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00 ?r DO 

OU 160379 >m 


Call No. ^[ O 3 n y & C Accession No. S^ S 

Author w> t_lo 

This book should be returned on or before the date last marked below. 


38 The Equilibria of Carbonic Acid and Its Salts Carbon dioxide and carbonic acid, sodium hydrogen 275 

carbonate and sodium carbonate, calcium hydrogen carbonate and calcium carbonate. Temporarily 
hard water. Limestone cave formation. 

E- Recapitulation of Problems on Ionic Equilibria A study assignment. lonization constants. Solu* 279 
bility products. 


39 The Negative Ions Qualitative Tests Sulfate ion, sulfite ion, sulfide ion, carbonate ion, chloride ion, 283 

bromide ion, iodide ion, phosphate ion, nitrate ion. 

F An Introduction to the Qualitative Analysis of the Metal Ions. Semimacro Techniques A study 280 

40 The Silver Group 293 

41 Precipitation with Hydrogen Sulflde. The Analysis of the Hydrogen Sulfide Group Hydrogen sulfide as 297 

a weak dibasic acid. The precipitation of metallic sulfides. Amphoteric sulfidcs. The oxidation of 

42 The Ammonium Sulfide Group Precipitation with ammonium sulfide. The hydrolysis of sulfides of 305 

weak bases. Buffer action. The oxidation of chromium and manganese. 

43 The Alkaline Earth and Alkali Groups Solubilities of alkaline earth salts. Precipitation with ammonium 315 

carbonate. The separation of magnesium. 

G The Qualitative Analysis of Cd", BI+++, AsO 3 , Sb"+, NI++, Co", and Sr++ A study assignment. 321 

44 The Analysis of General Unknown Inorganic Substances. The Solution of Solids Methods for the 323 

solution of difficultly soluble solids. The use of acids, oxidizing agents, reducing agents, complex 
ions, metathesis, and fusion processes. 


I The Measurement of Physical Quantities The use of dimensions, significant figures, experimental errors. 320 

II Tables of Data Logarithms, the metric system of units, various units and conversion factors, the vapor 333 
pressure of water at different temperatures, temperature corrections for barometer readings, con- 
centration of desk acid and base solutions, the volatility of acids, replacement scries of some metals, 
the solubility of some common gases, general solubility rules for common salts and bases, the rela- 
tive concentration of ions in 0.1 F solutions of electrolytes, the color changes and pH intervals of 
some important indicators, oxidation- reduction potentials, equilibrium constants, solubility products, 
table of ionic valences. (The periodic table and international atomic weights are given on the inside 


A review of chemical arithmetic and fundamental mathematical operations is given in Study Assignments A and B. 

A review of the rules of chemical nomenclature is given in connection with Experiments 8 and 18. 

A Teacher's Manual is available to all instructors. This provides lists of supplies needed, of reagents and solutions, 
with directions for their preparation, and numerous suggestions to the instructor regarding individual experiments. 

A Preface to the Studenl 

This laboratory manual has been designed to cor- 
relate with the presentation of the subject in Coir 
lege Chemistry, An Introductory Textbook of General 
Chemistry , by Linus Pauling. References to chap- 
ters in this text will be found at the head of each 

True knowledge and understanding in any field 
of learning depend on the careful observation and 
correlation of facts, and on logical deductions from 
them. In chemistry, these "facts" are the experi- 
mental observations of the way substances behave 
with respect to one another under various condi- 
tions, both in nature and in the laboratory. Out of 
such observations by scientists have come the 
"theories" and the "principles" which you are 
studying in your textbook. It is important that, in- 
sofar as time and reasonable laboratory facilities 
permit, you have the opportunity to experience 
this "scientific method" in your own study of 
chemistry. Your laboratory work is the central core of 
your chemistry course. 

Here are a few suggestions which will help to give 
you a good start in this course: 

(1) Use your ingenuity and common sense. While 
laboratory directions usually are quite specific, 
there is always abundant opportunity for clean-cut, 
logical, original, and imaginative thinking. These 
are qualities a corporation's chemical-laboratory 
director is seeking. The routine thinker will be as- 
signed to the routine jobs. Ingeniousness is as im- 
portant to you in the study of chemistry as in any 
other activity. 

(2) Broaden your outlook by becoming familiar 
with the introductory literature of chemistry early 
in your course. Tabular data on the properties of 
substances are given in handbooks, such as: Hand- 
book of Chemistry and Physics, Charles D. Hodge- 
man, editor in chief, Chemical Rubber Publish- 
ing Company, Cleveland, Ohio, and Handbook of 
Chemistry, N. A. Lange, Handbook Publishers, 
Inc., Sandusky, Ohio. Consult your instructor as to 
the many excellent textbooks of general chemistry 
and qualitative analysis which are available. A 
thorough, general reference work is A Comprehen- 
sive Treatise on Inorganic and Theoretical Chemis- 

try, 3. W. Mellor, Longmans, Green and Company, 
New York, 1932-1937. 

(3) Familiarize yourself with the experiment you 
are to perform before you come to the laboratory. 
Study the "Review of Fundamental Concepts" 
thoroughly, and read the "Experimental Proced- 
ure." In certain experiments, you will find some 
"Preliminary Work" to complete before the labora- 
tory period. Laboratory work is of much greater 
value and you can make more efficient use of your 
time if you understand beforehand what you are 
trying to do. 

(4) Note beforehand any extra equipment re- 
quired from the stockroom, and obtain all of it at 

(5) You will gain self-reliance by working alone 
in the laboratory, except when directed to do other- 

(6) The "Report Sheets" are not merely a place 
to "put down the answers." They are intended as a 
stimulus and guide to your thinking. It is important 
that you think through and fill in these reports as 
you perform the experiments. Keep the reports, after 
they have been corrected by the instructor, in regu- 
lar order. They thus become a more important part 
of this book. You will be graded in this course, 
however, not so much by your entries in any formal 
report sheet, as by what you know about the princi- 
ples underlying the experiment. 

(7) Scientists learn much by discussion with one 
another. Since the reports are not "tests" or "ex- 
aminations," you can often profit by discussion 
with your classmates but not by copying from 
them. The very keystone of all science rests first of 
all on integrity. You will also profit by frequent 
reference to your textbook while you are working 
in the laboratory. (Books are generally even more 
complete and reliable as sources of information 
than are your classmates!) , 

(8) The "Safety Precautions" and the "Labora- 
tory Rules" (see pp. vii and ix) will help you to 
safeguard not only your interests, but those of 
other members of the laboratory class as well. 
Know these precautions and rules and obey them. 





NEVER point a test tube of boiling liquid at your neighbor it may bump. 

Safely Precautions 

(1) The laboratory is a place for serious work. denly formed bubble of vapor may eject the con- 

Maintain a wholesome, businesslike attitude at all 
times. Do not attempt unauthorized experi- 
ments. Accidents and trouble will be avoided by 
following this rule. This is for the safety of the 
whole class. 

(2) Any accident involving even a minor injury 
should be reported to the instructor at once. 
Beware of hot glass: glass cools very slowly and 
may be very hot without appearing so. 

tents violently and dangerously. 

(4) When diluting sulfuric acid, pour the acu 
slowly and carefully into the water with constant 
stirring. Never add the miter to the acid. So mucl 
heat is liberated on solution that steam may b< 
formed with almost explosive violence. 

(5) Neutralize spilled acid or base as follows 
Acid on clothing use dilute ammonium hy- 

(3) Do not point your test tube at your neigh- Base on clothing -use dilute acetic acid, fol 
bor or yourself when heating substances. A sud- lowed by ammonium hydroxide. 

NEVER pour water into a strong 
acid the heat generated may 
spatter the mixture or break the glass. 

ALWAYS pour a strong acid into 

water slowly, and stir constantly. 




NEVER, force <x thistle tube or funnel 
into a stopper by rets ping the 
large end. Use the stem and 
twist as you push. 

ALWAYS wrap your hands in a towel 
when putting a glass tube into a 
stopper. Moisten with water and 
insert with a twisting motion. 

Acid or base on the desk top or floor wash off 
with plenty of water. Solid crude sodium bicar- 
bonate may be used to neutralize large amounts of 
either acid or base, and then the mixture is washed 
off with water. 

Corrosive liquids on the skin use plenty of 
water. Consult the instructor. 

(6) When putting glass tubing through a rubber 
stopper, lubricate the tube and stopper first with 
water (unless moisture must be avoided in the ex- 
periment). To insert the tubing, hold it with a 
cloth near the end to be inserted and use a twisting 
motion. This applies also to the insertion of ther- 
mometers and thistle and funnel tubes. (If you 
hold a thistle tube by the "thistle" end, it is very 
likely to break when inserted in the stopper.) 

(7) Never taste a chemical or solution unless 
directed to do so. (Poisonous substances are not 
always so labeled in the chemistry laboratory.) 
When directed to taste a solution, touch a drop of 
it, suspended on a stirring rod, to the tongue; then 
wash out the mouth with water. If any chemical is 
accidentally swallowed, see your instructor. 

(8) When observing the odor of any liquid, do 
not put your face directly over the container. Some- 
times a heated liquid will suddenly form, even 
when no longer being heated, a bubble of steam 
(called "bumping' 1 ). The contents might thus be 
discharged into your face. Fan a little of the vapor 

toward you by sweeping your hand over the top of 
the test tube or beaker. 

(9) In any experiments in which poisonous or 
otherwise objectionable gases or vapors are dis- 
charged, perform the operations under the HOOD. 
This provides suction to remove any such gases or 

(10) To protect your clothing from corrosive 
chemicals, always wear a laboratory apron when 
doing experimental work. (Aprons are cheaper 
than clothing.) 

ALWAYS smell a substance by 
wafting its odor gently toward 
your face. 

Laboratory Rules 

(1) At the close of each laboratory period, leave 
your glassware clean and dry. Wash and wipe off 
the desk top, so it is not left spotted and dirty. 

(2) Throw all solids to be discarded into the 
waste crocks. Wastepaper belongs in the waste- 
baskets. Never throw matches, litmus paper, or any 
solid, insoluble chemicals into the sink. Liquids are 
emptied into the sinks and washed down with 
water, because acids, and salts of copper, silver, 
and mercury are corrosive to plumbing, particu- 
larly if it is made of lead. 

(3) The reagent bottles on the side shelves 
should not be carried to your desk. Clean test tubes 
or beakers should be used for carrying liquids, and 
beakers, watch glasses, or small squares of paper 
for carrying solids. 

(4) Read the label twice before taking anything 
from a bottle. 

(5) Use as little reagent as is convenient to per- 
form your experiment. Two or three milliliters are 
usually sufficient in test tube experiments. 

(6) Never return unused chemicals to the stock 
bottles. You may make a mistake from which 
another student's experiments will suffer. 

(7) Do not insert your own pipettes or medicine 
droppers into the reagent bottles. Pour out the 
solution instead. This will avoid any possible con- 
tamination of the stock solution. 

(8) Do not lay the stopper of a bottle down, as 
impurities may be picked up and thus contaminate 
the solution when the stopper is returned Hold the 
stopper as illustrated in Figure i-6 or i-7. 

(9) Graduated cylinders and bottles are not to 
be heated, because they break too easily. (See 
below.) Likewise test tubes are likely to break if 
heated above the level of the liquid in them, and 
liquid is then splashed over the hot glass. Evapor- 
ating dishes and crucibles may be heated red hot 
if desired. Do not heat any piece of apparatus too 
suddenly at first, or it may break. Apply the flame 
intermittently at first until the vessel is hot. 

NEVER heat <x graduated 

cylinder or bottle. 

Laboratory Manipulations 

The following suggestions on common laboratory the hottest portion of the flame which is thus al- 
procedures will assist in a proper orientation to lowed to spread about them. If placed down in the 
your work. Read these now, and refer to them cold inner cone, consisting of unburned gas, the 
again in succeeding experiments, as you encounter objects are not heated effectively. 

The Meker burner (or the similar modern 
Fisher burner) is designed to give a concentrated, 
very hot flame (Fig. i-2). For maximum tempera- 

the first use of the various operations. 

A. Laboratory Burners 

The Bunsen burner, used for most laboratory 
heating, produces a cone-shaped flame, as illus- 
trated (Fig. i-1). Ordinary beakers, crucibles, and 

ture, have the gas on full pressure, and with the 
air vents open adjust the needle valve at the base 

other objects to be heated are placed just above to give a short blue flame of many short cones about 


region I 

*' ^_^ 1 1 

violet \ 


Hottest part 
of flam& 

Rotating sleeve 

for regulating 

"" the air supply 




rises off 
the burner 

This gets 
very hot 

If the flame U 
like this, open the 
air regulator. 

' flame 

If the f lame is like 
this, turn the gas off 
a moment and then 
light it again. 

If the flame is 
Pike this, turn 
down the gas. 


Pale blue 

Bunsen burner 

with correct flame. 

A flame .spreader on the top 
of a Bunsen burner makes a 
flat flame like this. 

Fia. i-1. The operation of a Bunsen burner 



Pale violet 

part of 

|part of 

Tiny blue cones 
reducing regio 

Needle valve 
for adjusting 
gas supply 


Needle valve 
for adjusting 
supply (jr^ 

^ ^S^s\ 

Pale violet 

Blue cone 

Needle valve 
for adjusting 
air supply 

and socket 
joint for di- 
recting blast 


Compressed air 

Meker burner 

Fisher burner Blast burner 

FIG. i-2. High temperature burners. 

0.5 cm high. The object to be heated is placed 
about 1 cm above the grid. 

The blast burner makes provision for mixing the 
gas with compressed air or oxygen gas and will 
give a very hot concentrated flame. It is particu- 
larly useful in glass-working operations. 

B. The Manipulation of Glass Tubing 

A few of the simpler and more frequently used 
procedures involved in the manipulation of glass 
tubing are presented in Figures i-3 and i-4. Study 
these carefully, and then construct a wash bottle 
as illustrated in Figure i-5. 

C. The Handling of Chemicals 

A few simple suggestions regarding the proper 
handling of solid and liquid chemicals are illus- 
trated for you in Figures i-6 and i-7. We repeat: 
Be considerate of others by always bringing your 
container to the reagent shelf to obtain a chemical. 
Do not take the bottle to your desk. Maintain the 
purity of the chemicals in the reagent bottles. Do 
not withdraw more than you need t and never re- 
turn any chemical to the bottle. Never contam- 
inate the stopper of a bottle by laying it down; 

hold it as illustrated. Do not insert your own medi- 
cine dropper into a reagent bottle or the medicine 
dropper from a reagent bottle down into your own 
test tube or solutions. (Fig. i-8.) 

Careful observance of these suggestions will pre- 
vent the spilling of chemicals and the contamina- 
tion of the stock bottles. If you do spill any chem- 
ical, clean it up completely, at once. A dirty 
laboratory is not conducive to good work. 

D. The Cleaning and Drying of Glassware 

Most vessels can be cleaned simply by washing 
them first with soap solution and then with tap 
water, and finally rinsing them with a spray of dis- 
tilled water from the wash bottle. Water wets a 
clean glass surface uniformly it does not stand 
in droplets over the surface. By observing this, 
you may tell when the glass is clean. (Fig. i-9.) 

Use your test-tube brush regularly, but be sure 
there are no exposed sharp metal points on the 
brush. These will scratch the inside of the test 
tube and cause it to break easily when it is heated. 
(This is a frequent but unsuspected cause of 
breakage.) Films adhering to the inside of flasks 
and bottles may often be removed by wetting the 



One stroke 
Don't "saw" 

Trim the * ^V^ 

edge of a \ 

^ broken graduate or 

First ^\' s \ large tube by stroking it 

With one stroKe scratch the tube with a piece of wire screen, 
with the edgre of a triangular file. 


Place the thumbs together oppo- 
site the scratch. 

After slowly introducing 
end into a Bunsen flame, 
the tube back and forth 
cut edges are rounded. 

Third - 

Pull and bend quickly. 

The cut After fire This has been 
end glazing heated too 

Fire glazing the end of a tube 

FIG. i-3. The manipulation of glass tubing. 



Roll the tube back and forth in a 
flat flame until it has become quite 

/? 'n ^ 

Remove from the flame and hold for 
a couple of seconds to let the heat 
become more uniform. 

Bend quickly to the desired shape 
and hold until it hardens. 

good bend 

Roll the tube in a Bonsen flame 
until it softens. Don't use aflame 

Allow the tube to become shorter 
as the walls thicKen to about twice 
their original thickness. 

Remove from the flame and after 
a moment pull until the softened 
region is as small as desired. 

Cut to length. 

Bad bends 

F/re glaze the tip. 

Fio. i-4. Further manipulations with glass tubing. 



Short rubber 
tube connection 

Constricted tip 

for blowing 
water out 
of tip 

Two hole 

rubber stopper 

250 ml 

Erlenmeyer flash 



This tube should 
come within Va cm 
of bottom. 

A wash bottle 

FIG. i-5. The construction of a wash bottle. 




Roll and tilt the bottle until some of 
the contents enters the inside of the 
glass stopper. 


Carefully remove the stopper so that 
some of the contents remains in it. 

_ / . 

Shovel out a little of the material 
on the spatula provided. 

Tap the spatula until the desired 
amount falls off. 

Second Method 



Tap the stopper with a pencil until 
the desired amount of material falls Roll and tilt the jar until enougl 

out. of the material falls out. 

First Method Third Method 

Handling Powders and Crystals ALWAYS replace the stopper. 

FIG. i-6. Transferring powders and crystals. 



First % 

Read the I Abel twice. 


Hold the stopper in and tilt the bottle 
until the contents wet the stopper. 


Moisten the inside of the neck and 
the lip with the wet stopper. 

NEVER set 
the stopper 

The moistened neclc and lip pre- 
vent the first drops from gushing out. 

Pour down a glass rod when 

Fourth '** ^[ ' When pouring from a beaker the 

Replace the stopper and withdraw stirring rod can be held in this 
it again with the back of the hand. manner. 

How to Remove a Stopper 

How to Pour a Liquid 

FIG. i-7. Transferring liquids. 



When putting 
the contents 
of a medicine 
dropper into 
a test tube 

and NOT 

Pro. i-8. The proper use of a medicine dropper. 

surface with dilute nitric acid followed by water. 
The use of cleaning solution (concentrated sul- 
furic and dichromic acids) or aqua regia (concen- 
trated nitric and hydrochloric acids) is seldom 
necessary and not always effective. These acids are 
good oxidizing agents but not universal solvents 
as many freshmen imagine. 

If it is necessary to dry the inside of flasks and 
similar vessels, they may be warmed over the Bun- 

Water spreads 
out smoothly on 
CLEAN glass 


stands in droplets 
on SOILED glas*. 

FIG. i-9. Clean and dirty glass ware. 

sen flame. A gentle stream of compressed air is 
then passed through a glass tube leading to the 
bottom of the vessel until it is dry. Graduated 
cylinders and heavy glassware such as bottles will 
break if heated. A warm not hot air jet, pre- 
pared by passing the air through a coil of coppei 
tubing heated over a burner, is most convenient, 
(Fig. i-10.) All equipment should be clean and dry 
when returned to the stockroom. 

E. The Separation of Precipitates 

Precipitates which form during the course of a 
chemical reaction are usually separated from the 


tubing / 

being dried 

Compressed \U 
air \ 

Bunsen burner 

FIG. i-10. A warm air drier. 



/ Fold again 

Fold and 
crease lightly 

About 10 

Open out 

corner I) He this 


Seal the 
edge of 
the filter 
against the 

Fill with 
water and 
let it run untif 
the air is washed' 
out of the .stem. 

When the air 
is out of the stem 
stop the flow and 
add the mixture 
to be filtered. 

The torn corner prevents air from 
leahing down the fold. 

The weight 
of this column 
of water has- 
tens filtration. 

The filtrate 
should run 
down the 
walls of 
the beaher: 


FIG. Ml. The process of filtration. 



'"mother liquor" in one of three ways: 

(1) Filtration. This is simply a process of 
straining the precipitate through a fine sieve the 
filter paper. With very fine grained precipitates, 
which tend to run through the filter, the separation 
is often aided by letting the mixture stand for some 
time, often with heating. This process is called di- 
gestion. This promotes the formation of larger 
crystals, with the resulting disappearance of the 
smaller ones, so that a clear filtrate is more easily 
obtained. The technique of filtration is well illus- 
trated by Figure i-11. Study it carefully. In wash- 
ing a precipitate, direct a stream of distilled water 
from your wash bottle around the top of the filter 
paper for a moment, and let all this water carrying 
off the impurities drain through the filter. The 
process may be repeated as often as necessary. 
Washing is more efficient when repeated with sev- 
eral amounts of water, than when carried out once 
with one larger amount. 

(2) Dccantalion. If the precipitate is quite 
dense, the mixture may be left standing until the 
precipitate has settled. The clear liquid is then 
"decanted," or poured off, leaving the precipitate 
in the original container. The precipitate may be 
washed by adding water, letting the solid settle, 
and again decanting. (Fig. i-12.) 

(3) Centrifugation. The use of a centrifuge, 
which is convenient and has now become very 
popular, substitutes centrifugal force for the force 
of gravity. It thus involves the same principles of 
separation as simple decantation. Figure i-13 illus- 
trates the operation of a centrifuge. The precipitate 
may be washed by mixing it with water and again 

F. Laboratory Balances 

A platform balance, or "Harvard trip scale," sen- 
sitive to about 0.1 gram, is used for crude weigh- 
ings. An agate, knife-edge, triple-beam balance, 
sensitive to 0.01 gram, is used for more precise 
work. (Fig. i-14 (a) and (b).) For most of the 

Position of sockets 
when handle b 
spun rapidly 

sockets contain 
tubes of 
liquid to be 

FIG. i-12. The separation of a heavy solid by decantation of 
the supernatant liquid. 

To prevent 
undue vibration 
the centrifuge tubes 
should be filled with 
equal amounts of liquid. 

FIG. i-13. The principle of the centrifuge. The rate of set- 
tling of a precipitate is increased many fold by the application 
of a centrifugal force 500 to 1000 times that of gravity. 



Pointer and scale 


Substance p/atform 

to 0.1 

for set- 
ting to 
nearest log 

Adjustment Adjustment weights 

weights Do not disturb 
Do not disturb 



10 beam 


and scale 

/ leveling 


FIG. i-14. (a) A platform balance, (b) An agate knife-edge triple-beam balance. 

Steel knife edges 
agate planes 

Adjustment weights 
Do not 


Leveling screw 

FIG. i-15. A pulp balance* 

quantitative experiments, either a pulp balance 
(Fig. i-15), or preferably an analytical type balance 
(Fig. i-16), capable of weighing to the nearest 
milligram, should be used. 

G. The Care and Use of the Balance 
Precautions and Balance Rules: 

(1) Handle the beam supports by using the cen- 
ter knob gently, so as not to jar the balance. 

(2) Always have the beam supports up in place 
when adding or removing any objects or weights 
from the pan, or when adjusting the rider. 

(3) Never put chemicals directly on the balance 
pans. Use a small beaker, or square of glazed paper. 

(4) Never weigh an object while it is hot, or even 
warm. Convection currents of air in the balance 
case will render the weighing meaningless. 

(5) Handle analytical weights only with the 
forceps provided, never with the hands. Do not 
borrow or lend weights. Each set is sufficient for 
any weighing up to 100 grams, when the largest 
weight in the set is 50 grams. 

(6) Keep the balance case closed when taking 
final readings on the pointer scale, and when you 
are through weighing. 



Sliding door in 
draft- proof case 

(Case cut away 
to show details) 

Attachment for 
adjusting rider 

Adjustment weigh 
Do not disturb 

Beam graduations 

Weight for 
Do not disturb 

Scale pans! 

x-\" ' 


and beam off their 




* ~ 

Turning this knob raises 
pads under the scale pans v Turning this 
to damp their swinging. knob raises the 

Forceps for 
handling weights 
Never use fingers 

Box of 

FIG. i-16. A balance for more precise weighing. Note the operation of the beam supports, and of the rider. 


(7) Request the instructor to make any neces- 
sary adjustments on the balance. 

Procedure in Weighing. We shall carry out 
weighings only to the nearest milligram (0.001 g). 
Your instructor will explain the exact procedure 
to be followed with the type of balance available 
in your laboratory. If an analytical balance, which 
is sensitive to 0.0001 g is used, it will not be neces- 
sary to take exact pointer-scale readings. In fact, 
unless the balance is in excellent condition, and 
accurately calibrated weights are used, and until 
you have had considerable experience in weighing, 
it is a waste of time to attempt closer weighing 
than 0.001 g. 

Carry out a weighing in the following steps: 

(1) Determine the Rest Point of the Balance. Be 
sure the balance pans are clean, the beam and pan 
supports and knife edges in proper position, with 
the rider off the beam, or at zero or center position. 
On releasing the beam and pan supports, cause the 
pointer to swing gently 3 to 7 scale divisions either 
side of the scale center, and note to the nearest 
half scale division, the point about which the 
pointer is oscillating (i.e. that point at which it 
would come to rest if there were no irregularity due 
to friction). This is the rest point, or point of refer- 
ence for the succeeding weighing. (If the balance 

beam does not swing freely and "die down" evenly 
or if the rest point is over 2 or 3 scale divisions from 
the scale center, call an assistant to make the neces- 
sary adjustments.) 

(2) Balancing the Object and Weights on the Pans. 
With the pans and beam fully supported, place the 
object to be weighed on the left pan. Add weights 
as needed to the right pan to restore the pointer 
fluctuations to a point almost, but not quite, to the 
initial rest point. In adding the weights, add first 
the largest denomination you think can be used, 
and release the beam support gently to ascertain 
that the weight is not too large. Add successively 
smaller denominations of weights until the weight 
has been determined within the nearest 10 milli- 
grams (i.e. with the pointer still fluctuating about 
a point slightly to the right of the rest point). 

(3) Adjusting the Rider. 1 Close the balance case 
so as to avoid convection currents in the air, for 
this final step in the weighing. The wire rider 
weighs 10 milligrams, hence by suitably placing 
this at successive 1 milligram intervals, and taking 
the rest point readings, adjustment may quickly be 
made so that the pointer swings coincide with the 
original rest point (within the nearest milligram). 

1 For balances which do not have a rider, follow the method as out- 
lined in paragraph 4 for the final adjustment. 

If the pointer swings 
from here to here 


If the pointer swings 

from here to here 


; f 
.;.- { 


K*st point of the 

empty balance 


6.2 >* 3,7 -7.4 + 7.O 

the rest point\ls at the rest poini^fo now 

+?.? -r 2 1.35. at - 0.4 *- * - -0.2 

The swinging should be The sum of the weights on 

stopped with the left hand the right scale pan plus the 

knob and the rider moved reading of the rider is now 

to the right. equal to the weight on the 

left pan plus a small error. 

a b - 


Rest point with i6.27g 

on right pan 
Rest point with ie.28g 
on right pan 

-1-3 b- 4 +3.5 -7.3 

% "* M -o* approximately 
The weight on the /eft pan id 
therefore about 10.274 g. 

FIG. i-17. (a) and (b). The use of pointer scale readings to determine the rest point, (c) The determination of the weight to 
0.001 g when the smallest weight used is 10 mg and the balance has no rider. 



(Note that if a 10-mg rider is used on a balance 
with a rider scale of only 5 scale divisions, each 
division equals 2 nig. See Fig. i-17ab.) 

(4) Using the Method of Pointer Scale Readings. 
For balances without a rider, the weight to the 
nearest milligram may be obtained as follows. Sup- 
pose the rest point as found in section (1) above is 
1 scale division to the right of the scale (+1, see 
Fig. i-17c). Suppose also that, when weighing the 
object, the scale reading with 16.270 g on the pan 
(totaling within 10 mg of the true weight), is +4 
scale divisions. Now add a 10-mg weight to the 
right pan, and again take the scale reading. Sup- 
pose it now reads 3.5 scale divisions. Note that 
adding 10 mg moved the scale-reading 7.5 scale 
divisions, while we only want to move it 3 scale 
divisions, (to +1). We therefore want to add 
3/7.5 X 10 mg, or 4 mg. The final weight is then 
16.274 g. 

(5) Counting the Weights. More errors are made 
in counting weights than in making the weighing 
adjustments. First count the weights on the pan 
and check by noting the empty places in the weight 
box. Make a record of each weight denomination 
you use, so that you can recheck the counting of 
the weights, if the results of the experiment indi- 
cate such an error may have been made. Keep and 
label this record. Total the result as illustrated, 
then enter the weight at once on your experiment 
report sheet. (Be sure that you record a 50-mg 
weight, for example, as 0.050 g, and not as 0.500 
g, when computing the total weight.) 

Example of totaling weights: 

10.000 g 

200 mg 

50 mg = 

20 mg = 

4 (rider) = 

16,274 g 

The weight of the object then, was 16.274 grams. 
Note that the milligram weights are finally ex- 
pressed as fractions of a gram, in the column to be 

H. Volumetric Measurements 

Approximate Volumes. The volume stamped 
with the trademark on beakers and flasks is only 

approximately correct. It can be used only where 
crude estimates of capacity will suffice. Frequently 
you are asked to add a small volume of from 1 to 
3 ml of a reagent. This can often be accomplished 
with sufficient precision by estimating the amount 
in a 10-cm test tube, which has a total capacity 
when filled of 8 to 10 ml. Likewise, you should 
know that a 15-cm X 2-cm test tube has a ca- 
pacity of about 30-ml. Use your graduated cylin- 
ders only when more precise measurements are 
necessary. It is not good practice to carry out 
chemical test reactions in graduated cylinders, as 
they cannot be heated, and are more difficult ot 
clean. Use test tubes for all such tests. 

Reading a Meniscus. The meniscus is illus- 
trated in Figure i-18. In all graduated ware, such 
as burettes, graduated cylinders, and volumetric 
flasks, always read the bottom of the meniscus. Be 

Always read 
the bottom of 
the meniscus 

This reads 
35.6 ml 

Eye level 

FIG. i-18. The proper method of reading a meniscus (curved 
surface of a liquid) in order to measure the volume of the 



sure the eye is on a horizontal with the level of the 
liquid, so as to avoid the error of parallax. Good 
and reproducible though not too bright lighting 
is essential for precise measurements, so that the 
meniscus is viewed under the same conditions in 
successive readings. Arrange the graduated cylin- 

der, or other vessel, so that you will look toward 
the light. White opaque paper back of the vessel 
sometimes helps, and a dark surface on the paper 
just a little below the level of the meniscus, or your 
finger held back of the vessel just below the menis- 
cus, helps to give a sharp definition. 



Vacuum shelf 


as to 
paper basket 

Crock for 



oards and 
drawers for apparatus 

Fio. i-19. A typical laboratory table arrangement* 







'290 ml 

250 ml 


^ / WWW' ^JZ^r: 

(Pyrex) Medicine droppe 
^ Funnel "^ 

FIG. i-20. Chemical laboratory apparatus glass and porcelain. 



Professional workers in every field keep their 
tools and equipment in the best possible working 
condition. Keep your equipment clean and con- 
veniently arranged in your locker, so that it is ready 
for use. 

Burette clamp, 

Clamp, utility* 

Ring stand 

with 4 inch rind 

Litmus paper, 
red and blue 



f jr -, Test tube 

Test tube // brushes 

holder fa 


Rubber stoppers 


Crucible tongs 

Mohr pinch 




Flame spreader 

FIG. i-21. Chemical Uxmtory apparatus accessories and implements. 

Rubber tubing 

Introductory Laboratory Techniques. 


Col/ege Chem/sfry, Chapter 1 

Experimental Procedure 

When beginning work in the chemistry lab- 
oratory you should do these four things: 

(1) Check the laboratory equipment in the 
locker assigned you. Refer to the drawings in 
Figures i-20 and i-21 on the preceding pages for the 
identification of any item with which you are not 
familiar. Ascertain that all items are present and 
examine them carefully to be sure they are in an 
acceptable condition. Hereafter, you will be re- 
sponsible for this equipment and will be charged 
for any breakage or shortage at the conclusion of 
the course. 

(2) Familiarize yourself with the "Safety Pre- 
cautions," "Laboratory Rules,** and "Laboratory 
Manipulations" as given on Pages vii to xxiv. 

(3) 1 Learn the elements of glass working, as 
directed and illustrated by the instructor, in ac- 
cordance with the suggestions on the manipula- 
tion of glass tubing, paragraph B, in the preceding 
section. Practice the fundamental operations, as 
shown in Figures i-3 and i-4, until you have at- 
tained a satisfactory proficiency, and then prepare 
a wash bottle in accordance with Figure i-5. Use 
a 250-ml Erlenmeyer or Florence flask for this pur- 
pose. An acceptable wash bottle will have the 
upper sections of the glass bends aligned in a 
straight line, cut to the designated lengths, and 
properly fire polished. Your instructor will indicate 
his approval by initialing your report sheet in the 
space provided. 

(4) Learn the metric system of measurements if 
you are not already familiar with it. See Table 
II, "The Metric System of Units," in Appendix 
II. Carry out the suggestions as given in the re- 
port sheet for this experiment for the measurement 
in metric units of the length, volume, or weight of 
various laboratory items, using the appropriate 
measuring device a meter stick, a graduated cyl- 
inder, or a suitable balance which is sensitive to 

1 In order to avoid congestion at the balances, half the class may 
be designated to do (4) on metric measurements* while the other 
learns glass working. 

0.1 g. You will thus gain a visual appreciation of 
metric dimensions. Further experience may be 

Hydrogen (H^ 

density o.o&9 

grarro per liter 

Oxygen (O.,) 


Mercury Wood 

Gasoline Water 


/ S -* / 2 

0.67* {J/cin i g/Cfn" 
Aluminum Iron Gold 


rTog/crn 3 7Stg/cm s io.3-g/cm 5 22.5 

FIG. 1-1. To illustrate the relationship of density to volume* 
The different size cubes represent the relative volumes of 
equal weights (about 0.03 g) of various materials, at 0C. 

gained by measurements of the density of various 
substances, as outlined in the next paragraph. 

Density. The determination of this important 
physical property requires measurements of two 


quantities: the mass, M, and the volume, V, of a 
given quantity of a substance. 1 The ratio of these 
two, or the mass per unit volume, is the density, 
that is, D = M/V. In the metric system, the result 
is expressed as grams per cubic centimeter, g/cm 8 . 
See also Figure 1-1. 

(a) Density of Liquids. After receiving instruc- 
tions on the method of weighing with the type of 
balance you are to use (also read paragraphs F, 
P. xix, and G, P. xx), weigh a clean, dry, 150-ml 
beaker to the nearest 0.1 g. (Enter this and all 
other data as you obtain it directly in the report 
sheet.) Place between 40 and 50 ml of the liquid 
to be used (water, carbon tetrachloride, or an un- 
known 2 provided by the instructor) in your gradu- 
ated cylinder. Read the volume to 0.1 ml (note the 
precautions in reading a meniscus, paragraph H, 
P. xxiii), and carefully transfer this liquid as com- 
pletely as possible to the beaker, lie-weigh the 
beaker and contents. From these data, calculate 
the density of the liquid. 

(b) Density of Solids. As an example of the 
technique to use with an irregularly shaped solid, 
determine the density of roll sulfur. Select about 
20 to 30 g of pieces of roll sulfur which have been 

1 It is important when making measurements of physical quantities, 
such as weight, length, or volume, that the measuring devices you use 
be consistent with the precision of measurement desired. Thus a pre- 
cision of 1% in the weighing of a 50-gram sample requires a balance 
which is accurate to only 0.5 gram (1% X 50 g = 0.5 g) ; for the same 
precision with a 1-gram sample you need a balance which is accurate 
to 0.01 gram. In this experiment, platform balances which read to 
0.1 gram and graduated cylinders which can be read to about 0.2 ml 
should not cause errors in the calculated density greater than about 


The percentage error in an experiment is calculated by dividing the 
actual error by the accepted value, and then multiplying the result 
by 100 to express it as percent, i.e. parts per hundred. 

See Appendix I for a more complete discussion of The Measurement 
of Physical Quantities. 

1 Suitable solutions of unknown density may be prepared by the 
instructor by dissolving inexpensive soluble salts such as CaCl 2 , 
NaCl, etc. in water. 

crushed small enough to be placed in your 50- oj 
100-ml graduated cylinder, but avoid unduly small 
pieces and any powder. Weigh the sulfur in a 
beaker, or evaporating dish, which has been pre- 
viously weighed. Place about 30 ml of water in the 
graduated cylinder and read the volume to 0.1 ml. 
Carefully add the pieces of sulfur, tilting the cyl- 
inder and sliding them in so as to avoid any loss 
of water by splashing. Tap the sides to dislodge air 
bubbles, and again read the volume. The increase 
will be the volume of the sulfur. See Figure 1-2. 


The difference 

bc-twccn these- 

readings is the 

"volume of these 

pieces of material. 

FIG. 1-2. Illustrating a method of measuring the volume of an 
irregularly shaped solid. 

Calculate the density. Other materials, as suitably 
shaped pieces of metal or metal shot, marble chips, 
etc., may be similarly used. 

If it is desired to determine the density of a 
regular shaped solid, as a rectangular block or a 
cylinder, the volume may be calculated from meas- 
urements of the length, breadth, and thickness, or 
the length and diameter, respectively. The weight 
may be obtained directly on the balance, and the 
density calculated. 

REPORT: Exp. 1 Name- 

Introductory Laboratory Techniques Date_ 


Locker Number- 

1. Glass Working 

Instructor's approval of wash bottle . . , . 

2. Metric Measurement of Length 

Learn to recognize approximate dimensions at a glance. Practice by measuring numerous objects about the 
laboratory. Record the dimensions of the following, to 0.1 cm or 1 mm. 

Large test tube: Length Diameter 

Small test tube: Length Diameter 

Large graduated cylinder: Length Diameter- 
Diameter of: Evaporating Dish Crucible 

Conversion factor. In order to determine the number of centimeters in an inch, measure the length of this page, 
in centimeters, and in inches. Express any fractional part of an inch (measured to the nearest sixteenth) as a decimal. 

Width of this page em in. 

Calculation of conversion factor (ratio of centimeters to inches) : 

Your value _ 

Ratio, from Appendix Table II ._ 

3* Metric Measurement of Volume 

Learn to recognize approximate volumes at a gVance. Practice by measuring various vessels in your desk equip- 
ment. In particular, learn to estimate the height of water in your small and large test tubes which corresponds to 
1 ml, 2 ml, 5 ml, and 10 ml volume. You will frequently estimate small volumes in this manner. Likewise, to dispel 
any false illusions you may have as to the precise volumes of beakers, flasks, etc., as indicated by the trademark, 
fill several of these level full with water, and determine the actual volume by repeated fillings into your graduated 

Capacity of test tubes: small large- 
Evaporating dish Crucible 

Beakers: "150" ml "400" ml "250" ml flask. 


4. Density of a Liquid (See directions in "Experimental Procedure") 
Liquid used 

Weight, beaker + liquid 
Weight, beaker . . . 

Weight, liquid 
Volume of liquid 
Density, calculated 1 . 
Density, from literature 
Percentage error 1 
1 Show method of calculation here: 

5. Density of a Solid 

Solid used Sul f ur 

Weight of solid 

Final volume 

Initial volume 

Volume of solid 

Density, calculated 1 

Density, from literature 

Percentage error 1 

1 Show method of calculation here: 

6. Questions 

(a) How would you modify the procedure as outlined, if you wished to determine the density of a solid which 
is soluble in water, for example, rock salt, or sugar? 

(b) Would your density determinations be more precise if you weighed your samples to 0.01 or 0.001 g, instead 
of only to 0.1 g, while other measurements were curried out as before? Explain. 

(c) If 20 g of sulfur is used in the above determination, what is the limiting precision of measurement when 
the weight is determined to a precision of 0.1 g. 

A Review off Fundamental Mathematical Operations. 

A Study Assignment 


If you are a typical beginning student of chem- 
istry, you are probably concerned about the rumors 
you have heard that chemistry requires a good 
background in mathematics. You should recognize 
very early that chemistry is one of the exact sci- 
ences, and it is indeed fortunate that we arc able 
to express many of its precise concepts, laws, and 
principles in the form of mathematical relation- 
ships. Once these relationships have been set up 
properly, however, you should be relieved to know 
that the mathematical operations required to 
simplify arid solve the equations arc mainly the 
arithmetical operations of multiplication, division, 
addition, subtraction, and handling fractions, and 
the simple algebraic operations required to solve 
first-degree and, occasionally, second-degree equa- 

The difficulties which students encounter with 
chemical problems are very rarely due to the 
inability to perform the mathematical operations; 
they stem, rather, from an incomplete understand- 
ing of the chemical laws or principles which form 
the basis of the mathematical set-up. In many 
cases the difficulty is simply due to the fact that the 
terms used in chemistry are new. For example, 
whereas a student would give a ready answer for a 
problem which asks how many eggs, at 72 cents 
per dozen, one could purchase for $1.56, he may 
be bewildered by an exactly analogous problem 
which asks how many milliliters of concentrated 
sulfuric acid, with density of 1.84 g/ml; should be 
used to provide 265 g of this substance for an 
experiment. Consequently, you should strive to 
obtain a thorough understanding of, and famili- 
arity with, all the terms and principles, so that you 
can reason out the proper set-up for the problem. 
You should always avoid solving problems by sub- 
stituting blindly in some formula which you have 
merely memorized and may not understand. 

A. How to Set Up Chemical Problems 

Although it is impossible to outline a general 
method for solving problems that is suitable to 

everyone, the following analysis of what is involved 
in problem-solving may be useful. 

1. Read the problem carefully. Be sure to find 
out the meaning of every term or concept in- 

2. Pick out the law(s), principle(s), or defini- 
tion (s) which may apply to the problem. Review 
these if necessary. 

3. Reason out the proper relationships between 
the numerical terms to obtain the mathematical 

4. As a check on your reasoning in the previous 
step, it is well to label each term with its dimen- 
sional unit(s). Factor out these units wherever 
possible to determine if the indicated calculations 
in the set-up will give the units desired in the 
answer. Units are multiplied or divided just as if 
they were numerical terms in this factor-tfw-unite 

5. The final step then involves carrying out the 
mathematical operations either by manual meth- 
ods or, preferably, with the aid of logarithms or a 
slide rule. Some suggestions on the use of these 
time-saving devices are given in the review which 

To illustrate the application of the above 
analysis, let us consider some problems dealing 
with an important physical property, the density 
of a substance. 

Illustrative Problem 1. A rectangular bar of 
iron has the dimensions 1.10 cm by 2.17 cm by 
6.41 cm. It weighs 121.7 g. What is the density of 

1. Density is defined as the ratio of the mass of a 
specimen to its volume. It should also be recog- 
nized that the volume of a rectangular bar may be 
found by multiplying its length by its w.'dth by its 

2. The definition of density may be expressed 
mathematically as 

density = 


3. The data of the problem are then analyzed 



and the mass of the bar placed in the numerator 
and the volume expression in the denominator: 

121.7 g 

density : 

1.10 cm X 2.17 cm X 6.41 cm 

4. Before carrying out the mathematical opera- 
tions, check the units in the set-up to be sure the 
units of density, g/cm 3 , will be obtained. 

5. When the arithmetic operations are carried 
out, the result is: 

121.7 g 

density ! 

15.5 cm 8 

7.85 g/cm a . 

Illustrative Problem 2. An experiment calls 
for 204 grams of mercury. What volume of liquid 
mercury should be measured out in a graduated 
cylinder to obtain the desired weight? The density 
of mercury at room temperature is 13.6 g/cm 8 . 

This time let us proceed by the "Monte Carlo" 
or "Las Vegas" method instead of the systematic 
analysis. As you see in Figure A-l, there are four 
possible ways of setting up the data given in this 
problem, and the student, Mr. M. I. Lucky, has 
one chance out of four to pick the right one. 

The alternative to this method of chance is, o 
course, to reason out a solution based on an under 
standing of the terms and concepts involved, 
followed by a factor-tlie-units check. Since density 
is the weight of one cubic centimeter, you should 
reason that by dividing the weight of mercury de- 
sired by its density you should obtain the number 
of cubic centimeters required: 

, 204 g 

volume =5 - - 
13.6 g/cm* 

A check of the units shows that carrying out this 
operation as indicated, g -f- g/cm 8 = g X cm s /g 
= cm 3 , gives the units of volume desired in the 
answer. Now the numerical answer may be safely 
obtained by dividing 204 by 13.6, which gives 
15.0 cm 3 as the volume of liquid mercury. In 
Figure A-2, another student, Mr. I. M. Shoor, has 
checked all the possible set-ups in Figure A-l by 
the factor-the-units method and has identified the 
proper one. 

Fio. A-l. One chance out of four. 

Fro. A-2, Factor-the-unita check. 


In summary, remember that the purpose of 
problem-solving in chemistry is to help you get a 
better working knowledge of the important laws, 
principles, and relationships which unify the many 
facts of the science. The crucial part of any prob- 
lem is the determination of the proper set-up or 
mathematical relationship. Your ability to do this 
will be limited only by lack of understanding of the 
meaning of the terms, concepts, laws, and prin- 
ciples involved. The mathematical operations 
necessary to solve the set-ups are usually quite 
simple, but in case you need a refresher the follow- 
ing review and exercises are included. 

B. How to Solve Algebraic Equations 

1. The same quantity may be added to or sub- 
tracted from each side of an equation without 
changing the equality. 

2. Each side of an equation may be multiplied 
or divided by the same quantity without changing 
the equality. 

3. First-degree equations, in which the exponent 
of the unknown quantity does not exceed one, may 
be solved by applying the above axioms. 

Example 1. 6x 4 2x + 16 Solve for x. 
Add 4 to each side: 6x 2x + 20 

Subtract 2x from each side: 4x 20 
Divide each side by 4: x 5 

Example 2. 


Solve for x. 

Multiply each side by 3b: 


Multiply each side by x: 3by : 
Divide each side by 2a: -T- ! 




4. Quadratic equations, in which the exponent of 
the unknown quantity does not exceed two, may 
be solved as follows: 

Example L 4x* 9 = Solve for JL 
Add 9 to each side and divide each side by 4, to obtain: 


X " 4 * 

Take the square root of each side : x = 

If a quantity x is a mass or volume or some other quantity 
which can have only a positive value, then the correct value 

Q O 

would be +~ ) and the root would be ignored. 

Example 8. 4x* + 4x * 3 Solve for x. 

This equation is similar to the general quadratic equation 

ax 2 + bx + c 

which may be solved by the method of completing the square 
to give the formula 

- b \/b $ 4ac 
x= _ 

To solve our example, we subtract 3 from each side: 

4x' + 4x -3=0. 
Then substitute the coefficients in the above formula: 

- 4 + 

- 4 8 


-7 or - 

C. Proportion and Graphical Relations 

A fraction expresses a division, that is, a ratio 
between the numerator and the denominator. Two 
such ratios (or fractions) which are equal to one 
another constitute a proportion. 

Example L Direct Proportion. The corresponding 
volumes and absolute temperatures of a given 
quantity of gas have a constant ratio, i.e. are 
directly proportional. This is Charles' Law. This 
fact may be expressed by the equations 

V - kT, or ^ = k. 

Since any two corresponding values of V/T are 
equal to k, we may write the proportion 

Vi V 3 . 4 Vi T! 

=---; or by transposing, ~- = -- 

ll 12 V 2 If 

If the temperature of 45 ml of hydrogen gas is de- 
creased from 20 C (293 K) to - 100 C (173 K), 
the decreased volume, Vi, is given by 

45 ml 293 

Multiplying both sides of this equation by 45 ml, 
we have 


26.6 ml H 2 . 

Calculate the volume at 100 C. (Answer, 57.3 ml.) 
Example 2. Inverse Proportion. The correspond- 
ing volumes and pressures of a given quantity of 
gas bear an inverse proportional relationship to one 
another. This fact is known as Boyle's Law, and 
may be expressed by the equations 

V* ~; or PV k. 



Since any two corresponding values of PV are 
equal to k, we may write the proportion 

PiVi = P 2 V 2 , or transposing, ~ = ~ 

Vj ri 

If the pressure of 45 ml of oxygen gas is increased 
from 1.0 atmosphere to 3.0 atmospheres, the de- 
creased volume, Vi, is given by 
Vi 1.0 atm 



3.0 atm 
1.0 atm 

X 45 ml = 15 ml O 2 . 

You may verify other corresponding values as 
follows: When P 6 = 5.0 atm, V 6 = 9.0 ml 2 , and 
when Pi = 0.50 atm, V 1 = 90 ml 2 . 

Scientists very frequently make a graph of their 
experimental data as a means of discovering funda- 
mental relationships, or of comparing their data 
with some known law. The extent to which their 
experimental values lie on a smooth curve is an 
indication of the precision of their measurements. 

The graph (Fig, A-3) for the data of Example 1, 
above, shows that a direct proportion plots as a 
straight line. Only two points are needed to fix the 
line, from which any other values then may be 
read. If the line were extrapolated to a volume 
of zero, it would cross the temperature axis at 
273 C. Of course, hydrogen would behave less 
like a perfect gas as the temperature is decreased, 
and would change to the liquid, and finally to the 

solid, state. By plotting precise experimental data 
for the volumes and temperatures, or pressures and 
temperatures, of a sample of hydrogen, the value of 
absolute zero can be determined. 

The graph (Fig. A-4) for the data of Example 2, 
above, shows that an inverse proportion plots as a 
hyperbola. A number of points are necessary to 
fix the curve. 

Figure A-5 shows a characteristic curve for the 
change in solubility of a salt with temperature. 
This is not a first order, or linear curve, $nd would 
have to be expressed by an equation in higher 
powers of x, where x represents the temperature. 
Usually a quadratic equation will give a close ap- 
proximation to the data. 

D. Exponents 

Very Large and Very Small Numbers. The student 
of chemistry finds that he uses many extremely 
large numbers, such as Avogadro's Number: 
602,300,000,000,000,000,000,000; and many ex- 
tremely small numbers, such as a sullidc ion con- 
centration of 0.00000000000000031 M. The reason 
obviously is the minute sizes of atoms and mole- 
cules, and the correspondingly enormous numbers 
of them which can exist even in a small space. 

Since such numbers arc awkward to handle, 
they are often expressed more conveniently by 
writing them as some simple number times ten 


f 40 - 

e 30- 

5 20- 

Hydrogcn (HJ 

too; 57. 3 ml 


i i i i | i r i 1 |l r i i J i i i T I i i 

> o* too' 

O i r*i r~j i ; i i I i | i | | i i i r 
-300 -200* -100 

Temperature (*C) 

FIG. A-3, The relation between the temperature and corresponding volume of a sample of a gas, such as hydrogen. A direct 
proportion, as represented by Charles' law, V * kT, plots as a straight line. 



Oxygen (OJ 

Pressure (atm) 

FIG. A-4. The relation between the pressure and corre- 
sponding volume of a sample of a gas, such as oxygen. The 
graph of an inverse proportion, as represented by Boyle's 
law, pv = k, is an hyperbola. 

raised to a power (the power is indicated by an 
exponent}. Thus, Avogadro's Number may be writ- 
ten 0.0023 X 10 24 , or 6.023 X 10 23 . The exponent 
23 means that 10 is raised to the 23rd power. The 
number 6.023 X 10 23 means that when the decimal 
point in 6.023 is moved 23 places to the right, the 
indicated number is obtained. Any number such as 
129,000 may be written in various forms, as, for 
example: 129 X 10 3 , 1.29 X 10 5 , or 0.129 X 1Q 6 . 

The number 0.00000000000000031 may be writ- 
ten as 3.1 X 10~ 10 , meaning that to obtain the 
number the decimal point in 3.1 must be moved 
16 places to the left. Likewise, any number such 
as 0.000736 may be written in a number of equiva- 
lent forms, as, for example: 7.36 X 10~ 4 , 736 
X 10- 6 , or 0.0736 X 10~ 2 . 

The Meaning of a Negative Exponent. In handling 
negative exponents, the student should note that 

IO- 4 = 

io- 3 

10< 10,000 
and so forth. In other words a number ex- 

pressed as a negative power of ten is equal to the 
reciprocal of ten raised to the corresponding posi- 
tive power, and vice versa. 

Multiplication and Division of Exponential 
Numbers. For numbers written in exponential no- 
tation, to multiply the numbers, add the exponents, 
and to divide the numbers, subtract the exponents. 
Other factors in the indicated product or quotient 
are to be multiplied or divided as usual. 


3 800 


2 600 


J 200 


NH 4 CI 

20 40 60 80 
Temperature (C) 


FIG. A-5. The change in solubility of ammonium chloride 
with temperature. Such a solubility graph is seldom linear, 
but can often be expressed by an equation of the binomial 
type: y = a + bx + ex 2 + . . . 

Examples of Multiplication: 

2 2 X 2 a X 2 7 = 2 2 + 3 + 7 = 2 12 , 

10 B X 1(T 2 = IO 3 , (Note that 5 + (-2) 

(2X 10 s ) (4 X 10 2 ) = 8X IO 5 . 

Examples of Division: 


io 7 /io 3 = io 7 ~ 3 = io 4 , 

10 2 /10- 6 = IO 7 , (Note that 2 - (-5) = 2 + 5 < 
(0 X 10 2 )/(2 X IO 3 ) = 3 X ID" 1 0.3. 


Powers of Exponential Numbers. The expo- 
nent is multiplied by the power desired. 


(IO 2 ) 3 = IO 2 X IO 2 X IO 2 = 16, 
(2 X IO 8 ) 4 = 16 X IO 12 . 

Roots of Exponential Numbers. The expo- 
nent is divided by the root desired. In case the ex- 
ponent cannot be divided evenly, the number must 
be changed to a form so that this can be done. 

= 0.0001. Likewise, IO 3 = Examples: 

The square root of 5 6 = (5 6 ) l/2 = 5 fi/2 = 5, 
The cubic root of 8 X IO 12 = (8 X 10 n ) l/a = 2 X IO 4 , 
The square root of 5 X 10~ 7 = (5 X 10~ T ) l/2 = 
(50 X IO- 8 ) 1/2 = 7.1 X 10- 4 . 

Addition and Subtraction of Exponential 
Numbers. The numbers first must be changed to 
a nonexponential form, or to the same power of ten. 


IO 2 + 10* = 100 + 1000 = 1100 = 1.1 X IO 3 , 
(6 X 1(T 2 ) - (2 X lO' 8 ) - (6 X 1(T 2 ) - (0.2 X 10"*) 
= 5.8 X IO- 2 = 0.058. 



E. Logarithms 

The common logarithm of a number is the power 
to which 10 must be raised to obtain the number; 
that is, a logarithm is an exponent and the pre- 
ceding rules for exponents are applicable. The 
logarithms of all numbers which are integral powers 
of 10 are whole numbers: 

Log of: 100, or 10 2 = 2 

10, or 10 1 1 

1, or 10 - 

Log of: 0.1, or 10" 1 = - 1 
0.01, or 10-* = - 2 
and so on. 

The logarithms of all intervening numbers, which 
are not integral powers of 10, are made up of two 
parts : a whole number, called the characteristic, and 
a decimal fraction, called the mantissa. A table of 
logarithms gives only the mantissas. 

The use of logarithms in the solution of problems 
shortens the labor of calculation, particularly in a 
series of multiplications and divisions. 

Example: Using a table of logarithms, find the 
logarithm of 174.5. 

The mantissa depends on the scries of digits, 
without regard to the decimal point. Mantissas 
are given as positive numbers in logarithm tables. 
Find the number opposite 17, in column 4 . . 2405 
Find the number in proportional parts 5, 

row 17 12 

The sum, with a decimal point prefixed, is 

the mantissa 2417 

The characteristic depends only on the decimal 
point in the number. For numbers greater than 1, 
it is a positive integer which is one less than the 
number of digits to the left of the decimal point. 
For numbers less than 1, it is a negative integer 
which is one more than the number of zeros im- 
mediately following the decimal point. In this ex- 
ample, the characteristic is 2. 

The log of 174.5 is therefore 2.2417. 

The log of 0.001745 is 3.2417. This means -3 
+ 0.2417, or written as one negative number, it 
means -2.7583. 

Example: Find the number corresponding to the 
logarithm 1.5280. 

In the table, the mantissa next lower than 5280 
is 5276, which is 4 units too small, and which 
corresponds to the digits 337. In the proportional 
parts columns of the same row 33, find 4, which is 

in proportional parts column 3. This is the four'' 
digit in the number. The sequence of digits is 33 o. 
Since the characteristic is 1, there are two digits 
to the left of the decimal point, and the number 
is 33.73. 1 

To multiply numbers, add their logarithms 
and find the antilogarithm (number correspond- 
ing to a logarithm) of this sum. To divide num- 
bers, subtract their logarithms and find the 
antilogarithm of this difference. In a series of con- 
secutive multiplications and divisions, it is con- 
venient to add the cologarithms of the divisors, 
instead of subtracting their logarithms. The colog- 
arithm of a number is found by subtracting its 
logarithm from 10, and then appending 10. For 
example, the logarithm of 174.5 is 2.2417. The 
cologarithm is 7.7583 - 10. 

Example: Find the volume at standard conditions 
of 253 ml of oxygen measured at 25 C (298 K) 
and 742 mm mercury pressure. 

Volume at C and 760 mm Hg 

298 760 mm 

log 253 - 2.4031 

log 273 - 2.4362 

log 742 - 2.8704 

log 298 2.4742, colog 298 - 7.5258 - 10 
log 760 - 2.8808, colog 760 7.1192 - 10 

22.3547 - 20 = 2.3547 

Antilog of 2.3547 = 226.3, or 226 ml volume at standard 

To find a given power of a number, multiply 
its logarithm by the power desired, and find the 
antilogarithm of this product. 

Example: Find the fifth power of 15, i.e., find 15 5 . 

The log of 15 is 1.1761. Multiplying by 5, we have 5.8805 
Antilog of 5.8805 759500, or 7.6 X 10 s = 15*. 

To find a given root of a number, divide its 
logarithm by the root desired, and find the anti- 
logarithm of this quotient. Note that if the char- 
acteristic is negative, the entire logarithm must be 
changed to a negative number before dividing by 
the root. 

i It is possible, by interpolation in the proportional parts columns, 
to determine the logarithm of 5-digit numbers, and vice versa. How- 
ever, most experimental work in this course is significant to three 
figures only, so that it is not even necessary to use the proportional 
parts columns at all. 



iple: Find the cube root of 0.00248. 

Log 0.00248 - 5.3945 - 3 + 0.3945 - - 2.6055 

i log -2.6055 - - 0.8685 - 1.1315 
Antilog 1.1315=0.1354, which is the cube root of 0.00248. 

(An alternate and perhaps simpler procedure is to trans- 
form the number 0.00248 to 2.48 X 10~, and then take the 
cube root of each of these factors separately.) 

F. The Slide Rule 

The slide rule scales which are used for multi- 
plication and division are graduated in lengths 
proportional to the logarithms of the digits from 
1 to 10, but the digits are placed on the scale, 

rather than the logarithms. Hence, to multiply 
numbers add the lengths, and to divide numbers 
subtract the lengths, by sliding the movable scale 
so that the lengths can be added or subtracted 
mechanically. The details of operation, whicli are 
quite easily mastered, are given in the manual 
accompanying the instrument. 

Since a course in chemistry necessitates a con- 
siderable amount of calculation, you are encour- 
aged to obtain and use a slide rule. An inexpensive 
10-inch rule is satisfactory. It may be read to 
about three significant figures, which is sufficiently 
precise for most calculations in this course. 

Drill Problems 

(You may refer to the answers to some of these problems, on P. 341, after solving them by yourself.) 

1. Express in the exponential form: 

(a) 1000 

(b) 200 

(c) 5,500 

(d) 0.1 

(e) 0.001 

(f) 0.004 

(g) 0,000000000003 
(h) 750,000,000,000 
(i) 2,006,000 

(j) 0.00204 

2. Change the decimal point in these expressions, 
without changing the value of the number, so that 
"%tre is one digit to the left of the decimal point. 

*** (a) 42.6 X 10* (e) 0.000465 X 10* 

(b) 41075 X 10~* (f) 6,023 X 10 2 ' 

(c) 0.375 X 10~ 4 (g) 30103 

(d) 0.07287 X 10 2 (h) 0.625 X 10 4 

3. Simplify these expressions as indicated, by 
multiplying, dividing, taking a power, extracting 
a root, adding, or subtracting. 

(a) 10* X 10 4 

(b) 10' X 10* X l<r 2 (k) 

(c) (4.6 X 10 4 ) - (3 X 10') 

(d) (5.42 X 10- 1 ) + (1.3 X 10 2 ) (1) 

(e) (10') 2 (10 2 )(10- 5 ) 

(f) 10V10 4 

(g) (10 2 )(10-')/(10->) 
(h) (2X10<)(4X10) 
(i) (2.5X10')* 

0) (25XiO') l 

(10')(16 X 10 6 ) 1 /' 

(3X10*)(2.3Xl(r 4 ) 

(x 2 )(y)(2x*) 

4. Express the following scientific laws as mathe- 
matical equations: 

(a) The pressure (P) of a gas, at constant volume 
(V), is directly proportional to its absolute tem- 
perature (T). (Solution: Stated as a direct propor- 
tion, we have: P T. Now by replacing the pro- 
portionality sign by an equal sign and a propor- 
tionality constant, we have the equation: 
t> - kT.) 

(b) The kinetic energy of a particle in motion is 
equal to one half the product of its mass times the 
square of its velocity. 

(c) The atomic number (Z) of an element is 
inversely proportional to the square root of the 
wave length (X) of its characteristic X-ray spectra. 

(d) Two electrically charged particles (ions) of 
opposite charge will attract one another with a 
force which is directly proportional to the product 
of their charges (ei and e 2 ), and inversely propor- 
tional to the square of the distance (d) between 

(e) At constant pressure, the volume of a gas 
changes by 1/273 of its volume at C, for each 
degree of temperature change. 

5 (a) At constant temperature, the masses (mi 
and m 2 ) and the squares of the average velocities 
(vi and v 2 ) of the molecules of two gases are in- 
versely proportional to one another. State as an 
equation, then use to solve the following problem. 

(b) If the average velocity of methane mole- 
cules (CH 4 ) is 4 X 10 6 cm/sec at a given tempera- 
ture, what will be the average velocity of sulfur 
dioxide molecules at this same temperature? 

6. Look up the logarithm of each 

(a) 146.8 

(b) 7.408 

(c) 0.003682 

(d) 50 

(e) O.G023 X 10 24 

(f) 1.8X 

7. Find the antilogarithms of which the follow- 
ing are the logarithms: 

(a) 2.4829 

(b) 1.0654 

(c) 4.5542 

(d) 9.8451 - 10 

(e) 6.7410 - 10 

(f) 18.6275-20 



8. Solve the following by logarithms, and then 
check your answers by the slide rule : 

(a) The volume of a sample of gas at standard 
conditions : 

(a) (6.5 X 10- 1 ) 2 

(b) (3.75 X 10 2 )' 

(c) (4.025) 2 (1.234) 

(d) (31 25) 1 " 

(e) (3.1 X 10 2 ) 3 (1.4 X 

(f) (0.02478) l/2 

(g) (8.1 X 10") 1 ' 2 

(h) (4.9 X 10~ 3 ) I/3 (3.0X 10 4 ) 

3050 ^ TG 
(b) The number of molecules in a drop of water: 

J!^!?LL x 0.6023 X 10 24 molecules/mole 
18 g/mole 

9. Use the simplest procedure practical to solve 
the following: 

10. Find the value of x in the following binomial 
expressions : 

(a) x 2 +6x- 12= 

(b) 2x 2 + 7x - 14 = 

(c) x 2 + (1.7 X 10- fi )x - (1.7 X 10~ 6 ) - 

(d) x(x + 0.01) = 3 X 10" 8 


The following is a 10-minute multiple-choice test 
which may be used to check your ability to handle 
some fundamental mathematical operations. Score 
each correct answer on part A as one point and on 
part B as two points. Select the one best answer by 
encircling one of the numbers at the right. 

Part A. Mathematical Operations 

1. d What does m equal? 


(1) d/v, (2) vd, (3) v/d, (4) d - v, (5) - 


2. x 2 = 4 + 4y. What does x equal? 
(l)4y-2, (2)2y-6, (3)4y+6, 
(4)2y+2, (5)2-2y 

2 1 

3 % . Reduce to a common denomi- 

x y 








x- y 

(5) 2y - x 



4. y (y b + a). Simplify. 
(1) 2y + b - a, (2) a - b, 
(3)2y-b+a, (4) b - a, (5)b + a 

5. x 2 9 = 16. What does x equal? 
(1)1, (2)13, (3)vT, (4)7, (5)5 


4 X 1Q 1 X 6 X KT 1 
3X 10* 

What does y 


(1) 0.8, (2) 8, (3) 80, (4) 800, (5) 800,000 

7. What is the square root of 16,900? 

(1) 13, (2) 43, (3) 130, (4) 430, (5) 1300 

8. What is 0.0096 divided by 0.06? 

(1) 16, (2) 1.6, (3) 0.16, (4) 0.016, 
(5) 0.0016 
0. What is 8% of 12? 

(1) 0.96, (2) 1.5, (3) 9.6, (4) 0.015, 
(5) 0.066 




10. 7 is 5% of what number? 
(1) 0.35, (2) 0.00714, (3) 14, 
(5) 140 


(4) 35, 

Part B. Setting Up Mathematical Relation- 

11. A table is half as wide as it is long. What 
is its area if x represents the width of the 

table? 12345 

(1) 3x, (2) x 2 , (3) 6x, (4) 2x 2 , (5) 4x 2 

12. If six-tenths of a ton of coal costs $8.40, 
what would a five-ton load of coal cost? 

Choose the proper set-up. 12345 

(1) $8.40 X 5 X 0.6, 


Q ^n 

ON I* 

W $8.40 ' 


. . U_7. TCV/ S\ V.VJ 


$8.40 X 5 


13. The density of mercury is 13.6 g/cm 8 . 
How much would 18 cubic centimeters 
of mercury weigh? 


> KW' 


"18" ' 

(3) 13.6 X 18, 



(5) none of the above. 

13.6 X 8 

14. The pressure of a gas in an automobile 
tire increases as the absolute tempera- 
ture is increased. The mathematical 
statement of this is: 

(1) P= Y> (2)P-kT, (3)T=~> 

(4) PT - k, (5) ~ = kT 

15. An automobile traveling at a velocity of 
y miles per hour consumes gasoline at 
the rate of x miles per gallon. How much 
gasoline is required to travel a distance 
of z miles at this rate of speed? 

(1) |- (2) yi (3) xyz, (4) j, 





Physical Properties of Substances. 


College Che mis fry, Chapters I, 2 

Review of Fundamental Concepts 

The Kinds of Matter 

Carefully study Figure 2-1, Page 14, and also re- 
view the definitions as stated in your text, as to 
the exact meanings of the following terms which 
classify and describe the kinds of matter: 

A substance all samples of which are identical 
in composition, as water, baking soda, and common 
salt. A mixture made up of more than one sub- 
stance. If the mixture is so intimate and homo- 
geneous that the several components are present 
as one phase, without separate boundaries, we 
speak of it as a solution. 

Properties the qualities or characteristics which 
distinguish one substance from another. Physical 
properties those which can be observed without 
changing the identity of the substance, as color, 
odor, density, melting and boiling points, crystal- 
line form, hardness, malleability, ductility, and 

thermal and electrical conductivities. Cfiemical 
properties those which describe the changes oc- 
curring when a substance is being transformed into 
one or more different substances. A test for a sub- 
stance a property sufficiently specific to identify 
it, particularly if interfering substances have been 

In this experiment we shall observe some of the 
physical properties which characterize the several 
types of substances, while Experiment 3 will in- 
troduce some of their chemical properties. The 
representative substances studied will include two 
metallic elements (copper and zinc), two non- 
metallic elements (sulfur and iodine), an acid 
(hydrochloric acid), a base (calcium hydroxide), 
and several salts (cupric carbonate, cupric nitrate, 
potassium chlorate, and others). 

Experimental Procedure 

Chemicals: Cu (turnings), Zn (mossy), S (roll), I 2 , IICI, 
Ca(OH)i (solid), CuC0 3 , Cu(NO 3 ) 2 , KC1O 3 , and other salts for 
crystallization, as NaCl, NaNO 3 , KNO 3 , CS 2 . 

Reread the laboratory rules on Page ix. Note 
again in Figures i-6 and i-7 the suggestions on 
handling and removing chemicals from the reagent 
bottles so as to avoid spilling them. Do not waste 
chemicals by removing from the bottle more than 
you need. Do not attempt experiments not called 
for without consulting the instructor (Safety Pre- 
caution No. 1). 

For each of the reactions which follow, fill in the 
table in the report sheet as you obtain the data. 
The chemical formula may be learned from the 
label on the bottle, or from your text. (Although 
you may not yet know the significance of these 
formulas, learn now to associate the appearance of 
the substance, its name, and correct formula.) 

Follow the suggestions below in observing each 
characteristic property for each substance in the 

1, Color and Physical State. Record accu- 

rately and concisely the color and general physical 
form and appearance of each substance. 

2. Odor and Taste. Smell cautiously by fanning 
the vapors toward the nose (Safety Precaution 
No. 8). Never taste chemicals except when di- 
rected to do so, and then only by touching the 
substance to the tongue. Do not taste copper salts 
as they are somewhat poisonous. The other sub- 
stances used in this experiment may be tasted, but 
first dilute the hydrochloric acid very much a few 
drops of acid in 5 ml of water. If the sour taste is 
insufficient to observe, add a little more acid. 

3. Density and Melting Point. The density of 
sulfur was determined in Experiment 1. No further 
experimental work is provided here. Note 
pare the values of the densities, and also of the 
melting points, which are given for your informa- 
tion in the report sheet. Note that some of the 
compounds decompose into simpler substances, 
rather than have a melting point of their own. A 
number of problems on density are included in the 
report sheet. 





All aamples of a, given substance 
, have the same composition. _ 


Each composed of one kind of 
atom i.e. of same atomic number 

Molecular combinations of more 
than one element 

METALS " malleable, ductile, 
good conductors of electricity 
And heat, form positive ions 

e.g. Sodium, 

Copper -"^ 


metallic and non-metallic 
eg. Silicon, 

Carbon t 


ACIDS 'taste sour, 

turn litmus red i^ 

Nitric acid, c 
Sulfuric acid 



conductors of heat and 
electricity, form negative ions 

<Lg. Oxygen, 

of definite 
of two or more 
elements are 

BASES-tarte brackish, o| 
2^ turn litmus blue *o 
I*Y e.g. Caustic Soda, 8 

SALTS- no common ^ S 
taste 2 | 

e.g. Alum, 5/lvcr nitmte, Jj ^ 
^ 5odium chloride S 5 


(cova)ent compounds) 

e.g. Ammonia (fas), 


Variable combinations 

of substances form 


both electrolyte* and non- 

e.g. Fats, Starch, 
Protein Alcohol, 

Aeneous mixtures of 
substances in varying* 

e.g. Air, 
Brass ; 

COLLOIDS -minute par- 
ticles of substances sus- 
pended in other substances 

e.g. Glue, India Ink, 

Egg white, 


aggregations of discrete par- 
ticles of substances or mixtures 

e.g. Wood, 


FIG. 2-1. The kinds of matter. 



4. Solubility in Water. Determine the relative 
solubility (designate only as soluble, fairly or 
slightly soluble, or insoluble) of each substance 
with which you are not previously familiar by shak- 
ing an amount equal to the size of a pea with about 
5 ml of distilled water in a clean 15-ml test tube 
for about 3 to 5 minutes* Do not waste distilled 
water by withdrawing more than you need. It is 
an expensive chemical. If unable to tell, by the 
disappearance of the solid or the color of the solu- 
tion, whether much or any of the substance has 
dissolved, proceed as follows: Filter the mixture 
(see instructions on filtration, Fig. i-11) into a 
clean evaporating dish or watch glass, and evapo- 
rate a few drops of it to dryness. A watch glass can 
only be used high above a small flame, or it may 
break. See Figure 2-2. The amount of residue in- 
dicates the relative solubility. 

Test material 1C 
dissolved in X 
water "^ 

\Vktch glass .? 

Wire gauze to 
distribute heat 

Do not bott. 

Low flame 

FIG. 2-2. The evaporation of a small amount of liquid in a 
watch glass to test for the presence of dissolved salts. 

5. Crystalline Form. Prepare crystals of as 
many of the substances below as your instructor 
directs. Observe, with a low-power magnifying 
glass, the characteristic habit of crystallization of 
each type of crystal (you may also observe crystals 
of different substances prepared by your class- 
mates), and describe each in your report sheet. If 
practical, include a sketch of the crystals. While 
observing the various crystals, be thinking of any 

supporting evidence of the atomic theory which 
the formation and growth of crystals suggests. 

(a) Rhombic Sulfur. Place about 2 g of crushed, 
roll sulfur 1 and about 5 ml of carbon disulfide in 
a 15-ml test tube. (Caution: Do not handle carbon 
disulfide within five feet of any flame\ its vapor, 
mixed with air, is very explosive.) Mix gently for 
about 5 minutes to promote solution. Fold a filter 
paper in the usual manner and hold this (without 
a funnel) directly over a watch glass to filter the 
mixture. Set the watch glass aside in a safe, quiet 
place, until the liquid has evaporated. Observe the 

(b) Monoclinic Sulfur. Sulfur also crystallizes, at 
an elevated temperature from the molten material, 
as monoclinic crystals. The equilibrium tempera- 
ture of the two allotropic forms is 95.5 C. Such a 
transition temperature is a definite physical prop- 
erty of a substance, analogous to the melting point, 
or boiling point. 

Prepare a cone of filter paper, and support it 
either in a funnel or a small beaker. Heat a 15-cm 
test tube, which has been filled two-thirds full of 
sulfur, slowly and uniformly so as not to superheat 
any portion of it. The sulfur will darken in color if 
it is superheated. This can be avoided by moving 
the test tube, held in a test tube clamp, in and out 
of the flame. When the sulfur is just melted it 
should be a light yellow, straw colored liquid. Now 
pour this into the filter cone previously prepared, 
and with a match stick in hand, watch it, while 
cooling, for the formation of long needle-shaped 
crystals. Just as the surface begins to solidify, 
break it open with the match stick, and quickly 
pour the remaining molten sulfur into a beaker of 
water. Let the filter cone cool, break it open, and 
observe the crystals. 

(c) Iodine. Iodine has a very appreciable vapor 
pressure at temperatures considerably below its 
melting point, so it may be easily sublimed (caused 
to crystallize directly from the vapor without first 
liquefying) from a warm surface to a cooler one. 
(Due to the higher cost of iodine, the instructor 
may designate a limited number of students to 
demonstrate this.) 

Put about 1 g of iodine in an evaporating dish, 
placed on a wire gauze and ring support. Over the 

1 Flowers of sulfur is not satisfactory for this purpose. 



evaporating dish, place a watch glass which is 
partially filled with water to act as a cool, con- 
densing surface. See Figure 2-3. Now, very gently 

Iodine crystals 
growing on bottom 
I of watch 

filled with 

Iodine being 
sublimed in 
bottom of 

FIQ. 2-3. Apparatus for the recrystallization of iodine by 

and with a small flame, warm the evaporating dish. 
Do not rush the process, as the growth of larger 
crystals is favored by slow growth with no dis- 
turbance of the vessel. When a sufficient time has 
elapsed, cool the vessel and observe the crystals, 
(d) Various Salts. Prepare a small amount of a 
concentrated solution by shaking about 3 g of the 
salt (sodium chloride, potassium chlorate, potas- 
sium nitrate, sodium nitrate, etc.) with about 5 ml 
of water in a test tube for about 5 minutes. Let any 
undissolved salt settle, and decant (Fig. i-12) the 
clear liquid on to a clean watch glass. Set aside in 
a quiet place overnight or until the next laboratory 
period, to permit the water to evaporate. Observe 
the crystals. (Ordinary table salt frequently con- 
tains well developed crystals which may be ob- 
served under low magnification. Rock salt, also, 
may contain excellent crystals. Optional: Some 
student may try crystallizing sodium chloride from 
a small amount of a 30% urea solution, instead of 
from water, to see if he can obtain the octahedral 
facial development of salt crystals.) 

REPORTs Exp. 2 

Physical Properties of Substances 



Locker Number- 




















Hydrochloric acid 1 


(No entry) 

(No entry) 

Calcium hydroxide 


dec. 580 
-H 2 

Cupric carbonate 



Cupric nitrate 


dec. 170 

Potassium chlorate 



i Hydrochloric acid is a solution of the gas hydrogen chloride, in water. The "dilute" laboratory reagent contains 21.9 g/100 ml of solution. 
Crystalline Form. Characterize and sketch the various crystals which you have observed. 

What supporting evidence of the atomic theory does the formation and growth of crystals suggest to you? 


Problems on Density 

Introductory note on the solution of problems. See Appendix I on the "Use of Dimensions" in the solution of 
problems, also for examples of the solution of density problems. 

In each problem label each quantity with its appropriate unit; set up the problem to indicate all mathematical 
operations, but omit the actual multiplication and division. The formula to be used in the solution is not sufficient. 
Actual quantities must be substituted for each symbol, as in the examples cited. The entire problem should be neat 
and legible. Record the answer in the space at the right. Good form is essential in your work. 

1. What is the density of a brass sample if 60 g of the coarse turnings, when placed in a 
graduated cylinder containing 10.3 ml of water, raise the level of the meniscus to a 
reading of 16.2 ml? (Consider that 1 ml = 1 cm 8 .) 

2. What is the weight in pounds of a cubic foot of mercury, whose density is 13.6 g/cm 8 ? 
(See Appendix II, Tables II and III, for any needed conversion factors. 

3. An experiment calls for 50 g of concentrated sulfuric acid, density 1.84 g/cm 8 . Suppose 
no balance is available and you decide to use a graduated cylinder. What volume 
should you use? 

4. The dilute nitric acid on the desk has a density of 1.19 g/cm 8 and is 32% l nitric acid, 
the remainder being water, 
(a) How many grams does 8.0 ml of this acid weigh? 

(b) What is the weight of pure nitric acid in this 8.0 nil? 

5. What volume of magnesium, density 1.74 g/cm 8 , would be equal in weight to 100 crn 8 
of lead, density 11.4 g/cm 8 ? 


_g/cm 8 


4 (a). 


1 The percent composition of solids and liquids is given as percent by weight unless percent by volume is specifically stated. The compo- 
sition of gases is usually expressed as percent by volume, however. 


Some Elementary Chemical Properties of Substances. 

Co//ege C/iem/tfry, Cfcapfm 3, 4 

Review of Fundamental Concepts 

This experiment continues the study of proper- 
ties of the same list of substances used in Experi- 
ment 2, this time from the standpoint of typical 
modes of chemical behavior, resulting in the forma- 
tion of new substances. 

Physical and Chemical Changes 
Taking Place on Heating 

When a substance is heated in the air, it may: 

(1) Change its physical state from solid to 
liquid, liquid to gas, or directly from solid to gas. 

(2) React chemically with a constituent of the 
air (usually oxygen) to produce one or more new 

(3) Decompose to produce simpler substances. 
Elements, of course, ordinarily cannot decompose. 1 
The products of a decomposition must contain all 
the elements originally present. None can be de- 
stroyed, and no additional elements can be present 
save possibly some constituent of the air with 
which there has been a reaction. 

The physical changes which accompany a given 
chemical change often serve to identify it. Thus, 
mercuric oxide (a red powder) decomposes to 
form mercury (a silvery metallic liquid) and oxygen 
(a colorless gas). On heating calcium carbonate 
(limestone), two simpler compounds result, cal- 
cium oxide (quicklime) and carbon dioxide (a 
colorless gas). Carbon (e.g., coke), heated in air, 
forms carbon dioxide, the oxygen coming from the 

Properties of and Tests for 
Some Common Gases 

Use the descriptions below to help you identify 
any gaseous products formed during the course of 
the experimental procedure that will follow these 

(1) Oxygen, O 2 : Colorless and odorless. Causes 
the glowing end of a wood splint to increase greatly 
in brightness or even burst into flame. This test 
may be made by thrusting the glowing end of a 
splint into the gas being tested. Only one other 

1 We are not considering here the unusual forces brought into play 
in modern research to smash atoms, by which different elements, with 
different properties, are formed. Such changes are not ''chemical 
changes*" They are called nuclear transformations. 

common gas, nitrous oxide, N 2 O, has similar prop- 

(2) Hydrogen, H^: Colorless and odorless. One 
of a number of gases which burn when ignited in 
the presence of air or explode when a mixture of 
the gas and air is ignited. With a lighted match 
held near the mouth of the tube, one can detect the 
presence of hydrogen. 

(3) Carbon dioxide, CO 2 : Colorless, odorless, with 
a slight acid or sour taste if in solution. Recognized 
by the milky precipitate of calcium carbonate it 
forms when exposed to calcium hydroxide solution 
(limewater). This may be observed by suspending 

Lime water 

If the evolved 
gas isCO 2 the 
limewatev will 
turn milky. 

FIQ. 3-1. The limewater test for carbon dioxide. 

over a. 
low flame 

a drop of limewater from a glass tube or rod held 
in the mouth of the test tube containing the sub- 
stance being tested, and noting whether it turns 
milky. (Touch the glass rod to some limewater in 
a test tube. Never put a rod, medicine dropper, etc. 
directly in a reagent bottle.) A better test is to 
connect the test tube containing the test material 
with a rubber stopper and bent delivery tube (see 
directions on glass bending, P. xix). Have the end 
of the delivery tube dipping in some limewater in 
a 10-cm test tube. On heating the substance, any 
C(>2 will be expelled into the limewater. Or, if 
testing for CCfe when an acid is added, disconnect 
the rubber stopper just long enough to add a little 
HC1, reconnect it at once, and if necessary warm 




a little to assist in expelling the evolved gas into 
the limewater. 

(4) Sulfur dioxide, SO 2 : A heavy, colorless gas 
with a sharp choking odor. 

(5) Nitrogen dioxide, NC^: A heavy, brown gas 
with a disagreeable odor. 

(6) Hydrogen sulfide, H 2 S: A colorless, poisonous 
gas, with a very characteristic foul odor. 

Chemicals: Cu (turnings), Zn (mossy), Zn (dust), S, I 2 , HC1, 
Ca(OH) 2 (solid), Ca(OH), (sat. sol.), CuCO,, Cu(NO,)* 
KC10,, Mg (ribbon). 

1. Reaction to Litmus. Test solutions of each 
of the soluble substances in the above list of chemi- 
cals with red and with blue litmus paper. Litmus 
turns red in an acid solution, and blue in a basic 
solution. (A blue color with iodine solution is due 
to starch in the paper. It is not basic.) 

2. The Effect of Heat on Certain Elements* 
Heat each of the following elements, as directed. 
In each case, note any physical changes, and also 
any evidence that a new substance may have been 

(a) Copper hold a small piece of copper sheet, 
or turnings, with the forceps at the top of the Bun- 
sen flame for a short time. 


FIG. 3-2. The ignition of sulfur in a crucible. 

(b) Sulfur heat a bit of sulfur the size of half a 
pea in a porcelain crucible, until it first melts, then 
catches fire as it becomes hotter. Note the odor 
cautiously. Continue heating until the sulfur is all 

Experimental Procedure 

burned out of crucible. Keep crucible under a hood. 
(c) Iodine heat one or two small crystals of 
iodine in a 15-cm test tube, keeping the upper part 
of the tube cool. 

3. The Decomposition of Certain Com- 
pounds by Heat. Place small samples each of 
cupric carbonate, cupric nitrate, and potassium 
chlorate in separate test tubes. Before you heat 
them, be prepared to test for any gases which you 
think may be formed. (See the previous discussion 
on properties of and tests for common gases, P. 19.) 
Note the changes occurring on heating, and iden- 
tify the products, both gaseous and solid. The 
white residue left on heating potassium chlorate 
contains no oxygen, and is potassium chloride. 

4. The Effect of an Acid on Various Sub- 
stances. Place small samples each of copper, zinc 
(use mossy zinc), sulfur, calcium hydroxide (solid), 
and cupric carbonate in separate test tubes. Be 
prepared to test for any gases you might anticipate. 
Add 3 to 5 ml of dilute hydrochloric acid to each. 
Note changes; identify gases, if any, that form. 

5. The Reaction of Zinc and Sulfur. 1 Mix 

thoroughly an estimated 2 grams of zinc dust 
(about 1 cm deep in a dry 10-cni test tube) with 
about twice its volume (1 gram) of powdered sul- 
fur. Place half of the mixture in a 15-cm test tube 
and add dilute hydrochloric acid. What gas is 
evolved? Place the remainder on a 7-cm square of 
asbestos paper, or in your evaporating dish, and 
ignite it as follows. Hold a 4-cm length of mag- 
nesium ribbon with your forceps; light this by 
heating the tip in the Bunsen burner, and at once 
touch this burning magnesium to the zinc and 
sulfur mixture. (Use reasonable care, and do not 
get too close, as it burns rapidly.) To gain addi- 
tional evidence as to any chemical change which 
has taken place between the zinc and sulfur, put 
this residue in a test tube, and add a little dilute 
hydrochloric acid. Note the odor. What is formed? 

'To avoid fumes in the laboratory instructors may prefer to 
demonstrate this. 

REPORT: Exp. 3 

Some Chemical Properties of 



Locker Number- 

1. Reaction to Litmus. List the substances whose solutions arc: Acidic- 



2. The Effect of Heat on Certain Elements. Note any changes in physical state, color, odor, and so forth, 
and name any new substances produced on heating: 



New Substances 




3. The Decomposition of Certain Compounds by Heat. Note any changes in physical state, color, odor, 
and so forth, and any tests performed on gaseous products in order to identify them, on heating the substance: 


Observations, Including Tests for A ny 
Gaseous Products Evolved 

New Substances 




4. The Effect of Acid on Various Substances. Note any observable effects, such as solution of the 
solid, color of the solution, tests performed, and results, on any gases evolved, when hydrochloric acid is added 
to the substance: 


Observations^ Including Tests Performed 

Identity of Gas t if Any 







5. The Reaction of Zinc and Sulfur. State clearly all the evidence in the experiment which leads you to 
believe that a chemical change did, or did not, take place when 
(1) the zinc and sulfur are mixed, before igniting, 

(2) the zinc and sulfur mixture was ignited. (Caution : Follow experimental directions.) 

Interpretation of Data 

It is important in drawing conclusions from experimental data that you consider all the facts observed, but also 
that you not generalize beyond the data. Let us consider a hypothetical case: 

Example. Six metals were treated with dilute hydrochloric acid. In every case a gas was evolved which, on 
mixing with air, could be exploded with a sharp report. 

Possible Generalizations: 

1 . Hydrogen can be formed by the action of hydroohlorir acid on certain metals. True. The behavior noted is 
characteristic of hydrogen. A sufficient number of cases is given to justify this conclusion. 

2. Every metal will displace hydrogen from hydrochloric acid. Insufficient evidence to judge. Not all metals were 
tried, under all possible conditions. 

3. When a metal is treated with any acid, hydrogen is liberated. Insufficient evidence. One cannot generalize 
about other acids which were not tried. 

4. The gas liberated when the above metals were treated with hydrochloric acid was carbon dioxide. False. 
Contrary to data. While no test for carbon dioxide was made, compounds of carbon were not even included in the 

For each of the following statements, circle the T if sufficient evidence is presented in Experiments 2 or 3 to 
justify the statement; circle the F if contrary to the data; or circle the I if insufficient evidence is given to decide. 

1. All metals when treated with an acid liberate hydrogen T I F 

2. Most metals unite with oxygen when heated in the air T I F 

3. All compounds may be decomposed by heating them T I F 

4. All compounds containing oxygen decompose to yield oxygen gas when heated. (The symbol 

for oxygen, O, with or without, a subscript, appears in the formula of any compound containing oxygen.) T I F 

5. The determination of any one specific property is always sufficient evidence to identify a 
substance T I F 

6. Calcium hydroxide is more soluble by reaction with hydrochloric acid solution than by solu- 
tion in water. (See Experiment 2.) T I F 

7. The aqueous solution of every substance not an acid or base is neutral to litmus T I F 

8. All salts are soluble T I F 

9. The disappearance of sulfur on heating always indicates its complete change into different 
substances T I F 

10. The decomposition of potassium chlorate by heat is a good example of one of its chemical 
properties T I F 

11. Copper oxide will dissolve in hydrochloric acid T I F 

12. Crystals of a substance are shaken with water, and some crystals remain after shaking. The 
substance is insoluble T I F 


The Packing of Atoms or Ions In Crystals. 


College Chemistry, Chapter 2 

Review of Fundamental Concepts 

The solid state of matter nearly always consists 
of a regular arrangement of atoms, molecules, or 
ions. If we represent each atom or group of atoms 
as a point, then the simplest configuration of 
points which has all the elements of symmetry of 
the whole crystal is called the lattice. The lattice 
may be looked on as the unit structure which, re- 
peated indefinitely, gives the crystal. The number 
of nearest neighbors of a particle in a crystal lattice 
is called the coordination number. 

Crystal Structure of the Elements. Most of 
the metallic elements crystallize in one of the 
following structures : 

A. Body-centered cubic packing, shown by the 
alkali metals Li, Na, K, lib, and Cs, and by some 
of the transition metals in groups Va and Via, such 
as V, Nb, Ta, <*-Cr, Mo, 0-W, and jS-U. 

B. Cubic closest packing , or face-centered cubic, 
shown by the noble metals Cu, Ag, and Au, some 
of the transition metals in group VIII, and a-Ca, 
Sr, and AL 

C. Hexagonal closest packing, shown by Mg, Be, 
and 7-Ca in Group II, and Ti and Hf in Group IVa. 

A substance may crystallize in more than one 
form, as indicated above by the Greek-letter pre- 
fixes, but in general each form is stable within a 
definite range of pressure and temperature. 

In the case of the non-metallic elements, niole- 
cules such as C1 2 , T?*, and S 8 constitute the building 
units of the crystals. 

The inert gases have an atomic lattice with 
either cubic closest packing, as A, Ne, and Kr, or 
hexagonal closest packing, as He. 

Crystal Structure of Compounds. Chemical 
compounds may be conveniently grouped into two 
broad classes: molecular crystals and ionic crys- 

Molecular crystals contain molecules such as 
water, H 2 O, carbon dioxide, CO 2 , hydrogen chlo- 
ride, HC1, arid sugar, CuHnOii, as building units. 
Here the atoms are held together by covalent 
bonds within the molecule, and the forces holding 

these neutral molecules together are relatively 

Ionic crystals contain positively charged par- 
ticles called cations or negatively charged particles 
called anions as the building units. The cation may 
be a positively charged atom of a metal, such as 
Na+, Ca+ +, or Ag+, or it may be a complex cation 
such as Cu(NH 8 )4++. The anion may be a nega- 
tively charged atom, such as Cl~, or a complex 
anion, such as NO 8 ~ or PtCl 6 . Most inorganic 
salts, such as sodium chloride, NaCl, magnesium 
bromide, MgBr 2 , potassium nitrate, KNO 3 , and 
barium sulfate, BaSO 4 , are members of this class. 
The structure of an ionic crystal depends on the 
relative sizes of the ions composing the crystal 
lattice, and it is possible to determine from geo- 
metrical considerations whether the number of 
nearest neighbors expected is three, four, six, or 
eight for a given ion. For example, if the ratio of 
the radius of the cation to that of the anion is 
greater than 0.73, eight anions can touch a given 
cation; if the ratio is less than 0.73 but greater 
than 0.41, only six anions can touch a given cation. 
The former case is typified by the lattice of the 
salt Cs+Cl- and the latter by the salt Na+CK 

It is the purpose of this experiment to help the 
student get some understanding of the various 
ways that spheres may be packed to form some of 
the typical crystal arrangements. It must be re- 
membered that when solid spheres are used to 
represent the relative sizes and shapes of atoms or 
ions, we are giving them a rigidity which they do 
not possess. The cloud of electrons surrounding the 
nucleus of an atom does not have a definite bound- 
ary, nor does its influence cease at a definite dis- 
tance from the nucleus. 

As the experiment is performed, it is well to 
remember the extent of the enlargement of the 
atoms or ions from their actual radii of 1 or 2 
Angstrdin units (10~* cm) to those of the spheres 
used in the models. Also bear in mind the large 
number of partides (Avogadro's number, 




6.023 X 10 23 ) actually involved in forming the 
gram-atomic weight of a metallic element or the 
gram formula weight of an ionic compound. 

It may also be profitable for you to review some 
of the simple geometric relationships involved in 
three-dimensional objects such as a cube and a 
sphere. For example, how many vertices, how 
many edges, how many faces does a cube have? 

What is the relationship between the edges of a 
cube and a face diagonal? What is the relationship 
between the body diagonal, the face diagonal, and 
the edge of a cube? How may a tetrahedron be 
constructed within a cube? What is the volume of 
a sphere in terms of its radius? Some simple three- 
dimensional sketches should help you to deter- 
mine the relationships desired. 

Experimental Procedure 

Special supplies: A set of cork balls (27 one-inch, 8 three- 
fourths-inch and 13 one-half-inch size); 36 pins which have 
had the heads cut off obliquely (may be conveniently stored 
in a cork) ; a small pair of pliers. 

1. A Study of Metallic Crystals 

Model A. Body-centered cubic packing (not 
closest packing). Construct the layers illustrated 
in Figure 4-1, using one-inch cork balls. Always 
insert the cut-off end of the pin into a ball with the 
aid of the pliers, and the other bull can then be 
easily attached to the sharpened end. Try to save 
wear and tear on the corks by inserting and pulling 

FIG. 4-1. Layers of atoms for the body-centered 
cubic arrangement. 

out the pins cleanly. The layers may be assembled 
without the use of pins by placing the single ball 
in the depression formed by the four balls of the 
first layer. Then place the third layer so that the 
four balls are directly over those in the first layer. 
As you study this model, you should realize that 
in an actual crystal there would be an infinite set 
of these unit cubes with common corners. With the 
model before you answer the questions in the report 

Model B. Face-centered cubic packing 
(closest cubic packing). Construct the layers illus- 
trated in Figure 4-2. Note that this time the balls 
are in actual contact with one another. Assemble 
the three layers without the use of pins. Place the 
second layer in a position so that the four balls all 
contact the center ball of layer one and fit between 
the outer balls. Then place the third layer so that 

its balls are directly over those of the first layer. 
Study this unit cell as you answer the questions 
in the report sheet. Save this model for comparison 
with the model for hexagonal closest packing. 

FIG. 4-2. Layers of atoms for the face-centered 
cubic arrangement. 

Model C. Hexagonal closest packing. Con- 
struct the layers illustrated in Figure 4-3. Place the 
first layer so that one of the vertices of the triangle 
is facing you. Then place the central ball of the 

FIG. 4-3. Layers of atoms for the hexagonal 
closest packing arrangement. 

second layer in the depression in the center of the 
first layer. Be sure that each ball of the first layer 
is in close contact with three balls of the second 
layer. Now place the third layer so that each of 
the three balls is directly over the corresponding 
ball of the first layer. Answer the questions in the 
report sheet relative to this model. 

Comparison of the two types of closest packing. 
Place Model B and Model C before you. Re- 
move the top layer of Model B. Rotate the top 
layer of Model C 60 to the right (or left) so that 
there is a layer of four balls forming a square facing 
you. Compare this view with that obtained by 



looking down at the bottom two layers of Model B. 
Place the top layer of Model B on the four balls 
facing you in Model C to see if you can construct a 
unit cell of cubic closest packing which resembles 
Model B tilted at an angle. 

2. A Study of Ionic Crystals 

Model D. The sodium chloride structure, 

which is one of the commonest ionic structures for 
salts which contain an equal number of positive 
ions and negative ions. Since the sodium ion, Na+, 
has a diameter of 1.90 A and the chloride ion, Cl~, 
has a diameter of 3.62 A, you may approximate the 
relative sizes of these ions by using one-half-inch 
balls for Na* ions and one-inch balls for Cl~ ions. 
Construct the layers illustrated in Figure 4-4. 

FIG. 4-4. Layers of atoms for the sodium 
chloride ionic arrangement. 

Arrange the three layers so that the sodium and 

chloride ions are alternately placed. Study this 
model, and answer the questions in the report 

Model E. The cesium chloride structure* 

The cesium ion, Cs+, has a diameter of 3.38 A 
compared with 3.62 A for the chloride ion; so you 
may approximate the relative sizes of these ions by 
using three-fourths-inch balls for the cesium ions 
and one-inch balls for the chloride ions. Construct 
the first and third layers as illustrated in Figure 
4-5. Place four loose three-fourths-inch balls in the 

FIG. 4-5. Layers of atoms for the cesium 
chloride ionic arrangement. 

depressions between the nine one-inch balls of the 
first layer. Then place the third layer directly over 
the first layer. Answer the questions in the report 
sheet relating to this model. 

REPORT! Exp. 4 

The Packing of Atoms and Ions 
in Crystals 

1. A Study of Metallic Crystals 

Model A. Body-centered cubic packing. 

Why is this structure called body -centered cubic? 



Locker Number^ 

What is the coordination number of an atom of sodium when packed in a crystal of this type? 


If this model were extended in all directions, what difference, if any, would there be in the relative positions 
of the corner spheres and those in the center? 

Note: In the calculations below, consider all distances to be measured between the centers 
of the balls. From a study of the right triangles involved in a cube, as illustrated, where a = the 
edge of the cube, f * the face diagonal, and b = the body diagonal, the following relationships 
should be apparent: b 2 = f 2 + a 2 , and since f 2 = a 2 + a a , it follows that b 2 = 3a 2 . 

In calculating the volumes, note that only one-eighth of each of the corner spheres is inside 
the cube which has as its corners the centers of the spheres. The volume of a sphere may be 
calculated from its radius by using the formula V = ^Trr*. 

What is the length of the body diagonal, b? inches, 

Calculate the length of an edge, a, and also the volume of the cube, a 8 : . . . a = 

Calculate the volume of the spheres inside the cell : 

What fraction of the total volume of the cube is actually occupied by the spheres in this 
type of packing? 

Model B. Face-centered cubic packing. 

Why is this type called face-centered cubic? 

What is the coordination number of a copper atom when packed in a crystal of this type? 

What is the length of the face diagonal, f ? inches, - 

Calculate the edge of this unit cell from the relation f 2 = a 2 + a 2 . Also calculate the 
volume of the cube: ' a 

a* - 

How many spheres fall wholly within this unit cell? 

Calculate the volume of the spheres inside the cell : _ 

What fraction of the total volume of the cube is actually occupied by the spheres in this 
type of packing? 





Could you distinguish between the corner spheres and those in the centers of the faces if the model were ex- 
panded in all directions? Explain. 


Model C. Hexagonal closest packing. 

How many spheres are in contact with any one sphere in this type of packing? .... 

How does the coordination number involved here compare with that in Model B with 
cubic closest packing? 

For the same radius of contact, how should the two types of closest packing compare in 

Make a brief statement comparing the packing in Model C with that in Model B. 

2. A Study of Ionic Crystals 

Model D. The sodium chloride lattice. 

Using the actual diameters of the ions in Angstrom units as given in the experimental 
procedure, calculate the radius ratio of sodium ion to chloride iori: 

How many chloride ions are in contact with any one sodium ion in this lattice? . 

How many sodium ions are in contact with any one chloride ion? 

In a crystal of table salt weighing 58.5 grams, how many sodium ions and how many 
chloride ions would be present? 

Molteii salts, such as fused sodium chloride, arc good conductors of electricity. How do 
you account for this? 

Using the actual diameters of the ions, record each of the following: 
The shortest distance between the center of a Na 4 * ion and a Cl~ ion. 

The shortest distance between two Na + ions __A 

The shortest distance between two Cl"" ions A. 

Model E. The cesium chloride lattice. 

Using the actual diameters of the ions as given in the experimental procedure, calculate 

the ratio of the radius of the Cs+ ion to that of the Cl~ ion. 

What is the coordination number of each of the ions in this type of lattice? . . For Cs+ = 

For Cl- = 
How do you account for the change in coordination number from Model D to Model E? 

Water solutions of salts are good conductors of electricity, whereas solutions of organic substances such as 
sugars are poor conductors. Explain. 


Substances in the Atmosphere. 

Co//ege Chemistry, ChaftT 

Review of Fundamental Concepts 

Look up, in your text, 1 the composition of the 
atmosphere. Note that it contains two gases in 
relatively large proportion, nitrogen and oxygen, 
three others in much lesser, but significant amounts, 
water vapor, argon, and carbon dioxide, and traces 
of several others. None of the constituents react 
readily with one another. Nitrogen and oxygen will 
combine in limited amount at extremely high tem- 

peratures, and water and carbon dioxide partially 
react in solution to form carbonic acid, H^COs. 

In the previous experiment, we have noted the 
reactivity of the oxygen in the air with a metal, 
copper, and a nonmetal, sulfur. In this experiment, 
we shall obtain qualitative evidetice of the presence 
of water, carbon dioxide, and nitrogen, and quan- 
titative evidence of the amount of oxygen in the 

Experimental Procedure 

Chemicals*: 0.1 F Ba(OH) 2 , CaCl 2 (4-mesh, anhydrous), Fe 
(filings), Mg (ribbon), 30% pyrogallol solution (freshly pre- 

1. Some Qualitative Observations 

(a) Water Vapor. Place several granules of an- 
hydrous calcium chloride on a watch glass and 
leave it exposed to the air. At the close of the 
period, or the next laboratory period, observe any 
change that has occurred. (The reasons for the ab- 
sorption of water by very soluble substances like 
calcium chloride will be considered when you study 
solutions. In extremely dry weather, calcium chlo- 
ride may absorb no water at all.) 

(b) Carbon Dioxide. Place about 15 ml of 0.1 F 
Ba(OH) 2 in a 500-ml flask. Stopper the flask, and 
shake it for a moment. Observe the change. (A 
barium hydroxide solution provides a more sensi- 
tive test for carbon dioxide than does the more 
commonly used limewater, or calcium hydroxide, 
solution, which you used in Experiment 3. The re- 
action is similar, and forms a precipitate of barium 
carbonate, BaCO 3 .) 

Just as a comparison, check on the relative 
amount of carbon dioxide in the breath, by open- 

1 Learn to use your text as a reference book. Consult the index to 
locate a topic which may be in advance of your present lecture assign- 

. * The "F ' preceding a formula indicates a solution of a specified 
concentration. Thus, a 0.1 F solution contains 0.1 of the gram formula 
weight (as defined in Experiment 10} per liter of solution. For the 
present, you only need to recognize that a 0.1 F solution is a mod- 
erately dilute solution of the substance whose formula is given. Units 
of concentration are discussed in Experiment 20. 

ing the flask, inhaling deeply, and then exhaling 
your breath through a piece of glass or rubber tub- 
ing, into the flask. Close and again shake the flask. 
Compare the relative amount of precipitate in the 
two cases. 

(c) Nitrogen. Place about 15 cm of magnesium 
ribbon, packed rather compactly, into a crucible, 
and cover it with the crucible lid. Heat this in- 
tensely with the Bunsen burner for about 15 
minutes, Cool the crucible, and replace the lid with 
a small watch glass, to the under side of which is 
attached a moist piece of red litmus paper. Lift 
this watch glass just enough to moisten the white 
residue with a few drops of water. Warm the cru- 
cible slightly. Note any change in the litmus. The 
reactions are as follows: Along with the reaction of 
magnesium to form magnesium oxide, some of the 
magnesium reacts, at high temperature, with nitro- 
gen in the air to form magnesium nitride, 

3Mg + N 2 

This, by reaction with water, forms ammonia 
(NHs) and the slightly soluble base magnesium 
hydroxide (Mg(OH) 2 ). Ammonia gas dissolves in 
water to form the weak, unstable base ammonium 

MgaN, + 6 H 2 O > 3 Mg(OH) 2 + 2 NH| 

It may be possible to detect the odor of ammonia 
gas above the moist white solid magnesium hy- 


Model C. Hexagonal closext^LEGE CHEMISTRY IN THE LABORATORY, No. 2 .Exp. 5 
How many spheres a~ - __-____ ___ 

How does th< 

cubic closest pr 

I . . *r$J Method. Absorption of Oxygen by Pyro- 

~)L Prepare the apparatus as sketched in Figure 

5-1. Remove the stopper and fill the flask, with the 

Equalize the levels inside t 

outside the flask / \ # 


Fio. 5-1. Apparatus for the absorption of oxygen from the air 
by pyrogallol solution. 

10-cm test tube still in it, with tap water. Tem- 
porarily disconnect the rubber tube and clamp, 
and re-insert the stopper and glass tube firmly into 
the flask so that no air bubbles are trapped. Re* 
move the stopper and accurately measure the 
volume of the water with your graduated cylinder, 
to obtain the capacity of the apparatus. Record 
this volume and all subsequent data, at once, in 
your report sheet. 

With the apparatus empty, but not dried, ac- 
curately measure about 20 ml of 6 F NaOH into 
the flask, and 5 to 6 ml of a 30% solution of pyro- 
gallol, 1 (also called pyrogallic acid) into the test 
tube. Record the exact volumes used. Now firmly 
close the flask with the stopper (there must be no 
leak), and re-attach the rubber tube. Avoid undue 
handling of the flask, as this would warm it above 
room temperature. When all is ready, tightly clamp 
the rubber tube just above the glass tube. Re- 
peatedly invert the apparatus, and shake it mod- 
erately, at intervals, to mix the contents. After 
about 15 minutes, during which time the oxygen 
should completely react chemically with the alka- 
line pyrogallol, invert the flask so that the rubber 
tube is completely under water in a large vessel, 
such as a pneumatic trough. Open the clamp to 
permit the entry of water, which will replace the 
volume of oxygen absorbed. With the flask cooled 
to room temperature, adjust the levels inside and 
outside the flask so they are equal, then close the 
clainp. (Again avoid handling that part of the 
flask which is exposed to the remaining air within.) 
This procedure will equalize the pressure so that 
all measurements are made under the same condi- 
tions. Place the flask upright at the table top. 

At this point in the procedure, test the character 
of the residual gas by removing the stopper just 
as you thrust a burning splint into the flask. Re- 
sults? With your graduated cylinder, carefully 
measure the volume of the brown liquid contained 
in the flask. From these data, calculate the volume 
of oxygen absorbed and the original volume of air, 
taking account of the volume of all liquids used, 
and the percentage of oxygen in the air. 

(b) Second Method. Reaction of Oxygen with Iron. 
Moisten the inside of your 50- or 100-ml graduated 
cylinder and sprinkle the walls with fine iron filings. 
Leave one side clear for observation. 2 Lower the 
inverted graduate, vertically, into a 400-ml beaker 
which is two-thirds filled with water, so that 
bubbles of air do not escape. Note that while the 
water level rises a little at once due to the in- 

1 A 30% solution means 30 g of solute dissolved in 70 g of water. 
This should be freshly prepared for class use, or you may use 2.0 g 
of pyrogallic acid dissolved in 5 nil of water. 

* A 100-ml cylinder is preferable, if available. If a 50-ml cylinder 
is used, place a rubber band about it before inverting it in the beaker, 
to assist you in marking the water level* as mentioned in the next 


creased pressure, the graduated cylinder is com- 
pletely filled, initially, with air at atmospheric 
pressure. Leave this undisturbed until the next 
laboratory period, during which interval the oxygen 
(and some water vapor) will react with the iron 
to form a hydrated iron oxide, or ordinary rust, 
Fe 2 O 3 -xH 2 O. See Figure 5-2. 

Without lifting the graduated cylinder out of 
the water, adjust the levels inside and outside so 
they are equal, and carefully read the volume of 
the residual gas. If the water level is below the 
graduation marks olume cannot be 

read directly nd carefully at 

tHe water 1* jme of residual 

md r iie cylinder, by 

iest graduation 

graduated cyl- 

ater needed to 

ad also to fill it 

te the percentage 

Fio. 5-2. The absorption of oxygen by iron filings. 

REPORT t Exp. 5 Name- 

Substances in the Atmosphere Z>a/*_ 


Locker Number, 

1. Some Qualitative Observations 

Describe briefly the qualitative evidence which you obtain for the presence in the air of: 

A. Water Vapor 

B. Carbon Dioxide 

C. Nitrogen 

2. Quantitative Determination of Oxygen in the Air 

A. First Method. Absorption of Oxygen by Pyrogallol 

1. 2. 

Volume of water required to fill flask with test tube in it ml _ ml 

Volume of GFNaOH ml 

Volume of pyrogallol solution ml 

Volume of all solutions in flask (after absorption) . . ml 

Volume of oxygen absorbed nil ml 

Volume of air originally in the stoppered flask . . ml ml 

Percentage of oxygen in the air ml ml 

Method of calculation: 

B. Second Method. Reaction of Oxygen with Iron 

1. 2. 
Volume of air originally in the graduated cylinder . ml 

Volume of residual gas in the graduated cylinder . . nil ml 

Volume of oxygen absorbed ml nil 

Percentage of oxygen in the air ml . ml 

Method of calculation: 



1. Judging from the equipment used, the technique employed, and the volumes measured, comment on the 
relative precision and accuracy of the two methods for the analysis of oxygen. 

2. What effect on the results would there be if: (1) All the measurements were carried out at a higher, but 
constant, temperature? (2) The temperature of the initial measurements of gas volume differed from that of the 
final measurements? 

3. The amount of water vapor in the air is variable, up to a maximum of 2.0% (at room temperature, 20 C). 
If dry air contains 20.99% oxygen, what would be the corresponding percentage of oxygen in this same air, when 
saturated with water vapor? (Hint: 20.99 parts of oxygen in 100 parts of dry air would correspond to 20.99 parts 
oxygen in 102 parts of moist air.) 

4. Calculate your percentage error 1 for each method of analysis for oxygen, assuming that the air you analyzed 
was saturated with water vapor, as the vessels were wet. 

(a) . 


5. The carbon dioxide in the air, 0.03%, is likewise absorbed, by the sodium hydroxide in the solution, in method 
(a) How much of an error in the calculation of the percentage of oxygen is caused by neglecting this factor? 

6. Summarize any physical and chemical properties of oxygen and of nitrogen which you have observed, or 
which it was necessary tp assume, in the performance of these experiments. 

7. List as many reasons as you can in support of the statement that the air is a mixture, rather than a com* 
pound, of its principal constituents. 

1 See Appendix I, Section C on Experimental Errors. 


The Preparation of Pure 
Substances by Chemical Changes. 


College Cfiemisfry, Chapter 4 

Review of Fundamental Concepts 

The Purity of Substances 

A really pure substance, that is one made up of 
only one kind of molecules, probably does not exist. 
Even when made with the most refined methods, 
traces of impurities are likely to be present in a sub- 
stance. For many purposes, impure substances are 
satisfactory and more economical, while in other 
cases chemicals of the purest grade obtainable are 

The principal grades used in science and industry 
are: "Commercial" or "Technical" (relatively 
crude, impure, and cheap); "Purified" (comprising 
an intermediate grade, often of a rather high degree 
of purity); "U.S.P." (meeting the standards of the 
United States Pharmacopoeia for the drug trade) ; 
"C.P." (the so-called chemically pure grade); and 
"Reagent" (comprising the purest and conse- 
quently most expensive grade, used in analytical 
and much scientific work). It is an important part 
of the training of a chemist to learn methods of 
purification and tests for purity. 

Methods of Purification 

The method of removing undesirable constitu- 
ents from an impure substance may be based on 
purely physical properties, such as differences in 
solubilities in suitable solvents, or differences in 
boiling points as in distillation procedures. Again, 
we may resort to methods in which one component 
is changed chemically into some other substance 
which has a different solubility, or boiling point, 
thus permitting a separation of the substance as a 
precipitate (solid substance), or perhaps as a vola- 
tile (gaseous) substance. 

Types of Chemical Change 

Chemical changes are characterized by the for- 
mation of one or more new substances, each with 
its own specific properties, and by the loss or gain 
of energy. Simple types of chemical changes are: 
(1) combination, in which two or more substances 
form a more complex one, e.g. the formation of 

copper sulfide from copper and sulfur; (2) decom- 
position, in which a more complex substance breaks 
down into two or more simpler ones, e.g. the de- 
composition of potassium chlorate (KC1O 3 ) into 
potassium chloride (KC1) and oxygen; (3) displace- 
ment or single replacement, in which one element dis- 
places another from its compound, e.g. the replace- 
ment of hydrogen in hydrochloric acid by zinc; (4) 
exchange or double replacement, in which two com- 
pounds interchange radicals to form two new com- 
pounds, e.g. silver nitrate (AgNO 3 ) and sodium 
chloride (NaCl) react to form silver chloride (AgCl) 
and sodium nitrate (NaNO 3 ). This last type of 
change usually occurs in solution, especially be- 
tween substances whose solutions are good conduc- 
tors of electricity. It is known that these substances 
are separated or dissociated in solution into elec- 
trically charged atoms or groups of atoms (radicals) 
called ions. Silver nitrate in solution consists of a 
mixture of silver ions and nitrate ions. Similarly a 
sodium chloride solution contains sodium ions and 
chloride ions. The formation of the precipitate of 
silver chloride consists of a combination of the 
silver ions and the chloride ions to form a compound 
that does not dissolve appreciably in water. The 
sodium ions and nitrate ions still remain in solu- 
tion. Unless the solution is evaporated to dryness, 
forming crystals of sodium nitrate, we really have 
only a simple combination. A discussion of ions and 
their properties is given in later experiments. 

Chemical Changes Occurring 
In this Experiment 

We shall start with a sample of a silver-copper 
alloy (such as one of the several alloys used in jew- 
elry manufacture) which has the approximate com- 
position of American coinage silver. By a succession 
of chemical and physical changes, we shall separate 
this so as to obtain pure silver and pure copper 
metals. In succeeding experiments we shall convert 
these pure metals quantitatively into pure com- 
pounds, and calculate the formulas of the com- 




pounds thus formed. The reactions will be carried 
out in the following steps: 

(1) Silver-copper alloy is dissolved with nitric 
acid (HNOs) to give a solution of silver nitrate 
(AgNO 3 ) and cupric nitrate (Cu(NO 3 ) 2 ). 

(2) This solution is treated with sulfuric acid 
(H 2 SO 4 ), evaporated to the point of the appearance 
of dense white fumes, and then diluted with water. 
The solution now contains silver sulfate (Ag2SO4) 
and cupric sulfate (CuSC^). 

(3) Copper metal is added to this solution, pre- 
cipitating the silver as the free metal, and forming 
additional cupric sulfate in solution. 

(4) Pure metallic copper is then obtained from 
this cupric sulfate solution by replacement with 
metallic zinc, in a manner similar to that of the 
replacement of silver above. 

These successive changes may be summarized as 


HN0 8 


Cu(NOs) a 

Ag2 SO 4 
CuSO 4 

CuSO 4 

Zn Cu 

""* ZnSO 4 

The experiment as a whole serves to illustrate the 
various types of chemical change, especially re- 
placement reactions of a more active metal for a 
less active metal from its salts. (See Table VIII, 
Appendix II.) 

Preliminary Work 

Fill in the data called for in Table A, "Physical 
Properties of Various Substances," of the report 
sheet. This will give you information helpful in 
understanding the experiment. 

Experimental Procedure 

Special supplies: Ag-Cu alloy. 1 

Chemicals: Copper wire, No. 18, 0.1 F Cu(NO 8 )*, 0.1F 
K4Fe(CN) 6 , 0.1 F AgNO 3 . Mossy zinc, 0.1 F Zn(NO 3 )2. 

1. Pure Silver from a Silver-Copper Alloy. 1 

Obtain a sample of silver-copper alloy weighing 
approximately 1.5 g. Using a balance sensitive to 
0.01 g (such as the agate, knife-edge, triple-beam 
balance), weigh the alloy carefully, recording the 
weight on the report sheet. (Get instructions on the 
use of the balance. Be sure the balance's knife-edge 
support is released and the balance properly ad- 
justed to zero, with all weight slides at zero posi- 
tion, before adding the alloy. A piece of metal may 
be weighed directly on the scale pan.) Place the 
weighed alloy in a 150-ml beaker and add 10 ml of 
dilute nitric acid (HNO 3 ). Caution: Do not spill 
nitric acid on your hands or clothing if this occurs 
flush well with water, neutralize acid on clothing with 
ammonium hydroxide. Place the beaker on a wire 
gauze supported on a ring stand under the hood, and 

1 Silver-copper alloy of the composition of a dime, or of various 
percentages in order to provide for "unknowns" if desired, may be 
obtained from dealers in the noble metals silver, gold, and platinum. 
The cost is only about three cents per student. 

1 Caution: Wash off any silver salt solutions spilled on the hands or 
clothing at once, otherwise a black stain of metallic silver will result. 
This is difficult to remove, but may be dissolved by treating the stain 
first with a few drops of tincture of iodine (to form Agl) and then with 
a saturated solution of sodium thiosulfate (Na&Qi), to remove excess 
iodine and dissolve the Agl. 

warm gently if needed, but only intermittently to 
maintain a moderate rate of solution of the metal. 
If a brown gas (what is it?) ceases to be evolved 
before all the metal dissolves, it may be necessary 
to add a little more nitric acid. After complete solu- 
tion of the metal, add 25 ml of dilute sulfuric acid, 
and continue the gentle heating in a manner to 
avoid any loss of solution by spattering, until the 
solution is evaporated to a syrupy liquid. This con- 
sists of concentrated sulfuric acid and dissolved 
salts. This point will be indicated by the very copi- 
ous emission of dense, white, choking fumes of 
sulfur trioxide (SO 3 ). The beaker may then be 
covered with a watch glass to prevent the further 
escape of these fumes into the room. Use caution, as 
concentrated sulfuric acid is very corrosive. Let Hie 
beaker cool, then add cautiously about 125 ml of dis- 
tilled water from your wash bottle. Heat the solu- 
tion quite hot, but do not boil it, and stir with a 
glass stirring rod to dissolve the salts. (Silver sul- 
fate, (Ag)2SO4, which is only slightly soluble at 
room temperature, is much more soluble in this 
strongly acid solution. In this case it is present 
largely as the acid salt, silver hydrogen sulfate, 
AgHSCX. This should dissolve readily on heating 
and stirring the solution.) 

Obtain a 30-cm length of about No. 18 copper 
wire, shape this into a very loose coil, and place in 
the warm solution of dissolved salts, leaving an end 



of the wire bent up out of the solution as a handle. 
Leave the copper wire quietly in the solution for 
ten to fifteen minutes, to observe the beautiful 
deposit. Then stir, or shake the wire gently, at in- 
tervals, to loosen the precipitated silver and to 

10 ml of 
HN0 3 

Silver -copper 
alloy sample 

Warm gently to 
start the reaction 
if necessary." 

Fio. 6-1. Dissolving the silver-copper alloy. 

expose fresh copper to the solution. While this re- 
action is proceeding, carry out the test tube reac- 
tions in the following paragraph, so as to learn 
suitable methods of qualitative tests which you will 
need to know in this experiment. 

Qualitative Tests for Silver, Copper, and Zinc Salts. 
We shall test solutions of silver, copper, and zinc 
salts, and mixtures of these salts, with three re- 
agents: hydrochloric acid (HC1), ammonium hy- 

droxide (NH 4 OH), and potassium ferrocyanide 
(K 4 Fe(CN)6), respectively. Refer now to the report 
sheet, Table B. Place 1-ml samples of the salt solu- 
tions, as specified in the left column of Table B, 
into 10-cm test tubes. Add a drop of dilute hydro- 
chloric acid to each, and note the results. Prepare 
another set of the salt solutions, and test each with 
1 ml of dilute ammonium hydroxide. Mix wvtf , and 
then note the results. (With cupric nitrate, am- 
monia forms a complex addition compound of the 
formula CuCNHsMNOa^, which contains the deep 

blue ion, Cu(NH 8 ) 4 ++ .) Test still a third set of the 
salt solutions with 1 ml of potassium ferrocyanide 
solution. In the silver nitrate salt mixtures be sure 
to use an excess of the potassium ferrocyanide, 
about 2 ml. State suitable methods (on the report 
sheet) for detecting the presence of silver, copper, 
and zinc salts in a solution. 

To return to our reaction, after the copper wire 
has been in the solution of the alloy metals for 
about a half hour, transfer a few drops of the solu- 
tion to a small test tube, and test this for silver ion. 
If a positive test is obtained, leave the wire in the 
solution for a longer period, until there is no more 
silver ion in the solution, this being shown by re- 
peated tests. 1 When the reaction is complete, the 
resulting solution is much richer in cupric sulfate 
solution than it was before. Remove the copper 
wire from the solution, and carefully shake loose 
all of the precipitated silver. Let the silver settle, 
and carefully decant (see Fig. i-12, P. xix) as much 
of the clear solution as you can into a clean beaker, 
taking care that no silver is transferred. Add 5 
ml of distilled water to the silver residue to wash 
it, mix this thoroughly, let it settle, and decant 
again, catching this first wash water in the cupric 
sulfate solution. Save this solution for paragraph 
2, which follows. Continue to wash the silver metal 
by decantation, using first a 5-ml portion of dilute 
ammonium hydroxide as the wash solution (to re- 
move any impurity of silver chloride), then using 
5-ml portions of distilled water, until the washings 
show no positive test for cupric ion when 5 drops 

1 The rate of reaction of silver sulfate solution with copper metal 
varies a great deal with the concentration of acid. It is quite slow in a 
neutral solution, requiring five to ten hours for completion. However 
with 25 ml of dilute sulfuric acid present in the 125 ml of solution, the 
replacement should be completed in 30 to 45 minutes. If your reaction 
is proceeding too slowly, you may add an additional 5 ml of sulfuric 



of ammonia are added to 1 ml of the wash water. 
Finally drain off the last wash water from the 
silver. Discard all wash solutions after the first. 

Heat a clean evaporating dish to dry it thor- 
oughly. Let it cool, and weigh the dish carefully on 
the balance. With a stirring rod, transfer all 
the silver to this dish, without loss, and heat 
it until it is thoroughly dry. Cool, and reweigh it. 
Calculate the percentage of silver in your alloy 

To check on the purity of your silver, take a very 
small amount of it, not over one twentieth of the 
total amount, dissolve it in a half milliliter of dilute 
nitric acid by warming it gently, and test it for 
cupric ion as above. (Be sure to add an excess of 
ammonia, so the solution smells of ammonia after 
mixing.) Show this test to your instructor, to- 
gether with the recorded results of the test, for his 
approval. Preserve your dry silver carefully in a 
covered vessel for use in Experiment 10. 

2. Pure Copper from a Cupric Sulfate Solu- 
tion. To your cupric sulfate solution from the 
above preparation of pure silver, add several grams 
of mossy zinc; (A large excess is not necessary.) 

Let this stand, with frequent stirring or shaking, 
until the blue color of the solution has disappeared. 
Test a few drops of this solution in a test tube with 
ammonia to see whether the copper is completely 
replaced. When a positive test for copper is no 
longer obtained, carefully decant the clear liquid, 
which may be discarded, leaving the solid ma- 
terial in the beaker. Pick out with your stirring rod 
any larger particles of zinc which remain, and add 
about 5 ml of dilute sulfuric acid, if necessary, to 
dissolve the remaining zinc. Does any copper dis- 
solve? When the reaction is complete, decant the 
solution, and wash the copper by decantation with 
5-ml portions of tap water, and finally with distilled 
water, until the last washing gives no test for zinc 
ion when 1 ml of potassium ferrocyanide is added 
to it. The washed copper may be dried at room 
temperature, but should not be heated, as it oxi- 
dizes readily when warm. 

To check on the purity of your copper, dissolve a 
very small portion of it in a half milliliter of warm 
dilute nitric acid, and test this for the presence of 
silver ion. Show this test to your instructor, to- 
gether with the recorded results of the test, for his 

REPORT: Exp. 6 

\ Preparation of Pure Substances 




Preliminary Work 

Locker Number^ 



The data for this table are to be obtained, before performing the experiment, from reference sources available, such as a 
table of physical constants of inorganic compounds in a handbook of chemistry. 



g/100 g 7/20, at 20C 



Cupric nitrate 
Cu(NO,) ,.3H,0 

Cupric oxide 

(No entry) 

Cupric sulfate 
CuSO 4 *5HsO 

(No entry) 

Potassium nitrate 

Silver chloride 

(No entry) 

Silver nitrate 

Silver oxide 
A g2 

(No entry) 

Silver sulfate 
A g2 S0 4 



Obtain these data experimentally, as directed. In the proper square, indicate any results obtained, such as "white cunly 
precipitate," "no effect," the color of any solution formed, etc. 

Test Solutions 

Acid (1 (trap) 

1 ml 


Pure silver nitrate solution 

Pure cupric nitrate solution 

Cupric nitrate solution and 1 drop 
silver nitrate 

Silver nitrate solution and 1 drop 
cupric nitrate 

Pure zinc nitrate solution 

Zinc nitrate solution and 1 drop 
cupric nitrate 


Utilizing the preceding results, name a reagent or reagents which will prove satisfactory when testing for the 
melal ions listed below, and give the name and color of any distinctive precipitates or solutions formed: 

Metal Ions 

Silver nitrate 
(silver ion) 

Cupric nitrate 
(cupric ion) 

Zinc nitrate 
(zinc ion) 

Reagent (9) 

Name and Color of Distinctive Product Formed 

1. Pure Silver from Silver-Copper Alloy 

Weight of silver-copper alloy taken . 
Weight of evaporating dish plus pure silver 
Weight of evaporating dish, empty and dry 
Weight of pure silver obtained . . 

Percentage of silver in the alloy 
Method of calculation: 

Test for purity of solid silver metal obtained: 


Observed result 
of test 


2. Pure Copper from Cupric Sulfate 

Test for purity of solid copper metal obtained: 


Observed result 
of test 


3. Chemical Changes Involved in This Experiment 

For each of the cases (a) to (i) on the next page, decide on the nature of the reaction, if any, which will occur. 
If you have had no previous chemistry, simply write the names of the products formed. At the right, indicate the 
type of the reaction: C for combination, D for decomposition, S for single replacement, and DR for double replace- 
ment. If you have studied chemistry, you may wish to write balanced equations for the reactions. Study Assign- 
ment A, Page 5, and Table XVI, Ionic Valences, in Appendix II, will help you to write the formulas correctly. 
As an example, we have given below both the names of the products and the equation for the solution of your alloy 
in nitric acid. This is a more complex, oxidation-reduction type of reaction. 

Example : Types of Reactions 

(x) Silver (or copper) is dissolved in nitric acid: 

Silver nitrate, nitric oxide, and water or 8 Ag + 4 HNO* - 


S AgNO* + NO + 

(A complex 
oxidativn reduction) 

Report on Exp. 6, Sheet 2 Name 

(a) Silver nitrate and sulfuric acid solutions are evaporated: 

(b) Hydrochloric acid is added to silver sulfate: 

(c) A copper wire is placed in silver sulfate solution: 

(d) Zinc metal is added to dilute sulfuric acid: 

(e) Copper metal is added to dilute sulfuric acid : 

(f) Zinc metal is added to cupric sulfate solution: 

(g) Potassium ferrocyanidc and cupric nitrate solutions are mixed: 

(h) Potassium fcrrocyanide and zinc nitrate solutions are mixed: 

(i) Copper metal is heated in air: 

(j) Silver oxide is heated : 

(k) Silver metal is heated in air: 

Types of Reaction* 

4. Replacement Series of the Metals 

Considering only the reactions in solution above which you marked SR (single 
replacement), make a list of the elements copper, zinc, silver, and hydrogen in such an 
order that each element comes above another that it replaces from solution. 

Interpretation of Data, and Problems 

In answering this section, use any of your data or Table A in this experiment, and also Table VIII in the 

1. (a) Predict two single replacement reactions which you did not try, but would expect to occur: 

(b) Predict a combination of a metal with a solution of a salt of another metal in which no reaction would 



2. Choose the method that you would use to separate the following pairs of substances. If a chemical change 
is required, encircle the C. If the separation is based on a difference in solubilities, encircle the S, or, if basec 1 
on a difference in boiling points, encircle the B. 

(a) Silver chloride and cupric sulfate ................. C S B 

(b) Water and silver nitrate .................... C S B 

(c) Zinc nitrate and cupric nitrate .................. C S B 

(d) Water and silver .......... ' ............ C S B 

(e) Zinc and copper ...................... C S B 

(f) Silver oxide arid silver nitrate .................. C S B 

(g) Cupric oxide from nitrogen dioxide and oxygen ............. C S B 

In case a chemical change takes place in any of the above, record the change involved in your separation: 

3. Three unknown solutions, which may contain salts of silver, zinc, and copper, but no other salts, are tested 
as indicated below. For each case, indicate what you know about the presence or absence of each metal ion in the 
solution. Encircle the P if the given metal is definitely proved present in the solution, A if it is definitely absent, 
and D if the test is insufficient to decide. 

(a) Unknown solution (a) is tested by adding ammonium hydroxide in excess, giving Silver PDA 
a clear, colorless solution. Zinc PDA 

Copper PDA 

(b) Unknown solution (b) is tested by adding a drop of hydrochloric acid, giving a Silver P D ^ 
white precipitate. Zinc P D ^ 

Copper PDA 

(c) Unknown solution (c) is tested by adding a drop of hydrochloric acid, with no Silver PDA 
noticeable effect, then adding a little potassium ferrocyanide, giving a white precipitate. Zinc PDA 

Copper PDA 

4. Cupric oxide contains 79.9% copper. What weight of American coinage silver, 
which is 90% silver, would be needed to produce, by suitable chemical reaction, 

10,0 grams of cupric oxide? ................. _______________ 

5. What weight of silver nitrate, which is 64% silver, could be produced from 
a U. S. dime weighing 2.5 grams? 


The Reduction of Metallic Oxides by Hydrogen 
and Other Reducing Agents 


Oxidation and Reduction 

As a first simple definition, we may say that a 
substance is oxidized when it combines with oxygen 
(or with some other nonmetallic element, as 
chlorine). The oxidizing agent for the reaction may 
be either free oxygen gas or some compound which 
supplies oxygen or another nonmetallic element. 
In general, when a substance is oxidized, the extent 
of its combination with nonmetallic elements is 

A substance is reduced when the extent to which 
it is combined with oxygen, or other nonmetallic 
elements, is decreased. The reducing agent is the 
substance for example, hydrogen gas or some me- 
tallic element which removes the nonmetallic 

Cof/ege Chem/sfry, Chapters 4, 6 

Review of Fundamental Concepts 

element by combining with it. A detailed study of 
oxidation and reduction is given in Experiment 17, 
and other subsequent experiments. Study Table 
VIII in Appendix II, and note the variation in 
the ease with which the oxides of different metals 
may be reduced. 

In this experiment, directions are given for the 
reduction of certain metallic oxides by the use of: 
(1) hydrogen gas, (2) your domestic fuel gas, and 
(3) the blowpipe, with a charcoal block. Use as 
many metallic oxides, and methods, as your in- 
structor directs. In observing your results, look for 
such characteristics as change in color and physical 
form, luster, malleability, hardness, magnetic prop- 
erties (in the case of iron), and ease of melting. 

Experimental Procedure 

Special supplier: Thistle tube, calcium chloride tube, blow- 
pipe, magnet. 

Chemicals: CaCla (anhydrous, 4-mesh), cotton, zinc (mossy), 
0.1 F CuSOi, charcoal stick, powdered charcoal, various metal- 
lic oxides for reduction, as CuO f Fe 2 O 3 , PbO, PbjO 4f PbOa, 
SnO 2 . 

1. Reduction with Hydrogen. Prepare the ap- 
paratus as illustrated in Figure 7-1. Place about 
2 to 4 g of the metallic oxide to be reduced in the 
pyrex test tube. Put about 15 g of mossy zinc, 
together with 10 ml of water, in the generating 
flask. Before generating hydrogen have your apparatus 
approved by the instructor, and have him initial the 
report sheet. He will not approve it later. 

When all is ready, pour 20 ml of dilute sulfuric 
acid down the thistle tube. Jf the gas is slc/w in 
generating, add 2 or 3 ml of cupric sulfate solution, 
as a catalyst. More acid may be added from time 
to time, so as to maintain a moderate rate of 
evolution of gas. (Wrap the generating flask in a 
towel to catch flying glass in case an explosion should 

Test for the purity of hydrogen gas throughout 
the system by inverting a 10-ml test tube over the 
exit tube to catch a tube full of the issuing gas, as 
illustrated in Figure 7-2. Repeat this test until the 
gas burns quietly, and then use this test tube of 
burning hydrogen to ignite the gas issuing from the 
capillary tip. As a safety precaution^ ignite the tip 
in no other way. 

Now heat the metallic oxide to a moderate tem- 
perature, maintaining a steady evolution of hy- 
drogen all the while by adding acid as needed. 
When the change is complete, warm the other part 
of the test tube to dry out all the water (where did 
it come from?), but do not burn the rubber stop- 
per. Remove the flame and permit the free metal 
to cool, still keeping the tube filled with hydrogen 
gas so that no air will enter to oxidize the metal. 

2. Reduction with Domestic Fuel Gas. Coal 
gas or a mixed gas usually contains free hydrogen 
and is an excellent reducing agent. Natural gas, 
which is largely methane (CEU), is not satisfac- 
tory. If your gas supply is suitable, place the oxide 
to be reduced in the test tube (Fig. 7-1), discon- 
nect the hydrogen generator and connect the gas 
jet by means of rubber tubing to the CaCla drying 
tube. Pass a slow stream of gas through the system, 
and ignite the excess at the exit tip. Heat the test 
tube to reduce the oxide to the free metal. 

3. Reduction by the Use of the Blowpipe. 
Hollow out a piece of charcoal near one end so as 
to make a depression about 1 cm wide and 5 to 7 
mm deep. Fill this with the metallic oxide to be 
reduced, and heat it with the reducing flame of the 
blowpipe, as illustrated in Figure 7-3. Your in- 
structor will demonstrate the proper technique, so 
as to maintain a continuous blowpipe jet. (Learn 
to maintain a positive air pressure, while inhaling 




Hydrogen flame 
Do not light until 
the air is out of 
the apparatus. 

Anhydrous Cad* grains 

Test tube tilted 
to prevent condensed 
water from running toward 
heated end. 


Metallic zinc and 

Flask should be wrapped 
in a towel because of 
the danger of an 

Collect some 

FIG. 7-1. Apparatus for the reduction of metallic oxide by hydrogen gas. 

at the same time.) Note that the oxide must be 
kept in the inner portion of the flame, in the pres- 
ence of the hot reducing gases which may consist 
both of the hot domestic gas and of carbon mon- 
oxide produced by the burning charcoal. Some- 
times, it is advisable to mix the metallic oxide with 
powdered charcoal before placing it on the char- 
coal block. 

Cheeks blown 
out like a 
bellows to give 


Carry it at 
two feet away from 
the rest of the apparatus 
ignite it. If the hydrogen 
pops and then burns quietly it is 
relatively pure. If it explodes with a sharp 
report ft is still mixed with air, 

FIG. 7-2. Testing for the purity of the hydrogen from the 

Low flame 

Charcoal stick 

FIG. 7-3. The reduction of a metallic oxide which is placed in 
a depression in a charcoal block, by the use of a blowpipe. 

REPORT: Exp. 7 Name 

Reduction of Metallic Oxides by 


Hydrogen and Other Reducing Section 


Locker Number 

Review Questions 

1. Write equations for the preparation of hydrogen gas by the following reactions: 

Sodium metal and water 

Steam and red hot iron 

Zinc and sulfuric acid 

Electrolysis of water . 

2. For the equations below, write in the spaces provided, the formulas of the substances designated: 

(a) FeO + CO * Fe + CO 2 

(b) 2 Al + 6 IIC1 * 2 A1C1 + 3 II 2 

For Equation (a) For Equation (b) 

The substance oxidized 

The substance reduced _ _ . 

The oxidizing agent ___ 

The reducing agent _ 

1. Reduction with Hydrogen 

Instructor's Approval of Apparatus .... 

1. List at least four physical properties of hydrogen which you observed in this experiment. 

2. Write the equation for the reaction which occurs in the explosion of impure hydrogen gas: 

3. Write equations for the reduction of each of the following oxides by hydrogen gas: 
Cupric oxide, CuO 

Ferric oxide, FeaOj 

Magnetic iron oxide, Fe 3 O 4 


4. List the metallic oxide or oxides which you reduced, and give all physical evidences which you observed, fc 
each case, that a reaction had taken place: 

2. Reduction by Other Reducing Agents 

1. Write equations for the reduction that takes place on heating stannic oxide, SnO 2 , with: 


Carbon monoxide 

2. List any metallic oxides which you reduced by the use of the blowpipe, and for each of these reaction; 
indicate all evidences of cheraical change which you observed: 

3. In the reactions of the blowpipe flame, (1) hot oxygen gas can react with the red hot charcoal, and (2) an 
carbon dioxide formed can likewise react with the red hot charcoal. Write the equations for these processes: 



Application of Principles 

Consider the results of this experiment, and Table VIII in the Appendix, in answering the following. 

1. Suggest three other metals, besides zinc, which might be satisfactorily used, by reaction with an acid, f< 
the preparation of hydrogen gas. 

2. List, in each category, two metals whose oxides: 
Can be reduced by heat alone, 

Cannot be reduced by heat but can by H 2 or CO, . 
Cannot be reduced by H 2 or CO but can by Al, . 
Cannot be reduced except by electrolysis, , . . 


The Oxides of the 

Elements and the Periodic Table. 


College Chemistry, Chapters 5, 6 

Review of Fundamental Concepts 

In this experiment, we shall prepare oxygen gas, 
and use this gas to prepare a number of repre- 
sentative oxides of various elements. We shall ob- 
serve the chemical character of the different oxides, 
by their reactions with water and by the formation 
of salts, and correlate the general character of the 
elements with their relative positions in the peri- 
odic table. 

Acids and Bases from Oxides 

In general, the oxide of an element (symbol El) 
reacts with water to form an addition compound in 
which there are hydroxide groups, 




(The number of OH groups, and the corresponding 
subscripts in the formulas, will, of course, vary 
with different elements.) The character of this 
hydroxide depends on the nature of the element to 
which the Oil groups are attached. The product 
may behave as an acid, 1 which is characterized in 
solution by the presence of hydrogen ions, H+. We, 
therefore, write the formula with the hydrogen 
atoms first, as H 2 E1O 2 . In solution, it ionizes ac- 
cording to the equation 

H 2 E10 2 

2 H+ + E10 2 . 

On the other hand, the compound may behave as 
a base, which is characterized, in solution, by the 
presence of hydroxide ions, OH~, so we write the 
formula with the hydroxide group together. In so- 
lution, it ionizes according to the equation 

E1(OH) 2 * E1++ + 2 QH-. 

An oxide which with water forms an acid is 
called an acidic oxide or acid anhydride. An oxide 
which with water forms a base is called a basic 
oxide or baMc anhydride. We may compute the 
formula of a basic or acidic anhydride which is re- 
lated to a given base or acid by subtracting water 

1 A number of important acids are not derived from oxides, bow- 
ever, as the binary hydrides of the nonmetals, i.e., HCl HBr, HsS, etc. 

so as to eliminate all hydrogen atoms, as for ex- 

Mg(OH),-H 2 vMgO 
2 HJ5O, - 3 H 2 O * B 2 O,. 


Salts are ionic-type compounds which consist of 
an electropositive radical (other than hydrogen 
ion) combined with an electronegative radical 
(other than hydroxide ion). An important method 
of formation of salts involves the reaction of either 
a base or basic oxide with either an acid or acidic 
oxide. Typical equations for such reactions are; 

2 NaOH + H 2 SO 4 * NajSO 4 + 2 H 2 O 
ZnO + H2SO 4 * ZnSO 4 + H t O 

Ca(OH) 2 + CO 2 K CaCO* + H 2 O 



H the reactions take place in solution, they more 
probably are of the first type, while if they take 
place in a molten mixture, free from water, they 
would, of course, involve the interaction of the 
basic and acidic oxides directly. In the study of 
the geological history of the earth's crust, the for- 
mation of many minerals is to be explained in 
terms of the crystallization of "salts" from the 
mixture of basic and acidic oxides present in the 
molten magma, as it cooled. The ceramic industry 
is based on such reactions, as are also certain 
metallurgical operations, for example, in the iron 
and aluminum industries. 

Salts may also be formed by the action of the 
free metal with an acid, in which case hydrogen 
gas, rather than water, is formed, as in the reaction 

Zn+H 2 SO 4 - 

ZnSO 4 +H,. 

Likewise, a salt is formed when a metal combines 
with a nonmetai, as in the reaction 

2Na+Cl, *- 

Chemical Formulas, Valence, Nomenclature 

There are many thousands of chemical sub- 
stances. However, their formulas and names in 




general correspond to simple rules, the application 
of which is not difficult for you to master. 

Binary compounds are those consisting of two 
elements. Their formulas are the simplest to build, 
and they may be named by remembering only the 
characteristic ending ide. In such compounds, the 
more electropositive (metallic) element is given 
first, followed by an abbreviation of the more 
electronegative (nonmetallic) element, then com- 
pleted by the ending ide. A few examples are : NaCl, 
sodium chloride; CaO, calcium oxide; MgBr 2 , 
magnesium bromide; H 2 S, hydrogen suliide; A1I 3 , 
aluminum iodide; Mg 3 N 2 , magnesium nitride; and 
K 2 O, potassium oxide. 

Ionic valence and the periodic table. The sub- 
scripts which you note in the above and in other 
formulas indicate the relative number of atoms of 
the different elements which are combined to- 
gether. The relative combining power of an ele- 
ment in a compound is called its valence. At this 
time we shall introduce only the ionic valence. 1 
This term implies that the atom, or radical, in the 
compound possesses a definite electrical charge. 
The metals become electropositive, and the non- 
metals become electronegative, in their binary 
compounds. The theoretical explanation of this 
positive and negative character will be presented 
later, when your background is better developed, 
but it will be to your advantage now to learn a 
few of the common valences, and to relate them to 
the position of the element in the periodic table. 
Note that in the first three groups the positive 
valence always corresponds to the periodic table 
group number. Thus we have a valence of + 1 for 
hydrogen and for the metals of group I in their 
compounds: H+, Li+, Na+, E>, Rb+, and Cs+; a 
valence of +2 for the metals of group II in their 
compounds: Be++, Mg++, Ca++ t Sr++, Ba++, and 
Ra++ ; and a valence of +3 for the metals of group 
III in their compounds: A1+ ++ , Sc+ + +, etc. 2 

The typical ionic valences of the nonmetals are, 
likewise, in accordance with their periodic group. 
In their binary compounds, we have a valence of 

1 The topic of valence, including other aspects of the problem, such 
as "covalerice" and "coordination number,** will be considered more 
thoroughly later. See College Chemistry, Chapters 10 and 11, and 
Experiments 15, 16, and 35. 

2 The character of boron is such that it is usually present in com- 
pounds as a part of an ion which possesses a negative charge, as in 
boric acid, HBO^ i.e. BOi . 

1 for the group VII elements: F~, Cl~, Br~, I" 
a valence of 2 for the group VI elements 
O , S , Se , Te ; and a valence of 3 fo 
the group V elements, N --- , P --- , As --- , etc. 
(Note that the negative valence is obtained b; 
subtracting the group number from eight.) 

Some elements, notably the transition metal 
(atomic numbers 22 to 32 and those below them ii 
the periodic table), show a variable valence. Ii 
these cases, the endings ons and ic are used t< 
show, respectively, the lower and the higher va 
lence. The more common instances of variabl 
valence are: 

Co++ Ni++ Cu+ Au+ Hg 2 ++ Sn+ + 



The names and formulas of a few correspondinj 

compounds are: 

ferrous bromide FeBr* mercurota oxide Hg 2 O 
ferric bromide FcBri mercuric oxide HgO 
cuprous sulfide Cu 2 S stannows chloride SnClj 
cupnc sulfide CuS stannic chloride SnCIi 

It will pay you to learn the names and formula, 
of a very few common oxygen acids, as this, in turn 
will teach you the valences of the correspondinj 
negative radicals. 4 Note that if the name of tin 
oxygen acid ends in ic 9 the name of the correspond 
ing negative radical ends in ate. 

nitric acid HNQi nitrate ion NOg*~ 

sulfurtc acid H 2 SO| sulfate ion SO 4 

carbonic acid HjCOj carbonate ion COs 

phosphoric acid H 3 PO 4 phosphate ion PO 4 

In general, acids and salts containing three differ 
ent elements (ternary compounds) are namec 
according to a different system than binary com 
pounds. This is considered in detail after Experi 
ment 18. 

To summarize the building of a formula: Note 
that one simply inserts such subscripts after eacl 
element or radical as will balance the total positive 
and total negative charges, as, for example 
(Na+)(Cl-) or NaCl, (Ca++)(Cl-) 2 or CaCl 2 

As we move from group VII to group V, the chemical bond forraec 
when these elements combine with other elements becomes less ioni< 
and more covalent in character. For the purpose of writing correct 
formulas, however, it is convenient to ascribe definite negative va 
lences to these elements. 

4 The term "radical" is applied to a group of atoms which oftei 
act as a unit in a chemical reaction. 



(Mg++)(N0 8 -) 2 or Mg(N0 8 ) 2 , (Ca++)(O~ ) or 
CaO, (Ca++)(SO 4 ) or CaSO 4 , and (Ba++), 
(P0 4 --- ) 2 or Ba 3 (P0 4 ) 2 . 

Some additional rules of nomenclature. Hydrox- 
ides are named according to the rule for binary 
compounds even though there are three elements 
present. The hydroxide ion (OH~) is a rather 
stable group of elements and is treated as though it 
were a single nonrnetal ion such as chloride (Cl~). 
For example, NaOH is sodium hydrox/c/e, and 
A1(OH) 3 is aluminum hydrox'wfe. 

The ion NH 4 + behaves very much like the 
alkali metal ions (Na+, K+, etc.) and is given the 
name "ammonium ion." It is treated exactly as 
though it were a simple metal ion. For example, 
NH 4 C1 is ammonium chloride, NH 4 OH is am- 
monium hydroxide, (NH 4 ) 2 S is ammonium sulfide. 

Some special mention should be made of binary 
compounds containing two nonmetallic elements. In 
these cases the more electronegative element is 
named last, and a prefix is attached to indicate 
the number of atoms of this element (rather than 
the ons and ic endings used in the case of metals) . 
Some examples arc: 

SO a sulfur dioxide N 2 O 3 nitrogen trioxide 

SOj sulfur trioxide NjC^ nitrogen pent-oxide 

Some compounds like water (H 2 Q), ammonia 
(NH 8 ), and phosphine (PH 3 ) possess common 
names which were given them before the nomen- 
clature of compounds was systematized. 

The hydrogen compounds of some of the nonmetals 
(HF, HC1, HBr, HI, H 2 S, H 2 Se, etc.) form acid 
solutions when dissolved in water, and when they 
are used in this sense they are njamed according to 
the following simple convention: 

HC1 (hydrogen chloride) becomes hydrochloric acid 
HBr (hydrogen bromide) becomes hydrohromic acid 
H 2 S (hydrogen sulfide) becomes hydrosulfuric acid 

The Writing of a Chemical Equation 

A chemical equation is simply the chemist's 
shorthand expression of a chemical reaction which 
takes place. Fundamentally, it expresses the quan- 
tities of the substances used and produced. All of 
the atoms included in the molecules of the re- 
actants must reappear in the molecules of the 
products, that is, an equation must balance. Thus, 
to write 

Zn + IIC1 * XnClj + H 2 

is false. It is necessary to use 2 HC1, since two 
atoms each of hydrogen and of chlorine are in- 
volved on the right side of the equation. We, 
therefore, write 

Zn + 2 HC1 * ZnClj + H 2 . 

In writing a correct, balanced equation, note 
these points: 

First be sure that the formulas of the sub- 
stances used, and of those produced, are correct, 
and then do not change the subscripts in order to 
balance the equation. 

Second place such coefficients in front of the 
formulas as are needed to have the same number 
of atoms of each element on each side of the equa- 
tion. This usually can be done by a simple in- 
spection. A good policy is to start with the most 
complicated formula. 

For example, when steam is passed over red 
hot iron, analysis of the oxide formed shows that 
it is Fe 3 O4, magnetic iron oxide. We, therefore, 
first write down the correct formulas of reactants 
and products: 

Fe + H 2 O * Fc 3 O 4 + H 2 (unbalanced). 

To balance it, start with the FeaO* noting that 3 
Fe and 4 H 2 will be required, and then 4 H a will 
be produced: 

3 Fe + 4 H 2 O * Fe 3 O 4 + 4 H 2 (balanced). 

Experimental Procedure 

Special supplies: Deflagrating spoon, 6 glass squares, pneu- 
matic trough, 6 250-ml wide mouth bottles. 

Chemicals: Ca (metal shavings), Cu (turnings), Fe (steel 
wool), Mg (ribbon), P (red), S, small pieces of charcoal, litmus 
solution, ECHO* MnO 2 (powdered), HjBO,, 60% solution of 
HC1O 4 , 

1. The Preparation of Oxygen. Assemble the 
apparatus as illustrated in Figure 8-1. Mix about 
10 g of potassium chlorate with 2 g of manganese 

dioxide. First place a very small sample of this 
mixture in the test tube and heat it enough to melt 
the potassium chlorate. If it decomposes smoothly, 
without obvious sparks and combustion, it is safe 
to use. 1 Let the test tube cool, then add the re- 

i if or g an i c materials are present, or if possibly the wrong chemicals 
are mixed, a dangerous explosion might result. It pays to treat 
potassium chlorate with due caution and respect. 



KCIOj and MnO 2 

Bottle being filled 
with oxygfen 

full of 

Bottle filled with water 
ready to be placed in 
the pneumatic trough 

FIG. 8-1. The laboratory preparation of oxygen gas. 

mainder of the mixture and connect the delivery 
tube. Fill six 250-ml wide mouth bottles with water 
and invert these in the pneumatic trough as needed. 
When all is ready, gently heat the potassium chlo- 
rate and manganese dioxide mixture just enough 
to maintain a moderate evolution of oxygen. First 
let a little of the gas escape into the air, to permit 
the generator to fill with pure oxygen, and then 
collect the six bottles of oxygen by displacement 
of water. Cover each with a glass square as soon as 
it is filled, and place it right side up on the table. 
Take the delivery tube out of the water before the 
generation of oxygen ceases or before you remove 
the flame. Why? 

2. The Preparation of Oxides. Prepare oxides 
of the following elements by burning them in 
oxygen gas. Keep the bottles covered as much as 
possible. Number or label each to avoid confusion. 
Immediately after each combustion, add 30 to 50 

ml of water, replace the cover, shake the bottle to 
dissolve the oxide formed, and set it aside for later 

Magnesium. Ignite a 10-cm length of magnesium 
ribbon, held with the crucible tongs, and at once 
thrust it into a bottle of oxygen. (Do not injure 
your eyes by looking directly at the brilliant light.) 

Calcium. Calcium metal is difficult to ignite, but 
burns very brilliantly. Use caution. Your instruc- 
tor may wish to demonstrate this. Prepare a deflag- 
rating spoon by lining it with a little asbestos paper 
or shreds, and ignite this to burn out any carbon. 
Heat a shaving of calcium metal, on the deflagrat- 
ing spoon, quite hot. At once add a 2-cm length of 
magnesium ribbon as a fuse, re-heat the calcium 
and ignite the magnesium, and then quickly thrust 
this into a bottle of oxygen. See Figure 8-2. 

As an alternative procedure, you may place a 
shaving of calcium in a crucible, and ignite it in the 



air at the maximum temperature over a Bunsen 
burner for 15 minutes, then wash out the product 
with water into a beaker. 

Iron. Add a little water to a bottle of oxygen, to 
form a protective layer on the bottom. Heat a 
little steel wool, held by the tongs, in the Bunsen 
flame to ignite it, and at once thrust it into this 

FIQ. 8-2. The use of a deflagrating spoon to burn a substance 
in oxygen. 

Carbon. Ignite a small piece of charcoal, held by 
the tongs or in a clean deflagrating spoon, and 
thrust the glowing charcoal into a bottle of oxygen. 

Phosphorus and Sulfur. For each of these, clean 
and re-line the deflagrating spoon, and ignite it to 
burn out any combustible residue. Add a bit of 
sulfur, or phosphorus (use red phosphorus, no more 
in volume than half a pea). Ignite each of these 
over the burner, then thrust them into bottles of 
oxygen gas. In each case, after the combustion dies 
down, re-heat the deflagrating spoon to burn out 
all remaining phosphorus or sulfur. 

3. Acids and Bases from Oxides. To each of 
the above bottles, in which the oxide of an element 
has been formed and to which water has been 
added to form a solution, add 1 to 2 ml of litmus 
solution. 1 Correlate the chemical character of the 

1 Ferric oxide or hydroxide if to insoluble that litmus ha& no effect 

reaction product with the position of the element 
in the periodic table. 

Oilier oxides. The above tests include representa- 
tive elements from groups 2, 4, 5, 6, and the transi- 
tion group 8, in the periodic table. To complete the 
series, let us examine an oxide or hydroxide from 
each of the other principal groups 1, 3, 7, and 
another example from group 5. (Why not include 
an element from group 0?) 

The normal alkali oxides, as Na 2 O, are difficult 
to obtain. Sodium peroxide forms when sodium 
burns in oxygen. The final reaction product with 
water, however, is the same as when Na 2 O reacts 
with water. The extra "peroxide oxygen" is liber- 
ated as free oxygen gas. Compare the equations 

NaaO + H 2 O 

2 NaOH 

2 Na 2 O 2 + 2 II 2 O > 4 NaOH + O 2 . 

Boil a very small amount of sodium peroxide in 5 
ml of water in a 15-cm test tube for a moment to 
complete the above reaction. Cool the solution, and 
test it with litmus. Also note the slippery, soapy 
feeling of such a caustic solution. Wash your fingers 

The oxide of boron, B 2 O 3 , is not readily available, 
and in some forms is very insoluble. Dissolve a 
small amount of boron hydroxide in 5 nil of hot 
water. Cool it, and test it with litmus. Decide 
which formula, B(OH) 8 , or H 3 BO 8 , is preferable. 
How is the substance usually named? 

To test the character of an oxide of nitrogen, con- 
nect a bent delivery tube to a 15-cm test tube 
which contains a small amount of metallic copper. 
Add 1 ml of water and 3 ml of concentrated IlNOj, 
then quickly close the tube, and bubble the evolved 
nitrogen dioxide through 5 ml of distilled water 
contained in a second test tube. Boil this solution 
a moment, cool it, and test it with litmus. The 
reaction is 

3 NO 2 + H*O * 2 HNO 8 + NO. 

The oxides of chlorine are all very unstable. Dis- 
solve several drops of a 60% solution of a hy- 
droxide of chlorine (whose formula may be written 
ClOa(OH) or HC1O 4 ), in 5 ml of water, and test the 
solution with litmus. Decide which of the two for- 
mulas you regard as the preferable one. 

on the solution. The effect may be somewhat feeble in one or two 
other cases, but should be sufficient to indicate the acidic or 
character of the hydroxide. 

Drill on Formulas and Nomenclature 

Note: Use this material, as needed, to help you learn how to write correct formulas and name compounds. Check your 
answers with your instructor, but they need not be handed in unless called for. 

1. Name the following: 



Fe 2 (S0 4 ) 3 - 



NC1 3 

PC1 6 

La(NO 3 ) a 

(NH 4 ) 2 S ______ 

S solution 


SnS 2 

Hg(NO,) 2 

- HgCh 

2. The spaces below represent portions of some of the main groups and periods of the Periodic Table. In the 
proper squares, write the* correct formulas for the chlorides, oxides, and sulfates of the elements of main groups I, 
II, and III, respectively. Likewise, write the formulas of the compounds of sodium, calcium, and aluminum with 
the elements of main groups VI arid VII. Two of the squares have been completed for you as examples. 

Group I 

Group II 

Group HI 

Group VI 

Group VII 



Li 2 O 
Li 2 S0 4 

(omit the aulfate) 



Na 2 S 




3. Give the valence (including the + or sign) for the italicized element or radical in each of the following 
formulas : 

NII 4 0//_ 

H 8 P04. 


4. From the valences which you found for the hypothetical elements X and Z above, what would be the for- 
mulas of the following? Fill in the proper subscripts: 

II , 

X ( 


,(P0 4 ) ( 


REPORT: Exp. 8 

The Oxides of the Elements Date 

and the Periodic Table 

Locker Number* 

1. The Preparation of Oxygen. Write the equation for the reaction by which you prepared oxygen. 

Compare (a) the heating of potassium chlorate alone, and (b) the heating of the same amount of potassium 
chlorate, mixed with some manganese dioxide, as to: 

(1) The relative temperatures at which decomposition readily takes place: 

(2) The relative amounts of oxygen which may be obtained: 

(3) Any changes taking place in the manganese dioxide: 

(4) The name applied to a substance used like manganese dioxide is used in this reaction: 

2. The Preparation of Oxides. Describe any changes occurring during the reaction of each element with 
oxygen, and any distinctive characteristics of the products formed. Write the equation for the reaction, in each ease. 








3. The Reaction of Oxides with Water. On the line corresponding to its periodic group, write the formula 
of each oxide (or hydroxide, if the oxide was not available) which you studied in this experiment. Indicate the 
reaction to litmus of its water solution, and write the equation for the formation of the acid or base. 

Formula of Reaction to Equation for Reaction, if Any 

Group Oxide Litmus 

II (two __ 


III . ^ 


V (two . 


VI . . 

VII . . 

VIII (transition) 

Comment on the acidic or basic character of the oxide of an element, as compared to its position in the periodic 

Application of Principles, Prom the position of the elements in the periodic table, write the formulas of 
three other oxides, which, in each case, would behave as: 

Acidic oxides Basic oxides 

Write the formulas of the anhydrides of: 

H 2 SO 3 Ba(OH), HNO 2 H 2 C(V 

H 3 P(>4 HNOa. 

Complete and balance the following equations: 

_MgO + H 2 S0 4 

+ CO, * 

-CatOH)* + 


The Chemistry of Nitrogen and Ammonia. 


College C/iem/sfry, Chop/en 5, 75 

Review of Fundamental Concepts 

We have noted already, in Experiment 5, that 
free nitrogen gas is quite inactive at room tem- 
perature. This is in accord with its relative abun- 
dance in the atmosphere, and in accord with the 
scarcity of nitrogen compounds. 

Impure nitrogen, which still contains the inert 
gases, principally argon, may be obtained from the 
air by any reagent which will remove the oxygen. 
This may be accomplished by the use of phos- 
phorus, pyrogallol or metallic iron as in Experi- 

ment 5, by the passage of air over hot metallic 
copper, or commercially by the fractionation of 
liquid air. Pure nitrogen, as we shall learn in this 
experiment, can be obtained by the chemical re- 
action of some of its compounds. We study am- 
monia this early in the course, because of its 
constant use as a laboratory reagent. The oxides 
and acids of nitrogen will be considered in Experi- 
ment 22. 

Experimental Procedure 

Special supplies: Three 250-ml wide mouth bottles, 3 glass 
plates, pneumatic trough, deflagrating spoon, asbestos fiber, 
wooden splints. 

Chemicals: NBUCl, NaNO 2 , Mg (ribbon), S, P (red), 
Ca(OH)fc CuO, litmus solution, 1 F NH 4 CL 

1. The Preparation of Pure Nitrogen. As- 
semble the apparatus as illustrated in Figure 9-1. 
Prepare to collect three bottles of the gas by dis- 

placement of water. Place 4 g of ammonium chlo- 
ride (NH 4 C1), 5 g of sodium nitrite (NaNO 2 ), and 
30 ml of water in the flask. 

Since the reaction is an exothermic one (releases 
energy as it proceeds), be prepared to loosen the 
clamp and immerse the generator in the pneumatic 
trough to cool it, in case the reaction becomes too 

Loosen clarnp and 
immerse flask in pneumatic 
trough if reaction 
becomes too rapid 

Worm slightly to 
tart the reaction 

FIG. 9-1. The preparation of pure nitrogen gas. 



rapid. Warm the flask just enough to start the 
reaction, then withdraw the flame. Permit the first 
gas generated (which will be mixed with the air of 
the flask) to escape, and then collect three bottles 
of nitrogen. 

Properties of Nitrogen. Observe the usual physi- 
cal properties of the gas, as color, odor, solubility 
in water, etc. Test the ability of nitrogen gas to 
burn or to support combustion by inserting a burn- 
ing splint into a bottle of the gas. Try, also, sub- 
stances which undergo a more vigorous reaction in 
the air, as a piece of magnesium ribbon, and a little 
sulfur or phosphorus on a deflagrating spoon which 
has been lined with asbestos. 

2. The Preparation of Ammonia, (a) Forma- 
tion. Place a mixture of several grams each of am- 
monium chloride and calcium hydroxide in a 15-cin 
test tube, and warm it. Note the odor. Also, expose 
a piece of moist, red litmus to the vapors. 
1 (6) Preparation. A more convenient method for 
our use is as illustrated in Figure 9-2. Assemble this 
apparatus. The generator flask contains about 30 
ml of concentrated ammonium hydroxide solution. 
The drying tube should be well filled with fresh, 
anhydrous, 4-mesh calcium chloride (CaCl 2 ), with 
loose plugs of cotton at either end. Note the rela- 
tive length of the delivery tubes in the collection 

bottles, and in the test tube. Place about a gram of 
cupric oxide in the test tube. In the receiving flask, 
place about 35 ml of water, and adjust the delivery 
tube so that it almost touches the water. 

To generate ammonia, gently warm the flask 
just enough to maintain a moderate rate of evolu- 
tion of bubbles in the final receiving flask. After 
the gas has been generating a minute or two, heat 
the cupric oxide in the test tube, with a second 
burner. When ammonia reaches the hot cupric 
oxide, two changes should be obvious to you. When 
the change in the cupric oxide is complete, let the 
test tube cool, but continue generating ammonia 
for a few minutes more to form a solution in the 
receiving flask, and then withdraw all heat and 
permit the operation to subside. 

3. Properties of Ammonia, (a) Reducing Ac- 
tion. You have already carried out this operation 
during the preparation of the gas. Record your 
observations of this in the report sheet. 

(b) Behavior on Burning. Disconnect the two 
bottles of ammonia from the collection system, and 
cover them at the same time with a glass plate so 
as to lose as little ammonia as possible. Thrust a 
burning splint into one bottle of ammonia. Note 
whether the gas is inflammable, and whether it will 
support combustion. 

Fio. 9-2. The preparation of ammonia gas* 



(c) Behavior with Water. Place about 300 ml of 
water in a 400-ml beaker, add 1 to 2 ml of litmus 
solution and 1 ml of dilute HC1 to turn the litmus 
red. Now immerse the second bottle of ammonia 
gas, mouth down, into the beaker of acidified 
water, and observe the results. 

Test 5 ml of the solution of ammonia gas in 
water (from the flask) with red litmus. Boil this 
solution a few minutes, then test the residual solu- 
tion again with red litmus. If necessary, repeat the 

In order to compare this behavior with that of 
a solution of hydrogen chloride, boil 5 ml of dilute 
hydrochloric acid solution as you did the ammonia, 
and test the residue each time with blue litmus. 

(d) Volatility. Place not over 1 g of ammonium 
chloride (NI^Cl) in an evaporating dish, and heat 
it strongly as long as any action occurs. 

Vary this procedure by placing a little solid 
NH 4 C1 in a dry 15-em test tube, and then place a 

loose plug of asbestos fiber about half way down 
the tube. Moisten a piece each of red and blue 
litmus paper, and attach them to the inside wall of 
the upper part of the test tube. Heat the NH 4 C1 
and explain any change in the litmus in terms of 
the relative rates of diffusion of the gases present. 
(e) Test for Ammonium Ion. To determine the 
conditions necessary for a positive test, place 1 ml 
of 1 F NTI 4 C1 in an evaporating dish, and cover 
this with a watch glass, to the under side of which 
is attached a moist piece of red litmus paper. Warm 
the dish slightly (do not boil the solution) to dis- 
cover whether ammonia gas will be liberated on 
warming ammonium salt solutions. After you ob- 
serve the result, add 1 ml of 6 F NaOII to the solu- 
tion, replace the watch glass, and watch for any 
change in the litmus. Warm again if necessary. 
(The NaOH solution is not volatile. Convince 
yourself of this by repeating the above test, with- 
out the NH 4 C1 present.) 

REPORT i Exp. 9 Name- 

The Chemistry of Nitrogen Date 

and Ammonia 

Locker Number- 

L The Preparation and Properties of Nitrogen 

Write the equation for the reaction by which you prepared nitrogen gas: 

Summarize your observations on the physical properties and general chemical behavior of nitrogen gas. 

Recall the reaction of nitrogen gas on very hot, active metals, by re-writing here the equation for its reaction 
with magnesium, and the equation for the reaction of the product with water. 

2, 3. The Preparation and Properties of Ammonia 

Write the equation for the reaction which occurs on wanning a mixture of solid ammonium chloride and calcium 

(1) Write the equation for the reaction of ammonia with cupric oxide. (The nitrogen is liberated as N gas.) 

In this reaction, which substance is the 
Reducing agent? Oxidizing agent ?_ 

(2) Does ammonia support combustion? Does ammonia burn?.. 

(3) When ammonia is dissolved in water, what evidence is there of any chemical action? 

What is the equation for the change that takes place: 

When ammonia dissolves in water? 

* When a solution of ammonia is boiled? 


Compare the behavior which you observe on boiling ammonia solution, and on boiling hydrogen chloride 

(4) Comment on the volatility of ammonium salts. 

Write the equation for the reaction that takes place when solid ammonium chloride is heated. 

Interpret the change in the litmus paper when solid ammonium chloride was heated in a test tube containing an 
asbestos fiber plug. 

(5) Compare the results of warming an ammonium chloride solution, first without, and then with, the addition 
of sodium hydroxide solution. 

What is the equation for this reaction? 

Outline the procedure you would follow in the testing of a commercial product, for example a fertilizer, for 
the presence of ammonium salts. 


Write the equation for: 

(1) The formation of white fumes when concentrated HC1 solution is brought in contact with ammonia gas. 

(2) The neutralization of ammonium hydroxide solution by hydrochloric acid. 

(3) The neutralization of ammonium hydroxide by sulfuric acid. 

"the Determination of the Formula 

of a Compound from Experimental Data. 


Review of Fundamental Concepts 

Units of Quantity as Used in Chemistry 

The terms, atomic weight and molecular 

weight, refer to the weights of the atoms and 
molecules of the various elements and compounds, 
respectively, relative to the mass of an oxygen 
atom which is arbitrarily taken as 16.0000. Be- 
cause atoms and molecules are the units in which 
chemical reactions occur, and since they are such 
minute particles that it is impossible to deal with 
them individually, for practical purposes we use 
larger unit quantities which contain equal numbers 
of atoms or molecules per unit. These are defined 
in the following paragraphs. 

The gram-atomic weight (usually referred to 
as a gram-atom) of an element is defined as a 
weight in grams numerically equal to the atomic 
weight of the element. Thus, 12 grams of carbon 
I (atomic weight 12) is one gram-atom of carbon. 
Likewise, a quantity of sulfur (atomic weight 32) 
which weighs 64 grams, is two gram-atoms of 
sulfur. A chemical symbol, as C, or S, repre- 
sents in a rather loose sense simply the name of 
the element. Very frequently the symbol is used 
to designate one atom of an element. In chemical 
calculations we must use a large enough unit to 
be practical, in which case a symbol represents one 
gram-atom of an element. Thus the symbol C 
stands for 12 grams of carbon, and O for 16 grams 
of oxygen. 

The gram-molecular weight, or mole, of a 
substance, likewise, is defined as a weight in grams 
numerically equal to the molecular weight of the 
substance, or is the sum of the gram-atomic 
weights of the constituent elements as represented 
by the formula of the substance. Thus one mole of 
carbon disulfide (1 CS 2 ) contains 1 C (12 grams) 
and 2 8 (2 X 32 grams) and weighs 76 grams. 

It is important to remember that one gram- 
atom of any element, or one mole of any compound, 
contains the same number of individual atoms or 
molecules, respectively. This is Avogadro's num- 
ber, 0.6023 X 10 24 particles. Thus 12 grains of 38gCS, 
carbon (1 C), 207.2 grams of lead (1 Pb), 76 grams 76 g/mole 

College Chemistry, Chapter 8 

of carbon disulfide (1 CS 2 ), and 18 grains of water 
(1H 2 O), each contains this same number of atoms 
or molecules, respectively. 

The gram-formula weight. In many sub- 
stances, such as the salts (NaCl, CuQ 2 for ex- 
ample), the units of crystal structure are not 
molecules, but are ions the same ions as those 
formed when the substance dissolves to form a 
solution. Strictly speaking, then, we may say there 
is no such thing as a particle of "NaCl," or "CuCl 2 " 
in which these formulas represent a single mole- 
cule. Such formulas express merely the simplest 
proportion of the constituent ions. The term "gram- 
molecular weight," or "mole," although frequently 
used in such cases, is thus a misnomer. Throughout 
this manual we shall use the term "gram-formula 
weight" abbreviated to gfw, to express the weight 
in grams as represented by the formula given. The 
term "mole" will be reserved for substances in 
which the molecular species is known and repre- 
sented by the formula used. 

Study Figure 10-1 to further clarify the meanings 
of the terms gram-atom, mole, and formula weight. 
Some examples will help you to learn to think in 
terms of chemical units of quantity. Remember 
that the gram-atomic weight means grams per gram- 
atom, and that the gram-molecular weight means 
grams per mole. Thus, the defining equations are: 

grams grams 

1 " 7^p *" (2) g mo wt " ~~ rr " 


For example, to calculate the weight of ten gram- 
atoms of sulfur, we transpose (1) thus, g at X g at 
wt = grams, and write 

10 g at S X 32 -- - 320 grams S. 

Conversely, for example, to calculate the number 
of moles in 38 grams of carbon disulfide, we trans- 

pose (2) thus, 

g inol wt 

moles, and write 

-0.600 moles CS* 


8.2 cm 

Helium (He) 
At. wt. 3 4.003 Volume = 23. 

MoLwt,- 32.00 Volume : 
1 mole = 32.oo d 

Carbon dioxide(CO a ) 
Mo), wt. 44.01 Volume 
1 mole - 44.01 tf 


Mercury (Hg) Water (H^O) Carbon disulfide(CS^) 

At wt. = 200,61 Density 13.546 Mol.wt.= is.oi6 Density = 1.00 MoLwt,^ 76.13 Density* 1.261 
1 gram atom * 200, aig 1 mole is.oieg 1 mole = it.isg 

3.00 cm 

4,35 cm 

@,m w* c 

Carbon (diamond) 
At.wi - ia.oi Density- 
Igram atom ^ 12.01 g 

FiG/10-l. Gram-atoms, moles, and formula weights. Although the weight in grams is very different for different substances, the 
number of atoms or molecules per gram-atom or mole is the same in all cases, namely the "Avogadro number." 

Salt (NaCl) Sulfur (S a ) 

formula wt.-jd.454 Den*ity~zi6S Molwt.=* 256.45 Penalty 

1 mole ~ 256.45 g 

The Calculation of the Formula from the Data 

While experimental data express the relative 
weights (in grams) of the constituent elements, a 
formula expresses the relative number of gram- 
atoms of each constituent element. Furthermore, 
this number of gram-atoms is always capable of 
being expressed as an integral ratio of small whole 
numbers. (This is a consequence of the atomic 
theory.) We need therefore only to calculate the 
number of gram-atomic weights of each element 
from the number of grams of each in any given 
amount of the compound, and find the simplest 
integral ratio of these. 

For example, in finding the formula of a chloride 
of copper, it was analyzed and 5,00 grams of it was 

foirnd to contain 2.35 grams of copper and 2.65 
grains of chlorine. We may calculate the respective 
number of gram-atoms of each as follows: 

2,35 g Cu 
63.5 g/g at 

0.0370 g at Cu 

0.0746 g at Cl. 

2.65 g Cl 
35.5 g/g at 

Dividing each of these numbers by the smaller 
one, we have a ratio of 1 Cu to 2.02 Cl. The simplest 
formula is, therefore, CuCl2. 

For some substances, where there is a definite 
molecule as the unit of structure of the substance, 
we can write a definite correct "molecular for- 
mula." The formula of hydrogen peroxide is 



not HO, because the molecular weight of this sub- 
stance has been determined experimentally, and 
has a value of 34, not 17. Note that the two for- 
mulas correspond to the same percentage composi- 
tion. The determination of molecular weights is 
considered in Experiment 13. 

In this experiment a weighed sample of silver is 
dissolved in nitric acid (UNO*) and precipitated as 
silver chloride. From the weight of silver and the 
weight of silver chloride the percentage composi- 
tion of the silver chloride may be calculated. The 
formula may then be calculated as in the case of 
CuCl 2 above. 

Experimental Precision vs. Significant Figures 

This experiment and the following one present 

the first strictly quantitative experiments, in 
which you are expected to carry out your weigh- 
ings with care and with as great a precision as the 
data justify. Before going ahead with the experi- 
mental procedure, read the section under "Labora- 
tory Manipulations" on The Care and Use of 
the Balance, page xx. Follow any special instruc- 
tions which your instructor gives you. 

When experimental data are recorded and when 
calculations involving these data are made, it is 
very important to record the answer in such a way 
as to convey the proper estimate of the precision 
of measurement. Study Section B in Appendix I on 
Significant Figures, and answer the questions in 
the report sheet on this point. 

Experimental Procedure 

Special supplies: Analytical- weights. 

Chemicals: Sample of pure silver (your preparation from 
Exp. 6 if pure). 

Heat, cool, and weigh (to 0.001 g if possible) a 
clean evaporating dish. (It is worth while to keep 
a record of the individual weight denominations 
used in each weighing, on a separate sheet, as a 
check against errors in counting weights.) Enter 
this weight in the report sheet, and then carefully 
transfer to the dish from 0.5 g to 1.0 g of pure dry 
silver metal. (Use your preparation from Experi- 
ment 6, if it is pure.) Again weigh the dish with its 
contents, and record this weight. Place the dish on 
a wire gauze supported on an iron ring. Add 3 ml 
of dilute nitric acid, and cover the dish with a 
watch glass. If action does not begin immediately, 
warm the dish momentarily, but avoid any loss of 
solution by spattering. When all the silver has 
been dissolved, remove the watch glass. Rinse the 
under side of the watch glass into the dish, by 
holding it edgewise over the dish, and spraying it 
with distilled water from your wash bottle, as illus- 
trated by Figure 10-2. 

Now add 3 to 5 ml of dilute hydrochloric acid 
to the solution, and evaporate the mixture very 
gently by placing the wire gauze with the dish on 
it at a height of two or three inches above a low 
(%") flame. Use great care to avoid any spattering 
of the precipitate. (See Fig. 10-3.) As the mixture 
approaches dryness, it may be wise to transfer the 

Rinse back and forth 
in this manner 

Rinse the 

bottom of the watch glass 
with distilled water 

from your wash bottle* 

Fio. 10-2. The technique of rinsing a watch glass. 

dish to the top of a beaker of boiling water (Fig. 
10-4) to complete the evaporation. When the pre- 
cipitate is apparently dry, again heat the dish 
cautiously directly on the wire gauze, wafting the 
flame back and forth under the dish until it is 
quite hot and thoroughly dry. Do not melt the 
residue, but be sure all vapors of water and acid 
have ceased to be evolved. Cool and weigh the dish 
and residue, being careful to avoid all unnecessary 
handling of the dish, as it will reabsorb moisture 



from the hands. Record the weight. From these 
data, calculate the formula of silver chloride, as 
outlined in the report sheet. 

Optional experiments, for students who have 
the time and interest to do further work. Some 
details 01' experimental procedure are left to the 
student to devise. Also, prepare a suitable report 
sheet for the entry of all data before they are 

lid, let the crucible cool, and examine the conten 
for vaporization of all excess sulfur (a black staw, 
will remain). Finally weigh the crucible and con- 
tents. Calculate the formula of the compound. 

(2) A chloride of lead. The same general proce- 
dure as for silver chloride may be used; use about 
0.5 g of lead, which has been cut in thin shavings 
to facilitate solution. After accurately weighing the 
dish, and the dish plus the lead, add about 10 ml of 

Heat the 
very gently. 

FIG. 10-3. The evaporation of liquid contained in an evap- 
orating dish. The liquid may simmer very gently, but you 
should avoid boiling it vigorously. 

When the 
dish is nearly 
dry, transfer it 
to the top of a 
beaker of boil- 
ing water. 

FIG. 10-4. The evaporation of a liquid over a water bath. 

(1) A sulfide of copper. First heat a clean cruci- 
ble, supported on a clay triangle, for a moment. 
Cool the crucible and weigh it accurately, without 
the cover. Add about 2 grams of copper, as wire 
or coarse turnings, and press this well down in the 
crucible. Weigh this accurately. Add 1 to 1.5 g of 
powdered sulfur. Avoid a large excess, as it takes 
longer to burn out. Now place the crucible, with 
lid in place, under the fume hood, and heat it 
moderately to form the compound and vaporize 
excess sulfur. Also play the flame around the sides 
and top of the crucible and lid. Do not remove the 
cover while the crucible is hot, as atmospheric oxida- 
tion would take place. Avoid heating longer than 
necessary. When sulfur ceases to burn around the 

coilc nitric acid. Cover the dish with a watch glass, 
and warm it occasionally as needed to promote 
solution of the metal, which is very slow. It may be 
necessary to leave this over night. When the metal 
is all dissolved (solid lead nitrate will remain), use 
a water bath (Fig. 10-4) to evaporate most of the 
acid. Then add 5 ml cone hydrochloric acid to con- 
vert the salt to lead chloride. Evaporate just to 
dryness, and again moisten the salt with a few 
drops of hydrochloric acid, and evaporate. Finally 
heat with the bare flame, very gently to avoid 
spattering, until all acid vapors are gone and the 
salt is thoroughly dry. Avoid a high temperature. 
Let the dish cool, and weigh it. Calculate the for- 
mula of the compound. 

REPORT: Exp. 10 

Percentage Composition and Formula 
of a Compound 



Locker Number, 

Data 1 



Weight of dish and silver chloride 



Weight of dish and silver 



Weight of dish, dried, at start 



Cram atomic weight of silver 2 



Gram atomic weight of chlorine 






Weight of silver chloride 



Weight of silver 



Weight of chlorine 



Number of gram atoms of silver in 
weight taken 4 



Number of gram atoms of chlorine in 
weight taken 



Simplest integral ratio between gram 
atoms of Ag and Cl (from your data) 

Formula of the compound 
(from your data) 

Theoretical % silver in compound, 
from its formula 



Experimental % silver in compound 
(from your data) 



Percentage error 




Approval of Instructor- 

1 Data are listed in the order most convenient for subtraction, rather than in the order in which they are obtained. Columns are provided 
for two samples, or for repetition of the experiment, as desired. 

2 An atomic weight table is given inside the cover of the manual. 

i Indicate the method of all calculations in the spaces provided, but do not carry out the actual multiplication, division, and so forth, 
here. As soon as you have finished your calculations, report the result to your instructor. 

4 Calculate the number of gram atoms to 3 significant figures, not just to "3 decimal places." 


Significant Figures, Problems 

"How far shall I carry out my figures?*' To learn the principles governing this common problem of the beginning science 
student study Appendix I, Sec. B, on "Significant Figures," and answer the following questions. 

Note: In general, in this course, calculate the answers to problems to three significant figures (slide-rule accuracy), except 
where the data given indicate that greater or less precision should be indicated in the answer. 

1. How many significant figures are there in each of the numbers (a) 35.6, (b) 3.56, 
(c) 742.0, (d) 7.0058, (e) 0.0058, (f) 0.6023 X 10 24 ? 

2. To a flask the following volumes of water were added: 4.27 ml, 101 ml, and 36.4 ml. 
The total volume of water should be recorded as: 

3. In finding the percentage of silver in silver chloride, the weight of silver was 0.552 gram. 
The weight of chlorine was 0.178 gram. The percentage of silver should be recorded as: 
(a) 70%, (b) 75%, (c) 75.6%, (d) 75.62%, (e) 75.616%, (f) 75.6164%. 

4. How many gram atoms in one pound (454 g) of 

5. How many moles in one pound (454 g) of 

gold? . . 

sugar, C 
water? . 

6. Methanol (wood alcohol) contains 37.5% carbon, 12.5% hydrogen, and 49.9% oxygen. 
What is its formula? Note: A 100-gram sample of methanol contains 37.5 g of carbon, 
12.5 grams of hydrogen, and 49.9 grams of oxygen. 

7. A compound of bismuth and chlorine is found to consist of 3.15 g of Bi and 1.60 g of Cl. 
What is the formula? 


1 (a) (b)_ 

(c) (d). 

(e) (f)_ 



The Atomic Weight off Magnesium. 


College Chemistry, Chapter 8 

Review of Fundamental Concepts 

Exact atomic weight determinations constitute 
some of the most fundamental of all scientific data. 
They are determined with the most painstaking 
care, with many elaborate precautions, and with 
the finest of balances and other equipment. As you 
know, they are based on the weight of a given 
element which combines with, or is equivalent to, 
one gram atom of oxygen, which is arbitrarily 
assigned a value of 16.0000 g. Since oxygen has a 
valence of two in most of its compounds with other 
elements, one half of its gram-atomic weight, 
namely 8.0000 g, is often taken as the standard of 
reference in determining the relative combining 
weights of other elements. The "combining weight" 
thus obtained, or some simple integral multiple of 
it, will be the true atomic weight of the given 

The factor by which this equivalent or combin- 
ing weight must be multiplied to give the correct 
atomic weight can be decided by several methods, 
as outlined in your text. At the present time the 
known position of the element in the periodic 
table, and its atomic number as experimentally 

determined by the Moseley method, give unmis- 
takable evidence of the approximate value of its 
atomic weight. In earlier days, as first discovered 
by Dulong and Petit, an experimental determina- 
tion of the specific heat of the element enabled 
chemists to obtain approximate atomic weight 

For some elements, direct measurements by 
comparison with oxygen are inaccurate or imprac- 
tical. For these, "secondary standards" other 
elements whose atomic weights have been most 
carefully obtained are used. The elements silver, 
chlorine, and bromine are such standards. Note 
that their atomic weights are known to Jive or six 
significant figures. 

The procedures of this experiment are quite 
simple. The precision you are able to attain will be 
limited primarily by your technique and by the 
limitation of the balance you use. If a balance 
equipped with a rider (which enables you to weigh 
to 0.001 g) is not available, estimate the nearest 
thousandth of a gram as near as you can by the 
technique illustrated by Figure i-17, Page xxii. 

Experimental Procedure 

Special supplies: Analytical weights. 

Chemicals: Mg ribbon. Other metals as needed in wire or 
thin sheet form, for optional experiments. 

Heat, cool, and weigh (to 0.001 g if possible) a 
clean crucible 1 and lid. (It is worth while to keep a 
record of the individual weight denominations 
used in each weighing, on a separate sheet, as a 
check against errors in counting weights.) Enter 
this weight in the report sheet. Now add to the 
crucible about 0.5 to 0.8 g (about 3 feet) of mag- 
nesium ribbon, which has been crumpled so as to 
fit into the lower part of the crucible and still 
allow good exposure to the air. Do not pack it 
down too tightly. Again weigh the crucible and lid 

1 Since this experiment causes damage to the glaze of the crucible, 
you may use old crucibles. Concentrated HC1 will afterwards remove 
any dark stain you may find. 

with its contents, and record the weight. Place the 
crucible on a clay triangle over the burner. Adjust 
the lid to allow a small amount of ventilation (sec 
Figure 11-1), and heat to oxidize the magnesium, 
'Avoid heating the crucible too rapidly, or with toe 
much ventilation, in order to prevent an appreci- 
able amount of "smoke" (magnesium oxide) from 
escaping. You can also regulate the rate of burning 
by using your crucible tongs to adjust the position 
of the crucible lid. 

Continue the heating until oxidation is about 
complete. This you can determine by lifting the lie 
slightly with your crucible tongs and seeing 
whether now the contents of the crucible do nol 
glow brightly. When you are sure, remove the lie 
to one side, to provide an abundance of air, anc 
continue the heating at the maximum temperature 




The equations for the reactions are: 
2Mg + 2 

3Mg + N, 

MftN, + 6 HOII + 3 Mg(OH), + 2 NH, 

Mg(OII) 2 HVIgO 

FKJ. 11-1. The rate of oxidation of magnesium may be con- 
trolled so as to avoid smoke, by regulating the amount of 
ventilation and the size of the flame. 

for another five minutes. 

After the crucible becomes cool, gently crush 
into a powder the product that has been formed, 
using the smooth end of a glass stirring rod. Make 
the powder lie compactly on the bottom of the 
crucible for thorough reheating in subsequent op- 
erations. Any powder adhering to the stirring rod 
may be tapped back into the crucible without loss. 
(A camel's hair brush is often used to loosen such 
powder from the rod. Your instructor may assist 
you with this operation if necessary.) 

Now moisten the product w r ith 10 to 15 drops of 
distilled water from a medicine dropper. This will 
change the magnesium nitride, some of which was 
formed at the same time, into magnesium hy- 
droxide. Warm this and note the odor of ammonia, 
which has also been formed. 

Finally, reheat the crucible with the lid fully in 
place, at first gently to avoid spattering, then at 
full heat (to give the crucible bottom a good red 
heat) for at least 5 to 10 minutes. This is neces- 
sary to achieve complete decomposition of the mag- 
nesium hydroxide. 

When the crucible, lid, and contents are com- 
pletely cool, weigh 1 them accurately and record the 

From these data you can calculate the combining 
weight of magnesium (the weight of the metal that 
combines with 8.0000 g of oxygen). From the 
known position of magnesium in Group II in the 
Periodic Table, and from its atomic number, you 
can decide by what integral factor you should 
multiply its combining weight to get the true 
atomic weight. 

Optional Experiment. The Atomic Weight of 
Tin. An oxide of tin may be formed by adding 
about 10 ml of concentrated nitric acid to a 
weighed sample of tin metal, in a weighed evapo- 
rating dish. After complete disintegration of the 
metal, evaporate the liquid very carefully (see 
Figures 10-2, 10-3, and 10-4 in the preceding ex- 
periment) so as to avoid spattering. Finally ignite 
the residue at the maximum temperature of the 
burner for at least 20 minutes, making certain that 
all portions of the dish become very hot, so that the 
metastannic acid first formed is completely decom- 
posed to the oxide. Cool the dish and weigh it. 

From these data you can calculate the combin- 
ing weight of tin (the weight of the metal that 
combines with 8.0000 g of oxygen). From the 
knowfi position of tin in the Periodic Table 
(Group IV elements characteristically have a 
valence in their compounds of two or four), and 
from its atomic number, you can decide by what 
integral factor you should multiply its combining 
weight to gel its true atomic weight. 

1 Weigh the MgO within about 20 minutes after it is completely 
cool, as it tends to regain moisture from the air and form Mg(OII)a 
again. If excessive time elapses, reheat and cool the crucible before 
weighing it. 

REPORT: Exp. 11 \ame- _ 

The Atomic Weight of Date 


Section __. _ . _ 

Locker \umbcr 

Data I 3 

Weight of crucible, lid, and magnesium oxide - - g , . _g 

Weight of crucible, lid, and magnesium g g 

Weight of crucible and lid ___ g g 


Weight of magnesium . .g . g 

Weight of oxygen ..__.__ g g 

Combining weight of magnesium, from data _ . ._..._. _. g . __g 

Atomic weight of magnesium, from data . - _____ g ________ _.. g 

Percentage error ,... .. __ _ . c { } _,_._, % 

Evaluation of Experimental Work 

1. A student completes his experiment a half hour before the close of the laboratory 
period. The first thing he should do is: (a) clean and lock his locker, (b) make his 
calculations and report his answer, (c) go home, (d) consult his neighbor's calculations. 

2. A student's magnesium oxide spatters on drying out the added water, and some is lost. 
He should: (a) estimate how much was lost and make allowance for it in his answer, 
(b) repeat the experiment with a new sample, (c) repeat the experiment with a new 
sample after devising a method to prevent spattering. 

3. A student weighs his magnesium to four significant figures, and the weight of the 
magnesium oxide obtained from it to two significant figures. His experimental results 
should be reliable to (a) 2, (b) 3, (c) 4, (d) 6 significant figures. 

4. Assuming that you used approximately 0.5 gram of magnesium, and that your balance 
is capable of weighing to 0.001 g, to how many significant figures are you justified in 
expressing your answer? Explain your answer: 

5. Suppose you did not bother to transform the impurity of magnesium nitride 

in your sample to magnesium oxide (MgO). Would this cause your value for the 
atomic weight of magnesium to be: (a) too low, (b) too high, (c) unaffected by 
the impurity? Explain your answer: 







1. Calculate the percentage composition of the elements in common salt, NaCl. (Use 
such atomic weight values as are necessary to give the correct answers to 3 significant 


2. Ethyl chloride is composed of 37.2% carbon, 7.8% hydrogen, and 55.0% chlorine. 
What is its simplest formula? 

3. A sample of an oxide of lead, which weighs 2.50 g, is reduced by heating in a stream of 
hydrogen gas, thus forming 2.27 g of lead. What is the simplest formula of this oxide? 

4. The hemoglobin of the blood, a protein, is known to contain 0.335% iron. There 
must be at least one atom of iron per molecule of the protein. On this basis, what must 
be the minimum weight of a mole of hemoglobin? 

5. (a) If the specific heat of element E has been determined and has a value of 0.0508 
cal/g deg, what is the approximate atomic weight of the element? (b) If 1.218 g of this 
element, when combined with oxygen, gives a total weight of 1.458 g, what is its 
combining weight (weight combined with 8.0000 g of oxygen) ? (c) What is its exact 
atomic weight? (d) What is the formula of the oxide? 

5 (a). 




The Molal Volume of a Gas. 


Co//g Chemistry, Chapter 9 

Review of Fundamental Concepts 

Avogadro's Law and the Molal Volume 

In order that you may understand the behavior 
of gases, we review two important facts that are 
related to each other. The first of these we consid- 
ered in Experiment 10, namely , that one mole of any 
substance contains the same number of molecules 
as a mole of any other substance. (This is inherent 
in the definition of a mole.) The second fact, which 
is a consequence of the kinetic theory of gases, and 
was first propounded by the Italian chemist 
Avogadro about 1811, states that equal volumes of 
all gases, measured under similar conditions of tem- 
perature and pressure, will contain the same num- 
ber of molecules. As a necessary corollary of these 
two facts, then, one mole of any gas will have a 
volume equal to that of one mole of any other gas 
measured under similar conditions. This volume is 
called the molal volume of a gas. 

The Molal Volume at Standard Conditions. The 
Perfect Gas 

To evaluate this volume in metric units, recall 
that the symbol O stands for 16.0000 grams of oxy- 
gen, and that one mole of oxygen gas, O 2 therefore 
weighs 32.0000 grams. As you will verify in this ex- 
periment, this has been found experimentally to 
have a molal volume of 22.4 liters at standard 
conditions (0C and 760 mm Hg pressure). Every 
other gas, likewise, will have a molal volume at 
standard conditions of 22.4 liters. 

Since all real gases deviate more or less from the 
behavior of a perfect gas, their respective molal 
volumes will vary slightly. Only in cases of perfect 
gases, were they to exist, would the molal volumes 
not vary. A perfect gas is that idealized gas in 
wl>ich the molecules would have no attractive forces 
whatever for one another, and likewise would be 
mere "points," without any significant volume. 
The standard molal volume for such a perfect gas 
has been calculated from measurements on real 
gases at very lowpressures to be 22.4140 liters. Most 
ordinary gases, unless they have a high molecular 
weight or are measured quite near their boiling 

point, have molal volumes which do not deviate 
more than about 1% from this value. 

The Calculation of Gas Volumes. The General 
Gas Law Equation 

It is much easier to measure the amount of a 
given gaseous substance by measuring its volume 
than by weighing it. Of course, this necessitates 
also the measurement of the temperature and the 
pressure for its volume is dependent on these vari- 
ables. The relationship is expressed by the General 
Gas Law Equation, which in its most general form 
is stated thus: 

PV - nRT. (1) 

In this, n is the number of moles of gas, of volume 

Fahrenheit Centigrade Kelvin 

F C C K 


212* - 

200 ' 


100 - 





- -373.16' 


r -273.18* 


_ -2oo r 

-400' - 


FIG. 12-1. A comparison of temperature scales. 






V, at the absolute temperature T and the pressure 
P. R is a proportionality constant called the "gas 
constant," and has the same value for all gases. It 
occurs in many fundamental equations of chem- 
istry and physics. 

In all cases where the same amount of a gas is 
involved, the value of nR will also be constant, and 
we may write 

PV = kT, or 




Now if Pi, Vi, and Ti represent the corresponding 
pressure, volume, and temperature of a given 
quantity of gas, and P 2 , 2, and T 2 represent some 
other set of related values of pressure, volume, and 
temperature (for example, standard conditions) 
for this same quantity of gas, then we may express 
the relation existing between them by 




This may be transposed to give 

Vi = V 2 X ~ X | f . 

2 1 I 

It will be noted that, in equations (3) or (4), if the 
temperature is constant (Ti = T 2 ), the inverse 
proportionality of pressure and volume (Boyle's 
Law) is expressed. Likewise for constant pressures 
(?! = P 2 ) the direct proportionality of volume and 
absolute temperature (Charles' Law) is expressed. 
(See also Study Assignment A, Sec. C, Page 7.) 
If any five of the quantities in equation (3) are 
known, the sixth can of course be calculated by 
simple algebraic moans. 

Many students and instructors prefer to reason 
out the relationships, rather than blindly "to fol- 
low a formula/' and to apply the pressure and tem- 
perature (corrective) factors according to whether 
the pressure change, and the temperature change, 
will cause an increase or a decrease in the volume. 
Either point of view gives the same result. An 
example: 300 ml of CO 2 gas, measured at 30C and 
780 mm Hg pressure, when changed to standard 
conditions, 0C and 760 mm Hg pressure, will have 
its volume decreased by decreasing temperature, 
and increased by decreasing pressure, in accordance 
with the equation 

The general form of the gas law equation (1), 
permits many direct calculations. One must know 
the value of the gas constant R. When V is given in 
milliliters, and P in atmospheres, R has the value 
82.0 ml atm per degree per mole. An example: 
What weight of chlorine gas, CU, would be con- 
tained in a 5.00 liter flask at 20C, and 600 mm 
(600/760 atm) pressure? Substituting in the 
equation (1) 

PV nRT, we have 

Tog atm X 5000 ml =* n X 82.0 ( 

X 293, or 

0.1 64 mole. 

And, in grams, 

0.164 mole X 71 C1 2 - 11.8 g CI* 

Aqueous Vapor Pressure, Dalian's Law of 
Partial Pressures 

When any gas in a closed container is collected 
over, or exposed to, liquid water, the water evapo- 
rates until a saturated vapor results, that is, until 
the opposing rates of evaporation and condensation 
of water molecules at the liquid surface reach a 
"balance." These gaseous water molecules exert 
their due share of the total gas pressure against the 
walls of the container. Thus if 3% of all the gas 
molecules are water, and 97% are oxygen mole- 
cules, then 3% of the total pressure is the vapor 
pressure of water, and 97% is the pressure due to 
the oxygen. Each gas exerts its own pressure regard- 
less of the presence of other gases. This is Dal tori's 
Law of Partial Pressures. Stated as an equation, 
we have 

Ptotai - PH,O + Po f or transposing, P 0a - Ptotal - 

To illustrate Dalton's Law, in Figure 12-2 we have 
shown in (a) a mixture of oxygen molecules and 
water vapor molecules. In (b), the water mole- 
cules have been removed, but all the oxygen gas 
is still present, in the same total volume. The pres- 
sure has thus been reduced by an amount equal to 
the vapor pressure of water. 



Experimental Procedure 

Special supplies: Analytical weights, 600-ml beaker, 100C 
thermometer, 500-ml graduated cylinder. 

Chemicals: prepared unknown mixture of KClOa, KC1 and 
MnO 2 ; or KC1O 3 , Mn0 2 . 

Prepare the apparatus as illustrated in Figure 
12-3. Note the details carefully. Test tube A must 
be clean and thoroughly dry. Fill the flask B with 
water, and have a small amount of water in the 
beaker D. Fill the rubber tube C with water by 
blowing in the rubber tube A momentarily. Syphon 
water back and forth through tube C by raising 
and lowering the beaker, to expel all air bubbles. 
Finally, with the flask nearly filled, but not quite 
to the top (water must not enter the short glass 
tube connected to the rubber tube A), close the 
clamp which has been placed on tube C near the 

Obtain an unknown mixture of KClOs, KC1, and 
MnO 2 from the instructor and at once record its 
code number on your report sheet. 1 By completely 
decomposing a weighed sample of this and measur- 
ing the volume of the oxygen liberated at a known 
temperature and pressure, the rnolal volume of 
oxygen, and also the percent of KClOa in the 
sample, may be determined. First weigh the empty 
test tube A, together with its connecting stopper 
and glass tube, as precisely as your balance per- 
mits, to the nearest milligram if possible. Then add 
from 1.5 to 4.0 g of the KC1O 3 mixture to the test 
tube. (Your instructor will suggest the approxi- 
mate amount for you to use, as you must not use 
an amount that will generate oxygen in excess of 
the capacity of the flask B.) Again weigh the test 
tube assembly. 

Connect the test tube and stopper to tube A, 
and make sure that your apparatus is air tight by 
opening the clamp on tube C, and noting that 
water does not flow out of the flask, even when the 
water levels are quite uneven. Have your appa- 
ratus approved by the instructor (see report sheet). 

1 If you are not using an unknown mixture, and are determining 
only the molal volume of O 2 , about 5g KCIQs may be dried by just 
melting it in test tube A (disconnected and held by a test tube holder, 
and heated throughout its entire length). Let this cool, then add a very 
small amount (10-15 mg) of dry MnO 2 . Weigh the lest tube and con- 
necting stopper and glass tube with the contents. As a safety 
precaution, it is wise to test a small amount of the KC10 3 , mixed 
with a little of the MnCfe you will use, as you did in Exp. 8, by melting 
it to see if it decomposes quietly without obvious combustion. 


Pressure of 
oxygen and 
water vapor 
equal* atmos- 
pheric press* 

The water 
vapor has 
been ab- 
sorbed by 
the CaCI* 

This equals 
the origin- 
al pressure 
of the water 

In this drawing the gas molecules are 
greatly exaggerated in arze. 

FIG. 12-2. Illustrating an application of Dalton's law of partial 

Equalize the pressures inside and outside the flask 
by raising the beaker until both water levels are 
the same. Water can syphon in or out of the flask 
until the internal and external pressures are equal. 
Close the clamp and discard the water in the 
beaker. Drain the beaker carefully and completely, 
but do not dry it. Replace the tube in the beaker, 
and open the clamp. During the heating that fol- 



Pyrex test 


tube C 

e oo ml beaker 
or sooml flash 

FIG. 12-3. Apparatus to determine the molal volume of oxygen. 

lows see that the end of the tube C is kept under 
the water in the beaker. 

Heat the KC1O 3 mixture gently at first, then 
more strongly, to maintain a moderate rate of 
evolution of oxygen gas. If white vapors appear in 
the tube, decrease the heating until this stops. 
Continue the heating as long as any oxygen gas is 
liberated, until there is no further transfer of water 
into the flask. (If you have prepared your own 
mixture, using excess pure KQOs and MnO 2 , stop 
the heating in time so that the water level does not 
go lower than 1 or 2 cm above the end of the tube 
in the flask.) Do not remove the tube from the 
beaker until the apparatus has cooled completely 
to room temperature and the internal and external 
pressures have been equalized. Also do not handle 
the flask in such a way that the gas contained will 
be warmed by the hands. Adjust the levels in the 
flask and beaker so they are equal, and then close 

the clamp on tube C. Obtain the final weight of the 
test tube, its contents, and its tube connection. 
Take the temperature of the oxygen by placing a 
thermometer directly in the gas. Measure the 
volume of oxygen by carefully measuring the water 
in the beaker, using a 500-ml graduated cylinder. 
The aqueous vapor pressure may be obtained from 
Table IV in Appendix II. Obtain the barometer 
reading for the day. This must be corrected for the 
unequal expansion of the mercury and the brass 
scale (Table V, Appendix II). The correction is to 
be subtracted from the direct barometer reading, 
and is necessitated by the fact that atmospheric 
pressures are always translated into terms of the 
barometer at C. 

Calculate from your data (a) the volume, at 
standard conditions, of 32 grams (1 mole) of 
oxygen gas, and (b) the percentage of KC1O 3 in 
the unknown sample issued to you. 

REPORT: Exp. 12 

The Mold Volume of a Gas 

Unknown sample number. 


Locker Number- 

Instructor's approval of apparatus. 




Weight of tube and contents before lieating 



Weight of tube and residue after heating 



Weight of empty tube 



Temperature of the oxygen 



Volume of oxygen collected 



Barometer reading (uncorrected at .... C) 



Aqueous vapor pressure at temperature of gas 






Weight of oxygen 



Temperature, absolute 



Barometric pressure, cor- 
rected for expansion of 
scale (total gas pressure) 



Pressure of oxygen alone, 
in the flask 



Volume of your oxygen 
at standard condition 



The molal volume of 



Percentage error 



Moles of your oxygen 



Holes of KClOg decomposed 



Weight of KClOj in sample 



Percent of KClOt in sample 




Exercises on Laboratory Technique 

1. Why should not the glass tube connected to rubber tube A extend to the bottom of the flask? What would 
happen at the conclusion of the heating if it did reach to the bottom? 

2. Why is it necessary that the other tube in the flask reach to the bottom? 

3. Should the end of the delivery tube C remain under water while the apparatus is cooling? Why? 

4. Why should the end of delivery tube C be under water when the pressure in the apparatus is equalized with 
atmospheric pressure? 

5. Why should the test tube be allowed to cool before the water volume is measured? 

6. What objection is there to weighing an object, 
such as the tube of potassium chlorate, on the balance 
while the tube is still warm? 

7. What purpose docs the clamp on tube C serve? 

8. If the water level in the flask were 10.3 cm above 
the water level in the beaker (x in Fig. 12-3), would the 
gas pressure in the flask be greater or less than atmos- 
pheric pressure? By how much would the pressures 
differ? (Express in mm of mercury. See the sketch and 

13.6 cm 


i cm 

FIG. 12-4. The mercury equivalent of a water column. Since 
the density of mercury is 13.6 times that of water, a mercury 
column will stand only 1/13.6 times as high as a water column 
which balances the same pressure. 


Rep. on Exp. 12> Sheet 2 


on Experimental Errors 

(Note: Study thoroughly the section in Appendix I on "Experimental Errors.") 

1. Calculate the percent of uncertainty of the data in your experiment, as follows: 
(a) the weight of oxygen (note that two weighings are required for each determination), 

(b) the volume of oxygen liberated, 

(c) the pressure of the oxygen (after correction for barometer scale expansion and for 
vapor pressure). 

2. From the answers to question 1, above, calculate the maximum percent of uncertainty 
of your experimental value for the molal volume of oxygen, due to these causes. 

3. Considering the number of significant figures in your weight, volume, and pressure of 
oxygen, respectively, to how many significant figures should your result for the molal 
volume be expressed? 




4. How does your percentage error in the experiment compare with the percent of uncertainty as calculated above? 
If your error is less, would you attribute this to chance, to more careful technique, or to some other cause? In 
general, what type of error (method, equipment, or observation) do you feel is responsible for any error in your 
result? Is a "blunder" indicated? 

5. List any other factors you can suggest that affect the amount of error in this experiment, which were not con- 
sidered above. 


Problems on the Gas Laws and Molal Volumes 

1. A pilot balloon filled with 100.0 liters of H 2 at a pressure of 740 mm Hg and 17C rises 
to a height of 14,000 ft, where the pressure is 370 mm, and the temperature is 33C. 
To what volume will the H2 expand? 

2. A liter of air at 29C and saturated with water vapor, is dried by bubbling it through 
cone. H 2 SO 4 . What is the volume of dry air, if the atmospheric pressure is 750 mm of Hg? 

3. Dry air is 78.0% N2 and 21.0% C>2. What is the partial pressure of each of these con- 
stituents when the total atmospheric pressure is 750 mm Hg? 

4. 6.0 liters of H2 at 80 cm Hg pressure is mixed with 8.0 liters of He at 75 cm Hg pressure, 
in a Id liter container. What is the partial pressure of each constituent, and what is the 
total pressure? 

5. A mixture of KC1O S and KC1, weighing 61.3 g was heated intensely, giving 5.60 liters 
of O 2 , at standard conditions. Write the equation for the reaction: 

How many moles of O 2 were liberated? How many formula weights, and grams of KClOs 
were present, and what was the percentage of KC1O 8 , in the mixture? 

6. All real gases deviate to some extent from the behavior of perfect gases. At standard 
conditions the density of O 2 is 0.0014290 g/ml, of H 2 is 0.00008988 g/ml, and of CO 2 is 
0.0019769 g/ml. Using these values, and the exact atomic weights, calculate the molal 
volume of each of these, in milliliters, to 5 significant figures 

(label and give units) 


The Molecular Weight of a Gas. 


College Chemisfry, Chapter 9 

Review of Fundamental Concepts 

Review the relationships between Avogadro's 
Law, the molal volume, and also the general gas 
law equation as developed in the last experiment. 
From these considerations we may state that the 
molecular weight of any gas is the weight in grams of 
22.4 liters of that gas at standard conditions. Like- 
wise the molecular weight of a volatile liquid or 
solid, whose vapor may be readily measured at 
known conditions and then calculated to standard 
conditions, can be determined by this same method. 

In this experiment we shall make direct measure- 
ments of the weight and corresponding volume of a 
given gas at the laboratory temperature and pres- 
sure. We shall first obtain the weight of the gas 
container, a 200 to 300-ml flask, filled with dry air. 
From a knowledge of the density of air and the 
volume of the flask, the weight of the contained 
air, and hence the weight of the empty (evacuated) 
flask, can then be calculated. The difference be* 

tween this weight, and the weight when the flask is 
filled with the gas under consideration, will of 
course give the weight of the volume of the gas 
contained in the flask. 

Various gases, such as carbon dioxide (CO 2 ), 
sulfur dioxide (802), ammonia (NHs), nitrogen 
(N 2 ), nitrous oxide (N 2 O), methane (CH 4 ), or 
acetylene (C 2 H 2 ), may be used in the experiment. 
When tanks of the compressed gas are available, 
they are convenient; but you will learn more chem- 
istry if you prepare your own sample of gas. You 
should study, in a general chemistry text, the 
properties and method of preparation of the gas to 
be used. Directions will be given here for the prep- 
aration of carbon dioxide. Sulfur dioxide may be 
prepared in a similar apparatus by substituting 
sodium bisulfite (NaHSO 3 ) in place of the marble 
chips, and warming the mixture slightly to drive 
the more soluble sulfur dioxide gas out of solution. 

Experimental Procedure 

Special supplies: Analytical weights, thistle tube, calcium 
chloride drying tube, 110C thermometer. 

CAmica^;CaCO 8 (marblechips)forCO b NaHSO a forSOibetc. 

The apparatus to be used is sketched in Figure 
13-1. First clean and thoroughly dry a 200-ml flask 
C, fitted with a one-hole rubber stopper and a 
glass tube extending very nearly to the bottom of 
the flask. This tube is connected by a section of 
rubber tubing to a drying tube B, filled with 4- 
mesh calcium chloride which is protected at each 
end by a loose plug of cotton. Leave the stopper in 
C loose so as to permit gas to escape, and connect 
the drying tube to the compressed air supply. Pass 
a very gentle stream of dried air through C for at 
least five minutes. If C has been warmed previously, 
and the air passed through while it is cooling, more 
complete drying will be assured. (If compressed 
air is not available, a two-hole stopper, with a short 
exit tube, may be used in C, and this connected to 
a water aspirator to suck air through B and C. In 
this case the second hole in the stopper should be 
plugged when making all weighings, so as to pre- 
vent undue diffusion of CO*.) 

In the meantime, prepare the carbon dioxide 
generator A, using a 200-ml Erienmeyer flask 
having a thistle tube reaching almost to the bot- 
tom, and having a short, right-angle exit tube, to 
be connected to the calcium chloride tube when 
ready to generate the gas. Place about 25 g of 
marble chips (CaCOa) in the generator, and add 
about 10 ml of water to cover the end of the thistle 
tube. Do not add acid until ready to generate the 
carbon dioxide. 

Disconnect the rubber tube from the flask C, and 
weigh this flask, including the stopper and glass 
tube, to an accuracy of 0.001 g. After this weighing, 
avoid all unnecessary handling of the flask. Hold it 
by the rim, and reattach it to the drying tube and 
CO* generator, placing it on a clean dry square of 
paper to avoid contamination by the table. Be 
sure the stopper is left loose so the displaced air 
and excess gas can escape between it and the flask. 
When ready add dilute hydrochloric acid (HC1) 
through the thistle tube, a little at a time as needed, 
so as to maintain a gentle evolution of CO 2 . (You 




Drying tube B filfed 
with 4 mesh Cad* 

This stopper 
very loose 

Marble chips 

FIG. 13-1. Apparatus to determine the molecular weight of a gas. 

may momentarily tighten the stopper in C and 
note the rate at which liquid backs up in the 
thistle tube, in order to estimate the rate at which 
gas is being generated.) 

Let the generator run for at least twenty minutes 
to displace all the air in C by CO 2 gas. Finally de- 
tach the rubber tube from C and push in the stopper 
firmly. Without undue handling, at once weigh the 
flask and contents. It is necessary to weigh at once 
to avoid undue diffusion of air into C through the 
glass tube. Again connect C to the drying tube, 
gently release the stopper in C and pass C(>2 
through it for an additional ten to fifteen minutes. 
Weigh the flask C as before. The two weights 
should agree if the flask was filled with COa the 
first time. 

Take the temperature of the gas with a thermom- 

eter placed in the gas shortly after the last weigh- 
ing, and obtain the barometer reading for the day. 
To measure the exact volume of the flask to the 
bottom of the stopper, fill the flask with water, re- 
place the stopper and glass tube, and wipe off 
excess water. Weigh this on the platform balance 
to the nearest gram. (This heavy a weight might 
injure an analytical balance, hence the more rugged 
platform balance is used. Furthermore, great pre- 
cision is not required for this weight. Why not?) 
How does the weight of water enable you to calcu- 
late its volume? Look up in a reference source the 
density of dry air at your laboratory conditions. 
Enter all data and calculations on the report sheet. 
From the volume of CO 2 , calculated to standard 
conditions, and the weight of this CO 2 , calculate 
its molecular weight (the weight in grams of the 
molal volume). 

REPORT: Exp. 13 

The Molecular Weight of a Gas 



Locker Number- 




Weight of flask and stopper, filled with CO 2 



Weight of flask and stopper, filled with water 



Weight of flask and stopper, filled with air 



Temperature of flask 



Barometer reading (uncorrected at C) 



Density of dry air at flask temperature and pressure 
Reference : 






Temperature, absolute 



Barometer reading, correction 



Volume of flask 1 



Volume of carbon dioxide 
at standard conditions 



Weight of air in flask 
at the start 



Weight of empty flask 
and stopper 



Weight of carbon dioxide 
contained in the flask 



Molecular weight of CO2 
(from your data) 

Percentage error 



1 Why can we neglect the weight of the air in the flask when obtaining the weight of the water (from which we calculate 
the volume of the flask), but cannot neglect it when calculating the weight of the carbon dioxide? Explain. 



1. Write the chemical equation for the reaction taking place in the preparation of carbon dioxide. 

2. How many gram formula weights are there in the 25.0 g of CaCOa used? 

3. How many liters of carbon dioxide could be generated at standard conditions by this 
weight of CaCO 3 ? 

4. What volume would this amount of carbon dioxide occupy at ordinary laboratory 
conditions? (Assume 27C and 740 mm.) 

5. A liter of ethyl chloride vapor weighs 2.87 g, recalculated to standard conditions. Find 
the approximate molecular weight. 

, What is the density relative to oxygen (ratio between the density of the gas and the 
density of oxygen) of a gas which has a gram molecular weight of 160 g? 

7. What is the weight of a liter of each of the following gases t at standard conditions: 
chlorine, CU; butane, C 4 Hio; phosgene, COC1 2 ? (Use no other data except atomic 
weights and the molal volume.) 

8. 100.0 g of water is vaporized at 100C and 740 mm pressure. What is the volume of the 
steam formed? (Hint: first convert the weight to moles.) 





The Solution of Problems Based on Equations. 

A Study Assignment 


This review is inserted at this time, after you 
have completed the study of weight relations in 
Experiments 10 to 13, in order that we may bring 
together and compare the methods for the solu- 
tion of problems involving the weights and volumes 
of reacting substances. In solving problems, the 
quantitative relationships inherent in an equation 
are most conveniently expressed in terms of the 
appropriate "chemical units" of quantity gram- 
atoms, formula weights, or moles. In the final 
answer, these may then be converted into units 
of weight (grams), or in the case of gases into units 
of volume (liters). This is illustrated as follows. 

When aluminum is dissolved in hydrochloric 
acid, the skeleton equation is 

Al + HC1 + A1CU -j- Hi (unbalanced). 

To balance this, note that the least common mul- 
tiple of the 3 Cl atoms and the 2 H atoms is 6, so 
that we need 6 HCL The balanced equation is, 

(balanced) 2 Al -f 6 HC1 *- 2 Aids + 3 H,. 

gfw, moles: 26 23 

grams: 2 X 27 6 X 36.5 2 X 135.5 3 X 2.016 

- 54 = 219 271 - 6.05 

liters (Applies only to gases at standard 3 X 22.4 

conditions): 67.2 

Under each formula, we have written its meaning 
in various units. Note that: 

(1) the coefficients give directly the number of 
gram-atoms, moles, or formula weights involved. 

(2) In the case of a gas, we may express the 
quantity involved either by weight (grams) or by 
volume (liters), whichever is called for in the 

The practical solution of problems involving the 
relative quantities of substances in chemical re- 
actions is best explained by several illustrative 

Example 1. Weight -Weight Relations. What 
weight of sulfuric acid would be needed to dis- 
solve 50.0 grams of aluminum? 

The equation is: 
2 Al -f 3 

Co //eg* Chem/sfry, Chapters 6, 9 

Al,(SOOi + 3 Hi 

The given weight, 50 g Al, is first expressed as gram- 

Then, from the equation, we note that 3 formula 
weights of H 2 SO 4 react with 2 gram-atoms of AL 
Therefore, we need 

| X 1.852 - 2.78 gfw H 2 SO 4 

Finally, we convert the formula weights of HaSO* 
back to grams: 

2.78 gfw X 98 -f- - 272 g HaSO* 

Example 2. Weight-Volume Relations. What 
weight of aluminum will be required to produce 
12.0 liters of hydrogen gas at standard conditions, 
by reaction with sulfuric acid? 
The equation is: 

2 Al -f 3 H 2 SO 4 - A1 2 (SO 4 ) 8 -f- 3 H, 
First express the 12.0 liters of hydrogen as moles: 
12.0 1 H 2 

22.4 I/mole 

0.536 mole H* 

From the equation, we note that 2 g at Al are 
needed to produce 3 moles of Ha. Therefore, we 

5 X 0.536 g at Al - 0.357 g at Al. 

Finally, we convert the gram-atoms of Al back to 

0.357 g at Al X 27-^ - 0.64 g Al. 

Examples. Weigkt-Volume-Gas Law Relations. 
What volume of hydrogen gas, measured at 27 C 
and 640 mm Hg pressure, will be produced by the 




reaction of 25.0 g of zinc with an excess of hydro- 
chloric acid? 
The equation is: 

Zn + 2 IIC1 * ZnCl 2 + H 2 

Therefore, as many moles of hydrogen will be pro- 
duced as there are gram-atoms of zinc: 

25<Qg / >U = 0.382 g at Zn, or also 0.382 mole H 2 . 
65.4 g/g at 

We may now use the general gas law equation, 
pv = nRT, to convert the 0.382 moles of gas (n 
in the equation) to liters at the conditions given. 
Note that when R = 0.082 liter atiu/mole deg, 
the pressure must be expressed in atmospheres, 
i.e., p = 640/760 = 0.842 atrn. Also, T = 273 + 
27 = 300 K. 
Transposing the equation, and substituting; 

nRT 0.382 mole X 0.082 liter atm/mole deg X 300K 

0.842 atm 


Example 4* Volume-Volume Relations. What 
volume of air, 20% oxygen, is needed to burn 10.0 
liters of butane gas, CJIio, completely to carbon 
dioxide and water? 
The equation is: 

2 C 4 H 10 + 13 O 2 * 8 COa + 10 H 2 O. 

The volumes of gases will be in the same ratio as 
the number of molecules, or moles, of the gases 
(Avogadro's Law), therefore, from the equation: 

2 liters C^io is equivalent to 13 liters O 2 (pure), 



10.0 liters C 4 Hi is equivalent to 10.0 X -75- = 65 liters O 2 . 


Finally, since the air is only 20% oxygen, the 
volume of air required will be 

05 liters/0.20 = 325 liters of air. 

To summarize, in solving problems based on 
chemical equations: 

Step 1. Write the balanced equation for the re- 
action concerned. 

Step 2. Read the problem carefully, taking note 
of the data given, and the items to be found. 

Step 3. Express the data given in the appro- 
priate chemical units of quantity (gram-atoms, 
formula weights, or moles) and from the equation 
find how many gram-atoms, formula weights, or 
moles of the desired substance this will give or 

Step 4- Finally, convert the latter back into the 
units of weight or volume asked for in the problem. 
When both substances are gases, the coefficients 
in the equation give directly the relative volumes* 

REPORT: Assignment B 

The Solution of Problems 
Based on Equations 


Locker y H m her 

Show the solution, in good order, below each problem. The first step is always to write the equation for the 
reaction. Express the answers to three significant figures. 

1. How many grams of zinc metal will react with three gram-formula weights of (a) 
hydrochloric acid, (b) sulfuric acid, (c) phosphoric acid? (Assume that all hydrogen 
atoms in each case react.) 

In Experiment 0, suppose that 1.30 g of silver was obtained by reaction of the silver 
sulfate solution with metallic copper. What weight of copper metal dissolved from the 
copper wire? 

3. How many formula weights, and how many grains, of calcium sulfate would be formed 
from 222 grams of calcium hydroxide, by reaction with sulfuric acid? 

4. What volume of carbon dioxide at standard conditions may be obtained by heating 
250 grams of limestone, CaCO 3 ? 

5. What volume of oxygen is needed to burn 25.0 liters of butane CJIio, completely 
earbon dioxide and water? 




6. 175 grams of calcium chloride is treated with excess silver nitrate solution. How many 
grain-formula weights of calcium chloride are there? How many grams of silver 
chloride will be produced? 


7. 50.0 grams of cupric sulfate crystals, CuSO 4 -5 H^0 f is dissolved in water and treated 
with excess aluminum metal. How many grams of aluminum will dissolve? How many 
grams of copper will precipitate? 


-g Cu 

8. 12.6 grams of sodium bicarbonate, NallCOa, reacts with dilute nitric acid to form 
water, carbon dioxide, and sodium nitrate. What volume of carbon dioxide, at 27 C 
and 700 mm Hg pressure, is liberated by the reaction? 

9. What volume of air, at 25 C and 740 mm Hg pressure, is required to burn a pint of 
gasoline completely to carbon dioxide and water? (Assume air is 20% Oj, gasoline is 
C 7 Hie, and that a pint of gasoline weighs 350 grama.) 

10. 135 grams of aluminum metal is dissolved in hydrochloric acid. The solution is then 
treated with sulfuric acid and evaporated, to change the aluminum chloride to alumi- 
num sulfate. Water and potassium sulfate are added, and the solution evaporated to 
form alum crystals, KA1(SO4V 12 H 2 O. What weights of (a) sulfuric acid, (b) potas- 
sium sulfate, are needed, and (c) what weight of alum crystals will be produced? (Note: 
Write the three equations, then, by a comparison of the relative number of gram- 
atoms and formula weights involved, solve directly for the desired substances. Avoid 
any unnecessary intermediate steps.) 


Group Relationships In the Periodic Table. 


Co/fege Cnem/sfry, Chapters 5, 6 

Review of Fundamental Concepts 

In Experiment 8, we studied the chemical char- 
acter of the oxides of a number of elements in re- 
lation to their positions in the periodic table. In 
this experiment, we shall continue this study, par- 
ticularly in relation to the variation in electro- 
positive and electronegative character as we move 
from clement to element vertically in a group. We 
identify increasing electropositive character with 

increasing basic tendencies of the compounds of 
the metals, and increasing electronegative char- 
acter with increasing acidic character of the com- 
pounds of the nonmetals. It will be of interest to 
compare these tendencies in the more active me- 
tallic groups I, II, and III, and in the very active 
nonmetallic group VII the halogens. 

Experimental Procedure 

Special supplies: Five 150 to 250-ml wide mouth bottles, 
5 glass squares, deflagrating spoon, asbestos pieces, tin foil, 
colored cotton cloth, colored flowers, newspaper, lead dish, 

Chemicals: Na, K, Ca, Mg, Al metals, CaF*, CCl* Br, 
water, Clj water, 0.1 F I* phenolphthalein indicator, 5% 
NaCIO (commercial brand), P, KBr, KI, NaCl, MnO* steel 
wool, Sb (powdered), 0.1 F KBr, 0.1 F EJ, 0.1 F NaCU 

I. The Behavior of the Metals of Groups I* 

II, and III. (1) Fill a 15-cm test tube with water, 
close the opening with your thumb, and invert the 
test tube into a 400-ml beaker half filled with 
water, so that no air enters the tube. Now wrap 
a small piece of sodium metal (size of a small pea) 
with a little tin foil. (Caution: Sodium metal and 
also potassium metal must be handled with extreme 
care. Use forceps or tongs. Never touch the metal 
with your hands. Observe all precautions which your 
instructor gives.) Punch several holes in the tin foil 
wrapped around the sodium. Hold this with tongs 
or forceps, and quickly insert it under the open 
end of the test tube. See Figure 14-1. When the 
reaction has subsided, remove the tube full of gas 
from the beaker, and at once ignite it. Also, test 
the solution in the beaker with phenolphthalein 
solution, or litmus* (Phenolphthalein is colorless in 
an acid, and red in a basic solution.) 

(2) Compare the relative activity of sodium and 
potassium metals with water, as follows. Place a 
watch glass over a beaker which is a third full of 
water. By means of forceps or tongs, drop a small 
piece of sodium (size of half a pea) into the water. 
Keep the beaker covered during the reaction, as a 

precaution. Repeat the experiment, using a small 
piece of potassium metal in place of the sodium. 
(3) Invert a test tube filled with water in a 

Fio. 14-1. The reaction of sodium metal with water. 

beaker of water. Drop a piece of calcium metal into 
the beaker and cover it with the test tube, so as 
to collect the evolved gas. Identify the gas, and 
also the solid which precipitates from the solution. 
Note the comparative rate of the reaction as com* 



pared with that of the other metals you used. Does 
magnesium react with cold water? 

(4) Compare the relative rates of reaction of 
small pieces of magnesium, calcium, and alumi- 
num, with 3 ml of dilute (6 F) hydrochloric acid 
mixed with an equal volume of water. 

2. The Behavior of Chlorine. Because of its 
very corrosive characteristics, caution must be exer- 

and H Cl 

FIG. 14-2. A simple method for the preparation of samples 
of chlorine gas. The escape of excess chlorine into the atmos- 
phere is prevented by absorption into NaOH solution. 

cised so as to avoid all unnecessary release of 
chlorine gas into the atmosphere. Work in the hood. 
(a) Preparation from Bleaching Solution. 1 Ar- 
range a 150 to 250-ml wide mouth bottle with a 
stopper and delivery tube as illustrated in Figure 
14-2. The delivery tube leads into a second bottle 
or flask containing a solution of about 20 ml of 
6 F NaOH with 50 ml of water, in order to absorb 
any excess chlorine. Place 20 nil of a standard 
brand of bleaching solution, which contains 5% 
sodium hypochlorite (NaCIO), and 5 ml of con- 
centrated hydrochloric acid, in the wide mouth 
bottle, and at once insert the stopper. Let this 
stand a few minutes to complete the reaction, for 
which the equation is 

2 HC1 + NaCIO >- C1 2 + NaCl + H 2 O. 

1 The preparation of chlorine by the usual laboratory method, 
namely, the oxidation of HC1 by MnOa, will be carried out in Experi- 
ment 18. Our purpose here is to obtain samples of Cl? gas in a simple 
manner, with a minimum contamination of the atmosphere. Your 
instructor may designate different students to perform the various 
tests for others to observe in order to decrease the amount of chlorine 
liberated In the laboratory. 

Quickly cover each bottle of chlorine gas with a 
glass square, as you remove the stopper. Prepare a 
bottle of gas for each of the tests of paragraph (b), 
immediately before you perform the test. Recharge 
the generator bottle with fresh reagents for each 
bottle of Ck which you prepare. The solutions used 
to prepare the chlorine may be left in the bottles. 
They will not interfere with the tests. 

Keep each bottle of chlorine which you use covered 
as completely as possible during the test. After com- 
pleting each observation, add 15-25 ml of the wash 
solution of NaOH from the second bottle to the 
covered bottle and mix it well. This will react with 
excess chlorine and thus avoid releasing it into the 
atmosphere when you clean the bottle. 

(b) Reactions of Chlorine. Test the behavior of 
chlorine with metals by sprinkling a little powdered 
antimony into a bottle of chlorine. The product is 
antimony trichloride (SbCls). Into a second bottle, 
insert a little steel wool which has been ignited by 
holding it momentarily in a flame. 

Try the reaction of chlorine with a nonmetal by 
inserting a bit of burning phosphorus into a third 
bottle of chlorine. The phosphorus should be con- 
tained in a deflagrating spoon which is lined with 
asbestos. Phosphorus trichloride (PCls) is formed 
as a vapor. (It is a liquid at room temperature.) 

Test the bleaching action of chlorine by insert- 
ing into a bottle of chlorine such materials as a 
moist strip of colored cotton cloth, a colored flower, 
a moist strip of paper with fountain pen ink on it, 
and a moist strip of newspaper with printer's ink 
on it. (The latter contains carbon black.) 

Prepare a glass jet by drawing down a piece of 
glass tubing (Fig. i-4, P. xiii). Connect this by rub- 
ber tubing to the gas supply, ignite the jet, and 
adjust the gas to give a one-inch flame. Now lower 
this burning gas into a bottle of chlorine gas. (For 
the purposes of this experiment, consider the gas 
to be methane (CHO* If it is coal gas, it will also 
contain considerable hydrogen gas.) By noting the 
appearance, and by exposing a piece of moist, blue 
litmus to the fumes, identify the products which 
arc formed. 

3. The Formation of Bromine and Iodine. 
Into a 250-ml flask, place 1 to 2 g of potassium 
bromide (KBr), a little manganese dioxide (MnO 2 ), 
and 1 to 2 ml of concentrated sulfuric acid. Warm 
this mixture just enough to note the product 



formed. (Bromine can be distilled from such a 
mixture and condensed to liquid bromine under 
cool water.) Observe the properties of a sample of 
liquid bromine, as exhibited by your instructor. 
Wash the contents of the flask down the drain with 
plenty of water. 

Place 1 to 2 g of potassium iodide (KI) and a 
little manganese dioxide in an evaporating dish. 
Add 1 to 2 ml of concentrated sulfuric acid and 
cover this with a watch glass which contains a 
little water (see also Fig. 2-3, P. 16). Warm the 
dish gently, to sublime the iodine onto the watch 

4. Replacement Reactions of the Free Halo- 
gens and Halide Salts. To 2 ml each of 0.1 F 
KBr and 0.1 F KI, add a little chlorine water (a 
saturated solution of chlorine gas in water). Now, 
to each of these samples, add 1 ml of carbon tet- 
rachloride, agitate the test tubes a moment, and 
observe the colors of the free halogen dissolved in 
the carbon tetrachloride. 

In a similar fashion, test the reactivity of 2 ml 
each of 0.1 F NaCl and 0.1 F KI, with a saturated 
solution of bromine in water. Use carbon tetra- 
chloride as before in each case, to help you decide 
whether a reaction has occurred. Likewise, test the 
reactivity of 2 nil each of 0.1 F NaCl and 0.1 F 
KBr by the addition of 1 ml of 0.1 F 1% solution. 

Summarize these results by comparing the re- 
activity of the free halogen elements in relation to 
their positions in the periodic table, group VII. 

5. The Hydrogen Halides. (a) Hydrogen Flu- 
oride. Etch a glass plate as follows. Coat the glass 
square on one side with a thin, evenly distributed 
film of paraffin, by warming the glass, and melting 
small bits of paraffin on the surface. When this is 
cool, write or sketch on the paraffined surface, 
using a pin or other pointed object. Now place 3 g 
of calcium fluoride (CaF 2 ) in a lead dish, moisten 
this with a few drops of concentrated sulfuric acid, 
and stir it with a match stick to make a paste. Place 
the glass square on the lead dish so as to expose 

the writing to the fumes. Leave this until the close 
of the period, or longer (at least an hour). Then 
remove the glass square, warm it and wipe off the 
paraffin. Wash out the lead dish. 

(b) Hydrogen Chloride. Place 2 to 3 g of sodium 
chloride and 1 to 2 ml of concentrated H 2 SO4 in a 
dry wide mouth bottle, and connect this, as in 
Figure 14-2, to absorb excess IIC1 into water. 
(Sodium hydroxide is not necessary.) Let this 
stand a few minutes to complete the reaction. You 
can tell when the bottle is filled with the gas by 
the fog which results when your moist breath is 
blown across the mouth of the bottle. Also test the 
gas around the bottle top with moist blue litmus. 
Note the odor and color of the gas. Invert the 
bottle of hydrogen chloride into a 400-ml beaker 
two-thirds filled with water, removing the glass 
square as you do so. 

(c) TJie Relative Ease of f Oxidation oj 'the Hydrogen 
Ilalides. To a very few crystals of potassium bro- 
mide, and of potassium iodide, in each of two 15-ml 
test tubes, add a few drops of concentrated H 2 SO4. 
Test the evolved gases with moist, blue litmus. 
Also, blow your breath across the mouth of the 
tubes. Warm the tubes slightly to obtain more 
definite evidence of the nature of the colored prod- 
ucts which are formed. Note any characteristic 

Both hydrogen bromide and hydrogen iodide are 
colorless. They are quite similar to hydrogen chlo- 
ride. Some of the free halogens are formed also, 
because concentrated H-jSCX, in addition to being 
an acid, is a good oxidizing agent. Typical equa- 
tions for these oxidation processes are 

2 KBr + 3 H 2 SO 4 > 2 KHSO 4 + Br, + H 2 SO 8 + HA 
8 KI + 9 H 2 SO 4 + 8 KHSO 4 + 4 I 2 + H 2 S + 4 HjO. 

Judging by the amounts of the free halogen formed 
in each case, compare the relative ease of oxidation 
of HC1, IIBr, and HI, and relate this to the posi- 
tion of the elements in group VII. 

REPORT: Exp. 14 Nam *~ 


Group Relationships in the 
Periodic Table Section. 

Locker Number- 

1. The Metals of Groups I, II, and III 

Write equations for the reactions with water of: 



What specific identification of the products formed hi the above reactions did you observe? 

Compare the relative activity of sodium and of potassium with water. What would you predict as to the re- 
activity with water of caesium? Of lithium? 

Judging by the relative activity of calcium and of magnesium with water, what would you predict as to the 
behavior of strontium or barium with water? 

Write equations for the reactions with dilute hydrochloric acid of: 



Predict the behavior of scandium metal with dilute hydrochloric acid. What would be the formula of its 

Summarize the comparative reactivity of the metals of group I, of group II, and of group III with water and 
with dilute acids. What would you predict as to the relative ease of oxidation of these metals on exposure to the air? 


2. The Behavior of Chlorine 

Summarize your observations, and write the equations for the reactions of chlorine with: 
Observations Equations 


Iron r 


Hydrogen gas also reacts readily with chlorine. 

Write the equation: 

3. The Formation of Bromine and Iodine 

Describe the results of warming (I) a mixture of potassium bromide, manganese dioxide, and sulfuric aci< 
(2) a mixture of potassium iodide, manganese dioxide, and sulfuric acid. Note the properties of the products forme 
in each case. 

What is the function of the manganese dioxide, in the above reactions? 

Complete and balance the equation for the above reaction of KBr (The reaction of KI is similar) : 
2 KBr + H 2 SO 4 + MnO a * KHSO* + MnSO 4 + + . 

4. Replacement Reactions of the Halogens and Halide Salts 

Describe the results, and write the equations for the reactions, if any, of solutions of CU, Br2, or L, with soli 
tions of NaCl, KBr, or KI, respectively: 

Observations Equations 

CU and KBr 

C1 8 and KI 

Br 2 and NaCl 

Br 2 and KI 

I, and NaCl 

I* and KBr_ 


Report on Exp. 14, Sheet 2 Name_ 

Comment on the relative activity of the halogen elements in relation to their positions in group VII of the 
periodic table. 

5. The Hydrogen Halides 

(a) Hydrogen Fluoride. Describe all your observations in connection with the preparation of hydrogen fluoride, 
and its use in etching glass. 

Equation for the preparation of HF:_ 

Complete the representative equation below for the etching of glass, of which calcium silicate is an important 

CaSiO, + HF + CaF 2 + + 

(b) Hydrogen Chloride. Describe all your observations of the properties of hydrogen chloride gas: 

Equation for the preparation of HCl*- 

(c) The Relative Ease of Oxidation of the Hydrogen Halides. As evidenced by the behavior of the halide salts 
with concentrated sulfuric acid, comment on the relative amounts of HF, HC1, HBr, and HI formed, as compared 
with the corresponding oxidation products Fa, Cla, Bra, and la. Correlate this with the position of the elements in 
the periodic table. 

Application of Principles 

1. Name, and give the location in the periodic table, of (a) the most electropositive element, and (b) the most 
electronegative element. 


2. Write the formulas (not equations) of the new substances, if any, which you predict would be formed when: 

(a) Francium metal is added to water . .. 

(b) Beryllium is treated with HC1 solution 

(c) Lithium metal and astatine are mixed 

(d) Rubidium bromide and iodine solutions are mixed 

(e) Fluorine gas is treated with calcium chloride 

(f) The gaseous product from (e) is bubbled into strontium bromide solution . . 

(g) Caesium bromide, manganese dioxide, and concentrated sulfuric acid are mixed 

(h) Sodium astatide is treated with concentrated sulfuric acid 

(i) Caesium metal is exposed to dry air 


The Periodic Classification of the Elements. 


A Study Assignment 
Review of Fundamental Concepts 

College Chem/sfry, Chapters 5, 6 

One of the most useful generalizations of in- 
organic chemistry is the Periodic Law, which 
states that the properties of the elements are 
periodic functions of their atomic numbers. The 
relationship can best be seen by arranging the 
elements in a certain way; this arrangement is 
called the Periodic Table. In order that you may 
appreciate more fully the kinds of information 
which are summarized by the periodic table, the 
following study assignment has been provided. The 
student should review carefully Chapters 5, 6 and 
9 of the text, which discuss the organization of the 
periodic table and contain numerous tables which 
summarize important properties of the elements by 

Special attention should be given to the char- 
acteristic properties of one of the common elements 
of each group and the relationship between its 
properties and those of the other congeners in the 
group. The increase in the metallic nature of the 
elements in Groups I and II from top to bottom, 
and the increase in the nonmetallic nature of the 
elements in Groups V, VI and VII from bottom 

to top are examples of vertical trends of impor- 

Horizontal trends which should be noted include 
the regular change in maximum oxidation number 
and the decreased basicity or increased acidity of 
the hydroxides of the elements, as one proceeds 
from left to right across the second and third 

The student should also know the location in the 
periodic table of the different types of elements, 
such as: the noble gases, the light active metals, 
the nonmetals, the transition metals, the metal- 
loids, the naturally radioactive metals and the rare 
earth metals. 

Before beginning this exercise, the student must 
realize that although the periodic table does sim- 
plify the task of learning what kinds of properties 
might be expected of the various types of elements, 
it also has its limitations. In order to obtain a broad 
knowledge of chemistry, a great number of specific 
facts must also be learned many of which cannot 
be completely coordinated by the periodic table or 
other unifying theories at this time* 


In the exercise which follows, the student is ex- 
pected to identify various elements and their loca- 
tion in the table from the description of char- 
acteristic properties and interrelationships within 
groups. Each group has been assigned a capital 
letter which is carried through in the fictitious 
symbols used for all the elements in a particular 
group. The subscript letters a, b, c, d, e and f are 
used to distinguish the individual members within 
the group but they have not been assigned in alpha- 
betical or any other special order. For instance, 
after reading the paragraph describing the family 
of elements whose symbols all begin with the capi- 
tal letter C, you will soon decide that the elements 
Ca, Cb, Cc, Cd, Ce, Cf are noble gases. By refer- 
ring to Table 5-2 in the text you will be able to 
determine the order in which they should be placed 

into Group O of the blank periodic table provided 
in the report sheet. 

The less familiar elements in the second and 
third rows of transition elements as well as most 
of the rare-earth metals have been omitted but a 
representative group of about fifty-seven elements 
has been included in the discussion. Identify each 
of them and locate them in their proper place in 
the blank periodic table on the report sheet. 

Discussion of the Properties of the Elements 

The element X occupies a unique place in the 
periodic table in that, strictly speaking, it belongs 
to no other group. It forms binary compounds with 
nonmetals in which its oxidation state is +1. It 
also forms binary compounds with metals in which 
its oxidation state is -1. It is the lightest of all 




gaseous elements. More compounds of this element 
are known than of any other element. 

The group of elements whose symbols all contain 
the capital letter C possesses electronic structures 
with extraordinary stability. They do not react to 
form chemical compounds readily. They are all 
gases at room temperature. The element Ce ranks 
as the second least dense of all gases and has the 
lowest melting point of all elements. It can only 
be solidified under pressure. Cb is naturally radio- 
active and is a member of the Uranium-Radium 
series of radioactive metals. Ca, because of its 
inertness and comparatively greater abundance, is 
used as the gas in incandescent light bulbs to cut 
down the rate of evaporation of the filaments. 
When an electric discharge is passed through a tube 
containing Cf at low pressure, a brilliant red light 
is emitted. Cb, Cd, and Cc occur in very much 
smaller amounts than the other members of this 
family. Cd has a melting point and boiling point 
which are intermediate to those of Cb and Cc. 

The elements to which the symbol D has been 
given are classified as active metals. They each 
decompose water readily with the evolution of 
hydrogen and the formation of hydroxides which 
are strong bases. They form ionic compounds with 
nonmetals in which they show an electro valence of 
+ 1. Da is the most abundant of the group with 
Db ranking second in occurrence. The other four 
elements are much more rare and of lesser im- 
portance. DC is the least dense of these metals and 
ranks along with Df at the top of the electromotive 
force series. De has been obtained in only very 
small quantities, is radioactive, and has been as- 
signed an official name rather recently. Db has a 
metallic radius less than that of Dd and Df but 
greater than that of DC and Da. Dd has a greater 
density than Db but is less dense than Df . 

In the group to which the symbol G has been 
assigned the oxidation states possible may vary 
from 3 in the hydrides to +5 in certain oxides. 
Ge and Ga are distinctly nonmetallic in nature but 
as one proceeds down the group the metallic prop- 
erties increase. Gc is sometimes classified as a 
metalloid since it possesses properties which are in- 
termediate to those of the nonmetals above it and 
those of the more metallic Gb and Gd below it. Ge 
is a relatively unreactive element at ordinary tem- 
peratures and makes up about four-fifths by vol- 

ume of the gases in the atmosphere. The strong 
ternary acid, HGeO 3 , is an illustration of a com- 
pound of Ge in its highest oxidation state. The 
element Ga forms the oxide Ga 2 O 5 when it is burned 
in air. This oxide reacts with water to produce the 
triprotic acid H 3 GaO 4 , which is a moderately strong 
acid. The element Ga exists in several allotropic 
forms in the solid state. It exists as a tetratomic 
molecule Ga 4 in the vapor state. Gd has physical 
properties which are mainly metallic, its chief use 
being as a component of low melting alloys. Its 
oxide Gd 2 O 3 is the anhydride of the predominately 
basic but slightly soluble hydroxide Gd(OH) 3 . Gb 
has a density, boiling point, and melting point 
which are intermediate with respect to those of its 
neighboring congeners Gc and Gd. 

The six elements to which the letter R has been 
assigned form compounds of the following types: 
oxides RO, hydroxides R(OH) 2 , carbonates RCO 3 
and sulfates RSO 4 . The metals of this family are 
less active than the alkali metals but are suffi- 
ciently active to necessitate their preparation by 
the electrolytic reduction of their fused chlorides. 
Rb and Rf react rather slowly with water at mod- 
erate temperatures, but as one proceeds down the 
group the metals react quite readily to displace 
hydrogen and form the hydroxides of the metals. 
Rd is the most abundant element of the group and 
is also the least dense. Re is a naturally radioactive 
element, a member of the Uranium 238 series. Rb 
has the lowest melting point of the group and is of 
strategic importance as one of the constituents of 
light weight alloys. Ra is located in the second long 
period of eighteen elements. Its chemistry is closely 
parallel to that of Rd and Re, the three elements 
composing one of Dobereiner's triads. The density 
and metallic radius of Ra is intermediate to those 
of Rd and Re. Due to the low solubility of the 
sulfate of Re, the Re 4 "" 1 " ion is used to test for the 
presence of sulfate ion in aqueous solutions. ReSO 4 
may be swallowed without poisonous effects and is 
used in obtaining X-ray photographs of the alimen- 
tary tract. Rf is used for making windows for X-ray 
tubes since X-rays readily penetrate elements of 
low atomic number. 

The family of elements to which the letter M has 
been assigned have as a member the element Ma 
which is the most electronegative and one of the 
most reactive of all the elements. The triad Me, 



Md, and Mb are closely related to each other in 
properties. They form gaseous binary compounds 
with hydrogen of the type HM, each of which is 
very soluble in water producing strong acids. Me, 
Md, and Mb exist in the gaseous, liquid, and solid 
states, respectively, at room temperature. The 
strength of the elements as oxidizing agents in- 
creases in the order Mb, Md, Me, to Ma, the latter 
element being the lowest one on the electromotive 
force series. The maximum oxidation number for 
this group is +7 and is illustrated by the ternary 
acids HMcO 4 and H 5 MbO 6 . Me is the least well 
known of this group and has just recently been 
given a name on the basis of its isolation from 
radioactive sources. 

Element EC is a nonmetal which shows oxida- 
tion states ranging from +4 to 4. Most of the 
chemical substances which constitute living matter 
contain this element. One of the allotropic forms 
of EC is the hardest substance known. By contrast, 
element Eb is the important constituent of a large 
number of minerals which make up most of the 
rocks in the inorganic realm of nature, and ranks 
as the second most abundant element in the earth's 
crust. Ea is usually classified as a metalloid, and 
as one proceeds down the group the metallic char- 
acter increases even more. Ed and Ee are quite 
dense, and their hydroxides in the bivalent states 
are predominately basic in nature. Different iso- 
topes of Ee are the end products of the various 
radioactive series. Ea is the least well-known of 
this group and was not known at the time of 
Mendeleef s attempt at a periodic classification. 
However, he was able to predict its properties quite 
accurately on the basis of its location in the table. 

The element Bd is the most abundant element 
in the earth's crust and is the most electronegative 
member of its group. Be forms compounds in which 
its oxidation state varies from 2 to +6. The acid 
H 2 BcO 4 is produced in greater quantities commer- 
cially than any other inorganic chemical. Although 
the group is classified as nonmetallic, there is an 
increase in the metallic nature with increasing 
atomic weight. For example, Be is a nonconductor, 
of electricity; one of the allotropic modifications of 
Be has a small but measurable conductivity; while 
Bb is classified as a semiconductor. The hydrides 
H 2 Bc, H 2 Be, and H 2 Bb, are gases. Ba is a radio- 
active element, a member of the Uranium-Radium 

series, and is the parent element of the end product 
of this series. 

The element Aa is the most abundant metallic 
element of the earth's crust. Its oxide, Aa^Oa, is 
the hardest of all naturally occurring substances 
except the diamond. Its hydroxide, Aa(OH)a, is in- 
soluble in water but is soluble in strong acids and 
also in strong bases. Ab is usually classified as a 
metalloid but resembles the nonmetallic elements 
in that its hydroxide H 3 AbO 3 is weakly acidic. 
Pyrex glass contains about 12% of the oxide Ab 2 Oa. 
The elements Ac, Ad, and Ae, are located in sub- 
group Illb, and there is an increase in metallic 
character in the order given. The principal oxida- 
tion number of these metals is +3, but Ae also 
forms some compounds iu which it exhibits a +1 
oxidation state. Ac has an unusually low melting 
point for a metal, melting just slightly above room 

Elements F, K, L, N, P, Q, J, W, and H are 
classified as transition metals and are discussed 
briefly in Chapter 24. Element K shows a maxi- 
mum oxidation state of +4, but also exhibits a 
fairly stable +3 and a less stable +2 state, K ranks 
as the seventh most abundant metal, and its 
oxide KO 2 has considerable commercial importance 
as a white pigment in paints. Element N shows 
oxidation states of +2, +3, +4 and +5. Element 
P exhibits the principal oxidation states +2, +3 
and +6 and got its name from the fact that its 
compounds are colored. L is a silver gray metal 
which is one of the most active of the transition 
metals, displacing hydrogen even from water at 
room temperature. It is located just below alumi- 
num in the electromotive force series. L shows a 
maximum oxidation state of +7. Elements F, Q, 
and W are members of a subgroup which, for his- 
torical reasons, is still singly numbered. The ele- 
ment F is greatly strengthened when alloyed with 
small amounts of carbon and moderate amounts of 
other transition elements. W has a slightly larger 
atomic weight than Q even though its atomic 
number is one less than Q. Element H only forms 
one series of compounds in which it is bipositive. 
It is a relatively low melting metal whose main in- 
dustrial use is to coat iron to protect it from rust- 
ing. It is located above hydrogen in the electro- 
motive force series. J shows oxidation states of +1 
and +2, and is the first element in the subgroup 


known as the noble metals because of their rela- radioactive series of elements bearing its name. One 

tive inertness. of the isotopes of Z and the artificially produced 

Elements S and Z are members of the Actinide element S undergo nuclear fission when exposed to 

series of rare earths. Z is the parent member of a slow neutrons. 

REPORT: Assignment C 

The Periodic Classification 
of the Elements 




Locker Number- 

By means of a periodic table, it is possible to organize the 101 known elements into seven horizontal rows or 
periods and into eighteen vertical groups or subgroups of elements. The periods consist of one very short period of 
two elements, two short periods of eight elements each, two long periods of eighteen elements each, one very long 
period of thirty-two elements and another very long period which is incomplete. The latter two periods are com- 
pressed into the table by writing the fourteen rare-earth metals and the uranium metals below the chart. For a 
further description of the table, see College Chemistry, Chapter 5. 

The increase of the maximum oxidation number of the elements from left to right across the table should be 
noted. Take the second period for example: Na+, Mg+ 2 , A1+ 8 , Si* 4 , P+ 6 , S+ 6 , C1+ 7 . Another horizontal trend is the 
decrease in basicity and increase in acidity of the hydroxides of the elements from Group I to Group VII. For 
example: NaOH is a strong base; Mg(OH)2 is basic; Al(OH)j is amphoteric; the hydroxide of silicon, SiO(OH)2, is 
a weak acid usually written H 2 SiO 8 ; PO(OH) 8 is a moderately strong acid (H 3 PO 4 ); SO a (OH) 2 is a strong acid 
(H 2 SO 4 ); and C1O 3 (OH) is a strong acid (HC1O 4 ). 

The increase of the metallic nature of the elements in Groups I and II from top to bottom, and the 
increase in the nonmetallic nature of the elements in Groups V-Va, VI -Via and VII -Vila from bottom to top 
are examples of some of the vertical trends which should be noted. 

Other trends in chemical and physical properties should be noted by referring to the numerous tables in Chap- 
ters 5 and 6 of College Chemistry, Second Edition. 

Group O 










































































































































La j| 




































> Th 










* Lanthanons 
4 Act-inons 































N P 













































! ri 






* Lanthanons 
$ Actinons 

















The Equivalent Weight of a Metal. Ionic Valence. 


College Cfiem/sfry, Chapters 6, 10 

Review of Fundamental Concepts 
Equivalent Weight 

The term equivalent weight, as the adjective 
"equivalent" implies, is used to designate the rela- 
tive amounts of substances which are chemically 
equivalent, that is, which just react with, or re- 
place, one another in chemical reactions. We define 
the equivalent weight of a substance in a given re- 
action as that weight of it, in grams, which reacts 
with or replaces one gram-atomic weight (1.008 g) 
of hydrogen. Thus at each point of the connecting 
lines in Figure 15-1 below, we have indicated the 
amounts of several elements which are equivalent 
to one another. 


56.2 <5 

student can deduce other possible equivalent rela- 
tionships from the diagram. 


l.oos g 




27.92 g or 13.62 gf 

FIG. 15-1. Some equivalent weights of several elements. 

56.2 g of cadmium, or 27.92 g of iron, is required 
by reaction with excess hydrochloric acid, to liber- 
ate 1.008 g of hydrogen. These are therefore the 
respective equivalent weights of cadmium and iron 
in these reactions. Likewise, in hydrochloric acid, 
1,008 g of hydrogen is combined with 35.46 g of 
chlorine, which is therefore one equivalent of this 
element. This amount of chlorine, 35.46 g, is like- 
wise the weight which is combined with 56.2 g of 
cadmium in cadmium chloride, and with 27*92 g of 
iron in ferrous chloride. Iron also forms another 
compound, ferric chloride, by direct reaction with 
chlorine, in which 35.46 g of chlorine is combined 
with 18.62 g of iron. The latter is therefore the 
equivalent weight of iron in this compound. The 

Equivalent Weight and Ionic Valence 

The equivalent weights of certain elements, e.g. 
sodium, potassium, chlorine, and bromine, are 
found to be equal to their atomic weights. One 
gram-atomic weight of sodium or potassium may 
replace one gram-atomic weight of hydrogen, while 
one gram-atomic weight of the nonmetallic ele- 
ments chlorine or bromine may unite with one 
gram-atomic weight of hydrogen. This means that 
these elements have a valence of one, (+1 for the 
two metals, and 1 for the two nonmetals) 1 . The 
equivalent weights of some other elements are 
simple fractional parts of their atomic weights. For 
example, the gram-atomic weight of cadmium is 
112.41 g, while its equivalent weight is 56.2 g. Its 
valence is therefore two (+2) in its compounds. 
The valence, then, indicates the number of gram 
equivalent weights in one gram-atomic weight of 
an element. 

T . . gram-atomic weight 

Ionic valence : : . . 

gram equivalent weight. 

The valence of an clement is often stated as indi- 
cating the number of atoms of hydrogen an atom 
of the element unites with or replaces in forming a 
given compound. The valence of an clement in a 
compound must always be a small whole number, 
as +], +3, +4, or 2. The valence of all uncom- 
bincd elements is always zero. (See Table XVI, 
Table of Ionic Valences, Appendix II.) 

Some elements, such as iron in the previous ex- 
amples, can have more than one equivalent weight, 
and correspondingly, will have more than one possi- 
ble valence. In ferrous chloride (FeQ 2 ) where the 
equivalent weight of iron, 27.92 g, is one half of its 
atomic weight, 55.85 g, its valence is +2, while in 
ferric chloride (FeCl 3 ) where the equivalent weight 
of iron is one third of 55.85, namely 18.62, the 
valence of iron is +3. 

1 We are using the term valence here in the sense of "ionic valence," 
in which metallic ions, or cations, carry a positive (-H) charge, and 
the nonmetallic ions, or anions, carry a negative ( ) charge. Review 
Experiment 8, on valence. 




Determination of the Equivalent Weight 
of a Metal 

The equivalent weight of an active metal, when 
it reacts with an acid, is easily determined. A 
typical example of such a reaction is: 

Cd+2HCl= CdCl 2 +H 2 . 

A weighed sample of the metal is allowed to react 
with the acid, and the hydrogen that is produced is 
collected. Its volume is measured at the laboratory 

temperature and pressure, and calculated to the 
volume at standard conditions. 

If we recall that the standard molal volume of a 
gas, as hydrogen, H 2 , is 22.4 liters, then the volume 
of one gram-atom of hydrogen (1 H) at standard 
conditions, is 11.2 liters, or 11,200 milliliters. For 
convenience in calculations, then, we may restate 
our definition thus : The equivalent weight of an active 
metal is that weight of it which will replace in an acid, 
11 $00 ml of hydrogen gas at standard conditions. 

Experimental Procedure 


in copper 



oy i 

of H 2 

this in cm 

FIG. 15-2. To illustrate the method of handling a 
gas-measuring tube. 

Special supplies: Analytical weights, 100 -ml gas-measuring 
tube, 100 C thermometer. Pieces of Al, Mg, Zn, cut to size. 

First heat 100 to 150 ml of tap water to boiling to 
remove dissolved air, and cool it under the water 
faucet to room temperature. Obtain a sample of the 
metal to be used (aluminum, zinc, or magnesium) 
which previously has been cut to such a size that 
not over 100 ml of hydrogen gas will be liberated. 
(Weights which will liberate about 80 to 95 ml of 
hydrogen are: 0,24 g Zn, 0.065 g Al, and 0.085 g 
Mg.) Clean the metal, if the surface is corroded, 
with emery cloth, and weigh it carefully to 0.001 g. 1 
Wrap the piece of metal with a 20-cm length of fine 
copper wire in all directions in such a way as to 
form a little cage enclosing the metal. Leave 5 to 
8 cm of the wire unwrapped, as a handle. (If a 
ribbon or wire form of the metal has been used, 
coil it rather tightly, and wrap the copper wire 
around the entire coil.) 

Clean a 100-ml gas-measuring tube, 2 and put into 
it the required amount of concentrated HC1 solu- 
tion. Because of differences in the activity of the 
mefals, it is necessary to vary the amount of acid 
accordingly. For magnesium use about 5 ml, and 
for aluminum or zinc use about 30 to 35 ml of con- 
centrated HC1. Next slowly and carefully fill the 
entire tube to the very top with the cooled, boiled 
water, so as to avoid undue mixing with the acid. 

1 To the Instructor: Samples in wire or ribbon form may be cut to 
exact length to give a known weight, threaded through a small card, 
and individually issued to students by code number. The corrected 
volume of hydrogen obtained may then be reported by the student 
before he is given the weight, as a check on his calculations and data. 

* If gas-measuring tubes are not available, you may, with slightly 
less precision, use a 100-ml graduated cylinder and a funnel (with the 
stem broken off) inverted in a beaker of water, as in Fig. 15-3. The 
graduated cylinder is filled with water. Add cone IIC1, a little at a 
time, as needed, through another funnel, to the bottom of the beaker, 
where it will react with the metal encased in copper wire. 



Insert the metal about two inches into the tube, 
leaving the end of the copper wire outside, and 
clamp it there by inserting a one or two-hole rubber 
stopper so that no air is entrapped in the tube. 
When all is ready, cover the end of the tube with 
your finger and invert it into a beaker partly filled 
with water. The acid, being more dense than water, 
stays down until inverted, and then quickly sinks 
and diffuses down the tube and reacts with the 
metal. The hydrogen liberated displaces the liquid 
from the tube. (Caution: If the reaction becomes 
too rapid, small hydrogen bubbles will escape from 
the bottom of the tube along with acid solution as 
it is expelled. In this case repeat the experiment, 
using less acid. If the reaction is too slow, use more 


FIG. 15-3. A simple apparatus for the measurement of the 
volume of hydrogen gas evolved. See footnote 2, p. 102. 

When the metal has completely dissolved, let the 
apparatus stand for a few minutes to allow it to 
cool to room temperature since heat is generated 
by the reaction. Any hydrogen bubbles adhering to 
the sides of the tube or copper wire may be freed by 
tapping the tube. If it is convenient, move the tube 
up or down so that the water in the tube is at the 
same level as the water in the beaker, but avoid 
warming the gas by contact of your hands with the 
tube. If this is not practical, measure the distance 
between the two water levels, using a metric rule, 

and recalculate this to its equivalent pressure in 
millimeters of mercury. Read and record the vol- 
ume of gas in the tube. Take the temperature of 
the gas by holding a thermometer close beside the 
tube. Obtain the barometer reading for the day. 

Calculate the weight of metal needed to liberate 
one gram-atom of hydrogen (11,200 ml at standard 
conditions). This is its equivalent weight. From 
this, calculate its valence. 

Alternate Procedure: 1 Set up the apparatus as 
sketched in Figure 15-4. The exit tube C from the 
test tube must not extend below the rubber stopper 
(so that gas will not be trapped in the test tube). 
The longer glass tube should extend very nearly to 
the bottom of the test tube, and may best termi- 
nate in a capillary tip bent over toward the wall of 
the tube. The flask E, 500 or 1000 ml capacity, is 
filled with water, and the flask F is partially filled. 
The siphon tube D, extending to the bottom of 
both flasks, is likewise filled with water. 

Place the carefully weighed sample of metal in 
the 15-cm test tube as indicated. It is well to have 
a fine copper wire wrapped in all directions about 
the metal, as a cage, and support this above the 
lower end of the glass tube, simply by leaving an 
extra length of the copper wire pointed downward. 
With the stopper in flask E loosened to permit air 
to escape, pour water into the funnel to completely 
fill the test tube and tubes B and C, and close 
clamp B. If flask E is not completely filled with 
water, raise flask F so that water siphons back into 
it, then push in the stopper in E tightly. 

When all air bubbles have been thus removed 
from the apparatus, release clamps B and D just 
enough to permit the water level in the funnel to 
fall just to the stem top, but no further, then close 
the clamp B. Now empty and drain, but do not 
dry, flask F, leaving tube D filled with water. Re- 
place the tube in the flask, leaving the clamp open. 
When all is ready, carefully measure 25.0 ml of 
cone HCI and place it in the funnel. Then release 
clamp B momentarily to permit a little acid to flow 
into the test tube and react with the metal at a 

1 This will give more experience with laboratory apparatus and 
techniques. It permits the use of larger samples, and other metals 
which react less rapidly with acid may be used, since the test tube may 
be warmed with a Bunsen burner to increase the rate, or the acid con* 
centration may be increased during the experiment. The metals may 
be issued as unknowns. Maximum weights for a 600-ml flask are 
about: Zn 1.10 g, Mg 0.40 g, Al 0.30 g, Fe 0.05 g. 





Nfcte the 
curved and 
pointed tip. 

Clamp here when (i) alt the sample has dis- 
solved, (2) all the evolved gas isinflaakE, (3) 
the water levels of flasksEand 
I have been adjust- 
ed to the 

Water level 

The end of this 
tube should not 
protrude below 
the stopper or it 
will trap das. 

Metal sample wrapped 
with fine copper wire 
should be above the 
end of the funnel 

5oo ml flasks 

Water .. 
displaced // 

FIG. 15-4. An alternate apparatus for the determirfation of the equivalent weight of a metal. 

moderate rate. Water will siphon from E to F in 
amount equivalent to the volume of hydrogen gen- 
erated, which collects in E. When the metal has all 
dissolved and the apparatus has stood ten minutes 
to attain temperature equilibrium, release clamp B 
enough to permit the acid level to fall to the funnel 
stem, but no further. Carefully measure 25.0 ml of 
water, add to the funnel, and again release the 
clamp to let the level fall exactly to the funnel 
stem, then close the clamp tightly. All gas should 
now be displaced from the test tube and connecting 
tubes into flask E. If not, repeat the addition of a 
measured amount of water. 

Adjust the levels in flasks E and F by raising or 
lowering one of them until they are even (do not 
warm the gas in E with your hands on the flask) ; 
then close clamp D tightly. Obtain the temperature 
of the hydrogen by removing the stopper enough to 
place a thermometer in flask E. Measure the vol- 
ume of the water in flask F by pouring it into a 
500-ml graduated cylinder. This volume minus the 
volumes of acid and water added to the funnel will 
be the volume of hydrogen generated. Obtain the 
barometer reading. Record all data on the report 
sheet, and calculate the equivalent weight of the 

REPORT: Exp. 15 

The Equivalent Weight of a 



Locker Number 




Weight of metal 



Volume of hydrogen 



Temperature of hydrogen 



Barometric reading (unconnected at .... C) 



Difference in water level inside and outside tube 



Aqueous vapor pressure at temperature of hydrogen 



Atomic weight of metal used 




Barometric pressure (cor- 
rected for scale expansion) 



Mercury equivalent of differ- 
ence of water levels 



Pressure of H 2 after correc- 
tion for difference in HjO 
levels and for vapor pressure 



Temperature, absolute 



Volume of dry hydrogen 
at standard conditions 



Equivalent weight of metal 
(from data) 



Valence of the metal 

Theoretical equivalent 
weight of metal 



Percentage error 



1. A certain metal has an equivalent weight of 68.7 g in a given reaction. Its valence is two 
in the compound formed. What is its atomic weight? 

2. The equivalent weight of a certain metal is found to be 29.68 g. Its atomic weight is 
118.70 g. What is the valence of the metal? 

3. An element of gram atomic weight 58.69 g forms a compound in which its valence is +3. 
What is its equivalent weight? In another compound it manifests a valence of +2. 
What is its equivalent weight in this case? 

4. 0.373 g of a certain metal replaces 250.0 ml of hydrogen, when collected over water at 
25C and 743 mm mercury pressure. What is the equivalent weight of the metal? 

5. How many gram atomic weights of hydrogen under standard conditions would be 
liberated from hydrochloric acid reacting with one gram atomic weight of each of the 
following metals: Al, Mg, Ca, K, Zn, Na? 

6. Calculate the weight of calcium which should be used with a 100-ml gas measuring tube, 
in order to liberate a convenient amount of hydrogen by reaction with an acid in an 
equivalent weight determination. Calculate on the basis of 85.0 ml of E^ at standard 



5. G at of H 2 from: 









Ionic and Covalent Compounds. Ionic Reactions 

Coif*?* Cfomi ft iy, Chapter! 10, ff 

Review of Fundamental Concepts 

Two Types of Chemical Bonds 

When an active metal reacts with an active non- 
metallic element, the metal becomes positively 
charged by the loss of one or more electrons to 
the nonmetal, which in turn becomes negatively 
charged. Each element tends to assume its most 
stable electronic configuration, with its outer elec- 
tron energy level saturated as in the noble gases. 
If we represent this outer level of electrons by 
dots, we may write, for example, 

Na. + -C1: *Na+:Ci:-. 

Such electrically charged atoms, or radicals, are 
called ions* The bonding force between such ions, 
which is due primarily to the attraction of unlike 
electrical charges, results in the formation of an 
ionic bond. 

On the other hand, when two elements of more 
similar electropositive or electronegative character 
react, they do so by the formation of a stable 
electron pair, which is mutually attracted to both 
of the atomic nuclei, as in the examples 

H. -J- H - >-H:H, orH* 

2 H. -f -O: * H:O: , or 


Such a bond is called a covalent bond. In the 
second example, since oxygen is somewhat more 
electronegative than hydrogen, the bond is some- 
what polar, or partially ionic, in character. 


Acids, bases, and salts either possess ionic bonds, 
or bonds which are quite polar, so that when these 
substances dissolve in water, the ions separate as 
independently moving particles, according to "ion- 
ization equations, 9 ' such as 

NaOH > Na+ + OH- f 

KjSO* + 2 K+ -f SO 4 . 

In the case of solid salts, the ions are already 

present as the structural units in the crystal, so 
that when the salt is melted and this crystal struc- 
ture broken down, the ions are free to move in- 
dependently. The ionization of acids in water 
solution is discussed in a later paragraph. 

Acids, bases, and salts are called electrolytes, 
because their water solutions conduct the electric 
current. Solutions of covalent compounds such as 
sugar (CwHraOn), are nonelectrolytea. 

The strong or active acids, bases, and most salts, 
ionize completely in dilute solution. In solutions of 
weak or slightly active acids and bases, a large part 
of the dissolved substance is present in molecular 
form, so that, although the total concentration may 
be high, the concentration of ions is low. This ac- 
counts for their slight activity. Relatively insolu- 
ble salts, although regarded as strong electrolytes 
in that all of the substance which is dissolved is 
present in ionic form, will likewise supply only a 
low concentration of ions to a solution. 

Ions and Electrical Conductivity 

Electricity, as carried by a metal wire, con- 
sists of a stream of electrons, which move along 
the wire from atom to atom. Electricity passing 
through a solution consists of a stream of negative 
ions moving toward the anode (positive electrode), 
and a stream of positive ions moving toward the 
cathode (negative electrode). In electrical conduc- 
tion through solutions, a chemical change occurs 
at each electrode. Electrons are taken from the 
cathode, and electrons are deposited on the anode, 
by the ions undergoing the change. 

These statements are illustrated by Figure 16-1. 
In this example, carbon electrodes dip into a rather 
concentrated solution of cupric chloride (CuCla), 
which contains cupric ions (Cu**) and chloride 
ions (Cl~). The positively charged cupric ions are 
attracted to the negative cathode and plate out as 
copper metal, 

Cu++ -f 2 0- + Cu. 

Each of the substances, copper (Cu) and cupric 
ion (Cu ++ ), has its own distinctive properties. Cu 




Carbon electrodes 


Negative electron* 

f/ow through wive Cu 4 *ion3 are at* Cficns Are at- 

65 through a tube traded by the tracted by the 

negative pole, positive pole. 

The battery force* 

electrons to go in 
one direction 
a pump. 

* t . ... - 

ions receive Cl ion* give up 

Dative electron*, negatn* electron* 

Metallic copper 
deposits on the 
negative electrode. 

Cl atoms combine 
to form bubbles of 
chlorine ga&(ClA 

FIG. 16-1. The ionization and electrical conductivity of a 
rather concentrated cupric chloride solution. There are twice 
as many Cl~ ions (valence 1) as Cu+ + ions (valence +2). 

is reddish brown in color and can exist alone; Cu 4 " 4 " 
is blue and is always mixed with an equivalent 
amount of some negatively charged ion. Cu and 
Cu ++ are different substances, differing in composi- 
tion by two electrons. 

Likewise, the negatively charged chloride ion, 
which is colorless and odorless, is attracted to the 
positive anode and gives up an electron to it, form- 
ing a neutral chlorine atom. 1 Two of these unite to 
form the greenish yellow, very corrosive, chlorine 

2 Cl- 

- C1 2 

2 -. 

Again, note that Cl" and C1 2 are different sub- 

Ions as Individual Substances 

Consider the common acids, with formulas HC1, 
HBr, HNO 3 , and II 2 SO 4 . As pure substances they 
are each distinctive, with individual characteristics. 
But now place them in water and the solution 
possesses a list of common properties the sour 
taste, the ability to dissolve active metals or to 
neutralize bases, and the ability to turn litmus red. 
These are the properties of hydrogen ion, written 
simply as H + . But, also, the solutions are different, 
possessing respectively the properties of chloride 
ion (Cl"~), bromide ion (Br~), nitrate ion (NO 3 ~), 
and sulfate ion (SO 4 ). Any solution of an electro- 
lyte is, thus, a mixture of at least two substances. 

The Hydration of Ions 

Since the hydrogen ion, as well as many others, 
is hydrated (combined with water), it is often 
written HaO" 1 ", called hydronium ion. (See Fig. 
16-2.) According to the older viewpoint, using the 
simple formula, H+, we write for the ionization of 
hydrochloric acid, 

HC1 * H+ + Cl- 

Using the hydrated formula, we write 
HC1 + H 2 * H S 0+ + C1-. 

The latter equation emphasizes the role of the 
solvent water in the ionization. The water molecule 
has a greater affinity for the simple hydrogen ion 

1 If a quite dilute solution of CuCU were used, we should find that, 
with such a low concentration of chloride ions, the electrode reaction 
would be different. Water molecules would then be decomposed tc 
supply the electrons needed, thus, 

2H a O 

- Oa + 4 H+ + 4 -. 



or proton (H+) than has the chloride ion (Cl~). 
Hence, the reaction takes place and two charged 
particles (ions) are produced. If hydrogen chloride 
gas is dissolved in some other solvent such as one 
of the many organic liquids, which are quite non- 
polar, we find that the proton (H 4 * ) is not attracted 
away from the chloride ion (Cl~~) by the solvent. 
As a result, hydrogen chloride is not an acid in 
such solvents. 

Most ions are hydrated more or less in water 
solution. Thus, cupric ion attracts four molecules 
of water, Cu(H2O)4+ + , and aluminum ion may at- 
tract six, A1(H2O) 6 +++ . These water molecules are 
held by covalent bonds. In this manual, for the 
sake of simplicity, we shall continue to use the 
simple unhydrated formula for most situations. 
You should recognize that whenever the formula 
H 4 " is written, the more precise IIsO" 1 " is implied. 

Ionic Equations 

Previously, we have written the equation for a 
neutralization reaction, for example, with mole- 
cular formulas, as if actual molecules were the re- 
acting particles, 

NaOH -fe HNOs + NaN0 3 + H 2 0. 
Since electrical conductivity data and the chemi- 
cal behavior of acids, bases, and salts indicate that 
their solutions consist of individual ions, it will be 
more nearly in accord with the facts to write the 
equation in terms of the principal substances (ions 
or molecules) actually present before, and after, 
the reaction. So we write 

Na+ + OH- -f H+ + N0 3 - > Na+ + NOr + H 2 O. 
This is called a total ionic equation. 

It is to be noted in this equation, that neither 
the Na + nor the N0 3 "~ have reacted both are 
present as separate particles after, as well as be- 
fore, the reaction. They may, therefore, be omitted, 
and the equation, which is now called a net ionic 
equation, becomes 

OH- + H+ > H 2 O. 

Such an equation focuses attention on the essential 
changes the disappearance of acid properties due 
to H 4 ", and of basic properties due to OH". The 
water formed is so slightly ionized that the con- 
centration of these ions is too low to show their 
characteristic properties. 

Ionic reactions occur whenever ions unite to form 
< weakly ionized or insoluble substances. In the neu- 

H/ ( 


Fio. 16-2. The formation of the hydronium ion by the attrac- 
tion of a simple hydrogen ion, or proton (H + ), to an unshared 
pair of electrons on the oxygen atom of a water molecule. 

tralization of nitric acid by sodium hydroxide, the 
net ionic equation shows that the only action oc- 
curring is the union of hydrogen ion and hydroxide 
ion to form water. The tendency for these ions to 
unite is the factor which causes the reaction to 
proceed. A similar reaction occurs when the ions of 
a weak acid or of a weak base unite as, for example, 

NH4+ + OH- * NH 4 OH. 

In the formation of the weak base ammonium 
hydroxide, the reaction is not as nearly complete 
as it is in the formation of water, for ammonium 
hydroxide has a far greater tendency to ionize than 
has water. A typical example of ions uniting to 
form an insoluble substance is expressed by the 

Ca++ + COj~ > CaCO,(*). 

(The (s) indicates that CaCO 3 is a solid.) It makes 
no difference in the reaction whether calcium ion 
(Ca ++) is obtained from calcium chloride, calcium 
nitrate, or any other soluble calcium salt. Car- 
bonate ion (CO 8 ) could be obtained equally well 
from sodium carbonate, ammonium carbonate, or 
any other soluble, well ionized carbonate. The 
chemical change, as recorded by the net ionic equa- 
tion, is the same in each case. 

It is also possible to change weakly ionized sub- 
stances into substances which ionize to an even lesser 
extent. When ammonium hydroxide (NEUOH) is 
neutralized by hydrochloric acid (HC1), the total 
ionie equation is 

NKUOH + H+ + C1- *NH4+ + Cl- + HjO. 
The net ionic equation is 

NHOH + H+ *NH4+ + HA 



Since ammonium hydroxide ionizes to a much 
greater extent than water, this reaction will pro- 
ceed. It, however, will not be as complete as a 
neutralization in which the acid and base are both 


*5o watt lamp 
with bulb 

Bottom view 
of apparatus 

no v R ll^^;-^ -'amp 

FIG. 16-3. An apparatus to compare the electrical conductiv- 
ities of various solutions. 

Again, the poorly ionized carbonic acid (H 2 CO 8 ) 
contains fewer carbonate ions (CO 8 ) in solution 
than does a saturated solution of the relatively 
insoluble calcium carbonate (CaCO 3 ), so the re- 

+ 2 H+ * Ca++ + H 2 CO 

will take place. Furthermore, the carbonic acid is 

not stable, but breaks down into water and carbon 

Unless there is a tendency for the ions in a solution 
to unite, no reaction occurs. Thus, if a solution of 
ammonium nitrate, NH 4 NO 8 , (containing NH 4 + 
and NOs"), and one of sodium chloride, Nad, 
(containing Na+ and Cl~~), are mixed, no reaction 
will occur; simply a mixture of the four kinds of 
ions will remain. The possible products, ammonium 
chloride, NH 4 C1, and sodium nitrate, NaNOs, are 
soluble, highly ionized salts; they, therefore, do not 

To summarize the technique of writing net ionic 
equations, note that this does not mean that all 
substances should always be written as ions. Poorly 
ionized, or slightly soluble substances, should be 
written as the un-ionized molecule. 

The Experimental Method 

By the measurement of the relative electrical 
conductivities of various pure substances, and their 
solutions, we shall obtain evidence concerning the 
presence and relative concentration of ions in these 
examples. We shall also trace the course of a num- 
ber of reactions by comparing the electrical con- 
ductivities of the reactants with those of the 
products formed. We shall interpret these results 
in terms of the net ionic equation for the reaction. 

Figure 16-3 shows a simple device for the com- 
parison of electrical conductivities. In this, the 
filament support wires from two 250-watt light 
bulbs, with the glass bulbs removed, are screwed 
into porcelain bases. These are connected in paral- 
lel with one another, and in series with a 10 or 15- 
watt light bulb, and with the 110- volt alternating 
current. Thus, we may compare the separate con- 
ductivities of two different solutions, tested one at 
a time, or we may test the combined conductivities 
of both solutions. 

Experimental Procedure 

Special supplies: conductivity apparatus. 

Chemicals: 17 F HCaHA, 6 F HC a HsO 2 , 0.1 F HC 3 H 8 O,, 
CAOH, 0.1 F NHUOH, 0.1 F Ba(OH) 8> CHe, CaCO, (marble 
chips). 0.1 F (XCJHA)* 0.1 F CuSO* CaH 6 (OH),, HC1 in 
benzene, 0.1 F HC1, KC10, 0.1 F KNO 8 , NaCl. 0.1 FNaCl, 
0.1 F NaOH, CuHaOn, 0.1 F HaSO 4f mossy Zn. 

Obtain an electrical conductivity apparatus of 
the type available in your laboratory, and deter- 
mine the relative conductivities of the pure sub- 

stances and solutions listed below. When testing 
solutions, use only 5 to 10 ml in a 10-cm test tube, 
a crucible, or a 60-ml beaker held at an angle. Do 
not waste solutions by using unnecessarily large 
amounts. Dip the conductivity wires in the solu- 
tion to a uniform depth of about 1 cm each time, 
and have them at a uniform distance apart. Be- 
tween each measurement, rinse the electrodes and 



dry them with a piece of filter paper. Caution: 
Disconnect the apparatus from the socket if 
the electrodes must be touched; keep your 
hands dry when using the apparatus. 

1. Electrolytes and Nonelectrolytes. By the 
above procedure, test the following, and classify 
each as a good, poor, or nonconductor of elec- 

(a) Pure Substances: (Save these for the next para- 
graph.) Water (distilled), alcohol 95% (C 2 H 6 OH), 
benzene (C 6 He), glycerine (C3H 5 (OH) 3 ), glacial or 
17F acetic acid (HC 2 H 3 O 2 ), solid sugar (CwHaOu), 
solid sodium chloride (NaCl), solid potassium 

chlorate (KCld). 

(b) Solutions: To each of the above, except the 
distilled water and the benzene, which is insoluble, 
add about 5 ml of distilled water, mix to effect 
solution, and again test these with the conductivity 

(c) Fused Salt: Melt a little KC1O 3 in a porcelain 
crucible, and test it, while molten, with the conduc- 
tivity apparatus. Wash the electrodes thoroughly 

(d) The Effect of the Solvent. After drying the 
electrodes, test a solution of HC1 gas (not hydro- 
chloric acid solution in water) dissolved in benzene 
or toluene. (This is already prepared for your use.) 
Compare the conductivity with that of aqueous 
HC1 as tested in the next paragraph. 

(e) A Comparison of Strong and Weak Acids and 
Bases. Determine the relative conductivity of 5 
to 10 ml each of 0.1 F HC1, 0.1 F HC 2 H 8 O 2 , 0.1 F 
NaOH, and 0.1 F NH 4 OH. Write equations to 
show the relative extent of ionization of each of 
the above substances. In doing this, use forward 
and reverse arrows. Vary the length of the arrow 
to indicate the desired tendency. Thus, - means 
that the ionization is practically complete; HP* 
means that the ionization takes place to only a 
moderate extent; and -*i means that the ioniza- 
tion is quite weak. 

Also, compare the above results of conductivity 
measurements with the physical and chemical prop- 
erties of strong and weak acids, as follows: 

1. Taste solutions of 0.1 F HC1 and 0.1 F 
HC 2 H 3 O 2 . 

2. Compare the relative rates of reaction of 0.1 F 
HC1, 0.1 F HCiHA, and HC1 in benzene, on (a) 
marble chips (CaCO 3 ) judge by the rate of evolu- 

tion of CO 2 gas, and (b) mossy zinc judge by the 
rate of evolution of Hk gas. 

2. Some Typical Ionic Reactions* Interpret 
the electricial conductivities of each of the follow- 
ing examples in terms of the solubilities and the 
extent of ionization of the substances involved. 

(a) A Mixture of Two Soluble Salts. Determine 
the conductivities of 10-ml portions of 0.1 F NaCl, 
of 0.1 F KNO 3 , and of their mixture. If the appa- 
ratus of Figure 16-3 is used, you can place the 0.1 F 
NaCl in contact with one electrode and the 0.1 F 
KNO 3 in contact with the other. Determine the 
total conductivity of the two solutions, separately, 
then mix them. Divide the mixture into the two 
containers and place these in contact with the 
electrodes. Note whether there is any change in 
the conductivity of the mixture as compared to 
that of the separate solutions. To what extent has 
a reaction taken place? 

(b) A Strong Acid with a Strong Base, to Form a 
Soluble Salt. Prepare 50 ml each of 0.01 F HC1 and 
0.01 F NaOH by diluting 5 ml of the 0.1 F solutions 
of each to 50 ml with distilled water, and mixing 
these well. Test the conductivities of these sepa- 
rate 0.01 F solutions, and of a mixture of equal 
volumes of them, as in paragraph (a). In preparing 
the mixture, add a drop of phenolphtkalein and 
carefully adjust the relative amounts to equiva- 
lence. How do you account for the slight decrease 
in the conductivity of the mixture as compared to 
that of the separate solutions? Write the total and 
the net ionic equations for the reaction. 

(c) A Weak Acid with a Weak Base, to Form a 
Soluble Salt. In a similar manner, mix equal vol- 
umes of the 0.1 F HC2lI,Q and the 0.1 F NH 4 OH 
solutions, and compare the brilliance of the light 
with that of either the acid or the base solution 
alone. Explain the effect by writing the total and 
the net ionic equations for the reaction. 

(d) A Strong Acid with a Strong Base, to Form an 
Insoluble Salt. As in paragraph (a), determine the 
conductivities of 0.1 F H 2 SO 4 and of 0.1 F Ba(OH)* 
solutions. Are these highly ionized? While the elec- 
trodes are dipping in the H 2 SO 4 solution, add a 
drop of phenolphthalein and then add the Ba(OH)2 
solution gradually, with stirring, until the exact 
equivalence point is reached as indicated by the 
change in color of the phenolphthalein. (With care, 
you can adjust the relative amounts of acid and 



base so that the light ceases to glow.) Explain, and 
write the total and net ionic equations for the re- 

(e) A Salt of a Strong Acid, with a Weak Acid, 
to Form an Insoluble Salt. First determine the con- 
ductivity of 0.1 F CuSO 4 solution, and of a solu- 
tion made by bubbling H 2 S gas 1 through 10 ml of 
distilled water, to saturate it. This gives an ap- 
proximately O.I F H 2 S solution. Then pass H^S 
gas directly into the 0.1 F CuSO 4 solution for two 

1 Consult your instructor as to the method of obtaining H2S gas 
in your laboratory. A cylinder of the compressed gas, or a Kipp gen- 
erator charged with ferrous sulfide sticks and 6 F IIC1, is frequently 
used. A commercial preparation of sulfur and a paraffin hydrocarbon, 
with asbestos added, is available from the Ilenga Co., 1833 Chestnut 
St., Philadelphia, Pa., or from various laboratory supply houses. On 
heating this substance in a test tube, H 2 S gas is liberated. Sec also 
Figure F-4, Page 292. 

or three minutes in order to precipitate the cupric 
sulfide, CuS, as completely as possible. Test the 
conductivity of the solution. Does the presence of 
the solid CuS materially contribute to the con- 
ductivity of the mixture? Why or why not? Write 
the total and the net ionic equations for the re- 
action, and explain any difference in conductivity 
in terms of the ions present. 

(f ) A Salt of a Weak Acid, with a Weak Acid 9 
to Form an Insoluble Salt. Determine the conduc- 
tivity of 0.1 F Cu(C2H 3 O 2 )2 solution, then saturate 
about 5 to 10 ml of this solution with H 2 S gas to 
precipitate the cupric sulfide completely, and again 
determine the conductivity of the solution. Write 
the total and the net ionic equations for the re- 
action. Explain the difference in conductivity in 
terms of the ions present. 

REPORT: Exp. 16 Name 

Ionic and Covolent Compounds. Date 

Ionic Reactions 

Locker Number- 

Review Questions 

The substance present in all acids which accounts for their common properties is: 

Four ways (composition or properties) in which this substance differs from hydrogen molecules are: 


What is the difference in composition between (a) a chloride ion and a chlorine atom? (b) a sodium ion and 
a sodium atom? 

1. Electrolytes and Nonelectrolytes. Indicate after each substance whether it is relatively a good, poor, or 

nonconductor of electricity : 

Distilled water Glycerine Solid NaCl 

Alcohol Glacial acetic acid Solid KClOs- 

Benzene Solid sugar Fused 

On the basis of your conductivity data of solutions, write beside the substances shown to be electrolytes, the 
formulas of any ions present in the solution, and beside the nonelectrolytes, the formulas of the molecular species 
in solution. 

Sugar__ . Glycerinej Ammonium hydroxide- 
Sodium chloride Sodium hydroxide Acetic acid 

Alcohol Potassium chlorate HC1 in water. 

HC1 in benzene- 

Can you suggest a reason for the difference in the behavior of HC1 dissolved in water, and in benzene? 

How do you explain the difference in the conductivity of solid KC1O> and of fused KC10? 


List the relative electrical conductivities of the acids and bases tested, and write the ionization equations for 
each, indicating the relative extent of ionization by the relative length of forward and reverse arrows: 

Conductivity Equation 


0.1 F NaOH 

0.1FNH 4 OIL 

Compare 0.1 F HC1 and 0.1 F HC 2 HsO 2 as to (1) taste, (2) reactivity with CaCO 8 , and with zinc: 

2. Some Typical Ionic Reactions, (a) A Mixture of Two Soluble Salts. The relative conductivities of the 
salt solutions tested are as follows: 

O.lFNaCl O.lFKNOs Mixture- 

The substances present before the solutions are mixed are 
(use ionic formula if ionized, molecular formula if un-ionized) : 

The substances present after the solutions are mixed are: 

Comment on the extent to which any chemical reaction has 
actually occurred: 

(b) A Strong Add with a Strong Base, to Form a Soluble Salt. The relative conductivities of the solutions tested 
are as follows: 

0.01 FHC1 0.01 F NaOH Mixture. 

The total ionic equation for the reaction is: 

The net ionic equation is: 

Interpret any change in conductivity of the solutions, before and after mixing, with the above equations: 

(c) A Weak Acid and a Weak Base* to Form a Soluble Salt. The relative conductivities of the solutions tested 
are as follows: 

0.1FHCtH 8 O 2 0.1FNH 4 OH Mixture. 

The total ionic equation for the reaction is: - 

The net ionic equation is: 

Interpret any change in conductivity of the solutions, before and after mixing, with the aouve equations: 


Report an Exp. 16, Sheet & No 

(d) A Strong Add with a Strong Base, to Form an Insoluble Salt. The relative conductivities of the solutions 
tested are as follows: 

0.1FH 2 SO 4 0.1FBa(OH) a Mixture . 

The solid substance formed by the reaction is: - 

The total ionic equation for the reaction is: - 

The net ionic equation is: _-_-___-___-___^^ 

Interpret any change in conductivity of the solutions, before and after mixing, in accord with the above 

(e) A Salt of a Strong Acid, with a Weak Acid, to Form an Insoluble Salt. The relative conductivities of the 
solutions tested are as follows: 

0.lFCuSO 4 0.1FH 2 S Mixture 

The total ionic equation for the reaction is: , , 

The net ionic equation is: __ 

Interpret any change in conductivity of the solution, before and after passing in II 2 S gas, in accord with the 
above equations: 

(f) A Salt of a Weak Acid, with a Weak Acid, to Form an Insoluble Salt. The relative conductivities of the 
solutions tested are as follows: 

0.1 F Cu(C2H 8 O 2 )2 Mixture with H 2 S_ 

The total ionic equation for the reaction is: : 

The net ionic equation is: - 

Interpret any change in conductivity of the solution, before and after passing in H 2 S gas, in accord with the 
above equations: 

Interpretation of Data 

1. Underline the formulas of the following substances which the experiment would lead you to believe would have 
many properties in common. (Other than similar electrical conductivity.) 

A. NaCl, HNO* C 2 H 5 OH, HQ|. 

B. HNOa, KNO,, NaaSO* Cud* 


2. Underline the formulas of the substances whose aqueous solutions are good conductors of electricity: 

NH 4 C 2 Ha0 2 , NH 4 OII, NaOH, CJ^OH),, HC 2 H 3 O 2 

3. Underline the formula of a substance which ionizes completely in solution, but for which it is impossible to 
prepare a solution with a high concentration of its ions: 

BaS0 4 , Na 2 S0 4 , BaCl 2 , NH 4 OH, C 2 H 6 OH 

4. Underline the formula of each substance which is very soluble in water, but whose water solution cannot contain 
a high concentration of its ions: 

BaSO 4 , Na 2 SO 4 , BaCl 2 , NH 4 OH, C 2 H 5 OH 

5. From data obtained from this experiment or from Tables X and XI in Appendix II, write, after the formula of 
each substance listed below, the formula or formulas of the principal substances found in the solution, and the 
molecular formula with a small (s) after it if the substance is insoluble. Four of the items are answered as examples. 

Sugar.. CiigQii BaSO 4 BaS0 4 (s) 

NH 4 OH 

HCN Ba(OH) 2 

Application of Principles 

Using information when needed from Tables X and XI in Appendix II, write in the space at the right the 
formulas of the principal ions or molecules found in the mixture when the substances listed at the left are allowed to 
react. If a solid molecular substance is produced, underline the formula. Assume that water is present so that 
soluble, strong electrolytes will be represented by ions, not molecules. Also assume that equivalent amounts are 
used. Note: Use coefficients preceding formulas only when expressing definite amounts of substances, but not when 
just writing the formulas. 

(1) Ba(OH) 2 and HC1 

(2) BaCl a and Na 2 COg (10) CuS and KNO^ 

(3) NaCl and AgNO 3 (11) KOH and HNO*- 

(4) NaC 2 H 3 O 2 and NaCl (12) Zn and CuSO 4 

(5) Fe(NO 3 )a and KOH (13) NH 4 C1 and Ba(OH) 2 

(6) ZnSO 4 and K 4 Fe(CN) (14) NII 4 OH and 

(7) Cu(C 2 H 3 O 2 ) 2 and 11,8 (15) NH 4 C1 and 

(8) Mg and H 2 SO 4 (16) CaCO* and HNOs- 


Oxidation -Reduction. 


College Ctam/sfry, Chapter 12 

Review of Fundamental Concepts 

Two Fundamental Types of Chemical Reactions 

Chemical processes may be divided into two 
general types. 

(1) The simpler of these types involves all those 
reactions in which there is no change in valence 
of any of the atoms taking part in the reaction. 
Such reactions include: The union of two ions to 
form a precipitate or slightly ionized substance, 
the solution of soluble substances to form the ions, 
and the formation of complex ions. Complex ions 
may be formed by the union of two kinds of ions 
or by the union of ions with neutral molecules. 
Familiar examples of these are the following: 

Ba++ + SO 4 ' > BaSO 4 , 

H+ -f OH- *HA 

NasSO 4 (solid) > 2 Na+ + SO 4 ", 
Ag+ -f 2 NH 3 > Ag(NH3) 2 +. 

(2) On the other hand, many reactions which 
we have studied involve changes in valence, or 
electrical charge, and are classed as oxidation-re- 
duction processes. These may involve such simple 
types as the replacement of one element in a com- 
pound by another, the direct union of two ele- 
ments, and the decomposition of a compound into 
its elements. Other types are more complex in na- 
ture, as illustrated by the last of the following ex- 
amples. Element Element 

Reaction Oxidized Reduced 

Cu + 2 Ag" 1 " > 2 Ag + Cu++ . copper silver 

2 112 + O 2 > 2 JI 2 O hydrogen oxygen 

2 HgO > 2 Hg -f O 2 ..... oxygen mercury 

2 Fc++ + Sn +"*-+ tin iron ' 

16 11+ -f 10 Cl~ + 2 MnO 4 - >- 
2 Mn++ -f- 5 C1 2 + 8 H 2 O . . chlorine manganese 

Note that in each case the ele- 
ment which is oxidized either in- 
creases in positive valence (copper, 
hydrogen, tin), or decreases in neg- 
ative valence (oxygen, chlorine). 
The substance which is reduced (sil- 
ver, oxygen, mercury, iron, manga- 
nese) undergoes the reverse change, 
decreasing in positive valence or 
increasing in negative valence. A 
graphical interpretation of these 
facts is given by Figure 17-1. 




Fio. 17-1. A 
change in oxida- 
tion number in- 
volves oxidation 
and reduction. 

Oxidation-Reduction and Electron Transfer 

According to the modern view of atomic struc- 
ture, when an element or simple ion undergoes a 
change in its ionic valence (that is, its electrical 
charge) it does so by gaining or losing electrons. 
An oxidation process always involves a loss of elec- 
trons. Correspondingly, a reduction process always 
involves a gain of electrons. In any chemical reaction 
where there is such a transfer of electrons, both 
oxidation and reduction must occur simultaneously. 
These processes may be written as separate "half- 
reactions." In the reaction of metallic copper with 
silver ion, the copper is oxidized to cupric ion with 
the resultant loss of two electrons: 

Cu > Cu++ -f 2 e~ (oxidation loss of electrons) 

At the same time, silver ions gain the electrons 
lost by the copper: 

2 Ag" 1 " + 2 e~~ * 2 Ag (reduction gain of electrons). 

FIG. 17-2. Oxidation-reduction and electron transfer. When 
chlorine, C^, reacts with magnesium metal, two electrons are 
transferred from one magnesium atom to two chlorine atoms. 




As another example, we may refer again to the 
last of the examples of oxidation-reduction given 
earlier in this section, namely, the rather compli- 
cated case of the reaction of Cl~~ in acid solution 
with MnO 4 ~. Here the 10 Cl~ are oxidized to 5 C1 2 , 

10 Cl~* "> 5 Cla + 10 e~ (oxidation loss of electrons), 
and the 2 Mn(>4~ are reduced to 2 Mn ++ , 

2 MnO 4 - + 16 H+ + 10 ~ + 2 Mn++ + 8 H 2 O (reduction 
gain of electrons). 

It is necessary to use 16 H + to react with the com- 
bined oxygen, giving 8 H 2 O. The total charges on 
the ions used are, 2 ( ) for the 2 MnO4~ and 
16 (+) for the 16 H+, or a net of 14 (+) charges 
on the left. On the right side of this reduction 
process, we have 4 (+) for the 2 Mn" 1 ""*". It is, 
therefore, necessary to write in 10 e~ on the left, 
in order to make this * 'half -reaction" balance elec- 
trically. We see, then, that the 10 Cl~~ lose the 
same number of electrons that the 2 MnO4~~ gain. 
These quantities are, therefore, equivalent from the 
standpoint of oxidation-reduction. Note that we 
have arrived at these equivalent quantities from 
the number of electrons lost and gained in the 
half-reaction processes, and have not found it nec- 
essary to make any assumptions as to the valence 
of individual atoms in the ions or molecules. 

The Oxidation State of an Element 

In the calculation of the valence change in oxi- 
dation-reduction reactions, it frequently is an ad- 
vantage to ascribe a definite oxidation number 
or oxidation state to each element in the com- 
pound. In binary compounds of metals with non- 
metals, this oxidation number is the same as the 
ionic valence of the elements concerned. Thus, we 
have an oxidation number of +2 for iron in Fe++, 
or FeCU, and of +1 for silver and 2 for oxygen 
in silver oxide, Ag 2 O, In ammonia, NH 8 , in which 
the bonds are quite covalent in type, the nitrogen 
is the more electronegative element, so we assign 
the oxidation number of 3 to nitrogen, and +1 
to hydrogen. 1 Any element in its free state, as zinc 
metal, Zn, or hydrogen gas, H 2 , has an oxidation 
number of zero. 

In compounds of three or more elements, the 
elements hydrogen, oxygen, and the alkali and 
alkaline earth metals, etc., are assigned their usual 
oxidation numbers. A third element in the com- 

l Sec Figure 11-11 in College Chemistry, on the dectronegativity 
Male of the elements. 

pound is then given such an oxidation number as 
will make the algebraic sum of the charges of all 
the elements in the neutral molecule add up to zero. 
Thus, in H^SC^ we have two positive charges due 
to the two hydrogen atoms, and eight negative 
charges due to the four oxygen atoms. The sulfur 
must then have a charge or oxidation state of +6. 
Similarly, the oxidation state of N in HNOs is +5, 
of P in H 3 PO 4 is +5 and of Cr in K 2 Cr 2 O 7 , is +6. 
Note in the last case that the sum of +2 (due to 
potassium) and 14 (due to oxygen) gives a total 
of 12, which is balanced by the opposite charges 
due to two Cr atoms, so the oxidation state of Cr 
is +6. 

The charge of an ion must be considered in the 
calculation of the oxidation state of a given ele- 
ment in the ion. Thus, in MnO4~~, there are eight 
negative charges due to oxygen. The oxidation 
state of Mn must therefore be +7, if the ion as a 
whole is to have a charge of 1. Likewise, in sul- 
fite ion (SO 3 ), we have six negative charges due 
to oxygen, so the oxidation state of sulfur must be 
+4, to account for the total ionic charge of 2. 

The assignment of a particular charge to an 
individual element in a compound or ion is quite 
arbitrary, since it is only in the extremely polar 
valence bonds that we are justified in assuming a 
complete transfer of an electron from one atom to 
another. In many cases, particularly those involv- 
ing carbon atoms, the bonding electron pair is 
shared by both atoms, so that the valence bond 
is more or less non-polar. However, in many cases, 
the assignment of a definite charge or valence num- 
ber to an element serves as a useful device in the 
classification of the different oxidation states of an 

The Relative Strength of Oxidizing 
and Reducing Agents 

The strength of an oxidizing agent depends on 
its tendency to gain additional electrons. Likewise, 
the less attraction a substance has for electrons 
which it already possesses, the stronger reducing 
agent it will be. Thus, silver ion is a stronger oxi- 
dizing agent than cupric ion because the reaction 

takes place as indicated, but not in the re- 
verse direction. That is, silver ion has a strong 
enough attraction for electrons to take them away 



from copper atoms. It is possible to arrange the 
various oxidizing and reducing agents as "oxida- 
tion-reduction couples" in a series according to 
their relative tendencies to gain or lose electrons. 
(See Table XIII, Appendix n, Oxidation-Reduc- 
tion Potentials.) 

Preliminary Extrcis* 

Before performing the experiment, complete the 
preliminary exercise in the report sheet, showing 
the substances oxidized and reduced in the several 
equations given, and the change in oxidation state 
of each. 

Experimental Procedure 

Chemicals: metal strips (about 5 mm x 15 mm) of Cu, Pb, 
Fe, Ag, Sn, Zn; saturated Br 2 water, CCU, saturated CU water, 
Cu turnings, 0.1 F Cu(NO 8 )2, 0.1 F I 2 , Fe filings, 0.1 F FeCU, 
0.1 F Pb(N0 8 X 0.1 F Hg(NOj) fc 0.1 F KBr, 0.1 F KJeCCN)* 
0,1 F KsFe(CN)a, 0.1 F KI, KMnO 4 cryst,, 0.1 F AgNO*, 
0,1 F SnCl 2 , 0.1 F Zn(NO 8 ) a . 

In this experiment, we shall determine qualita- 
tively in the laboratory the relative position of a 
limited number of oxidation-reduction couples in 
the potential series. We shall start with only three 
metals and their ions, and in subsequent sections 
gradually include more couples until we have con- 
sidered thirteen in all. 

1. A Simple Potential Series for the Metals* 
(a) Place a small strip of copper in 3 ml of 0.1 F 
zinc nitrate solution, and a small strip of zinc in 
3 ml of 0.1 F cupric nitrate solution. Which is the 
stronger reducing agent, copper or zinc? Now try 
a small strip of lead (sandpaper it if necessary) in 
cupric nitrate and also in zinc nitrate solutions. 
Try also copper and zinc strips in lead nitrate solu- 
tion. Write equations wherever reactions occur. 
Compare the strength of lead as a reducing agent 
with that of copper and of zinc. Prepare a table 
by arranging the symbols of these three metals in 
a vertical column, placing the strongest reducing 
agent at the top, and the weakest at the bottom. 
To the right of each metal, make a dash, followed 
by the symbol for the oxidized form of the metal, 
thus Me -Me ++ . You now have a brief "oxida- 
tion-reduction potential series." Which is the 
stronger oxidizing agent, lead ion or cupric ion? 
Why do the lead pipes used in draining laboratory 
sinks frequently corrode and develop leaks? Be 
specific in your answers. 

2. Other Metals in the Potential Series. Now 
expand your series to include the couples, H 2 H+, 
Sn-Sn++, Fe-Fe++, Ag-Ag+ and Hg-Hg2++. 
To do this, try various combinations of the sand- 
papered metal strips with 3-ml portions of the ni- 
trates or chlorides of other metals. Try only the 
combinations necessary for you to determine the 

position of each Me Me++ couple in the series. 
Use only ferrous, not feme, salts. Try only mer- 
curow*, not mercuric, nitrate on silver and copper 
metals. Free mercury need not be used. In cases 
of doubt as to reaction, rub the metal surface 
with your finger to note any deposit formed. You 
may not be able to determine the relative position 
of iron and zinc, since such electropositive metals 
do not plate out readily in water solution. Zinc is 
the stronger reducing agent. Likewise, tin and lead 
are so nearly the same that it is difficult to get a 
replacement. Tin is a slightly stronger reducing 
agent than lead. In determining the position of the 
Ha H+ couple, use hydrochloric acid. Nitric acid 
cannot be used here because nitrate ion in acid 
solution is a much stronger oxidizing agent than 
hydrogen ion and the results therefore would be 
confusing. Write the net ionic equations for all 
cases in which you obtained a reaction. Arrange 
your potential series, including these couples with 
the three from the preceding paragraph, first on 
scratch paper. When you have decided on the right 
order, enter them on the report sheet. Name thd 
strongest oxidizing agent and the strongest re- 
ducing agent thus far included in your series. 

3. The Oxidizing Power of the Halogens* 
Review the results of the replacement reactions of 
the free halogens and halide ions, as carried out 
in Experiment 14, paragraph (4), If you did not 
do this, carry out that procedure now. Re-write 
the equations for ail reactions which occurred, and 
interpret the results by arranging the halogens and 
their ions in a potential series. Again place the 
strongest reducing agent (Is this the free element 
or the ion?) at the top and at the left side of the 
couple. The strongest oxidizing agent will be at the 
bottom, on the right. 

4. The Ferrous Ion Ferric Ion Couple. De- 
termine whether ferric ion is a stronger or a weaker 
oxidizing agent than iodine or bromine, by adding 
1 ml of 0.1 F FeCU to 2 ml each of 0.1 F KBr and 



of 0.1 F KI. Add a little carbon tetrachloride and 
shake to note the formation of any free halogen. 
The presence of any ferrous ion in the solution may 
be tested by adding a little potassium ferricyanide 
solution, which gives the deep blue ferrous ferri- 
cyanide Fe 3 (Fe(CN) 6 )2 precipitate. Write equa- 
tions for any cases in which the ferric ion is re- 
duced; note that it does not go to metallic iron. 
Place the Fe + + Fe+++ couple in its proper place 
in your potential series of the halogens. 

5. The Manganous Ion Permanganate Ion 
Couple. Try the action of a little potassium per- 
manganate (add several small crystals of the solid) 
in 3 ml of 6F HC1. Warm, and note the odor. (See 
previous discussion for this equation.) Place the 
Mn+ + MnO 4 ~ couple in your potential series. 

6. The Reaction of the Halogens with 
Metals. Are the halogens strong enough oxidizing 
agents to oxidize the metals to the corresponding 
metal ions? To decide this, add bromine water to 
a little fine iron filings in a test tube. Try also the 
action of fine copper turnings with bromine water. 
Shake each mixture a few minutes, decant the 
solutions into clean test tubes, boil a moment to 
remove all free bromine, and test for the presence 
of bromide ion by adding a few drops of silver ni- 
trate solution. To test for the presence of ferric ion 
and also of cupric ion, add a little potassium ferro- 
cyanicle reagent to each solution. Ferric ion gives a 
deep blue ferric ferrocyanide Fe4[Fe(CN)6] 3 pre- 
cipitate, while cupric ion gives the maroon cupric 
ferrocyanide Cu 2 [Fe(CN) 6 ]. It is also possible to 
oxidize silver or mercury metals by means of the 
halogens. It should be noted, however, that when 
a metal such as silver is reacted on by a halogen, a 

precipitate, as Agl, is formed, which keeps the 
Ag+ at a low concentration. Under these conditions 
(low Ag+ concentration), silver metal is a stronger 
reducing agent than when it is in the presence of IF 
Ag + . Why? As a matter of fact the Ag Ag+ and 
the Hg Hg 2 ++ couples are below the I~ 1 2 and 
the Fe ++ Fe"*" 4 "*" couples but above the Br~ Br 2 

7. Summary. You should now be able to com- 
bine the two separate potential series you have de- 
veloped into one general oxidation-reduction po- 
tential series, which shows the relative tendencies 
of the various elements and ions studied to lose 
electrons. Note that each couple is written so that 
the change reading from left to right represents 
the loss of electrons. Along the left side of your 
list, write "Reducing Agents," and along the right 
side write "Oxidizing Agents." Which end of the 
left column represents the strongest (S) reducing 
agents? Which the weakest (W) ? Label these posi- 
tions with an S and W. Which end of the right 
column represents the strongest oxidizing agent? 
Which the weakest? Label these with an S and W. 

Such a table as this may be expanded to include 
many more oxidation-reduction couples, and is 
useful in predicting the course of many reactions. 
(Table XIII, App. II.) Any reducing agent (on 
the left) will have the possibility of reacting with 
any oxidizing agent (on the right) which is stronger 
than it is, that is, which is lower down in the series. 
In some cases, while there is a definite tendency 
for a reaction to take place, the rate of the reaction 
is extremely slow. Again, the concentration of the 
ions in a solution has a definite effect on the 
tendency for reaction. This factor will be con- 
sidered in Experiment 19. 

REPORT: Exp. 17 





Locker Number- 

Preliminary Exercise 

As a review, in the following equations for familiar reactions, underline once the reducing agent, and twicq the 
oxidizing agent. Indicate at the right the number of electrons gained or lost per atom for each element concerned. 
If there is no change, simply write in "no valence change." 



Per Atom 


Per Atom 

Fe + S * FeS 

2Fe + 3Cl 2 ^2FeCl 8 

2 Al + 3 HaSO* * 3 H 2 + A1 2 (SO 4 ) 8 

Hg++ + H 2 S * HgS + 2 H+ 

Ca + 2 H 2 > Ca(OH) 2 + H a 

3 Ca++ + 2 P0 4 > Ca 3 (P0 4 ) 2 

PbO + 2 H+ > Pb++ + H 2 

PbO-f H 2 ^Pb + H 2 

Cu -f 4 HNO 3 * Cu(NO 3 ) 2 + 2 NO 2 + 2 H 2 O 

NH 4 ^ + OH- * NH S + H 2 

CaC0 3 + 2 HC1 >- CaCl 2 + II 2 O + CO 2 

4 Zn + 5 H 2 SO 4 * H 2 S + 4 ZnSO 4 + 4 H 2 O 

1. A Simple Potential Series for the Metals. Write net ionic equations for any reactions taking place be- 
tween the metals and metal ions listed below (indicate any cases of no action). 

copper and zinc ion 
zinc and cupric ion 
lead and cupric ion 
lead and zinc ion 
copper and lead ion 
zinc and lead ion 

Which metal is the stronger reducing agent, copper or zinc? 

lead or copper? 

lead or zinc? . 

Construct an oxidation-reduction potential series for lead, copper, and zinc and 
their ions in the space at the right, following the directions in the experimental procedure. 

Which is the strongest oxidizing agent, lead ion, cupric ion, or zinc ion? 

Discuss a frequent cause of corrosion of lead plumbing in chemistry laboratory table drain pipes: 

2. Other Metals in the Potential Series. Write net ionic equations for the reactions performed in deciding 
on the relative position of iron, tin, silver, mercury, and hydrogen, and their ions in the potential scries (this time 
omit listing cases where no action occurs) : 


Construct a potential series for lead, copper, zinc, iron, tin, silver, mercury, and 
hydrogen in the space at the right, in accordance with your experimental results. 

The strongest oxidizing agent in this series is 
The strongest reducing agent in this series is 

3* The Oxidizing Power of the Halogens. The net ionic equation for the laboratory method of preparinj 
chlorine from HC1, with MnO 2 , is: * 

In this reaction the oxidizing agent is 

and the reducing agent is 

Write net ionic equations for the reactions of the free halogens with the halide ions, in accordance with you 
experimental data: 

chlorine and bromide ion, . 
chlorine and iodide ion, . 
bromine and iodide ion, . 
iodine and bromide ion, , 


Report an Exp. 17, Sheet * 


Construct a potential series for the three common halogens and their ions in the 3 
space at the right, arranging it as directed in the experimental procedure. 

The strongest oxidizing agent in this series is . 

The strongest reducing agent in this series is 

If fluorine were included in the series, the F~ Fi couple 
would be located 

4. The Ferrous Ion-Ferric Ion Couple. Write net ionic equations for the reaction of Fe+++ with the halide 
ions, in accordance with your experimental results. 

Ferric ion with iodide ion, 
Ferric ion with bromide ion, 

The Fe++ Fe+++ couple should therefore be placed between the- 

5. The Manganous Ion-Permanganate Ion Couple. Note your observed 
result on mixing MnOr" with CI~ in acid solution: K 

The equation for this reaction is given in the preceding discussion. Rewrite it here: 

In this reaction, the oxidizing agent is , the reducing agent is . 

Construct a potential series, including the three halogens, the Fe++ Fe+++ 
couple, and the Mn++ MnOr couple, in the space at the right, in accordance with 
your observations. 

6. The Reaction of the Halogens with Metals: Write net ionic equations 7 
showing the results of your experiments on the reaction of bromine water with iron 
and with copper. 

7. Summary. Construct an oxidation-reduction potential series for all thirteen 
couples studied in this experiment. Write "Oxidizing agents" and "Reducing agents" 
along the proper sides of the table. Indicate the position of the strongest and weakest 
oxidizing agents and reducing agents by placing (8) and (W) beside the formulas for 
these substances. 


.couple and the 

Application of Principles 

(Refer to Table XIII, Appendix II, to answer the following.) 

1. After each of the following, write an O if it may be used as an oxidizing agent, an R if it may be used as a 
reducing agent, and both O and R if it may be used as either, depending on the substance with which it reacts. 
Also write the formula of the reduced form, the oxidized form, or both, as the case may be. 

Al( )( ) Br 2 ( )( ) II 2 S( )( ) 

Mn0 2 ( )( ) - Mg( )( ) --- -- Fe++( 
H+( )( ) -- HC1( )( ) - 

2. Indicate the following as true (+) or false ( ) statements: 

(a) Manganese metal can dissolve in dilute IIC1 ................ ( ) 

(b) An acid solution of stannous nitrate is a stable solution ............. ( ) 

(c) Sn metal will reduce Sn+++ + to Sn++ .................. ( ) 

(d) H 2 O 2 in acid solution can oxidize Br~~ to Br 2 ................ ( ) 

(e) Ag metal will dissolve in IINOa, liberating H 2 gas .............. ( ) 

(f) Gold may be dissolved by 1 F HNO 8 .................. ( ) 

(g) Oxygen from the air can oxidize Fe++ to Fe+++ ............... ( ) 

3. Name a substance which can 

(a) reduce Fe+++ to Fe++, but cannot reduce Sn++++ to Sn++ ..... ___ . _ . 

(b) oxidize Al to A1+++, but cannot oxidize Ag to Ag 4 " ........ . _ _ ___ _ ___ 

(c) reduce Sn++ to Sn, but cannot reduce Co + + to Co ........ _____________ 

(d) oxidize Fe to Fe++, but cannot oxidize Fe++ to Fe+++ ....... ______ ____ 


Write a thoughtful and reasonably complete list of the most important principles and viewpoints you have 
gained from this experiment: 


Common Oxidizing and Reducing Agents. 

The Balancing off Oxidation-Reduction Equations. 


College Chemistry, Chapter 12 

A Study Assignment 

Charts of Oxidation States for Common 

Much useful information about the behavior of 
oxidizing and reducing agents, under various condi- 
tions, can be summarized in the form of charts. 
Such charts for some common elements are pre- 
sented below, arid are repeated again in later 
experiments when these elements are studied in 
greater detail. These charts, and the comment 
given with them, will help you to predict the prob- 
able changes in oxidation state which will occur 
in a particular reaction. Note that the oxidation 
state is given just before each formula in the chart. 
Let us illustrate by commenting briefly on the 
charts for sulfur and oxygen compounds. 

It should be obvious from the chart on sulfur 
compounds that, since H 2 S represents the lowest 
possible oxidation state, it can act only as a re- 
ducing agent in oxidation -reduction processes, in 
which case it can be oxidized to free sulfur, or to 
some higher state, as a sulfate, depending on the 
conditions and on the strength of the oxidizing 
agent. Sulfuric acid (HsSCh), representing the 
highest oxidation state, can act only as an oxidizing 

agent, its reduction products being any of the 
lower states of sulfur. On the other hand, since 
sulfurous acid (H 2 SO 8 ) represents an intermediate 
state of sulfur, it may act either as a reducing 
agent with substances which can take on electrons, 
as chlorate ion (ClOs~), or as an oxidizing agent 
with substances which can lose electrons, such as a 
metal like zinc. 

Hydrogen peroxide and the perox-ides are impor- 
tant oxidizing agents, both commercially and in the 
laboratory. Note, according to the chart, that 
when hydrogen peroxide acts as an oxidizing agent 
it is reduced to water, or in basic solution to hydrox- 
ide ion. It is oxidized to free oxygen only when it is 
acting as a reducing agent, in the presence of 
stronger oxidizing agents. The instability of hydro- 
gen peroxide, especially in the presence of certain 
catalysts, is due to this ability of one molecule to 
oxidize another molecule of the same substance 
(auto-oxidation-reduction). The half-reactions cor- 
responding to these statements are: 

2 H+ + 1I 2 2 + 2 e~ > 2 H,O (reduction) 
H 2 Oj * 2 H+ + Of - 2 e~ (oxidation) 

2 H 2 O 2 > 2 H 2 O -j- O 2 (auto-oxidation-reduction) 

Sulfur compounds: 

+6 S0 3 , H 2 S0 4 , SOr 
+4 SO 2 , HjSOs, SO- 

ave +2 S 2 O, 




Oxygen compounds 9 peroxides: 


-1 H 2 2 , HOr 
-2 HAOH- 

Concentrated acid is a strong oxidizing agent. 

Active either as oxidizing or reducing agent. 

Thiosulfatc ion. Decomposes to S and H 2 SO| in acid solution. Oxidized to 

(tetrathionate ion) by free I 2 . 

Strong reducing agent, usually oxidized to S. 

Active as an oxidizing or as a reducing agent. 

Chlorine compounds: 

+7 (Cl 2 Or), HClOfc C1O 4 ~ C1 2 O 7 is unstable. HC1O 4 is a very strong oxidizing agent. Reduced to Cl"". 

+5 HClOj, C1O|- Strong oxidizing agent. Reduced to Cl~. 

+4 ClO a Unstable, explosive. 
+3 HClOj, ClOr 

H-l CIA HCIO, CIO- Good oxidizing agent. Reduced to Cl~. 

C1 2 Good oxidizing agent. Reduced to Cl~. 
-1 Cl~ 






Nitrogen compounds: 

+5 N 2 6 , HN0 8 , NOr 

+4 N0 2 , (N 2 4 ) 

+3 (N 2 0,, HN0 2 ), NOr 

+2 NO 

+1 NiO 


-3 NH,, NH 4 OH, NH4+ 

Chromium compounds: 

Acidic Basic 

+6 Cr 2 O 7 CrO 4 
+3 Cr+++ Cr(OH)r 
+2 Cr++ 


Manganese compounds: 
+7 MnOr 

+6 MnO 4 

+4 MnO 2 , MnO(OH) 2 

+3 Mn+++Mn(OH), 

+2 Mn++ 


Strong oxidizing agent, usually reduced to NO 2 in concentrated acid, or to NO in 

dilute acid. With strong reducing agent may go to NH|. 

A heavy brown gas. 

N 2 O 3 and HNO 2 are unstable; nitrites are fairly stable. Active as oxidizing or as 

reducing agent. 

Oxidized by the air to NO 2 . 

Supports combustion quite vigorously. 

Strong oxidizing agents. Bichromate ion is orange, chromate ion is yellow. 
Amphoteric. Chromic ion is green to violet. Chromic hydroxide complex ion is green. 
An uncommon ion, because it is such a strong reducing agent that it reduces water 
to hydrogen gas. 
The metal. 

Permanganate ion, purple. Strong oxidizing agent, reduced to Mn ++ in acid solu- 
tion, or to MnO 2 (sometimes to MnO 4 ) in neutral or basic solution. 
Manganate ion, green. Easily reduced to manganese dioxide. 
Brown as precipitated from solution. 
MH+++ j s unstable, gives Mn+ + and MnO 2 . 

Colorless in solution, pale pink as solid manganous salts. Mn(OH) 2 is oxidized by 
air to Mn(OH),. 
The metal. 

Some common reducing agents: 

a) The metals oxidized to their positive ions, 

Sn oxidized to Sn+ + 
Zn oxidized to Zn+ + 

b) Ions in which the metal has another higher 
oxidation state, as: 

Sn++ oxidized to Sn+ +++ 
Fe+ + -oxidized to Fe+++ 
Hg 2 ++ oxidized to Hg++ 

c) Carbon and organic compounds may be oxi- 
dized to other organic compounds, or to C0 2 and 
H 2 O, aa: 

C (coke, much used in industry) oxidized to CO or to CO a 
CjHjOH (alcohol)-may be oxidized to HC 2 H 8 Oa (acetic acid) 

Thus: CJH.OH + H 8 O > HC,H,O 2 + 4 H+ + 4 e~ 
HCHO (formakiehyde)--oxidized to HCH0 2 (formic acid) 

Thus: HCHO + H,O * HCHO, + 2 H+ + 2 e~. 

The Balancing of Oxidation-Reduction 

Since the fundamental event in an oxidation- 
reduction reaction consists in a transfer of elec- 

trons, it is obvious that in balancing the equation 
for such a reaction one must take such relative 
amounts of the reactants that the electrons sup- 
plied by the oxidation process are all used by the 
reduction process. There are several techniques for 
doing this, differing in the mechanics of the opera- 
tion, but all involving the same principle. 

The Half-Reaction Method 

Separate half -reactions, or electron reactions, are 
first written for the oxidation and for the reduc- 
tion processes. In developing these, one may first 
determine the number of electrons required from 
the change in oxidation number, then insert H+ 
(or OH- if the solution is basic) to balance the 
charges, and finally add H 2 O to balance the atoms. 
For an example of this technique, study the oxida- 
tion of FeSOi by KC1O as given in your text, 
College Chemistry, Chapter 12. 

The reverse process is sometimes used. By this 
technique, first balance the atoms in the half- 
reaction by inserting H+ and HjO as needed, and 
then insert as many electrons as needed to balance 



the charges. Study the following examples. 

Example 1. Let us consider the oxidation of 
sodium sulfite (Na a SO) by potassium dichromate 
(K 2 Cr 2 O 7 ) in an acid solution. Reference to the 
charts of oxidation states for sulfur compounds 
and for chromium compounds will indicate that 
the sulfite ion would be oxidized to sulfate ion, and 
that the dichromate ion would be reduced to chro- 
mic ion (Cr+++), as would be evidenced by the 
green color of the solution. We may write first a 
partial equation including only the sulfite ion 
(80s ) and its oxidation product, 

Step(l) SO," *S0 4 

We need another oxygen atom on the left, which 
will be supplied by water, and then we write the 
hydrogen as 2 H+ on the right. (Note that we keep 
the oxygen with a valence of 2 and the hydrogen 
with a valence of +1 on both sides of the equation, 
since they are not the substances oxidized and 
reduced.) Our partial equation then becomes 

Step (f) H a O + SO," * SO 4 ~ + 2 H+ 

We still need to balance the charges so that they 
are the same on both sides of the equation. To do 
this we add two electrons on the right side of the 

Step (5) H 2 + SO, + S0 4 + 2 H+ + 2 e~ (a) 

which completes the oxidation half-reaction, and 
emphasizes the fact that an oxidation process in- 
volves the loss of electrons. 

The reduction of dichromate ion to chromic ion 
may be expressed similarly. The several steps are: 

Step (1) Cr 2 7 * 2 Cr+++ 

then balance the hydrogen and oxygen by insert- 
ing 14 H+ to react with the seven oxygen atoms 
to form 7 H 2 O, 

+ 14 H+ > 2 CT+++ + 7 H 2 O 

Step (0) 

Finally, balance the charges. As above written, 
we have 12 positive charges on the left and 6 pos- 
itive charges on the right, or a net charge of 6+ 
on the left. We, therefore, need 6 electrons on the 
left side to complete the reduction half-reaction, 

Step (S) Cr a O 7 + 14H+ + 6~ *2Cr++++ 7H,O(b) 

This emphasizes the fact that a reduction involves 
the gain of electrons. Note that the requirement of 

6 electrons corresponds with the change in oxida- 
tion state of chromium from +6 to +3, so that 
the two chromium atoms decrease by a total of 6 
charges. However it was not necessary to assume 
these oxidation states in order to balance the re- 
duction half-reaction properly. 

Finally, we may combine (a) and (b) in such a 
way as to balance the electrons gained against 
those lost, since free electrons never appear in the 
final equation. To do this, we multiply (a) by 3, 
and add algebraically to (b), 

Balanced equation: 

3 Hrf)+ 3 SO, *3 SO 4 + 6 H++ 6 *~ (a) X 3 
Cr a O T + 14 H+ + 6 e~ + 2 Cr+++ + 7 H,0 (b) 
Cr 2 O 7 +3 SO, +8 H+ >*2 Cr++++3 SO 4 ~ +4 H f O 

You should always check your results to see that 
the equation balances both as to atoms and as to 

Example . The spontaneous decomposition of 
aqua regia results from the slow oxidation of chlo- 
ride ion by nitrate ion in a strongly acid solution. 
This time we shall give only the completed half- 
reactions involved. See if you can develop them, 
on a piece of scratch paper, by either or both of the 
above techniques. For the half-reactions we have 





Now multiply (a) by 3 and (b) by 2 to give the 
same number of electrons in each case, and then 

Balanced equation: 
6C1- *"3Cl 2 

-2NO+4H 3 



Example 8, A case in basic solution. If the same 
reaction studied in Example 1 is carried out in 
basic solution, we proceed as follows. Write the re- 
ducing agent and its oxidized form as before: 

8tep(l) SO," *S0 4 ~ 

But now the reaction must be balanced in terms of 
H 2 O and OH" (since H> is not available), so we 
add 2 OH" on the left (twice as much as the re- 

1 Nitric oxide, NO, will combine with free chlorine, Cl* to give 
nttrosyl chloride, so the final equation may be written to include 
2 NOC1 + 2 CU, instead of 2 NO -h 3 Cl* 



quired oxygen) and then water on the right to 
balance the atoms: 1 

Step (f ) 2 OH- + SO," * S0 4 ~ + H 2 O 

Finally add electrons to balance the charges, as 

Step (3) 2 OH" + SO, > SO 4 ~ + H,O + 2 e~ (a) 

For the reduction of the dichromate ion, note, 
in the chart on chromium compounds, that in 
basic solution this will be present as chromate ion, 
CrO 4 , and when reduced to the trivalcnt state 
will be present in basic solution as the hydroxide 
complex ion, Cr(OII 4 )~. We therefore write: 

Step (1) Cr0 4 " > Cr(OH) 4 ~ 

The oxygen now balances, but there are four 
extra hydrogen atoms on the right; so, by adding 
4 OH- on the right to react with these, and 4 H 2 O 
on the left, we have achieved a balance of atoms: 

Step (2) 4 II 2 + Cr0 4 + Cr(OH) 4 - + 4 OH~ 
Finally, add electrons to balance the charges: 
Step (3) 3 *-+4 II 2 0+Cr04~ >Cr(OH) 4 -+4 OH- (b) 

We may now combine the oxidation half -reaction 
(a) with the reduction half-reaction (b) so as to 
eliminate electrons. 

Balanced equation: 
60H-+3SO, *3 
6 g-+8 H,0+2 CrQ 4 ~ 

4 +3H 2 0+6<r- (a) X 3 
2 Cr(OH) 4 -+8 OH- (b) X 2 

3SQT" +2CrO 4 

+2Cr(OH) 4 -+20H- 

The Oxidation Number Method 

This technique differs from the preceding one 
principally in that both the oxidizing and the re- 
ducing agents, and their respective products, are 
written as a single preliminary equation. The rela- 
tive amounts of each are then computed from the 
changes in oxidation numbers. The total charge on 
each side of the equation is then balanced by the 
addition of H+ (or OH~ in a basic solution), and 
the atoms are balanced by the addition of H 2 O. 
Study the following example. 

Example 1. Let us consider again the oxidation 

1 For cases in which it is easy to assign definite oxidation states, it 
may be simpler for you to note, as in this case, that sulfur changes 
from oxidation state -f 4 in SQg to 4-6 in SO 4 , and therefore 2 e~ 
are needed on the right. Finally, then you can balance the charges by 
adding 2 OH* on the left and H 2 on the right to balance the atoms. 

of sodium sulfite by potassium dichromate in an 
acid solution: 

Step (1) Write down the principal substances 
used and produced: 

a + CryO," * S0 4 + Cr+++ 

Step (2) Note the change in oxidation number, 
i.e., the number of electrons lost and gained. These 
may be indicated by brackets, thus: 


i + 

+4 +6 



.loses 2 0-_ 

I gains 3 e~~ per Cr 

Step (3) Take such amounts of oxidizing and oi 
reducing agents as will make the number of elec- 
trons lost equal those gained. In this case, since 
each S loses 2 e~ and each Cr gains 3 e~ 9 we need 
3 S to 2 Cr, or 

3 SOj + Cr 2 7 + 3 S0 4 + 2 Cr+++ 
Step (4) Note the total charges on each side of 
the equation, in this case 8 negative charges on the 
left and charges on the right. Then add II + so 
as to balance the charges, and insert H 2 O to bal- 
ance the atoms. In this case we need 8 H+ on the 
left and 4 H 2 on the right. The final equation is 
then : 

Balanced equation: 

3 S0 8 +Cr 2 7 +8 H+ K* SO 4 "+2 O++++4 H 2 O 

Example 2. If this same reaction were being car- 
ried out in basic solution, the chromium would be 
present as chromate ion (CrO 4 ) before the re- 
action and as chromic hydroxide complex ion 
(Cr(OH) 4 -) afterward. (See the chart for chro- 
mium compounds.) The steps in balancing the equa- 
tions are: 

'SO^- +Cr(OH) 4 - 

+6 +3 

Step (1) SO 3 + CrO 4 
8tep(&) SOr~+CrO~- 

+4 +6 

I I 

I loses 2 9- 

I gains 30 -_ 

Step (3) 3 SQi~ + 2 CrO 4 ~ > 3 SO 4 ~ + 2 Cr(OH) 4 ~ 

Step (4) Since we have 10 negative charges on 
the left and 8 negative charges on the right, we 
need 2 OH~ on the right, and 5 H 2 on the left: 

Balanced equation: 

3 SO* + 2 CrO 4 " + 5 H 2 O > 

3 S0 4 ~ + 2 Cr(OH) 4 ~ + 2 OH". 

REPORT; Study Assignment D Name 

Common Oxidizing and Date 

Reducing Agents . 

Locker Number* 

Application of Principles 

I. What is the change in oxidation number (give number of electrons gained or lost, per atom; e.g., 3 e~ gained) 

C1 2 *NaClO . . _ HsSOs *S . . . _ HA 

CrA~ *CrO 4 . KMnO 4 *MnO 2 . Nil, 

II. Give the formula of a product (derived from the first-named substance) which may be formed when: 

1. HNO 2 is treated with an oxidizing agent . 

2. IINO 2 is treated with a reducing agent 

3. HzS is treated with H 2 SO 

4. Hot, cone H 2 SO4 reacts with carbon 

5. HA is treated with FeSO 4 and II 2 SO 4 

6. HA is treated with KMnO 4 and H 2 SO 4 . 

7. MnO 2 is treated with cone HC1 

III. Write the half -reaction equation for the oxidation of: 

1. Cl- to CIO- (acid) 

2. NO 2 ~ to NO 8 - (acid) 

3. SO| to SO 4 (basic) 

4. Cr+++ to CrA" (acid) 

5. C 2 H 6 OH to CHCHO (acid) 

IV. Write the half -reaction equation for the reduction of: 

1. N0 3 - to NH 4 + (acid) 

2. SOr~ to S (acid) 

3. MnO 4 ~ to MnO 2 (basic) 

4. C1O 4 - to C1 2 (acid) 

5. SOi~ to S (basic) 


V. Using the method you prefer, write balanced net ionic equations for the reactions of the following substances. 
(Show your method in each case.) 

1. Cla gas is bubbled into H-zS solution, forming a yellow precipitate. 

2. Al metal is used to reduce FeCl 3 to FeCl 2 . 

3. 1I2S gas is passed into concentrated IINOs, giving a brown gas. 

4. Bleaching solution, NaCIO, is added to an acid solution of FeSO 4 . 

5. SOj gas is bubbled into an acid solution of KMnOi. 

6. SOa is added to a basic solution of KMnOi. 

7. Ag metal is dissolved by dilute UNO*. 

8. Sn++ is added to an acidified Cr 2 O7 solution, turning it green. 

9. H20a and NaOH are added to Mn+ + solution, giving a brown precipitate. 


The Preparation and Properties of 

Some Oxygen Compounds of the Halogens. 


Co//ege Chemistry, Chapter 13 

Review of Fundamental Concepts 


Halogen compounds: 

(C1 2 O 7 ), HC1O4, C1O 4 - Cl2O 7 is unstable. HC1O 4 is a very strong oxidizing agent. Reduced to Cl* 
Strong oxidizing agent. Reduced to Cl~. 
Unstable, explosive. 

Good oxidizing agent. Reduced to Cl~. 
Good oxidizing agent. Reduced to Cl~, 


(C1 2 7 ), 


cio 4 - 





C1O 2 

r +3 

HC1O 2 , 


1 + 1 





1 _! 


The oxygen compounds of the halogen elements 
chlorine, bromine and iodine are important 
substances whose preparation and properties fur- 
nish further interesting illustrations of the princi- 
ples of oxidation-reduction. Due to its great elec- 
tronegativity (see text), fluorine exists only in 
the oxidation states 1 and 0. The other halo- 
gens, however, are capable of existence in a range 
of oxidation states extending from 1 to +7. The 
student should acquaint himself with the names 
and formulas of the oxides, acids, and anions of the 

halogens in each of the oxidation states. 

In this experiment, we shall follow the element 
chlorine through changes in its oxidation state 
from 1 to to +1 to +5 by a series of successive 
reactions. We shall also perform some experiments 
which illustrate the ability of a lighter halogen to 
increase the oxidation number of a heavier halogen 
while the lighter halogen is undergoing a decrease 
in its oxidation number. This is in consequence of 
increasing electronegativity of the halogens in the 
order I, Br, Cl, F. 

Experimental Procedure 

Chemicals: MnQ& KOH, CCl^ starch iodide paper, solu- 
tions of 0.1 F KI, 0.1 F KBr. 

1. Preparation of Chlorine, Potassium Chlo- 
rate and Sodium Hypochlorite. Before begin- 
ning the experiment, turn to the report sheet and 
write the equations for the reactions which you are 
to observe; namely, the oxidation of chloride ion in 
an acid solution by MnCV, the reaction of chlorine 
with cold sodium hydroxide solution; the reaction 
of an excess of chlorine with a hot solution of po- 
tassium hydroxide. 

Set up the apparatus as illustrated. After you 
have obtained the instructor's approval on the ar- 
rangement of the apparatus, place the following 
chemicals in their respective containers. Into the 
Erlenmeyer flask, put approximately 0.1 gram-for- 
mula weights (gfw) of MnO 2 . Place a loose plug 
of cotton in the bulb B of the drying tube, which 
serves to catch any spray from the chlorine gener- 
ator. Dissolve about 0.06 gfw of solid KOH in 15 
ml of water and put this solution into the 15-cm 
test tube in a beaker, C. Dilute 5 ml of the 6 F 

NaOH stock solution with 20 ml of distilled water 
and place this solution in test tube D, which is sur- 
rounded by an ice bath. Heat the water bath in C 
almost to boiling. Add about 75 ml of 6 F HC1 
through the thistle tube and warm flask A gently 
until chlorine gas is being produced. Keep the 
water in C almost at the boiling point by heating 
it intermittently. Use the same burner to heat flask 
A gently from time to time, but do not let the re- 
action mixture boil. 

After the air has been displaced from the appa- 
ratus, note how the chlorine gas-bubbles decrease 
in size as they are absorbed by reaction with the 
basic solutions. Continue the heating until the 
chlorine gas is no longer appreciably absorbed in 
the cold NaOH solution. Allow the system to cool, 
and when flask A is cool enough to be handled, dis- 
connect the drying tube and replace it with a 
right-angled tube which is long enough to dip into 
a beaker of cold water. Allow the generator to re- 
main this way until the end of the period, when it 
should be completely cooled and the contents may 




25 ml Na OH 


Fio. 18*1. The preparation of chlorine, potassium chlorate and 
sodium hypochlorite. 

be washed down the sink with water. This should 
prevent any appreciable amount of chlorine from 
escaping into the laboratory. 

2. The Crystallization of Potassium Chlo- 
rate. Transfer the solution in the test tube in C to 

a small beaker and evaporate it to about half its 
volume. Cool this solution in the ice bath until a 
crop of crystals has formed. Separate the crystals 
from the supernatant solution by filtering. Rinse 
any crystals remaining in the beaker into the fun- 
nel with a few ml of ice water. Transfer the crystals 
to a small beaker and dissolve them in as small an 
amount of hot water as possible (5 to 10 ml should 
suffice). Cool the solution in the ice bath; separate 
the crystals by filtration and dry them between 
several thicknesses of filter paper. Refer to the 
solubility chart, Fig. 18-4 of the text, to deter- 
mine the relative solubilities of KC1 and KC1O 3 in 
water at different temperatures. 

While the KC1O 3 crystals are drying, perform 
the following experiments with the NaOCl solution 
in the test tube at D. 

3. The Chemical Properties of the Hypo- 
chlorite Ion. Test the bleaching properties of 
hypochlorite ion by putting several drops of the 
solution in D on pieces of red and blue litmus 

To test the oxidizing strength of the hypochlorite 
ion, put 3 ml of 0.1 F KI and 1 ml of CC1 4 in a 
10-cm test tube and add one or two drops of the 
NaOCl solution. Shake the test tube and note the 
color of the CCU layer. Write the equation for this 
reaction. Now add about 1 ml more of the NaOCl 
solution to the same test tube, shake it again and 
note any change in the CCU layer. Write the equa- 
tion for this reaction. Are these results in agree- 
ment with the general principle stated in the text? 

Before beginning the next experiment, refer to 
Appendix II, Table XIII for the relative oxida- 
tion potentials of the CIO" ion in basic and in acid 
solutions. Record these on the report sheet. Now 
repeat the experiment of the preceding paragraph 
using 0.1 F KBr in place of the KI solution. If no 
results are obtained with the basic solution, acidify 
with several drops of 3 F I^SO*. Write the equa- 
tion for the reaction in the acid solution. (Some- 
times the freshly prepared NaOCl solution contains 
some dissolved chlorine, which of course displaces 
Br 2 from the KBr solution, even when alkaline.) 

The greater ease with which I" is oxidized to I 2 , 
and thence on to IO 8 ~ by the hypochlorite ion in 
alkaline solution may be utilized to differentiate 



between the I" and Br~ ions when they are present 
in the same solution. Put 1 ml each of 0.1 F KI and 
0.1 F KBr and CC1 4 into a test tube. Add one or 
two drops of the alkaline NaOCl solution and shake 
the tube. Note the color of the CC1 4 layer. Now 
add 1 to 2 ml of NaOCl and shake the tube again 
until the color in the CC1 4 layer has disappeared. 
Then acidify the solution with several drops of 3 
F H 2 SO4 and shake the test tube again. Note the 
color in the CC1 4 layer now. Explain the results in 
the report sheet. 

4. The Chemical Properties of Potassium 
Chlorate. Test the dry crystals of KC1O 3 prepared 
in 2, by heating about half of the crystals in a small 
Pyrex test tube. If the KClOa is kept just at its 
melting point, note that very little visible decom- 
position takes place. (See text for the reaction.) 
Now add a small amount of MnO 2 to the test tube 
and note any difference in the rate and type of 
decomposition which takes place. Write the equa- 
tion for this reaction. 

Treat a few crystals of KC1O 3 with about 1 ml of 
12 F HC1 in a small test tube. Test for the product 
of this reaction with moistened starch-iodide paper. 
Write the equation for the reaction which occurs 
in the test tube and also for the one which takes 
place in the starch-iodide paper. 

Dissolve the remainder of the KC1O 3 in about 10 
ml of water. Add 5 ml of this solution to a test tube 
containing 1 ml of 0.1 F KI and 1 ml CC1 4 . Repeat 
this test, but acidify the second solution with 5 
drops of 3 F H 2 SO 4 . Shake the test tubes and ob- 
serve any signs of a reaction. Allow the tests to 
stand until the next period and then shake the test 
tubes and note any oxidation of iodide ion. Note 
the relative positions of the chlorate ion and the 
hypochloritc ion in the oxidation-reduction poten- 
tial series in Appendix II. One may account for 
the difference in results noted in 3, with the hypo- 
chloritc ion, by relative rates of the reactions rather 
than by any significant difference in the oxidation 

Fundamental Rules of Nomenclature off Inorganic Compounds 

Optional drill for students who have not yet mastered the elementary rules of nomenclature. 

In Experiment 8 the naming of binary com- 
pounds was discussed. Now, with the introduction 
of the variety of oxyhalogen compounds in Experi- 
ment 18, a more complete presentation of the rules 
of nomenclature is appropriate, 

Binary Compounds 

Rule 1. Binary, or two-element, ionic com- 
pounds are named by the following general system: 

Thus SO 2 is sulfur dioxide; SO* is sulfur trioxide. 

NOj is nitrogen dioxide; N^Os is nitrogen pentoxide. 
CO is carbon monoxide; CCU is carbon tetrachloride. 

Rule 3. Binary acids are named by the following 
general system: 

hydro nonmetal (abbrev.) ~ic acid 
Thus HC1 is hydrochloric acid. 
1X28 is hydrosulfuric acid. 
HI is hydroiodic acid. 

metal nonmetal (abbrev.) -ide 

Thus NaCl is sodium chloride MgBr 2 is magnesium bromide 
CaO is calcium oxide KOH is potassium hydroxide 

If the metal has a variable valence, the suffix ovs 
is used to denote the lower valence state and the 
suffix ic the higher valence state. 

Thus FeCl 2 is ferrous chloride; FeCl 3 is ferric chloride. 
is mercurous oxide; HgO is mercuric oxide. 

Rule 2. Binary covalent compounds use the 
same ending, ide, but in case of variable valence 
the Greek prefixes mono, di, tri, etc. are used to 
denote the number of atoms of the more negative 

Ternary Compounds 

Most of the nonmetals form compounds with 
oxygen and hydrogen which are acidic in nature. 
These are called oxy acids; examples are HaBO*, 
H 2 CO 8 , HNO 3 , H 2 SQ 4 , and HC1O 3 . Some transition 
metals in their higher oxidation states also form 
hydroxides which are oxyacids e.g., H 2 CrO 4 and 
H 2 ]VInO 4 . In many cases there may be a series of 
oxyacids, each one containing the same nonmetal 
in a different oxidation state e.g., HC1O, HC1Q 2 
HClOs, HC1O 4 . In order to distinguish between 
these acids and their respective salts, the following 
system of suffixes and prefixes has been devised: 




Rule 6. In some cases there may be several 

state of Cl Acid 


oxyacids of an element in the same oxidation state. 

+ 1 HC10 
+3 HC10 2 

hypocliloTous acid 
chlorous acid 


The common acid of such a series is given the prefix 

+5 HClOs 

chloric acid 

ortho. The acid which contains one less water 

+7 HC10 4 

perchloric acid 

molecule in its formula is called the meta acid. 



The acid whose formula may be derived by re- 

+1 NaCIO 

sodium Aypochlonta 

moving one molecule of water from two molecules 

+3 NaC10 a 
+5 NaCIO, 

sodium chlorite 
sodium chlorate 

of the ortho acid usually has the prefix pyro. When 

+7 NaC10 4 

sodium perchlorate 

no prefix appears, the ortho, or common, acid is 

^^ t t 

f* * t Ij. *j.l 

assumed to be the one intended. 

Rule 4. Oxyacids with suffix ic form salts with 
suffix ate. Oxyacids with suffix ous form salts with 
suffix ite. Oxyacids with prefix hypo and suffix ous 
form salts with prefix hypo 1 and suffix ite. Oxy- 
acids with prefix per and suffix ic form salts with 
prefix per and suffix ate. 

Rule 5. In a series of oxyacids, the ic acid is the 
one in which the nonmetal has the oxidation num- 
ber corresponding to its group number in the 
periodic table, with the exception of the halogens, 
whose ic acids have the oxidation state +5. 

Examples: HsP0 4 orJAophosphoric H 8 BO 8 orthoboric 
HP0 8 metaphosphoric HB0 2 metaboric 

H 2 S0 4 

H 2 SiC>3 metasilicic 


Examples of ic acids are: 

Group III IV V 

II 2 CO, HN0 8 
H 4 Si0 4 H 8 P0 4 
H 8 As0 4 



H 2 S0 4 HC10, 
H 2 Se0 4 HBrO, 
HeTeO, HIO, 

Some ic acids formed by metals are H 2 SnOs, H 2 ZnOj, HVOi, 
H,CrO 4l H 2 Mo0 4) and H 2 Mn0 4 . 

Rule 7. In forming salts of polyprotic acids 
such as H 2 S0 4 or H 8 P0 4 , it is possible to replace 
one or more of the hydrogen atoms by metal ions. 
To differentiate between these salts, several sys- 
tems of nomenclature are in use; in the illustra* 
tions below the first name given is usually pre- 

NaH 2 P0 4 monosodium phosphate, primary sodium phosphate 
Na 2 HP0 4 disodium phosphate, secondary sodium phosphate 
Na*P0 4 trisodium phosphate, tertiary sodium phosphate 
NaHS0 4 sodium acid sulfate, sodium hydrogen sulfate, 

sodium bisulfate 
sodium sulfate, normal sodium sulfate 

Optional Drill on Nomenclature 

To the student: When you answer the following, write both 
the formula of a substance and its corresponding name side 
by side, in a simple outline or chart form. This will be an aid 
to your understanding and memory. 

1. Name the following binary compounds: C1 3 0, 
Hg 2 0, HgO, SnCl 2 , SnCl 4 , FeS, Fe 2 S 8 , N 2 4 , SiF 4 , 
PC1 8 , PCI,, Mn0 2 . 

2. Write the formulas of the following: cuprous 
oxide, nitrogen trichloride, mercuric sulfide, carbon 
monoxide, silicon dioxide, ferrous hydroxide, phos- 
phorus pentoxide, chlorine heptoxide, arsenious 
oxide, mercuric chloride. 

3. In the first of four vertical columns copy the 
following formulas of acids: HF, H 2 C0 8 , HN0 2 , 

1 The prefix hypo is from the Greek meaning under or less than, 
and the prefix per is Greek for above or more than. 

HNOa, H 2 S0 3 , H 2 S0 4 , HBrO, HBrO a , HBr0 3 , 
HBr0 4 . In the second column write the name of 
each acid, paying attention to the correct prefixes 
and suffixes. In the third and fourth columns write, 
respectively, the formula of the corresponding 
sodium salt, and its correct name. 

4. Write the names of the following ortho, meta, 
or pyro acids and salts: H 8 As0 8 , HsAs0 4 , HAs0 2 , 
HAs0 8 , Pb 8 (As0 4 ) 2 , Na 8 As0 8 , KAs0 8 , Ca(As0 2 ) 2 . 

5. Write the formulas of the following: ferrous 
sulfate, ferric sulfate, ferric phosphate, ferrous 
phosphate, potassium sulfite, potassium hydrogen 
sulfite, magnesium hydrogen sulfite. 

6. Name these: Na 2 S, Ba(HS) 2 , BaS0 8 , AgHS0 4 , 
KH 2 P0 4 , K 2 HP0 4 , Mg 3 P 8 7 , Pb(OH)Cl, NalO, 
KBrO,, KC1Q 2 , Ba(C10) 2 . 

REPORTs Exp. 18 

The Preparation and Properties of Some 
Oxygen Compounds of the Halogens 



Locker Number- 

Preliminary Exercise: 

Fill in the following chart with the names and formulas of the oxygen acids of chlorine, bromine, and iodine. Do 
the same for the anions corresponding to these acids. 






1. The Preparation of Chlorine, Potassium Chlorate, Sodium Hypochlorite. 

Equation for the reaction of MnO 2 with 6 F HC1: 

Equation for reaction of C1 2 and cold NaOH solution L. 

Auto-oxidation-reduction of hypochlorite ion in hot KOH solution :_ 

2. Crystallization of KC1O 3 . 

Compare the solubility of KC1 and KC1O 8 in water at C. and at 100 C._ 

On the basis of the above data explain the reasons for the steps involved in the crystallization and recrystalliza- 
tion of KC1O 3 . 

What is main commercial use of NaOCl?_ 

3. Properties of the Hypochlorite Ion. 

Result of bleaching test on litmus paper L, 


How is NaOCl made on a commercial scale?. 

Balanced equation for reaction of basic solution of OC1~ (few drops) on 0.1 F KI solution- 

Equation for reaction of excess OC1~~ on I 2 produced above: 

Equation for reaction of OC1~ in acid solution on 0.1 F KBr solution: 

Explain the basis of the test by which 1 iboth I~ and Br~ may be determined by the use of the OC1" 

Properties of Potassium Chlorate, 

Equation for reaction which takes place if KClOa is kept just at its melting point: 

Equation for decomposition of KC1O 3 in presence of MnO 2 :_ 

How do you account for the decomposition taking place at a lower temperature in the presence of 

Equation for reaction of KC1O 3 and 12 F HC1:. 

Equation for reaction of product of previous reaction with starch-iodide paper: 

Reaction of solution of KCIO 3 on 0.1 F KI: 

Record the relative oxidation-reduction potentials for the various couples involved in this experiment: 

How do you account for the difference in the results with ClOa" and CIO"" as oxidizing agents? 

How do the electronegativities of the halogens affect the relative stabilities of the various oxidation levels of 
these elements? 


The Production of an Electric Current 

By Means of Oxidation -Reduction Reactions. 


Col/ege Chemistry, Chapter 12 

Review of Fundamental Concepts 

The Nature of Electrical Cells 

We have observed that whenever an oxidation- 
reduction reaction occurs, there is a transfer of 
electrons from the substance oxidized to the sub- 
stance reduced. Thus, when zinc is oxidized by 
cupric ion, the zinc atom loses two electrons and 
the cupric ion gains two electrons. We may express 
this as two separate half-reactions: 

Zn > Zn++ + 2e~ (oxidation) 

Cu++ + 2 e~ > Cu (reduction) 

The sum of these two half-reactions of course gives 
the total equation for the process 

Zn + Cu++ + Zn++ + Cu. 

This total equation does not contain any free elec- 
trons for, obviously, the electrons lost by the zinc 
are all gained by the cupric ion. An electrical cell, 
or battery, is simply a convenient mechanical de- 
vice for carrying out an oxidation-reduction re- 
action in such a way as to transfer the electrons 
through a wire rather than by actual contact of 
the oxidizing agent with the reducing agent. The 
chemical reactions taking place at the separate 
poles of the battery are simply the half-reactions 
above described. The oxidation half-reaction takes 
place at the negative pole, since electrons are lib- 
erated there, and the reduction half-reaction takes 
place at the positive pole. See Figure 19-1. 

Cell Voltage 

The volt is the unit of electrical potential, or 
driving force. It furnishes a measure of the work 
done when a unit electrical charge is transferred 
from one substance to another. The voltage of a cell, 
sometimes called its electromotive force, is thus a 
quantitative value expressing the tendency of the 
chemical reaction occurring in the cell to take place. 
This voltage, of course, depends on the strength of 
the oxidizing and reducing agent used. In other 
words, if the oxidizing agent has a very great 
affinity for electrons as compared to the tendency 
of the reducing agent to hold these electrons, the 
electrical potential, or voltage, will be corrrespond- 
ingly large. 


not shown 

Fro. 19-1. A simple electric cell which transforms the 
energy liberated by a chemical reaction into electrical energy. 
The electrical current in the solution consists of sulfate ions 
(not shown) moving from right to left, as well as of the positive 
zinc and cupric ions moving to the right as shown. 

Standard Electrode Potentials 

The total voltage of the cell is a combination of 
the separate voltages due to the oxidation half- 
reaction and the reduction half-reaction involved. 
Since we cannot measure the voltage due to a 
single couple, it is customary to calculate the volt- 
age of all oxidation-reduction couples in terms of 
the total voltage which would be given by a cell 
in which the couple under consideration is com- 
bined with the H 2 - H+ couple. The H 2 - H+ couple 
is arbitrarily assigned a potential of zero, so that 
the total voltage is then ascribed to the other 




couple. For example, if a cell is measured involving 
the reaction of zinc in an acid, we have the half- 
reactions, with the potentials for each, as follows: 


Zn++ + 2 e" 


+ 0.763 volts 

Zn + 2 H+ > Zn++ + H 2 SE = + 0.703 volts 

This value, + 0.762, is called the standard electrode 
potential 1 for the Zn Zn++ couple. See Figure 
19-2. In all such data, the values refer to the volt- 


Escaping H* 

Glass shield 



Fio. 19-2. Hydrogen gas, adsorbed on the platinum elec- 
trode, acts as a hydrogen electrode. When this is immersed in 
1 M H+, and coupled with zinc in 1 M Zn++, the meter reads 
0.762 volts. 

ages obtained when all substances in solution are 
at a concentration of one molar, all gases at one 
atmosphere pressure, and the temperature at a 
fixed, convenient value, usually 25 C. (See Table 
XIII, Appendix II.) 

To obtain the voltage for any given cell, we cal- 
culate the algebraic difference of the two oxidation* 

1 There it a lack of uniformity among authors regarding the sign 
of the potential. We shall follow the convention of calling the potential 
of couples of the active metals, that is, those above the Hi H+ 
couple, positive (+) and those below it negative (). 

reduction couples concerned. For example, we may 
make a battery utilizing the strong reducing agent 
zinc, reacting with the strong oxidizing agent chlo- 
rine, according to the reaction 

Zn + C1 2 > Zn++ + 2 C1-. 

A zinc rod is placed in 1 F zinc chloride, and a 
platinum electrode in 1 F potassium chloride satu- 
rated with chlorine gas. A salt bridge connects the 
solutions. The platinum electrode is inactive, and 
serves merely to conduct electrons from the solu- 
tion as chlorine is reduced to chloride ion. The two 
couples, with the corresponding half-reactions and 
voltages as given in Table XIII, are: 

Zn-Zn++, i.e. Zn 
Cl- - C1 2 , i.e. 2 C 

Zn++ + 2 e~ 
- C1 2 + 2 e~ 

E=+0.7G3 volts 
a -1.3595 

(subtracting) Zn-f C1 2 *-Zn++ + 2 Cl~, SE = 2 . 122 volts 

Since the standard electrode potentials are always 
tabulated in the direction of loss of electrons, i.e., 
oxidation, it was necessary to subtract the second 
equation from the first in order to obtain the cor- 
rect cell reaction. (We could have obtained the 
same result by reversing the second equation and 
also the sign of its voltage, thus, Clz + 2 e~ > 
2C1-, E = +1.3595, and adding.) 

Electrode Potentials and the Principle of 
Le Chatelier 

We have observed that the tendency for a re- 
action to take place is measured by the voltage 
created when the reaction takes place in an electric 
cell. Thus, the electromotive force of 2.122 volts 
created by the cell in the preceding paragraph, for 
the reaction 

is indicative of the behavior of quite a strong re- 
ducing agent, Zn, with a strong oxidizing agent, 
Cl2. There is also some tendency for the reverse re- 
action to occur. However, zinc ion (Zn+ + ) is quite 
a weak oxidizing agent, and chloride ion (Cl~) is 
a weak reducing agent; hence this reverse tendency, 
as indicated by the short reverse arrow in the 
equation, is not very great. A reaction, in which the 
forward and the reverse processes have attained a 
balance, and are taking place at equal but oppos- 
ing rates, is said to be in chemical equilibrium. 

According to the principle expressed by Le 
Chatelier's Theorem, and in accord with observed 
fact, an increase in the concentration (or pressure) 







FIG. 19-3. Electrolysis and counter electromotive force. The products of electrolysis in (a) will create an electric cell which 

operates in the reverse direction, as in (b). 

of chlorine gas (Cla) will increase the rate of re- 
action with zinc, and consequently will increase the 
voltage of the cell. Conversely, an increase in the 
concentration of zinc ion or of chloride ion will 
favor the reverse process and therefore decrease 
the voltage, 1 

Again, in the forward reaction of zinc with chlo- 
rine, heat is evolved. Therefore, if the cell is oper- 
ated at a higher temperature, say 50 C, the voltage 
will be less, since this higher temperature will re- 
press the evolution of heat, and the tendency for 
the forward reaction will be decreased. 

1 While the potential values in Table XIII are all calculated for 
1 M concentration, or 1 atmosphere pressure in the case of gases, we 
may calculate the corresponding potential, , for a given reaction, 
at other conditions from the equation 

where E is the standard electrode potential, n is the number of 
electrons gained by the oxidizing agent in the reaction equation, and 
Q is the product of the concentrations (molarity of solutes, gas pres- 
sure in atmosphere) of the products, divided by the product of the 
concentrations of the reacting substances. 

Example: In the reaction tibove we found that for 

-KWlatm) : 

lZu++(lM) -f 2 Cl- 

E-2.12 volt. 

To calculate the corresponding voltage, if the Cl a were at 4 atm 
pressure, and the Zn 4 " 1 " were 0.01 M, we would have 

E - E - 

- 2.12 + 0.08 - 2.20 volt. 


The electrolysis of a solution involves just the 
reverse process to that discussed thus far in this 
experiment. In this case, the electric current itself 
constitutes the oxidizing and the reducing agent. 
The negative electrode in the solution (the cath- 
ode), with its excess of electrons, is the reducing 
agent. The positive electrode (the anode) is, cor- 
respondingly, the oxidizing agent. Note, in Figure 
19-3a, that at the negative electrode, the excess of 
electrons results in the electrode reaction 

2 0~ + Cu++ >- Cu (reduction) 

At the positive electrode, the deficiency of elec- 
trons results in the removal of electrons from the 
chloride ions 

2 Cl- > C1 2 + 2 e~ (oxidation) 

Study the electrode reactions in the commercial 
processes for the manufacture of sodium, sodium 
hydroxide, chlorine, aluminum, etc., as described 
in your text, College Chemistry Chapter 14. 

In any process of electrolysis, the substances pro- 
duced at the electrodes, together with their cor- 
responding ions, constitute oxidation-reduction 
couples. These, in turn, act as an electric cell to 
set up a counter electromotive force. (See Fig. 
19-3b.) It is necessary, therefore, in order for any 



electrolysis to occur, to apply a greater voltage 
than this internal counter electromotive force. 1 

Faraday's Law 

The amount of oxidation and reduction which 
takes place of course depends on the amount of 
electricity passed through the solution. The unit 
quantity of electricity is the coulomb. The unit of 
electric current is the ampere. There is a current 
of one ampere when one coulomb per second passes 
a given point in the circuit. That is: 


amperes ; 


The relationship between the quantity of elec- 
tricity and the amount of chemical action produced 
by it, called Faraday's Law, states that a flow of 
96,500 coulombs (one Faraday) of electricity will 
liberate one chemical equivalent of a substance. This 
is the amount of substance corresponding to the 
gain or loss of one "equivalent of electrons." Since 
there are 0.6023 X 10 24 atoms (Avogadro's Num- 
ber) in one gram-atom of an element, and since 
one electron is liberated when a silver atom be- 
comes a silver ion (Ag+), two electrons when a 
copper atom becomes a cupric ion (Cu++), and 
so forth, we may say that one Faraday of electricity 
is equivalent to the transfer of 0.6023 X 10 24 elec- 

Electrolytic Separation of Substances 

In any process of electrolysis, the negative elec- 

trode (acting as a reducing agent) will react most 
easily and completely with the strongest oxidizing 
agent in contact with it. Thus, in a mixture of gold, 
silver, and copper salts, since auric ion is a stronger 
oxidizing agent than silver ion, and this in turn is 
stronger than cupric ion, we can plate out first 
gold, then silver, and finally copper, as the voltage 
is successively increased. As long as the voltage is 
kept below the reduction or decomposition poten- 
tial for the remaining ions, only the more active 
one can plate out. 2 

Likewise, at the positive electrode (acting as an 
oxidizing agent), the most active reducing agent 
present will be liberated first. Thus, in a mixture 
of the halide ions, since iodide ion is the strongest 
reducing agent, then bromide ion, and finally 
chloride ion, the free halogen elements would be 
liberated, on electrolysis, in the same order first 
iodine, next bromine, and then chlorine, as the 
voltage is successively increased. If the voltage is 
only slightly more than enough to liberate iodine, 
no bromine or chlorine will be produced. 

In the commercial preparation of chlorine, it is 
necessary to use a concentrated chloride ion solu- 
tion, because at low concentrations the chloride ion 
is a weaker reducing agent than the water present, 
in which case oxygen gas from the water, rather 
than chlorine gas, is liberated. At intermediate 
concentrations, or if the current density is too high 
to allow time for the chloride ion to migrate to the 
electrode, both oxygen and chlorine will be liber- 

Experimental Procedure 

Special supplies: voltmeter (1000 ohms/ volt) for class use; 
two dry cells connected in series, with 8-inch leads of bare 
copper wire to use as electrodes; 10-cm U-tube; electrode 
strips of Cu, Zn, Pb, and Sn, about 1 X 10 cm; very narrow 
strips or wire (1 mm X 8 cm) of Zn, Sn, and Cu, and also ft 
few nails, for paragraph 5. 

Chemicals: 1 F NH 4 NO,, 0.5 F CuS0 4 , 0.5 F Pb(N0 8 ) 2 , 
0.1 F KI, 2 F Na 2 S, 0.5 F SnCl 2 , 0.5 F ZnSQ 4 . 

1. A Simple Daniell Cell. Place a strip of sand- 
papered copper inetal in a 100 or 150-ml beaker 

1 This discussion has not taken into account an empirical factor 
called the "overvoltage." This varies with the electrolysis reaction, 
the electrodes, and the conditions of electrolysis. It is frequently a 
polarization effect. To obtain any effective electrolysis, the applied 
voltage must be greater than the counter electromotive force plus 
the overvoltage, 

which contains about 35 ml of 0.5 F CuS0 4 solu- 
tion. (Do not waste solution by using an excessive 
amount.) The metal electrode may be bent so as 
to lie across the bottom of the beaker and should 
extend above the top of the beaker. Prepare a 
similar "half-cell" with a strip of zinc metal in a 
0.5 F ZnSO 4 solution. The solution should be at 
the same height in both beakers to avoid any 
siphoning action. Connect the two beakers by a 
"salt bridge" consisting of an inverted U-tube 
which is filled with zinc sulfate solution and which 

2 If a voltage sufficiently high to plate out copper is applied initially, 
and if the current density is too high to allow time for the more active 
ions to migrate to the electrode, those of lesser* activity, such as Ag 4 
and Cu ++ , will also plate out with the gold. 



has the ends plugged with cotton wet with the 
zinc sulfate solution. The voltage produced by this 
cell may then be measured by connecting the metal 
strips by means of copper wire to a high resistance 
voltmeter. 1 What constitutes the electric current in 
the wire? In the solution? Complete the diagram 
of this cell on the report sheet, labeling it as di- 
rected to show the various features and character- 
istics of the cell. Keep the cell for parts 2 and 3. 

2. Standard Electrode Potentials. On the re- 
port sheet, rewrite the oxidation-reduction poten- 
tial series for the couples studied in the last experi- 
ment and, at the right of each couple, list in another 
column its "standard electrode potential" as ob- 
tained from Table XIII, App. II, or from any 
handbook, text, or other reference source. Now cal- 
culate the potential which your Daniell cell should 
have, by algebraically subtracting (why do you 
subtract?) the voltage for the Cu Cu++ couple 
from that for the Zn Zn + + couple. 2 

For the experimental part of this section, set up 
cells similar to the Daniell cell, except use the 
Pb Pb++ couple. First connect it with the Zn 
Zn+ + couple, and then with the Cu Cu+ + couple, 
reading the voltages and noting the polarity in 
each case. For these cells, use a U-tube salt bridge 
of 0.5 F NH 4 NO 3 , to avoid forming insoluble salts. 
(Cells using other couples, such as the Sn 811++ 
couple in combination with those already studied, 
may be tried, if you have time. The Ag -Ag+ 
couple, which is too expensive for class use, may be 
demonstrated by the instructor.) Compare the 
voltage obtained in each case with the calculated 
values from the standard electrode potentials. 

1 An ordinary voltmeter which has about 250 ohm/volt will give 
a low reading because of the high internal resistance of cells con- 
structed as suggested here. Voltmeters with a resistance of about 
1000 ohm/volt, as the Weston Model 301, are quite satisfactory. 
These (one for each 10-15 students) may be located conveniently at 
a wall shelf for all to use. 

A demonstration cell which has a low internal resistance and, 
therefore, gives a more accurate voltage measurement, may be made 
by placing the zinc strip and zinc sulfate solution in an un glazed 
porcelain cup which is supported in the cupric sulfate solution. Ionic 
contact thus is made through the porous porcelain, so that a salt 
bridge is unnecessary. This type of cell, using very small unglazed 
porcelain cups, may be used by the entire class, if desired. 

* In these cells, we have used 0.5 F solutions, rather than 1 F on 
which the standard electrode potentials are based. This does not 
affect the equilibrium potential in these cases, since the concentra- 
tions of both product and reactant have been changed alike. See 
footnote 1, P. 139, and paragraph 3, this page. 

3. The Effect of Concentration. In order to 
see how changes in concentration affect the volt- 
age, again set up the Daniell cell (Zn Zn++ with 
Cu Cu++). After reading the voltage, add an ex- 
cess of sodium sulfide (about 10 ml of 2 F Na a S) 
to the Cu++ half-cell, at the same time diluting the 
Zn++ solution with an equal volume of water. Mix, 
and read the voltage. If there is no material change, 
add a little more Na 2 S. Explain the results. Will 
the presence of solid CuS affect the voltage? 

4. The Electric Current as an Oxidizing and 
Reducing Agent. Dip the ends of the copper wires 
leading from two or more dry cells connected in 
series, into a small beaker containing 10 ml of 
0.1 F KI solution and three drops of phenolphthal- 
ein. Explain the formation of the yellow color at 
one of the electrodes. Which one? Has this elec- 
trode acted as an oxidizing or a reducing agent? 
Write the equation for this half -reaction. The red 
color at the other electrode indicates the formation 
of what substance? What gas is being liberated? 
The reason for this is as follows. At this electrode 
(which one?), the most easily reduced substance 
will, of course, be the one to react. This is hydrogen 
ion from the water, not potassium ion, so instead 
of liberating potassium metal, the half -reaction 
here is 

2 e-+ 2 H 2 O 

2 OH~. 

As shown in the discussion, the products of the 
electrolysis will act as an electric cell, generating a 
counter electromotive force which tends to work 
against the applied voltage. From Table XIII, we 
see that the standard electrode potential for the 
H 2 HzCKlO- 7 M 11+) couple, corresponding to the 
reverse of the half-reaction above, is +0.414 volts. 
From this, subtract algebraically the standard po- 
tential for the I- 1 2 couple, to get the minimum 
external voltage necessary to apply to cause any 
oxidation-reduction for this solution. Would a 
single Daniell cell be satisfactory to electrolyze a 
potassium iodide solution? (The student should re- 
member that standard electrode potentials are 
based on solutions of definite concentration, one 
molar unless otherwise stated. Since these condi- 
tions are not maintained during the electrolysis, 
particularly with respect to the hydroxide ion 
around the hydrogen electrode, the calculated volt- 
ages are only a rough approximation.) 


5, The Corrosion of Metals. 1 This phenom- of these a small narrow strip of tin, around another 

enon is closely related to oxidation-reduction and a copper wire, and around the third a small, nar- 

its application in electric cells. Select three bright row strip of zinc. Place the nails in separate test 

iron nails or other pieces of iron. Wrap around one tubes and partially cover them with water. Let 

be set up advantageously as a demonstration in the them Stand Over night or longer. Explain the re- 

laboratory for all to observe. suits in terms of the action of an electric cell. 

Drill on Products of Electrolysis and on Cells 

Note: Here is some supplementary drill material for you to work on after completion of your regular Report Sheets. This 
need not be handed in unless called for. 

I, In each of the following cases, indicate, in the spaces provided at the right, all the substances (ions or mole- 
cules) which will be liberated,/^ at each electrode when electrolysis takes place. Assume inert electrodes, such as 
carbon rods, and a low current density. 

At the Positive At the Negative 

Electrode Electrode 

1. Concentrated CuCl 2 solution . . . . 

2. Fused NaCl 

3. Concentrated NaCl solution 

4. Dilute NaCl or Na 2 SQ 4 solution . . . . 

5. Dilute H 2 S0 4 solution 

6. A solution of CuS0 4 and NiS0 4 . . . . 

7. A concentrated solution of FcBr 2 and SnCU 

II. In the commercial preparation of aluminum, purified A1 2 3 , obtained from bauxite, is dissolved in molte:i 
cryolite, NasAlFe, and the mixture is elcctrolyzed with carbon electrodes. List all the ions present in the molten 
mixture, write equations for the electrode reactions, and account for the fact that the other ions present are not 
liberated as the free elements under these conditions. 

III. Make simple diagrammatic sketches of the lead storage battery and of the dry cell (see your text for any 
needed information). Show all essential parts electrolytes (a "dry cell" isn't dry) and electrodes, with the charge 
induced on each. Write equations for the half -reactions at each electrode, with the voltage for each (from Table 
XIII in Appendix II), and combine these to get the total cell reaction and total voltage. Compare with your com- 
mon knowledge of this and with the value given in your text. (Use another paper for these sketches.) 

REPORT: Exp. 19 

The Production of on Electric 
Current by Means of Oxidation- 
Reduction Reactions 

1. A Simple Daniell Cell. Complete the diagram 
at the right by writing in the formulas for the substances 
present in various parts of the cell. Indicate the charge 
at each electrode by (+) and ( ) signs. 

What constitutes an electric current in a wire? 

Indicate the direction of these particles by an arrow 
placed on the wire. 

What constitutes an electric current in a solution? 

Indicate the direction of the movement of these 
particles by placing their formulas on the proper arrow 
below the sketch. 

The half-reaction taking place at the negative electrode is 

The half-reaction taking place at the positive electrode is 

The total cell reaction is 


Locker Number- 

2. Standard Electrode Potentials. In the space at the right, rewrite tins 
potential series for the couples studied in Experiment 17, page 123 and insert the 
standard electrode potential for each, as obtained from Table XIII, App. II. 

Calculate the potential for your Daniell cell, as directed: 

What voltage did you obtain experimentally for the Daniell cell? 1 

1 The values in Table XIII are obtained by measurements in a reversible manner, with very small 
currents, using a potentiometer, so that the internal resistance of the cells does not tend to decrease the 
voltage reading, as is the case when a voltmeter is used. 


Potential Series 



In the space below, write the equation for the cell reaction for each cell that you tried. Indicate which metal is 
the positive and which the negative electrode, and give the experimental and the calculated voltages in each case. 

Cell Reaction 





3. The Effect of Concentration. Write the net ionic equation for the reaction taking place when Na2S is 
added to the copper sulfate solution: 

Explain the effect of this addition of Na2S on the cell voltage : 

How would you adjust the concentrations of the Cu + + and of the Zn + + in the Daniell cell in order to get the 
maximum voltage possible for this cell? 

4. The Electric Current as an Oxidizing and Reducing Agent. At the right sketch a diagram for the 
electrolysis of a potassium iodide solution, indicating the 
cathode, the anode, the direction the various ions move in 
the solution, and the products formed at each electrode. 
(Make a neat sketch.) 


The yellow color at one electrode is due to the formation 

This appeared at which electrode? 


Report on Exp. 19. Sheet 2 Name- 

this electrode the oxidizing agent or the reducing agent?_ 

The equation for the half-reaction taking place at this electrode is: 

The equation for the half-reaction taking place at the other electrode, which liberates Ha gas (see experiment 
directions), is: 

The net ionic equation for the electrolysis of a potassium iodide solution is therefore: 

Calculate the minimum external voltage needed to cause oxidation-reduction of a 1 F KI solution: 


Would a single Daniell cell be satisfactory to bring about this oxidation-reduction? Explain. (See experimen 

5. The Corrosion of Metals* The observed results on the rusting of iron in contact with other metals is in 
dicated below: 

Iron with copper 

Iron with zinc 

Iron with tin 

Explanation of these results : 

Why does a "tin" can (iron plated with tin) rust more easily than a galvanized pipe (iron plated with zinc] 
when exposed to weathering? 

Why do we use cans of tin plate, rather than galvanized iron, for preserving fruit and other food products? 

Application of Principles 

1. At the right make a sketch for a cell using the Br~ 
Br 2 and the Zn Zn++ couples. Indicate the behavior of all 
parts of the cell, as you did for the Daniell cell. 

2. Write the equations for the electrode reactions, and 
for the total cell reaction: 

3. Calculate the potential to be expected if all ions are at 1 M concentration, and the solution is saturated with 
Br 2 . 

. volts 

4. Heat is evolved when zinc reacts with bromine. Would the potential for this cell be greater or smaller if the 
cell were heated? Explain. 

5. a) In commercial silver plating the article to be plated, e.g. a teaspoon, is connected to which electrode of 
the battery? 

b) Is this electrode the oxidizing or the reducing agent? 

c) To plate a surface of silver 0.1 mm thick on a knife (estimated 75 cm 2 surface) would require about 7.00 grams 
of silver. How long would it take to plate out this much silver, using a 0.400 ampere current? 


d) How long would it take to plate out 4.00 grams of gold from an auric chloride solution on this same knife, using 
0.400 ampere? 



The Concentration of Solutions. 

The Fractional Crystallization of Salt! 


Co//ege Chemistry, Chapter IB 

Review of Fundamental Concepts 

Units of Concentration as Used in Chemistry 

Since the concentration of the solutions we use 
is an important factor to be considered in this and 
in many succeeding experiments, we shall intro- 
duce a study of this concept at this time. It is, 
first of all, important to distinguish between units 
of quantity and units of concentration. The term 
concentration refers to the amount of dissolved 
substance in a given amount or volume of solution, 
that is 

amount of solute 

concentration = 


For example, if we dissolve 34.2 g of sugar 
(Ci 2 H 2 2On) in enough water to make 100 ml of 
solution, we will have quite a concentrated solu- 
tion which will taste distinctly sweet. But if we 
dissolve 34.2 g of sugar in enough water to make 
10 liters of solution, we will have used the same 
amount of sugar, but the concentration will be much 
less than before, and the solution will hardly taste 
sweet at all. 

In chemical practice, it simplifies calculations to 
express the concentration of solutions in terms of 
the fundamental chemical units of quantity, as 

Formality. A solution which contains one gram- 
formula weight of solute dissolved in water and 
diluted to a total volume of one liter (not added to 
one liter of water) is called a one formal solution, 
designated 1 F. One might dissolve a half formula 
weight in a half liter, or ten formula weights in ten 
liters, and likewise have a. IF solution. The defin- 
ing equation is 

__ .. formula weights of solute _ gfw 

Formality = * , or F^r^r r- 

liters of solution V(hters) 

Molarity. Likewise, a solution which contains 
one mole of substance per liter of solution is desig- 
nated a one molar (1 M ) solution. The defining 
equation is 

- . . moles of solute ,_ moles 

Molarity = , or If rrr^ -. 

* liters of solution V(htera) 

Note that the terms "formula weight*' and "for- 
mality" are inclusive terms, and include "mole' 1 
and "molarity," respectively. While these latter 
terms are often used in the same sense as formula 
weight and formality, we shall restrict their use to 
cases where a definite molecular or ionic species is 
represented by the formula used. Thus, we speak 
of a 1 F (not 1 M ) NaCl solution, since the formula 
NaCl does not represent a definite molecule which 
is present as such either in the solid crystal or in 
solution. Since the solution is regarded as com- 
pletely ionized, we may speak of a 1 F NaCl solu- 
tion as containing 1 M Na + and 1 M Cl"~. 

Several examples will help you to understand 
the meaning of the above defining equations. 

Example 1. What are the concentrations of the 
sugar solutions mentioned above, which contain 
34.2 g Ci2H 2 2O u in 100 nil and in 10 liters of solu- 
tion, respectively? 

Since the solution contains molecules of the com- 
position CiaH^Ou, we can designate this as molar- 
ity. Formality would be equally correct, but less 
specific. First we shall express 34.2 g as moles, 

342 g/mole 
From the defining equation, we write 

M - 

moles 0.1 mole 
V ~ 0.1 liter 

1 M f or the first solution, 


0.1 mole 
10 liters 

0.01 M for the second solution. 

Example 2. How many gram-formula weights are 
there, and what is the weight of HzSOi in grams, 
in 250 ml of 0.300 F H 2 SO 4 ? 

Transposing the defining equation F gf w/ V, we 

gfw _ F X V 0.300 ^ X 0.250 liter * 0.075 gfw H f SO 4 . 
And to express this quantity in grams, we write 
0.075 gfw X 98 -- 7.35 g j 



Example 3. What volume of 0.300 F H 2 SO 4 will 
be required to react with 4.00 g NaOH? 

4.00 g NaOH . <AA . XT _ 
7-7^-7-. - 0.100 gfw NaOH. 
40.0 g/gfw * 

By the equation for the reaction 

2 NaOH + H 2 SO< * Na 2 SO 4 + 2 H 2 O 

it is evident that 2 gfw NaOH will require 1 gfw 
H 2 SO 4 , therefore 0.1 gfw NaOH will require 0.050 
gfw H 2 SO 4 . From the defining equation F = gf w/ V, 
we have by transposing, 

- - 167 1 or 167 ml of - 300 F H ' so <- 

n /i 

r 0.300 gfw/1 

Some 0for ?7ntte of Concentration. Other ways in 
which the concentration of solutions may be desig- 
nated are the following. Percent by weight refers 
to parts of solute per one hundred parts of solution. 
Thus, a 5% solution is one which contains 5 g of 
solute to every 95 g of solvent. Sometimes, as in 
tables of solubility data in the reference literature, 
the concentration is expressed as grains of solute 
per 100 g of solvent. Again, in cases where the 
relative number of molecules of solute and solvent 
are fundamental to the theory, concentration is 
expressed as mole fraction, or as weight formal- 
ity or weight molarity. These two latter terms 
refer to the number of formula weights or moles, 
respectively, per 1000 g of solvent (rather than per 
liter of solution). Normality as a unit of concen- 
tration will be introduced in Experiment 27. 

Saturated Solutions 

A saturated solution is one in which an equi- 
librium exists between the rate at which mole- 
cules or ions in solution are depositing on the sur- 
face of the solid crystals and the rate at which 
molecules or ions are leaving the crystal surface 
and mixing with the solvent molecules in solution. 
At equilibrium these opposing rates are equal, so 
that the concentration of the solution remains con- 
stant. Any factor which disturbs this system, such 
as the evaporation of solvent or the cooling of the 
solution, may cause additional crystals to deposit. 

When soluble salts dissolve in water, they are 
present as separate ions in solution (Exp. 16). Any 
given positive ion, or cation, may combine with 
any given negative ion, or anion, to form a crystal- 
line salt whenever the concentration of ions in solu- 
tion is great enough to exceed the solubility of the 

salt in question. Thus from a solution containing 
sodium ion (Na+), potassium ion (K + ), chloride 
ion (Cl~~), and nitrate ion (NOs"), it is possible 
to form four crystalline salts, with the formulas 
NaCl, NaNO 3 , KC1, and KNO 3 . The particular salt 
which crystallizes out first will depend on the tem- 
perature and relative concentration of the several 
ions in the solution. The deposits of soluble salts 
which occur in nature, as those at Stassfurt, Ger- 



O 2O 40 6O 60 100 1 2O 

Temperature *C 

FIG. 20-1. The principles of fractional crystallization, as illus- 
trated by the solubility curves for two substances A and B. 

many, and those at Searles Lake in California, 
have been formed by such fractional crystalliza- 
tion from solution, as large bodies of salt-bearing 
water evaporated. The following discussion will 
help you to understand how such fractional crystal- 
lization may occur. 

The Theory of Fractional Crystallization 

Study the graph, Figure 20-1, which shows the 
saturated solution curves for two typical sub- 
stances. The solubility of A increases rapidly with 
increasing temperature, while that of B increases 
only slightly. Note the different situations in dif- 
ferent regions of the graph. In area (1), below both 
curves, the solution is unsaturated with respect to 
both A and B. In area (2) it is unsaturated with 
respect to A, saturated with respect to B, and con- 
tains excess solid B. What is the situation in area 
(3)? In area (4)? 

Now suppose we start from a point w, at which 



the solution will contain about 14 g each of A and 
B per 100 g of water, at the boiling point. Let us 
cool the solution. The point moves horizontally to 
the left. At n (about 50 C), the saturation point 
of A is reached and solid A begins to separate. 
When the solution has cooled to 20 C, approxi- 
mately 7 grams of A per 100 grams of water will 
have crystallized. 

Again start as before from point m but, instead 
of cooling the solution, evaporate it at the boiling 
point until half the water is removed. Now the 
point moves directly upward. At point o, solid B 
begins to separate, so that at point p, when half 
the water is evaporated, we have about 28 grams 
of A per 100 grams of water. We also have a sat- 
urated solution of B and about 8 grams of solid B. 
Now filter off the solid and cool the filtrate (mov- 
ing to the left from both p and o, on the diagram). 
A little additional solid B will at once begin to 
separate and when the solution is cooled to about 
80 C (point q), solid A will also separate. When 
the mixture is cooled to 10 C, we will have a large 
amount of A (about 24 grams per 100 grams of 
water) and only a small amount of B (about 4 
grams) crystallized out. We have, thus, achieved 
quite a separation. How could this mixture of A 
and B be dissolved and recrystallized so as to ob- 
tain pure A? 

Preliminary Exercise 

Before the laboratory period, make a neat and 
accurate graph of the solubility, in grams per 100 

grams of water, of each of the four salts given in 
Table 20-1 below, at temperatures from C to 
100 C. Arrange the temperature scale to extend 
well across the horizontal axis, with a suitable scale 

TABLE 20-1. 


(Solubility in grams of solute per 100 grams of water. Data 
from Scidell, Solubilities of Inorganic and Organic Compounds, 
Vol. 1, P. 747, 834, 1218, 1275. D. Van Nostrand Co., N. Y. 


10 C 

20 C 

30 C 


80 C 

100 C 









NaNO 3 
























of concentration on the vertical axis. Make all four 
curves on the same graph paper, and draw a smooth 
curve through the points for each substance. Label 
each curve with the formula for the salt. From a 
study of the theory of fractional crystallization as 
given above, and from your graph, decide what 
effect temperature and concentration will have in 
determining which particular salt will crystallize 
out first under various conditions. (Answer the four 
questions in the report sheet bearing on this ques- 

Experimental Procedure 

Special supplies: Buechner funnel and filter flask to fit, with 
pressure tubing connection. Nichroine wire and 2 squares of 
cobalt glass for flame tests. Ice. Thermometer. 

Chemicals: Chile saltpeter NaNO 8 , crude KC1, FeSO 4 , 0.1 F 
AgN0 8 . 

We shall utilize the same method for preparing 
potassium nitrate that is extensively used in chem- 
ical industry, namely by fractional crystallization 
from crude Chile saltpeter, NaNO 8 , and crude po- 
tassium chloride, KC1. The sodium chloride ob- 
tained as a by-product will also be purified. 

1. Preparation of Potassium Nitrate and 
Sodium Chloride. Calculate the weights of % of 
a grain-formula weight each of crude NaNO 8 and 
KC1. (Do not use the reagent grade chemicals on 
the reagent shelf.) Place these weights of the 
NaNO a and KC1, together with 140 ml of tap 

water, in a 400-ml beaker, and heat the mixture 
almost to boiling to dissolve the salts. Filter while 
hot with the aid of vacuum, using either technique 
as illustrated in Figure 20-2, to remove any in- 
soluble foreign matter. Transfer the hot filtrate to 
a 250-ml beaker and chill it, with stirring, to 5 C 
or lower, by immersing the beaker in an ice bath 
in some suitable container. Clean and rinse the 
funnel and filter flask. (Cool the fuftnel if it is still 
warm.) Quickly filter the cold KNO 3 crystals and 
press out all excess moisture with another filter 
paper laid over the crystals (Fig. 20-3). Set these 
crystals aside for later purification. 

By reference to your graph, estimate the weight 
of KNO 8 , which still remains dissolved in the fil- 
trate. (Remember that 140 ml of water was used.) 



A filter paper laid flat covers 


the holes in 
the bottom 
of the Buch* 
ner funnel. 

Ti[ter paper (12.5 cm) 

5 cm gauze 

The g 

the tip. 

Fold again ^. 
^-^ >s 

Fro. 20-2. Vacuum filtration technique: (a) using a Buechner funnel, (b) using an ordinary funnel with 

gauze folded in the filter paper to strengthen it. 


Lay a filter paper on 
top of the crystals. -* 

Press out excess 
moisture with the 
bottom of a small 
beaker while the pump 
is still running. 

Flo. 20-3. Pressing out the excess moisture from the crystals. 

To obtain a second crop of KNO 3 crystals, place 
the filtrate in a beaker and boil it down rapidly. 
What solid separates as the solution is concen- 
trated? As the boiling continues, the mixture be- 
gins to bump violently. This may be minimized as 
follows. Seal one end of a 20-cm glass tube. Leave 
the other end open with sharp edges. Use this, 
open end down, to stir the mixture continuously 
(Fig. 20-4). 

Continue the evaporation until the volume has 
been reduced to a half or a third of its original 
volume, as estimated by its height in the beaker, 
and until quite a large amount of salt has sepa- 
rated. Do not reduce the volume to less than a 
third of its original volume, however. Filter this 
mixture, while boiling hot, by vacuum filtration. 
Press out all the mother liquor you can and at 
once transfer the filtrate to a small beaker. Rinse 
the flask with a volume of water about one tenth 
of the volume of the filtrate (estimate it rather 
carefully), and add this to the filtrate. Why is this 
done? Save this crude NaCl for later purification, 
if time is available. 

Heat the filtrate to redissolve any salts which 


This end must >r 
be sealed. / 


'Small bubbles 
of steam rise 
from the sharp 
edges of the 
freshly cut tube. 

Fio. 20-4. The use of a dosed tube to prevent excessive bump- 
ing as sodium chloride is crystallized by evaporation. 

have separated and then chill the filtrate to 5 C or 
less by immersing the beaker in ice, as before, to 
obtain a second crop of KNOs crystals. Filter, and 
combine these crystals with the first batch. They 
may be weighed damp, and the dry weight esti- 
mated as about 90% of the damp weight. Or if you 
have time before recrystallizing, they may be 
spread out on a clean paper in your locker to air 

A colored flame 
indicates a 

Clean the 
wire by alter- 
nately dipping 
in concentrated 
HCI and heating 
to incandescence. 

dry. Record the weight of the crude KNOi you 
have prepared. 

2. Tests for Purity. Both of your crude salts 
will need to be purified by recrystallization. Be- 
fore you do this, learn the following qualitative 
tests and use them to check on the purity of your 
salts as you recrysallize them. 

Flame Teste for Sodium Ion and Potassium Ion. 
Obtain a nichrome wire and two pieces of cobalt 
glass. Clean the wire by alternately heating it in- 
tensely in the flame and dipping it in a little con- 
centrated HCI in a small test tube. Continue this 
treatment until the wire does not impart a definite 
color to the flame. Then moisten a very small 
pinch of the salt to be tested and touch the ni- 
chrome wire to it. (Be sure the wire touches noth- 
ing else after it has been cleaned.) Place the wire 
in the edge of a low Bunsen flame to heat it to in- 
candescence. A strong yellow flame indicates so- 
dium, while a fainter violet flame, which does not 
last very long, indicates potassium. With a mixture 
of sodium and potassium, the intense yellow flame 
completely masks the fainter violet flame. The 
sodium flame may be screened out by placing two 

Observe the color of 
the flame when the wire is 
heated in the edge of the 
Bunsen flame. 

look at 
V/the flame 
against a 
dark back- 

Dip the clean 
wire into the sub- 
stance which has 
been moistened 
with distilled 

Use two 
layers of 
cobalt glass 
to observe the 
pale violet pot* 
assium flame if 
sodium {s also 

FIG. 20-5. The flame test for sodium and potassium: (a) Cleaning the wire, (b) Observing the flame colorations. 



thicknesses of cobalt glass in front of the flame. 
The shorter wave-length potassium flame is trans- 
mitted, and may be seen through the glass; how- 
ever, it lasts only a short time (see Fig. 20-5). 

Test for Chloride Ion. 1 Dissolve a very small 
amount (0.01 g) of the salt to be tested in 2 ml of 
water, acidify with 2 drops of 6 F HNO 3 , and then 
add a drop of 0.1 F AgNO 3 . A milky precipitate of 
AgCl indicates the presence of chloride ion. 

Test for Nitrate Ion. 1 Dissolve 0.01 g of the salt 
to be tested in 2 ml of water and add this to 2 ml 
of saturated FeSCX solution. (Since FeSO 4 solution 
is easily oxidized, prepare your own solution by 
adding a small amount of the solid to a little water 
in a test tube and mixing.) Incline the test tube 
containing your test sample and FeSO* at quite an 
angle and very gently add a little concentrated 
H 2 SO 4 so that it runs down and underlies the other 
solution. A characteristic brown ring at the junc- 
ture of the two layers indicates the presence of a 

3. The Purification of Potassium Nitrate. 
Before recrystallizing, test a very small amount 
(half the size of a pea) for chloride ion. Saw thu 
testy as well as subsequent tests for Cl~, so they 
may be compared and any increase in purity thus 
observed. To recrystallize your KNO 3 , place the 
salt in a 150-ml beaker, add as many milliliters of 
distilled water as you have grams of dry KNO 3 
crystals, and warm the mixture to completely dis- 
solve the crystals. Cool the solution in an ice bath 
to 5 C or less, and filter the crystals while they 
are cold by vacuum filtration. Press down the 

1 Both the chloride and the nitrate ion tests are discussed more 
fully in Experiment 39. See Figure 39-1 for an illustration of the 
brown ring test for nitrate ion. 

crystals and suck out all the moisture you can. 
Again test the purity of your KNO 8 crystals by 
the chloride test. It will probably be necessary for 
you to recrystallize your KNO 3 a third time, or 
possibly a fourth time, before you get it reasonably 
pure, so that it fails to give a significant chloride 

Test a small amount of the final salt also for 
nitrate ion, and by the flame test for potassium ion 
and for freedom from sodium ion. The sodium 
flame test is exceedingly sensitive, and responds to 
the slightest amount of impurity. 

Air-dry the crystals by spreading them out on a 
watch glass, or clean paper. Weigh the crystals and 
compute the percent of pure KNO 3 recovered from 
the crude salt. Hand in your crystals, labeled with 
your name, the formula of the salt, and a note as 
to the extent of any impurities found, as your 
laboratory instructor may direct. 

4. The Purification of Sodium Chloride. 
(Optional.) Test a small sample (size of half a pea) 
of your crude NaCl for K+ and for NO 3 ". The salt 
may be recrystallized by first dissolving it in the 
minimum amount of boiling water (use about 3 nil 
of H 2 O to each gram of salt), and then by boiling 
the solution rapidly to concentrate it to one third 
of its volume. The addition of concentrated HC1 
at this point (use a volume about one tenth that 
of the mixture) will cause some additional precipi- 
tation of NaCl, due to the increase in the Cl~" con- 
centration. Filter the crystallized salt by vacuum 
filtration. The salt may be dried by heating it, with 
stirring, in an evaporating dish. A second crop of 
crystals of lesser purity can be obtained by further 
concentration of the filtrate. Test the recrystallized 
samples for purity. 

REPORT: Exp. 20 p ate 

Fractional Crystallization Section*. 

Locker Number 

Preliminary Exercise 

The graph of salt solubilities you have made should be handed in with this report. Use this graph to help you 
answer the following: 

a. If a solution containing equal amounts of the four ions, Na+, K+, Cl~ t and NOj~, is evaporated at the boiling 
point, which salt will crystallize out first? Explain. 

b. If this hot mixture is now filtered, and the filtrate cooled with ice, which salt will crystallize out in largest amount? 

c. Will any of the same salt that crystallized out at the boiling point (question a above) separate out along with this 
second salt as the solution cools? Explain. 

d. What effect would the addition of a little water to the hot filtrate (in question b) have on the purity of the crystals 
which form when the solution is cooled? Explain. 

Approval by instructor of the graph, and of the answers to the questions above 

1* Preparation and Purification of Potassium Nitrate 

Weight of crude salts taken : NaNO 8 KC1 - 

My estimate of the weight of KNO 8 which still remains in the filtrate after the first 
crystallization of KNOg (based on graph, and volume of water used) g 

Weight of dry crude KNOs obtained (both batches) g 

Theoretical weight of KNOa, based on the weight of crude salts taken. Show calcu- 

Percent yield of crude KNOa 1 % 

1 In commercial saltpeter plants, it IB possible to obtain nearly a 100% yield of KNO^ because it b then practical to recirculate the filtrate* 
still containing some KNOi with the richer liquors, thus finally obtaining nearly all the KNO, in crystalline form. 


2. Purification of Potassium Nitrate 

After successive recrystallizations, the relative amounts of AgCl formed on testing for Cl~ was as follows: 

1st crystallization 2nd recrystallization 3rd recrystallization 

Tests of the final purified KNOg showed the following results: 
For Na+ For K+ For NO 8 - 

Why does successive recrystallization of a salt such as KNO 8 increase its purity? 

Weight of pure Percent of your crude KNO 8 

dry KNO 3 obtained g recovered as the pure salt % 

3. Purification of Sodium Chloride 

If you recrystallized your NaCl, comment on the relative freedom from K+ and from NO 8 "~ (1) of the crude 
salt, (2) of the first crop of crystals on recrystallization, and (3) of a second crop of crystals, if obtained. 

Why is the recrystallization less efficient as to yield than in the case of the recrystallization of KNO*? 


1. Calculate the gross profit one could make in the manufacture .of a ton of crude KNO 3 from Chile saltpeter 
(NaNOs) and muriate of potash (KC1), if a 100% theoretical yield could be obtained. (Use these commercial 
prices: KNO* $200 per ton, NaNO 3 $50 per ton, and KC1 $30 per ton. Neglect the by-product formed, NaCl.) 

2. Repeat the calculations of problem 1 on the basis of your actual percent yield of crude KNOs (i.e. before 
recrystallization). Hint: Note that the lower your percent yield, the larger the weights of NaNO 3 and KC1 are 
which you must use to obtain a ton of 

Report on Exp. 90, Sheet 2 Name, 

Exercises on the Units of Concentration 

In each of these cases, indicate clearly the method of solution, including the proper use of dimensions. 
1. Complete the items called for in each of the following: 

Formula Number 

Solution Weight of gfw Formality 

(a) 20.0 g NaOH in 250 ml solution 

(b) 30.3 g KNO 3 in 500 ml solution 

(c) 245 g H 2 SO 4 in 3000 ml solution 

2. What amount of the solute is present in each of the following solutions? 

(a) 500 ml of 4 F HC1 gfw 


(b) 750 ml of 0.2 F HCI gfw 


(c) 750 ml of 0.2 F CuHaaOii (sugar) gfw 


3. It is desired to measure out 5.0 g of each of the solutes in the following solutions. What volume of each solution 
should be used? 

(a) 3 F NaC 2 H 3 O 2 ml 

(b)0.1FCuSO ml 

(c) 3 F HNOa ml 


4. Given a solution of a substance, as indicated in the first column, with certain other data given in succeeding 
columns. From this data calculate the corresponding values to fill in the blank spaces. (Thus, for the first case 
given 9.1 g of HC1 in 500 ml of solution, we may calculate thus: we have 9.1 g/36.5 g per gfw = 0.25 gfw; since then 
are 0.25 gfw in 500 ml, the concentration is 0.25 gfw/0.5 1 = 0.5 F.) 





of gfw 

Weight of 
Solute (g) 








0.20 ' 













5. Concentrated nitric acid has a density of about 1.42 g/ml, and is about 16.0 F. Express this concentration in 
each of the following ways: 

(a) as percent composition 

(b) as g/100 g of solution 

(c) as g/100 g of solvent 

(d) as weight formality 

6. Concentrated hydrochloric acid contains about 37.0% HC1 and has a density of 1.184 g/ml. What is the for- 
mality of this solution? 

7. Concentrated sulfuric acid is about 18.0 F and contains 96.0% [2804. What is the density of this solution? 


The Chemistry of Sulfur and Its Compounds. 


College Chemistry, Chapter 14 

Review of Fundamental Concepts 

Sulfur compounds: 

+6 S0 3 , H 2 S0 4 , SO 4 ~ 
+4 S0 2 , H 2 SO 3 , SO 4 
o r ave -f2 S 2 O, 


-2 H 2 S, S 

Concentrated acid is a strong oxidizing agent. 

Active either as oxidizing or reducing agent. 

Thiosulfate ion. Decomposes to S and HjiSOj in acid solution. Oxidized to SiOe""" 

(tetrathionate ion) by free I*. 

Strong reducing agent, usually oxidized to S. 

In this experiment we shall study the physical 
and chemical properties of the element sulfur and 
its more important compounds. The chart given 
above summarizes the various oxidation states 
which are illustrated by sulfur compounds. 

As you perform the experiments outlined, keep 
constantly before you the structural formulas of 
the compounds involved by reference to the text- 
book. Recall that sulfur, a sixth group element, 

may be represented by the electron dot formula 

: S By gaining two electrons from metallic ele- 

ments, it forms sulfides in which the sulfide ion, 
S , has attained the stable configuration of an 
inert gas. The oxidation state is also -2 in the 
covalent compound H 2 S, where the sulfur atom 
shares electrons with two hydrogen atoms. 

By sharing its electrons with more negative ele- 
ments such as oxygen, sulfur attains positive oxida- 
tion states. In sulfur dioxide, SO 2 , four of the sulfur 
electrons are involved in forming covalent bonds 

with oxygen. Sulfur dioxide is the anhydride of the 
weak acid, H^SOa, which forms salts with bases in 
which the ions, HSO 8 ~ or SO 3 are present. 

In sulfur trioxide, all six of the sulfur electrons 
are involved in forming bonds with oxygen atoms. 
This oxide is the anhydride of sulfuric acid, H2SO4, 
which is one of the most important industrial 
inorganic chemicals. Salts of this acid contain the 
ions, HSC>4~~ or SO 4 , which are further illustra- 
tions of sulfur in the + 6 oxidation state. 

In elementary sulfur, the sulfur molecule at or- 
dinary temperatures contains eight atoms arranged 
in a puckered ring. These rings are oriented to form 
rhombic crystals at ordinary temperatures, but 
above 95.5 C., the transition temperature, these 
molecules rearrange themselves to form mono- 
clinic crystals. Above its melting point, sulfur un- 
dergoes some interesting changes in properties with 
increasing temperature which can be accounted for 
in terms of changes in the structure of the sulfur 

Experimental Procedure 

Special supplies: Thermometer, thistle tube. 

Chemicals: Crushed roll sulfur, iron filings, powdered mag- 
nesium, mossy zinc, copper turnings, KMnOi> NasSOs, 0.1 F 
solutions of NaCl, CaCl 2 , Zn(NO 8 ) 2 , Cd(NO 3 ) 2 , Cu(NO 3 ) 2 , 
SbCl 3 , Pb(N0 8 ) 2 , AgN0 8 , Hg 2 (N0 3 ) 2 , BaCl 2 , Ba(OH) 2 , 2 F 
Na 2 S, Bromine water, 1 F Na 2 SO 4 . 

1. Oxidation State Zero, Elementary Sulfur 

(a) Forms of solid sulfur. Recall the differences 
in appearance and properties of the two crystalline 
modifications of sulfur prepared in Experiment 2, 
Part 5, and record this data in the report sheet. 

(b) Liquid sulfur. Fill a small test tube about 
one-third full of crushed sulfur and heat it very 
slowly over a small flame until the sulfur just melts. 

Raise the temperature gradually and note the color 
and fluidity of the sulfur from time to time. Con- 
tinue the heating until the sulfur boils. Pour the 
molten sulfur into a beaker of cold water and ex- 
amine the product. Test its solubility in 3 ml of 
carbon disulfide. (Caution: Handle CSt at least 
five feet from any flame.) 

2. Oxidation State 2. Sulfides and Hydrogen 
Sulfide (Optional) 

(a) Preparation of a sulfide. Mix approximately 
3.5 g of iron filings with 2 g of crushed sulfur in a 
crucible supported on a triangle. Place a lid on the 




crucible and heat it with a Bunsen burner until the 
reaction begins. Remove the lid and burner oc- 
casionally to note whether the reaction continues 
with the evolution of heat. Continue the heating 
to burn off any excess sulfur and allow the crucible 



a 50 ml 

H 2 

4Om) HfcO^ 
FIG. 21-1. The preparation of sulfur dioxide. 

to cool. Write the equation for the reaction. Place 
a small piece of the compound into 3 ml of 6 F 
HC1 in a small test tube and cautiously note the 
odor of the gas evolved. 

(b) Hydrogen sulfde as a precipitating reagent. 
Many metallic sulfides are difficultly soluble in 
water or in acids, and advantage is taken of this 
fact in the separation of groups of metals in Quali- 
tative Analysis. (See Study Assign. 6.) Using a 

source of hydrogen sulfide available in your labora 
tory, (see footnote in Exp. 16, P. 112) saturate I 
ml of each of the following solutions with H 2 S gas 
0.1 F CuSO 4 , 0.1 F SbCla, 0.1 F Cd(N0 8 ) 2 , 0.1 1 
Zn(NO,) 2 , 0.1 F CaCl 2 and 0.1 F NaCl. Note ii 
which cases insoluble sulfides are formed. Recorc 
the formula and color of each of the precipitates 

(c) Hydrogen sulfide as a reducing agent. Fron 
its position in the chart of oxidation states, ii 
should be obvious that H 2 S can act only as a re 
ducing agent, in which case it may be oxidized t< 
free sulfur or to higher oxidation states depending 
on the strength of the oxidizing agent used. Satur 
ate each of the following solutions with H^S: 5 m 
each of warm 3 F HNO 3 , of 0.1 F I 2 , a freshly pre 
pared solution of H 2 SO 3 made by adding a fev 
crystals of Na-jSOs and a drop of 6 F H 2 SO 4 to 5 m 
of distilled water. In each case note the product 
of the reaction and write balanced equations foi 
the oxidation-reduction reactions. 

(d) Polysulfides. Add a small pinch of powderec 
sulfur to 5 ml of a 2 F Na2S solution. Warm th< 
mixture gently for a few minutes as you stir th< 
powdered sulfur to wet it. Account for the resuli 
obtained by drawing electron-dot formulas for th< 
new ion or ions formed. Acidify the solution witl 
6F HC1 and note the products formed. How is thi: 
reaction analogous to the decomposition of hydro 
gen peroxide? 

3. Oxidation State +4. Sulfur Dioxide, Sul- 
furous Acid, Sulfite Ion 

(a) Preparation of sulfur dioxide. In the metal 
lurgical industry, large quantities of sulfur dioxide 
are produced as a by-product during the roasting 
(oxidation) of metallic sulfides. It may be pre 
pared in the laboratory by burning sulfur (recal 
Exp. 8), by adding a strong, nonvolatile acid sucl 
as sulfuric to a solution containing sulfite ions, 01 
by the reduction of hot concentrated suifuric acic 
by certain metals. Set up the generator shown it 
Figure 21-1. Place about 40 ml of distilled watei 
in the 50-ml flask and dissolve about 15 g of Na^SO 
in 25 ml of hot water in the 250-ml flask. Add 5 m 
portions of 3 F H 2 SO 4 , as needed to maintain con- 
tinuous evolution of the gas in order to saturate 
the water in the small flask. Remove the deliver) 
tube from the flask and collect a 10-cm test tube 
of the gas by the displacement of air. Determine 



for yourself whether this should be done by the 
upward or downward displacement of air, by esti- 
mating its density from its molecular weight and 
comparing with that of air. Place your thumb over 
the mouth of the test tube and invert the tube in 
a 400-ml beaker of water. Note any change in the 
level of water in the tube as you swirl it gently 
under the water and estimate the solubility of SO* 
in water. 

(b) Chemical properties ofsulfurous add. Test the 
acidity of the saturated solution of SOa with a piece 
of neutral litmus paper. Place 5 ml of the solution 
in a small test tube, and add a small pinch of 
powdered magnesium. Note the result. 

Prepare a very dilute solution of potassium per- 
manganate by adding a minute crystal of KMnOi 
to 5 ml of distilled water, acidified with a drop of 
3 F H 2 SO 4 . Add some of the SO 2 solution, drop by 
drop, until a color change takes place. What hap- 
pens to the MnOi"" and HSO 8 ~ ions during this 

Repeat the above experiment, using 3 ml of 
saturated bromine water instead of KMnOi. How 
do you account for the decolorizing of the bromine 
solution? Test for the presence of sulfate ions in 
this solution by adding 2 ml of 0.1 F BaCl 2 . Is the 
precipitate formed soluble in 6 F HC1? 

To 10 ml of the saturated sulfur dioxide solution 
in a small beaker add 0.1 F Ba(OH) 2 solution until 
a precipitate forms. Is this precipitate soluble in 
6 F HC1 ? Compare with result obtained above and 
write equations for reactions which will account 
for the results obtained. 

4. Oxidation State +6* Sulfuric Acid, Sulfates 

(a) Preparation of sulfuric add by a method an- 
alagous to the lead chamber process. Refer to the 
textbook for the reactions involved in the lead 
chamber process and keep them before you as you 
perform this experiment. Identify the various re- 
actions with the steps in the commercial process 
and in the simplified version described below. 

Set up the apparatus shown in Figure 21-2, Place 
about 10 g of copper turnings and 25 ml of 18 F 
H 2 SO 4 in the flask fitted with the thistle tube. Place 
about 10 drops (0.5 ml) of 16 F HN0 3 in the other 
reaction flask. Fill the test tube about half-full of 
3 F NaOH which will serve to neutralize the excess 
gases (acidic anhydrides) which might otherwise 


Fio. 21-2. Apparatus to demonstrate the reactions of the 
lead chamber process for the manufacture of sulfuric acid. 

escape into the room. Heat the reaction flask gently 
until some of the nitric acid decomposes to form 
some of the oxides of nitrogen necessary for the 
process. Now heat the flask containing the copper 
and sulfuric acid gently until the SOa produced 
reacts with the gases in the reaction flask to a point 
where the brown color disappears. What reaction 
has taken place? Can you see any crystals of nitro- 
sylsulfuric acid on the sides of the flask? Allow the 



apparatus to cool a little, and then open the re- 
action flask and wash down the sides with 5 to 10 
drops of water delivered from a medicine dropper. 
Note the reappearance of the brown gases. What 
reaction accounts for this? Now reassemble the 
apparatus, allow more SO 2 to react with the brown 
gases until the reaction flask is again colorless, and 
then wash down the sides of the flask with a few 
drops of water. This process is repeated over and 
over in the commercial process, some of the react- 
ions taking place in the "lead chambers," others 
in the Gay-Lussac tower and the Glover tower. 

Pour out the reaction product into a test tube 
and test the solution for the presence of sulfuric 
acid by adding 2 or 3 ml of 0.1 F BaCl 2 to precipi- 
tate any sulfate ions present. 

(b) Physical properties of sulfuric acid. Carefully 
measure 40 ml of distilled water into your large 
graduated cylinder and 10 ml of 18 F H 2 SO 4 into 
your small graduated cylinder. Take the tempera- 
ture of the water, and then add slowly the con- 
centrated sulfuric acid to the water. Stir the solu- 
tion obtained in the large cylinder with the ther- 
mometer and note the highest temperature reached. 
From the change of temperature and the combined 
weight of the water plus sulfuric acid, calculate the 
number of calories evolved. 

Allow the solution to come to room temperature, 
and then read the volume of the solution and com- 
pare this with the sum of the volumes of the two 
ingredients mixed. What is the percent change in 
volume due to mixing? 

Place a few drops of concentrated sulfuric acid 
on a few crystals of sugar in a small evaporating 

dish. Repeat using a small piece of filter paper. To 
what property of concentrated sulfuric acid do you 
attribute the results obtained? 

(c) Chemical properties of sulfuric acid. Compare 
the action of concentrated and dilute sulfuric acid 
with that of concentrated and dilute hydrochloric 
acid on metals above and below hydrogen in the 
electromotive force series, by carrying out the fol- 
lowing experiment. Place about 5 ml of each of the 
acids, 18 F H 2 S0 4 , 3 F H 2 SO 4 , 12 F HC1, 6 F HC1, 
into 10-cm test tubes and drop a small piece of 
mossy zinc into each. Heat gently, if necessary, to 
initiate a reaction. Note cautiously the odor of the 
gases evolved and write the equations for the re- 
actions. Repeat the experiment using small pieces 
of copper instead of zinc. How do you account for 
any differences in results obtained ? 

Investigate the oxidizing strength of concen- 
trated sulfuric acid by adding 1 ml of 18 F II 2 SO 4 
to 1 g of each of the following salts in 10-cm test 
tubes : NaCl, NaBr and Nal. Heat each test tube 
gently and note cautiously the odor, color and 
acidity of the gases evolved. In which cases are 
oxidation-reduction reactions involved? Note the 
relative positions of the halide ions in the electro- 
motive force? series (Table XIII, Appendix II). 

(d) Solubility of sulfates. Add 1 ml of 1 F Na 2 SO 4 
to 3 ml of each of the following solutions in sepa- 
rate 10-cm test tubes: 0.1 F Ca(NO 3 ) 2 , Zn(NO 3 ) 2 , 
Cd(N0 3 ) 2 ,Ba(N0 3 ) 2 ,A g N0 8 ,Pb(N0 3 ) 2 ,Hg 2 (N0 3 ) 2 . 
Note the cases in which insoluble sulfates are 
formed and write a solubility rule summarizing 
your results. 

REPORT: Exp, 21 

The Chemistry of Sulfur and Date 

Mts Compounds 

rt . 


Locker Number^ 

1. Oxidation State Zero. Elementary Sulfur 

(a) Comparison of the physical properties of the allotropic forms of solid sulfur as recalled from Experiment 2: 

(b) Changes in color and fluidity of molten sulfur with increasing temperature: 

Structural formulas of the two forms of liquid sulfur which are proposed to explain the above result: 
Description of the plastic sulfur obtained: 

Its solubility in CS2 Is plastic sulfur a form of liquid or solid sulfur?.. 

2. Oxidation State -2. Sulfides and Hydrogen Sulfide (Optional) 

(a) Equation for reaction between iron and sulfur: 

Was the reaction exothermic or endothermic ? 

Reaction of hydrochloric acid on ferrous sulfide:- 

(b) Formula and color of insoluble sulfides produced by saturating solutions with H 2 S : Cu+ + 

, Cd++ ___ , Zn++ 

, Na+ 

(c) Equations for the oxidation of H 2 S by each of the following: 
By 3 F HNO 8 : 
By 0.1 F I 2 r 
By H 2 SO 3 : 

(d) The proposed structure of the ion formed when sulfur dissolves in aqueous sodium sulfide solution: 
Equation for reaction of IIC1 on polysulfide solution: 


How is this analogous to the decomposition of hydrogen peroxide? 

3. Oxidation State +4. Sulfur dioxide, Sulfurous Acid, Sulfite Ion 

(a). Write balanced equations for the reactions for the production of sulfur dioxide: 

By burning sulfur : _ 

By roasting a sulfide ore (ZnS) _ 
By acidifying a sulfite* 

By reducing sulfuric acid with Coppers 

Density of SO 2 at standard conditions, estimated from its molecular weight and the molal volume: 

Density of air at standard conditions is 1.29 g/liter. Sample of SO 2 was collected by the 

displacement of air. 

The extent of solubility of 862 gas in water was. 
(b). The reaction of SO a and water: 

The acidity of this solution was demonstrated by its reaction with powder magnesium: 

The oxidation of HSO~ ions by acid permanganate: 

Saturated bromine water oxidized the HSOa~~ ions as follows: 

The precipitate produced by the addition of BaCl 2 solution to the resulting solution above: 

Is it soluble in 6 F HC1? 

The reaction of the aqueous SO 2 solution with 0.1 F Ba(OH) 2 solution: 
The precipitate obtained ^ dissolve in dilute HCL 

Report an Exp. 81, Sheet 9 Name- 

4. Oxidation State +6* Sulfurtc Acid, Sulfates (Optional) 

(a). Reactions given in the text for the lead chamber process for the commercial production 

Describe the stages in the laboratory experiment corresponding to each of the above reactions. 

Are the oxides of nitrogen used up in this process? What role do they play ?_ 

Was the test for sulfate ions in the reaction vessel positive? 

Write the equation for this test: 

(b). The change in temperature noted in mixing sulfuric acid with water. 

The weight of the water used . (Density of water 1.0 approx.] 

The weight of sulfuric acid . (Density of sulfuric acid 1.8] 

Calories evolved = grams of solution X temperature increase X specific heat. Calories = 

(Specific heat may be taken as 1.) 

The change in volume on mixing was . The percent change 

The effect of sulfuric acid on carbohydrates such as sugar and filter pn-p^r? 
What property of sulfuric acid is illustrated here? . 

(c). Fill in the following table to indicate the products formed in each case. Mark no reactions with an X. 

Zinc Copper 
Conc.HCl .... 

Dilute HC1 .... . 

Cone. H 2 SO 4 .... 

Dilute H 2 SO 4 . . . 

How do you account for any differences in results obtained? 


Write the equations for the reactions obtained by adding cone HsSO* to each of the salts: 




Account for any differences observed in terms of the relative oxidation strength of cone sulfuric and the position 
if the halides in the electromotive force series. 

(d). A solubility rule summarizing the solubility of sulfates of the common metals: 


The Chemistry of the Compounds of Nitrogen. 

Co//ege Chemistry, Chapter 15 

Nitrogen compounds: 

+5 N 2 6 , HNOs, NO,- 

+4 N0 2 , (N 2 4 ) 
+3 (N 2 8 , HN0 2 ), 

+2 NO 

Review of Fundamental Concepts 

N 2 
-3 NH,, NH 4 OH, 

Strong oxidizing agent, usually reduced to NO 2 in concentrated acid, or to NO in 

dilute acid. With strong reducing agent may go to NH 5 . 

A heavy brown gas. 

N 2 O 3 and UNO* are unstable; nitrites are fairly stable. Active as oxidizing or as 

reducing agent. 

Oxidized by the air to NO 2 . 

Supports combustion quite vigorously. 

Nitrogen, the first element in Group V, is an im- 
portant nonmetal which forms compounds illus- 
trating all the oxidation states from 3 to +5. 
The chart above summarizes the more important 
compounds to be considered in this experiment. 

In Experiment Q the properties of elementary 
nitrogen and ammonia were considered. You should 
recall the remarkable stability of the zero oxidation 
state where the nitrogen molecule has the struc- 
ture, :N:::N:, in which the triple bond seems to 
provide a special degree of stability to the mole- 
cule. You should also review the properties of the 

3 oxidation state in which the covalent molecule, 

ammonia, has the structure H:N:H, which readily 


adds a hydrogen ion to form the ammonium ion. 

When nitrogen combines with the more electro- 
negative element oxygen, it forms compounds 
where nitrogen has positive oxidation states. The 
chemistry of the +2 and +4 states is of impor- 
tance, since NO and NO 2 are involved as inter- 
mediates in the commercial production of nitric 

acid by the oxidation of ammonia or of nitrogen. 
In the +4 state the reversible reaction 2NO 2 +3z 
N 2 O 4 , between the colorless gas N 2 O 4 and the brown 
gas NO 2 , furnishes an opportunity to apply the 
Principle of Le Chatelier to this system in chemi- 
cal equilibrium. Special attention should be given 
to the +5 oxidation state as illustrated by the 
very important compound, nitric acid. This acid is 
not only a strong acid in the sense that its aqueous 
solutions contain high concentrations of hydrogen 
ions, but it also is a strong oxidizing agent due to 
the presence of the nitrate (NOg"") ions. It is capa- 
ble of oxidizing some metals below hydrogen in the 
electromotive force series as well as many other 
inorganic and organic substances. It may be re- 
duced to any one of the lower oxidation states of 
nitrogen depending on the concentration of the 
nitric acid and the strength of the reducing agent. 
The student should be sure to review the methods 
for balancing oxidation-reduction equations since 
the chemistry of nitrogen and its compounds is 
principally concerned with changes in oxidation 

Chemicals: Cu (turnings), Zn (mossy), Mg (ribbon), Pb 
(powdered), Sn (mossy), Al (turnings), NHJMO*, NaNOj, 
Pb(NO,)i NaaOfe Bromine water, 0.1 F XL 

1. Preparation of Nitrous Oxide. Oxidation 
State +1 (Optional). Assemble the apparatus 
illustrated in Fig. 22-1, and prepare to collect two 
15-cm test tubes of gas by displacement of water. 
Since nitrous oxide is less soluble in hot water than 
in cold, the water displaced in the beaker may be 

Experimental Procedure 

warm. Place about 3 g of NH 4 NO 3 into the clean, 
dry 15-cm Pyrex test tube which is to serve as the 
generator. Heat the tube very gently but steadily 
with a small flame so that the gas is generated at 
the rate of one bubble per second. (Cautions 
Ammonium nitrate may explode if heated too rapidly. 
Especially avoid overheating near the top of the 
molten nitrate. If for any reason you need to add 
more ammonium nitrate to make more gas, allow 




the tube to cool, and clean and dry it before re- 
charging it with fresh ammonium nitrate.) After 
the air has been displaced from the generator, col- 
lect two test tubes of the gas. Remove the delivery 
tube from the water to avoid sucking back water 
into the generator. What is the equation for the 
decomposition of NI^NOa? Note the color and 
odor of the gas remaining in the generator. Test 

Fio. 22-1. Apparatus for the preparation of the oxides of 
nitrogen. For NjO, the generator test tube contains NH 4 NO 8 ; 
for NO, it contains copper turnings and 6 F HNO 3 . 

one tube of the gas for its ability to support com- 
bustion by inserting a glowing splint. Save the 
other test tube of N 2 O for a test with nitric oxide 
in the next section. Look up the electronic struc- 
ture of nitrous oxide and suggest how this might 
account for its activity as an oxidizing agent. 

2. The Preparation of Nitric Oxide. Oxida- 
tion State +2: Clean out the generator used in the 
preceding section and place about 3 g of copper 
turnings into the tube. Prepare to collect four 
15-cm test tubes of nitric oxide by displacement of 
water. Add 15 ml of 6 F HNO 3 to the generator, re- 
place the delivery tube connection, and warm the 
tube gently to initiate the reaction. After the air 
has been displaced from the apparatus and the gas 
bubbling through is colorless, collect three test 
tubes full of the gas and the fourth tube about half- 

full. What is the reaction for the reduction of dilute 
nitric acid by copper? 

Test one tube of nitric oxide with a glowing 
splint to see if it will support combustion. Note 
the colored gas produced when the tube was ex- 
posed to the air. Write the equation for the reac- 
tion which accounts for this change. 

Take the test tube of nitrous oxide saved in 
Part 1 and place it quickly mouth to mouth with a 
tube of nitric oxide, so as to avoid undue exposure 
to the air. Is there any evidence of a reaction 
similar to that between nitric oxide and oxygen of 
the air? If you were asked to determine whether a 
test tube contained oxygen or nitrous oxide, could 
you use the glowing splint test? What test could 
be used? 

Test the third tube of nitric oxide for its solubil- 
ity in water by swirling the test tube with its mouth 
under the water to allow contact of fresh water 
with the gas in the tube. Note if the water level 
in the tube rises. Now take the tube out of the 
water for a few seconds and allow the oxygen of 
the air to react with the gas as shown by the for- 
mation of a brown gas. Invert the test tube under 
the water and swirl it to note the solubility of the 
brown gas. Write the equation for the reaction 
taking place when this gas dissolves in water. 

Mark the level of water in the fourth test tube, 
which is about half-full of nitric oxide, with a wax 
pencil or a gummed label. Set up the small oxygen 
generator shown in Figure 22-2. Place about 2 g of 
sodium peroxide (Na^) into the dry 15-cm Pyrex 
test tube and draw up a few ml of water into the 
medicine dropper. Sodium peroxide reacts vigor- 
ously with water to produce oxygen and sodium 
hydroxide, so allow only a drop of water to fall on 
the peroxide whenever the flow of oxygen becomes 
too slow. When the air has been displaced from 
the generator, place the delivery tube under the 
marked test tube and allow 8 to 10 bubbles of 
oxygen to enter. Note whether the level of the 
water is lowered by the additional oxygen. Recall 
that in the balanced equation for the reaction that 
is taking place, two volumes of nitric oxide react 
with one volume of oxygen to produce two volumes 
of nitrogen dioxide. Now swirl the test tube with its 
mouth under the water and note what happens to 
the water level as the NO 2 reacts with the water. 
Allow more oxygen to enter the tube until the 



gases again turn brown, note the water level, and 
again allow the gases to dissolve in water. Repeat 
the process until the water level approaches the 
top of the tube. Remember that excess oxygen is 
not soluble in water. What substances are present 
in the water solution in the test tube? Apply a 
simple test to verify your answer. What part of one 

Fio. 22-2. Generator to prepare a small amount of oxygen 
by the reaction of water on NajO 2 ; the reaction of this oxygen 
with NO. 

of the commercial processes for the production of 
nitric acid does this experiment illustrate? 

3. Nitrites, Nitrous Acid. Oxidation State 
+3 (Optional)* Auto-oxidation-reduction of ni- 
trogen dioxide in an alkaline solution produces the 
nitrite ion as well aa nitrate ion according to the 
reaction 2 NQa + 2 OHr 5= NO a ~ + NO 8 ~ + 
H 2 O. However, in order to prepare a sample of a 
nitrite relatively free from nitrate, we shall reduce 
sodium nitrate by heating it with metallic lead. 

Mix about 2 g of NaNO 3 with 5 g of powdered 
lead in a dry test tube and heat the mixture quite 
strongly for two or three minutes. Do not heat to 
the point where brown fumes of NO* are evolved. 
Note what happens to the powdered lead, and 
write the equation for the reaction. Cool the tube 
and dissolve the sodium nitrite produced by add- 
ing 10 ml of water and warming the tube gently. 
Filter the solution and acidify with a ml of 6 F 

HC1. Note if there is any evidence for the de- 
composition of the nitrous acid produced, and write 
the equation for any reaction. Retain the solution 
for the following test. 

Since the nitrite ion is exhibiting an intermediate 
oxidation state, it may be oxidized to the nitrate 
ion by some strong oxidizing agents, but in other 
instances it may be reduced to nitric oxide by suffi- 
ciently strong reducing agents. To illustrate this, 
place 5 ml of 0.1 F KI into a 10-cm test tube and 
5 ml of bromine water into another test tube. Add 
the acidified sodium nitrite solution drop by drop 
to each of the tubes, mix, and note the products of 
the reactions. Write balanced oxidation-reduction 
equations for the reactions. 

4. Nitrogen Dioxide, Nitrogen Tetroxide. 
Oxidation State +4* At ordinary temperatures, 
nitrogen dioxide (NOa), a red-brown gas, exists in 
equilibrium with nitrogen tetroxide (N 2 O 4 ), a color- 
less gas. This mixture of gases can be prepared by 
the reduction of concentrated nitric acid by copper 
or by the thermal decomposition of lead nitrate. 
Write the equation for the latter reaction which 
we shall use. Set up the apparatus shown in Figure 
22-3a. Place about 5 g of Pb(NO 8 ) 2 in the dry 
15~cm test tube. Have two pinch clamps ready to 
place on the pieces of rubber tubing on each end 
of the 30-cm piece of glass tubing. Heat the lead 
nitrate gently until decomposition takes place and 
the brown gas has displaced the air in the appa- 
ratus. When the brown gas has been coming out of 
the end of the glass tube for about a minute, at- 
tach the pinch clamps and quickly disconnect the 
tube from the generator. 

We now have a closed system containing the 
mixture of NO 2 and N 2 O 4 gases in chemical equilib- 
rium. In order to study the effect of heat on this 
reversible reaction, place the closed tube in an up- 
right condition in a 400-ml beaker containing cold 
water so that the lower third of the tube will be 
kept cool. (See Figure 22-3b.) Allow several min- 
utes for the gases in the tube to cool, then with a 
small flame heat the upper third of the tube by 
fanning the flame back and forth a few times* 
(Avoid overheating at this point or appreciable 
amounts of NO a may decompose to the colorless 
gas NO and thus complicate the interpretation of 
your results.) View the gases in the tube against 
a piece of white paper and note whether there i* 



Pb(N0 3 ) 2 


FIG. 22-3. (a) The preparation of nitrogen dioxide and nitrogen tctroxide. (b) The effect of temperature on the 

dissociation of 

more of the colored gas, NC>2 in the cooler portion 
or in the heated portion of the tube. Write the 
equation for the reversible reaction and include the 
term heat on the proper side of the reaction to ac- 
count for the result obtained. The shift in the 
equilibrium, in this case caused by the addition of 
heat, is an illustration of the application of an im- 
portant principle, Le Chatelier's principle, which 
will be further applied in the next experiment. 
Look it up now in the text and note how it applies 
to the present experiment. 

5. Nitric Acid, Nitrates. Oxidation State +5. 
Many reactions involving the reduction of nitric 
acid and the decomposition of nitrates have been 
studied already in the preceding parts of this ex- 
periment. We shall illustrate the oxidizing prop- 
erties of nitric acid further by varying the con- 
centration of the acid and the strength of the 
reducing agents. Pay particular attention to the 
oxidation state of the reduction products and what 
factors influence the extent of reduction. 

Add about 3 ml of 16 F HNO 8 to a small piece 
of copper in a 10-cm test tube. Place about 5 ml 
of 1 F HNO, into each of two test tubes. Place a 

small piece of Zn into one and a small strip of Mg 
ribbon into the other. Note the products of each 
of the reactions. Pour the solution resulting from 
the reaction on the zinc into a small beaker, add 
about 5 ml of 6F NaOH and cover the beaker with 
a watch glass which has a small piece of moistened 
red litmus paper on the under side. Warm the 
beaker gently and note any change in color of the 
litmus paper. 

Recall the product obtained when 6 F HNO 8 
reacted with copper in Part 2 of this experiment. 
Summarize your findings concerning nitric acid as 
an oxidizing agent in the report sheet. 

Some metals such as arsenic, antimony and tin 
are oxidized to insoluble oxides by hot concentrated 
nitric acid. Put a small piece of tin into about 5 ml 
of 16 F HNOa and note the reaction products. 

A few metals, such as aluminum or chromium, 
are rendered passive or inactive in nitric acid. 
Cover a small piece of Al with 5 ml of 16 F HNO 8 
and watch for any reaction. 

Are nitrate salts of metals generally soluble in 
water? Make a one sentence solubility rule which 
Holds for most nitrates. 

REPORT* Exp. 22 Name- 

The Chemistry of the Compounds Date 

of Nitrogen c ,. 

** Sectwn- 

Locker Number- 

1. Oxidation State +! Preparation and Properties of Nitrous Oxide (Optional) 

The electron-dot structures proposed for N2O: 

Equation for the decomposition of NH 4 NO 3 : 

Properties of N2O: color , odor , ability to support combustion 

How does the position of the oxygen atom in the N2O molecule affect its activity as an oxidizing agent? 

2. Oxidation State +2. Preparation and Properties of Nitric Oxide 

The electron-dot structures proposed for NO: 

Why is nitric oxide called an odd molecule? 
Equation for reaction of dilute nitric acid on copper: 

Properties of NO: color , odor , ability to support combustion. 

What is the colored gas formed when NO is exposed to air 

Equation for this reaction : 

Does nitrous oxide give up its oxygen to react with NO?_ 

To what extent is NO soluble in water? Is NO 2 soluble in water ?_ 

How do you account for this? .. 

Equation for the reaction of NO 2 with H 2 O: 

Equation for the production of oxygen from Na 2 O2 and 

Describe the changes taking place in the test tube half-full of NO gas as oxygen was added: color of gases 
. t change in water level .. Account for these observations: 


When test tube was swirled under water, the change in water level noted was , color of 

gases became Account for these observations: 

What results were obtained when the above process was repeated? 

What substances are present in the aqueous solution in the test tube at the end of the process? 

What part of which commercial process for the production of nitric acid does this part of the experiment illus- 

3. Oxidation State +3. Nitrites and Nitrous Acid (Optional) 

Equation for the production of NaNO2 by the reduction of NaNOa with Pb: 

What is the yellow solid formed in the tube? . 

When the NaNO2 was leached out with water and acidified with dilute HC1 the reaction which took place was: 

The half-reaction for the reduction of nitrite ion in acid solution: 

The half-reaction for the oxidation of nitrite ion in acid solution: 

Was bromine water (Br 2 ) a sufficiently strong oxidizing agent to oxidize nitrite ion? -- What is 

your evidence? _ . What happened to the Tlr t ?! 

Equation : - 

Was the nitrite ion able to oxidize the iodide ion in the acid anlnfinn? . What happened to 

the iodide ion? - . Equation: - ' 

If you were given two white crystalline salts, one of which was KI and the other KBr, how would you apply 
the principle of selective oxidation illustrated by this experiment to determine which salt was KI? 


Report on Exp. 88, Sheet 2 Name. 

4. Oxidation State +4. The Equilibrium Between NO* and N*O 4 

Equation for the preparation of the NOa, N*O 4 mixture by the decomposition of Pb(NOs)*: 

The electron-dot structures of NOa and of 

Which gas is colored? 

Describe the changes which take place in the closed tube when heat is applied to the upper third of the tube: 
Write the equation for the reversible reaction involved in the equilibrium system under study here: 

Which way was this equilibrium shifted by the application of heat?_ 

In this reaction there are 15,000 calories of heat involved per mole of N 2 C>4. On which side of the equation above 

should the heat term appear with a positive sign? . 

What is the Principle of Le Chatelier and how does it apply to the interpretation of this experiment? 

5. Oxidation State +5. Nitric Acid and Nitrates 

From the various experiments carried out in this experiment which involve the use of nitric acid as an oxidizing 
agent, write balanced equations illustrating its reduction to the +4, +2, and 3 oxidation states. 

From +5 to +4: 

From +5 to +2:. 
From +5 to 3:_ 

What was the main product of the reaction when Mg, a very strong reducing agent, reacts with very dilute 

nitric acid? 

Summarize the results obtained by making a statement concerning the influence of the strength of the reducing 
agent and the dilution of the acid on the reduction products. 

Equation for reaction of concentrated nitric acid on tin: 

How do you explain the action of concentrated nitric acid on aluminum? 
A solubility rule concerning the solubility of most metallic nitrates: 


The Rate of Decomposition 
of Sodium Hypochlorite 


College Chem/sfry, Chapter 19 

Review of Fundamental Concepts 

A study of the factors that affect the rate at 
which a reaction proceeds is both of commercial 
importance and of great theoretical interest. Even 
though the energy relations of a reaction are such 
that one would predict that it should take place 
readily, it may take place so slowly as to be com- 
mercially impractical. The theoretical chemist is 
interested in what must be done to make the atoms, 
molecules, or ions collide in an effective manner so 
that new combinations will result at a desirable 

Most reactions with which you have had some 
experience have been practically instantaneous be- 
cause they have usually involved the collision of 
ions in an aqueous solution. Thus, when Ag+ and 
Cl~ ions collide, new crystalline grains of the 
slightly soluble substance AgCl are produced. 
Many reactions between covalent molecules, how- 
ever, take place very slowly. For example, the key 
reaction in the production of fertilizers and explo- 
sives, N 2 + 3 H 2 += 2 NH 8 + 21,880 cal, was 
made commercially feasible only after years of re- 
search revealed the appropriate catalyst and the 
conditions of temperature, pressure, and concen- 
tration which would favor the production of am- 
monia at a reasonable rate. 

Reactions do not take place upon every collision 
between molecules, and only the more energetic 
molecules are involved in the effective collisions. 
Some of the factors which determine both the 

number of energetic molecules and the chances for 
their collision are the concentration of molecules, 
their temperature and pressure, and the presence 
of a catalyst. 

The term "catalyze" means to loosen. Theo- 
retical chemists suggest that in the case of contact 
catalysis the reacting molecules are adsorbed on the 
surface of the solid catalyst and are subjected to 
strain which makes them more susceptible to the 
desired breaking of chemical bonds. In other in- 
stances the catalyst may serve as an intermediate 
by reacting with one of the substances to form an 
intermediate compound, which in turn reacts 
readily with the other reactant to form the desired 
substance and regenerate the catalyst. 

The reaction which we shall study is the decom- 
position of sodium hypochlorite in a slightly basic 
solution (commercial bleaching solution), cata- 
lyzed by the presence of an oxide of cobalt. 1 

The net reaction is: 2 OQ- * O 2 + 2 CK 
The rate may be followed by allowing the oxygen 
formed to displace water and by measuring the 
volume collected at regular intervals of time. In 
order to standardize the effect of the catalyst and 
also minimize any complicating side reactions, it is 
necessary to add the same amount of cobalt nitrate 
solution in each of the trials. Only a small portion 
of the sodium hypochlorite reacts with the cobalt 
ion to produce the solid cobalt oxide catalyst. 

Experimental Procedure 

Special supplies: a watch with a second hand, thermometer. 
Chemicals: 0.17 F Co(NO 8 ) 2 (5%), and 5.25% NaOCl 
(commercial bleaching solution). 

1. Set up the apparatus illustrated in Figure 
23-1. Fill the Florence flask with water, and insert 
its stopper. Fill the delivery tube to the graduated 
cylinder by blowing into the tube connected to the 
length of rubber tubing. Close the pinch clamp. 
Measure carefully 25 ml of sodium hypochlorite 
solution in a graduated cylinder (or pipette) and 

pour it into the Erlenmeyer flask. Place exactly 
5 ml of the cobalt nitrate solution in a 10-cm test 
tube, and insert it carefully in the reaction flask. 
Adjust the glass tubing in the stopper so that the 
test tube is held loosely in the position shown. 
Remove the pinch clamp. If your apparatus is 

1 The formula of the oxide of cobalt is not known with certainty, 
but it is thought to be CojOj, which is probably formed by the reaction 
2 Co++ 4- OC1- -f- 2 H*0 - CojOt + 4 H+ + Cl~. Cod hai also 
been suggested in the literature. 




25 ml 
i-^ solution 

FlQ. 23-1. Apparatus to measure the rate of decomposition of sodium hypochlorite. 

tight, only a drop or two of water will drop into 
the graduated cylinder. Have your data sheet 
ready. Observe the time as you quickly tip over 
the reaction flask so as to mix the cobalt nitrate 
solution with the hypochlorite solution. Hold the 
flask at its neck, and shake it constantly with a 
gentle swinging motion. Record the volume of 
water displaced every 30 seconds until a total vol- 
ume of about 100 ml has been reached. Plot the 
results on a graph, using time as the abscissa and 
volume of oxygen as the ordinate. 

2. Repeat the procedure at a temperature ap- 
proximately 10 C above room temperature. The 
solutions may be conveniently warmed by placing 
the Erlenmeyer in a larger beaker containing 
water at about 20 to 30 above room temperature. 
When the desired temperature has been reached, 
the Erlenmeyer should be removed from the water 
bath and connected to the apparatus as before. 

3. (Optional.) If time permits, repeat the ex- 
periment at a temperature approximately 10 C 
below room temperature. Use chipped ice and 
water in the water bath this time. 

4. To determine the effect of changing the con- 
centration, repeat the experiment at room tem- 
perature, but add 30 ml of water to the hypochlo- 
rite solution in the flask so as to effect a twofold 
dilution in the final volume after mixing. 

5. (Optional.) If time permits, the experiment 
may be repeated at room temperature with a 
fourfold dilution by adding 90 ml of water to the 
hypochlorite solution. 

Plot all results neatly on the same graph, and 
label each run. Note the variations in rate which 
resulted from the changes in temperature and in 
concentration. Make your comparisons as quanti- 
tative as possible in the discussion requested on the 
report sheet. 

REPORTi Exp. 23 

The Rate off Decomposition 
of Sodium Hypochlorlte 

Summary of Data 



Locker Number- 


Volume of oxygen produced (ml) 

Room temp. 






































Graphical representation of data. Label axes and each curve. 

1. Discuss in as quantitative a manner as possible the effect of changes in temperature upon the rate of this 

2. How do you account for this in theoretical terms? 

3. Give a reasonably quantitative discussion of the effect of changes in concentration upon the rate of the 

4. How do you account for this in theoretical terms? 

5. What is the probable role played by the cobalt oxide catalyst in the mechanism of the decomposition 


Reversible Reactions and Equilibrium. 


Co//ege Ctam/ffry, Chapter 19 

Review of Fundamental Concepts 

The Reversibility of Chemical Reactions 

A chemical reaction in which a slightly ionized 
or a slightly soluble substance is formed, or in 
which a large amount of energy is released, will 
generally continue until one or more of the react- 
ing substances is effectively all used. Many reac- 
tions, however, even if allowed to proceed for a 
very long time, do not go to completion. This will 
be the case if the products formed remain in con- 
tact or are intimately mixed with one another in a 
reactive condition, so that they may reform the 
initial substances again. Thus, hydrogen gas and 
iodine vapor at 400 C will interact quite readily 
to form hydrogen iodide. But, likewise, under the 
same conditions, hydrogen iodide molecules inter- 
act to re-form hydrogen and iodine. This is there- 
fore a reversible reaction. These opposing tenden- 
cies may be indicated by the use of double arrows 
in the equation, thus, 

HI T i i > f> TTT 
2 ~r 2 < r & Hi 

Chemical Equilibrium 

Let us carry out the above opposing reactions 
as follows. In one bulb, we place one mole each of 
hydrogen and iodine. In a second similar bulb, we 
place two moles (the equivalent amount) of hydro- 
gen iodide. The first bulb shows the violet color of 
iodine vapor; the second is colorless. We shall now 
heat both bulbs at 400 C for a long time, after 
which we observe that each bulb has attained the 
same intensity of violet color. Samples withdrawn 
for analysis show the same concentration of free 
iodine in each bulb, and likewise the same con- 
centrations of hydrogen and of hydrogen iodide. 
Additional heating produces no further changes. 

This does not mean that reaction has ceased, but 
only that the rate at which hydrogen and iodine 
are interacting to form hydrogen iodide is exactly 
counterbalanced by an equal rate of dissociation of 
hydrogen iodide into hydrogen and iodine. A re- 
action in which the resultants are being changed back 
into the reactants at the same rate as the reactants are 
forming the resultants is said to be in equilibrium. 

Reaction Rates and Le Chatelier's Principle 

The rate of a chemical reaction is affected by (1) 

the temperature, (2) the concentration of the re- 
actants, and (3) the presence of a catalyst. In this 
experiment, we shall study only the qualitative 
effect of changes in the concentration of the re- 
actants and products on the equilibrium. 1 

These concentration effects may be most simply 
interpreted if we keep in mind the kinetic picture 
of matter as made up of moving molecules. The 
rate of formation of hydrogen iodide, according to 
the preceding equation, must be proportional to 
the number of effective collisions of hydrogen and 
iodine molecules. (See Fig. 24-1.) If we add more 
hydrogen gas to the container, thereby increasing 
the concentration of the hydrogen molecules, we 
will have more collisions per second of hydrogen 
and iodine molecules and will thereby increase the 
rate of the forward reaction. This will momentarily 
"unbalance" the equilibrium until the concentra- 
tion of hydrogen iodide molecules has increased 
enough so that their rate of decomposition (the 
reverse reaction) has increased to the point where 
the opposing processes are taking place at equal 
rates. The over-all effect of adding more hydrogen 
gas is thus to create a new equilibrium, in which 
we have more hydrogen, less iodine, and more hy- 
drogen iodide than in the original case. The equilib- 
rium has been "shifted to the right." 

Conversely, if some of the iodine is removed 
from the container, there will be fewer collisions 
per second of hydrogen and iodine molecules, re- 
sulting in a slower forward reaction. The reverse 
reaction will predominate until a new equilibrium 
is established, with less hydrogen iodide, more hy- 
drogen, and somewhat less iodine, than in the 
original mixture. These relationships are in accord 
with the Law of Le Chatelier, that a system in 
equilibrium tends to shift in such a direction as to 
restore in so far as possible, the effect of any stress 
(change in concentration of one of the substances) 
which has been applied to the system. 

The common ion effect is a special case of the 
application of the law of chemical equilibrium to 
ionization reactions* For example, in a solution ot 

1 We shall consider the quantitative aspects of the law of chemical 
equilibrium in Experiments 86 and 37. 




H 2 + I a 




FIG. 24-1 The formation of hydrogen iodide, and the reverse process (its dissociation into hydrogen and iodine), are occurring 

simultaneously. This results in a chemical equilibrium. 

the weak base ammonium hydroxide, we have the 


If we add some ammonium chloride, NH 4 C1, am- 
monium sulfate, (NH 4 )SO 4 , or any other soluble 
ammonium salt, we will thereby increase the con- 
centration of ammonium ion. There will then be 
more collisions per second between ammonium ions 
and hydroxide ions than before, the equilibrium 
will be shifted to the left, and the hydroxide ion 
concentration will be decreased. The ammonium 
ion, since it is common to both the ammonium 

hydroxide and the ammonium salt added, is called 

a "common ion". 

In the same way, salts that are only very slightly 
soluble can be made even less soluble by increasing 
the concentration of a common ion. Thus, in the 

AgCl (solid) + Ag++Cl-, 

silver chloride is less soluble in any solution to 
which additional silver ion, or chloride ion, has 
been added, than it is in pure water. A higher con- 
centration of either silver ion, Ag+, or chloride ion, 
Cl~, tends to shift the equilibrium to the left. 

Experimental Procedure 

Chemicals: 6 F HC,H,O 2 , 0.25 F (NHOaCaQ* 0.1 F Bad* 
0.1 F CaCl a , 0.1 F FeCl,, 0.5 F HjCA, 0.1 F KCI, 0.1 F 
KCNS, 0.1 F AgNO,, 3 F NaC,H,O 2 , sat. (5.4 F) NaCl, 
1 F NadSO 4 . 

1. A Test for the Completeness of a Reac- 
tion. We shall prepare solid CaSO 4 and solid 
BaSO 4 from equivalent amounts of their respec- 
tive ions. The equilibrium equations are 


Ca++ + S0 4 " ^^ CaS0 4 (solid), 
Ba++ + SO,"- :=* BaSOi (solid). 

Consider the following questions while you are 
doing the experiments. Are these reactions com- 
plete or is an equilibrium established between the 

solid and the ions when there is still a detectable 
amount of the ions present? Which of the ions, 
Ca++ or Ba++, precipitates SO 4 more com- 

Place 1.0 ml of 1 F Na^Ch in each of two test 
tubes, measuring each sample carefully with your 
10-ml graduated cylinder. To one sample add about 
15 ml' of 0.1 F CaCl a , and to the other add about 
15 ml of 0.1 F BaCU. (It is necessary to have a 
slight excess of CaCl 2 or BaCla in each case.) Heat 
each tube to boiling, and let it stand, preferably 
for at least a half hour. (Meanwhile proceed with 
paragraph 2.) Test each mixture for completeness 
of precipitation by adding a drop of CaCla solu- 
tion to the CaSO 4 mixture, and a drop of BaCla 
solution to the BaSO 4 mixture. Add more reagent 



Direct the 
stream around 
the edge of 
the filter 

Puncture the 
filter paper. 

Wash the pre- 
cipitate into 
a test tube* 

FIG. 24-2. The technique of puncturing the filter paper and washing the precipitate into a test tube. 

if needed. If there is no further precipitation, filter 
or centrifuge 1 each mixture. 

Now to the CaSO 4 filtrate, add 2 to 3 ml of 0.1 F 
BaCl 2 . Note the results. To the BaSO 4 filtrate, add 
about 2 to 3 ml of 0.1 F CaCl 2 . Note the results. 
Questions. From these results which of the above 
equilibrium reactions, for the precipitation of 
CaSO 4 or of BaSO 4 , took place more completely? 
What relationship does the residual concentration 
of SO 4 in the filtrate have to the solubilities of 
the salts? Which is more soluble, CaSO 4 or BaSO 4 ? 

1 If the precipitates have settled completely the filtrates may be 
decanted from the solids, or they may be placed in 10-cm test tubes 
and eentrifuged. Obtain instructions before using the centrifuge. It 
is essential that solutions are well balanced before centrifuging. If 
you filter the BaSO 4 mixture, a special fine grained filter paper is 
necessary, as the precipitate IB very fine grained and tends to run 
through the filter paper. 

2. The Shifting of an Equilibrium. The 
Common Ion Effect 

A. A Saturated Silver Acetate Solution. Prepare 
some freshly precipitated silver acetate by mixing 
10 ml of 3 F NaCaHA with 25 ml of 0.1 F AgNO,. 
Let the mixture stand a few minutes to complete 
the precipitation, then filter it. Drain the precipi- 
tate and rinse it with not over 2 ml of distilled 
water, and again let this drain. Place a clean 15-cm 
test tube under the funnel, puncture the filter with 
a stirring rod, and wash the precipitate into the test 
tube with a small stream of water from the wash 
bottle; but do not use over 10 ml (^J of a test tube) 
of water. Warm this mixture slightly (it should not 
be hot, however,) and shake it gently for at least 
ten minutes to establish the equilibrium of a satu- 
rated solution according to the equation 
(aotid) *=t Ag+ + C&Or. 



Filter this solution through a dry filter, retain- 
ing most of the residue in the original test tube 
(save this), and divide the filtrate into two 4 to 
6-ml portions in small test tubes. Add to these two 
samples, respectively, about 6 to 10 drops of 3 F 
NaC 2 H 3 2 , and 6 F HC 2 II 8 2 . Observe the results, 
which may require several minutes to be notice- 
able. Explain, in detail, on the basis of the above 
equilibrium equation. 

To the test tube containing the residual solid 
AgCJIsOu, add 1 to 2 ml water, and then several 
drops of 6 F HN0 3 . Explain the results. 

B. A Saturated Sodium Chloride Solution. Meas- 
ure out 10 nil of a saturated solution of NaCl. Cal- 
culate the concentration of the ions, assuming 
complete ionization. (Solubility of NaCl: 358 g 
/1000gH 2 O, or 358 g/1358 g solution. Density: 
1.200 g/ml. Verify this as 5.4 F NaCl.) Add a few 
ml of concentrated IIC1 (12 F). 

NaCl (solid) HF: Na+ + Cl~ 

How do the concentrations of Cl~ in a saturated 
solution of NaCl and in 12 F HC1 compare? What 
is the common ion in HC1 and NaCl? Explain the 
action occurring on the addition of 12 F HC1 to 
saturated NaCl in terms of the above equilibrium 

C. The Formation of a Complex Ion. The follow- 
ing experiment differs from the two preceding ones 
in the fact that the product formed on the union 
of the ions is not insoluble, but is a slightly ionized, 
soluble, complex ion. 1 

(dark red). 

Add 10 ml of 0.1 F FeCl 3 to 10 ml of 0.1 F KCNS, 
and dilute the solution until a medium red color is 
obtained. About 100 to 125 ml of tap water will be 
required. Place 10-ml portions of this solution in 
each of four test tubes. Add 10-ml portions of the 
following reagents, respectively, to these four 

1. H 2 O 

2. 0.1 F FeCla 

3. 0.1 F KCNS 

4. 0.1 F KC1 

Predict the effect of each addition on the color 
of the solution before making the test. Record the 

1 See article by Bent and French, J.A.C.S. 63, 568, on the formula 
for this complex ion. The ion undoubtedly is hydrated, and may vary 
in composition from Fe(II 2 Q) ft CNS++ to Fe(CNS). . 

results of the tests as redder or lighter. An in- 
crease of redness indicates an increase of the con- 
centration of what substance? What shift in the 
equilibrium does the formation of this substance 
indicate? What shift in the equilibrium does a de- 
crease of redness indicate? 

3. Application of the Law of Chemical 
Equilibrium to Analytical Procedure : In quali- 
tative analysis, Ca++ is usually precipitated for 
identification as calcium oxalate, 

Ca++ + C 2 4 ~ 

CaCA (solid). 

What are the conditions needed to make the pre- 
cipitation as complete as possible? The Ca++ con- 
centration cannot be varied in the experiment as 
it is the ion whose presence is being tested for in 
the unknown. The completeness of the reaction, 
then, depends on the adjustment of the concentra- 
tion of the C 2 O 4 . From which can the highest 
concentration of C 2 4 be obtained, from a solu- 
ble salt, e.g. (NH 4 )2C 2 O4, or from the moderately 
weak acid, H 2 C 2 O 4 ? The ionization equations are 

(NH 4 )2C 2 O 4 
H 2 C 2 O 4 

2 NH 4 + + C 2 O 4 ~ , and 
HC 2 4 ~ 


What effect would the presence of H+ in the 
solution have on these equilibria? What would be 
the effect of the presence of OH~? Should Ca++ be 
precipitated as CaC 2 O4 in an acidic or in a basic 
solution? Test out your reasoning by making the 
following tests. 

Mix 10 ml of 0.1 F CaCl 2 with 10 ml of distilled 
water, and divide into two 15-cm test tubes. To 
one add 1 ml of 0.5 F H 2 C 2 4 , and to the other 
2 ml of 0.25 F (NH 4 ) 2 C 2 O 4 . Compare the results. 
To the H 2 C 2 O 4 mixture, add 1 to 2 ml of 6 F HC1, 
and mix. Explain the results. Now to this same 
solution add a slight excess of 6 F NH 4 OH, and 
mix. Determine whether the precipitate may be 
Ca(OH) 2 by adding a little 6 F NH 4 OH to some 
diluted CaCl 2 solution. What is the precipitate? 
Explain all these results in terms of the equilibrium 
equations above. 

Would the acidity of the solution be of more im- 
portance in the precipitation of the salt of a strong 
acid or in the precipitation of the salt of a weak 

REPORT: Exp. 24 


Reversible Reactions and 
Equilibrium Section - 

Locker Number- 

1. A Test for the Completeness of a Reaction 

(a) Was there in this experiment sufficient SO 4 left in solution after precipitation with an equivalent amount of 

Ca++ to precipitate with Ba++? 

(b) Was there sufficient SO 4 left in solution after precipitation with an equivalent amount of Ba+ + to precipitate 

(c) Which filtrate, from the CaS0 4 , or BaS0 4 , has the higher concentration of SO 4 ? 

(d) Which salt, CaSO 4 or BaSO 4 , is the more soluble, according to your experimental 

(e) What are the solubilities (formalities) of CaSO 4 and of BaSO 4 at room temperature? 

( CaSCV 
Reference source: < 

( BaSO,- 

(f) Which ion, Ca++ or Ba++, precipitates SO 4 more completely? 

(g) What additional information would you need to find whether the reaction, Ba* 1 "* + SO 4 3P* BaSO 4 , is more 
complete than the reaction of SO 4 with any other metallic ion? 

2. The Shifting of an Equilibrium. The Common Ion Effect 

A. Saturated Silver Acetate Solution. Rewrite the equation for the equilibrium present in a saturated solution 
of silver acetate: 

State whether a precipitate was formed when each of the following was added to the above equilibrium solution: 

3 F NaC 2 H 3 O 2 6 F HC 2 H 8 O* 

What effect would you predict if 3 F AgNOs (which is too expensive to use) were added to the above equilibrium 

Explain concisely, but in some detail, why silver acetate was precipitated in some cases, but not in others. 


What happened when 6 F HNO* was added to the residual solid silver acetate? Explain in terms of the equi- 
librium equation. 

B. A Saturated Sodium Chloride Solution. 

What is the "common ion" in solutions of NaCl and HC1? 

The Cl~ concentration of the saturated NaCl is approximately 

Is, then, the concentration of Cl~ increased or decreased by adding concentrated HC1?_ 

Explain the action occurring on the addition of concentrated HC1 to saturated NaCl, in terms of the equilibrium 
equation, NaCl (solid) + Na+ + C1-. 

C. The Formation of a Complex Ion. Compare the colors of the solutions of the tubes 2 to 4 with tube 1, in each 


(2) FeCU 

(3) KCNS 

(4) KC1 . 

What two ions are most effective in increasing the red color, (as compared with tube 1)? 

Interpret the above results in terms of the equilibrium equation, Fe+++ + CNS~ ^F* Fe(CNS)++. 


Report on Exp. 4, Sheet 


3. Application of the Law of Chemical Equilibrium to Analytical Procedure 

Substances Mixed 

Observations (Indicate the Relative Amount of Precipitate) 

(1) CaCU + H,C,0 4 

(2) CaCU + (NH 4 )2C 2 4 

(3) Results of (1) + HC1 

(4) Results of (3) + excess NH 4 OH 

(5) CaClj + NH 4 OH 

(a) Does the CaC 2 O 4 precipitate more completely in an acidic or in a basic solution? 

(b) Does a solution of (NH 4 ) 2 C 2 O 4 , or a solution of H 2 C 2 O 4 of equivalent concentration, give a higher concentratior 
ofC 2 4 ?Why? 

(c) By use of the equilibrium equation, 

H+ + HC 2 4 ~ 

show how the concentration 

H+ + C 2 4 , 

is affected by the H + concentration of the solution. 

(d) How does the increasing of the OH~ concentration of the solution affect the CaO 4 concentration in an oxalic 
acid solution? What was the precipitate, formed on addition of NH 4 OH to the acid solution of Cad* and oxalk 


(e) How was the equilibrium reaction, 

Ca++ + C 2 O 4 :?r* CaC 2 O 4 (solid) 
affected by an increase of H+ concentration? Explain. 

(f) Is it more important to control the H+ concentration in the precipitation of the salt of a weak acid or of th 
salt of a strong acid? Why? 

Application of Principles 

Predict whether the following insoluble salts would be dissolved by the addition of a strong acid such as HNOj 
In each case where solution would take place, write the net ionic equation for any reaction occurring. (See Table XI 
Appendix II.) 


Would They Dissolve? 

Equation, if Any 




PbSO 4 



Mg,(P0 4 ), 

Equilibria Involving Volatile and Insoluble 
Substances. The Transformation of Salts. 


Co//ege C/iem/sfry, Chapters 13, 14, 15 

Review of Fundamental Concepts 

This continues the study of chemical equilib- 
rium with reference to the formation of volatile 
and insoluble substances (see Tables VII and X, 
Appendix II) and shows how this may be used in 
the conversion of one salt to another or in the re- 
moval- of undesired ions. It is often necessary in 
analytical procedures to eliminate an ion from a 
solution before making a test. 

The Conversion of a Salt of a Volatile Acid to a 
Salt of a Non-Volatile Acid 

Suppose you have a solution of sodium nitrate 
which you wish to convert to sodium sulfatc. Sul- 
fate ion must be provided from some source, and 
at the same time a positive ion must be provided 
which can be removed with the nitrate ion. The 
latter could be removed, if it formed an insoluble 
or a volatile compound. Since there are no insoluble 
nitrates, the precipitation method is not available. 
Nitric acid has a low boiling point. If, then, sul- 
furic acid, which provides both hydrogen ion and 
sulfate ion and has a high boiling point, is used, the 
equation for the reaction is 

2 NaNO 3 + H 2 S0 4 >- Na 2 SO 4 + 2 HNO 3 . 

The evaporation of the solution will vaporize the 
nitric acid and leave the sodium sulfate as a solid 
residue. 1 

The Conversion of a Salt of a Volatile Acid to a 
Salt of Another Volatile Acid 

If you wish to convert a chloride to a nitrate, or 
vice versa, another principle is involved. While to 
some extent during an evaporation, the reaction 

NaCl + HNO 3 > NaNO* + HC1 

takes place, we also have an oxidation-reduction 
reaction, in which both chloride ions and nitrate 
ions are destroyed. 

4 H+ + 3 Cl- + NO 3 ~ Cl, + NOC1 + 2 H 2 O. 

1 With an excess of sulfuric acid, the principal product will be 
sodium hydrogen sulfate NaHSO 4 . 

If an excess of nitric acid is added, the chloride 
ion is eventually all destroyed. To obtain a com- 
plete transformation, repeated trials are necessary. 
Each time nitric acid is added, and the solution 
evaporated to dry ness, the conversion becomes 
more complete. 

The Conversion of a Salt of a Non-Volatile Acid 
to a Salt of a Volatile Acid 

In converting a sulfate to a chloride, the method 
of treating the sulfate with hydrochloric acid and 
evaporating the solution is unsuccessful, for ori 
evaporating the solution, the hydrochloric acid, 
and not the sulfuric acid, evaporates. Two other 
methods are available. (1) Barium chloride solu- 
tion in equivalent amount may be added, the 
barium sulfate filtered out, and the remaining 
sodium chloride then evaporated to crystals, 

Na 2 S0 4 + BaCl 2 * BaSO 4 (solid) + 2 NaCl. 

(2) The metal ion may be precipitated out as a 
hydroxide, or carbonate, filtered, and the precipi- 
tate redissolved in hydrochloric acid. The solution 
is then evaporated to crystals. 

CuS0 4 + 2 NaOH - 
Cu(OH),+ 2HCl- 

Cu(OH) 2 (solid) + 
CuCl 2 +2H 2 O. 

The Choice of Reagents and Test of Completion 

Note that in the examples given for all three 
cases, no substances were added that would remain 
in the solution with the product after the reaction 
was complete. The by-product was always either 
a precipitate which could be filtered out, or a 
volatile substance which could be boiled off. The 
product, if in solution, could be crystallized out as 
a pure substance. 

To see if conversion is complete, a test is made 
on the product for the negative ion that supposedly 
has been completely removed. 




Experimental Procedure 

Chemicals: 0.1 F BaCl 2 , FeSO 4 7H 2 O, 0.1 F AgNO,, NaCl, 
NaNO,, NajSO*, 1 F 

1. The Conversion of a Salt of a Volatile 
Acid to the Salt of a Non- Volatile Acid. To 

about 1 gram of NaNO 3 , add 5 ml of 3 F HtSO 4 . 
Evaporate the solution to dryness. Dissolve the 
residue in 5 to 10 ml of water. Test a small portion 
of this solution for the presence of NO 8 ~ to see if 
the conversion is complete. 1 If not complete, add a 
small excess of H^SCX, and repeat the process. 

2. The Conversion of a Salt of a Volatile 
Acid to the Salt of Another Volatile Acid. To 
about 1 gram of sodium chloride, add 5 ml of 6 F 
HNOf. Evaporate just to dryness and test a small 
portion of the residue for Cl~. 1 If a Cl~ test is ob- 
tained, repeat the process with a second 5-ml por- 
tion of HNOa. If a chloride test is still obtained, 
repeat the process a third time. Is there any dif- 

1 See Experiment 20 for details of the tests for NO 3 - and for Cl~. 

ference in the intensities of successive chloride 

3. The Conversion of a Salt of a Non- Volatile 
Acid to the Salt of a Volatile Acid. To about 1 
gram of Na 2 SO 4 add 5 ml of 6 F HC1, and evapo- 
rate the solution to dryness. Test a small portion 
of the residue for sulfate ion. Repeat the process, 
and again test the residue for sulfate ion. Is this a 
good method for converting Na 2 SO 4 to NaCl? 

To 2 ml of 1 F Na 2 SO 4 , add an equivalent 
amount, 20 ml, of 0.1 F BaCl 2 solution. The solid 
BaSO 4 may be removed by filtration through a fine 
filter paper (obtained from the stockroom) ; or if a 
centrifuge is available, the mixture may be centri- 
fuged. Obtain instructions, and be sure the centri- 
fuge tube is carefully balanced. Evaporate the 
filtrate to crystallization and test the crystals 
formed for SO 4 . Which conversion is more com- 
plete, with HC1, or with BaCl 2 ? Explain the dif- 
ference in behavior. 


REPORT: Ezp. 25 

Equilibria Involving Volatile and Date 

Insoluble Substances. The Trans- Section 

formation of Salts j^, Number _ 

1. The Conversion of a Salt of a Volatile Acid to the Salt of a Non-Volatile Acid 

To what extent can you convert NaNO 3 to NaaSOi by evaporating a mixture of NaNO and H a SO 4 ? 

List the principal substances present in separate solutions of NaNO 3 and 

Write the equation for the reaction, if any, which occurs on mixing solutions of NaNOa and EUSOi at room 

What is the net ionic equation for the reaction, if any, which occurs when this mixture is evaporated to crystal- 

Why should one expect the conversion of NaNO 3 to Na 2 SO4 to be successful under the conditions used in the 

Why is the negative test for N0 3 ~ in the product a better test of the completion of the reaction than is a posi- 
tive test for SO 4 ? 

2. The Conversion of the Salt of a Volatile Acid to the Salt of Another Volatile Acid 

To what extent can you convert NaCl to NaNO 3 by evaporating a mixture of NaCl and HNO 3 ? Is repeated 
treatment with HNO 8 more effective in converting the salts? Explain fully. 

Would you expect to be able to convert NaNO 3 to NaCl by a similar process? Explain fully. 

3. The Conversion of a Salt of a Non-Volatile Acid to the Salt of a Volatile Acid 

To what extent can you convert Na a SO 4 to NaCl by evaporating a mixture of Na 2 SO 4 and HC1? Is repeated 
treatment with HC1 more effective in converting the salts? Explain fully. 


Was the conversion of Na-iSCh to NaCl more successful when using BaCl 2 than when using HC1? Explain. 
What would be the disadvantage of using an excess of BaCU in order to make the reaction more nearly complete? 

Application of Principles 

1. In making each of the following transformations, indicate the reagent(s) and procedures necessary to form 
the final pure salt, in solid form, from the first named substance. Also write the net ionic equations. 

Na 2 SO, to NaCi 

MgCl 2 to Mg(OH), 

Mg(OH) 2 to MgCl, 

Bad* to Ba,(PO 4 ) 2 

K^O 4 to KC1 

2. Solutions of the following are mixed. In each case indicate the molecular formulas of any new substances 
produced. Also, at the right, give the reason for the reaction, if any, thus: no action, precipitate, weak acid, volatile 
product, etc. If necessary to heat the mixture to form the new substance, indicate this fact. 

(a) Sodium carbonate and 

sulfuric acid . . . 

(b) Silver nitrate and 

aluminum chloride 

(c) Potassium hydroxide and 

ammonium nitrate 

(d) Magnesium chloride and 

sulfuric acid 

(e) Potassium nitrate and 

ferric chloride . 

(f) Sodium sulfate and 

barium chloride . 

(g) Nitric acid and - P* 

sodium acetate , . 

(h) Phosphoric acid and 

potassium carbonate . 

(i) Nitric acid and calcium 
hydroxide (solid) . 


Strong and Weak Acids and Bases; Indicator* 

Review of Fundamental Concepts 


College Chemistry, Chapter* 79, 20 

The adjectives, strong and weak, as applied to 
acids, bases, or salts, refer to their relative degrees 
of ionization. A strong acid is "strong" that is, 
very reactive as an acid because of the high con- 
centration of hydrogen ion (H + ) that it contains 
in solution. A weak acid, although able to neutral- 
ize as much base as an equivalent amount of a 
strong acid, is "weak" because only a small pro- 
portion of its molecules are dissociated into hydro- 
gen ion and the anion. It is, therefore, less reactive 
as an acid. Correspondingly, the strength of a base 
depends on the extent to which its molecules are 
dissociated into the cation and hydroxide ion. (See 
Table XI, The Relative Concentration of Ions in 
0.1 F Solutions of Electrolytes, Appendix II.) 

Measuring H+ and OH~ Concentrations by 
Means of Indicators 

An indicator is a complex organic compound the 
bolor of which, when in aqueous solution, depends 
on the relative concentration of hydrogen and hy- 
droxide ions in the solution. With different in- 
dicators, this color change takes place at very dif- 
ferent relative hydrogen and hydroxide ion con- 
centrations. Thus, methyl violet exhibits its color 
change in a moderately acid solution, methyl orange 
in a slightly acid solution, phenolphthalein in a 
slightly basic solution, and indigo carmine in a 
solution which has a considerable hydroxide ion 
concentration. (See Table XII, The Color Change 
and pH Interval of Some Important Indicators, 
in Appendix II.) 

In this experiment, we shall first prepare a series 
of solutions of known hydrogen ion and hydroxide 
ion concentration. This may be done by making 
tenfold dilutions of 0.1 F solutions of the strong 
acid hydrochloric acid, and of the strong base 
sodium hydroxide, respectively. Since these sub- 
stances are thought to be completely ionized in 
dilute solution, the molarity of the hydrogen ion 
and of the hydroxide ion, respectively, will be the 
same as the formality of the corresponding acid or 
base solution. Solutions of extreme dilution are re- 
Squired, since the ionization of water is so slight that 

in pure water the concentrations of hydrogen ion 
and hydroxide ion are each only about 0.0000001 
M, or 10~ 7 M. 1 

We shall use this series of solutions as standards, 
to learn the colors of several indicators and the 
concentration range through which a change of 
color takes place with each indicator. Also, using 
this method, we shall determine the approximate 
degree of ionization of acetic acid and of ammon- 
ium hydroxide, each at two different concentra- 

It should be noted that the hydrogen ion or 
hydroxide ion concentration in a solution can be 
estimated closely by one indicator only if the con- 
centration lies within the color change range of 
that indicator. As an example, we may consider 
Congo red, which gives a red color in any solution 
which contains 0.00001 M H+ or less, an inter- 
mediate red-blue at 0.0001 M H+, and a blue color 
at 0.001 M H 4 " or greater. If the color is inter- 
mediate between that of the 0.0001 M H + and the 
0.001 M H+ solutions, we might prepare a series 
of standards in between these limits and make care- 
ful color comparisons with our sample. In the 
absence of such additional standards, we may pro- 
ceed as follows. If, for example, we judge the color 
to be about two-thirds of the way from the color 
of the 0.0001 M H + standard toward the color of 
the 0.001 M H + standard, we would estimate the 
sample to contain about 0.0007 M H 4 ". The series 
of expanded numbers given below may help you to 
express the intermediate values correctly. 

lO" 4 .... 0.0001 M H+ (red-blue to Congo red) 
. . 0.0005 

0.0007 (2/3 toward the blue) 
. . 0.0010 M H+ (blue to Congo red) 

1 If you are not familiar with exponential notation, review the dis- 
cussion on "Exponents" in Study Assignment A. The pH system of 
recording hydrogen ion concentration is explained in Experiment 34, 





Note especially that half way between 10" 4 and 
10~* is not lO" 3 * 8 , since exponential numbers are 
logarithmic, not linear. (The log of 0.0005 is -3.3. 
Thus, 0.0005 - 10--.) 

Dilution of Solutions 

When diluting a solution of known concentra- 

tion, the concentration after dilution will be in- 
versely proportional to the volumes before and 
after dilution. Thus, if we dilute 2 ml of 6 F HC1 
to 200 ml, the concentration will be 

Experimental Procedure 

Chemicals: 1 F HCJIA, 0.1 F HC^O* 1 F NH 4 C 2 HA, 
1 F NH 4 C1, 1 F NH 4 OH, 0.1 F NH 4 OH, 0.1 F HC1, H+ solu- 
tions from 10~ 4 to 10~ 7 M H+, OH~ solutions from 10~ f 
to 10- 7 M OH- 1 F NaCaHjOz, 1 F NaCl, 0.1 F NaOH. 
Indicator solutions as follows: Methyl violet, methyl orange, 
phenolphthalein, alizarin yellow R, and indigo carmine. 

1. The Preparation of Solutions of Known 
H+ Concentration. Prepare solutions of 0.01 M 
H+, and 0.001 M H+, by dilution of 0.1 F HC1, as 
follows. First, thoroughly clean two 100 or 200-ml 
beakers or flasks. Rinse these with tap water, then, 
with not over 5 ml of distilled water by spraying 
from your wash bottle. By means of your grad- 
uated cylinder, carefully measure 5.0 ml of 0.1 F 
HCl from the stock bottle and add distilled water 
to make a total volume of 50.0 ml. Pour this back 
and forth between the graduated cylinder and one 
of the clean flasks several times to mix it thor- 
oughly. What is now the H+ concentration of this 
solution? Leave 5.0 ml (carefully measured) of the 
solution in the graduated cylinder and make a 
second tenfold dilution by filling carefully to the 
50.0-ml mark with distilled water. Mix as before 
and transfer to a second flask. The more dilute 
solutions from 10 ~ 4 M to 10 ~ 7 M H+ respectively 
are already prepared for your use. 1 

2. The Colors of Indicators in Acid Solu- 
tions. Now prepare a series of clean 10-cm test 
tubes containing 5 ml each of the acid solutions 
ranging from 10 -* to 10 ~ 7 M H+. Add to each test 
tube one drop of methyl violet indicator. Stir each 
and note the colors 2 which are characteristic of the 
given H+ concentrations. 

1 These solutions of very low H + concentration are "buffered" so- 
lutions. That is, the substances present in the solution act to keep 
the H+ concentration in the solution very nearly constant, even 
when some additional acid or base is added. The carbon dioxide 
from the air reacts with the water, and form* too much H+ to main- 
tain a known concentration, without these buffers. 

1 The methyl violet in the stronger acid solutions will fade on long 

Save the tubes covering the color change range 
(three or four tubes) and label each. You will need 
them in sections 3 and 4. Now prepare another 
series of tubes containing 5 ml each of the acid 
solutions from 10 ~~ 1 M to 10 ~~ 7 M H + , and test each 
with one drop of methyl orange indicator. Stir each 
and note the colors which are characteristic of the 
given H + concentrations. Again save the tubes 
covering the color change, labelling each, for use 
in sections 3 and 4. For each indicator, learn the 
range of hydrogen ion concentration through which 
it changes color, and the respective colors. You will 
need to fix the appearance and color in your mind 
as well as possible, for any descriptive names used 
in your notes will help only partially. Whenever 
you want to make accurate measurements with in- ( 
dicators, it is necessary to prepare a comparison 
series with the given indicator and make actual 
color comparisons. The amount of indicator used, 
of course, has some effect on the intensity and ap- 
pearance of the color. The indicators are prepared 
so that one drop in 5 ml should be ample. An excess 
should be avoided. 

3. The H+ Concentration of an Unknown 
Solution. Take a clean test tube, labelled with 
your name, to your laboratory instructor and ob- 
tain a solution of unknown H+. Test portions of 
this solution with methyl violet and methyl orange. 
Report the H+ concentration of your unknown so- 
lution to your instructor at once (see report sheet). 

4. The Degree of lonization of Acetic Acid* 
Obtain a 5-ml portion of 1 F HC2H 3 O 2 , and test it 
with one drop of methyl violet indicator. Hake as 
accurate an estimate of the H+ concentration in 
the solution as possible, by comparison with your 
standards. If the color lies between the colors of 
two of your standards, try to estimate the H* 
concentration in between the values represented by 
these standards. What would be the color of methyl 
orange with 1 F HQjHA? Try it if you wish. What 



Compare indicator colors by 
looking lengthwise through the 
tubes against & white background. 

Label the H* concentration 
of each tube* 

FIG. 26-1. Correct technique for the observation and comparison of indicator colors. 

is the concentration of acetate ions in this same 
solution? Of un-ionized ac6tic acid molecules? 
What fraction of all the acetic acid molecules put 
in solution are broken apart into ions? (Express as 
percent ionization). 

Test also 5-ml samples of 0.1 F HC 2 H 8 O 8 with 
both methyl violet and methyl orange to estimate 
the H+ concentration. Calculate the C 2 H 3 O2~ con- 
centration and the concentration of un-ionized 
HC2HaO2. Calculate the percent ionized and com- 
pare with the percent ionized for a 1 F solution. 

5. The Preparation of Solutions of Known 
OH~ Concentration. Prepare from the stock 
0.1 F NaOH solution a 50-ml sample of 0.01 F 
NaOH by tenfold dilution as in section 1 for strong 
acids. Be careful to rinse all vessels used. Keep the 
solutions stoppered in order to prevent the absorp- 
tion of carbon dioxide from the air. As with the 
acid solutions, the more dilute basic solutions, from 
10~~ 8 to 10~ 7 M OH~", are already prepared for 
your use. They are likewise buffered solutions. The 
absorption of carbon dioxide from the air into an 
unbuffered basic solution would be even more rapid 
and troublesome than in the case of dilute acid 

6. The Colors of Indicators in Basic Solu- 
tions* This time we shall use three different indi- 
cators, namely, indigo carmine, alizarin yellow B 

and phenolphthalein, to learn to identify the hy- 
droxide ion concentration of solutions. Prepare a 
series of 10-cm test tubes containing 5 ml each of 
the alkaline solutions ranging from 10""" 1 to 10""* 
M OH"". Test each with two drops of indigo 
carmine indicator. 1 Stir each and note the colors 
and, in particular, the range of the color change. 
Keep three or four tubes which cover the range 
of the color change and label each for use in section 
7. Likewise, determine the colors and color range 
with alizarin yellow R, and with phenolphthalein, 
respectively. Keep the tubes covering the color 
change in each case for use in section 7. 

7* The Degree of Ionization of Ammonium 
Hydroxide* Test 5-ml portions of 1 F and of 0.1 F 
NKUOH with indicators to determine the hydroxide 
ion concentration in each case. What is the concen- 
tration of NH 4 +, and of undissociated NH 4 OH, in 
each case? Calculate the percent ionized in each 

8. Reactions Involving Weak Acids and 
Weak Bases. To 5 ml of 0.1 F HC1 add a drop of 
methyl violet indicator, and then gradually add 
1 to 2 ml of 1 F NaC JI 8 O 2 solution. Try the same 
experiment again except this time use 1 F NaCl in 

1 Indigo carmine indicator does not keep very well. If it ia not a 
deep blue color it must be discarded and a fresh solution prepared. 



place of the 1 F NaC 2 H 8 O 2 . Also try 1 F NH 4 C 2 H 8 2 
instead of 1 F NaC 2 H 3 O 2 . Explain the results and 
write the net ionic equation for any reaction taking 

To 5 ml of 0.1 F NaOH solution add a drop of 
alizarin yellow R and then gradually add 1 to 2 ml 
of 1 F NH 4 C1 solution. Try the same experiment 
again, except use 1 F NaCl in place of the 1 F 
NH 4 C1. Also try 1 F NH 4 C 2 H 3 O 2 instead of 1 F 
NH 4 C1. Explain the results and write the net ionic 
equation for any reaction taking place. 

9. The Electrical Conductivity Method for 
Determining the Degree of lonization. As an 
optional alternate or additional procedure, the 
same apparatus used in Experiment 16 may be 
used to compare the conductivity of 0.1 F HC 2 H 3 2 
with that of a very dilute solution of hydrochloric 
acid, prepared as follows. Fill the 10-ml graduated 
cylinder exactly to the mark with 0.1 F HC1, then 
with the aid of a medicine dropper add this solution 
drop by drop to 100 ml of distilled water, stirring 
after each addition, until this solution has the same 
conductivity as that of the 0.1 F HC 2 H 3 O 2 . The 
two solutions can be compared by raising first one, 
then the other, alternately, to make contact with 
one of the pairs of electrodes. Be sure the electrodes 
are the same distance apart, and that you immerse 
them to the same depth. Make electrical connec- 
tions only momentarily, as continuous passage of 
the current will heat the solution and invalidate 
the results. 

Note the precise volume of 0.1 F HC1 used (first 
returning any excess from the medicine dropper to 
the graduated cylinder). From this volume, and 
the volume of diluted HC1, calculate the concen- 
tration of H + in the diluted HC1, assuming com- 
plete ionization. If we assume that the 0.1 F 
contains the same H+ concentration as 

the diluted HC1, 1 we may calculate the approxi- 
mate degree of ionization of the 0.1 F HC^HaC^. 

Now replace the 0.1 F HC 2 H 3 O 2 by 1 F HC 2 H 3 O 2j 
and continue to add 0.1 F HC1 from the graduated 
cylinder to the diluted HC1 solution until its con- 
ductivity equals that of the 1 F HC 2 H 3 O 2 . Calcu- 
late the degree of ionization as before. 

The degree of ionization of 0.1 F NH 4 OH and of 
1 F NH 4 OH may be determined in an analogous 
manner. For these, add a measured volume of 0.1 F 
NaOH to 100 ml of distilled water until the con- 
ductivity is equal to that of the solution being 
tested. 2 Calculate the results as in the preceding 

10. Review Questions on lonization and 
Ionic Substances. It is very much worth while 
to form the habit of thinking of the various strong 
and weak acids, bases, and salts, in terms of the 
actual molecular or ionic species largely present in 
their aqueous solutions. Study the examples as 
given in the section of the report sheet headed 
Application of Principles, then calculate the con- 
centration of the principal substances present, in 

the problems which follow. j 

_ J 

1 Strictly speaking, the ionic concentrations are not quite equal 
when the conductivities are equal The equivalent conductances at 
25 C of the separate ions, in reciprocal ohms, are: H + 350, C 2 H3O2~ 
40.8, Cl~ 75.5 (data from Lange, Handbook of Chemistry, 7th ed., 
F. 1417). The conductivities of equal concentrations of the ions of HC1 
and of HC 2 H 8 O 2 will be in the ratio of (350-|-75.5)/(35Q-f 40.8), 
which equals 1.09 to 1. Therefore, you may correct the calculated 
value of the concentration of H+ in the acetic acid by multiplying 
by the factor 1.09. 

2 Here, also, we may correct our calculations for the slight differ- 
ences in the conductivities of sodium ion and ammonium ion. The 
equivalent conductances at 25 C are: NH 4 + 74.5, Na+ 50.9, and OH~ 
192. The conductivities of equal concentrations of the ions of NaOH 
and of NH4OH will be in the ratio of (50.9-f-192)/(74.5-f 192), which 
equals 0.91. Therefore you may obtain a corrected value for the OH" 
concentration in the NH 4 OH solutions, by multiplying by the factor 

REPORT: Exp. 26 

Strong and Weak Acids 
and Bases, 



Locker Number- 

1. The Preparation of Solutions of Known H + Concentration. The hydrogen ion concentration in my 
solution prepared by diluting 5.0 ml of 0.1 F HC1 to a volume of 50.0 ml is 


2. The Colors of Indicators in Acid Solutions. The indicator colors obtained with various concentrations 
of hydrogen ion in strong acids are as follows: 


Methyl Violet 

Methyl Orange 

10- 1 M H+ 

10- 2 M H+ 

10~ 3 M H+ 

10- 4 lf H + 

10- 6 MH + 

10- 6 J/H + 

10~ 7 M H+ 

3. The H+ Concentration of an Unknown Solution. My unknown H+ solution No._ 
following indicator colors: 

gave the 

Methyl violet- 

Methyl orange. 

The hydrogen ion concentration is therefore: 
Instructor's approval of report on unknown . 

4. The Degree of lonization of Acetic Acid. The color of acetic acid solutions with indicators, and the esti- 
mated molal concentrations of the substances present are as follows : 







Concent ration 




1 F HC 2 H 3 O 2 





Q.1FHC 2 H 3 2 





The calculation of the percent ionization is as follows: 

Does the degree of ionization of weak electrolytes for example, HC2H 8 O2 increase or decrease or remain the 

same, as the dilution is increased? (This is a general rule.) 


5. and 6. The Preparation of Solutions of Known OH**" Concentration. The Colors of Indicators in 
Basic Solutions. The indicator colors obtained with various concentrations of hydroxide ion in strong bases are as 


Indigo Carmine 

Alizarin Yellow R 


10- 1 M OH- 

IO- 2 M OH- 

IO- 8 M OH- 

IO* 4 M OH- 

IO- 8 M OH- 

IO" 6 M OH- 

IO" 7 M OH- 

7. The Degree of lonization of Ammonium Hydroxide. The color of ammonium hydroxide solutions 
with indicators, and the estimated molal concentrations of the substances present are as follows : 



Yellow R 









1 F NH 4 OH 





O.I F NH 4 OH 





The calculation of the percent ionization is as follows: 

8. Reactions Involving Weak Acids and Weak Bases. The results obtained are indicated in the following 

Solutions Mixed 

Indicator Used and Color CJiange 

5 ml HC1 with NaCjH,O 2 added 

5 ml HC1 with NaCl added 

5 ml HC1 with NH 4 C 2 H 8 O 2 added 

5 ml NaOH with NH 4 Cl added 

5 ml NaOH with NaCl added 

6 ml NaOH with NH 4 C 2 HsO 2 added 



ttepon on &xp. # 

Explain in words any differences in the effects of the salts on HC1, and write the net ionic equation for any 
reaction occurring: 

Explain in words any differences in the effects of the salts on NaOH, and write the net ionic equation for any 
reaction occurring: 

9. The Electrical Conductivity Method for the Determination of the Degree of lontzatton. 

0.1 F HCJhOt 

1 F HCJltO* 

0.1 F NHtOH 

1 F NH<OH 

Volume of 0.1 F HC1 (or NaOH for the NH<OH 
solutions) added to 100 ml H 2 O 





Calculated molarity of the H+ (or OH~) 





Degree of lonization 





The calculation of the percent ionization is as follows; 

Again compare the degree of ionization and the concentration of the solutions: 

Application of Principles 

In interpreting a solution as to the actual form (ions or molecules) in which the dissolved substances aw 
present, the following points should be noted: 

(1) Are the substances present largely in ionized form, or as molecules? 

(2) Do any of the ions or molecules react with one another? How many moles of reactants used, and of product* 
formed, are there? (Determine from the equation for the reaction.) 

(3) In calculating the final concentration of the substances in solution, has there been any dilution effect, either by 
adding water, or due to the mixing of two or more solutions ? 

Examples: What is present, and at what concentration in: 

(I)O.IFHCI? 0.1 M H+ 0.1 M Cl~ 

(2) a mixture of 1 gfw of HC1, and 0.5 gfw of NaCl, in 250 ml of solution? 

There will be 1 mole H+, 1.5 mole Cl~, and 0.5 mole Na+, in 250 ml (J liter) of final solution. The concentra- 
tion in moles per liter will therefore be four times the number of moles, or ... 4 M H+, 6 M Cl~, 2 M Na"* 


(3) a mixture of 300 ml of 1 F NH 4 OH and 200 ml of 2 F HN0 3 ? 

First figure the number of moles of each substance before reaction, then from the equation 

NH 4 OH + H+ 5P^ NH 4 + + H 2 O 

1 mole 1 mole 1 mole 

0.3 mole 0.3 mole 0.3 mole 

calculate the number of moles of each substance produced, or remaining in excess. In this case 0.3 mole of NH 4 OH 
will require 0.3 mole of II +. Since 0.4 mole of H + was available, 0.1 mole of H + will remain in excess. Finally cal- 
culate the concentrations. The following form is useful to summarize the calculations: 

300mllFNH 4 OH 
200ml2FHN0 8 

Final volume: 500 ml 

before reaction 

0.3 mole NH 4 OH 
0.4 mole H + 
0.4 mole NO 3 - 

after reaction 

0.3 mole NH 4 + 
0.1 moleH + 
0.4 mole N(V 


0.6 M NH 4 + 
0.2 M H+ 
0.8 M NO 8 - 

Problems. For each of the following, list at the right the principal substances present. If in solution, give the 
molarity; and if a solid or a gas, give the number of formula weights or moles formed. Indicate small amounts by 
"Trace." Any volume changes due to the formation of water may be neglected. In items (5) to (8) inclusive show 
your method, in the space below each item, and also give the net ionic equation for any reaction. 

(1) 0.01 F Ba(OH)2 (assume complete ionization) 

(2) A mixture of 1 gfw Na2SO 4 and 2 gfw K^SQi in a liter of solution 

(4) 0.1 gfw NH 4 OH in 100 ml of solution __ 

(5) A mixture of 0.1 gfw HC1 and 0.1 gfw Ba(OH) a in a liter of solution .... 

(6) A mixture of 250 ml of 0.1 F NH 4 OH and 250 ml of 0.1 F HC 2 H 3 O 2 (neglect 

(7) One liter of 0.5 F CuS<X saturated with H 2 S gas 

(8) 2.4 g (0.1 g at) of magnesium metal dissolved in 250 ml of 2 F HC1 


The Titration of Acids and Bases. 


Co/feg* Chemistry, Chapter 20 

Review of Fundamental Concepts 

The Equivalent Weights of Acids and Bases 

In the neutralization of an acid with a base, one 
hydrogen atom (either as the free ion, H+ or as a 
reactive hydrogen atom in the molecule of a weak 
acid) reacts with one hydroxide group (as free ion 
or as radical in the molecule) to form one molecule 
of water. These amounts are therefore equivalent. 
Thus we may calculate the equivalent weights of 
acids and of bases by dividing the respective for- 
mula weights by the number of hydrogen atoms or 
hydroxide groups in the formula, provided that all 
these take part in the reaction under consideration. 
Consider the reactions 

HaSO 4 + Ca(OH) 2 
H 2 S0 4 + 2 NaOH 
H 2 SO 4 + NaOH 

-* GaSO 4 + 2 H 2 O (1) 

> Na 2 SO 4 + 2 H 2 O (2) 

- NaHS0 4 + H 2 (3) 

In equations (1) and (2), the equivalent weight 
of H 2 SO4 is one-half of its formula weight (^ of 
98 g, or 49 g), since both hydrogen atoms are 
neutralized. The equivalent weight of Ca(OH) 2 is 
half of its formula weight (J^ of 74 g, or 37 g), 
while the equivalent weight of NaOH is its formula 
weight (40 g). In equation (3), only one of the 
two hydrogen atoms in H 2 SO4 has been neutral- 
ized by the one formula weight of NaOH, which is 
obviously one equivalent. The equivalent weight 
of H 2 SO 4 in this reaction is, therefore, the same as 
its formula weight, 98 g. 

Sometimes chemists refer to the equivalent 
weight of an acid or base without reference to a 
particular reaction, in which case they generally 
infer the maximum number of equivalents per for- 
mula weight. 

Normality as a Unit of Concentration 

We shall define a new unit of concentration, 
which is related to molarity and formality in the 
same way that the equivalent weight is related to 
the mole and the formula weight. The normality of 
a solution expresses the number of equivalents of 
solute per liter of solution. The defining equation is 


equivalents of solute 
liters of solution 

orN - 

V (liters) 

One advantage in using this unit is that equal 
volumes of solutions of the same normality are 
equivalent chemically and just react with one 
another. This is not always true when concentra- 
tions are expressed in molarity or formality, since 
reactions do not always take place one formula 
weight to one formula weight. The following ex- 
amples will illustrate the meaning of normality. 
Study also Figure 27-1 to review the distinction 
between a unit of quantity, such as the equivalent 
weight, and the corresponding unit of concentra- 
tion normality. 

Example 1. How many equivalents, and how 
many grams of NaOH, are there in 200 ml of 0.300 
From the defining equation, N = equiv/V (liters), 

equiv - # X V - 0.300 X 0.200 liter - 0.0600 equiv. 

And to express this as grams, 

40 o NftOH 

0.0600 equiv X g - 2.40 g NaOH. 


Example 2. What is the normality of a solution 
which contains 1.11 g Ca(OH)a dissolved in 2.00 
liters of solution? First express 1.11 g Ca(OH)2 as 
equivalents. The formula weight is 74.0 g. Since 
there are two replaceable OH groups, the equiva- 
lent weight is 74.0/2 or 37.0 g, and the number of 
equivalents are 

Q *' ,- g . - 0.0300 equiv. 
37.0 g/equiv ^ 

From the defining equation, we may then calculate 
the normality: 

- '>* 

Example 3. What is the normality of 3 F H 2 SO 4 , 
when used in the usual manner where both hydro- 
gen atoms in the formula react? 

3 gf w HaSO 4 2 
liter gfw 


or 6 N 




Na 4 


'"* ("> ' A 

** V** ,,:, ; -J 

.*r... % 


I equiv in 1 liter 2 equiv in 2 liters 0.5 equiv in 0.5 liter 

The above solutions contain different amounts of solute, but are at the same concentration, namely 1 N, calculated thus: 

. T ,. A 1 equiv 2 equiv 0.5 equiv , .. 

Normality ~- -^ -^rFT~ 1 ^V 

1 liter 2 liters 0.5 liter 

2 equiv in 1 liter 4 equiv in 2 liters 1 equiv in 0.5 liter 

These three lower solutions, likewise, contain different amounts of solute, twice as much, respectively, as above, and are 
each 2 N in concentration; 

VT i A 2 equiv 

Nonnahty=T __ 

4 equiv 1 equiv 
2 liters ** 0.5 liter " 

FIG. 27-1. To illustrate the meaning of normality as a unit of concentration. In this drawing, 1 equivalent 
is represented by 8 Na+ ions and 8 OH~" ions, at an enormous scale. 

Tftratfon and Standard Solutions 

The pro cess of titration consists of the gradual 
addition of a solution of known concentration to 
a measured volume of a solution of unknown con- 
centration, or to a weighed amount of sample dis- 
solved in water, until the same number of equivalents 
of each substance has been used. This point is de- 
tected by means of a suitable indicator. Phenol- 
phthalein, which is red in basic solution and color- 
less in acid solution, is often used for acid-base 
titrations. The end-point, or point at which a per- 

manent color change occurs after thorough stirring 
of the solution, is very sharp. Only a drop, or frac- 
tion of a drop, of excess solution will bring about 
the color change. Volumetric analysis can there- 
fore be performed with almost the same precision 
as the gravimetric process of weighing. The solu- 
tion of known concentration which is used as the 
reference standard is called a standard solution. 

The relative concentrations of the two solutions, 
and consequently the normality of the unknown 
solution, can be determined from the relative vol- 



umes of the solutions used. Thus, if two solutions, 
(1) and (2), are titrated to the end-point, we have 
the fundamental relationship, 

equivalents of solution (1) * equivalents of solution (2) 

and by the defining equation of normality, namely 
N = equiv/V, we note that the equivalents of 
solution (1) = NiVi, and of solution (2) = NzVt. 

NiVi = MY* or transposing, -~ = ~. 

The relative volumes required thus will be in- 
versely proportional to the normalities of the solu- 

It should be noted that while the number of 

grams of reacting substances are seldom equal and 
the number of moles of reacting substances are 
only sometimes equal, the number of equivalents of 
reacting substances are always equal by definition* 
For this reason, concentrations are expressed in 
normalities in the equation, NiV\ = NjV 2 . If any 
three of the quantities are known, the fourth can 
be calculated. 

Example. What is the normality of a sodium 
hydroxide solution, if 43.25 ml of it are required to 
neutralize 50.00 ml of 0.500 N oxalic acid solution ? 
From the equation NiVi = A^Vs, by transposing, 

V a 

0.500 N X 50.00 ml 

0.578 N NaOH. 

Experimental Procedure 

Special supplies: analytical weights, 250-ml volumetric flask, 
2 burettes, 1 burette clamp, double. 
Chemicals: oxalic acid crystals. 

1, The Preparation of a Standard Acid Solu- 
tion. In this experiment, we shall prepare as a 
standard solution, 250 ml of 0.5 N oxalic acid. 
This substance is readily obtained as pure crystals 
of definite composition, of the formula E^^O^ 
2H 2 O. The water of crystallization is a part of the 
substance as weighed and must, of course, be in- 
cluded in your calculation of the equivalent weight. 
The hydrogen of the water of crystallization does 
not contribute to the acidity of the solution, so that 
from the formula above, there are two replaceable 
hydrogen atoms per mole. 

Calculate the number of grams of oxalic acid 
crystals needed to prepare 250 ml of 0.5 N solu- 
tion. (Note: First find the number of equivalents 
needed, then the weight.) Check your weight cal- 
culation with the instructor to be sure you have 
made no error. Weigh an amount of oxalic acid 
crystals close to this amount, recording the exact 
weight. Do not waste time at the balance attempt- 
ing to adjust the amount to the exact weight cal- 
culated. You can calculate the exact normality 
later, if you know the exact weight used, by divid- 
ing the number of equivalents weighed out by the 
volume of the solution, 0.250 liter. 

Place a clean 250-ml volumetric flask on a clean 
sheet of paper on your desk and, with the aid of 
a stirring rod, very carefully transfer the weighed 

crystals, without loss, to the volumetric flask* 
Rinse the beaker and stirring rod with distilled 
water, and add these rinsings to the flask. Now add 
distilled water to the flask until it is about % full, 
and mix gently by rotating the flask until solution 
is complete. Add more water until the bottom of 
the meniscus of the solution comes to the gradua- 
tion mark. Be careful not to go past the mark. The 
last water may be added dropwise with a medicine 
dropper. Stopper the flask and mix the contents 
very thoroughly for several minutes by repeatedly 
inverting and rotating the flask. Hereafter, do noth- 
ing that will alter the concentration of this stand- 
ard solution, such as adding it to a wet vessel, or 
permitting evaporation of the solution. Calculate 
the normality of your standard oxalic acid solu- 
tion. Put the solution just prepared into a clean 
dry flask or bottle. Stopper this container to pre- 
vent any evaporation. Label it with the name of 
the substance, concentration, date, and your ini- 
tials. Save the solution, for it may be used in the 
following experiment also. 

2. The Preparation of an NaOH Solution 
for Standardization. Dilute 40 nil of the ap- 
proximately 6 N desk reagent of NaOH with 440 
ml of water. (Desk reagents are not made up with 
quantitative accuracy to the indicated strength, 
hence your diluted solution will only be approxi- 
mately the normality desired, and must be stand- 
ardized against your oxalic acid.) Place your basic 
solution in a bottle, close with a rubber stopper, 



filled with 
sodium hy 

filled with 
oxalic acid 

for waste 

To release the 
liquid pinch the / 
tube just above / 
the glass bead. 

Give the liquid 
a rotary motion 
before starting 
to titrate. 

Rinse the walls 
of the flask 

air bubbles 
from the 
tip Hke 

the rubber 
tube --*. 

the drop 
the tip. 

Near the 
end point 
the trail of 
color from 
each drop 
is quite 

FIG. 27-2. Techniques in the process of titration. 



and mix very thoroughly by repeatedly inverting 
and rotating the solution. Label the bottle and save 
it for use in the next two experiments. Calculate 
the approximate normality of your NaOH. 

3. The Titration of Your NaOH Solution. 
Clean two burettes thoroughly. Soap solution and 
a long burette brush may be used, if needed. Water 
should flow freely, without forming droplets, over 
the inside surface of the burette. After rinsing 
thoroughly with tap water and draining, rinse one 
burette with a 5-ml portion of your oxalic acid 
solution. Reject the rinsings. Rinse a second time, 
again rejecting the rinsings by running them 
through the burette tip. Fill the burette with the 
oxalic acid solution, and make sure that there are 
no air bubbles in the tip of the burette. 1 The 
burette has a rubber delivery tube containing a 
glass bead valve. This is operated by placing the 
fingers on the tube just slightly above the bead and 
squeezing the tube so as to make a channel in the 
tubing around the bead, through which the liquid 
from the burette may flow. (Do not press the rub- 
ber tube below the bead, as this would cause it to 
act as a medicine dropper, expelling liquid, and 
sucking in an air bubble. This would cause an in- 
correct reading.) If the burette continues to drip 
after releasing the pressure, return it to the stock- 
room and have the rubber tubing replaced. 

Fill the second burette with your NaOH solu- 
tion, taking the same precautions as before in re- 
gard to rinsing it. Adjust the level of the liquid in 
both burettes so that it is at, or slightly below, the 

1 Study Figure 27-2 to learn various convenient techniques in the 
nrocess of titration. 

zero mark, catching the discharged solution in a 
beaker. Touch off the drop which adheres to the 
tip at the side of the beaker. When you read this 
initial burette volume and also the final volume, 
estimate the burette reading to the nearest tenth 
or fifth of the smallest division. With a little prac- 
tice, you should be able to read a burette to 0.02 ml. 

Perform the titration as follows. Run about 20 
ml of the oxalic acid solution into a clean, rinsed 
200 to 300-mi Erlenmeyer flask. Add 2 drops of 
phenolphthalein indicator. Now run in NaOH solu- 
tion, fairly rapidly at first, stirring by a rotary 
motion of the flask. As the end-point is approached, 
which is indicated by the fact that the pink color 
does not disappear so rapidly and trails around 
the solution as it is rotated, add the NaOH solu- 
tion slowly, drop by drop, until the final drop, 
after stirring, leaves a slight pink color which per- 
sists in the solution at least 15 seconds. If you have 
gone past the end-point, add a few drops of oxalic 
acid, and again approach the end-point. The last 
drops adhering to both burette tips may be touched 
off in making the final adjustment. The wash bottle 
will be found convenient in rinsing down the walls 
of the flask. Read and record the final volume of 
solution in each burette. Make two additional ti- 
trations of your solutions. Calculate the normality 
of your NaOH solution, and obtain the instructor's 
approval. If concordant results are not obtained, 
additional titrations may be made. 

Save both the unused oxalic acid and sodium 
hydroxide solutions for the next experiment. They 
should be kept stoppered. 

Supplementary Drill 

Note: Use this material as needed, after you have completed the report sheet, to provide additional drill on calculations 
involving quantity and concentration of solutions. It need not be handed in unless called for. Show your method in each case. 

1. How many grams of each of the following are required to just react with 200 ml of 0.35 N HNO? 

(a) Mg(OH) 2 . . . g (c) Na 2 COi .... g 

(b) CaO .... g (d) NaHCOi .... g 

2. What is the concentration of each of the following? (Note that dilution does not change the number of equiva- 
lents of solute. Use ViNi V 2 N 2 , or ViFi - VjF 2 .) 

(a) 25 ml of 0.4 N HC1, diluted to 500 ml volume N 

(b) 10 ml of 3 F H 2 SO 4> diluted to 15 ml volume F 

(c) 10 g of NaOH, dissolved in 250 ml H 2 and evaporated to 100 ml final volume N 

3. To what volume must each of the following be diluted to give the concentration called for? 

(a) 20 ml of 6 N HN0 3 , to give a 0.1 Absolution ml 

(b) 38 ml of cone (18 F) H 2 SO 4 to give a 0.2 F solution ml 

(c) 36 ml of cone (18 F) H 2 SQ 4 to give a 0.2 N solution ml 

(d) 100 ml of glacial acetic acid (17 F) to make it 6.0 AT ml 

4. 60.0 ml of 0.2 N NaOH is mixed with 40.0 ml of 0.05 N HC1. What is the concen- 
tration of the remaining QH~? (Final volume is 100 ml.) N 

5. 1.00 g of Ca metal is dissolved in 200 ml of 1.00 N HC1. What is the concentration 

of the remaining 11+ after complete solution of the metal? N 


The Equivalent Weight of a 

Solid Acid. Other Volumetric Analyse! 


Co//e*e Cfombfry, ttapter 20 

Review of Fundamental Concepts 

This experiment makes further use of principles 
developed in the preceding experiment. Make sure 
that you know: 

(1) How to find the number of equivalents of 
the solute in a given volume of solution of known 

(2) The relationship between the weight, equiv- 
alent weight, and the number of equivalents for a 
given quantity of a substance, 

(3) The relationship between the numbers of 
equivalents of substances that react. That is, if you 
have 2 equivalents of NaOH, how many equiva- 
lents of HaSOi would be required for neutraliza- 

These relationships are all summarized by the 
series of equations: 

no. of equiv 

equiv wt 

Experimental Procedure 

Special supplies: analytical weights, 2 burettes, vial of un- 
known solid acid. Bring your own samples of vinegar, citrus 
fruits or juices, and baking soda. 

1. Equivalent Weight of a Solid Acid. Obtain 
from the assistant a numbered vial containing a 
sample of a solid unknown acid. Record this num- 
ber on your report sheet at once. Weigh out three 
1 to 2 g samples of the acid, as precisely as you can, 
into 250-ml Erlenmeyer flasks. Follow the sug- 
gestions of Figure 28-1. The second weight for the 
first sample becomes the first weight for the second 

sample, and so forth, so that only four weighings 
are needed for three samples. Record each weight 
as obtained, and number the flasks to avoid con- 
fusion. Dissolve the samples by adding to each 
flask about 50 ml of distilled water. The exact 
amount of water is not important; more may be 
added if needed to dissolve the sample. 1 

1 A difficultly soluble sample may be allowed to dissolve during the 
titration, as the base is added. Proceed slowly when near the end- 
point, stirring constantly, and be sure the solution is complete before 
you cease titrating. 


Tap out 
some of the 


Weigh the vial and 
its contents to the 
nearest o.ooi g. 


Weigh again to 
find out how much 
was removed. 

Fio. 28-1. Steps in weighing a sample from a test tube which is used a* a weighing bottle. 




Clean, rinse, and fill a burette with your stand- 
ardized NaOH solution. 1 Titrate each sample of 
acid by running the NaOH solution directly into 
the flask in which the sample was dissolved, using 
two drops of phenolphthalein in each flask as an 
indicator. Titrate to the appearance of the first 
faint, permanent pink color, after stirring. If you 
pass the end-point slightly in this first titration, 
you can calculate quickly the approximate volume 
needed for successive titrations from the first vol- 
ume and the relative weights of the samples. Then 
add the first part of the necessary volume of NaOH 
rapidly, and approach the end-point slowly by 
dropwise additions. From the data obtained, cal- 
culate the equivalent weight of the unknown acid. 

2. Other Volumetric Analyses. Note: No 
report sheet for this part has been provided. Your 
report will be judged by the care with which you prepare a 
suitable form for the entry of all needed data in good 
logical order, and for the extension of all calculations. Do 
this after studying the procedure, but before performing 
the experiment. Carry out one or more of these pro- 
cedures, as directed by the instructor, and as time per- 

(a) Determination of Acetic Acid in Vinegar. 
Rinse and fill a burette with a light-colored sample 
of vinegar. (A highly colored vinegar will need to 
be treated first with adsorbent charcoal to remove 
the coloring, and filtered through a dry filter.) Fill 
a second burette with your standard sodium hy- 
droxide solution, as usual, and titrate the two so- 
lutions, using phenolphthalein indicator. Check 
the results by a second titration. 

From the average density of vinegar, 1.005 g/ml, 
calculate the weight of a liter of the vinegar. The 
weight of the solute in a liter of the solution can 
be obtained from your experimentally determined 
normality and the equivalent weight of the solute. 
Assume that all the acidity of the vinegar is due 
to acetic acid (HC2H 3 O 2 ). How does your result 

1 In case of any doubt as to the normality of your NaOH solution 
prepared in Experiment 27, you may restandardize it against 20-ml 
portions of a standardized H 2 SO 4 solution provided by the instructor 
for this purpose. Use phenoiphthaleiu as the indicator. 

compare with the legal requirement, which specifies 
that vinegar must contain not less than 4% acetic 

(b) Determination of Citric Acid in Citrus Fruit. 
Squeeze out about 100 ml of lemon, orange, or 
grapefruit juice. Determine its density by means 
of a hydrometer or by carefully weighing on the 
agate bearing triple beam balance, a carefully 
measured volume, 50 or 100 ml. Place the juice in 
one burette, and your standard NaOH solution in 
a second burette. Titrate the solutions, using 5 
drops of phenolphthalein indicator. If orange juice 
is used, keep a sample out for comparison, as the 
first appearance of the red color of the indicator 
is rather difficult to detect in such a highly colored 

Calculate the acidity of the citrus juice, as per- 
cent citric acid in the sample. The formula for 
citric acid is HsCeHoO?. 

(c) Determination of the Purity of Baking Soda. 
Weigh out accurately two 1.5 g samples of baking 
soda. Place in Erlenmeyer flasks and dissolve each 
separately in water. Obtain about 100 ml of a stand- 
ardized H2$O 4 solution of about 0.5 N strength 
from the instructor and fill a burette with it in the 
usual manner. 2 Add 2 drops of methyl orange in- 
dicator to each solution and titrate to the end- 
point. A dilute solution of methyl orange in a basic 
or neutral solution is straw yellow. In an acid, it 
is pink. The end-point is considered the point at 
which the solution begins to turn from yellow to 
orange (not yet pink); that is, when a slight 
amount of pink color appears with the yellow. 

Write the equation for the reaction of sodium 
bicarbonate, NaHCO 3 , with H 2 SO 4 . From the vol- 
ume of the standard H 2 SO4 used, the equivalents, 
and hence the grams, of NaHCOa may be calcu- 
lated. Express the result as percent of NaHCOs in 
the baking soda. 

* A weak base such as baking soda cannot be titrated with a weak 
acid such as your oxalic acid solution, hence the use of a sulfuric 
acid solution. When a weak base is neutralized by a weak acid, the 
color change of an indicator is so gradual that quantitatively ac- 
curate results cannot be obtained. 

REPORTS Exp. 28 

The Equivalent Weight of a Solid 
Acid, Other Volumetric Analyses 



Locker Number^ 

The number of my unknown acid sample is- 


Sample Number 




Sample vial, 1st weight 




2nd weight 

Weight of sample 




NaOH burette, 2nd reading , 




1st reading 

Volume of NaOH 




Normality of standard NaOH N 





Equiv of NaOH to titrate 
the solid acid 




Equiv of solid acid in the 




Weight of one equivalent of 
solid acid 




Average valu 



Report your value to the instructor, who will give 
you the formula of your unknown solid acid . 

Calculate your percentage error. (This is a check 
of your accuracy, in both Exp. 27 and Exp. 28.) 



1. 0.613 g of an acid required the addition of 11.70 ml of 0.382 N NaOH to neutralize the 
acid completely. What is the equivalent weight of the acid? 


2. 1.24 g of a solid basic substance is dissolved in 50.0 ml of water and titrated with 40.2 
ml of 0.642 N HC1. What is the weight of one equivalent of the basic substance? 

3. A solid acid has an equivalent weight of 59.0 g. Calculate the weight of a sample of this 
acid which would be titrated with 45.0 ml of 0.516 N NaOH. 

4. How many ml of 0.643 N NaOH are required to titrate 0.784 g of an acid whose equiva- 
lent weight is 63.0 g? 

5. 0.454 g of an acid of equivalent weight 137 g, requires 39.2 ml of an NaOH solution for 
titration. What is the normality of the NaOH solution? 


Report an Exp. 8, Sheet 


6. 45.0 ml of an NaOH solution required 40.0 ml of a 0.0725 N HsSO 4 solution for titra- 
tion; 0.462 g of a solid acid required 35.0 ml of the NaOH solution for neutralization. 
What is the equivalent weight of the solid acid? 

7. Given a solution of a substance, as indicated in the 1st column, with certain other data given in succeeding 
columns. From this data calculate the corresponding values to fill in the blank spaces. 







No. of 

No. of 






Q (OH), 





H,P0 4 









Exercises on Experimental Errors 

(Study Appendix I, Experimental Errors.) 

1. Underline the correct answer in the following: 

(a) A good analyst 

(1) has no need to check his work. 

(2) consistently runs checks on important experiments. 

(b) A student in titrating ran out of his standardized NaOH solution. To continue the titration, he should 

(1) use the 6 N NaOH solution from the reagent bottle on the desk as a standard. 

(2) borrow a neighboring student's standard NaOH solution. 

(3) make up some more NaOH solution and standardize it. 


2* Why is the volume in which the solid sample is dissolved not included in your calculations? 

3. A number of ways of carrying through the experiment just completed have been suggested by students. Three of 
the methods suggested are: 

(a) One larger quantity of the unknown acid could be weighed. This could be dissolved in water and made up to 
a known volume, and shaken, well. Measured portions of the solution could then be titrated. 

(b) The method you have used, to weigh three samples separately and titrate each separately, could be 

(c) The weights of the three samples could be added together, and the three volumes of solution added together, 
thus reducing the calculations to one computation instead of three. 

From the above descriptions of the three suggested methods, underline the correct answer in the following: 

(1) The method that would best show up any lack in ability to titrate correctly would be: 

(a) (b) (c) (None of these) 

(2) The method (s) that would best show up any possible error in the student's technique in the entire experi- 
ment would be: 

(a) (b) (c) (None of these) 

(3) The method (s) that would fail to show up any error either in weighing or in titration would be: 

(a) (b) (c) (None of these) 

4. Underline the correct answer in the following: 

(a) A student titrated three 20.00 ml portions of standardized EbSC^ solution with his NaOH solution. The 
volumes of NaOH solution used in the three trials were: 31.85 ml, 31.90 ml, and 31.85 ml. These values 
indicate a high degree of 

(1) accuracy, 

(2) precision. 

(b) A student determined the value of the equivalent weight of a solid acid as 157.2 grams. The true value was 
157.5 grams. This indicates a high degree of 

(1) accuracy, 

(2) precision. 

5. Calculate the percentage error involved in the above case of the solid acid, question 4(b) 


Some Chemical Properties 

of Chromium and Manganese. 


College C/iemisfry, Chapter 29 

Review of Fundamental Concepts 


o r 

I I 
I I 

Chromium compounds: 

Acidic Basic 
+6 O 2 O7 CrO 4 
+3 Cr+++ Cr(OH) 4 - 
H-2 Cr++ 


Manganese compounds: 
+7 MnO 4 ~ 

+6 MnO 4 ~ 

4-4 MnO 2 , MnO(OH)i 

+3 Mn+++, Mn(OH), 

+2 Mn++ 


Strong oxidizing agents. Bichromate ion is orange, cbromate ion is yellow. 
Amphoteric. Chromic ion is green to violet. Chromic hydroxide complex ion Is green. 
An uncommon ion, because it is such a strong reducing agent that it reduces water 
to hydrogen gas. 
The metal. 

Permanganate ion, purple. Strong oxidizing agent, reduced to Mn 4 "* hi acid solu- 
tion, or to MnO2 (sometimes to MnO 4 ) in neutral or basic solution. 
Manganate ion, green. Easily reduced to manganese dioxide. 
Brown as precipitated from solution. 
Mn 4 * ++ is unstable, gives Mn"* + and MnOa. 

Colorless in solution, pale pink as solid inanganous salts. Mn(OH)s is oxidized by 
air to Mn(OH) a . 
The metal. 

In the previous experiments on oxidation-re- 
duction, we have seen that during oxidation a sub- 
stance is changed from a lower oxidation state to 
a higher oxidation state with the attendant loss of 

Cr + + M> + ++ +0~(an oxidation half-reaction). 

During reduction, a substance is changed from a 
higher oxidation state to a lower oxidation state 
with the attendant gain of electrons: 

Mn +4 "+20~" *-Mn(a reduction half -reaction). 

In order to give you some actual experience with 
reactions involving changes in oxidation states, we 
shall consider the chemistry of two of the transi- 
tion metals, chromium and manganese. These ele- 
ments are located in the first long period of the 

periodic table, chromium in Group VI and man- 
ganese in Group VII. They are capable of forming 
bipositive and terpositive ions as typical metals 
do, and, in addition, tend to coordinate with oxy- 
gen to form oxides or negative complex ions in 
which they have higher oxidation states. 

Before beginning the experiment, the student 
should acquaint himself with the principal oxida- 
tion states of chromium and manganese and with 
the formulas, names, and colors of the ions or 
compounds which exist. This information may be 
found in College Chemistry and also in the 
charts above. Since most of the reactions which will 
be observed are concerned with changes in oxida- 
tion states, the student should also review the 
methods available for balancing oxidation-reduc- 
tion reactions. 

Experimental Procedure 

Chemicals: chromium metal, manganese metal (small chips). 
CrOs, MnO 2 , (NH^OzQj, KOH, KClOj, NajSOj, NaBiO 8 , 
KMnO 4 , powdered Zn, FeS(V7H 2 O, 0.1 F Mn(NO 8 )2, 
0.1 F CrCNO*)*, 2 F NaA 0.1 FBaCk 0.1 F HgCU, 0.1 F 
CuSOi, 0.1 F KBr, 0.1 F KI, 1.0 F K 2 CrO4, HA 3% 
solution, Cl* water, C<X 

The Chemistry of Chromium 

1. Oxidation State CV; Properties of Metallic 

Select a small piece of metallic chromium about 
the size of the head of a pin. Note some of its 
physical properties and also refer to the text 
for additional characteristics. List these on the 
record sheet. 

Determine the relative activity of the metal by 
performing the following experiments. Cover the 
metal with about 1 ml of 6 F HNO 8 in a 10-cm test 




tube. Warm the test tube gently and watch for any 

Pour off the dilute nitric acid and wash the 
metal several times with water. Decant the wash 
water and add about 3 ml of 6 F HC1. Warm 
gently for several minutes, tap the tube, and watch 
for the rather sudden beginning of a reaction. Test 
the gas evolved for inflammability. The colored 
ion first formed is the chromous ion, but this is 
readily oxidized to chromic ion by exposure to air. 
The green chromic ion formed is quite probably a 
coordination complex which includes one or two 
chlorides substituted for some of the water mole- 
cules in the hydrated complex ion. 

Pour off the green solution and save it for Part 
2. Wash the remaining piece of metal with water 
several times. Cover it with distilled water and 
warm it to see if chromium will replace hydrogen 
from cold or hot water. (See Appendix II, Table 

2. Oxidation State +2; The Chromous Ion 

Add a small amount of powdered zinc to the 
hydrochloric acid solution of chromium obtained 
above. Warm it gently if necessary to initiate the 
reaction, then stopper it with a loose plug of cot- 
ton. After the reaction has proceeded for several 
minutes, withdraw some of the solution and add it 
quickly to a test tube containing 2 or 3 ml of 0.1 F 
HgCl2. Observe carefully the product which is 
formed. Add another portion of the solution con- 
taining the chromous ion to a test tube containing 
2 or 3 ml of 1.0 F CuSO 4 and observe the result. 
When metallic chromium is dissolved in dilute 
sulfuric acid, the chromous ion first formed is not 
so readily oxidized as it is in the hydrochloric acid 
solution, and a blue solution containing some 
chromous ion results. In order to dissolve chrom- 
ium in sulfuric acid, it is usually necessary first to 
remove the "oxide film" with dilute HC1. After 
the metal has started to react with the HC1, it 
should be washed off with several portions of water 
and then covered with 5 ml of 3 F H 2 SO 4 . Warm 
the test tube, if necessary. Test the blue solution 
obtained for chromous ion by adding portions of 
it to solutions of HgC^ and CuSC>4 as directed in 
the preceding paragraph. 
Summarize your results by writing equations for 

the reactions and make a statement concerning 
the properties of the chromous ion. 

3. Oxidation State +3; The Chromic Ion and 
Chromic Oxide. 

The chromic ion, Cr" 1 " + +, forms many types of 
coordination complexes involving water, chloride 
ions, or hydroxide ions whose colors vary from 
violet to green. Note the color of the stock solu- 
tion of 0.1 F Cr(N0 8 ) 8 and compare it with the 
color of the chromic chloride solution you pre- 
pared in Part 1. How do you account for the dif- 

To 3 ml of 0.1 F chromic nitrate, add 6 F NH 4 OH 
drop by drop and observe the precipitate formed. 
Add more NH 4 OH to determine whether this pre- 
cipitate is soluble in excess ammonia solution. Does 
the chromic ion form a stable complex in which 
ammonia molecules are coordinated? (Text, Chap. 
22, Table 22-1.) 

To 3 ml of 0.1 F chromic nitrate, add 6 F NaOH 
drop by drop until a precipitate is formed. How 
does it compare with the one formed with NH 4 OH ? 
Now add several drops more of NaOH until the 
precipitate disappears. What ion of chromium is 
now present in the solution? What is the name 
given to hydroxides or oxides which dissolve either 
in acids to form cations, or in bases to form anions, 
as nonmetals do? Write equations for the reac- 

To the solution obtained above with excess 
NaOH, add 3% hydrogen peroxide solution drop 
by drop until a color change is noted. What new 
ion is formed by the oxidation of chromite ion in 
an alkaline solution? Write a balanced equation 
for the reaction, 

As an illustration of a change in oxidation state 
which is just the reverse of that in the previous 
paragraph, place a small, conical pile of powdered 
ammonium dichromate (about 1 cm high) in a 
small evaporating dish and ignite it by touching 
the tip of the cone with a small Bunsen flame. What 
is the powder which is formed? What are its com- 
mercial uses? Write an equation for the re- 

4. Oxidation State +6; Chromium Trioxide 9 
Chromates, Dichromates. 

Place about 0.1 g of CrOa in a 10-cm test tube 
and add about 5 ml of water. Test the solution 



formed with litmus paper. Is CrOs the anhydride 
of an acid or a base? Is your answer in conformity 
with the general rules about the acidic and basic 
properties of ,metallic oxides and the oxidation 
state of the metal? Write an equation for the 
reaction. Is this an oxidation-reduction reaction? 

Add 6 F NaOH drop by drop to the solution 
just obtained until there i? a color change. What 
new ion is now present? Write the equation for this 
reaction. Now add 6 F HC1 drop by drop until the 
color changes again. What ion is now present? How 
is this reaction related to the one just written? 

Add 10 drops of 1.0 F K 2 Cr0 4 to 5 ml of water 
and then add several drops of 0.1 F BaCb Write 
the equation for this reaction. Repeat the experi- 
ment, but this time add 5 drops of 6 F HC1 to 
the diluted chromate solution before adding the 
barium ion. How do you explain the difference in 
the result obtained this time? Now add 5 to 10 
drops of 6 F NaOH to the last test and note the 
result obtained. How does this series of experi- 
ments illustrate the principle of Le Chatelier? 

The dichromate ion in acid solution is a rela- 
tively good oxidizing agent. (Appendix II, Table 
XIII.) To illustrate this property, add 1 ml of 
1.0 F K 2 Cr0 4 and 1 ml of 3 F H 2 SO 4 to 5 ml of 
water and add a few crystals of FeSO 4 7H 2 0. 
Repeat the experiment, this time using a small 
amount of solid Na^SOa instead of FeSO 4 as the 
reducing agent. Write the oxidation half-reaction 
and the reduction half-reactions separately. How 
many electrons per mole are gained by the di- 
chromate ion when it is reduced? 

Add 1 ml of 1.0 F K 2 O0 4 and 1 ml of 3 F H 2 S0 4 
to 5 ml of water and then add some 3% H 2 O 2 drop 
by drop. The color formed is thought to be due to 
a peroxychromic acid whose exact formula is un- 
certain. This constitutes a rather useful test for 
the dichromate ion. 

The Chemistry of Manganese 

1. Oxidation State Mn; Properties of Metallic 

Select a small piece of metallic manganese about 
the side of the head of a pin. Note some of its 
physical properties and also refer to the text 

for additional characteristics. List these on the 
record sheet. 

Determine the relative activity of the metal by 
performing the following experiments. Place the 
piece of metal in a 10-cm test tube and add about 
5 ml of 6 F HC1. Note the relative vigor of the re- 
action and test the gas evolved for inflammability* 
Write the equation for the reaction. 

Pour off the hydrochloric acid, and wash the 
remaining piece of metal with several portions of 
water. Cover the metal with distilled water and 
watch carefully for any reaction of manganese with 
cold water. Warm the test tube gently and note 
any difference in the rate of reaction. Write the 
equation for this reaction. 

Pour off the water and add 5 ml of 6 F HNO 8 . Is 
manganese passive in this acid? 

2. Oxidation State +2; The Manganous Ion. 

To 3 ml of 0.1 F Mn(NO 3 ) 2 , add a drop of 2 F 
Na 2 S. Note the color of the precipitate and write 
the equation for the reaction. 

Add about 5 drops of 6 F NaOH to 3 ml of 0.1 F 
Mn(NO 3 ) 2 and note the color of the precipitate as 
it is first formed. Note also the darkening of the 
precipitate at the surface of the suspension. Now 
add an excess of NaOH to determine whether the 
precipitate is amphotcric. Write equation for the 
initial reaction and explain why the precipitate 

To 5 drops of 0.1 F Mn(N0 8 ) 2 in 5 ml of water, 
add 1 ml of 6 F HNO 3 and a small amount of solid 
sodium bismuthate (NaBiO 8 ). The pink color pro- 
duced is due to the MnO 4 ~ ion. Refer to Appendix 
II, Table XIII and note the relative positions df 
the Mn++ MnO 4 ~ (H+) couple and the Bi+++ 
BiOs"" (H + ) couple. (The latter couple is listed as 
BiO+ Bi 2 O 6 (H+).) Write the half-reactions in- 
volved in this experiment and combine them into 
a single, balanced equation. 

3. Oxidation State +3; The Manganic Jon, 

The brown manganic compound, MnO(OH)* 
which was formed by the oxidation of Mn(OH)2, 1ft 
Part 2, and the oxide Mn 2 Oa are the only relatively 
stable compounds containing Mn in the oxidation 
state, Mn+ 3 . Other salts are unimportant or Ufi* 
stable, and will not be considered. 



4. Oxidation Slot* +4 Manganese Dioxide* MnO*. 
Manganese dioxide, which constitutes the main 

source of manganese in nature, is the only im- 
portant compound of quadri-positive manganese. 
It may be reduced by carbon to produce the metal. 
It may serve as an oxidizing agent, in which case 
it is reduced to manganous ion. 

Place a small amount of powdered MnO 2 (about 
one-fourth the size of a pea) in the bottom of a 
10-cm test tube. Add about 5 ml of 12 F HC1 and 
cautiously observe the product of the reaction. 
Write the equation. 

To another test tube containing a small amount 
of MnO 2 add about 5 ml of 0.1 F KBr and acidify 
with 1 ml of 18 F H 2 SO 4 . Warm gently if neces- 
sary to initiate the reaction. Then cool, and add 
1 ml of CCU. Shake, and note the color of the 
CCU phase. Write the equation for the reaction. 

Repeat the experiment using 5 ml of 0.1 F KI 
instead of the KBr solution. Identify the product 
and write the equation for the reaction. 

Cover a very small amount of MnO 2 with about 
3 ml of distilled water which is acidified with 1 
ml of 6F HNO 8 . Now add some 3% H 2 O 2 solution, 
about a ml at a time and warm gently until the 
MnO 2 has disappeared. In this instance, hydrogen 
peroxide is acting as a reducing agent in the acid 
solution, and manganous ion and oxygen are the 
products of the reaction. (See Fig. 42-1 in this 
manual for a chart relating the oxidation states of 
manganese and the action of H 2 O 2 in solutions of 
different acidity or basicity.) Write the balanced 
equation for the reaction. 

5. Oxidation State +6; The Manganate Ion, 

Put approximately 3 g of solid KOH and 2 g of 
tCClOa in a crucible, which is placed in a triangle 
dn an iron ring. Heat the mixture gently until it 
fuses, then add slowly from a spatula about 2.5 
grams of powdered MnO 2 . Keep the mixture fused 
for several minutes; then allow it to cool. Put the 
cold crucible into a 250-ml beaker and cover it 
with distilled water. Write the balanced equation 
for the reaction. 

The manganate ion is stable in the alkaline solu- 
tion, but upon acidification, undergoes auto-oxida- 
tion-reduction to the +7 and +4 oxidation states. 
Decant some of the green solution obtained above 
and acidify 5 ml of it with several drops of 3 F 
H 2 SO 4 . Observe the products of this reaction. Now 
make this same solution basic again by adding 
several drops of 6 F NaOH. Note the change in 
color. Write the equation for the reversible reaction 
which is involved here and explain the changes ob- 
served on the basis of the principle of Le Chatelier. 

To another 5-ml portion of the green manganate 
solution, add about 2 ml of chlorine water, and 
note the results. Write the equation for this re- 

6. Oxidation State +7; The Permanganate lon t 

Potassium permanganate is p*obably the most 
important compound of manganese, since it is one 
of the stronger oxidizing agents and is very con- 
venient in analytical determinations where volu- 
metric oxidation-reduction reactions can be used. 
(See Exp. 30, in which you will determine the 
amount of a reducing agent by titration with a 
standard KMnO 4 solution.) 

Prepare an approximately 0.01 F solution of 
KMnO* by dissolving approximately 0.15 g of 
KMnOi in 100 ml of distilled water. Place about 
3 ml of this solution into each of four test tubes. 
Acidify the first with 1 ml of 3 F H 2 SO 4 , add 2 
drops of 6 F NaOH to the second, add 3 ml of 6 F 
NaOH to the third, and leave the fourth neutral. 
Now, with a spatula, add a small amount of a re- 
ducing agent such as sodium sulfite (Na2SO 8 ) to 
each of the test tubes and observe the color pro- 
duced in each case. Write the equations for the re- 
actions. How many electrons per mole are gained 
by the permanganate ion when it is reduced in an 
acid solution? How many per mole in a slightly 
basic solution? How many per mole in a strongly 
basic solution? 


REPORTt Exp. 29 


Some Chemical Properties 

of Chromium and Manganeste Sechon 

Locker Number 

The Chemistry of Chromium 

1. Physical properties of Cr._ 

Behavior with dil. HCL 

Behavior with dil. HN(V 

Behavior with water- 

Statement concerning activity of Cr_ 

2. Reactions by which the Cr++ may be produced: 

Zn and Cr+++ 

Cr and H+ (H 2 SO 4 )- 

Reactions of the chromous ion : 
With HgCl 2 

With CuSO4- 

Summary of properties of Cr"*" 1 " ion. 

3. Color and formula of the chromic ion in solutions of: 

Chromic chloride 

Chromic nitrate 

Reactions of the chromic ion: 
With dil. NH 4 OH 

With excess NH 4 OH Explanation- 

With dil. NaOH. 

With excess NaOH- 

Action of HjOs on chromite ion in basic solution 
Oxidation half-reaction 

Reduction half-reaction. 
Balanced net reaction 


Products of the decomposition of ammonium dichromate: 
Reaction . _ _ 

I. Litmus test water solution of CrO 3 Type anhydride. 


Is reaction oxid.-red.? Why?_ 

Effect of NaOH on color of solution Ion formed- 


Effect of HC1 on above solution color Ion formed 

Reaction - 

Color and formula of reaction product formed when Ba++ is added to aqueous 
Acid K 2 CrO4 Basic K 2 CrO4 

Explanation of the above series of reversible reactions in terms of the principle of Le Chatelier_ 

Oxidizing properties of Cr 2 O 7 in acid solution: 
With ferrous sulfate; half-reactions: 

Oxidation : 


Balanced net reaction. 

With sodium sulfite; half-reactions: 


Balanced net reaction- 

Observed result of effect of H 2 O 2 on acid Cr 2 O 7 - 
!h em is try of Manganese 

Physical properties of Mn 

Behavior with dil. HC1 


Behavior with cold water With hot water_ 

Reaction . 

Behavior with dil. HNOa Is Mn passive? 

Reactions of the manganous ion : 

With Na 2 S ^ 

Color of precipitate formed., 

With NaOH - . __ 

Color of initial precipitate formed 

Effect of excess NaOH Conclusion 

Explanation of darkening of precipitate 


Report on Exp. 29, Sheet 2 Name- 

Oxidation of Mn+ + by NaBiOs in acid solution; half-reactions: 


Balanced net reaction- 

3. Formulas of stable compounds of manganic ion.- 

4. MnO2 as an oxidizing agent in acid solution: 

Reaction with HC1 

Reaction with Br~_ 
Reaction with I~~ 

Reaction with H 2 O 2 ; balance using half-reactions: 
Oxidation - 


Balanced net reaction : 

5. Preparation of the manganate ion by oxidation of MnO 2 with KC1O 3 in KOH. 
Reaction , 

Change noted when MnO4 was acidified. 

Change when solution is made basic again^ 

Explanation of latter two reversible reactions in terms of the principle of Le Chatelier 

Reaction of chlorine water with manganate ion: 

6. The reduction of permanganate ion in solutions of different acidity or basicity: Write reduction half-reactions. 

In acid solution- 

In neutral solution- 

In weakly basic solution 
In strongly basic solution^ 

O1 T 

Titrations Involving 

Oxidizing and Reducing Agents. 


Co/lege Chemistry, Chapter* 29, 12 

Review of Fundamental Concepts 

The Equivalent Weight in Oxidation-Reduction 

The equivalent weight of a substance under- 
going oxidation or reduction is the same as the 
"chemical equivalent" which was discussed in the 
section on Faraday's Law, Experiment 19. Study 
this section again. We may calculate the equivalent 
weight of an oxidizing or reducing agent by divid- 
ing its formula weight by the number of electrons 
gained or lost, as indicated by the half-reaction 
equation for the process taking place. For example, 
if potassium bichromate (K 2 Cr 2 O 7 ) is used ' 
dize ferrous sulf ate (FeSO 4 -7H 2 O) to ferric si. 
the half -reactions are 

Cr 2 O 7 -f 14 H^ + 6 e- - 

Fe+ + * Fe+++ + 1 e~, 

7 H2O + 2 Cr+++ f 

and the corresponding weights are 

Formula weights: 

K 2 Cr 2 O 7 = 294.22 

Equivalent weights: 

K 2 Cr 2 7 


FeSO 4 -7H 2 O - 278.01 


FeS(V7H 2 O 

- 278.01. 

Note that we have used the formula weight of the 
salt, rather than simply of the ion taking part in 
the reaction, since if one wished to weigh the sub- 
stances, he would use the solid salts. 

Normal Solutions 

A normal oxidizing solution, or a normal reduc- 
ing solution, is one which contains one equivalent 
(defined as above) per liter of solution. 


Indicators for oxidation-reduction titrations have 
not been developed as extensively as they have 
for acid-base titrations. In principle, they would 
need to be substances which undergo a color change 
when they are oxidized or reduced, and of such a 
nature that their oxidation-reduction potential is in 
the proper range for the particular titration being 
performed. In many cases, as in the present experi- 
ment, the substance being titrated undergoes a 
sufficiently sharp color change so that no other in- 
dicator is needed. 

!u metric Oxidation-Reduction Processes 

n this experiment, we shall prepare a potassium 
rmanganate solution by weighing out approxi- 
mately the weight desired and dissolving this in 
che requisite amount of water. This solution will 
be standardized by titrating it against a sodium 
oxalate solution of known concentration. Finally, 
the equivalent weight of an unknown reducing 
substance will be determined by weighing out 
small portions and titrating these with the stand- 
ard potassium permanganate solution. 

You may perform an additional optional experi- 
ment, if time permits, by determining the amount 
of calcium in the laboratory tap water. This is 
done by precipitating the calcium as calcium oxar 
late (CaC 2 O 4 ), filtering, and dissolving this in di- 
lute sulfuric acid, and titrating against standard 
potassium permanganate. 

Experimental Procedure 

Special supplies: 2 burettes, analytical weights. 
Chemicals: KMnO 4 , a standard NaaCjA solution, 0.5 F 
vials of unknown reducing substances. 

1. Standard Potassium Permanganate So- 
lution. Calculate the weight of KM nO 4 needed to 
make up 500 ml of an approximately 0,02 F KMnOi 
solution. After approval of your calculation, weigh 
just this amount on the triple beam balance to 

0.02 g, dissolve in about 100 ml of boiling dis- 
tilled water, cool, and dilute to 500 ml with dis- 
tilled water. The solution cannot be made up to 
an exact standard due to impurities present. A 
graduated cylinder is, therefore, sufficiently ac- 
curate in measuring these volumes. Mix the solu- 
tion well by pouring from one container to another 
several times; place in a clean labeled bottle. 




Now standardize this solution by titration 
against a standard sodium oxalate solution (con- 
taining 6.5 to 7.0 g/1), provided by the instructor. 
He will report to you the exact weight, in grams, 
of sodium oxalate, Na 2 C 2 O 4 , per liter, from which 
you may calculate the normality of the solution as 
a reducing agent. Carry out the titration as fol- 
lows. Fill two clean burettes with the sodium 
oxalate and potassium permanganate solutions, 
respectively, having first rinsed each burette with 
a little of the solution. Into a clean 250-ml Erlen- 
meyer flask put about 50 to 75 ml of distilled water 
and 30 ml of 3 F H 2 SO 4 . Now run in about 20 ml 
of the sodium oxalate solution, heat the contents 
of the flask almost to boiling, and titrate this, 
while hot, with the potassium permanganate solu- 
tion, to the first permanent faint pink color which 
remains after stirring. 1 At least three titrations 
should be performed, or until concordant results 
are obtained. 

The oxalate ion, C 2 O 4 , is oxidized to CO 2 g 
Write the half-reaction equation for this proct 
calculate the equivalent weight of Na 2 C 2 O 4 and t 
normality of the standard sodium oxalate solution 
you used. You can then calculate the normality, 
as an oxidizing agent, of your potassium per- 
manganate solution, based on the inverse rela- 
tionship between normality and volume, i.e. 
ViNi = V 2 N 2 . From this normality, and the fact 
that your solution was made up to be about 0.02 F, 
calculate the number of equivalents per formula 
weight of KMnO 4 in this reaction. Verify this rela- 
tionship by writing the half-reaction equation for 
the reduction of MnO 4 ~ (Mn++ is formed) in acid 
solution. Now write the net ionic equation for the 
whole oxidation-reduction process. 

2* The Equivalent Weight of an Unknown 
Solid Reducing Agent. Obtain from the stock- 
room a vial of unknown reducing agent. Record 

1 A delay in the decolonisation of the potassium permanganate 
color at the start of the titration is due to an interesting autocatalytic 
effect. Until a sufficient amount of manganous ion (Mn+ +) has been 
formed to catalyze the reaction, the rate is very slow. 

the number on your report sheet. Using this vial aa 
a weighing bottle, weigh out approximately 0.2 
samples to 0.001 g into 400 or 500-ml beakers or 
flasks, as you did with the Na 2 C 2 O 4 . Add 75 ml 
of distilled water, 30 ml of 3 F H 2 SO 4 , heat to 
boiling, and titrate as before with your KMnO 4 . 
Calculate the equivalent weight of your sample, 
and report the value at once to your instructor. 

3. Calcium in Tap Water. (Optional.) The 
hardness of water is due largely to the presence 
of calcium and magnesium salts. Using a graduate, 
measure a 250-ml portion of water to be examined, 
and place in a 400-ml beaker. Add 1 ml of 6 F HC1 
and evaporate to a volume of about 50 ml. Add 
2 drops of methyl orange indicator, and then am- 
monium hydroxide drop by drop until the acid is 
neutralized. If a precipitate appears, just dissolve 
it by adding a drop or so of dilute HC1. The solu- 
tio^ ill now be faintly acid. Heat to boiling and 
jwly about 25 to 30 ml of 0.5 F oxalic acid 
ion. Make alkaline with NH 4 OH, boil for a 

w minutes, and allow the precipitate to settle. 

Filter the precipitated CaC 2 O 4 . If the filtrate is 
not clear, heat it again and let it stand longer to 
coagulate into coarser particles. Wash the filte] 
paper repeatedly with a stream of water from your 
wash bottle, until 20 ml of the wash water will not 
decolorize a drop of 0.1 N KMnO 4 . Puncture the 
tip of the filter paper, place a 400-ml beaker under 
the funnel, and pour about 50 ml of 1 F H 2 SO 4 in 
a very gentle stream all over the surface of the 
paper, to dissolve the precipitate. Rinse the filter 
with 50 ml of hot water, adding it in small portions. 
Heat the solution in the beaker to boiling and 
titrate with your standardized KMnO 4 solution. 
Write equations for the reactions taking place, and 
calculate the relationship between moles of Ca++ 
present and equivalents of KMnO 4 used. Report 
your results in parts per million, which is the cus- 
tomary manner of reporting water analyses. (Parts 
per million is the same as milligrams per liter.) 
Prepare your own report form for this part of the 
experiment, showing equations, data, and calcu- 
lated values in a neat, orderly arrangement. 

REPORT: Exp. 30 

Titratlons Involving Oxidizing 
and Reducing Agents 



Locker Number^ 

L* Standard Potassium Permanganate Solution 

Calculated weight of KMnO 4 required for 500 ml of 0.02 F solution: 

Weight of Na 2 C 2 O4 per liter, in standard solution used, as reported by the 





Na 2 C 2 O4 burette, 2nd reading 




1st reading 

Volume of Na 2 C 2 O4 solution used 




KMn(>4 burette, 2nd reading 




1st reading 

Volume of KMnOi solution used 




Write the half-reaction equation for the oxidation of oxalate ion, C 2 O4 , to carbon dioxide gas: 

Calculate the equivalent weight of Na 2 C 2 O4 and the normality of the Na 2 C 2 O4 solution, based on this equation and 
on the grams per liter: 







of KMnO 4 




Average value N 

From this normality of the KMnO 4 and the approximate formality (0.02 F) as the solution was prepared, 
calculate the number of equivalents per formula weight of KMn04, in this reaction : 

equiv per gfw 

Verify this by writing the half -reaction equation for the reduction of MnO 4 " to Mn++: 

Combine the two half -reactions to give the net ionic equation for the reaction of permanganate ion with oxalate 
ion in acid solution: 


2, The Equivalent Weight of an Unknown Solid Reducing Agent 

Number of unknown sample. 





Initial weight, vial + unknown 

Final weight, vial + unknown 

Weight of unknown 

Final burette reading (KMnO 4 ) 

Initial burette reading 

Volume of N KMnOi used 





Equivalents of KMnC>4 
(and of unknown) in 
the sample titrated 

Weight of one equivalent 
of unknown 


Average value 


Instructor's report: Your unknown is_ 

Percent error 

3. Calcium in Tap Water (Optional) 

Prepare your own report form for this part of the experiment, showing equations, data, and calculated values in a neat, 
orderly arrangement. 


Report on Exp. SO, Sheet 2 Name 

Application of Principles 

1. What is the equivalent weight of the following: 

a) K 2 Cr 2 O 7 (reduced to Cr+++) .... 

b) Na2SOa (oxidized to a sulfate) 

c) KC1O 3 (reduced to a chloride) 

2. Two formula weights of an acid solution of ferric chloride are mixed with a solution 
containing four formula weights of sodium iodide. How many moles, and grams, of 
iodine, I 2 , will be produced? 

3. What substances are present, and if in solution at what concentration, in the following: 

a) one formula weight each of stannous chloride and of ferric chloride are mixed in 
water to make ten liters of solution. (First write the equation.) 

b) 0.5 mole of chlorine gas, CU, is passed into one liter of a solution containing one 
formula weight each of potassium iodide and of potassium bromide. 


Soirfe Elementary Experiments 
in Organic Chemistry 


College Chemistry, Chapters 30, 7 

Review of Fundamental Concepts 

Organic chemistry is concerned with the chemis- 
try of the compounds of carbon. In the early 
history of chemistry it was thought that these 
aaturally-occurring compounds could not be 
made in the laboratory but required the vital in- 
fluence of the living organism, hence the name 
"Organic" was applied to this class of compounds. 

The element carbon has four valence electrons 
and when it combines with other elements it shares 
its electrons and forms four covalent bonds. The 
elements usually combined with carbon are: hydro- 
gen, oxygen, and nitrogen, and less frequently, 
sulfur, phosphorus, and the halogens. Since carbon 
atoms have a great tendency to combine with other 
carbon atoms to form ring structures or chains of 
carbon atoms, very complex molecules of high 
molecular weight are possible. The great variety 
of structural arrangements which result makes 
possible the existence of virtually an infinite num- 
ber of different compounds. To date nearly a half- 
million organic compounds have been isolated from 
natural sources or have been synthesized in the 

One of the simplifying principles which helps to 
systematize the study of organic chemistry is the 
fact that organic compounds may be classified into 
different types, according to the nature of the 
functional group present. For instance, the groups 
-OH, -CHO, or -COOH each impart a char- 
acteristic reactivity to the molecules containing 
such a group. All alcohols contain the OH group 
in addition to the carbon-hydrogen portion of the 
molecule as is seen by the following illustrations: 
methyl alcohol is CH 3 OH, ethyl alcohol is CjiH 6 OH 
and amyl alcohol is C 6 H U OH. The general for- 
mula, ROH, is assigned to all alcohols containing 
one hydroxyl group. The letter R stands for the 
carbon-hydrogen radical to which the functional 
group is attached. 

A summary of some of the types of organic com- 
pounds, together with their general formulas and 
a common illustration is given below. 

Alcohol R OH 
Ether R O R 

Methyl alcohol CH,OH 
Diethyl ether (C 2 H 6 )aO 

Aldehyde R C=O Formaldehyde HCHO 


Ketone R C R Acetone (CH 8 ) 2 CO 


Acid R C OH Acetic acid CH 8 COOH 


Ester R C OR 
Amine R NH 2 
H O 

Ethyl acetate 
Butyl amine 

Amino R C C OH Glycine 
acid | 

NH 2 

CH 2 (NH 2 )COOH 

Compounds containing only carbon and hydro- 
gen are called hydrocarbons and are generally 
classified into two series. The aliphatic series con- 
tains the carbon atoms linked together in open 
chains, while in the aromatic series the carbon 
atoms are linked together to form closed rings. In 
the aliphatic series, the hydrocarbon is said to be 
saturated when all four carbon valences are being 
satisfied by other carbon or hydrogen atoms. Com- 
pounds formed when there are two or more hydro- 
gen atoms less than that required for saturation 
of the carbon valences are called unsaturated 
compounds. Some illustrations of the various 
types of hydrocarbons are given below* 




H H 

Paraffin series I | 
(saturated) C n H2n+i Ethane H C C H 
1 1 
H H 

As hydrogen atoms are successively replaced bj 
oxygen atoms, the oxidation number of the carbon 
atom substituted increases as shown below for the 
simplest hydrocarbon. 

Ethylene series 

(unsaturated) C n 

Hta Ethylene H C==C H 

-4 -2 Zero +2 +4 

1 1 


H H 

1 1 1 1 

Acetylene series 


(unsaturated) CJ 

Elzn-j Acetylene HC^C H 

I 1 

H H 


methane methyl formal- formic carbon 

A mm nt in S#T*I*> 


alcohol dehyde acid dioxide 

(cyclic, unsaturated) 

Benzene HC 







By adding C units to increase the length of 

the open chain compounds or to substitute in the 
side chain it is possible to obtain a very great num- 
ber of hydrocarbons. In the paraffin series, for 
example, the gases CH 4 , C 2 H 6 , C 3 H 8 , C 4 Hio continue 
up the homologous series by CH 2 increments to 

the liquids C 6 Hi 2 , C 6 Hi 4 Ci 6 H 3 4 until 

finally solid paraffins with as high as 60 or more 
atoms of carbon per molecule have been reported. 
The alcohols, aldehydes, acids, and other types 
of compounds mentioned before, may be considered 
to be derivatives of the hydrocarbons in which the 
functional group or groups have replaced one or 
more of the hydrogen atoms in the hydrocarbon. 

Thus, when methane undergoes complete com- 
bustion the carbon atom changes from the -4 to 
the +4 oxidation state. When methanol undergoes 
mild oxidation, formaldehyde may be the resulting 
product, whereas more strenuous oxidation condi- 
tions may produce formic acid or carbon dioxide. 

Reactions between covalent organic molecules 
generally take place rather slowly in contrast with 
the rapid reactions which characterize ionic in- 
organic compounds. For instance, when an alcohol 
and organic acid react to form an ester, very little 
reaction takes place until the mixture is heated for 
some time in the presence of a dehydrating agent 
to remove the water formed and shift the equilib- 
rium to the right as in the following equation: 


B/-OH + R COOH 3= H 8 O + R COOR'. 

In the experiments which follow, we shall at- 
tempt to survey some of the properties of the vari- 
ous types of hydrocarbons and their derivatives, 
We shall also illustrate the relative reactivity oi 
some organic compounds. 

Experimental Procedure 

Chemicals: CaC, kerosene, paraffin, benzene, naphthalene, 
methyl alcohol, ethyl alcohol, amyl alcohol, formaldehyde 
solution, Brs in CCU, Cu wire, soda lime, NaCaHaC^, salicylic 
acid, potassium dichromate, CCU 0.1 F AgNOi, glacial acetic 

1* Properties of Some Hydrocarbons 

(a) Saturated Hydrocarbons, Methane or Paraffin 
Series. Using the natural gas supply in the labora- 
tory as a source of methane and ethane, bubble 
some of the gas slowly through 3 ml of Br 2 solution 
in CC1 4 for several minutes. Is the bromine decolor- 
ized ? Collect a test tube of the gas by displacement 
of water. Is the gas soluble in water? Ignite the 
gas in the test tube and note the flame produced. 

Put 1 ml of Hmewater into the test tube and shake 
it with the products of combustion. What gaseous 
product does this test indicate is present? Write 
the equation for the combustion of ethane in an 
excess of oxygen. 

What are the molecular formulas of the hydro- 
carbons present in gasoline and in kerosene? Test 
3 ml of kerosene with 1 ml of the Br 2 in CCU solu- 
tion to determine whether thqre are present any 
unsaturated compounds which would "add on" the 
Br and decolorize the reagent. Note also whether 
the kerosene is miscible with the CCU. Place about 
5 drops of kerosene in a watch glass and ignite it 



with a small flame. Note the type of flame pro- 

Place a small fragment of paraffin the size of a 
small pea in a test tube and add 3 ml of the Br a 
in CCli solution. Does the paraffin dissolve in the 
CC1 4 ? Is the bromine decolorized? Ignite a small 
piece of paraffin on a watch glass. Does it have a 
high ignition temperature? If you have any diffi- 
culty igniting it place a small piece of string in the 
melted paraffin and use it as a wick. Note the flame 

(b) Unsaturated Hydrocarbons, OpenChain Type. 
Of the many types possible here we shall use 
acetylene, C 2 H2, with the structure, H C=C H, 
as an illustration of this type compound. Set up 
the small generator shown in Fig. 22-2 and place 
a small lump of calcium carbide, CaC 2 , in the dry 
test tube. Draw in a few milliliters of water into 
the dropper tube. Collect a 15-cm test tube of 
acetylene by the displacement of water. Allow the 
rest of the gas generated to bubble through about 
3 ml of the Br 2 solution in CC1 4 . Caution: acetylene 
is explosive when mixed with air in the proper 
proportions, so keep the generator away from open 
flames. When the generator has become inactive, 
test the combustibility of the tube of acetylene 
you have collected, taking the precaution to wrap 
the test tube with a towel before holding its mouth 
to the flame. How did the flame compare with that 
produced when methane burned ? 

(c) Aromatic Hydrocarbons, Benzene, Naphtha- 
lene. Test 3 ml of benezene with 1 ml of the solu- 
tion of Br 2 in CCU. Is the benzene miscible with 
the CCU? Is the Br 2 solution decolorized? Ignite 
about 5 drops of benzene on a watch glass, and 
note the type of flame produced. 

Repeat the above tests using a small amount of 
naphthalene about the size of a pea. What is the 
structure of the napthalene molecule? 

2. Alcohols, R OH. Write the structural for- 
mula for the following alcohols: methyl alcohol, 
ethyl alcohol, and amyl alcohol. Place 1 ml of each 
of these alcohols into separate test tubes and note 
their odor. Add 1 ml of water to each to determine 
the solubility in a polar solvent. Repeat the solu- 
bility test using the nonpolar solvent, CC1 4 . Did 
the length of the hydrocarbon chain have any 
effect on the solubility tests you observed? What 

could you predict about the solubility of glycerol, 
a trihydroxy alcohol with the formula CaHeCOH)^ 
in water and in a hydrocarbon? Answer the same 
question for the alcohol CieHasOH. 

3. Aldehydes, R CHO. Write the structural 
formula for formaldehyde. What is the oxidation 
number of the carbon atom in an aldehyde? What 
compound would be produced when formaldehyde 
is oxidized to the next higher oxidation state? The 
Fehling test for aldehydes consists in the oxidation 
of an aldehyde by the use of an alkaline solution 
containing the cupric-tartrate complex ion. The 
latter oxidizing agent is thereby reduced to the 
insoluble cuprous oxide, Cu 2 O. Perform this test 
as follows: Mix 5 drops of each of the Fehling solu- 
tions, A and B, with one ml of water. Warm to 
60 C in a water bath and add a few drops of the 
formaldehyde solution. Note the character of the 
precipitate formed. This test is also used to detect 
the presence of a reducing aldehyde group which 
is one of the characteristic groups of certain sugars. 

Make a spiral of copper wire by inserting a 10- 
inch piece of medium gauge copper wire into a cbrk 
to serve as a handle and wind the other end around 
a glass rod to make a spiral. Place 1 ml of methyl 
alcohol in a small test tube. Heat the spiral to red 
heat in a flame and plunge it into the methyl 
alcohol. Repeat this procedure several times. Can 
you recognize the odor in the test tube? Note that 
the wire becomes dark when oxidized in the flame 
but turns bright again when dipped in the alcohol. 
Write the equation for the reaction for the oxida- 
tion of CH 3 OH by CuO. 

4. Organic Acids, R COOH. Organic acids 
may be prepared from their salts by adding a non- 
volatile acid and distilling over the volatile acids. 
Place about 1 g of NaC 2 H 3 O 2 in a test tube and add 
1 ml of 18 F H 2 S04 and warm gently. Note cau- 
tiously the odor of the vapors and their effect on 
moistened neutral litmus paper. Is oxidation-re- 
duction involved here? What is the oxidation state 
of the carbon atom in the carboxyl group, COOH ? 

Write the structural formulas of ethyl alcohol, 
the aldehyde, and the acid which would be pro- 
duced by the oxidation of the alcohol. Dissolve 1 g 
of K 2 Cr 2 O 7 in 5 ml of 3 F H 2 SO 4 . Add 5 drops of 
ethyl alcohol cautiously. Note the change in color 
of the oxidizing agent. Note the odor of the vapors, 
cautiously. A pearlike odor indicates the presence 



of acetaldehyde. Now warm the tube gently and 
test the vapors with moistened neutral litmus 
paper for any acid vapors. To what is the green 
color of the solution due ? Write the balanced oxida- 
tion-reduction equations for this reaction either 
to the formation of the aldehyde or to the acid. 

5. Esters, R COOR'. Esters are produced by 
the reaction of organic acids and alcohols in the 
presence of concentrated sulfuric acid. 

Prepare some ethyl acetate by warming gently 
a mixture of 1 ml ethyl alcohol, 1 ml glacial acetic 
acid and 10 drops of 18 F H 2 SO 4 . After about three 
minutes pour the reaction mixture into a small 
beaker and note the odor. 

Repeat the procedure using amyl alcohol and 
acetic acid and note the odor. What does it re- 

Repeat again, this time using 1 g of the solid 
salicylic acid and methyl alcohol. What is the odor 
noted here? 

Write the equation for the preparation of ethyl 
acetate. Why is sulfuric acid necessary? Compare 
this reaction with that taking place between NaOH 
and HC1 in aqueous solution as to type of molecules 
involved and reaction rate. 

6. Tests for Halogens in Organic Molecules* 

We shall use carbon tetrachloride, CCU, as an 
illustrative compound for the following tests. 

The Beilstein test is performed by heating the 
copper wire spiral used in Part 3 and plunging it 
into the compound containing a substituted halo- 
gen. When the wire is returned to the flame a green 
flame results from the volatilization of the copper 

Another test for the presence of a halogen atom 
in an organic molecule illustrates the conditions 
necessary to cause the precipitation of AgCl when 
the ionic compound AgNO 3 reacts with a covalent 
molecule such as CCU. First, add 5 drops of CCU 
to 1 ml of 0.1 F AgNO 3 and shake the test tube. Is 
there any sign of reaction? Now, place about 0.2 g 
of soda lime in a clean 15-cm test tube. Heat to 
redness and add 2 or 3 drops of CC1 4 . Heat the soda 
lime again and add the CC1 4 as before. Cool the 
tube and acidify with 8 to 10 nil of 6F HNO 3 . Now 
add 1 ml of 0.1 F AgNO 3 and note the result. Com- 
pare the procedure required above, with the re- 
action of NaCl and AgNO* in an aqueous solution, 
How do you account for this difference in reactiv- 

REPORT: Exp. 31 

Some Elementary Experiments 
in Organic Chemistry 



Locker Number- 

1. Properties of Hydrocarbons 

Fill in the data called for in this table either from information obtainable in the text or experimental determina- 

Hydrocarbon Solubility in Reaction with Type of 

Name Formula Type Nonpolar Solvent Br% in CClt Flame 







Equation for the combustion of ethane : 

Equation for the preparation of acetylene: 

Compare the ignition temperatures of the various saturated hydrocarbons tested. 

How does the ratio of C to H in the molecule affect the amount of carbon in the flame? 

If the Br 2 in CCU is decolorized what does this indicate concerning the nature of the hydrocarbon? 

Which of the hydrocarbons tested above showed a reaction with Br 2 under the conditions used here? 

Equation for the reaction with Br 2 :- 

2. Properties of Some Alcohols 

Alcohol Structural 

Name Formula 

Methyl alcohol 


Solubility in 
Polar Solvent 

Solubility in 
Nonpolar Solvent 

Ethyl alcohol 
Amyl alcohol 

What effect does length of the hydrocarbon chain have on the solubility in different type solvents? 


What do you predict concerning the solubility of glycerol in water 
in a hydrocarbon solvent - ? 

What do you predict concerning the solubility of CieHgaOH in water - 
in a hydrocarbon solvent - ? 

3. Properties of Aldehydes, R CHO 

Structural Formula Oxidation Number of Carbon 

Methyl alcohol . . _ - . 

Formaldehyde . . - - , - 

Formic acid . . . - - 

Describe the appearance of the Fehling test. 

What is the precipitate formed? . What is the nature of the oxidizing agent 

present in Fehling solution? 

In the hot copper wire-methyl alcohol test, what was the odor observed? , 

Why did the copper wire become bright on plunging it into the methyl alcohol ?_ 
Write the equation for the reaction of methyl alcohol and hot cupric oxide: 

4. Organic Acids, R COOH 

Write the equation for the preparation of acetic acid by the action of 18 F H2SO4 on sodium acetate: 

Is oxidation-reduction involved here?_ 

The structural formula of, and the oxidation number of the substituted carbon in: 

Ethyl alcohol 

The aldehyde produced by oxidation . 

The acid produced by oxidation . . 

Write the balanced oxidation-reduction reaction for the effect of acid dichromate solution on ethyl alcohol: 

Report on Exp. 81, Sheet $ Name 

5. Organic Eaters, R COOR' 

Write the equation for the preparation of ethyl acetate: 

Describe the odor of each of the esters produced. 

What is the role of sulfuric acid in the reactions involved here? 

Compare the conditions and rate of this reaction with an acid-base reaction. 

6. Tests for Halogens in Organic Compounds 

Describe the Beilstein test for halogen compounds. 

Does aqueous AgNOa react with carbon tetrachloride?- 

Describe the reaction necessary to obtain a halogen test with CCU. 

How do you account for the difference in reactivity of organic halides and inorganic haiides? 


Some Elementary Experiments in Biochemistry. 

Review of Fundamental Concepts 


Cof/ege Che mi's fry, Chapter 31 

Three classes of organic compounds, carbohy- 
drates, proteins and fats, are important constit- 
uents of foods. A major portion of the field of 
biochemstry is concerned with the important role 
these compounds play in building body tissues and 
in supplying energy to the body. Since these com- 
pounds are very complex in nature, careful atten- 
tion should be given to the discussion of their 
structure and properties in the text and in the sec- 
tion which follows. 

Carbohydrates, C x (H 2 O) y . The carbohydrates 
are produced by plants by the process of photo- 
synthesis, in which carbon dioxide from the air 
and water from the soil combine in the presence of 
chlorophyll and sunlight to produce compounds 
containing carbon, hydrogen and oxygen. The fi- 
brous part of the plant is composed of cellulose 
which is a polysaccharide which consists of pos- 
sibly 200 units of C 6 Hi O 5 . The starch produced by 
many plants is believed to contain about 30 units 
of the simple saccharide molecule (CeHioOe^o-I^O. 
The disaccharides are composed of two molecules 
of rnonosaccharides linked together. Examples of 
these are sucrose, maltose and lactose, which vary 
slightly in structure but have the common formula 
C^H^On. The monosaccharides with the formula 
C 6 IIi2O 6 contain either an aldehyde or ketone group 
in their structure which is capable of reducing cer- 
tain solutions of metal ions. One such reagent is 
Fehling solution, which, when reduced to red cu- 
prous oxide, constitutes a characteristic test for 
glucose, fructose, and other reducing sugars such 
as maltose and lactose. Other carbohydrates which 
do not reduce Fehling solution, such as sucrose, 
starch, and even cellulose, will form the simple re- 
ducing sugars when hydrolyzed by boiling with 
dilute acids or bases. During digestion of foods rich 
in starches such as rice, potatoes or bread, the 
starch is hydrolyzed to maltose, a disaccharide, and 
finally to glucose. Enzymes present in the saliva, 
ptyalin and maltose, catalyze this reaction so that 
it takes place with ease at body temperature. The 
glucose is carried by the blood to all parts of the 

body where it may be oxidized to carbon dioxide 
and water, accompanied by the liberation of energy 
for muscular movement and heat. Some of the 
glucose may be stored in the liver and muscles as 
glycogen, a polysaccharide composed of glucose 

Proteins, Complex Nitrogenous Substances. 
The elementary composition of proteins averages 
approximately 52% carbon, 7% hydrogen, 23% 
oxygen, 16% nitrogen, to 3% sulfur, to 0.8% 
iodine, and often other elements such as phosphorus, 
iron and copper are present. The molecular weights 
of proteins will range from 30,000 to possibly sev- 
eral million. The fundamental building units are 
amino acids, CHR(NH2)COOH, which are com- 
bined by the union of the NH 2 group of one acid 
with the COOH group of another acid to form 


a peptide linkage or bond, N . This 


linkage is responsible for the color produced in the 
Biuret test. Due to the presence of both an acid 
group (COOH) and a basic group (NH 2 ), the pro- 
tein molecule is amphoteric in nature and will form 
a protein salt with an acid or a base. When heated, 
the polypeptide chains are uncoiled from their 
characteristic structure as native protein and 
thereby lose many of their characteristic properties. 
They are said to be denatured. 

During digestion, enzymes, such as pepsin and 
rennin in the stomach and trypsin and erepsin in 
the small intestine, catalyze the hydrolysis of pro- 
teins into more simple molecules, and finally into 
amino acids. The amino acids pass through the 
walls of the intestines into the blood stream and 
are distributed to all the tissues of the body where 
they are used to renew tissue which is constantly 
being broken down in the life process. Excess pro- 
teins may be converted to carbohydrates or may 
be oxidized to carbon dioxide, water, urea, and 




ammonium compounds, accompanied by the liber- 
ation of energy. 

Proteins are produced by both plants and ani- 
mals but the primary source may be considered to 
be from plants, since the proteins in such sub- 
stances as meat, eggs and milk are produced by 
animals fed on vegetable protein. Such foods as 
wheat flour, corn flour, oat meal, beans, peas and 
nuts are important foods containing plant pro- 

Fats, Fatty Acid Esters of Glycerol. Fats are 
produced in plants and animals by synthesis from 
carbohydrates. Foods rich in fats include butter, 
lard, fat meas, cheese, nuts, and vegetable oils. 
The elements present are carbon, hydrogen, and 
oxygen which are combined as esters of various 
fatty acids with the tri-hydroxy alcohol, glycerol. 
One of the fats present in olive oil is the oleic acid 
(CuH 2B COOH) ester of glycerol (CiHi(OH),), 
whose structure is 

H O 

H C O C Ci B H 
I O 

H -C O 

I O 


H C O C CieHa 


Chemicals: Absorbent cotton, acetone, d-glucose, dried egg 
albumin, ethyl alcohol, Fehling solutions A and B, lead acetate 
paper, Methyl orange indicator, Molisch reagent (alpha naph- 
thol in alcohol), olive oil, phenolphthalein indicator, KHSO^ 
soap chips, soda lime, starch, sucrose, whole milk, 0.1 F 
CuSO* 0.1 F I 2 , 0.1 F PbCCAO,)* 0.1 F HgClj. 

1. Carbohydrates: Glucose, Sucrose, Starch, 
Cellulose. Prepare solutions of glucose and sucrose 
by dissolving 2 g of each in 20 ml of water. Prepare 
a colloidal suspension of starch by adding a paste 
made by wetting 1 g of starch with 5 ml of cold 
Vater to 50 ml of boiling water and stirring for one 

(a) The Molisch Test for Carbohydrates. This test 
is given by all types of carbohydrates, regardless 
of their complexity. Add five drops of the alcoholic 
solution of alpha-naphthol to 3 ml of each of the 
prepared solutions of carbohydrates. Allow about 
2 ml of 18 F HjzSO* to run down the side of each 

When this molecule is hydrolyzed, it is split into 
glycerol and the fatty acid. When this is done in 
the presence of an alkali such as NaOH the process 
is called saponification, and the sodium salt of the 
fatty acid produced is a soap. 

When a fat is heated strongly in the presence of 
a dehydrating agent such as KHSO 4 , the glycerol 
portion of the molecule is dehydrated to form the 
unsaturated aldehyde, acrolein CH 2 = CH CHO, 
which has the peculiar odor of burnt grease. This 
is a standard test for fats. 

During digestion, fats are emulsified and are 
hydrolyzed in the stomach with the aid of the 
enzyme lipase, and in the alkaline medium of the 
small intestines by the enzyme steapsin. The fatty 
acids and glycerol pass through the intestinal mem- 
branes, recombine to form fats, and pass into the 
blood where they are carried to all parts of the 
body. They may be oxidized to carbon dioxide and 
water to furnish heat and energy or they may be 
stored as fatty tissue to give form to the body and 
to support the internal organs. 

In this experiment we shall observe some of the 
characteristic tests used to identify carbohydrates, 
proteins, and fats, and then apply these tests to 
the various components separated from the com- 
mon food: milk. 


test tube and form a layer at the bottom. Do not 
mix. Watch for the appearance of a characteristic 
color at the contact zone after a short time. 

(b) The Fehling Test for Sugars with Reducing 
Groups. This test is given by all monosaccharides 
and those disaccharides which have an aldehyde or 
ketone group which is not involved in the di- 
saccharide linkage. When the Fehling solutions A 
and B are mixed, we have an oxidizing agent con- 
sisting of the complex cupric tartrate ion in an 
alkaline medium which can be reduced to the in- 
soluble cuprous oxide. 

Add 10 drops of each of the Fehling solutions A 
and B to 3 ml of water. Mix and divide the solu- 
tion between three test tubes. Place the tubes in a 
beaker containing water at about 60 C. Add 2 ml 
of each of the carbohydrate solutions and allow to 



stand for several minutes. Which of the carbohy- 
drates contained a reducing group? 

Place a test tube containing 5 ml of the sucrose 
solution and 1 ml of 12 F HC1 in a boiling water 
bath, and allow the sucrose to hydrolyze for about 
ten minutes. Neutralize the acid solution with 2 
ml of 6 F NaOH and make the Fehling test as out- 
lined before with some freshly prepared Fehling 
solution. How do you account for the results now 
obtained ? 

(c) The Hydrolysis of Starch. In order to determine 
when the complex starch molecule has been broken 
down into simpler molecules and finally into glu- 
cose, we shall employ the iodine test for starch and 
the Fehling test for glucose. Prepare the iodine test 
solution by diluting 2 drops of 0.1 F 12 solution 
with 1 ml of water. Place one ml of the prepared 
starch solution on a small watch glass which is 
resting on a piece of white paper. Add a drop of 
the dilute iodine solution and note the color of the 
iodo-starch complex. Repeat the test with one ml 
of glucose solution, and note the result. 

Dilute 25 ml of the starch solution with 100 ml 
of distilled water. Add 15 ml of 12 F HC1 and boil 
for one minute. With a medicine dropper transfer 
about 1 ml of the solution to a small test tube, 
neutralize with a few drops of 6 F NaOH and add 
one drop of the iodine solution. View the color 
against a white background. Continue to boil the 
solution and withdraw samples at one minute in- 
tervals until no color is obtained with the iodine 
solution. Now test 2 ml of the hydrolyzed starch 
solution, neutralized with NaOH, with 1 ml of 
freshly mixed Fehling solution. What has happened 
to the starch molecule during hydrolysis? How is 
corn syrup, used for candy making, produced com- 

(d) (OptionalExperiment) . The Enzymatic Hydrol- 
ysis of Starch by Ptyalin from Saliva. This experi- 
ment will enable you to note the relative ease with 
which biochemical reactions take place in the 
presence of enzymes, as compared to the rather 
strenuous conditions of temperature and acid con- 
concentration required in the preceding experi- 

Make a fresh suspension of starch as directed 
before. Place about 25 ml of the suspension in a 
50-ml beaker which is suspended in a slightly larger 
beaker containing water at body temperature 

(37 C.) In order to follow the course of the hy- 
drolysis, have 5 test tubes ready, numbered 1 to 5 
containing 1 ml each of Fehling solution and a 
similar number containing 1 drop each of dilute 
iodine solution. Note the time and add about 25 
drops of saliva with a medicine dropper to the 
starch solution. At two-minute intervals, remove a 
few drops of the solution and add to the iodine 
solution. Also add 1 ml of the starch to the Fehling 
solution. When the starch-iodine test has turned 
from blue to red to colorless, the starch is hy- 
drolyzed. Now heat the test tubes containing the 
Fehling solution to 60 C in a water bath. Note 
the test tube in which the reducing sugar first ap- 
peared. If the hydrolysis is not complete in the 
time allotted, repeat the experiment using double 
the amount of saliva. 

(e) Hydrolysis of Cellulose. In order to hydrolyze 
the complex cellulose molecule even more strenuous 
conditions must be applied. Add 1 ml of 18 F 
H 2 SO 4 to a small wad of absorbent cotton on a 
small watch glass and stir with a glass rod for 
about three minutes. Pour this solution into 20 ml 
of water in a small beaker and boil for 10 minutes. 
Add water from time to time if necessary to replace 
evaporated water. Remove about 2 ml of the solu- 
tion, neutralize with a few drops of 6 F NaOH and 
add 1 ml of Fehling solution. Proceed with the 
boiling until a positive Fehling test is obtained. 
How does the complexity of the cellulose molecule 
compare with that of starch? 

2. Proteins, Their Characteristic Tests and 
Properties. We shall use egg albumin, molecular 
weight 33,400, as our illustration of a protein sub- 

(a) Analysis of Elements Present in Proteins. Heat 
about 0.2 g of dried egg albumin with 1 g of soda 
lime in a small test tube. Hold a moistened red 
litmus paper at the mouth of the tube and note 
any change in color. Heat the tube strongly until 
a charred residue remains. Allow the tube to cool, 
and then add about 5 ml of 6 F HC1 to the residue. 
Hold a small piece of filter paper, moistened with 
0.1 F lead acetate solution, over the mouth of the 
tube. If a dark stain of PbS is produced, it is due 
to H 2 S gas evolved as a result of a series of re- 
actions from the sulfur in the protein. What evi- 
dence for the presence of carbon, hydrogen and 
oxygen did you observe during the above tests? 



(b) Color Tests for Proteins. Dissolve about 0.5 g 
of dried egg albumin in 50 ml distilled water for 
use in the following tests. 

The Biuret test indicates the presence of the 
peptide linkage C N , which is present in all 

O H 

proteins. Prepare a very dilute solution of copper 
sulfate by diluting 2 drops of the 0.1 F CuSO 4 re- 
agent with 1 ml of water. Now add 1 ml of 6 F 
NaOH to 1 ml of the egg albumin solution. Mix 
and add 2 drops of the diluted CuSO 4 solution. 
Watch for a characteristic color to develop. 

The Xanthoproteic test is given by most pro- 
teins. To 1 ml of egg albumin solution add 5 drops 
of 16 F HNO 3 . Heat the mixture for a few minutes. 
Note the color produced. Allow the tube to cool 
and cautiously add an excess of 15 F NEUOH. Note 
the intensification of color which results. Place a 
drop of 16 F HNO 3 on your fingernail, wash it off 
with water and apply a drop of NH 4 OH. Repeat 
the test using a piece of wool yarn or cloth. The 
student often inadvertently makes the Xanthpro- 
teic test when he spills nitric acid on his skin or on 
wool and silk clothing. 

(c) Coagulation of Proteins. Add one drop of 6 F 
acetic acid to 3 ml of egg albumin solution and 
heat the test tube to boiling for several minutes. 
What change in the structure of the protein mole- 
cule is proposed to account for the change ob- 
served? Is the coagulated protein soluble in water? 
To 3-ml samples of egg albumin in each of three 
test tubes, add a few drops of 0.1 F lead acetate to 
one, of 0.1 F HgCl 2 to the second and of 0.1 F 
CuSO 4 to the third. Note the precipitates formed 
in each case, and comment on the effect these salts 
might have if taken internally. Can you suggest 
why raw egg whites are recommended as an anti- 
dote for mercury poisoning? 

(d) Amphoteric Nature of Proteins. Dilute 1 drop 
of 6 F HC1 with 30 drops of water. Add one drop 
of this diluted acid and 1 drop of methyl orange 
indicator to 3 ml of water in a 15-cm test tube. The 
solution should have a red color, a pH. of about 3. 
Mix in about 5 ml of egg albumin solution and note 
the color now. Methyl orange turns yellow when 
the pU is increased. 

Prepare some dilute sodium hydroxide by add- 
ing one drop of 6 F NaOH to 30 drops of water. 

Add one drop of the dilute base and a drop of 
phenolphthalein indicator to 3 ml of water in a 
15-cm test tube. Stir in 5 ml of egg albumin solu- 
tion and note any change in the color of the in- 
dicator. Explain with suitable equations the re- 
actions which occurred when the protein solution 
was added to the acid and to the base. Could it 
be said that proteins act as buffers? 

3. Fats, Their Properties and Characteristic 

(a) The Solubility of Fats. Place 5 drops of olive 
oil, the glyceryl ester of olcic acid, into each of 
three 10-cm test tubes. To the first tube add 5 ml 
of water, to the second 5 ml of acetone and to the 
third add 5 ml of water to which has been added 
a pinch of soap chips. Shake each tube well and 
allow to stand for a few minutes. Observe whether 
solution or emulsification has taken place. 

(b) The Saponification of Fats. Add 2 ml of olive 
oil and 1 g of NaOH to 20 ml of ethyl alcohol in a 
50-ml Erlenmeyer flask and heat on a water bath 
maintained at about 75 C for 15 minutes. Cool 
the solution in a beaker of cold water. Remove a 
small bit of the solid product (about the size of a 
pea) with a stirring rod and place it in a 15-cra 
test tube two-thirds full of distilled water. Place a 
pipette into the test tube and blow gently through 
it. What is the nature of the substance dissolved 
in the tube? 

Dissolve most of the remainder of the solid in 
the flask in some distilled water and place the solu- 
tion in a 15-cm test tube. Acidify with a few drops 
of 18 F H 2 SO 4 . Mix with a stirring rod and note 
the oily material which collects on top of the test 
tube. Dip a piece of neutral litmus paper into the 
oily layer and note whether it is neutral like the 
original oil or acid. What is the oily material? 
Write equations for the saponification of the olive 
oil and also for the reaction that takes place upon 
acidification of the saponified product. 

(c) The Acrolein Test for Fats. In a dry Pyrex test 
tube place about 1 g of KHSO 4 and 5 drops of olive 
oil. Heat the test tube rather strongly for a few 
minutes, and then allow it to cool before noting 
the characteristic odor of acrolein. Do not put your 
nose aver the tube while it is hot. Write the equation 
for the reaction by which the unsaturated alde- 
hyde, acrolein, is produced by the loss of two moles 



of water by the glycerol portion of the fat mole- 
cule. Will this test work for all fats? 

4. Analysis of the Various Food Constit- 
uents Present in Milk. 

(a) Precipitation of Protein Material. Dilute 20 
ml of fresh, whole milk with an equal amount of 
water in a 50-ml Erlenmeyer flask. Warm to about 
50 C. Dilute 1 ml of glacial acetic acid with 10 
ml of water and add this dilute acid drop by drop 
to the warm milk. Shake the flask after each addi- 
tion and look for the appearance of curds of casein. 
About 10 to 20 drops usually are sufficient; avoid 
an excess of acetic acid. Allow the casein to settle 
and decant the clear liquid through a filter paper 
in a funnel. Save the filtrate for Part (c). Transfer 
the casein to the funnel, wash it with water, and 
let it drain. Remove the casein to a pad of several 
pieces of filter paper and press it reasonably dry. 

Apply the Biuret and the Xanthoproteic tests 
to small, portions of the casein by following the 
directions given earlier in this experiment. 

(b) Extraction of Fat. Place the remainder of the 
casein in a small beaker and work 5 ml of acetone 

into the coagulated material for several minutes. 
Decant off the acetone into a small evaporating 
dish. Extract the casein with another 5-ml portion 
of acetone and add this solution to the previous 
extract. Evaporate away the acetone over a beaker 
of boiling water. Note the appearance of the resi- 
due and the nature of the spot produced when some 
filter paper is touched to it. There will not be 
enough fat extracted to run the Acrolein test, but 
if some cream off the whole milk is available it may 
be tested in this way. 

(c) Test for Carbohydrates. Heat the filtrate ob- 
tained in (a) to boiling for a few minutes. Filter out 
the additional proteins coagulated by heating. Cool 
the filtrate and add about 2 ml of 0.1 F lead acetate 
to precipitate out other proteins. Filter again and 
test the filtrate for the presence of milk sugar, 
lactose, which is a reducing disaccharide, by apply- 
ing the Fehling solution test to 1 or 2 ml of the 
clear solution. If you wish, you may allow the 
filtrate to evaporate until the next laboratory 
period and note the crystals of lactose which re- 

REPORT: Exp. 32 

Some Elementary Experiments 
in Biochemistry fi-faL _ 

Locker Number- 

1. Carbohydrates 

(a) Describe the appearance of the Molisch test. 

This test was given by the following carbohydrates tested: 

(b) The Fehling test consists of the reduction of the complex ion in an alkaline medium 

to the insoluble . The group responsible for the reduction is . 

The carbohydrate (s) which gave a positive test . What is the molecular formula of 

glucose , of sucrose , of starch ? 

How do you account for the positive Fehling test obtained after sucrose was boiled in an acid solution? 

(c) The Hydrolysis of Starch. 

Describe the appearance of the test obtained when dilute iodine solution was added to a starch solution. 

_, to glucose . What changes in color were noted in the iodine- 

starch test as the hydrolysis of the starch proceeded ?_ 

What happened to the starch molecule during hydrolysis? 

Account for the Fehling test obtained after the hydrolysis was complete* 

(d) (Optional Experiment). The Enzymatic Hydrolysis of Starch. 

How do you account for the relative ease of hydrolysis of starch under the conditions of this experiment 

compared to those in Part (c) ? 

What is known about the chemical nature of enzymes? 

How are carbohydrates digested in the body? 

(e) The Hydrolysis of Cellidose. 

How does the complexity of the cellulose molecule compare with that of starch? 

Was the Fehling test positive after the treatment with concentrated sulfuric acid and boiling? 

What happened to the cellulose molecule during the boiling treatment? 


Can the digestive system of man hydrolyze and use cellulose?- 
Can animals use cellulose as food? . 

2. Proteins 

(a) Elementary Analysis. 

What elements were shown to be present by the reaction of soda lime on egg albumin?- 

How are these elements present in the protein molecule? 

What are the basic building units which combine to make up the complex protein molecules? 
What was the basis of the test for sulfur? 
Write the formulas for two amino acids containing sulfur. 

What evidence for the presence of carbon, hydrogen and oxygen did you observe during the above tests? 
(b) Describe the appearance of the Biuret test __ 

Is this test given by all proteins ? Why ?_ 

The Xanthoproteic test consists in the production of a color when the protein is 

treated with concentrated nitric acid. When treated with concentrated ammonium hydroxide the color 

(c) Coagulation of Proteins. 

Describe the change which took place when the slightly acidified protein solution was heated to boiling. 

What change in the structure of the protein molecule is proposed to account for the change observed? 
Is the protein now soluble in water? . 

Describe the results obtained when the salts of heavy metals were added to tlje egg albumin solution. 

Are these salts toxic when taken internally? . What is the antidote usually 

prescribed? . Why? 

(d) Amphoteric Nature of Proteins. 

Refer to Table XII, Appendix II, on the colors of indicators, and determine the effect of adding the protein 

solution to a) the slightly acid solution and b) the slightly 

basic solution . 

Explain the results obtained by writing suitable equations for the reaction which took place in each case. 

Protein acting as a base: 

Protein acting as an acid:__ _ 

Do proteins act as buffers in the body? Explain; 


Report on Exp. 32, Sheet 2 Name.. 

3. Fats 

What is the structural formula for the fat olein, which is the major component of the olive oil used in this 

(a) Solubility. Record the case(s) in which the olive oil dissolved, , in which emulsifi- 

cation took place . 

(b) The Saponification of Fats. What is the meaning of the term 
The equation for the saponification of olive oil by NaOH: 

What were the properties of the aqueous solution of the product obtained by saponification? 

When the water solution was acidified what was the oily product formed? . Write 

the equation for the reaction of acid on the saponified product: 

(c) The Acrolein Test. Describe the odor of acrolein 

Is this test applicable to all fats ? , 

Write the equation for the reaction by which acrolein is produced by dehydration of the glycerol portion 
of the fat molecule: 

4. Analysis of milk. 

(a) Protein Constituents. Describe the nature of the material which precipitated out from the slightly acidified milk. 

What is the material called ? . What are some of its 

uses ? 

Describe the appearance of the Biuret test , 

and of the Xanthoproteic test- 

(b) Extraction of Fats. Did the acetone extract any of the fats which were present in the coagulated protein 

fraction ? , 

How did you recognize the residue obtained upon the evaporation of the acetone as a fat? 

If a sample of cream or butterfat was available did you obtain a positive acrolein test? 

(c) Carbohydrates Present in Milk. What was the nature of the material precipitated by boiling the filtrate ob- 
tained in Part (a)? ; by the lead acetate solution? 

Did the resulting clear solution give a positive Fehling test? 

What is the sugar present in milk? _. 


thermochemistry. The Heat of Neutralization. 

Review of Fundamental Concepts 


Co/fege Chemistry, Chapter 2$ 

Chemical thermodynamics is the study of 
the amounts of energy involved during chemical 
changes. Such questions as how much electrical 
energy can be produced by a certain amount of 
chemical change taking place in an electrical cell 
or battery are considered. In other instances it is 
concerned with the amount of energy in the form 
of work that can be done by chemical means. In 
the present experiment we shall be concerned with 
the amount of energy in the form of heat which is 
involved during a chemical reaction. This part of 
thermodynamic chemistry is called thermochem- 

Reactions which involve the evolution of heat 
are called exothermic reactions, while those which 
absorb heat are said to be endotherraic reactions. 
The general term, heat of reaction, may be classi- 
fied into several more specific categories. For in- 
stance, the heat of combustion is the amount of 
heat evolved when one mole of a combustible sub- 
stance, such as carbon or methane, undergoes com- 
bustion. The heat of formation of a compound is 

the amount of heat involved when one mole of the 
compound is formed directly from the elements 
composing it. The heats of vaporization, fusion and 
sublimation are concerned with changes in state. 
The heat of neutralization, which we shall measure 
in this experiment, is defined as the amount of 
heat evolved when one mole of water is produced 
by the reaction of an acid and a base. 

The heats of reaction are usually expressed in 
the units: calories per mole. They are experiment- 
ally determined by allowing the reaction to take 
place in an insulated device called a calorimeter. 
The heat involved is determined by noting the 
change in temperature in the water surrounding 
the reaction chamber. Since the heat capacity 
(specific heat) of water is one calorie per gram per 
degree centigrade, the heat involved can be ob- 
tained by simply multiplying the weight of the 
water present by the change in temperature. A 
correction for the heat which is absorbed by the 
calorimeter apparatus must also be taken into ac- 
count in the calculations. 

Experimental Procedure 

Special supplies: A thermometer, preferably one graduated 
to 0.1 or 0.2 C. Less precise results will be obtained with a 
thermometer with 1 graduations. 

Chemical*: 1 F NaOH, 1 F HC1, 1 F HNO,, 1 F HCaH.0* 
1 F KOH. 

To measure the heat of neutralization we shall 
allow solutions containing equivalent amounts of 
acid and base to react in a simple calorimeter con- 
sisting of a 250-ml beaker inside a slightly larger 
400-ml beaker. Some loosely crumpled filter paper 
should be placed in the bottom of the larger beaker, 
and a few strips of paper may be inserted between 
the beakers to insulate the sides. A square of card- 
board, with a small hole in the middle for the 
thermometer, may serve as a lid. 

In order to determine the correction to be made 
for the heat lost to the calorimeter and thermome- 
ter, the following determination, run under similar 
conditions to those to be used later, should be made 

in duplicate. Obtain about 250 ml of distilled water 
and adjust its temperature to that of the room or 
very slightly below room temperature. Carefully 
measure exactly 100 ml of this water into the 
calorimeter, using a clean graduated cylinder. Ob- 
tain another 250 ml of distilled water and heat it 
to approximately 10 C higher than room tern* 
perature. Measure out exactly 100 ml into the 
graduated cylinder and determine its temperature 
as accurately as possible, estimating the nearest 
0.1 C. Transfer the thermometer to the calori- 
meter and read the temperature of this water 
again. Note the time, then add the measured warm 
water to the calorimeter, and stir thoroughly with 
the thermometer. Read the temperature at the end 
of each minute for five minutes. Plot on some graph 
paper the temperatures to the nearest 0.1 C as 
ordinates against the time in minutes as abscissa* 




Extrapolate the curve connecting the plotted points 
back to zero time to obtain the highest tempera- 
ture attained immediately following the mixing. 
Calculate the heat lost by the warm water by 
multiplying its decrease in temperature by its 
weight, 100 g. (Assume the density of water to be 
1.0 and that its specific heat is 1.0.) Calculate the 
heat gained by the cooler water by multiplying its 
increase in temperature by its weight, 100 g. The 
difference between the heat lost and that gained by 
the water is the heat gained by the calorimeter and 
thermometer. This number of calories, divided by 
the change in temperature of the calorimeter is 
called the water equivalent of the apparatus. Re- 
peat this determination to check your results. 

The following example is given to illustrate the 
use of the water equivalent in a determination: 
How much heat was liberated by a reaction taking 
place in a calorimeter, whose water equivalent is 4 
calories per degree, if the temperature of 100 ml 
of solution was raised 6.5 C by the reaction? 
Calories gained by the water = 100 X 6.5. (As- 
sume density of solution and its specific heat are 
each 1.0.) To this must be added the heat lost to 
the calorimeter: the water equivalent 4 cal/degree 
X 6.5 = 26 cal. Thus the heat involved in this 
reaction was 650 cal + 26 cal or 676 cal. 

1. The Determination of the Heat of Neu- 
tralization of a Strong Base by a Strong Acid 

Prepare to run duplicate determinations first 
with 100 ml each of 1 F NaOH and 1 F HC1 and 
then use a different pair: 1 F KOH and 1 F HNO 8 . 
Be sure that the temperature of each solution is 
at, or very nearly at, room temperature so as to 
avoid errors caused by mixing solutions at different 
temperatures. Use a clean, dry graduated cylinder 
to measure out exactly 100 ml of the solution of 
base into the calorimeter. Measure exactly 100 ml 
of the acid solution and accurately determine the 
temperature of each solution to the nearest 0.1 C. 

Note the time and pour the acid into the base and 
stir thoroughly. Read the temperature at one- 
minute intervals for five minutes. Plot the temper- 
atures against time and extrapolate to zero time so 
as to obtain the highest temperature reached when 

Record the data for both pairs of strong bases 
and acids, and calculate the heat of neutralization 
per mole of water produced. 

Write the net ionic reaction for the neutraliza- 
tion of a strong acid by a strong base. The accepted 
value for the heat of neutralization of strong acids 
and strong bases is 13,700 calories. How do you 
account for the fact that this value is essentially 
the same for all strong acid strong base neutrali- 

2. The Heat of Neutralization of a Weak 
Acid by a Strong Base 

Following the same procedure as outlined above, 
determine the heat of neutralization of the strong 
base, NaOH and the weak acid, acetic acid. Record 
the data obtained in the report sheet. 

How do you account for the different value ob- 
tained in this experiment? Write the equation for 
the net reaction involved here. How does it differ 
from the one for the neutralization of a strong base 
and strong acid? 

What value for the heat of neutralization would 
you expect from the reaction of ammonium hy- 
droxide and hydrochloric acid ? Explain the reason 
for your answer. 

Consider the following data for another weak 
acid, hydrocyanic acid, and its neutralization by 
sodium hydroxide. The heat of neutralization is 
only 2,900 calories per mole. For a strong acid 
strong base the value is 13,700 calories per mole. 
Ho^v many calories must be involved in the dis- 
sociation process of the weak acid ? Write equations 
for the three processes involved and include the 
heat term appropriate to each. 

REPORT: Exp. 33 

Thermochemistry, the Heat of 



Locker Number- 

Determination of the Water Equivalent of the Calorimeter 


Weight of water at lower temperature g 

Weight of water at higher temperature , g 

Lower temperature C 

Higher temperature __ _ C 

Final temperature after mixing , C 

(1) Heat lost by the warm water cal 

(2) Heat gained by the cold water __cal 

Heat lost to calorimeter (1) (2) _ cal 

Water equivalent of the calorimeter cal/C 

The Heat of Neutralization of Hydrochloric Acid and Sodium Hydroxide 

Weight of 1 F acid solution g 

Weight of 1 F base solution ._ _.. g 

Original temperature of solutions _ .. _ _ C 

Final temperature after reaction -.._-___ C 

Calories absorbed by aqueous solution 1 cal 

Calories absorbed by apparatus cai 

Total calories evolved *cal 

Number of moles of water formed moles 

Heat of neutralization cal/ mole 

Accepted value cal/mole 




_cul/ C 




1 Assume a specific heat of 1 calorie per gram per degree for these aqueous solutions, and also for those in the data on the next page, 


The Heat of Neutralization of Nitric Acid by Potassium Hydroxide 

1 t 

Weight of 1 F acid solution ......... __ - g - - g 

Weight of 1 F base solution ......... _ - g - g 

Original temperature of solutions ........ - C _ - C 

Final temperature after reaction ........ - - C - C 

Calories absorbed by aqueous solution 1 ...... --- cal - cal 

Calories absorbed by apparatus ........ - _cal -- cal 

Total calories evolved ........... - cal - cal 

Number of moles of water formed ....... - moles . - moles 

Heat of neutralization ........... - cal/mole . - cal/mole 

The net ionic equation for the neutralization of a strong acid by a strong base: 

How do you account for the fact that the heat of neutralization is essentially the same for all strong acids and 
strong bases? 

The Heat of Neutralization of Acetic Acid by Sodium Hydroxide 

1 t 

Weight of 1 F acid solution ......... _ g . - g 

Weight of 1 F base solution ......... , - g - g 

Original temperature of solutions ........ _ - C ----- C 

Final temperature after reaction ........ ____ C __ ----- ~_ --- C 

Calories absorbed by aqueous solution 1 ...... ____ cal ---- cal 

Calories absorbed by apparatus ........ _ __ cal -- cal 

Total calories evolved ........... __ cal -- _cal 

Number moles of water f owned ........ ___ moles _ moles 

Heat of neutralization ...... ..... _ cal/mole _ cal/mole 

How do you account for the different value obtained in this experiment? 

Write the equation for the net ionic reaction involved here:. 

How does it differ from that written for a strong acid strong base? 

What value for the heat of neutralization of ammonium hydroxide and hydrochloric acid would you expect? 


How do you account for the low value for the heat of neutralization of hydrocyanic acid? 

Write equations for the three processes involved and include the heat term appropriate to eacht 

1 See footnote on the proceeding page. 



Basic to the Qualitative Analysis Procedure 

To the Student: 

The study of qualitative analysis is not a separate subject; it is an integral 
part of the general chemistry course. As you undertake the study of the proce- 
dures of the scheme of qualitative analysis used in Experiments 39-44, you will 
bring into active service all of the important principles which you have learned 
so far. 

It is particularly important that you fully understand the principles relating 
to the equilibria of ion* in solution. The next group of experiments (E*p. 34-38 
and Study Assignment E) are presented as an integral part of the qualitative 
analysis section of this manual so that you can study these important principles 
first in separate experiments. 

For example, studies of the equilibria involving the ions of water and the 
ions of weak acids and bases should help you to understand how the concentra- 
tion of hydrogen ion (pH), or of other ions, such as hydroxide, sulfide, or car- 
bonate ions, may be calculated or controlled as is necessary in the group separa- 
tions. A study of the solubility product principle should make clear the factors 
which regulate the formation and dissolution of slightly soluble substances. 
Acquaintance with hydrolysis, buffer action, amphoteric substances, complex 
ions, and equilibrium constants such as K. p or Ki will make the directions in the 
qualitative scheme take on real meaning. Without this background the analytical 
procedure would mean little more to you, so far as chemical training is concerned, 
than would a set of recipes in a cookbook. 

Types of Equilibria The Ions 
of Water, pH, Hydrolysis 


College Chemistry, Chapter 20 

Review of Fundamental Concepts 

The Ions of Water 

A summary of the experimental observations re- 
garding the ionization of water should emphasize 
these two opposite viewpoints: 

(1) Hydrogen ions and hydroxide ions react with 
each other quite completely, as evidenced by the 
various titration experiments involving acids and 
bases, and as interpreted by the following net ionic 
equation for the reaction taking place when a 
strong acid reacts with a strong base, 

H++OH- *HA 

(2) The electrical conductivity of very pure 
water, while too small for any but the most refined 
measurements, indicates a slight but definite dis- 
sociation of the water molecule into hydrogen and 
hydroxide ions, 

H 8 >-H++OH-. 

Since these two opposing tendencies must be going 
on all the time, we have in any sample of water, 
pure or impure, the equilibrium represented by the 
reversible reaction: 

In water, when we have no other ions present 
which will unite with either the hydrogen ions or 
the hydroxide ions, the concentration of the ions, 
at 25 C, from conductivity data, is: 

concentration of hydrogen ion, (H+) 10~ 7 M 
concentration of hydroxide ion, (OH) - * 10~ T M . 

The Equilibrium Constant Expression 
for the Water Reaction 

The equilibrium constant for the water reaction 
may be written: 

or (H+XOH-) 


since the quantity, (H 8 O), is practically a con- 
stant, Ki(H 2 O) will also be a constant, K, and the 
expression may be written, for any aqueous solu- 
tion, at 25 C, 

(H+)(OH-) - K - I(T M . 

The evaluation of the constant, K, as 10 ~ 14 is ob- 
tained by substituting the actual concentrations of 
the ions in pure water in this equation. However, 
the same constant, of course, applies in any situa- 
tion where water is present* in pure water, and ia 




aqueous solutions of acids, bases, or salts. It is, 
therefore, possible to calculate the hydrogen ion 
concentration in a basic solution and the hydroxide 
ion concentration in an acid solution. Thus, in a 
0.01 F hydrochloric acid solution, the hydroxide 
ion concentration is 10~ 12 M 9 since 10~ 2 M 11+ 
X 10- 12 M OH- equals the constant 10~ 14 . 

The pH of a Solution 

Chemists frequently designate the relative con- 
centrations of hydrogen ion and hydroxide ion in a 
solution, especially between the range from 
1 M H+ (10- 14 MOH-) to 1 MQII- (10- 14 M H+), 
by a different system of notation, based on the 
concentration of hydrogen ion alone. This is called 
the pH of the solution and is defined by the equa- 

pll - log (H+). 

The meaning of pH, which, you should note, is a 
logarithmic function, is illustrated by the following 
selected corresponding values: 

IMH+orlQ MH+ 10-"MQH- pH 

10r* M H+ 10- 10 3/OH- pH = 4 

10-' MR+ 10~ 7 M OH- ?II=* 7 

10- u JtfH+ 10-' J/OH- pH=ll 

10- 14 MH+ 10 JfOH-orlMOH~ pH 14 

Note that in any such cases, when the hydrogen 
ion concentration is a simple power of 10, the pH is 
the exponent of 10 with the minus sign dropped. 

To calculate the pH for intermediate values 
of the hydrogen ion concentration, which cannot 
be expressed as simple, integral powers of 10, we 
simply look up the logarithm of the hydrogen ion 
concentration and change the sign. Be careful to 
include both mantissa and characteristic in the 
expression for the logarithm of the hydrogen ion 
concentration. Note the following: 

Example: What is the pH of a 1 F acetic acid 
solution, in which the hydrogen ion concentration 
is 0.004 M or 4 X 10~ 3 

pH - log (H+) - - log (4 X 10-') 

- (log 4+ log 10-') 

= - (0.60 - 3) - - (-2.4) = 2.4 

The pH is 2.4; that is, the number 4 X Itf" 1 1(T 2 *. 

To calculate the hydrogen ion concentra- 
tion from the pH, which is the reverse of the 
above calculation, note the following cases: 

Example 1: What is the hydrogen ion concen- 
tration of a solution which has a pll of 4.3? 

If the pH is 4.3, the H+ concentration is 10~ 43 M. This 
exponent, or logarithm, may be broken into the positive 
mantissa and the negative characteristic thus: 

io-< 3 = 10 7 ~' = 10 7 x io- 6 

= 5.0 X KT 6 (The antilog of 0.7 is 5.0.) 
The H+ concentration is 5.0 X 10~ 5 M, which is the same 
as 1CT 4 3 M . 

Example 2: What is the hydroxide ion concen- 
tration of a solution which has a pH of 11.7? 

This is a basic solution, but we first calculate the hydrogen 
ion concentration as in Example 1 : 

If the pH is 11.7, the H+ concentration is KT 11 - 7 1/. 

lOrii.7 = lO-3- 12 =5 10 3 X 10~ u 

= 2.0 X 10~ 12 (Antilog of 0.3 is 2.0.) 

The H+ concentration is 2.0 X 10~ 12 M. We use this to 
calculate the OH" concentration, from the ionization equi- 
librium for water: 

(OH-) = 

(H+XOH-) = 10-", or 

!?!!!=: 1 x icr 14 

(H+) 2 X 10~ 12 

0.5 X IO- 2 or 5.0 X 10~ 3 M OH". 

The Hydrolysis of a Salt 

The word "hydrolysis" means "reaction with 
water." Certain salts and a number of organic com- 
pounds hydrolyze. The hydrolysis of a salt is fre- 
quently spoken of as the reversal of the neutraliza- 
tion process. Hydrolysis always involves a shift to 
the right of the reaction 

This reaction will be shifted to the right whenever 
salts of weak acids or bases are present, resulting 
in a solution which will be acidic if the OH"" con- 
centration has been reduced by the formation of a 
weak base, basic if the II + concentration has been 
reduced by the formation of a weak acid, or still 
neutral if both H+ and OH~ ions have reacted with 
the ions of the salt to the same extent. 

As an example of hydrolysis, let us study the be- 
havior of a 0.1 F sodium cyanide solution, NaCN. 
Such a solution is distinctly red to phenolphtha- 
lein, showing an excess of hydroxide ions. These 
hydroxide ions have resulted because the cyanide 
ions in the solution react with hydrogen ions from 
the water to form the weak acid HCN. More 
water then ionizes to maintain the equilibrium, 
thus creating an excess of hydroxide ions: 




V CN- 


Note that there is a competition of OH~ ions for 
H+ ions to form HOH, and also of CN~~ ions for 
H+ ions to form HCN. The ultimate position of 
the equilibrium will depend on how poorly the 
HCN is ionized as compared with the water. The 
above may be more conveniently represented by 
the net ionic equation for the hydrolysis: 

CN- + HOH -+ HCN + OH~. 

Note that water itself always enters into the hy- 
drolysis equation, not just the H+ ion (or OH~ ion 
in other examples). Since there is no tendency for 
Na+ ions to react with OH" ions, Na+ ion is not a 
part of the hydrolysis reaction. Note also that the 
hydrolysis equation is just the reverse of the equa- 
tion for the neutralization of the weak acid HCN 
by the strong base NaOH. 

Polyvalent ions of weak bases or weak acids will 
hydrolyze in steps. For example, in a solution of 
sodium carbonate, which is distinctly basic, the 
carbonate ion will hydrolyze in steps as follows : 

CO 8 + HOH : 
HC0 3 ~ + HOH . 

: HCOr + OH- 
: H 2 CO 3 + OH- 

In such cases the OH- formed by the first step 
represses the second step so that it occurs to a very 
slight extent, and no CO 2 results from the very 
small amount of H 2 CO 3 formed. 

Similarly, the hydrolysis of A1+++, Fe+++, 
Cu++, etc., results in an acid solution, but usually 
no precipitate of A1(OH) 8 , Fe(OH) a or Cu(OH) 2 is 
formed unless some ion is present which uses up 
the H+ ion formed and thus promotes the hy- 
drolysis to completion. In the preparation of 
reagent shelf solutions of ferric salts, stannous and 
stannic salts, and some others, additional H+ ion 
is added to reverse the hydrolysis so that the 
metallic hydroxides will not be precipitated. 

The Degree of Hydrolysis 

As explained above, the extent of hydrolysis de- 
pends on the formation of a weak acid, a weak 
base, or both, by the reaction of the ions of the 
salt with water. If only a weak acid or a weak base 
is formed, we may determine the extent of hy- 

drolysis by measuring the OH" ion or H+ ion 
concentration by means of indicators. 

For example, a 0.1 F Na 2 CO 8 solution is red to 
phenolphthalein and gives an orange-red color 
with alizarin yellow R, such that we may estimate 
the OH- ion concentration as about 0.004 M. As 
discussed above, the equation is 

CO, + HOH + HCOr + OH". 

Now if this reaction had proceeded to completion, 
the 0.1 M CO 8 " would have produced 0.1 M OH~. 
Since only 0.004 M OH~ resulted, the fraction of 
the salt hydrolyzed is 

0.004 M 

0.1 M 

0.04, or 4%. 

If both a weak base and a weak acid are pro- 
duced, the solution may remain about neutral, but 
hydrolysis will always occur to a greater extent 
than for either the positive ion or the negative ion 
separately. This is due to the effect of each reaction 
on the other. As an extreme example, if Fe+++ ion 
and CO 3 ion are mixed, the H+ ions and OH" 
ions produced by each separate hydrolysis will 
neutralize one another, so that each hydrolysis 
will take place to a greater extent, in fact through 
all the possible steps of hydrolysis. As a result, 
Fe(OH) 3 , not Fe 2 (C0 3 ) 8 , is precipitated, and CO, 
gas is evolved. The equations for these reactions 

Fe+++ + 3 HOH + Fe(OH), + 3 H+ 
CO," + 2 HOH -+ H 2 CO, + 2 OH~ 

Combining these by neutralization of the H+ and 

2 Fc++++3 CO," +6 HOH K2 Fe(OH),+3 H 2 CO, 


3 H 2 0+3 CO,. 

Preliminary Work 

Before starting the experimental procedure, com- 
plete the Indicator Chart in the report sheet. 
Record on the chart the colors of the indicators 
which you have studied in previous experiments: 
methyl violet, methyl orange, phenolphthalein, 
alizarin yellow R, and indigo carmine. Record these 
by inserting at each pH position the color (abbre- 
viate: B = blue, and so forth) given by that indi- 
cator at that pH. The chart is completed for methyl 
red as an example. 



Experimental Procedure 

Chemicals: Al (powdered), Al 2 (SO 4 )s or alum, NH|C|HA 
1 F NH 4 C 2 H 3 O 2 , (NHOaCO*. NH 4 C1, 1 F NH 4 C1, H+ and 
OH- solutions from 10~ M H+ to 10~* M OH", i.e. pU 4 to 
pH 11, 1 F NaC 2 H 8 Oj, NaHCO,, 1 F Na,CO,, 1 F NaCl, 

S (powdered), and the indicator solutions: methyl violet, 
methyl orange, methyl red, brom-cresol purple, brom-thymol 
blue, phenolphthalein, alizarin yellow R, and indigo carmine. 

1. Indicators that Change Color Near the 
Neutral Point. To complete the chart, informa- 
tion should be obtained on indicators which change 
color in the range from pH 5 to pH 8. Use the pre- 
pared H+ and OH~ solutions covering this range to 
determine the color change which the indicators 
brom-cresol purple, brom-thymol blue, and phenol 
red undergo in solutions of varying pH. For each 
indicator prepare a set of clean 10-cm test tubes, 
containing respectively 5 ml of the standard solu- 
tions varying from pH 5 to pH 8. To each test tube 
of one set add two drops of brom-cresol purple; 
mix this, and note the particular pH range in 
which the color change is most marked. Do the 
same for the other two indicators. Record the re- 
sults on the report sheet on the line corresponding 
to the indicator used. 1 Have the instructor ap- 
prove your chart before continuing with the ex- 
periment, so that any errors may be corrected. 

2. The Hydrolysis of Salts. The solutions to 
be tested for hydrolysis are: 1 F NaCJIaOj, 1 F 
NaCl, 1 F NH 4 C1, 1 F NH 4 C 2 H8O 2 , and 1 FNa*CO 8 . 
Test 5-ml portions of these salt solutions 2 with one 
or more indicators which change in color in the pH 
range, 5 to 9. If a solution is out of the range of 
these indicators, use other indicators that change 
in a still more acidic or basic solution, as your 
evidence dictates. In this manner, determine the 
approximate pH of each solution as accurately as 
you can. Record your results on the report sheet. 
In the case of sodium acetate, the hydrolysis is 
due to the formation of what poorly ionized sub- 
stance? What ion from the salt must, therefore, be 
involved in the hydrolysis? Write the equation for 
the net ionic reaction which takes place in the 
hydrolysis of a sodium acetate solution. Remember 
that, although H+ is used up in the reaction, it is 

1 You may compare your results, both for these indicators and 
for those used previously, with the data in Table XII, Appendix II. 

* Traces of impurities in the 1 F NaCl may cause incorrect results. 
Test a solution of a gram of reagent grade NaCi in 20 ml of distilled 

partially replaced by further ionization of water, 
so that the main substance used up is water rather 
than H+. Also write equations for any hydrolysis 
which takes place in the other solutions tested. 
3. The Degree of Hydrolysis* 

A. Salts of a Strong Base and a Weak Acid. In the 
hydrolysis of sodium carbonate, the net ionic re- 
action is 

CO 8 + HOH ^= HCOr + OH-. 

What would have been the hydroxide ion concen- 
tration, if the hydrolysis of the 1 F Na 2 C0 8 solution 
had proceeded to completion? By comparing the 
actual OH~ concentration just obtained by experi- 
ment with this hypothetical value for complete 
hydrolysis, calculate the percent of the salt which 
did hydrolyze. In a similar manner, calculate the 
percent of the sodium acetate which hydrolyzed in 
the 1 F NaC 2 H 8 O2. Carbonic acid, the acid in a 
"soda," is a much weaker acid than acetic acid, 
the acid in vinegar. Make a general statement re- 
lating the degree of hydrolysis with the strength 
of the ocid (or base) formed by the hydrolysis re- 

B. Salts of a Weak Base With Strong and With 
Weak Acids. 

(1) Hydrolysis of Ammonium Salts. Remove the 
stopper and smell (use caution) the bottles of solid 
ammonium chloride, ammonium acetate, and am- 
monium carbonate on the laboratory shelf. There is 
enough water adsorbed on each of the apparently 
dry salts to make hydrolysis possible. Explain the 
result, remembering that NH 4 OH is unstable and 
decomposes readily into NH and HjO. 

(2) Hydrolysis of Aluminum Sulfide. The in- 
structor will prepare a quantity of aluminum sul- 
fide as a class demonstration, using powdered 
aluminum and powdered sulfur in the proportion of 
about 4 g of Al to 10 g of S. The reagents are mixed 
thoroughly and transferred to a clay crucible which 
is placed in an extremely well ventilated place free 
from fire hazard, preferably outdoors. The mixture 
is then ignited by means of a fuse of magnesium 
ribbon. When cool, the product, Al 2 Si, is broken 
into small pieces with a hammer or in an iron 

Take a small piece of this aluminum sulfide, and 



ribbon fuse 

Clay crucible 
filled with a 
mixture of 
sulfur and 

This demonstration 
should be made 

a wise pre- 

FIG. 34-1. The preparation of aluminum sulfide. 

add it to 10 or 15 ml of water in a small beaker or 
test tube. Note any evidence of the formation of a 
gas or precipitate. Write the equation for this hy- 
drolysis reaction, remembering that the salt is 

present this time at the beginning as a solid, not 
as ions. What would you conclude about the rela- 
tive degree of hydrolysis in any case where the acid 
and base formed are both extremely weak, or 
where the products are both removed from the 
sphere of action by precipitation or gas evolution? 

4. An Application of Hydrolysis. Practically 
all baking powders consist of a mixture of sodium 
hydrogen carbonate with some solid substance 
which furnishes H+ when water is added. In alum 
baking powders, this H+ is furnished by the hy- 
drolysis of alum, KA1(SO 4 V12 H 2 O. 

Place about 3 grains of solid alum or of alumi- 
num sulfate in one dry beaker, and 3 grams of solid 
sodium hydrogen carbonate in a second dry beaker. 
Mix a little of each of these together in a third 
dry beaker. Is there a reaction? Now add 10 ml of 
water to each beaker and note the results. Test 
samples of the two separate salt solutions with 
litmus paper and suitable indicators to determ'ne 
their approximate pH. 1 Finally mix the two salt 
solutions. Explain the results and write net equa- 
tions for all reactions involved. 

1 The slight ionization of HCO S ~ as a very weak acid is counter* 
balanced by its hydrolysis, so that a hydrogen carbonate solution if 
slightly basic. On boiling, it changes to COt and CO: 

HCOr <"-*- H+ -f CO* (ioniaation) 
HCOr 4- HUH +i OH- + HiCOi (hydrolysis) 
2 HCOr > COr~ -1- HO + CO* (on boiling). 


REPORT* Exp. 34 

Types of Equilibria Ions 
of Water, pH, Hydrolysis 


Locker Number^ 

Preliminary Work, and 1. Indicators that Turn Near the Neutral Point. 

Complete the chart below, as directed. 

acidic neutral 




KT 1 


lO' 3 

10 4 



10' 7 


10' 9 

i<r' ? 




10" 4 









tO 7 


H>- 5 

10' 4 





? i i i t.g $ i s 8 K.H..J! a a....? 

Methyl violet 

Methyl orange 

Methyl red 

R R R R R 


Brom-cresol purple 

JBrom- thymol blue 

Phenol red 


Alizarin yellow R 

Indigo carmine 


1 2 3 4 5 6 7 8 9 10 11 12 13 14 

Abbreviations: B = blue, V = violet, BV = blue violet, P = purple, R = red, O = orange, Y = yellow, G green 
C colorless. Colored crayons may be used, if desired, to fill in lightly the actual colors obtained at various pll. The regioi 
showing the color change may be bracketed for emphasis. 

!. The Hydrolysis of Salts. 


Indicators Used 





1 M NaC 2 H 8 2 

1 M NaCl 



1 If Na 2 CO, 


In the case of sodium acetate the hydrolysis is due to the formation of: 
The ion from the sodium acetate involved in the hydrolysis is therefore: 

The net ionic equations for the hydrolysis, if any, of the various salt solutions are given below: 

Sodium acetate 

Sodium chloride 

Ammonium chloride 

Ammonium acetate 

Sodium carbonate 

3. The Degree of Hydrolysis 

A. Soft* of a Strong Base and a Weak Acid 

If a 1 F sodium carbonate solution hydrolyzed completely according to the 
equation given, the concentration of the hydroxide ion would be: M 

A comparison of the actual hydroxide ion concentration with this hypothetical value 
indicates that the percent of the salt hydrolyzed in this 1 F sodium carbonate solution is* % 

Method of this calculation: 

Similarly the percent of sodium acetate hydrolyzed in a 1 F sodium acetate solution is : % 

Method of this calculation : 

The general principle relating the degree of hydrolysis and the strength of the acid (or base) formed by the 
hydrolysis reaction is as follows: 

Considering both the positive ions and negative ions involved, which of the two 
salt solutions is hydrolyzed most, 1 F sodium acetate, or 1 F ammonium acetate? . _ 



Report on Exp. 84, Sheet 2 

B. Salts of a weak base with strong and weak acids. 

(1) Hydrolysis of ammonium salts. 

My results of smelling the three ammonium salts indicate that the one giving the 
(a) strongest odor of ammonia is 

(b) next strongest odor of ammonia is 

(c) weakest odor of ammonia is 

My explanation of the reason for this order is as follows: 

(2) Hydrolysis of aluminum sulfide. 

The equation for the preparation of aluminum sulfide is: 

The net ionic equation for the hydrolysis of this substance is: 

List below all the specific factors which you can think of which cause this reaction to go practically to com- 

4. An Application of Hydrolysis 

Approximate pH. of solutions of A^CSOJa , and NaHCOs- 

The net ionic equation for the hydrolysis reaction involved in the AlaCSO^a solution is: 

Combine this equation with the equation for the neutralization of the sodium bicarbonate: 

H+ + HCO 8 ~ ^r^ HaO + CO 2 , 
so as to eliminate H+ and give the final net ionic equation for the whole process: 


Application of Principles 

(Refer. to Table XI in the Appendix, if necessary.) 
1. For each of the salts listed, indicate in general whether its water solution would be acidic, basic, or neutral: 



NaN0 3 _ 

2. Consider the following list of salt solutions: 

1 F KNO 8 , 1 F Na 2 S, 1 F NH 4 C1, 1 F NII 4 C2ll3O 2 , 1 F KC 2 II 3 O 2 

Indicate which solution has the characteristic called for: 
The highest H+ The lowest degree 

concentration . of hydrolysis ________ 

The highest OH~ The two solutions 

concentration ___ most nearly neutral 

The highest degree 

of hydrolysis . _ . _ 

3. List the solutions given below on the lines at the right, in order of increasing pH 

(highest 11+ concentration at the top) : 

1 F Na 2 C0 3 1 F (NHOaSOi 

1 F NH 4 C 2 H 8 O 2 1 F IIC1 . 

1 F NaOH 1 F (NH 4 ) 2 CO 3 

4. Experiment: Some magnesium ribbon is placed in a warm solution of aluminum sulfate. Bubbles of a gas 
are formed. 

Write two net ionic equations to explain __ 

what you think occurs: 

5. Write equations for the hydrolysis of 
ferric ion in steps: 

6. Consider the two compounds NaHCOs and NaIISO 4 . One of these gives an acid reaction in solution, the 
other a slightly basic reaction. Explain these facts, both by words and by net ionic equations: 

7. A 0.5 F solution of sodium cyanide, NaCN, has a pll of about 12. Write the hydrolysis equation, and calcu- 
late the degree of hydrolysis. 


Types off Equlllbrla-Sllghtly Soluble 
Substances. Coordination Compounds. 


Co//ege Chemistry, Chapters 21, 22 

Review of Fundamental Concepts 

Thus far, we have studied the equilibria which 
are established between the undissociated mole- 
cule and its ions in the case of pure water, in solu- 
tions of weak acids, of weak bases, and of their 
salts. We also have considered the equilibrium be- 
tween a slightly soluble solid substance and its ions 
in solution. In this experiment, we shall review 
these principles in an experimental study of the 
solubility of metallic hydroxides. We shall also in- 
troduce a new type of equilibrium which involves 
the formation of "complex ions," as discussed in 
the following paragraphs. 

Coordination Compounds and Complex Ions 

Most of the common ions, for which we usually 
write simple formulas, as Cu++, Zn++, A1+ ++, 
H+, Cl~, and so forth, in reality are hydrated, both 
in solution, and in many crystals. In such hydrated 
ions, a number of water molecules are grouped in 
a stable configuration about the central ion. The 
more complex ions permit a greater number of 
water molecules to be grouped about them. Several 
examples are: H 8 O+ or H(H 2 O)+, Cu(H 2 O) 4 ++, 
A1(H 2 O) 8 ++- 1 -, and Fe(H 2 O) 6 + + + . The formation 
and stability of such hydrated ions are related to 
the fact that the water molecule possesses un- 
shared pairs of electrons and is quite polar (see 





FIG. 35-1. Three ways to represent the polar character of 
the water molecule. The hydrogen to oxygen to hydrogen bond 
is not linear, but forms an angle of about 105. The oxygen side 
of the molecule is more negative than the hydrogen side. 

Fig. 35-1). Water molecules thus tend to group 
about a cation so that their unshared pairs of elec- 
trons form or complete a noble gas electron struc- 

These hydrates constitute one example of the 
type of substances which are called coordination 
compounds. Other neutral, but polar, molecules, 
such as ammonia, NH a , and also a number of ions, 

as OH-, C1-, CN-, S 2 O 8 , and C 2 O 4 ~, likewise 
can form similar, very stable, coordination group- 
ings about a central ion. These coordination com- 
pounds may form by the replacement of the water 
molecules from the hydrated ion by these other 
molecules or ions which are present in the solution 
at high concentration, to form a more stable bond. 
The resulting coordination compound may be a 
positively or negatively charged ion (a complex 
ion), or it may be a neutral molecule, depending 
on the number and kind of coordinating groups at- 
tached to the central ion. A few examples are: 
NH 4 + or H(NH 3 )+, Cu(NH,) 4 ++, AuCl 4 -, 
HgCl 4 , Fe(CN). , and Ag(CN),-. Such com- 
plex ions may be diagrammed or illustrated in 
various ways, as shown in Figure 35-2. 

The ammonia complex ions 1 are formed in the 
presence of a high concentration of ammonium 
hydroxide. In this latter solution, we have several 
substances in equilibrium, as represented by the 

NH, + H 2 O qpfc NH 4 OH +. NH 4 + + OH". 

We shall study these ammonia complexes in this 
experiment and show how we may decide which 
substance in the above equilibrium is responsible 
for the formation of the complex. 

Hydroxide Complex Ions, or Amphoterlc 

The hydroxides of most metals are insoluble in 
water, so that when sodium hydroxide, for ex- 
ample, is added to a metal ion in solution, as lead 
ion, Pb + +, we get a precipitate, 

Pb++ + 2 OH~ ^=>- Pb(OH) 2 . 

It would be expected that excess hydroxide ion 
would give more complete precipitation. Instead, 
the precipitate dissolves. This -is explained by the 
tendency of lead ion to form a coordination com- 
pound with additional hydroxide ion, 

Pb(OH), + 2 OH- '?= Pb(OH) 4 . 
This may be written as a replacement of the water 

1 See Pauling, College Chemistry, page 489, for a table of important 
ammonia complex ions. 




f<m^ law 


V>V. 1 '.;.V^'jff 

,-.'' : '"- : % 


H H 


H,N NH 3 

I i 

N-C Fc^-C-N 


Fc(CM) 6 " 

Fio. 35-2. Diagrams showing the spatial arrangement of the coordinating groups about a central ion in the formation of 
a complex ion. In the bracketed diagrams (immediately above the conventional formulas in each case) the lines represent a 
bonding pair of electrons. Above these, the groups are represented at the points of tetrahedra or octahedra, or in the case of 
Cu(NH) 4 ' t " ( " as a square coplanar structure. In these, the atomic diameters have been reduced to one-third their correct size 
M compared to the interatomic distances, to show their relative spatial positions. The upper figures show these in their proper 
relative dimensions. 

molecule in the hydrated formula, as follows: 1 




_H,0 OH 

or with 

+ OH- 




H 2 O OH 


\ / 


Simple Ion 


Hydroxide Complex Ion 

Pb(OH) 2 

Pb(OH)4 plumbite ion 


Zn(OH) 2 

Zn(OH) 4 , zincate ion 



A1(OH)4~, aluminate ion 



Cr(OH)4~, chromite ion 


Sn(OH) a 

Sn(OH) 4 , stannite ion 


Sn(OH) 4 

Sn(OH)e 9 stannate ion 

A number of metal ions, but not all, behave in 
a similar manner. Examples of these are indicated 
in the following table. 

1 There is some uncertainty as to whether Pb(OH) t ~, and also 
Sn(OH)t-, will coordinate further with hydroxide ion to form 
Pb(OH),T"t and Sn(OH) 4 ~. 

These hydroxide complex ions frequently have 
been written in the anhydrous form: PbOa , 
Zn0 2 , A1O 2 -, CrO a ~, SnO t ~, and SnO s ~. 
(Such formulas may be derived from the hydroxide 
complex ion formulas simply by subtracting 2 or 3 



The reactions to form these hydroxide complex 
ions are entirely reversible. If an acid is added to 
the above, we have first the re-precipitation of the 

Pb(OH) 4 + 2 H+ :=+ Pb(OH), + 2 EM). 

With an excess of acid, this dissolves as the simple 
metal ion, 

Pb(OH) a + 2 H+ :?T* Pb++ + 2 HjO. 
Such metallic hydroxides, which may be dissolved 

by an excess of either a strong acid, or a strong 
base, are called amphoteric hydroxides. For ex* 
ample, with aluminum hydroxide, we have, with 
excess acid: 

Al(OH), + 3 H+ :j=* A1+++ + 3 HA; 
or with excess base: 

Al(OH), + OH- :=+ A1(OH) 4 - 

We shall study the nature of this type of equilib- 
rium experimentally. 

Experimental Procedure 

Chemwh: I F NH 4 C1, 1 F NH 4 OH, 0.1 F CaCl 2 , Ca(OH) 2 
solid, 0.02 F Ca(OH)* CuS<V5HA 0.1 F CuSO* 0.1 F 
FeCU, 0.1 F Mg(NO,)* 0.1 F AgNO,, 0.1 F Zn(NO,)* 

1. Slightly Soluble Metallic Hydroxides. The 

slightly soluble metallic hydroxides give sufficient 
OH" concentration in solution to carry out con- 
venient test tube reactions and, thus, determine 
experimentally the relative concentrations of OH"* 
in their saturated solutions, as follows: 

(a) To 2 or 3 g of solid Ca(OH) 2 in a small 
beaker, add 5 ml of 1 F NH 4 C1. Warm the mixture. 
The solid Ca(OH)2 will, of course, dissolve until a 
saturated solution is formed. Is there any evidence 
that this can react with 1 F NH 4 C1? What can you 
conclude as to the relative concentration of OH~ 
in saturated Ca(OH) 2 and 1 F NH 4 OH? As a 
further check on your reasoning, test 3 ml samples 
of saturated Ca(OH) 2 (limewater), and 1FNH 4 QH 
with indigo carmine indicator, to discover which 
has the higher OH" concentration, 

(b) Would you expect to get a precipitate of 
Ca(OH) 2 when 0.1 F CaCl 2 and 1 F NH 4 OH are 
mixed? Try it. Try also 6 ml of 0.1 F CaCl a with 
1 ml of 6 F NaOH. Note the results. 

(c) Try 2 ml of 0.1 F FeCU with 2 ml of 1 F 
NH 4 OH. Which hydroxide furnishes more OH"" to 
a solution, saturated Fe(OH) 8 or 1 F NH 4 OH? 

(d) Prepare two test tubes, one containing 2 ml 
of 1 F NH 4 OH + 1 ml of 1 F NH 4 C1, and the other 
containing 2 ml of 1 F NH 4 OH + 1 ml of H 2 O. 
To each of these, add 1 ml of 0.1 F Mg(NO 8 ) 2 . 
Write net ionic equations for any reactions and ex- 
plain the difference in the two cases. Compare the 
OH" concentration in each of the three cases, 
saturated Mg(OH) 2 , 1 F NHiOH, and 1 F NH^OH 
containing NH 4 C1. 

(e) To the mixture of NBUOH and NBUC1 used 
in (d) f now add a few drops of 0.1 F FeCl*. Was 

there still an appreciable concentration of OH"" in 
the mixture? 

Summarize all your results in this section by 
arranging all the substances studied in the order 
of decreasing OH" concentration, as provided in 
the table in the report sheet. 

2. The Formation of Insoluble Hydroxides 
and Oxides. When a high concentration of OH"" 
is added to a metallic ion, either the metallic hy- 
droxide or the corresponding oxide may be pre- 
cipitated. This depends on the relative insolubility 
of the hydroxide as compared with the oxide, and 
on the rate of the decomposition of the hydroxide 
into the oxide and water. In some cases, the pre- 
cipitate is of variable and rather indefinite com- 
position, as in the case of ferric hydroxide above, 
which probably is more properly written Fe 2 O* 
xH 2 O, and is sometimes called a hydrous oxide. 

(a) To 3 ml of 0.1 F CuSO 4 in a large test tube, 
add 1 ml of 6 F NaOH. Note the results at room 
temperature, then heat the solution and note any 
changes. (Caution: Be very careful never to point 
a test tube containing hot alkali or hot acid toward 
anyone.) The black precipitate formed is cupric 

(b) When OH" is added to silver ion at room 
temperature, the oxide, not the hydroxide, pre- 
cipitates. Try it with 1 ml of 0.1 F AgNO, and a 
drop of 6 F NaOH. 

3. The Formation of Complex Ions. To 3 ml 
of 0,1 F CuSO 4 , add a drop of 6 F NH 4 OH. Note 
the result. Now continue to add NH 4 OH a little 
at a time, with shaking, until a distinct change 
has taken place. Is this contrary to the law of Le 
Chatelier? Since it is obvious that the cupric hy- 
droxide dissolved while the OH" concentration was 
increasing, how must the Cu+ + concentration have 
changed? Did it increase or decrease? 

To learn which of the substances present in 



ammonium hydroxide solution is responsible for 
the change, try adding 1 ml of 1 F NH 4 C1 to 1 ml 
of 0.1 F CuSO 4 . Try also the effect of ammonia gas 
by placing several crystals of solid cupric sulfate 
in a dry beaker. Also place in the beaker, but not 
touching the CuSO 4 crystals, a piece of filter paper 
moistened with concentrated ammonia solution. 
Cover with a watch glass. Observe whether the 
ammonia gas liberated from the solution has any 
effect on the crystals. What do you consider to be 
responsible for the dark blue color, NH 3 or NH 4 +? 
Write an equation to show the formation of this 
new substance, when excess ammonium hydroxide 
solution is added to a cupric salt solution. (To 
assist you in the writing of correct formulas for 
this type of complex ion, note the following rule: 
The number of moles coordinated with one mole 
of the ion usually is twice the valence of the ion. 
In this case, the formula is Cu(NH 3 ) 4 " f "" l "0 

To 1 ml of cupric ammonia complex ion solu- 
tion, add 6 F HNO 3 in excess. Explain the result 
and write the equation. 

4. The Formation of Amphoteric Hydrox- 
ides. To a 3-ml sample of 0.1 F Zn(NO 3 ) 2 , add a 
drop of 6 F NaOH. Note the result. Now continue 
adding NaOH, with shaking, as long as a change 
occurs (2 to 3 ml may be needed). Is this result 
contrary to the law of Le Chatelier? Has the con- 
centration of Zn + + increased or decreased as more 
OH~~ is added? Write separate equations for each 

of the two steps of the reaction. Now to this 
strongly basic solution, add 6 F HC1 a little at a 
time. What is the precipitate? Continue to add an 
excess of HC1. What are the principal substances 
in the solution now? Write equations for these re- 

5. Reaction of Zinc Ion with Ammonia. 
When ammonium hydroxide solution is added to 
Zn++, zinc hydroxide first forms, then dissolves 
when excess ammonia is added. Does the zinc 
hydroxide dissolve as zincate ion Zn(OH) 4 , due 
to the excess base added, or does it dissolve as 
Zn(NH 3 ) 4 ++, due to the NH 3 molecules added? 
To answer this question take 5 ml of 0.1 F 
Zn(NO 3 ) 2 , and add 6 F NaOH drop by drop, with 
shaking, until the Zn(OH) 2 first precipitated has 
almost, but not completely, redissolved. Avoid an 
undue excess of NaOH solution. Find the OH"" 
concentration in this solution by dividing it into 
two portions. Test one with alizarin yellow indica- 
tor, and the other with indigo carmine. 

Now to a second 5-ml sample of 0.1 F Zn(NO 3 ) 2 , 
add 6 F NH 4 OH, drop by drop, until the precipitate 
of Zn(OH) 2 just about dissolves. Again find the 
OH~ concentrations, this time using phenolphthal- 
ein, and alizarin yellow indicators. What is your 
conclusion as to the possibility of forming zincate 
ion, Zn(OH) 4 , by adding ammonium hydroxide 
to a zinc salt solution? Explain. 

REPORT: Exp. 35 

Types of Equilibria Slightly 
Soluble Substances. Coordination 

1. Slightly Soluble Metallic Hydroxides 

(a) The equilibrium equation for a saturated solution 
of Ca(OH) 2 , with excess solid, is : 

The experimental evidence that Ca(OH) 2 (s) can 
react with NH 4 C1 solution is: 



Locker Number- 

The net ionic equation (above reaction) is: 

The reason this reaction will take place as indicated is: 

Color with indigo carmine: Ca(OH) 2 _ 

-, NH 4 OHL 


has the higher OH" concentration. 

(b) to (e) The net ionic equation for any reaction in each of the following mixtures is: 

CaCl 2 andNH 4 OH . 

CaCUandNaOH . 

FeCI 3 and NH 4 GH 

NH 4 OH, NH 4 C1, and Mg(NO 3 ) 2 . . 

NH 4 OH and Mg(NO 8 ) 2 

NH 4 OH, NH 4 C1, and FeCl 8 .... 

From the above experimental evidence, list in the chart below, the following hydroxide solutions, in the order 
of decreasing OH"- concentration: Saturated Fe(OH) 8 , saturated Ca(OH) 2 , saturated Mg(OH) 2 , 1 F NaOH, 1 F 
NH 4 OH, I F NH 4 OH+NH 4 C1. Optional honor problem: Verify the order you have written down, by calculating the 
OH- concentration in each case. Obtain any data needed from a handbook, and from Experiment 26 for the NH 4 OH 
solution. Omit the NH 4 OH+NH 4 C1 mixture, as you have not yet studied the methods for calculating this case. 

List of Hydroxides 1 


Method of Calculating the OH~ Concentration 



1 Note that this list of hydroxides is arranged in such an order that any substance lower down would tend to be formed by the reaction 
of its positive ion with any substance above it in the list. Compare with the equations you have written. 


2. The Formation of Insoluble Hydroxides and Oxides* 

Write net ionic equations for the reactions indicated, and give the color of any precipitates formed: 

CuSO 4 and NaOH (cold) _ L_ 

CuSO 4 and NaOH (hot) 

AgNO 8 and NaOH 

3* The Formation of Complex Ions. 

Write the net ionic equation for the reaction of CuSO 4 , in excess, with a small amount of NH4OH: 

From this reaction and from Le Chatelier's theory, would you expect the addition of excess NH 4 OH to forte 
more, or less, solid Cu(OH)2? 

From the actual result, how must the Cu+ + concentration have changed as more ammonium hydroxide is added 
What is the result of adding NH 4 C1 to CuSO 4 ? 

What is the result of adding NH 8 gas to CuSO 4 crystals? 

The equation for the reaction of excess ammonium hydroxide solution on copper sulfate solution is: 

What result did you observe, and what is the net ionic equation, for the reaction of HNO 3 on the producl 
formed above? 

4. The Formation of Amphoterlc Hydroxides. 

Write the net ionic equation for the reaction of Zn(NOj)2 in excess, with a small amount of NaOH: 

From this equation, and Le Chatelier's theory, would you expect the addition of excess OH~ to form more, 01 
less, solid Zn(OH) 2 ? 

From the actual result, how must the Zn ++ concentration have changed as more NaOH is added: 


Report on Exp. $6, Shed 


Explain in your own words why the Zn(OH)t solid dissolves with excess OH~~, and write the net ionic equation 
for the reaction: 



What is the result when a moderate amount of HCI is added to this strongly basic solution? 


What further change takes place when an excess of HCI is added? 
Observation : 

Equation : 

5. Reaction of Zn++ with Ammonium Hydroxide Solution* 


Indicator Color 

OH~ Concentration 

Zn(NO*)* + NaOH to 
almost dissolve ppt. 

Alizarin yellow R 

Indigo carmine 

Zn(NO 8 )2 + NH 4 OH to 
almost dissolve ppt. 


Alizarin yellow R 

What proof does this experiment give as to the coordination compound [Zn(OH)i or Zn(NHt)4++\ which 
Zn+ + forms with excess NH 4 OH solution? State your reasons clearly. 

The net ionic equation for the reaction of Zn++ with excess NH 4 OH solution is: 

Interpretation of Data 

1. Will calcium hydroxide, Ca(OH)2, precipitate when saturated barium hydroxide, Ba(OH)2, is mixed with 
an equal volume of 1 F CaCl 2 ? The solubility of barium hydroxide is 3.9 g per 100 g of H 2 O at 20C. See also Table 
XI in the Appendix. (Show calculations, and explain.) 

2. When ammonium hydroxide is added to zinc nitrate solution, a white precipitate forms, which dissolves on 
the addition of excess ammonia; but when ammonium hydroxide is added to a mixture of zinc nitrate and ammonium 
nitrate, no precipitate forms at any time. Can you suggest an explanation of this difference in behavior? 

3. Suppose you are given the following experimentally observed facts regarding the reactions of silver ion : 

(1) Silver ion reacts with chloride ion to give white AgCl, 

(2) Silver ion reacts with ammonia to form a quite stable complex ion, Ag(NHs)2 + , 

(3) A black suspension of solid silver oxide, Ag2O, shaken with sodium chloride solution, changes to white AgCl, 

(4) Solid silver chloride will dissolve when ammonia is added, but solid silver iodide does not dissolve under 
these conditions. 

Write equations for any net reactions in each of the above cases : 




Arrange each of the substances, AgCl, Agl, Ag 2 O, 
and Ag(NH 8 )2+ in such an order that their solutions 
with water would give a successively decreasing con- 
centration of Ag+ . 


The lonization Constant of a Weak Acid. 


Co//ege Chemistry, Chapter* 19, 20 

Review of Fundamental Concepts 

The Quantitative Treatment of Chemical 

In this experiment an introduction to the quan- 
titative study of equilibrium will be given through 
a study of the reaction 

HC 2 H,O, 

H+ + CJEM>t- 

It is a fundamental concept in all equilibrium situa- 
tions, as stated in the preceding experiment, that 
the rates of the forward and of the reverse reactions 
are equal. 

In reactions such as this example (where the 
coefficient of each constituent in the equation is 
one), the rate of each reaction, forward and re- 
verse, is proportional to the concentration of the 
reacting molecules or ions. 1 Thus we have, for the 
forward reaction, 

HC 2 H 3 2 -^H+ -f C 2 H 3 O 2 -, Ratd = k^ 
and for the reverse reaction, 
HC 2 H 3 O 2 7-H+ + C 2 H S O 2 -, Rate 2 = k 2 (H+) (C 2 H 3 O 2 -). 

The symbols ki and k 2 are the proportionality con- 
stants, or the specific reaction rates. They express 
the constant ratio between the rates of the reaction 
and the concentrations of the substances involved. 
Thus, in this case, ki is small, since the tendency 
for acetic acid to ionize is not very great. If the 
concentration of acetic acid, represented by the 
bracketed symbol (HC 2 H 3 O 2 ), is doubled, ratei 
will be doubled. Similarly, we find that k 2 is rela- 
tively large, since the tendency for hydrogen ion 
and acetate ion to unite is great. Rate 2 will be 
proportional, both to the concentration of hydro- 
gen ion, written (H + ), and to the concentration of 
acetate ion, written (C 2 H a O 2 ~). Thus, if we double 
the hydrogen ion concentration, and triple the 
acetate ion concentration, rate 2 will be increased 
sixfold. It should be remembered that ki and k 2 
do not change as the concentrations are varied. 

1 The more general case, where the coefficients of the several 
formulas in the equation are not always one, is considered in Experi- 
ment 37 and Study Assignment F. 

They are constants, characteristic of the reactions 

The relative concentrations of the reactants and 
products will automatically be adjusted as the 
rates of the forward and of the reverse reactions 
correspondingly change, until these rates are equal, 
that is, until 

Rate 2 - Ratei. 

The processes do not stop, but there is no further 
change in the concentration of any of the con- 
stituents, since each is being formed as fast as it is 
being used up. If we equate the concentration ex- 
pressions, corresponding to rate 2 and ratei, we 

k 2 (H+) (C 2 H,0 2 -) - ki (HC 2 H 8 2 ). 

Now transposing the variable concentration fac- 
tors to one side of the equation, and the constants 
to the other, we have, 

(H+) (C a HQ 2 -) k, 

~ "" 

Since ki and k 2 are both constant, their quotient K 
will also be constant. It is called the equilibrium 
constant, or in this case the ionization constant. 2 

While it is often impossible to measure ki and 
k 2 separately, since the speeds of the reactions of 
many ionic reactions are immeasurably great, the 
ratio, ki/k 2 , or K, can be measured easily. A small 
value of K (ki smaller than k 2 ), means that the 
forward reaction is slow, and consequently the 
equilibrium mixture is largely present as reactants. 
Conversely, a large value of K means that the 
products of the reaction are present at relatively 
high concentration. 

Since the numerical values of ki and k 2 are not 
affected by variations in the concentrations of hy- 
drogen ion, acetate ion, or acetic acid molecules, 
their ratio, ki/k 2 , or K, will also be unaffected by 
changes in these concentrations. Whether the con- 
centrations of these substances are varied by dilu- 
tion, by the addition of sodium acetate to supply 

* Equilibrium constant expressions are customarily written so that 
the products of the reaction (as the equation is written), are in the 
numerator, and the reactants are in the denominator. 




acetate ion, or by the addition of hydrochloric acid 
to supply hydrogen ion, the same numerical rela- 
tionship between K and the molar concentrations 
of the reacting substances in the equilibrium holds: 

(HC 2 HgO a ) 

Although a change in the equilibrium concentration 
of one of the three substances forces the other two 
concentrations to change, the numerical ^alue of K 
remains unchanged. This fact makes an ionization 
constant or equilibrium constant a very useful tool 
when it is desired to calculate the equilibrium con- 
centrations without actually measuring them ex- 

In this experiment, the ionization constant of 
acetic acid will be determined. We shall measure 

the hydrogen ion concentration by means of in- 
dicators. In pure acetic acid solutions, the acetate 
ion concentration will be equal to the hydrogen ion 
concentration. The concentration of un-ionized 
acetic acid molecules (molarity of HC 2 H 3 O 2 ) can 
be calculated by subtracting the measured molarity 
of the hydrogen ion from the formality of acetic 
acid in the solution, since one acetic acid molecule 
is used up for each hydrogen ion formed. 

We shall also vary the ratio of hydrogen ion to 
acetate ion by adding excess sodium acetate. In 
this latter situation, the concentration of acetate 
ion and of un-ionized acetic acid may be calculated 
from the formalities of the acid and of the salt in 
the solution, as suggested in the experimental pro* 

Experimental Procedure 

Chemicals: 1 F HCaH^, 0.1 F HCJW)* 1 F 
or NaC2H3O2 crystals. Methyl violet, methyl orange, and 
methyl red indicators. H+ solutions from 10 -* to 10 ~ M. 

1. The effect of a Common Ion. To 10 ml of 
1 F HC 2 H 3 02 colored with a drop of methyl orange 
indicator, add a little 1 F NaC 2 H 3 2 , or a few 
crystals of NaC 2 H 3 2 . Interpret the change in 
terms of the equilibrium equation. 
2. The Ionization Constant of Acetic Acid 

A. Recall from your data for Experiment 26, the 
H" 1 " concentration of 1 F HC 2 H 3 O 2 . If necessary, 
determine this again, using methyl violet indicator, 
as in that experiment. Calculate the molarity of 
C2H 3 O2~~ and of HC 2 H 3 O 2 as described above. Sub- 
stituting the values found in the equilibrium con- 
stant expression, find the numerical value of K for 
1 F HC 2 H 3 2 . Tabulate and record the values in 
the chart in the report sheet. 

B. Using the value for the H + concentration in 
0.1 F HC 2 H 3 O 2 (from Exp. 26), calculate the molar- 
ity of C 2 H 3 2 -, of HC 2 H 3 2> and the value of K 
for 0.1 F HC 2 H 3 2 . Tabulate and record these 

C. Add 25.0 ml of 1 F NaCiHA to 10.0 ml of 
1 F HC 2 H 3 2 , and mix well. To a 5-ml portion 
of this in a 10-cm test tube, add a drop of methyl 
red indicator and estimate the H + concentration. 
To do this, compare the color with color compari- 
son standards made by adding a drop of the indi- 

cator to each of 5-ml portions of the standard acid 
buffer solutions of hydrogen ion concentration 
10~ 4 M H+, 10~ 6 M H+, and 10~ 6 M H+. 1 Save 
these color standards for use in part D. Calculate 
the molarity of the C 2 H 3 G 2 ~ and of the HC 2 H 3 2 . 
(Assume that NaC 2 H 3 2 is completely ionized, and 
that all of the C 2 H 3 O 2 ~ comes from the NaC 2 H 3 O2. 
The amount of C 2 H 3 2 ~ from the slightly ionized 
HC 2 H 3 2 is negligible in comparison to that from 
the NaC 2 H 3 Q 2 . These assumptions simplify the 
calculations and do not introduce an error as large 
as the experimental error.) Calculate K for this 

D. Dilute 10 ml of the solution used in C with 
40 ml of distilled water and mix well. Again de- 
termine the H+ concentration experimentally, us- 
ing methyl red indicator. Calculate K for this 
diluted mixture. 

E. Mix 4.0 ml of 1 F NaC 2 H 3 O 2 with 22.0 ml of 
1 F HC2H 3 O 2 . This time use methyl orange in- 
dicator to measure the H + concentration. Compare 
the color with acid solutions of the concentrations: 
10-' M H+ 10- 4 M H+, and 10~ 6 M H+. Calcu- 
late the value of K for this mixture. 

i The color of methyl red indicator changes appreciably toward the 
red when the H+ concentration increases only very slightly beyond 
10~ 5 M. It is almost completely transformed to the red color at 
1Q-*- 1 or 0.00003 if H+. 

REPORT: Exp. 36 


The lonizotion Constant 
3f a Weak Acid Section - 

Locker N umber . 

I. The Effect of a Common Ion* 

The color of methyl orange in 1 F HC 2 H 8 O2 is 

The color of methyl orange in 1 F HC 2 H 8 O 2 to which NaC 2 H 8 2 has been added is 

Give a brief explanation of this change in color in terms of the equilibrium equation 
HC*H 8 2 + H+ 

2. The lonization Constant of Acetic Acid. 

Enter all data, as calculated below, in the chart on the following page and calculate the value of K for each 

A. Recall the data from Experiment 26 on the molarity of H+, CiH 8 Or, and HC 2 HO 2> in 1 F HC*HsOi. 

B. Recall the data from Experiment 26 on the molarity of H+, C|H|CV% and HC 2 H 8 O 2 , in 0.1 F HCjHjOf. 

C. In the mixture of 25.0 ml of 1 F NaC 2 HO 2 + 10.0 ml of 1 F HC 2 H 8 O 2> the formality of each substance is: 



D. In the mixture of 10 ml of mixture C above plus 40 ml of distilled water, the formality of each substance is: 

F NaCaHjO* 


E. In the mixture of 4.0 ml of 1 F NaC2H 8 O 2 + 22.0 ml of 1 F HC^A, the formality of each substance is: 

_ F NaCaHaOt 


Fill in the values in the following table, for the five solutions studied. In the space provided below the table, 
show the method of calculation of the constant, K, for each trial: 


Formality of 

Formality of 

Molar it y 

Molar it y 



1 F HC,HgO 2 

0.1 F HC 2 H 3 2 

Accepted value of K for HC 2 H 3 O2, from the literature 

Method of calculation of K for each trial : 


1. The H+ concentration of 0.1 F hydrofluoric acid, HF, is 0.0086 M. Write the equilibrium equation for its 
ionization, the equilibrium constant expression, and solve for the value of K. 

2. The OH- concentration of 0.01 F ammonium hydroxide, NH 4 OH, is 4.2 X 10~ 4 M . Write the equilibrium 
equation for its ionization, the equilibrium constant expression, and solve for the value of K. 



The Solubility Product off Slightly Soluble Salts. 


College Chemittry, Chapter 2! 

Review of Fundamental Concepts 

This experiment extends our quantitative study 
of chemical equilibrium to the case of saturated 
solutions. Such equilibria are established every 
time we form a precipitate of a slightly soluble salt. 
The study of such equilibria is very important, for 
it enables us to calculate the experimental condi- 
tions (concentration, acidity, etc.)> to cause, or to 
prevent the formation of a given precipitate. In 
Experiments 24 and 36 we learned the essential 
characteristics of chemical equilibrium. Let us re- 
capitulate these briefly. 

Chemical Equilibrium 

When a process has reached a state of equilib- 
rium, we have the following three conditions : 

(1) A dynamic situation that is, the forward 
and the reverse processes continue to take place. 
The action seems to have stopped only because the 
relative amounts of the various reactants and prod- 
ucts do not change. 

(2) A "balance" so that the rate of formation 
of the products is just equal to the rate at which 
they are being decomposed again to give the initial 
substances. Representative equations for equilibria 
which we have studied are: 

H+ + 


=+ NH 4 + + C,H,O,- + 

NH40H + 

(neutralization and hydrolysis) 
(solid) + Ag+ + C 2 H 3 O a - 
(solubility of salts) 

Chemists frequently generalize the idea of an equi- 
librium process by the equation 



(3) A definite mathematical relationship which 
enables one to calculate the concentrations of the 
various substances present under any desired situ- 
ation. This relationship, the law of chemical equi- 
librium, is treated in considerable detail in general 
chemistry texts. We shall review the essential ideas 
of this law briefly in the following section. 

The Equilibrium Constant 

In the generalized reaction 

A + BqpC - 


the rate of reaction of the substance A with the 
substance B will depend on the number of collisions 
between molecules of A and molecules of B, and 
this will be proportional to the concentration of 
each of the reactants. Thus, the rate, or speed, 
of the forward reaction may be expressed by the 

Si oo (A) (B), or Si - k! (A) (B), 

where the bracketed symbols represent the con- 
centration in moles per liter, and 1^ is the propor- 
tionality constant. Similarly, the speed of the 
opposing or reverse reaction is expressed by the 

82 oo (C) (D), or S, - k, (C) (D). 

Changes in concentration will take place until the 
forward and reverse reaction rates become equal, 
that is, until 


Therefore, when equilibrium has been attained, 
we have the relation 

k, (C) (D) - ki (A) (B). 

or rearranging terms, 

(C) (D) Id 

By custom, we always write the products in the 
numerator, and the reactants in the denominator. 
If more than one molecule of a given substance 
is involved in the equation for a given reaction, 
that is, if 1 A reacts with 2 B to form the mole- 
cule AB 2 , it is evident that the number of col- 
lisions per second which occur in the formation of 
the intermediate substance AB will be proportional 
to the concentrations of A and of B. The rate of 
formation of AB 2 must then depend on the chance 
of collision of this intermediate AB with molecules 
of B a second time. The rate of formation of AB 2 
will therefore be proportional to the first power of 




A, and to the square of B. The equilibrium constant 
expression for the equilibrium 

A + 2 B + AB, 
is, therefore, 

I \ M-A 



(A) (B)' 

In general, the concentration of a substance must be 
raised to the same power as its coefficient in the 
equation for the reaction. 1 

FIG. 37-1. Equilibria at the crystal surface in a saturated 
silver chloride solution. Silver ions and chloride ions are 
leaving the crystal surface and returning to it at equal, oppos- 
ing rates. 

The Solubility Product Law 

With slightly soluble substances, we have the 
equilibrium of a solid salt with its ions in solution. 
In this case, the equilibrium constant expression 
may be expressed in a simplified form called the 
solubility product law. For example, we may write 
AA(solid) 4Zfc 2 A+++ + 3 B . 

The equilibrium constant expression for this would 

(AM) l 

However, the factor (A 2 B S ) is practically a con- 

1 For a discussion of reaction rates and for a more thorough de- 
velopment of the general equation for the equilibrium constant, see 
ftbo Chapter 19 in your text. 

stant, for its "activity," which depends on its rate 
of solution per unit surface, is practically constant. 
Wfc therefore have the simplified expression 

(A+++) 2 (B ) ki (AzBa) - K. (At saturation). 

The solubility product law thus states that in a 
saturated solution of a slightly soluble substance, 
the product of the concentration of the ions of the 
salt (each raised to its proper power, corresponding 
to its subscript in the molecular formula) is a con- 
stant. A salt is less soluble in the presence of an 
excess of either of its ions than it is in pure water. 
This so-called common ion effect was observed qual- 
itatively in Experiment 24, paragraph 2A, for a 
saturated solution of silver acetate. 

The Effect of High ionic Concentration 
on Equilibrium 

The law of chemical equilibrium applies with 
quantitative precision only in cases where we have 
a low total concentration of ions in the solution. In 
a solution which contains many ions, each charged 
particle mutually affects every other nearby 
charged particle, with the result that the ions are 
thus retarded to some extent in their movements. 
This is sometimes called a "drag effect." As a con- 
sequence, the ions are less "active," and behave as 
if they were at a lower concentration than would 
be the case if they were largely surrounded by 
neutral molecules. 

A high concentration of any ions in the solution 
therefore affects the solubility of a slightly soluble 
salt, and, in general, tends to make it more soluble. 
Thus, as in this experiment, while sodium nitrate 
contains no ions in common with those of silver 
acetate, we find that the latter is somewhat more 
soluble in a sodium nitrate solution than in pure 
water. Even in a strong sodium acetate solution, 
where the common ion will cause a decrease in 
solubility of silver acetate, this decrease is not as 
great as the solubility product law would predict. 
This effect was formerly called the "salt effect/* 
and is now explained by the Debye-Hiickel theory, 
which applies the law of chemical equilibrium to 
strong electrolytes by substituting a corrected con- 
centration, known as the "activity," for the molar- 
ity. This takes account of the electric charge and 
the concentration of all ions in the solution. 


The formation of a saturated solution involves 



the establishment of an equilibrium between the 
ions in solution and the crystal surface of the solid. 
It sometimes happens that, although the concen- 
tration of the ions gives an ion-product somewhat 
greater than the solubility product for that sub- 
stance, no precipitate forms. This is explained by 
the fact that there is no solid present to which 
the ions may attach themselves and also because 
the formation of the first bit of solid crystal is a 
more difficult process than the further growth of 
the crystal. The difficulty usually can be overcome 
by shaking, or by "seeding the solution with a 
small crystal of the solid substance. In this experi- 
ment the methods used are favorable for the exist- 
ence of supersaturated solutions, and it will be in- 
structive to study their behavior. 

Experimental Method 

We shall test the solubility product relationship 
quantitatively for the same salt used before, 
namely silver acetate. As an optional experiment, 
you may also test the relationship for a saturated 
solution of lead chloride (PbCla). This will give 
opportunity to show the necessity of squaring the 
chloride ion concentration in order to obtain a 

We shall mix measured volumes of 0.2 formal 

solutions of silver nitrate and of sodium acetate 
in several different proportions. The concentrations 
of silver ion and of acetate ion have been arbi- 
trarily chosen so that, in the mixture which results, 
their product, (Ag+) X (CjHaOa-), is greater than 
the solubility product for silver acetate. Conse- 
quently, solid silver acetate will separate out until 
the above ion-product is reduced to the value it 
always has when the equilibrium 

AgC 2 HA (solid) + Ag+ + C 2 H,Or 

is established. We shall then determine the silver 
ion concentration in this solution by titration with 
0.1 N potassium thiocyanate (KCNS) solution, 
using a ferric alum indicator. In this titration, the 
very insoluble silver thiocyanate (AgCNS) is pre- 
cipitated. As soon as the end-point is reached and 
an excess of thiocyanate ion (CNS~~) can remain 
in solution, the blood red ferric thiocyanate ion 
(FeCNS++) is formed. If we know the silver ion 
concentration at equilibrium, and the total num- 
ber of moles of silver ion and of acetate ion mixed 
in a known volume at the start, we can calculate 
the acetate ion concentration, and hence the solu- 
bility product constant 


Experimental Procedure 

Special supplies: 1 burette. 

Chemicals: carefully prepared solutions of 0.2 F AgNO*, 
0.2 F NaC 2 HA, 0.2 F Pb(NO 3 ) 2 , and 0.4 F NI^Cl. Standard- 
ized solutions of 0.1 N KCNS, 0.1 N AgNO 3 . Indicators of 
ferric alum, and 1 F 

1. The Solubility Product of Silver Acetate* 

Obtain about 130 ml each of the specially prepared 
solutions of 0.2 F AgNO 3 , and 0.2 F NaC 2 H,O 2 , in 
dry beakers or dry flasks. By careful measure- 
ment with a graduated cylinder, prepare the four 
mixtures listed below. In each case, place the solu- 
tions in dry, numbered 250-ml flasks. Stopper these 
to avoid evaporation, especially if they are to be 
left overnight. 


1 Use 20.0 ml 0.200 F AgNOs + 40.0 ml 0.200 F NaQH.O, 

2 30.0 * u 4-30.0 * * 

3 * 36.0 * * * 4-25.0 * * 

4 - 40.0 * * * + 20.0 

The mixtures, which contain an excess of Ag+ 
and C 2 H 3 O a ~, may remain supersaturated for a 

while. If necessary, they may be seeded with a 
small crystal of solid silver acetate to induce crys- 
tallization. Mix these, or shake gently at intervals 
for at least 30 minutes after the precipitation ap~ 
pears to be complete, in order to give time for the 
establishment of the equilibrium. A longer time is 
desirable. If convenient, it is well to let these 
mixtures stand overnight, although this is not 

Clean a burette, rinse it with tap water, then 
with a 5-ml portion of distilled water, and finally 
with two 5-ml portions of a standardized 0.100 JV 
potassium thiocyanate solution (KCNS), letting 
some run through the tip. Finally, fill the burette 
with the KCNS solution. Use a dry filter and fun- 
nel to filter your mixture No. 1 into a dry 250-ml 
beaker. Accurately measure out 25.0 ml of the 
filtrate, using a 50-ml graduate which you have 
rinsed with a little of the solution. Place this solu- 
tion in a beaker which contains a stirring rod, f Sr 



the titration. Add 1 ml of a saturated ferric alum 
solution as indicator, and 1 ml of 6 F HNO 3 . If 
the red color due to the hydrolysis of Fe+ ++ is not 
dispelled completely, add a little more HNO 3 . Ti- 
trate the solution to the first permanent appearance 
of red color due to the formation of Fe(CNS) + +. 
You may need to add another drop or two after 
you have apparently reached the end-point, as the 
Ag+ ion is partially absorbed in the precipitated 
AgCNS, and is slow in reacting. Repeat the titra- 
tion with a second 25.0-ml sample from this same 
filtrate if your results seem at all uncertain. Filter 
and analyze the other three mixtures for the con- 
centration of Ag+ in a similar manner. You can 
have the next mixture filtering while titrating the 
preceding sample. 

Calculations. Find the concentration of acetate 
ion in each mixture by calculating successively the 
various items called for from (a) to (g) under the 
calculated data section in the report sheet. Study 
through these items first, so that you understand 
what you are calculating, and why each step is 
needed. Finally, calculate the solubility product, 
K AgC|Hl (v by multiplying the experimentally de- 
termined Ag~*~ concentration by the calculated 
C2HsO2"" concentration. Carry out all calculations 
(a) to (g) to three significant figures, and the final 
solubility product, K AgCfHl0a , to two significant 

2. The Solubility Product of Lead Chloride* 
(Optional.) This experiment provides a more rigor- 
out test of the solubility product law, than the 
AgC 2 Hs02 equilibrium does, since here we have 
2 Cl~~ to 1 Pb++ and, therefore, the constant in- 
volves the square of the Cl~" concentration. The 
determination may be carried out by the same gen- 
eral method as was used for the silver acetate. 
Obtain about 130 ml each of the specially prepared 
0.2 F Pb(NO 8 ) 2 and OAF NH 4 C1 solutions, in dry 
beakers. Use these to prepare the following care- 
fully measured mixtures: 


1 Use 20.0 ml Q.200FPb(NQ,) 2 -f 40.0 ml 0.400F NH 4 C1 -f 20.0 mlH s O l 

2 25.0 * u -f 35.0 " * " +20.0 " 

3 30.0 * * +30.0 * * * +20.0 * 

4 35.0 * * * +25.0 " * +20.0 * 

* Do not add the 20 ml of distilled water until after a precipitate 
starts to form, thus avoiding the tendency to form a rather stable 
supersaturated solution in some cases* 

The mixtures (80 ml in all) must be kept in 
stoppered flasks to avoid evaporation, if they are 
to be left overnight. Shake them gently, at in- 
tervals, for at least 30 minutes after the precipita- 
tion appears to be complete, to allow time for the 
establishment of equilibrium. 

This time we shall determine the concentration 
of the Cl- by titration with a standard 0.1000 N 
AgNO 3 solution. Clean, rinse, and fill a burette 
with this standard AgNO 8 solution. Filter the mix- 
tures, using a dry filter paper and dry receiving 
vessel. Measure 25-ml portions of the filtrates care- 
fully with a graduate, and titrate with the AgNO 8 
solution. The indicator to be used this time is a 1 F 
K 2 CrOi solution. The first indicator added will, of 
course, precipitate the very insoluble PbCrO 4 , so 
it is necessary to use enough indicator so that about 
1 ml will be present after the Pb" 1 "* is all precipi- 
tated. This will require about 1.5 ml of 1 F K 2 CrO 4 
for mixture No. 1, increasing to 5 ml for mixture 
No. 4. In the titration AgCl precipitates until the 
end-point is reached, when, as soon as excess Ag+ 
appears, the slightly more soluble Ag 2 CrO 4 sepa- 
rates as a red precipitate. The end-point is then 
the first permanent appearance of a slightly red- 
dish cast to the yellow mixture. If a momentary 
red color does not appear where the AgNOs from 
the burette strikes the solution, insufficient indi- 
cator is present. 

Calculations. Prepare your own report sheet, in 
neat order, for this part of the experiment. You 
can model it after the form given for silver acetate. 
However, note that we are determining the Cl"" 
directly by titration this time, not the positive ion. 
Also note that the number of moles of solid PbCl 2 
which precipitates out will be one-half the differ- 
ence between the total moles of chloride in the 80- 
ml mixture and the moles of Cl"" (dissolved) in the 
solution. The total moles of Jb*b+ + (dissolved) can 
then be calculated as the difference between the 
total moles of Pb* 4 " in the mixture and the moles 
of PbCl 2 precipitated out. From this the concen- 
tration of Pb++, and the solubility product, (Pb++ ) 
X (Cl"") 2 , can be calculated. As a comparison, 
show also on your report sheet the calculation for 
the simple ion-product, (Pb++) X (Cl~). Which 
calculation shows the best constant? 

REPORT: Exp, 37 

Equilibria of Slightly Soluble Salts 

1. The Solubility Product of Silver Acetate 



Locker Number- 






Volume of 0.200 F AgNO 8 (ml) 





Volume of 0.200 F NaC 2 H 8 O 2 (ml) 





Total volume of the mixture (ml) 





Volume of mixture titrated 

Volume 0.100 N KCNS required for the titration 


a) Total moles Ag+ added in 
preparing each mixture 

b) Total moles C 2 H 3 O 2 - added in 
preparing each mixture 

c) Concentration of Ag + at 
equilibrium with solid AgCaHsC^ 

d) Total moles Ag+ (dissolved) in 
total volume of mixture 

e) Total moles of solid AgC 2 HOa 
precipitated out. (a-d) 

f) Total moles C 2 HO 2 - (dissolved) 
in total mixture, (b-e) 

g) Concentration of C 2 H 8 O 2 "" at 
equilibrium with solid AgC^jO-j 

h) The solubility-product, 
K AgC ,H30, - (Ag+) X (C 2 Hrf) a -) 

' ">how details of calculation for the first mixture only, where there is not space for all 



1. Look up the solubilities of silver acetate and of lead chloride at 20 C, in grams per 100 ml of water, in a reference 
source, and record the values below. 

Reference source' 

(a) The solubility of silver acetate, at 20 C, is: 
Calculate its solubility product. 1 

(b) The solubility of lead chloride, at 20 C, is: 
Calculate its solubility product. 







2. How do these values compare with your own experimental values? If your values are slightly larger, can you 
suggest a reason? 

3. Why, in making a qualitative test based on the formation of a precipitate, should one generally allow some time, 
and shake the test solution, before deciding that the substance tested for is absent? 

1 Remember that concentrations are always expressed in mole* per liter in calculating solubility products. Why? See also Examples 1 and 
2, Page 280. 

274 * 

The Equilibria off Carbonic Acid 
and Its Salts 


Co/feg* Chemistry, Chapter* 2f, 7 

Review off Fundamental Concepts 

This experiment will help you to understand the 
principles involved in the behavior of a typical 
weak, dibasic acid, such as carbonic acid, and its 
salts, and will enable you to apply these principles 
to similar situations. Our study will be both experi- 
mental and theoretical from the equilibrium equa- 
tions. We shall determine the principal ionic and 
molecular substances present in solutions of COt 
and H 2 CO 3 , NaHCO 8 and Na 2 CO 8 , and Ca(HCO 8 ), 
and CaCO 8 . 

These are all important substances. Carbon 
dioxide and carbonic acid are important in the 
beverage industry and in pH control; sodium 
hydrogen carbonate and sodium carbonate are im- 
portant as baking soda and washing soda respec- 
tively, and as cheap alkaline chemicals for indus- 
trial use; calcium hydrogen carbonate and calcium 
carbonate are important in hard water and its 
softening, in the formation of limestone caves and 
other limestone deposits, in the lime and cement 
industries, and as an alkaline flux in metallurgical 

As a basis for this study, we shall give below the 
fundamental equilibrium relationships involved. 

C0 t (aq) + H^O qpfc H.CO, (aq). 

The position of this equilibrium in solution is diffi- 
cult to measure, and is not important here, for the 
total number of moles of solute* whether as CO 
or as HiCOi, is the same. In the following treat- 
ment all the dissolved carbon dioxide is written as 
carbonic acid* 

As a weak acid, carbonic acid ionizes in steps, as 

(1) H,C0 4 + H+ + HCO,- 

(2) HCOi- 

More detailed comment on these equations is 
given in the experimental procedure. 

+ + CO," 

Experimental Procedure 

Special supplies: thistle tube* 

Chemicals: CaCO, (marble chips), NaHCOi solid, Ca(OH) s 
(sat. sol.), 0.1 F CaCl 2 , 1 F NaHCO, (freshly prepared), 
1 F Na^CO,, 0.1 F NaOH, pH 8 and pH 11 buffer solutions, 
and indicator solutions: methyl orange, bromthymol blue, 
phenol red, phenolphthalein, alizarin yellow R, and indigo 

1. Carbonic acid. Prepare a simple carbon diox- 
ide generator, similar to that illustrated in Fig. 
13-1. The Erlenmeyer flask contains about 5 grams 
of marble chips and 10 ml H*O, and is fitted with a 
thistle tube and a gas delivery tube to which are 
attached a length of rubber tubing and a 15-cm 
length of glass tubing. Add concentrated HC1 
through the thistle tube, a few milliliters at a time, 
as needed, when you wish to generate CO* gas. 

Prepare a saturated CO 2 solution by bubbling 
the gas through about 20 ml of distilled water in a 
15-cm test tube. We shall determine the approx- 
imate relative concentrations of H|CO$, H+, 

HCOi~, and CO in this solution, as follows: 

(a) From the solubility data in Appendix II, 
Table IX, calculate the total formality of dis- 
solved COi. Record this all as formality of H f CO 8 . 
(Recall that 1 mole of CO, gas at 20 C would 
equal 22,400 ml X 293/273 - 24,000 ml.) 

(b) Determine the approximate H+ concentra- 
tion in this saturated solution by testing a 5-ml 
portion with methyl orange indicator. 

(c) Add 3 ml of 0.1 F CaCU to a 3-ml portion of 
the saturated CO) solution. If you are not satisfied 
with the result, bubble more CO f into the solution 
to saturate it. From this result, and by calculation 
from the solubility product for CaCO, (Table XV 
in Appendix II), estimate the maximum COi 
concentration in saturated CO* solution. (It is 
actually considerably less than this.) From the 
results of paragraphs (a) and (b), and from the 
ionization equations for H,CO, how will the 




HCO 8 ~ concentration compare with that of H+ 
and of CO,? 

2. Sodium hydrogen carbonate. As a salt, this 
substance will be completely ionized in solution 
into Na+ and HCO 8 ~. The HCO 8 ~ can ionize 
further into H+ and CO* , according to equation 
2 for the ionization of H 2 CO 3 , and it also can 
hydrolyze, because H 2 CO 8 is a weak acid, as experi- 
mentally determined in paragraph (a) above. 

Determine the approximate pH of a freshly pre- 
pared 1 F NaHCO 8 solution, by testing 5-ml por- 
tions with bromthymol blue, phenol red, and 
phenolphthalein indicators (or recall the results of 
Experiment 34, paragraph 4). You may compare 
the colors with pR 8 buffer solution in case of any 
doubt. Now utilize this pH value and the K 2 value 
given above for the second step in the ionization of 
HaCOs, to calculate the approximate ratio of 
(CO,~) to (HCO 8 ~) in 1 F NaHCO 3 . Also esti- 
mate the actual approximate concentrations of 
HCO 8 - and CO, in 1 F NaHCO 3 . Would you 
expect this solution to precipitate CaCO 8 when 
added to 0.1 F CaCl 2 ? Try it. (Note the relative 
amount of any precipitate.) 

3. Sodium carbonate. This salt will, of course, be 
completely ionized into Na + and CO 8 . The 
CO 3 will be partially hydrolyzed to HCO 8 ~, be- 
cause HCO 8 ~ is a very weak acid, as indicated by 
the second ionization equation for H 2 CO 8 . In a 
solution as basic as this, further hydrolysis to 
H 2 CO 8 is practically negligible. 

Determine the approximate pH of a 1 F Na f CO 3 
solution, using alizarin yellow R and indigo car- 
mine indicators (or recall the result of Experiment 
34, paragraph 2), In case of any doubt, compare the 
colors with those obtained from a pR 11 buffer 
and from a pH. 12 standard prepared by 10-fold 
dilution of a little 0.1 F NaOH. Use this pH value 
and the K 2 value for the ionization of HCO 8 ~, to 
estimate the ratio of (CO 8 ~ -) to (HCO 8 ~), and the 
approximate actual concentration of each, in 
1 F Na 2 CO 8 . How would you expect the amount of 
CaCO 8 precipitate obtained on mixing 0.1 F CaCl t 
and 1 F NaCOa to compare with that obtained 

above with 0.1 F CaCl 2 and 1 F NaHCO 8 ? Try it, 
using comparable volumes of solutions. 
4. The instability of HCO*- solutions. 

(a) Boil about 10 ml of 1 F NaHCO 8 solution 
for several minutes. Cool the solution, and again 
determine the approximate pH, using phenol- 
phthalein and alizarin yellow R indicators. Com- 
pare this result with that of paragraph 2 above, 
and explain any change by writing an equation 
which combines the ionization equation and the 
hydrolysis equation for HCO 8 ~. Consider also the 
volatility of H 2 CO 8 . 

(b) Temporarily hard water, and limestone cave 
formation. Bubble CO 2 gas from your CO 2 gener- 
ator through about 10 ml of limewater contained 
in a 15-cm test tube. Continue this operation as 
long as any change occurs. Explain fully, by words 
and by equations: first, why a precipitate forms in 
this case, when no precipitate resulted from the 
solution of CO 2 in CaCl 2 solution [paragraph l(c) 
above]; second, why the precipitate rcdissolves as 
the solution becomes saturated with CO 2 . 

Divide this "temporarily hard water" which you 
have just prepared into two portions. Boil one por- 
tion for several minutes, or until a change occurs; 
to the other portion add an equal volume of lime- 
water. For each case, explain fully, by words and 
by equations, the reasons for the changes which 
you observe. 

(c) Soda ash. Place 2 or 3 grams of solid NaHCO 8 
in a test tube which has been fitted with a rubber 
stopper and a gas delivery tube. Arrange the de- 
livery tube so it is immersed in some limewater 
contained in a test tube. Heat the NaHCO 3 quite 
hot for a short time. 1 What gas is evolved? Dissolve 
in distilled water some of the solid which remains, 
and test it with phenolphthalein indicator. Ex- 
plain all changes which have occurred and also the 
significance of the name "soda ash." Much soda 
ash is manufactured this way by the Solvay soda 

1 Remove the delivery tube from the limewater before the heating 
is discontinued. 

REPORTS Exp. 38 

The Equilibria of Carbonic Acid Date 
and Its Salts 


Locker Number _ 

1. Carbonic Acid 

(a) Solubility of CO 2 at 20 C (Appendix II, Table IX) 1/100 1 

Calculation of this as formality: 

F HjCOt 

(b) Concentration of hydrogen ion in saturated CO 2 solution: 

Methyl orange color M H+ 

(c) Observed results of adding 0.1 F CaCl 2 to saturated CO 2 solution . . . 

Maximum concentration of COa in saturated CO 2 solution, based on the 

above observation, and the solubility product of CaCOji 

M CO," 

Approximate concentration of HCOs~ (indicate reasoning): 

_jf HCOr 

2. Sodium Hydrogen Carbonate 

Concentration of hydrogen ion and of hydroxide ion 
in 1 F NaHCOa (record indicator colors) : M II +, M OH~ 

Rewrite the equation, and the equilibrium constant expression, for the second step in the ionization of HaCO,: 

Concentration of carbonate ion and of hydrogen carbonate ion in 1 F NaliCQ, 
(use this K 2 expression and the above H+ concentration): 

M CO, 



Comment on your ability to precipitate CaCO, (and indicate the relative amount of precipitate) from 0.1 F 
CaCl 2 and 1 F NaHCO,: 

3. Sodium Carbonate 

Concentration of hydrogen ion and of hydroxide ion 
in 1 F Na 2 CO (record indicator colors): _ M H+ f _ M OH~ 

Concentration of carbonate ion and of hydrogen carbonate ion in 1 F Na 2 CO, 
(use this H+ concentration and the K 2 Expression): 

_ __ M CO, 

_____ M HCO,~ 


Comment on your ability to precipitate CaCOi (indicate the relative amount of any precipitate) from 0.1 
CaCl 2 and 1 F Na 2 CO 8 : 

Summarize the data of this experiment 
by placing, at the appropriate place in the 
chart at the right, the formulas H 2 CO 8 , H+, 
OH-, HCOr, CO 8 , and Na+, for each 
of the solutions indicated. 


1 F XaHCOt 

1 F NotCOi 









4. The Instability of HCO|~ 

(a) Approximate concentration of hydrogen ion and of hydroxide ion in 1 F NaHCO 8 after boiling the solution 
(record indicator colors) : 


M OH- 

Explain the above behavior, both by words and by the equations for the ionization and hydrolysis of HCO 8 ~: 

(b) Temporarily hard water. Limestone cave formation. Explain, both by words and by equations, all the results 
you observe as the limewater solution is saturated with CO 2 : 

Account for the difference in behavior of CO 2 with Ca(OH) 2 solution and with CaCl 2 solution: 

Explain, both by words and by equations, the results obtained when the "temporarily hard water" 
is boiled: 

is treated with additional Ca(OH) 2 : 

Correlate the above results with the solution of limestone rock and the formation of stalactites in limestone 

(c) Soda ash. Explain, both by words and by equations, all the results you observe when solid NaHCO 8 is 
heated. Account also for the name "soda ash": 


Recapitulation of Problems 
on Ionic Equilibria 

Cof/ege CJiemistiy, Chapters 19, 20, 21 

A Study Assignment 

After studying Experiment 36 on the Equilib- 
rium Constant and Experiment 37 on the Solubil- 
ity Product, you may have gained the impression 
that here we have a means of determining by 
mathematical calculation the exact concentrations 
of the several substances present in equilibrium 
with one another. In any solution containing a 
considerable concentration of ions, this may be far 
from the truth. Furthermore, it is not always easy 
to determine the exact concentration of a particu- 
lar molecular or ionic species because of hydro- 
lysis, complex ion formation, amphoteric behavior, 
and similar phenomena. 

The calculated concentrations, based on the 
mathematical equations for the equilibrium con- 
stant and solubility product, give idealized results 
which might be expected if the interionic attrac- 
tions were negligible, and if there were no other 
complicating equilibria. With very dilute solutions 
one obtains quite satisfactory agreement between 
the experimental and the calculated values. 

The mathematical equations would be exact if 
we used "activities" instead of concentrations. The 
activity is obtained by multiplying the concen- 
tration by a corrective factor, called the "activity 
coefficient/* This latter factor is not constant for 
a given ion but depends on the total concentration 
of all the ions in the solution, and on their valence 
type. In this manual we shall simplify our study 
of the principles involved by using the simple con- 
centrations of the ions. 

In spite of these limitations, the quantitative 
study of equilibrium conditions by means of the 
theoretical equilibrium constant expressions serves 
as a very valuable tool in chemical reasoning. You 
will do well to become familiar with the methods 
involved in such calculations. 

Typical Problems Involving lonization Constants 

Example 1. A 0.01 F formic acid (HCHO 2 ) solu- 
tion is found by electrical conductivity measure- 
ments to be 13.2% ionized. What is its ionization 

The equilibrium expressions are 


: H+ -f CHOr, and K - 

(H+) (CHOQ. 


The hydrogen ion concentration, (H+), will be 
13.2% of 0.01 F, or 0.00132 M H+. The formate 
ion concentration, (CHO*-), will also be 0.00132 M 
CHO 2 ~". The un-ionized formic acid concentration, 
(HCHO 2 ) will be 0.01 - 0.00132 - 0.00868 M 
HCHO 2 . Therefore 

(H+) (CH0 2 -) (0.00132) (0.00132) 
"" (HCHO*) " (0.00868) * ' 

Example S. If the ionization constant for formic 
acid is 2 X 10 ~ 4 , what is the hydrogen ion con- 
centration, and what is the degree of ionization, 
of a 0.1 F formic acid solution? 

(H+) (CHO,-) 

(HCHO,) * A * 

Let x - (H+) - (CHO,-). Then 0.1 - x 


Substituting in the formula, we write 

This is a quadratic equation. Our solution will be 
much simplified if we note that the value of x will 
be small as compared to the total concentration 
(0.1), so that (0.1 x) is approximately equal to 
(0.1). Making this simplification, we have 

<LM 2 x 10 -<, 

x 2 X 10- 6 - 20 X 10-*, 
x 4.5 X 10-*, or 0.0045 M H+. 
If the 0.1 F HCHO 2 were completely ionized, the 
hydrogen ion concentration would have been 0.1 M 
H+, so that the fraction ionized is 


0.045, or 4.5%. 

Comparing this value with the 13.2% ionization 
in 0.01 F formic acid, we see that the equilibrium 
constant expression requires that the degree of ion- 
ization increase with the dilution. This fact was 
observed experimentally in Experiment 26. ! 




Example 3. Calculate the hydrogen ion concen- 
tration in a buffer mixture which contains 0.1 
formula weight of formic acid (HCHO 2 ) and 0.2 
formula weight of sodium formate (NaCHO 2 ) in 
a liter of solution. 

, (H+) (CHOr) 

+ CHO 2 -, and- 

2 X 10-*. 

2 X 10~ 4 . 

2 X 

(HCH0 2 ) 

Let x - (H+), then (x + 0.2) = (CHQr), and 
(0.1 - x) - (HCH0 2 ). 

Note, in the above expression for the formate ion 
concentration, that x moles come from the ioniza- 
tion of the formic acid, and 0.2 mole comes from 
the 100% ionization of the sodium formate. 
Substituting these values in the equilibrium con- 
stant expression, 

(H+) (CHOr) = (x) (x + 0.2) 
(HCHO 2 ) (0.1 x) 

Now, due to the common ion effect, (H + ) will 
be even smaller than it is in a formic acid solution 
alone, so we may neglect the x in (x + 0.2), and 
also in (0.1 x). The expression then simplifies to 
(x) (0.2) 

x = 1 X 10-< M H+. 

Thus, we see that the addition of 0.2 formula 
weight of sodium formate per liter to the 0.1 F 
formic acid solution reduces the hydrogen ion con- 
centration from 0.0045 M to 0.0001 M H+. 

Problems Involving Ionization Constants 1 

Note. Be systematic and use a good form in 
setting up and in solving problems. This is just as 
important as getting the right answer. 

1. Calculate the ionization constant for each of 
the following from the data given: 

(a) 0.1 F HNO 2 is 6.5% ionized. 

(b) 0.01 F NH 4 OH is 4.15% ionized. 

(c) A 1 F solution of H 8 PO 4 has a H+ concen- 
tration of about 0.083 M. Assuming that this H+ 
all comes from the first stage of ionization, cal- 
culate KI. 

2. Use the ionization constants given in Table 
XIV, App. II, to calculate the H+ concentration 
in each of the following: 

(a) 0.5 F H,BO, (eye wash solution) 

1 Some of these problems, as well as some of those in the following 
set, may advantageously be left for assignment until similar situations 
Are encountered in the qualitative procedures, Experiments 39-44. 

(b) 0.1 F H 2 CO 8 (a carbonated beverage at the 
soda fountain. Use KI only.) 

(c) Vinegar. Assume it is 5% HC 2 H 3 O 2 and has 
a density of approximately 1.0. Calculate the mo- 
larity first. 

3. Use the ionization constant of NH 4 OH giver 
in the table to calculate the OH- concentration 
and also the degree of ionization, for: 

(a) 1 F NH 4 OH 

(b) 0.1 F NH 4 OH 

(c) 0.01 F NH 4 OH 

4. (a) 15 grams of acetic acid is dissolved tc 
make a liter of solution. What is the H+ concen 
tration, and the C 2 H 8 2 - concentration? 

(b) 15 grams of sodium acetate is dissolved ir 
the solution formed in problem 4(a). Now what it 
the H+ concentration, and the C 2 H 8 O2"" concen 
tration? (Assume no change in volume.) 

5. 4 grams of NH 4 C1 is added to 100 ml of 0.1 / 
NH 4 OH solution. What is the OH~ concentration! 
(Special problem: Calculate the pR of this solution 
See Exp. 34.) 

6. It is desired to make a solution having a H" 1 
concentration of 10 ~ 5 M. How many grams o\ 
sodium acetate must be added to 250 ml of 0.1 f 
HC 2 H 3 O 2 to do this? 

7. 400 ml of 0.5 F HC 2 H 3 O 2 and 600 ml of 0.2 f 
NaOH are mixed. What is the resulting H + con- 
centration? (Hint: First calculate the formality oi 
the HC 2 H 3 O 2 , and of the NaC 2 H 3 O 2 formed, in 
the final mixture.) 

Typical Problems Involving Solubility Product! 

Example 1. The solubility of silver bromide al 
room temperature is about 1.2 X 10 ~ 8 grams pei 
100 ml H 2 O. Calculate its solubility product, (Ag+; 
(Br-) K AgBr . 

From the formula weight of AgBr, 188, we maj 
first calculate its formality: 

03 " 

Since we regard salts as 100% ionized, thus 
AgBr * Ag+ + Br-, 

(Ag+) = (Br-) - 6.4 X 10" 7 M. 
The solubility product is, therefore, 

(Ag+) (Br-) - (6.4 X 10~ 7 ) (6.4 X 1Q- 7 ) 

- 40 X 10-" - 4 X 10- 18 - KAgBr. 

Example 8. Consider the case of a substance o 

different valence type, such as Ca(OH) 2 . For this 



the solubility is 0.161 grams per 100 ml H 2 O. Let 
us calculate the solubility product. 1 

We first calculate the formality of the saturated 


74 g/gfw 

Noting that the ionization of this would give one 
Ca++, but two OH"-, 

Ca(OH) 2 + Ca++ + 2 OH-, 
we have 

(Ca++) - 0.0218 M 

(OH~) - 2 X 0.0218 M - 0.0436 M 

and the solubility product would be calculated thus: 

(Ca++) (OH-) 2 = (0.0218) (0.0436) 2 = 4 X 10-*. 
Note that to get the (OH""), we double the total 
formality of the compound, and then we square 
the (OH~) according to the solubility product 
formula. We do not double the (OH"), and then 
square that. 

Example 3. Suppose we have given the solubility 
product of Mg(OH) 2 as 6 X 1Q- 12 . How many 
grams of Mg(OH) 2 will dissolve in 100 nil of 

(Mg++) (OH-) 2 = 6 X 1Q- 12 . 
Let x = (Mg++) in the saturated solution, 
then2x = (OH-). 

Substituting in the solubility product expression, 

(x) (2x) 2 - 6 X 10-* 
4x' = 6 X 10~ 12 
x 8 1.5 X 10- 18 
x 1.14 X 10~ 4 M Mg++ or 1.14 X 

10-< F Mg(OH) 2 . 
The weight of Mg(OH) 2 = (1.14 X 10~ 4 X 58) g/liter, or 

7 X 10~ 3 g/liter 

- 7 X 10" 4 grams Mg(OH) 2 per 100 

Note: The above example is a somewhat com- 
plicated type. For a simple univalent salt, such as 
AgCl, letting x = (Ag+) (Cl~), we would 
simply have x 2 = K^. 

Example 4- How many grams of SrSO 4 will dis- 
solve in 5.0 liters of 0.01 F SrCl 2 solution? 

1 For compounds of polyvalent ions such as Ca +4 ", and Mg"*"** in 
Example 3, it is likely that at least the intermediate ion Ca(OH) " 
is poorly ionized, so that the equation, Ca(OH)i -Z Ca++ + 2 
OH", does not completely describe the equilibrium. Likewise, the 
activity of the ions is considerably less than their concentration. For 
reasons of simplicity, such factors are neglected here, and the cal- 
culated solubility products are larger than they should be. 

From the Table of Solubility Products, 

(Sr++) (S0 4 ) - 2.8 X 10-'. 

Let x the formula weights of SrSO4 which dis- 
solve, then x = (SO 4 ),and (0.01 + x) - (Sr++). 
Substituting in the solubility product expression, 
we have 

(0.01 + x) (x) 2.8 X 10-'. 

Since the quantity (0.01 + x) does not differ 
materially from 0.01, we write as an approxima- 

(0.01) (x) 2.8 X 10-', 

(x) = 2.8 X 10-' F SrS0 4 . 
The weight of SrSO 4 in 

5.0 liters - (2.8 X 1<T X 183.6) g/1 X 5.01 

- 0.026 g SrSO 4 . 

Example 5. (Theory of Fractional Crystalliza- 
tion). To a solution which contains Cl~ and Br~ 
each at a concentration of 0.1 M , add silver nitrate 
solution. Which salt will crystallize first, and under 
what conditions can the other salt crystallize? 

By noting the relative size of the solubility 

(Ag+) (C1-) - 1.6 X 10-" 
and (Ag+) (Br-) = 4 X !<)- 

it is obvious that only AgBr solid will separate 
first. The (Ag+) cannot rise high enough to pre- 
cipitate AgCl, as long as considerable Br~ remains 
in solution. For both precipitates to co-exist in 
equilibrium at the same time, the (Ag+) would be 
the same for both (since there is only one solution), 
We may thus divide one solubility product ex- 
pression by the other, cancelling out (Ag" 4 " ) : 

(Ag+) (C1-) (CQ _ 1.6 X 10-" ^ 
(Ag+) (Br-) (Br-) 4 X 10" 1 * *"' 
Thus, the chloride ion concentration will be 400 
times that of the bromide ion. If the (Cl~) were 
0.1 M, no AgCl would precipitate until the (Br*~) 
were reduced to 

which thus allows the silver ion concentration to 
increase sufficiently to precipitate AgCL 

Problems Involving Solubility Products 8 

8. Calculate the solubility products for each of 
the following difficultly soluble salts, from the 

1 Some lack of agreement between your answers and the solubility 
products or solubility data given in the literature is accounted for in 
part by the fact that your calculations are based on concentrations, 
rather than activities* 



solubility in grams per 100 ml of water as given 

(a) AgCl 1.8 X 10- 4 (c) Pbl* 6.4 X 10-* 

(b) PbSO 4 4.1 X 10~ (d) Ag,CrO 4 2.6 X 10- 1 

9. Calculate the solubility of each of the follow- 
ing, in grams per 100 ml of water. See Table XV, 
Solubility Products, Appendix II, for the necessary 

(a) Agl (c) BaS0 4 

(b)A g2 S0 4 (d)Fe(OH). 

10. How many grams of silver acetate will dis- 
solve in 

(a) 100 ml of 0.1 F NaC 2 H 3 O 2 ? 

(b) 100 ml of 0.1 F Ca(C 2 H 3 O 2 )2? 

11. If solid sodium sulfide, Na 2 S, is added gradu- 
ally to a solution containing 0.01 mole of Pb++ 
and 0.01 mole of Zn+ + per liter, 

(a) What is the sulfide ion concentration when 
solid PbS begins to precipitate? 

(b) What is the sulfide ion concentration when 
solid ZnS begins to precipitate? 

(c) What is the concentration of the remaining 
Pb+ + when the ZnS begins to precipitate? 

12. How many formula weights of BaSO 4 will 
dissolve in a liter of 

(a) pure water (d) 0.1 F NaCl 

(b) 0.01 F Na 2 SO 4 (e) 10~ 6 F Na 2 SO 4 (Use 

(c) 0.1 F BaCl 2 the quadratic equation 

in this case. Why?) 

13. Given a 0.1 F MgSO 4 solution, calculate the 
hydroxide ion concentration necessary to pre- 
cipitate Mg(OH)2. Also calculate the pR of the 

14. What concentration of oxalate ion, C 2 O 4 , 
is necessary to precipitate CaC 2 O 4 from a saturated 
solution of CaSO 4 ? 

15. A liter of solution contains 0.1 M Ag 4 " and 
0.1 M Ba + +. If a strong K a CrO 4 solution is added 
drop by drop, which will precipitate first, Ag 2 CrQ 4 
or BaCrOi? Show calculations. 

16. Suppose the above solution, question 15, con- 
tained Ag+ and Ba++, each at 0.001 M concen- 
tration. Now, which will precipitate first, AgaCrO 4 
or BaCrO 4 , as K 2 CrO 4 solution is added? Show cal- 

Problems Involving Both lonization Constants 
and Solubility Products 

17. A solution contains 1 F NH 4 OH and 1 F 
NH 4 C1. What is the maximum concentration of 
Mg++ which could be present in such a solution 
without precipitating Mg(OH) 2 ? (Hint: Calcu- 
late the (OH~) first.) 

18. How many grams of NH^Cl should be added 
to 50 ml of 0.2 F NH 4 OH to just prevent precipita- 
tion when this solution is added to an equal volume 
of 0.2FMgCl 2 ? 

19. A solution containing cadmium chloride, 
CdQ 2 , is saturated with H 2 S, and the precipitated 
CdS is filtered out. The concentration of H 4 " is 
found to be 0.5 M. What is the concentration of 
Cd+ + which remains in the solution? (Hint: Multi- 
ply the equilibrium constant expressions for the 
first and second stages of ionization of H 2 S to- 
gether, to get the relation for the equilibrium of 
H 2 S = 2 H+ + S . See Experiment 41, for a 
fuller discussion of this. Calculate the sulfide ion 
concentration, and finally the cadmium ion con- 

20. A solution is prepared by dissolving 6.3 
grams of oxalic acid crystals (0.05 gfw) in 100 
ml of 2 F HC1. How many grams of solid CaCl 2 
could be dissolved in this solution without pre- 
cipitating calcium oxalate? (See the hint in Prob. 
19 above.) 

21. To what value must the concentration of the 
hydrogen ion in a 0.1 F PeCl 2 solution be adjusted, 
in order to just prevent the precipitation of ferrous 
sulfide (FeS) when hydrogen sulfide gas is passed 
into the solution? (See the hint in Prob. 19.) 

The Negative Ions Qualitative Tests. 


Co/fege Chemistry, Chapters ? 3, 14, 15, 16 

Review of Fundamental Concepts 

As a background for your study of the behavior 
of, and of tests for, the common negative ions and 
their acids and salts, the Appendix II will provide 
useful information, with which you should become 
thoroughly familiar: Table XI, on the relative 
strength of various acids; Table VII, on the rela- 
tive volatility of various acids; and Table X, on 
the general solubility rules for salts and hydrox- 

Typical Reactions of Common Anions 

The identification of negative ions in a sample 
generally depends on the establishment of one or 
more of the several types of equilibria which we 
have studied. These involve the formation of weak 
acids, volatile acids, insoluble salts, or complex 
ions. While we have tested for several of these ions 
in previous experiments, we must now consider the 
principles on which these tests are based. 

Sulfate Ion. This ion is the anion of a strong 
add: it forms characteristic precipitates with cer- 
tain metal ions, as with barium ion: 

Ba+++S0 4 ~ : 

BaSO 4 

Other anions also form insoluble precipitates with 
barium ion: 

Ba++ + CO, :<F* BaCO, 

Ba++ + S0,~ qp^ BaSO, 

3 Ba++ + 2 PO 4 --- :*r* Ba,(PO 4 ),. 

However, these are salts of weak adds and would 
all dissolve, or fail to precipitate, in an acid solu- 
tion. (See the chart on the next page.) In the pres- 
ence of excess hydrogen ion, the concentration of 
the free anion is reduced to such a low value that 
the solubility equilibrium is reversed, and the pre- 
cipitate dissolves or fails to form in the first place. 
We may therefore use barium ion in an acid solu- 
tion as a test reagent for the presence of sulfate ion. 
Sulfite Ion, Sulfide Ion, and Carbonate Ion. 
In each of these anions of weak, volatile adds, the 
free acid is formed by the addition of a strong acid: 

CO, + 2 H+:?T>" H,CO,^=:COt (g) + H|O 
SO, + 2 H+ r* HSO*qFSO, (9) + H,O 

The odors of sulfur dioxide (SO*) and hydrogen 
sulfide (HjS) are usually sufficient identification. 
Sulfite ion (SO ) may be oxidized to sulfate ion 
(SO 4 ) by bromine water (Br 8 ) or hydrogen 
peroxide (H 2 O t )> and then tested as the sulfate ion 
is tested. Sulfide ion (S ) may be further con- 
firmed by placing it in contact with a lead salt 
solution on filter paper, thus forming a black de- 
posit of lead sulfide (PbS). 

In your study of the oxidation states of sulfur 
(Study Assignment D and Experiment 21), you 
have learned that sulfite ion, which is intermediate 
in its oxidation state, can be reduced to free sulfur 
by sulfide ion, which in turn is oxidized to sulfur. 
The equation is 

SO," + 2 S + 6 H+ 

3 S + 3 

These ions, therefore, are not likely to be found to- 
gether in the same solution, particularly if His add. 
Likewise, separate solutions of these ions are rather 
unstable because of their ease of oxidation by at- 
mospheric oxygen: 

2 SO, + 2 

In the carbonate ion test, the evolution of a 
colorless, almost odorless gas when an acid is 
added to the sample is an indication, but not proof, 
that the gas is carbon dioxide. The sample must 
be reprecipitated as an insoluble carbonate saljt, 
such as calcium carbonate (CaCO,). When the 
slightly soluble carbon dioxide gas is bubbled into 
a neutral solution containing caldum ion, the weak 
acid which is formed, carbonic acid (HjCOi), 

COt + H,0 ^=fc H,CO, +=t 2 H+ + CO,, 

does not furnish a high enough concentration of 
carbonate ion to precipitate calcium carbonate. 
What would be the effect on the carbonate ion 
concentration if the carbon dioxide gas were 
bubbled into an alkaline solution instead of into 
the neutral one? This is the reason for using lime- 
water, Ca(OH)t, although the hydroxide is only 
slightly soluble, rather than some soluble calcium 



salt, in the carbonate ion test. The over-all net 
ionic equation is 

CO, + Ca++ + 2 OH- HP* CaCO, + 

Sulfite ion, if also present in the unknown, will 
interfere with the carbonate ion test, as it will 
liberate sulfur dioxide gas (SO*) along with the 
carbon dioxide gas when the sample is acidified, 
and will form a calcium sulfite (CaSO,) precipitate 
when the gases are absorbed in the limewater. 
(Sulfide ion, S , would not interfere, as CaS is 
soluble.) To overcome this, it is necessary first to 
oxidize any sulfite ion to sulfate ion, and thus pre- 
vent the formation of any sulfur dioxide gas. 

Chloride Ion, Bromide Ion, and Iodide Ion. 
These ions are, like the sulfate ion, the anions of 
strong acids. Their salts with silver ion are in- 
soluble, whereas the silver salts of most other 
anions, which are also insoluble in neutral solution, 
will dissolve in acid solution. Silver sulfide (Ag 2 S), 
however, is so very insoluble that the presence of 
hydrogen ion does not reduce the sulfide ion con- 
centration sufficiently to dissolve it. 

We may test for the presence of chloride ion in 
the mixed precipitate of AgCl, AgBr, and Agl by 
dissolving the AgCl from the still more insoluble 
AgBr and Agl by adding ammonium hydroxide 
solution, thus forming the complex ion Ag(NH,)t+ 

and Cl~. When the filtrate is again acidified, the 
AgCl is reprecipitated: 

AgCl + 2 NH 3 ^z Ag(NH 3 ) 2 + + d- 
Ag(NH,) 2 + + Cl- + 2 H+ > AgCl + 2 NH 4 +. 

It is thus possible, by carefully controlling the 
NH 4 OH and Ag+ concentrations, to redissolve the 
AgCl without appreciably redissolving the more 
insoluble AgBr and Agl. Furthermore, AgCl is 
white, AgBr is cream-colored, and Agl is light 

The distinctive differences which enable us to 
separate and test for bromide ion and iodide ion in 
the presence of chloride ion depend on differences 
in their ease of oxidation. Iodide ion is easily oxi- 
dized to iodine (I 2 ) by adding ferric ion (Fe+++), 
which does not affect the others. After adding car- 
bon tetrachloride (CC1 4 ) to identify (purple color) 
and remove the iodine, we may next oxidize the 
bromide ion with potassium permanganate or 
chlorine water solutions, absorb the bromine (Br 2 ) 
formed in carbon tetrachloride, and identify it by 
the brown color produced. 

Phosphate Ion. This is the anion of a moder- 
ately weak, nonvolatile acid, phosphoric acid 
(HsPO 4 ). It is tested by the typical method of 
forming a characteristic precipitate. The reagent 
used is ammonium molybdate solution, to which 





Pb ++ 





C1-, Br~, 

AgCl, white; AgBr, cream; 
Agl, yellow. All insoluble 
in HNOi. AgCl soluble, 
AgBr slightly soluble, and 
Agl insoluble, in NH 4 OH. 



PbClj and PbBr 2 , white, 
soluble in hot water. 
PbI 2 , yellow, slightly 
soluble in hot water. 

so 4 

Moderately soluble 

BaSO 4 , white, insoluble 
in HNO, 

CaSO 4 , white, slightly 

PbSO 4 , white, insoluble 
in HNO, 


Ag 2 SO,, white, soluble in 
NH 4 OH and in HNO, 

BaSO,. white, soluble in 

CaSO,, white, soluble in 

PbSO 8 , white, soluble in 


AgsS, black, soluble in hot, 
cone HNO, 



PbS, black, soluble in 
HN0 3 

P0 4 

Ag*PO 4 , yellow, soluble hi 
NH 4 OH and in HNO, 

Ba,(PO 4 ) 2 , white, soluble 
in HNO, 

Ca,(PO 4 ) 2 , white, soluble 
in HNO, 

Pb,(PO 4 ) 2 , white, soluble 
in HNO, 


AgsCOfc white, soluble in 
NH4OH and in HNO, 

BaCO,, white, soluble in 

CaCO,, white, soluble in 

PbCO,, white, soluble in 



an excess of ammonium ion is added in order to 
shift the equilibrium further to the right. An acid 
solution is necessary. The equation is 

3 NH 4 + + 12 MoO 4 ~ - 

21 H+ qi 
(NH 4 )3P0 4 12 MoO, + 12 H 2 O. 

The yellow precipitate is a mixed salt, ammonium 
phosphomolybdate. Sulfide ion interferes with this 
test but may be removed first by acidifying the 
solution with HC1 and boiling it. 

Nitrate Ion. In this case, since all nitrates are 
soluble, we cannot use a precipitation method as 
the test. Two other facts are utilized. First, nitrate 
ion is a good oxidizing agent in an acid solution 

when a reducing agent such as ferrous ion (Fe++) 
is added: 

4H++ NOr+3Fe++ *3 Fe+++-f NO + 2HA 

Second, the nitric oxide (NO) formed readily 
unites with excess ferrous ion present to form a 
brown complex ion: 

NO + Fe+ + + Fe(NO) ++. 

It is essential that an excess of ferrous ion be used, 
or the test will fail. 

Summary. The preceding chart may be of 
assistance to you in summarizing the behavior of 
the above negative ions with the metallic ions: 
silver ion, barium ion, calcium ion, and lead ion. 

Experimental Procedure 

Chemicals: 0.5 F (NH 4 )2MoO 4 , 0.1 F BaCl s , saturated Br t 
water, saturated C1 2 water, 0.02 F Ca(OH) 2 , 0.1 F FeCli, 
FeS0 4 7 H 2 O, 3% H 2 O 2 , Pb(C 2 H 3 O 2 ) 2 paper, 0.1 F KBr, 
0.1 FKI, 0.1 FKN0 3 , Ag(C 2 Ha0 2 ), 0.1 F AgNO,, 1 FNa 2 CO,, 
0.1 F Nad, 1 F KII 2 P0 4 , 1 F Na 2 SO 4 , 0.1 F Na 2 SO, 
(freshly prepared), 0.1 F Na 2 S (freshly prepared). 

First familiarize yourself with the test proce- 
dures by performing tests on known solutions of 
the ions to be tested: S , SO 4 , SO 8 , CO 8 , 

C1-, Br-, I-, PO 4 > and NO 8 ~. Then obtain two 

unknown solutions from the instructor, and per- 
form analyses on these for each of the ions. Use a 
fresh portion of the unknown for each test. 

Test for Sulfide Ion. To about 2 ml of the 
test solution add a slight excess of 6 F HC1. Note 
any odor of H 2 S, and, as a more sensitive test, place 
a piece of moistened lead acetate paper over the 
mouth of the test tube. Heat the tube gently. A 
darkening of the paper indicates S in the orig- 
inal solution. (If S is found to be present, S0 8 
cannot be present [why?], and the SO* test may 
then be omitted. Note also the modification of the 
PO 4 test when S is present.) 

Test for Sulfate Ion. To 2 ml of the test 
solution add 6 F HC1 drop by drop until the solu- 
tion is slightly acid. Then add 1 ml of 0.1 F BaCl, 
solution, or more as needed to complete the precip- 
itation. A white precipitate of BaSO 4 proves 
SO 4 \ (Save for the sulfite test.) 

Test for Sulfite Ion. If the solution from the 
previous sulfate ion test had a sharp odor of SO* 
when it was made acid with HC1, SO* is present. 

If in doubt, filter or centrifuge the solution to ob- 
tain a clear filtrate, add a drop or more of 0.1 F 
BaCl 2 to be sure all SO 4 was precipitated, and if 
necessary add more BaCl 2 ; then refilter or re- 
centrifuge. To the clear solution add 1 to 2 ml of 
bromine water to oxidize any 80s to SO 4 . A 
second white precipitate of BaSO 4 now proves the 
presence of S0 8 . 

Test for Carbonate. Fit a 15-cni test tube with 
a 1-hole rubber stopper and bent delivery tube 
(see Fig. 3-1). Place about 3 ml of the test solution 
in this test tube. If sulfite ion (SO* ) is present in 
the unknown, add 1 ml of 3% H 2 O 2 to oxidize it to 
sulfate ion. Now insert the delivery tube into some 
clear limewater, Ca(OH)j, in another test tube. 
When ready, remove the stopper just enough to 
add a little 6 F HC1 to the test solution. Immedi- 
ately close the stopper again, and heat the tube 
gently to boiling to drive any CO* gas into the 
limewater. Be careful not to let any of the boiling 
liquid escape through the delivery tube into the 
limewater. A white precipitate in the limewater in- 
dicates C0 8 or HCOj- in the test solution. 

Test for Chloride Ion. To a 2-ml portion of 
the test solution add a few drops of 6 F HNOs, 
as needed, to make the solution slightly acid. 
(Test with litmus paper.) Any sulfide ion present 
may be removed by boiling the solution a moment. 
The free sulfur formed does not interfere. Add 1 ml 
of 0.1 F AgNO 8 . (No precipitate here proves the 
absence of Q-, Br-, or I~.) Centrifuge the mix* 


ture. (See Fig. F-l, in the following study assign- 
ment, on the use of the centrifuge.) Test the cleat* 
filtrate with a drop of 0.1 F AgNO, for complete 
precipitation. If necessary, centrifuge again. Dis- 
card the filtrate. Wash the precipitate (see Fig. 
F-2 on washing precipitates) with distilled water to 
remove excess acid and silver ion. Now to this 
precipitate in the test tube add 3 ml of distilled 
water, four drops of 6 F NH 4 OH, and a half ml of 
0.1 F AgNOi. (The proportions are important, as 
we wish to dissolve only the AgCl from any mix- 
ture of AgCl, AgBr, Agl, and Ag 2 S.) Ag(NH,) 2 + 
and Cl" will form. Shake the mixture well. Centri- 
fuge. Transfer the clear solution to a clean test 
tube, and acidify with 6 F HNO 8 . A white precipi- 
tate of AgCl confirms Cl~. 

Test for Iodide Ion. To 2 ml of the test solu- 
tion add 6 F HC1 to make the solution acid. If 
S or SOs is present, boil the solution to remove 
the ion. Add 1 ml of 0.1 F FeCl a to oxidize any I- 
to It. (Br~ is not oxidized by Fe+++.) Add 1 ml of 
carbon tetrachloride (CC1 4 ), and agitate the mix- 
ture. A purple color proves I~. (Save this for the 
Br- test.) 

Test for Bromide Ion. // no I~ was present, 
add 2 ml of chlorine water to the above I~ test 
mixture, and agitate it. A brown color in the CC1 4 
layer proves Br". If I~ was present, separate, by 
means of a medicine dropper, as much as possible 
of the solution above the CC1 4 layer which con- 
tains the It, and place it in a clean test tube. Again 
extract any remaining I 2 by adding 1 ml of CC1 4 , 
agitating the mixture, and separating the solution. 
The solution may be boiled a moment to remove 
any remaining trace of I 2 . Then add 2 ml of chlo- 
rine water and 1 ml of CCU, and agitate the mix- 
tare. A brown color proves Br*. 

Test for Phosphate Ion. First mix about 1 ml 
of 0.5 F (NH 4 )MoO 4 reagent with 1 ml of 6 F 
HNOj. (If a white precipitate forms here, dissolve 
it by making the solution basic with NH 4 OH, and 
then re-acidify with HNO$.) If S has been found 
in the unknown, first make a 2-ml sample of it 
distinctly acid with HC1, boil it a moment to re- 
move all H S S, then add this, or a 2-ml sample of the 
original unknown if no S is present, to the above 
dear molybdate solution. A yellow precipitate of 
(NH 4 )iPCV12 Mod, appearing at once or after 
wanning a few minutes to about 40 C, proves 

PO4 "". 

Test for Nitrate Ion. // Br- and I~ are absent, 
use 2 ml of test solution acidified with 3 F H 2 SO 4 . 
Add 1 ml of freshly prepared saturated FeSO 4 . In- 
cline the test tube at about a 45 angle, and pour 
about 1 ml of concentrated H 2 SO 4 slowly down the 
side of the test tube. Be careful to avoid undue 
mixing. A brown ring of Fe(NO)++ at the interface 

Concentrated HgSQ. 

Ferrous sulfate solution 
containing test sample 

Formation of 
thin brown layer or 
ring indicates the 
presence of nitrate 
or nitrite ions. 

Observe the 
ring against 
a white back- 

FIG. 39-1. The nitrate test. 

of the two liquids proves NO~, A faint test may be 
observed more easily by holding the test tube 
against white paper and looking toward the light. 
(See Fig. 39-1.) 

// JBr~ and 7~ are present, free Br t or Ij may form 
at the interface with the concentrated HjSOi and 
invalidate the test. In this case, place a fresh 2-ml 
sample of the test solution in a clean mortar, add 
about 10 mg of solid silver acetate, and grind the 
mixture together for a couple of minutes to raetath- 
esue it to solid AgBr and Agl, Decant the liquid 
into a test tube, and centrifuge. Treat the clear 
solution with FeSO 4 and cone H 2 SO 4 , as above. 

Name . .-' L - , ^ 

REPORT: Exp. 39 

Date , 

The Negotive Ions 
L Qualitative Tests Section 

Locker Number 

1. Review Questions 

(1) In the tests for Cl" and SOi , explain fully why the addition of acid will dissolve other insoluble silver 
salts and barium salts, respectively, which would interfere with the tests if the solutions were neutral. (See the 
chart in the preceding discussion.) 

(2) Suppose a white precipitate is obtained on adding BaCU reagent to a neutral unknown solution. What 
might the precipitate be? Write as many formulas as you can of possible substances, considering both positive and 
negative ions as given in the preceding chart. 

(3) A moderate test for S0 4 is to be expected whenever you have SO 8 or S in an unknown solution, 
even when no SCh has been placed in the solution. Why? 

(4) Why cannot both S and SO 8 be present in the same solution? 

(5) Why is an excess of Fe++ necessary in the test for N0a~, if the test is to be successful? 

(6) Suppose you have two solutions, one a carbonate, C0 , and the other a bicarbonate, HCO 8 ~. Give specific 
procedures by which you could distinguish between the two. (Hint: Consider the pH of the solution.) 

2. Report of Unknown Solutions 

Write in the correct formula for each ion found, for each unknown. (The ions, if added at all in these unknowns, 
are present in more than traces.) 

Unknown No _ - Ions found . 

Unknown No. Ions found. 


3. Supplementary Drill 

(1) Give the formulas of all new characteristic molecules or ions formed when solutions of the following are mixed. 
(Indicate any cases of no action.) 

Cl-% Br~ t I-, Fe+++_ . Ag+, Cl~, NH 4 OH (excess) 

Cl- Br~, I- f C1 2 water Ag+, I- NH 4 OH (excess) 

Fe++, Br 2 water- S0 2 bubbled in Ca(OH) 2 . 

SO 8 % HzO (acid)_ . CO 2 bubbled in CaCl 2 . 

S", SO," (acid) . Pb(C 2 H 8 O 2 ) 2 , H^S gas 

(2) An unknown solution is to be tested for the ions, or their derivatives, 1 listed at the right. The following 
tests are performed on separate portions of the unknown: 

(a) Phenolphthalein added to a test portion gives a red color. (2) (3) 

(b) When a test portion is warmed, red litmus held in the mouth of the test tube turns blue. 

(c) When a test portion is acidified with HC1, and lead acetate paper is held in the mouth 4 
of the test tube, the paper turns black. 


(d) A test portion is acidified with HC1, boiled, cooled, then C1 2 water and CC1 4 added. 
After shaking the mixture, the CC1 4 layer is reddish-brown. 


(e) Addition of dilute H 2 SO 4 in excess to a test portion results in a white precipitate. 

On the basis of these tests, considered as a whole, mark, in the column headed 
(2), a plus sign (+) for each ion or its derivative, in the original solution, which is ^1~ 
definitely present, a minus sign ( ) if the ion is definitely absent, or a question 
mark (?) if there is no evidence to prove the ion present or absent.* Br~ 


(3) A second unknown solution is to be tested for the same list of ions given SO 4 
above. The following tests are performed: 

(a) The solution is basic to phenolphthalein, and smells of ammonia. ^ 

(b) The addition of excess HNOi to a test portion results in the formation of a white NO 3 ~ 

co 8 

(c) If the mixture from (b) above is filtered, and ammonium molybdate reagent added to 

the clear filtrate, a yellow precipitate results. PQ 4 

or a 

On the basis of these tests, considered as a whole, mark, in the column headed (3), a plus sign, a minus sign, 
question mark for each of the ions in the list, as you did for the preceding unknown solution. 

i B y "derivative" ia meant, for example, PO 4 or H 8 PO 4 , Ag+ or Ag(NHs) f +, NH 4 + or NH 4 OH, etc. 

Take account of the insolubility of certain salts. For example, if an acid solution is known to contain Fb++ then S0 4 , S ; etc., 
cannot be present. 


An Introduction to the 

Qualitative Analysis off the Metal Ions. 


A Study Assignment 

Qualitative analysis is concerned with the identi- 
fication of the particular substances present in a 
given sample of material. In the analysis of in- 
organic substances, this involves separate analyses 
for the metallic constituents or cations, and for the 
non-metallic constituents or anions. We have given 
a brief consideration of the latter in the preceding 
experiment. We now must turn our attention, for 
the following series of experiments, to the analyt- 
ical procedures by which we separate and identify 
the cations. 

You should recognize at the outset that the 
methods we shall use are not the procedures gener- 
ally adopted in industrial laboratories. In the 
latter, the investigator usually is concerned with 
the determination of the presence or absence of 
only a very few metals in a given sample. Further- 
more, the problem often is simplified because the 
nature and source of the sample make it unneces- 
sary to consider many metals. The methods used 
in industry naturally involve the shortest and 
most direct procedures that will give the informa- 
tion desired. Such observations as the ignition of 
the material to form characteristic decomposition 
products, the use of specific organic reagents for 
the various metal ions, or the microscopic study 

of crystalline form, may be used. Frequently, the 
identification and the quantitative determination 
as well are based on purely physical properties, 
such as the measurement of characteristic spectral 
lines. Such specialized tools as the spectrophotom- 
eter, which measures the absorption of various 
wavelengths of light in passing through a solution 
of the sample, and the mass spectrograph, which 
measures the masses of the various kinds of mole- 
cules in the sample, have recently come into wide- 
spread use. These latter tools are particularly use- 
ful in the analysis of organic mixtures. 

Grouping the Metal Ions for Qualitative 

In a systematic analysis, a solution of the un- 
known sample is prepared, using nitric acid or 
other reagents, if needed. Then, by the successive 
addition of specific reagents, insoluble salts of the 
metal ions are precipitated in groups. Each group 
can be subdivided further and tested for each 
metal ion in that group. The most commonly used 
scheme of analysis involves the separation into 
major groups of the twenty-four ions usually con- 
sidered, as indicated in the accompanying outline. 

As would be expected, there is some correlation 


Nitric Acid Solution of the Sample 
Reagent : HC1 or NH 4 C1 

Hg 2 Cl 2 
PbCl 2 (partly) 

(Group 1) 

Solution (Ions of Groups 2-5) 
Reagent: H S S in 0.3 M H+ 

Sb 2 S, 

(Group &) 

Solution (Ions of Groups 3-5) 
Reagent: (NH^jS in NH 4 OH solution 

A1(OH) 5 

(Group 8) 

Solution (Ions of Groups ^-6) 
Reagent: (NH 4 ),CO 8 + NH 4 CI 


(Group 4) 


(Group 5) 




between the groups of elements as precipitated for 
qualitative analysis and the groups which con- 
stitute the families of elements in the periodic 
table. Periodic sub-group 1 is represented by silver 
and mercury in qualitative Group 1. The alkaline 
earths, as precipitated by ammonium carbonate, 
constitute qualitative Group 4. With some change 
in procedure, magnesium would also precipitate 
here. The alkali metals (periodic group 1), nearly 
all of whose salts are soluble, constitute qualita- 
tive group 5. 

It will be profitable, if time is available, for you 
to prepare a solution containing one ion from each 
of these groups, for example Ag+, Cu++, Zn++, 
Ca++, and K+, and by the successive addition of 
the group reagents given in the scheme, to remove 
each ion in turn as a precipitate until only potas- 
sium ion remains. The instructor may wish to 
demonstrate this. 

In this course, we shall study a limited number 
of these ions, including all the ions of qualitative 
Groups 1 and 5, but omitting several ions from 
Groups 2, 3, and 4. No fundamental principles of 
qualitative separation will be omitted, however. 

Instructors who wish to consider all the ions 
usually studied in a course in qualitative analysis, 
as outlined in the preceding chart on the Separa- 
tion of the Metal Ions into Groups, are referred to 
the following Study Assignment G for the inclu- 
sion of the optional ions. 

Laboratory Techniques for Qualitative Analysis 

The volumes of solutions used in these qualita- 
tive procedures are rather small and approach the 
semimicro scale of work. While the experimental 
operations are, in general, similar to those we have 
been using, the reminders and suggestions which 
follow and the accompanying figures will help you 
to attain better technique, to save time, and to be 
more accurate in your analyses. 

The oft-repeated admonition keep your desk 
neat and orderly will pay big dividends. Keep your 
working space on the table top and in front of the 
sink clear of unnecessary equipment. Keep dirty 
test tubes and other articles in one place. Clean 
and rinse these at the first opportunity, so that a 
stock of clean equipment is always ready. Label 
any solutions which are to be kept some time. Do 

A smalt 
motor driven 


diagram to 
show how 
it worhs. 


test tubes of 
liquid to be 
These swing 
to the dot- 
ted posit jo 
when the in- 
strument is 

shield cut 
away to show 
rubber pad 

Typical centrifuge 

Note Opposite pairs of tubes most 
be balanced .by filling them with equal 
amounts of liquid or the centrifuge 
will vibrate excessively. 

FIG. F-l. The centrifuge and its use. 

your thinking and keep your laboratory record up 
to date as you work. 

The Separation of Precipitates. Either the centri- 


fuge (see Fig. F-l) or the vacuum filtration tech- 
nique (see Fig. 20-2b) may be used. The former 
process is faster (about 2 minutes centrifuging time 
usually is sufficient) and has been specified in most 
instances throughout these procedures. Filtration 
may be substituted if centrifuges are not available. 
When you filter a small volume (1 to 5 ml) of 
liquid, use a small filter paper so that all the liquid 
is not lost by absorption in the paper. Note the 
detailed techniques for each process, as illustrated. 
In particular, be sure you know how to wash a pre- 
cipitate properly, as this generally is essential in 
qualitative work (see Fig. F-2). 

The Evaporation of a Solution. When you boil 
down a solution in a 10-cm test tube, it is very 
easy to "boil it over" or suddenly eject some of 
the contents, especially if there is much precipitate 
present. With more than 2 ml of liquid, it usually 
is advisable to transfer the sample to a 15-cm test 
tube, and shake it sidewise while heating it (see 
Fig. F-3). If more than 10 ml of liquid is being 
evaporated, use a small beaker, casserole, or evap- 
orating dish. 

The Preparation and Use of Hydrogen Sulfide. 
Because of the frequency with which we use this 
gas, it is advisable to make the best possible prepa- 
ration for handling it. If hydrogen sulfide is not 
piped directly into a well-ventilated hood for your 
use, generate it by one of the methods illustrated 
in Figure F-4, as directed by your instructor. A 
ready-prepared commercial mixture, called "Aitch- 
tu-ess," 1 is available for method (a). Always work 
in a hood, as hydrogen sulfide is very poison- 
ous. A pronounced odor of the gas in the 
laboratory can and must be prevented. 

Thioacetamide, CH 8 CSNHj, may be used to 
generate hydrogen sulfide, without contaminating 
the atmosphere. The reagent is rather expensive. 
To use this method, add 5 to 15 drops of 1 F 
CHjCSNHj, as needed, directly to the test solu- 
tion, after adjustment of the H+ ion concen- 
tration. Heat the test tube in a small beaker of 
hot water to promote hydrolysis, according to the 

CH,CSNH, + 2 H,O + H 2 S 

C 2 H,Or. 

The precipitate must be washed by 
removing the liquid, refilling and mix* 
ing with distilled water, and ccntri- 
fuging again. This should be repeat* 
ed several times. 

The liquid may be removed either 
by decanting or with a drawn down 

medicine dropper 

FIG. F-2. How to wash a precipitate which 
has been ccntrifuged. 

Wave bach and 
forth to prevent 
the liquid from 
boiling violently. 

1 "Aitch-tu-esa" is supplied by The Hengar Co., 1833 Chestnut 
St., Philadelphia, Pa., or by various laboratory supply houses. 

FIG. F-3. This will improve your technique when you 
evaporate a solution in a test tube. 



Prepared source of 
called "Aitch-tu-ess" 

To stop the flow 
remove the flame. 

Loose cotton 

the material 
is exhausted 

//N Heat move the flame 

f'/s)j gently up the tube. 

Cotton filter to catch FeCI^ spray 
Release pinch clamp 
and leave stopper loose 
until the air is 
flushed out of 
the test tube 

of FeS 

then close and 
let the solution 
saturate with H a S. 

Remove this tube and 
rinse off solutions when 




in i5 cm test tube 

through with generator. 

FIG. F-4. Methods for the generation of hydrogen sulfide gas for individual use: (a) by heating a mixture of sulfur, paraffin, 
and asbestos, (b) by the reaction of HC1 on FeS in a small automatic generator. 

The Silver Group. 


Co/tog* Chemistry, Chapters 26, 26 

Review of Fundamental Concepts 

The precipitating reagent for the group is chlo- 
ride ion in an acid solution. Only three of all the 
twenty-four metal ions form a precipitate on the 
addition of this reagent, namely silver ion (Ag+), 
mercurous ion (Hg^" 1 "), and lead ion (Pb++). 
The mercurous ion is not the simple ion, Hg+, but 
consists of two atoms held together by a covalent 
bond, with a double positive charge. Since lead 
chloride is somewhat soluble, it is not completely 
precipitated here, and hence lead ion appears also 
in Group 2. 

Ammonium hydroxide solution gives different 
results with each of the three metal ions, as shown 
by the preliminary experiments which follow. The 
behavior of mercury salts is peculiar and needs 
some explanation. Mercury exhibits three oxida- 
tion states, zero in free mercury, Hg, +1 in mer- 
curous ion, Hg2 ++ , and +2 in mercuric ion, Hg 4 " +. 
In some situations, the intermediate mercurous 

ion, Hg 2 ++, is unstable, part of the mercury being 
reduced to the metal, and part oxidized to Hg++, 

HH Hg + Hg++ 

When ammonia is added to mercuric chloride solu- 
tion, a white "ammonolysis" product, HgNHaCl, 
is formed. This is analogous to the partial hydroly- 
sis of HgCl 2 to form a basic salt. Compare the two 



HOH * Hg(OH)Cl + HC1, 
HNH 2 * Hg(NH a )Cl + HCL 

If mercunws chloride is treated with ammonia, 
part of it is oxidized to the white mercuric amino- 
chloride above, while part is reduced to black 
finely divided mercury: 

Hg 2 Cl a + 2 NH, + Hg(NH 2 )Cl + Hg 

The mixed precipitates appear jet black. 

Experimental Procedure 

Chemical*: 1 FNH 4 OH, 0,1 FPb(NO,) 2 , O.I FHgCI 2 , 0.05 F 
Hg 2 (NO 3 ) 2 , 0.1 F KBr, 1 F K 2 CrO 4 , 0.1 F KI, 0.1 F AgNO 8 , 
0.1 F NaCI, 1 F Na 2 S 2 O 8 . 

A. Typical Reactions of the Silver Group 

1. Solubility of the Chlorides. Prepare a sample 
of each of the three chlorides by adding 2 ml of 
0.1 F NaCI to 2 nil of test solutions of AgNO a , 
Hg*(NQOi, Pb(NO 3 ) 2 . If a precipitate fails to 
form in any of these (why?), add 1 ml of 6 F HCL 
Let the precipitates settle, decant and discard the 
supernatant liquid. Add to each about 2 ml of 
distilled water. Heat each nearly to boiling and 
shake. Results? Cool under the cold water tap. 
Results? Save for the next paragraph. 

2. Behavior with Ammonia. To each of the three 
precipitates, add 1 ml of 6F NH 4 OH. Shake to see 
if they will dissolve. Compare the behavior of the 
Hg2Cl2 precipitate with ammonia with that of 
HgCl 2 . To do this, add 1 ml of 6 F NH 4 OH to 1 
ml of 0.1 F HgCl 2 solution. Divide the ammoniacal 
solution of AgCl into two unequal portions. To the 
smaller, add a slight excess of 6 F HNO 8 . Result? 

3. Relative Solubility of Silver Salts. To the larger 
portion of the ammoniacal AgCl solution, add 1 
ml of 0.1 F KBr. Result? Now, to this add 1 ml 
of 1 F Na&Oi (sodium thiosulfate) and shake. 

The complex ion formed is Ag (8203)2 "". (This is 
the reaction of "hypo" in fixing the developed 
film in photography.) To this solution, add a little 
0.1 F KL Compare the results with the solubility 
data for AgCl, AgBr, and Agl. 

4. Stability of Ag(NH*\+. The formation and 
dissociation of silver ammonia complex ion is 
governed by the equilibrium 

Ag(NHa) 9 + + Ag+ + 2 NH S , 

and the corresponding equilibrium constant expres- 

sion is 


(Ag(NH,),+) ~ 

We can get a fairly accurate value of this constant 
and therefore of the stability of this complex ion 
by the following simple experiment. 

Place 3.0 ml of 0.1 F AgNOa (measure it ac- 
curately in a 10-ml graduate) in a 15-cm test tube. 
Add 3.0 ml (also carefully measured) of 1 F 
NH 4 OH. Now, prepare some 0.02 F NaCI by dilut- 
ing 2.0 ml of 0.1 F solution to 10.0 ml in your 10-ml 
graduate. Mix thoroughly, note the exact volume, 
and then add this from a medicine dropper, about 
a half milliliter at first and then drop by drop until 
a very faint permanent milky precipitate of AgCl 





To the solution to be tested, add 6 F HC1. 
Filter. Wash the precipitate. 

Res. ll l AgCl, HfrCh, PbCl* 
Pour boiling water over the precipitate on the filter. 

Sol. 11 
Cations of 
Groups 2-5. 
Save for 
Group 2 

Sol. 12 Pb++ 
Add K 2 CrO 4 
Yellow PbCrO 4 
proves Pb++. 

Res. 12 AgCl, Hg z Cl 2 
Add NH 4 OH on the filter. 

Res. 13. 
Hg, HgNH 2 Cl 
Black, proves 

Hg2 ++. 

Sol. 13. 
Ag(NH 8 ) 2 + Cl- 
Acidify, HNO 8 
White AgCl, 
proves Ag+. 

1 The first number in a residue or solution 
sequence in the group. 

remains after shaking. Return any excess NaCl 
from the medicine dropper to the graduate, and 
note the exact volume of the 0.02 F NaCl used. 
From these data, calculate the concentrations of 
AgCNH*^" 4 ", Ag+, and NH 3 in the final mixture, 
and the value of K, as outlined in the report sheet. 
B. Analysis of the Silver Group (Group 1) 

For each group, we shall present first a brief 
outline of the method of separation, which will en- 
able you to visualize and learn the general steps 
in the analysis. This is followed by the "Pro- 
cedure," which gives detailed directions for the 
actual analysis. (Note: You first should prepare a 
"known** solution containing the three ions and 
analyze it to gain practice in technique, and then 
analyze one or two unknown solutions supplied by 
the instructor.) 

Procedure for the Analysis of the Silver Group 
Ag+ Hg 2 ++, Pb++: 

Precipitation of the Group. To 4 ml of the solu- 
tion to be tested, add a half ml of 6 F HC1 (or 
about 10 drops added by a medicine dropper) and 
mix well for a minute or two. (No precipitation 
indicates absence of Ag+, Hg2" l " H , and of much 
Pb + +.) Filter 2 into a 15-cm test tube, and test the 
filtrate for complete precipitation with 1 drop of 
HCL After the liquid has all drained through the 
filter, spray the precipitate (Res.ll) with 1-3 ml 
of distilled water from your wash bottle to wash 
it Let drain. Reserve the filtrate (label it Sll) for 
the analysis of Groups 2-5 if you are analyzing for 

1 It probably is simpler in this experiment to use the usual filtra- 
tion procedure (vacuum is unnecessary), rather than the centrifuge, 
although the centrifuge may be used, if desired, with a corresponding 
modification of the directions which follow. 

indicates the group, the second number indicates 

these. If not, discard it. 

Test for Lead Ion. Heat 10 - 15 ml of water to 
boiling, and pour 3-4 ml of this boiling water over 
the residue Rll on the filter, catching the hot 
filtrate S12 in a small test tube. (Wash the re- 
maining residue by pouring 10 ml of boiling water 
in portions over it, discarding these washings.) Add 
several drops of 1 F KsCrCX to the filtrate. A yellow 
precipitate indicates Pb" 1 ""*". 

Test for Mercurous Ion. To residue R12 on the 
filter, add 1 ml of 6 F NH 4 OH and then 2 ml of 
water, 3 collecting the filtrate S13 in a small test 
tube. A black residue on the filter indicates Hg2 4 " 1 ". 4 

Test for Silver Ion. If the filtrate S13 is not per- 
fectly clear, it may be due to colloidal Pb(OH)Cl 
coming through the filter. In this case, refilter the 
solution as often as necessary through the same 
filter, until a perfectly clear filtrate is obtained. 
Then, acidify the filtrate with a little 6 F HNOj 
and mix. (If in doubt, test for acidity with litmus.) 
A white precipitate, AgCl, proves Ag + . 

3 Small volumes may be estimated by noting that a 10-cm test 
tube holds about 8 or 9 ml, and using the appropriate portion. Measure, 
by means of a 10-ml graduate, 1 ml, then 2 ml, of H 2 O into a 10-cm 
test tube. Remember the resulting height of liquid as a guide foi 
future use, as directions frequently call for 1 or 2 ml of a reagent. 

4 If a large amount of Hg* ++ was present (heavy black precipitate)* 
a small amount of Ag + in the unknown may escape detection in the 
regular Ag + test, due to the fact that free Hg will reduce Ag+ to 
Ag, thus retaining it with this black residue, instead of leaving it in 
solution as Ag(NH) 3 +. 

This silver may be detected by opening out the filter on a watch 
glass, and pouring over the residue a mixture of 1 ml of 6 F HNOi 
and 1 ml of 12 F HC1, catching the solution in a small beaker or test 
tube. Evaporate off most, but not all the acid* and dilute with 3 ml 
HA adding a drop of 6 F HCl if no white precipitate of AgCl forma 
without it. Further identification may be made by noting if the sus- 
pected AgCl dissolves on the addition of excess NH<OH. 



Locker Number- 

REPORT: Exp. 40 

The Silver Group 

A. Typical Reactions of the Silver Group , 

1, 2. In the proper space in the following chart, note the characteristic behavior, if any (solubility, color, etc.) 
and write the formula (s) of all new substances formed by the reaction: 




Hot water 

NH 4 OH (Excess) 

1 Look up the relative solubilities to help you decide whether a reaction takes place between PbCla and NT^OH. 

3. Write net ionic equations for the successive reactions that take place when AgCl is treated with ammonium 
hydroxide, and then each product, in turti, is treated with potassium bromide, sodium thiosulfate, and potassium 
iodide. Also look up the solubilities of the silver halides and list these. 

Explain and correlate these reactions with the solubility data given: 

Solubility (g/100gH 8 Q); 
AgCl - 


4. The stability of silver ammonia complex ion> 

Indicate your calculations for each step, in the spaces provided. 

a) Volume of 0.02 F NaCl used 

b) Total volume of the final mixture 

c) Concentration of Ag(NHs)2+ 

(Assume all the silver to be present as this complex ion, neglecting the small 
amount of free Ag+ formed) - 

d) Concentration of Cl"~ 

(Neglect any trace of Cl~ removed as AgCl) - 

e) Concentration of Ag+ 

(Use the value for Cl~ obtained above, and the solubility product relationship, 
- 1.6 X 10~ l .) - 


f) Concentration of free NHs 

(Assume that 1 F NH 4 OH is 1 M in NHs. First calculate the NHs concentration 
as if none combined with Ag+, then subtract from this twice the concentration 
of Ag(NH 8 )2+ found above.) _ 

g) Use the values found in e), f), and c) to calculate the value of the constant for 
silver ammonia complex ion : _ 

(Ag(NH 8 ) 2 +) 

h) Accepted value for this constant in the literature (See the Appendix, Table XIV.)- 

B. Analysis of the Silver Group 

Summary, Unknown No 

Unknown No 

Ions found- 
Ions found- 

Indicate, for each unknown, each essential step of the procedure, identifying each succeeding stage in the 
analysis by its Sol. or lies. No. 


Reagent (t?) 




Precipitation With Hydrogen Sulfide. 

The Analysis off the Hydrogen Sulfide Group. 



Fia. 41-1. The hydrogen sulfide molecule, the hydrosulfide ion, 
and the sulfide ion. 

Hydrogen Sulfide a Weak Dibasic Acid 

When hydrogen sulfide is bubbled into water at 
ordinary atmospheric pressure, it dissolves to give 
a saturated solution which is about 0.1 F H 2 S. This 
behaves as a weak acid, and since there are two 
replaceable hydrogen atoms we may represent its 
ionization in steps, thus: 

H2S -fdfc H+ + HS- (1) 

HS- + H+ + S-- (2) 

The extent of the ionization of hydrosulfide ion, 
step (2) above, is much less than that of step (1), 
since it involves the removal of a positive hydrogen 
ion from the negative particle HS~~, while the first 
hydrogen ion only has to be separated from a 
neutral molecule, H 2 S. 

The corresponding equilibrium constant expres- 
sions for these stepwise ionizations are 

Coffoge Chtm/ifry, Chapters 26, 28 

Review of Fundamental Concepts 

i- (1.1 X 10~ T ) (1.0 X 10-") ~ 1.1 X 10-* 


Furthermore, in any solution saturated with hy- 
drogen sulfide at atmospheric pressure, the con- 
centration of molecular H 2 S remains sensibly con- 
stant, and equals about 0.1 M. We may, therefore, 
simplify equation (5) thus: 


" K - 


Note that K 2 is about 100,000,000 times smaller 
than Ki, corresponding to the much smaller extent 
of ionization in step (2). The expression (H+) 
refers to the same value, of course, in both equa- 
tions (3) and (4), since both equilibria are present 
in the same solution. Most of the H+ comes from 
the first stage. Likewise the HS~ concentration 
(from the first stage) is about equal to the H+ con- 
centration and is not nearly so small as the S 
concentration which results from the second stage. 
If we multiply equation (3) by equation (4), can- 
celling out the common factor (HS~), we get 

or (H+)*(S ) - l.l X 10-. (6) 

This is a very important relationship, as it shows 
us, for instance, that if the hydrogen ion concen- 
tration in a solution is increased tenfold, the sulfide 
ion concentration in the solution is thereby re- 
duced a hundredfold. That is, the sulfide ion con- 
centration in a solution saturated with hydrogen 
sulfide gas is inversely proportional to the square 
of the hydrogen ion concentration in that solution. 
We may thus control the maximum sulfide ion 
concentration in a solution by controlling the hy- 
drogen ion concentration. 

For example, in a 10 ~ 2 M H+ solution, such as 
we may have in an acetic acid solution, we can 
calculate the maximum sulfide ion concentration 
resulting when hydrogen sulfide is bubbled into the 
solution, by substituting in equation (6) above: 

(10-*) 8 (S ) 1.1 X 10"* 

iiv in-* 1 
(S ) - fa - 1.1 X 10- M. 

Similarly, in a 0.3 M H+ solution, such as is used 
when precipitating the hydrogen sulfide group in 
acid solution, we may calculate the maximum sul- 
fide ion concentration, as follows: 
(0.3) J (S ) - 1.1 X 10- 
1.1 X 10- 

CS") - 


- 1.2 X 10- M. 

Note that in these two examples increasing the 
hydrogen ion concentration 30 times (from 10 ~* to 
0.3 M) has resulted in a decrease of about 1000- 
fold in the sulfide ion concentration. What effect 
does this large difference in sulfide ion concentra- 
tion have on the ability of hydrogen sulfide to 




precipitate a metallic sulfide? The next section will 
discuss this point. 

The Precipitation of Metallic Sulfides 

In a saturated solution of cupric sulfide, we have 
the equilibrium 

CuS-fiCu + + + S . (7) 

According to the solubility product principle, the 
product of the concentrations of the ions in the 
solution cannot be greater than a certain value, 

(Cu++) (S ) Kcus - 10- 40 . (8) 

The amount of cupric ion which can remain in 
solution depends, therefore, on the sulfide ion con- 
centration, and as we have seen, this in turn de- 
pends on the hydrogen ion concentration. For a 
0.3 M H+ solution, the sulfide ion concentration 
cannot be greater than 1.2 X 10 ~ 22 M. (See the 
above examples.) From this, we may calculate the 
maximum cupric ion concentration which could 
remain in such a solution. Substituting in equa- 
tion (8), we have 

(Cu++)(1.2 X 10- 22 ) = 10-* 
in -40 

(Cu+<f) " 


(approximately) J 

The precipitation of cupric ion from such a solu- 
tion is, therefore, practically complete. 

Compare this with the somewhat more soluble 
ferrous sulfide, for which, in a saturated solution 

(re**) (S ) - 10- 22 . 

Again, calculating for a 0.3 M H+ solution, in 
which (S") = 1.2 X 10~ 22 , 
(Fe++) (1.2 X 10- 22 ) * 10-" 


(Fe++) - j^-^^-^ 1 if (approximately). 2 

A high concentration of ferrous ion, thus, can re- 
main in solution without being precipitated at all. 
However, if the solution is made basic with ain- 

1 The solubility products of the metallic sulfides in many cases arc 
not known more precisely than to the nearest power of 10. Such 
calculations as this, and those which follow, should be regarded as 
only approximate. They do serve, however, to show the principles 
on which the separation of metallic sulfides is based. 

The corresponding calculation with zinc ion (solubility product 
for ZnS ID** 4 ) would indicate that only 0.01 M Zn+ + could remain 
in ft 0.3 M H+ solution saturated with H*S. ZnS does not precipitate 
readily, however, from quite concentrated Zn++ solutions under these 
conditions, due, in part at least, to the tendency of ZnS to delay 
precipitation and form supersaturated solutions. Zn+ + , therefore, is 
not precipitated with the HaS group, but precipitates with the am- 
monium sulfide group when the solution is made basic. 

monium hydroxide, in which the fiulfide ion con- 
centration is about 10 ~ 7 M when treated with 
hydrogen sulfide, we calculate 

(Fe++) (10- 7 ) - 10-* 

(Fe++) - 10~ 18 M. 

Thus, ferrous ion will be almost completely pre- 
cipitated in a slightly basic solution, but not at all 
in a 0.3 M H+ solution. 

The very insoluble metallic sulfides are precipi- 
tated along with cupric sulfide in the hydrogen 
sulfide group, while those which are slightly more 
soluble may be precipitated along with ferrous sul- 
fide by ammonium sulfide. The readily soluble 
sulfides will not be precipitated in either case. This 
is the basic principle of the separation of Groups 2 
and 3 in the qualitative scheme. 

The Separation of the Hydrogen Sulflde Group 
into Sub-Groups. Amphoteric Sulfides. 

As we have just seen, the hydrogen sulfide group 
includes all those sulfides which are so insoluble 
as to be almost completely precipitated in 0.3 M 
H + solution by hydrogen sulfide. This includes 
eight of the twenty-four metal ions commonly 
studied. For convenience, these may be subdivided 
into two smaller groups for individual tests of the 
ions. This is accomplished by the treatment of the 
mixed sulfide precipitate with a high sulfide ion 
concentration. Using a sodium sulfide solution, we 
find that the sulfides of mercury, arsenic, anti- 
mony, and tin dissolve as the corresponding sulfo- 
salts. (If ammonium sulfide, which gives a lower 
sulfide ion concentration due to hydrolysis, is used, 
the mercuric sulfide remains undissolved with the 
"copper" sub-group.) The theory of such separa- 
tions is as follows. 

Just as the hydroxides of certain metals are 
amphoteric, dissolving in an excess of hydroxide 
ion to form hydroxide complex ions (see Exp. 35), 
so also the sulfides of some of the metals will dis- 
solve in a high concentration of sulfide ion to form 
sulfide complex ions. The addition of only a slight 
excess of hydroxide ion, or of sulfide ion, to a 
stannic salt solution forms the expected precipi- 

+ 4 OH" ^ Sn(OH)4, 8 
2 S" 3r* SnS>. 

3 An ion with such a high positive charge as -|-4 will have a very 
strong tendency to coordinate with any negative ions in solution, 
and will seldom, if even be present as the simple ton. In ft hydro- 



with a high concentration of these ions, the 
precipitates redissolve: 

Sn(OH) 4 + 2 OH- 
SnSa + S~ + 

The addition of hydrogen ion to the above 
strongly basic solutions will decrease the concen- 
tration of hydroxide ion, or of sulfide ion, and 
reprecipitate stannic hydroxide, or stannic sulfide, 
respectively. In a high concentration of hydro- 
chloric acid, both precipitates redissolve again, due 
not only to the action of the hydrogen ion in form- 
ing water and hydrogen sulfide, respectively, but 
also to the high chloride ion concentration, which 
forms the complex ion SnCl 6 . 

A very interesting situation is the fact that 
stannic sulfide dissolves readily in a sulfide ion 
solution, while stannous sulfide will not dissolve. 
It should be remembered that sulfide ion is a base 1 
and stannic sulfide is acting as an acid as it dis- 
solves. Stannous sulfide is not a sufficiently strong 
acid to react with sulfide ion. This behavior is in 
accord with the more acidic character of an element 
in its higher oxidation state. Likewise, nitrous acid 
(HNO 2 ) and sulfurous acid (H 2 SO 3 ) are weak 
acids, while the corresponding acids in which the 
elements are more highly oxidized, nitric acid and 
sulfuric acid, are strong acids. It is for this reason 
that, in the analytical procedure, an oxidizing 
agent, hydrogen peroxide, is added to oxidize any 
tin to stannic ion before adding sulfide ion. 

The Oxidation of Hydrogen Sulfide and Metallic 

Thus far, we have dealt with hydrogen sulfide 
only as a weak acid. It contains sulfur in its lowest 
oxidation state, namely minus 2, and is a good 
reducing agent, reacting readily with any strong 

oxidizing agents present. The sulfide is nearly al- 
ways oxidized to free sulfur, together with some 
sulfate ion which forms by further oxidation. 

Nitric acid, or nitrates in acid solution, are 
nearly always present in the solution to be tested 
for Group 2 ions. If the solution is quite dilute, 
and not too hot, these nitrates do not interfere as 
oxidizing agents. In stronger solution, especially 
when heated, sulfur will always form when hydro- 
gen sulfide gas is passed into the solution. The 
reaction is 

3 HjS + 2 H+ + 2 NOr * 3 S + 2 NO + 4 HA 

This also explains why nitric acid is a better 
solvent than hydrochloric acid for most metallic 
sulfides. To dissolve cupric sulfide, we must de- 
crease the concentration of either the cupric ion 
or the sulfide ion, so as to shift the equilibrium 

CuS += Cu++ + 8~ 

to the right. The sulfide ion concentration is de- 
creased more completely by oxidizing it to free 
sulfur with nitric acid, than it is by forming the 
weak acid hydrogen sulfide with hydrochloric acid. 
The reaction is 
3 CuS+8 H++2 NO,- >- 3 Cu+++3 S+2 NO+4 HfO. 

Some few sulfides, particularly mercuric sulfide, 
are so extremely insoluble that they will not dis- 
solve even in concentrated nitric acid. Hot aqua 
regia, however, is able to dissolve mercuric sulfide 
quite readily. This reagent owes its powerful sol- 
vent action to the fact that, in addition to the 
oxidizing action of nitric acid, there is a high con- 
centration of chloride ion which readily forms com- 
plex ions with a number of metallic ions. With 
mercuric sulfide, we have the reaction 

HgS + 4 H+ + 2 NOr + 4 Cl" * 

HgCU" + S + 2NO, + 

Experimental Procedure 

Chemicals: 6 F HCHO 2 , 3 F NI^CjHjQz, saturated Br 2 
water, 0.1 F Cu(N0 8 ) 2 , 0.1 F Fed,, 3% HaO* 30% HA, 
Fe (1-cm wire pieces), 0.1 F Pb(NO 3 ) 2 , 0.1 F HgCl 2 , 0.1 F 
Hg(NO*)t, 1 F K 2 CK)4, 0.1 F K^CN)* 1 F NajSO* 2 F 
NaHS, 0.1 F SnCU, 0.1 F SnCl* 0.1 F Zn (NO,)*. 

chloric acid solution, stannic salts form the chloride complex ion* 
SnCU~~", so that the reactions with hydroxide or sulfide ions really 
constitute replacements: 

find* 4- 4 OH- ^r* Sn(OH) 4 + 6 C1-, 

Suds" + 2 S :?=* Sn& + 6 d~. 

1 This is in accord with the definition of a base as any substance 
which can react with protons, H+. Thus, sulfide ion, carbonate ion, 
acetate ion, etc., as well as hydroxide ion, are bases. 

A. Typical Reactions of the Hydrogen Sulfide 

1. The Precipitation of Metallic Sulfides. Dilute 
3 ml of 6 F HC1 to 9 ml with distilled water and 
mix, to make a 2 M H+ solution. Dilute 1 ml of this 
to 10 ml to make a 0.2 M solution. Use 1 ml of 
this latter solution to prepare 10 ml of 0.02 M H* f 
and 10 ml of 0.002 M H+, by successive tenfold 
dilutions* Mix 2 ml of each of these with an equal 
volume of 0.1 M Cu++ to give four solutions vary* 
ing in H + concentration from 1 M to 0.001 M. 




Sol. 11 Cations of Groups 2-6 
Adjust the H+ concentration to 0.3 M . Warm, and saturate with H 2 S gas. Centrifuge. Wash residue. 

Res. 21 HgS, PbS, CuS, SnS, SnS* S 

Add 6 F NaOH, H-jO. Warm and agitate. Cool, if dark residue remains add 
3% HaO* Agitate. Heat to decompose excess HzOj. Add 2 F NaHS re- 
agent. Warm and mix for 2-3 minutes. Dilute with H2O. Centrifuge. 

Sol. 21 Cations of Groups 3-5. 
At once boil out H^, and save for Group 
3 analysis 

Wash any dark residue with 2 nil H 2 O containing NaHS and NaOH. Discard washings. 

Res. 22 PbS, CuS, (HgS?) 

Add 3 F HNO 3 , heat almost to boiling for 2 minutes only. 
Centrifuge. (Combine any dark residue with Res. 24.) 
To the filtrate add 3 F H 2 SO 4 ; evaporate to dense white 
fumes. Cool. Add H^O (caution). Centrifuge. 

Res. 23 PbS0 4 
Dissolve in 
Add 1 F K 2 CrO 4 . Yel- 
low precipitate of 
PbCrOi proves lead. 

Sol. 23 Cu++ 
Make basic 

with NH 4 OH. 

Blue Cu(NH 8 )<++ indicates 
copper. To confirm, acidify 
with HC 2 H 3 O 2 . Add 0.1 F 
K4Fe(CN)e. Maroon precipi- 
tate of CujFe(CN)e, proves 

Sol. 22 HgSr*, SnS 9 ~ 

Add 3 F H 2 SO 4 to just neutralize the solution, noting the 
volume, and add ^ volume in excess. Boil to remove 
HgS. Cool. Centrifuge, discarding filtrate. Treat the 
residue (HgS, SnS* S) with 6 F NaOH, H 2 O and special 
30% H 2 O 2 . Warm, then boil to decompose excess H 2 O 2 . 
(If yellow color of S 2 persists, repeat H 2 2 treatment.) 

Res. n HgS 

Add aqua regia, heat, 
evaporate, add H 2 O. To 
the clear solution, add 
0.1 F SnCl 2 . Hg 2 Cl 2 
(white), or Hg (black), 
proves mercury. 

Acidify with 6 F HC1. Add Fe 
wire. Evaporate almost to 
dryiicss. Add H 2 O. Decant. 
To the clear solution add 0.1 F 
HgCl 2 . Hg 2 Cl 2 (white), proves 

Prepare a similar set of solutions using 0.1 M Zn+ +. 
Pass H 2 S gas (see Fig. F-4) into each of the eight 
solutions to saturate them with the gas. Note the 
formation of any precipitate, in the chart provided 
in the report sheet, and explain the results in the 
light of the differences in solubility and of the 
ionization of H 2 S. 

2. The Oxidation of Hydrogen Sulfide. Into 2-ml 
samples each of warm 6 F HNO 3 , of Br 2 water, and 
of 0.1 F FeCl 3 acidified with about 5 drops of 6 F 
HC1, pass H 2 S gas. Note, and explain any changes 
in color, or formation of precipitates. 

3. The Solution of the Sulfides in Acid. First pre- 
pare a little CuS and HgS by passing H 2 S into 1-ml 
samples of 0.1 M Cu++ and 0.1 M Hg++ solutions. 
Add an equal volume of 6 F HNO 3 to each to give 
3 F solutions, and heat just about to boiling for 
two minutes only. Results? If either precipitate 
fails to dissolve, let it settle, decant, and to the 
residue add 10 drops of 12 F HC1 and 5 drops of 
16 F HNO 3 . Warm, and note results. 

4. Amphoteric Sulfides. Prepare a little CuS and 
SnS 2 by adding 2 drops of 2 F NaHS solution to 
1-ml samples of 0.1 M Cu++ and 0.1 M Sn++++ 
solutions. Let the precipitate settle, decant, and to 
the residue add 1 ml of 2 F NaHS and 0.5 ml of 
6 F NaOH. Warm, if necessary, and note any dif- 

ference in behavior of the two precipitates. To the 
one which dissolved, now add 3 F H 2 SO 4 drop by 
drop to just make it .acid. Results? 

5. The Behavior of Lead Salts. Prepare a little 
PbSO 4 precipitate by mixing 1 ml of 0.1 FPb(NO 3 ) 2 
and 2 drops of 1 F Na 2 SO 4 . Divide the mixture in 
two test tubes. To one, add 1 ml of 6 F HC 2 H 3 O 2 . 
To the other, add 1 ml of 3 F NH 4 C 2 H 3 O 2 . Note 
and explain the results. To the one which dissolved, 
add a few drops of 1 F K 2 CrO 4 . Results? 

6. The Reduction of Mercuric Ion by Stannous Chlo- 
ride. To 1 ml of 0.1 F HgCl 2 , add several drops of 
0.1 F SnCl 2 . Then, add 2 ml of additional SnCl 2 , 
Let stand and observe the change. The black color- 
ation is due to free mercury. 

B. Analysis of the Hydrogen Sulfide Group* 
(Group 2) 

We shall include only the ions Cu" 4 "*, Pb+ + , 
Hg++, Sn*" 1 " and Sn" 1 "** 4 " in our study of this 
group. For practice, analyze a known solution con- 
taining these ions by the following procedure. 
Then, obtain an unknown from the instructor. If 
you have time, analyze an unknown containing 
ions of both Groups 1 and 2 for both groups. 
Procedure for the Analysis of the Hydrogen 

To 4 ml of the solution to be analyzed (or Sol. 



11 from the silver group, if a general unknown), 
add 6 F NH 4 OH, or 6 F HC1, drop by drop, as 
needed, to adjust to 0.3 M II +. (This acidity may 
be determined with methyl violet paper, which 
may be prepared by soaking filter paper with 
methyl violet indicator solution and drying. Touch 
a drop of your solution on a stirring rod to the 
indicator paper. This may be matched with the 
blue-green color produced by a drop of 0.3 F HCL) 
An alternate method of adjusting to 0.3 M H+ is 
to neutralize 1 your solution to the nearest drop by 
drop-wise addition of NHUOH and HC1, then add 
1 drop of 6 F HC1 for each milliliter of solution. 
(Not valid if anions of weak acids are present.) 

Precipitation of the Group. Heat the solution, and 
saturate it with H 2 S gas by slowly bubbling the 
gas into the solution for several minutes. Be sure 
the gas delivery tube, which should preferably be 
drawn down to a coarse capillary, is clean. To 
clean it, immerse it in warm 6 F HC1 in a 10-cin 
test tube and rinse. Filter, or centrifuge, the mix- 
ture. (Get instructions on the use of the centrifuge. 
The solution in a 5-ml centrifuge tube or 10-cm 
test tube must always be carefully balanced with 
an equal amount of water in a similar tube placed 
opposite in the centrifuge. A larger amount of so- 
lution may be divided equally in two similar tubes 
to centrifuge it.) Decant the clear solution care- 
fully into a clean test tube and test for complete 
precipitation 2 by again bubbling in BUS for a mo- 
ment. If the solution is to be saved for further 
analysis, boil it for a moment to expel BUS com- 
pletely, label it S21, and set it aside for Group 3. 
Wash the precipitate (R21) by adding 1 to 2 ml 
of water. Centrifuge again and discard the wash 

Separation of the Copper and Tin Sub-Groups. To 
Res. 21, add 1 ml of 6 F NaOH, and dilute to 2 ml 
with water. Warm and agitate the mixture gently 
for a moment. Cool it, and, if a dark residue re- 

1 In using litmus paper to test for neutrality, it is best not to put 
the paper in the solution, as the absorption of ions renders it less 
sensitive, and also because the litmus dissolves in the solution. Rather, 
touch a drop of the solution, held on the end of a stirring rod, to the 
litmus paper. 

8 In case a very large amount of precipitate has formed, the solu- 
tion may have become too acid due to a reaction of the type: 
M++ + H*S * MS + 2 H+. 

In this case, boil the solution to remove H?S, cool it, and readjust 
to 0.3 M H+ before passing in HsS a second time. 

mains (which might be stannous sulfide), add 1 ml 
of 3% hydrogen peroxide to oxidize any tin to the 
stannic state. Agitate again, and finally heat to 
decompose any excess H 2 O 2 . 8 Now, add 2 ml of 2 F 
NaHS reagent. Warm and mix for several minutes, 
then centrifuge. Save the filtrate (S22) for analysis 
for Hg++ and Sn" 1 "* 4 " 1 ". Wash any dark residue 
(R22) with 2 ml H 2 O containing several drops of 
2 F NaHS and a drop of 6 F NaOH, and discard 
the washings. 

Test for Lead Ion. To Res. 22, add 1 ml of H 2 O 
and an equal volume of 6F HN0 3 , and heat almost 
to boiling, with shaking, for 2 minutes, but no 
longer. Centrifuge, and if a dark residue remains 
(which might possibly be HgS), 4 save it and com- 
bine it later with Res. 24. To the filtrate, contained 
in a small beaker, add 2 ml of 3 F H2SO4, and 
evaporate until almost dry and very dense white 
choking fumes of S0 8 are evolved. (Hood: The 
excessive escape of these fumes into the room can 
be prevented by covering the beaker with a watch 
glass, after the evaporation. The purpose of this 
heating is to remove all HNO 8> in which lead sulf ate 
is more soluble.) Cool, then cautiously add 3 ml 
H 2 O. Cool, and let stand 5 minutes. A white pre- 
cipitate, settling readily, is probably PbSCV Cen- 
trifuge, and decant the clear solution (S23) and 
save for the cupric ion test. Wash the white pre- 
cipitate (R23) with a little water, let settle or 
centrifuge, and discard the washings. To the resi- 
due, add 1 ml of 3 F NH4C 2 H 3 2 and warm to dis- 
solve any PbSO 4 . Add 1 ml of 1 F K 2 CrO 4 . A yellow 
precipitate proves lead. 

Test for Cupric Ion. To Sol. 23, add 6 F NH 4 OH 
to make it distinctly basic. A deep blue color, 
Cu(NH8)4" l " f " indicates copper. This is quite dis- 
tinctive, but if desired Cu+ + may be confirmed by 
acidifying with 6 F HC 2 H 8 O 2 , and adding 0.1 F 
KtFe(CN)6. A maroon coloration or precipitate of 
Cu 2 Fe(CN)o proves copper. 

Precipitation of the Tin Group. To the filtrate 
(S22) which is strongly basic and contains much 
S , add 3 F H 2 SO 4 drop by drop just enough to 
neutralize it, testing with litmus and noting the 

9 A black residue, possibly PbS or CuS, may remain at this point. 

4 Since HgS is not too readily dissolved by the previous NaHS 
treatment, it is wise to test a dark residue here for Hg. However, this 
dark residue may be simply a trace of CuS or PbS, encrusted over 
with free S formed by oxidation, and hence rendered insoluble. 



volume of acid used. (Mix it well, but be cautious, 
or the H 2 S may be evolved too rapidly. Work 
under the hood. As long as H 2 S is evolved rapidly 
on the addition of each drop of H2SO4, the solu- 
tion is still basic.) When the solution is neutral, 
add about ]/& this volume of H 2 SO 4 , in excess. 1 
Centrifuge, or let settle, and discard the solution. 
A dark or colored residue indicates the presence of 
the tin group. If the residue is white or very pale 
yellow, it is sulfur, formed by the decomposition 
of the S 2 present: 

2 + 2 

' H2S + S. 

Test for Mercuric Ion. Treat any colored residue 
formed above with 1 ml H 2 O, heat to boiling to 
expel H 2 S, add 0.5 ml 6 F NaOH, and 0.5 ml of 
30% (not 3%) H 2 O 2 . 2 Transfer this to a 15-cm 

1 Since HSO<~ is a weak acid (K = 1.2 X 10~'), a slight excess of 
H*SO 4 reacts with some of the SO 4 formed 

H,SO 4 -I- SO 4 HH*- 2 HSO<r, 

so that the acidity is repressed by a buffering action. With H 
excess acid, the SO 4 and HSO 4 ~ concentrations will be about 
equal, so that 

- " X 


- 12 X 

This is not sufficient H+ to dissolve the SnSa precipitate. See the 
explanation of buffer action in Experiment 32. 

* Any SnSi will dissolve as Sn(OH)e" and S~ . This S~ , unless 
destroyed by oxidation with ILO 2 , might dissolve some HgS, which 
is otherwise insoluble in OH~, and thus render the separation in- 

test tube and boil it to decompose excess H 2 O 2 . S 
If the yellow color of S 2 persists, cool, and re*- 
peat the H 2 O 2 treatment. Centrifuge, and save the 
filtrate, Sol. 24, for the Sn++++ test. 

Confirm mercury by dissolving the residue (B24), 
to which has been added any dark residue from 
the treatment of Res. 22, in a little aqua regia 
made by mixing 1 ml 12 F HC1 with 0.5 ml 16 F 
HNO 8 . Heat this mixture and evaporate almost, 
but not quite, to dryness. Add 2 ml of H 2 O, and 
filter or centrifuge, if necessary. The solution must 
be perfectly clear. Add 2-3 drops of 0.1 F SnCl 2 . 
A white silky precipitate slowly turning dark when 
an excess of about 1 ml of SnCl 2 is added, proves 

Test for Tin. Acidify Sol. 24 with 6F HC1, using 
a slight excess of acid so that any Sn" 1 "** 4 " will 
remain in solution. Add a 1-cm piece of iron wire 
and evaporate the solution until only 1 or 2 drops 
of liquid remain. The iron reduces any Sn++++ to 
Sn+ +. Add 1 ml of H^O. Decant this clear solution 
into a clean test tube, and add 5 to 10 drops of 
0.1 F HgCl 2 . A white precipitate of Hg 2 Cl 2 , or 
black Hg, proves tin. 

8 Shake the tube while heating it to minimize the chance of loss of 
material by the rapid evolution of oxygen. Continue heating until 
the excessive frothing subsides and the mixture boils gently, and on 
removal from the flame, no more oxygen gas is evolved. 

REPORT: Exp. 41 

The Hydrogen Sulfide Group 

A* Typical Reactions of the Hydrogen Sulfide Group 

1. Precipitation of Metallic Sulfides. 



Locker Number* 

Cu+ + 



0.1 M H + 

O.OLJf H+ 

0.001 M 11+ 

Explanation of the above results : 

2. The Oxidation of Hydrogen Stdfide. 

Write net ionic equations for the reactions of H^S with each of the three substances, HNOa, Br2, Fed 3 in acid 
solution. Note any changes in the appearance of the solutions. 


List the two different types of chemical reactions exhibited by hydrogen sulfide (in pars. 1 and 2). 



8. Solution of Sulfides in Acid. 

Observed results of any action of 3 F HNO 3 on CuS and HgS: 

f The equation is_ 

If HC1, instead of HNOa, is added to CuS, the equation is 

Why does a mixture of HC1 and HNOa exert a greater solvent action on HgS than does HNOa alone? 

The equation is_ 


Amphoteric Svlfides. 
The net ionic equation for the reaction of NaOH and NaHS is:_ 

Explain the difference in behavior of CuS and SnS2 with S , including any equations needed: 

What happens on acidifying the resulting solution? Write equation. 

6. Behavior of Lead Salts. 

Observed results on treating PbSCh with 

(1) HC 2 Il3O2 (2) 

What conclusion can you draw from this as to the solubility and extent of ionization of 

The net ionic equation is_ 

When K2Ci<>4 is added, the net ionic equation for the reaction is_ 

6. The Reduction of Mercuric Ion by Stannous Chloride. 

Write net ionic equations for the reaction of mercuric chloride with (1) a small amount of SnCU, (2) an excess 
of SnCU, recording also the color of any precipitates formed: 


B. Analysis of the Hydrogen Sulfide Group 1 

Summary, Unknown No Ions found- 

Unknown No. Ions found- 

1 For these, and each succeeding group, prepare on a separate sheet (with your name at the top) a form similar to that used for the silver 
group, as on the report sheet for Experiment 40, and hand in with your experiment report. 


The Ammonium Sulfide Group. 


College Chemistry f Oiapfen 26, 27, 2, 29 

Review of Fundamental Concepts 

Precipitation with Ammonium Sulflde 

An ammonium sulfide solution is rather difficult 
to keep, since such a high concentration of sulfide 
ion is readily oxidized by the air to free sulfur. 
This dissolves in the ammonium sulfide solution to 
form a yellow ammonium polysulfide solution, 

S + S >- S 2 ~ . 

This latter ion gives a bothersome precipitate or 
colloid of free sulfur every time the solution is 

S 2 ~ + 2 H+ > H 2 S + S. 

We therefore prepare the ammonium sulfide re- 
agent as we use it, by making the test solution 
basic with NH 4 OH, and then passing in II 2 S gas, 

2 NH 4 OH + HzS > 2 NH 4 + 4- S~ + 2 H 2 O. 
Actually, the reaction does not proceed much past 
the first stage of neutralization 

NH 4 OH 4- H 2 S > NH 4 + + HS~ + H 2 O, 
since both ammonium hydroxide and hydrogen 
sulfide are so poorly ionized that ammonium 
sulfide, (NH^S, is largely hydrolyzed to ammon- 
ium hydrogen sulfide, NHJIS. The net ionic 
equation for the precipitation of a metallic sulfide 
in ammonia solution, therefore, is principally 

NH 4 OH + Zn++ + HS- > ZnS + NH 4 + + H 2 O. 
However, for such reactions, we shall usually write 
the simpler equation 

Zn++ + S + ZnS. 

The Hydrolysis of Sulfldes of Weak Bases 

The precipitation of sulfides has been discussed 
in the preceding experiment. The student will 
notice, however, that two of the ions in this group 
are precipitated as hydroxides, not as sulfides, on 
the addition of ammonium sulfide solution. This 
fact is due to the large extent to which salts of 
weak bases and weak acids hydrolyze. (Review 
Exp. 34 on hydrolysis.) Aluminum hydroxide is 
such a weak base that when ammonium sulfide is 
added to A1+ ++ solution, we have the reactions 

2 A1+++ + 6 HOH 
3S" + 6 HOH 

2 Al(OH) t + 6 H+ 
+ 6 OH" 

2 A1+++ + 3 S + 6 HOH * 2 A1(OH), + 3 HjS 
The hydrolysis of both the aluminum ion and the 

sulfide ion is shifted completely to the right, since 
the products of the reaction, hydrogen ion and 
hydroxide ion, neutralize one another, and also 
because aluminum hydroxide separates as a pre- 
cipitate and hydrogen sulfide is evolved as a gas. 
Chromic ion, Cr+++, behaves in a similar fashion. 
Likewise, the addition of carbonate ion to a metal- 
lic ion whose hydroxide is a weak base, as alumi- 
num ion or ferric ion, results in the precipitation 
of a hydroxide instead of a carbonate, inasmuch 
as carbonic acid is very poorly ionized. 

Buffer Action 

Frequently, in chemical processes, it is desirable 
to maintain the pH, or relative acidity and basicity 
of a solution, at an approximately constant value, 
even when considerable quantities of base or acid 
are added. In life processes, the pH of the blood 
must be kept practically constant at a value of 
about 7.4. This is accomplished by the bicarbonate 
and acid phosphate ions HCO 3 ~, HPO 4 , and 
H2PC>4~, which are present. In general, such con- 
trol of the pR of a solution is attained by having 
present a weak acid together with a salt of that 
acid, or a weak base together with a salt of the 
base. Such a combination, which resists change in 
pH, is called a buffer mixture. 

In the precipitation of the ammonium sulfide 
group from Solution 21, ammonium hydroxide is 
added to make the solution basic before passing in 
hydrogen sulfide gas. It is important, however, 
that the solution does not become too basic, or 
magnesium ion from the alkaline earth group may 
precipitate as magnesium hydroxide. Also, a strong 
ammonium hydroxide solution has a hydroxide 
ion concentration almost high enough to redissolve 
aluminum hydroxide as the amphoteric hydroxide 
complex ion, A1(OH) 4 ~. These situations are 
avoided by having present a moderate concentra- 
tion of ammonium ion (added as ammonium 
chloride), which maintains the hydroxide ion con- 
centration at a value somewhat less than that of 
ammonium hydroxide alone. The buffer action may 
be explained as follows. 




In a mixture of ammonium hydroxide and am- 
monium ion, with a strong acid added, the hydro- 
gen ion is reacted on by the ammonium hydroxide 
present; or if a strong base is added, the hydroxide 
ion is reacted on by the ammonium ion present: 

NH 4 QH + H+ * NKU+ + H 2 O, 
or NH 4 + + OH- >-NH 4 OH. 

The mixture thus does not change its pH. greatly 
with the addition of either acid or base. This is 
further explained by the equilibrium expressions 
for the ionization of ammonium hydroxide: 

+ OH- 

v i Q v in-j& 

TNH 4 OH) -K- 1.8X10*. 

If we rewrite this in the following manner, 
X (OH-) = 1.8 X 10~ 5 , 

we see that the hydroxide ion concentration de- 
pends on the ratio of ammonium salt, i.e. NH 4 +, to 
free ammonium hydroxide, NH 4 OH. So long as 
there is a considerable amount of both salt and 
base present, the ratio will not change greatly on 
the addition of moderate amounts of acid or base. 
The hydroxide ion concentration will be corre- 
spondingly constant. 

We shall also make use of another buffer, a 
mixture of sodium sulfate and sodium hydrogen 
sulfate. The hydrogen sulfate ion is a weak acid, as 
is represented by its equilibrium expressions : 

(H+)(SOr-) io xlo - 

(HSOr) - K - L2X1 

The addition of hydrogen ion shifts the above 
equilibrium to the left, and the addition of hydrox- 
ide ion shifts it to the right. However, so long as 
there is a considerable amount of both sulfate ion 
and hydrogen sulfate ion present, the ratio of 
(SO4 )/(HSO 4 ~) does not change rapidly on the 
addition of acid or base, and the hydrogen ion 
concentration, as shown by the above equation, 
should remain about 10 ~ 2 M. 1 

In Experiment 41 we used this buffer to prevent 
the re-solution of the stannic sulfide formed when 
SnS 3 -- is acidified (footnote 1, p. 302). We shall 
use it here to prevent the solution of zinc sulfide 
while dissolving all other sulfides and hydroxides 
of the group. 

1 Actually, in the buffer mixture which we use in this experiment, 
0.5 F Na2SO 4 and 0.5 F NaHSO 4 , the pH is nearer A value of one. 
This fact is to be explained by the effect of such high concentrations 
on the activities of the ions present. 

Oxidation States of Chromium and Manganese 

While these elements are metals, they quite read- 
ily coordinate with oxygen to form oxides or nega- 
tive complex ions in which they have higher oxida- 
tion states. These are very important substances. 
(Consult your textbook and review the charts in 
Study Assignment D on principal oxidation states 
of chromium and manganese, and the amphoteric 
character of chromium hydroxide. Also review the 
chart on the oxidation states of oxygen so as to 
understand the behavior of hydrogen peroxide. We 
shall use this substance both as a powerful oxidizing 
agent and, under different conditions, as a reducing 
agent. You should know all these facts well.) 

We make use of the great change in properties 
of the ions of chromium and manganese with 
change in oxidation state in order to separate and 
test for these elements. The properties of chromic 
ion (Cr+++) are very similar to those of the triva- 
lent ions of aluminum (A1+++) and iron (Fe+++). 
However, when the chromium is oxidized to chro- 
mate ion (CrO 4 ), it behaves very differently and 
is readily separated and tested. The oxidation is 
accomplished by hydrogen peroxide in a basic so- 
lution. 2 The peroxide ion in basic solution is HQ 2 -. 

a The pH of the solution has a great deal to do with the action of 
hydrogen peroxide on the ions of chromium and manganese. While 
hydrogen peroxide is not nearly so strong an oxidizing agent in basic 
solution, the potentials for the oxidation of chromic hydroxide and 
manganous hydroxide to chromate ion and manganese dioxide, re- 
spectively, are also much lower, so the oxidation takes place readily. 
Compare the oxidation potentials from La timer, Oxidation Potentials 
(Second Edition), Prentice-Hall, 1952. 

Acid Solution Basic Solution 

H 2 Oz -Oa -0.67 volt Cr(OH), -CrO 4 +0.13 volt 

Mn++ -MnOz -1.23 HO*- -Ot +0.076 

CP+-H- -Cr 2 O 7 -1.33 Mn(OH) 2 -Mn(OH)* -0.1 

Mn++ -&nQ- -1.51 Mn(OH) 8 -MnO* -0.50 

MnOa -MnQ 4 - -1.695 MnOj -MnO 4 " -0.60 

H0 -HaO* -1.77 HaO -HOT -0.88 

In acid solution, CrzOr is reduced by H^a (potential difference 
1.33 - 0.67 or 0.66 volt) to Cr+++ While the HA is also a very 
powerful oxidizing agent, and could certainly oxidize Cr" M " f to 
CrjO? (potential difference 1.77 L33 or 0.44 volt), the former 
reaction predominates. When Cr+ ++ is oxidized to CrO 4 in the basic 
solution, it is therefore necessary to destroy any excess HsO* com- 
pletely before the solution is acidified, in order to prevent the subse- 
quent reduction of the CrjO? back to Cr" 1 "*"*" again. 

Likewise, while HaOa is sufficiently powerful, either in acid or in 
basic solution, to oxidize Mn++ or Mn(OH)2 to MnO 4 ~, in acid 
solution HjOj does not react on Mn+ + because it more rapidly reduces 
MnO 4 ~ or MnO 2 to Mn + + and liberates O 2 . In basic solution, the 
peroxide ion reduces Mn0 4 ~ to the very insoluble MnO% but no 
further. Also, the basic peroxide ion oxidizes Mn"*" * to MnOj, but no 



The excess hydrogen peroxide is readily decom- 
posed into water and oxygen gas by boiling the 
solution. (Note that no oxygen gas is formed, how- 
ever, by the hydrogen peroxide which is used to 
oxidize the chromic ion*) 

Any manganous ion which is present in the 
above treatment with hydrogen peroxide in basic 
solution, is precipitated by oxidation to manganese 
dioxide (MnO 2 or more probably a hydrated form 
MnO(OH)2). After the separation of the man- 
genese dioxide precipitate, it is redissolved by 
hydrogen peroxide in a nitric acid solution. In this 
case, the manganese dioxide is reduced to man- 
ganous ion by the hydrogen peroxide, which, in 
turn, is oxidized to oxygen. Finally, the very char- 
acteristic purple permanganate ion (MnO4~) is 
produced by adding the powerful oxidizing agent, 
sodium bismuthate (NaBiOa), 1 which is reduced to 


Chemicals: 1 F HC 2 Hs0 2 , Aluminon reagent, 0.1 F Al(NOj)t, 
1 F NH4C 2 H 3 2 , 3 F NI^Cl, saturated Br, water, 0.1 F 
Cr(NO 8 ),, 0.1 F FeCls, FeSO 4 '7H 2 O, 3% HA, 30% ftO* 
Pe (filings), 0.1 F Pb(C2HjO 2 ) 2 , 0.1 F Mn(NO 8 )j, 0.1 F 
K 8 Fe(CN) 6 , 0.1 F IQFeCCN)* 0.1 F KCNS, NaHCO, (pow- 
dered), NaBiO (solid), 1 F Na*CO 8 , 1 F NaHSQ 4 1 F NasSO*, 
0.1 F Zn(NOs) 2 . 

A. Typical Reactions of the Ammonium 
Sulfide Group 

1. Behavior of Al+++,Fe+++, Cr+++, Mn++, and 
Z n ++ w ith Typical Reagents. In the report sheet, 
is a chart form on which you may enter pertinent 
data as to any reaction of the above ions with 
NH 4 OH, NaOH, NaiCOa, H 2 S, and (NH 4 ) 2 S, re- 
spectively. For any reactions with which you are 
already thoroughly familiar, fill in the chart with- 
out wasting time to repeat the experiment. For 
each reagent, take about 2-ml samples of 0.1 F 
solutions of the five ions, and add first only 1 to 3 
drops of the reagent. Where an excess is called for, 
add more (up to 2 ml if needed) to the same solu- 
tion to determine if this produces a different kind 
of product. Review Experiment 35 in this con- 
nection on the formation of complex ions and 
amphoteric hydroxides. Before testing with B^S 
gas, add a drop of 6 F HC1 to the 2-ml samples of 
each ion. Then, to test with (NHOA make these 

the trivalent bismuth ion (Bi++ +). These changes 
in the oxidation state of manganese are indicated 
in Figure 42-1. 

c, 4 


i The commercial product here called aodium bi&muthate, NaBiO* 
is probably quite complex, consisting essentially of BiO4 (bismuthyl 
bismuthate, BiO BiO 3 ) as the oxidizing agent, which in strong acid 
easily oxidizes Mn++ to MnOi"". 

-I *; -, *-r 

! HA ! H a (X ! NaBiO, ' 
! (NaOH) ! (HNOj) ! (HNQ,) I 

FIG. 42-1. Changes in oxidation state in the separation of 


same solutions slightly basic with 6 F NHiOH and 
pass in more EtS if necessary. 

If you cannot decide the nature of a given pre- 
cipitate, for example, whether the reaction of so- 
dium carbonate with ferric ion gives a carbonate 
or a hydroxide, devise a simple experiment to test 
the point. (In this case, note the evolution of a 
gas. After washing the precipitate, what reagent 
would you add to cause effervescence of COa, if a 
carbonate were present?) 

2. Red Lake Formation with Aluminum Hydrox* 
ide. The light flocculent precipitate which forms 
when ammonium hydroxide is added to aluminum 
ion is often difficult to observe* The dye called 
"aluminon" (ammonium aurin tricarboxylate), if 
added to the mixture, is adsorbed by the precipi- 
tate to form a characteristic red "lake" which 
renders the identification of the aluminum hydrox- 
ide easier. To 1 ml of water, add a few drops of 
0.1 M A1+++ solution. Add 2 drops of aluminon 
reagent, ai^d then 3 to 5 drops of 6 F NHUOH to 
precipitate the Al(OH)s. Note the color. Now make 
the solution acid with about 0.5 ml 6 F HC1, and 
again make just basic by drop wise addition of 6 F 
NH 4 OH. After letting the mixture stand a minute, 
note the color, both of the precipitate (which now 
should be red), and of the solution. A good red 
color is not obtained if the solution is too basic. 
Explain how the above treatment guarantees a 



very slightly basic solution. 

3. Reactions of Fe++ andFe+++. You will need 
to make your own solution of ferrous ion, since 
such solutions are too easily oxidized by the air to 
be very stable. For this purpose, dissolve a very 
small amount of ferrous sulfate, or ferrous am- 
monium sulfate crystals in 6 to 8 ml of distilled 
water, for use in the following experiments. 

a. Tests for Fe++ and Fe+++. To very dilute 
samples of each ion (1 to 2 drops in 1 ml H 2 O), 
add several drops of 0.1 F K4Fe(CN) 6 solution. 
Repeat, using 0.1 F K 3 Fe(CN) 6 solution. Note, in 
particular, the cases which give dark blue pre- 
cipitates or colors. Which reagent, then, would you 
use to test for Fe++? For Fe+++? (Note that we 
get the intense blue colors when opposite valences 
are present, i.e., ferrous ion with ferricyanide ion, 
and vice versa.) 

To 1 ml of very dilute samples of each ion, add 
1 ml of 0.1 F KCNS. Was your ferrous ion solu- 
tion entirely free from ferric ion? 

b. Changes in the Oxidation State of Iron. Add 
several drops of 6 F NaOH to a 2-ml sample of the 
Fe** solution. Result? Divide the mixture into two 
portions. To one, add a few drops of 3% H 2 O2 and 
mix for a moment. Result? Pour the other portion 
of this mixture over a filter paper and expose to 
the air for 10 to 15 minutes. Result? 

To 1 ml of Fe++ solution, add 1 ml of Br 2 water, 
boil this to remove any excess free bromine, cool 
the solution and test it for Fe" 1 " 4 " 4 ". 

Shake 2 ml of 0.1 F FeCl 8 with some iron filings 
for about 5 minutes. Filter, or decant, and test the 
resulting solution for Fe 4 " "*". 

4. Higher Oxidation States of Chromium and Man- 

a. Oxidation with Hydrogen Peroxide. To 1-ml 
samples of 0.1 F Cr(NO 3 ) 3 and of 0.1 F Mn(NO 8 ) 2 
in 15-cm test tubes, add 2 drops of 30% H 2 O 2 , then 
5 drops of 6F NaOH. Mix, and after a moment boil 
the mixture to decompose excess H-zOa. 1 Results? 
(Save the solutions for parts b and c.) 

b. Chromates and Dichromates. To the yellow 
Na 2 CrO4 solution prepared above, add 3 F H 2 S0 4 
to make it acid. Result? Now again make it basic 
with 6 F NaOH. Result? 

c. Manganese Dioxide, Manganous and Perman- 

> Any HiO* which is not decomposed will reduce the CrO 4 back 
to O+++ when the solution is acidified. See also Footnote 2, p. 306. 

ganate Ions. Divide the manganese dioxide mixture 
prepared above into two portions. To one, add 1 ml 
of 6 F HN0 8 ; and to the other, add 1 ml of 6 F 
HC1. Warm each. Compare and explain the results. 
To the one which did not dissolve, add 3 drops of 
3% H 2 O 2 , and mix. Explain the result. Boil the 
solution to decompose the excess H2O 2 , then cool 
it and add a small amount (a fourth the volume 
of a pea) of solid sodium bismuthate, NaBiO*. 
Result? What would happen if you add 1 ml of 
6 FHC1 to this? Try it. 
5. A Study of Buffer Action 

a. Ammonium Acetate as a Buffer. Since acetic 
acid and ammonium hydroxide have about the 
same value for their respective ionization con- 
stants, a solution of ammonium acetate will be 
practically neutral, and have the same pH as 
water. Place 5 ml of 1 F NH 4 C 2 H 8 O 2 in a 15-cm 
test tube; add 5 ml of distilled water in a second 
15-cm test tube. Add 2 drops of methyl orange 
indicator to each. Prepare a little 1 F HC1 by a 
sixfold dilution of the*6 F HC1, mix well, and fill 
your 10-ml graduate to the mark. Now to the 5-ml 
sample of water, add first a drop of the HO, and 
more if needed, until the red methyl orange end- 
point is reached, i.e., until the neutral water has 
been changed to about 10 ~ 3 M in H + . What 
volume of the 1 F HC1 was needed to do this ? Now 
repeat the experiment using the 5-ml sample of 

1 F NH4C 2 H 3 Q 2 instead of water. What volume of 
1 F HC1 was required this time to make the neutral 
solution 1G~ 8 M in H+? 

Let us also test the buffering action of ammon- 
ium acetate against the addition of a base. Prepare 

2 more 5-ml samples of H 2 O and of 1 F NH 4 C 2 H 3 O 2i 
respectively. To each, add 2 drops of indigo car- 
mine indicator. (Alizarin yellow R may be used 
instead.) Prepare a little 1 F NaOH by sixfold 
dilution of the 6 F NaOH. Add a drop of this, and 
more if needed, to the water sample until the 
indicator color changes, (about 10~ 2 M OH"). 
Also add the base, from a 10-ml graduate, a little 
at a time to the ammonium acetate sample, until 
the hydrbxide ion concentration has been increased 
to 10 ~ 2 M OH~. Note the respective volumes of 
1 F NaOH needed. Explain why the ammonium 
acetate solution is able to neutralize both acids 
and bases. 

b. Sodium Hydrogen Carbonate as a Buffer. As an 



instance of the use of an acid salt of a weak acid as 
a buffer, prepare a dilute sodium hydrogen carbon- 
ate solution by dissolving an estimated half gram 
of solid NaHCOa in 10 ml of distilled water. Divide 
into two 5-ml portions. To one add 2 drops of 
indigo carmine, then 1 F NaOH a little at a time 
until the indicator color change is reached (10~ 2 M 
OH-). To the other add 2 drops of methyl orange, 
then 1 F HC1 until red methyl orange color appears 
permanently throughout the liquid (10~ 8 M H+). 
Compare the volumes of base and acid added with 
the similar data for water. Explain how sodium 
i hydrogen carbonate acts as a buffer. 

c. Control of the Precipitation of Sulfides by Buf- 
fer Action. Dissolve 1 or 2 small crystals of ferrous 
sulfate in 2 ml of 0.1 F Zn(NO 3 ) 2 . Acidify this by 
adding about 0.5 ml of 1 F HC 2 H 8 02, and saturate 
the solution with H 2 S gas. After noting the color 
of the precipitate which forms, make the solution 
just basic by adding 3 to 5 drops of 6 F NH 4 OH. 
Explain the result in terms of the relative solubili- 
ties of zinc sulfide and ferrous sulfide. Centrifuge 
the mixture, and discard the solution. To the pre- 
cipitate, add 1 ml of 1 F Na 2 SO 4 , and then 1 ml 
of 1 F NaHSO 4 . Shake the mixture, and explain 
the result. 

B. The Analysis of the Ammonium Sulfide 

For practice, analyze a known solution contain- 
ing A1+++, Fe+++, Zn++ Cr+++, and Mn++, by 
the following procedure. Then, obtain two un- 
knowns from the instructor for analysis. If time 
permits, an unknown containing ions of both 
Groups 2 and 3 may be analyzed for both groups. 
This will give practice in the separation of the 
sulfides based on their differences in solubility in 
acid and basic solution. 

Procedure for the Analysis of the Ammonium 
Sulfide Group, A1+++, Fe++, Fe+++, Cr+++, 
Cr 2 O 7 , Mn++, MnOr, Zn++. 

To 4 ml of the solution to be analyzed (or Solu- 
tion 21 from the H 2 S group, if a general unknown), 
add 2 ml of 3 F NH 4 C1 or 1 ml of 6 F HC1, and 
then add just enough 6 F NH 4 OH to neutralize it, 
and about 0.5 ml in excess. 1 The absence of a pre- 
cipitate here (it will flocculate more on standing 
for several minutes) indicates the absence of which 
ions? Without filtering off any precipitate, saturate 
the solution thoroughly with H 2 S gas. Again add 

1 Any Fc + 4 " 1 ", C^O?""", or MnO 4 ~ would have been reduced to 
Fe++ Cr+++ and Mn++, respectively, by the HaS treatment in acid 
solution, when Group 2 was precipitated from a general unknown. 

If phosphate ion were present in the unknown, the ions of the alka^ 
line earth group of metals (Ca++, Sr++, Ba++, Mg++) would pre- 
cipitate here as phosphates, when the solution is made basic. This 
complication will not be introduced into our limited study of quali- 
tative analysis. 


Sol. 21 Cations of Groups 3-5. 
Add 3 F NH 4 C1, or 6 F HC1, then 6 F NH 4 OH to make basic. Saturate with HjS gas. Add NH 4 OH. Warm. Centrifuge, 

and wash the residue. 

Res. 31 Al(OH)s, Cr(0tf), FeS, Fe* t MnS, ZnS. 
Add 1 F Na 2 SO4, 1 F NaHSO 4 , as a buffer. Agitate the 
mixture. Centrifuge, and wash any residue. White to 
gray residue indicates zinc. 

Sol. 81 Cations of Groups 4-6. 
At once add HC1 and boil out 

Save lor Groups 4-5 

Res. 82 ZnS. 
To confirm, add 6 F HC1, 
evaporate almost to dry- 
ness. Add 1 F Na2SO 4 , 
and 1 F NaHS0 4 . Add 
lFNH 4 C 2 H 3 O 2 .Passin 
HjS. White ZnS proves 

Sol. 82 Al++\ Cr+++, Fe++, 
Boil to remove HjS. Cool. Add 6-8 drops 30% Hj0 2 , and add this to HaO + NaOH. Let 
stand. Boil to decompose excess H 2 O 2 . Centrifuge and wash the residue. 

Res. S3 Fe(OH)t, Mn0 2 . 
Add 6 F HNO, and HjO* Boil. 
Two portions: 
(1) Test for F*++\ 
Add NH4OH, HC1, then 0.1 F 
KCNS. A red color of 
FeCNS++ proves iron. 
(*) Test for Mn + +. 
Add NaBiOj, Purple Mn0 4 ~ 
proves manganese. 

Sol. S3 Al(OH)<~, CrOC". 

Acidify, 5 drops excess HC1, then NH 4 OH. FloccuJent 
precipitate indicates aluminum. Centrifuge and wash 

Res. 34 Al(OH)i. 
To confirm, add excess HC1, 
then aluminon reagent. 
Add NH 4 OH. Flocculent 
precipitate, Al(OH)a, dyed 
red, proves aluminum. 

Sol. 84 CrOc~. 
Acidify with 1 


.a*. Yel- 
low precipitate of PbCrC>4 
proves chromium. 



0.5 ml of 6 F NH4OH and warm the mixture 
slightly. Centrifuge the mixture. (Divide into two 
tubes to centrifuge it if there is too large a volume 
for one tube.) Save the solution (S31), acidifying 
it with HC1 and boiling out HjS at once, if analyses 
for Groups 4-5 are to be made. Otherwise, discard 
it. Wash the residue (R31) with 5 ml of distilled 
water and centrifuge. Discard the washings. Do 
not let Residue 31 dry; continue at once to the 
separatation of zinc ion. 

Separation and Test for Zinc Ion. To Residue 31, 
add first 2 ml of 1 F NaaSCh, and then 2 ml of 1 F 
NaHSO 4 . Agitate the mixture for 2 to 3 minutes. 
Centrifuge it, and save the solution (S32). Wash 
and centrifuge any residue (R32). 1 Add 5 drops 
of 6 F HC1 to the solid, and evaporate this just to 
dryness, (Avoid overheating.) Redissolve the resi- 
due with 1 ml each of 1 F NasSO^ 1 F NaHSO 4 , 
and 1 F NH^HaOa. Saturate this with H 2 S. Addi- 
tional NH^HaOa may be added, drop by drop, to 
promote precipitation. A white precipitate of ZnS 
proves zinc. 

Separation of the Iron and Aluminum Sub-groups. 
Boil Solution 32 to remove all H 2 S gas. Cool it, 
add 6 to 8 drops of special 30% H 2 2 , and add 
this mixture to a 15-cm test tube containing a 
solution of 2 ml of water and 0.5 ml of 6 F NaOH. 2 
Mix the solution and let it stand for 1 to 2 minutes, 
then boil it cautiously. (Recall Footnote 3, P. 302, 
as to the precautions to be observed to prevent 
loss of material when decomposing H 2 O 2 .) Centri- 
fuge, and save the solution (S33) for the tests for 
aluminum and chromium. 

Test for Iron (ferrous or Ferric Ion). Treat Resi- 
due 33 with 1 ml of 6 F HNO 3 . If solution is not 
complete, 8 add 3 drops of 3% H 2 O 2 to reduce any 
MnO 2 to Mn+ +, and boil the solution to decom- 
pose excess H 2 O 2 . Cool the solution, dilute it to 2 

1 The dark color may be due to some undissolved iron sulfide. 
If cobalt and nickel salts were present, black CoS and NiS would 
remain here, due to the slow rate at which they dissolve at this pH. 

* The order of addition of reagents is important as it assures an 
adequate mixing with the H 2 2 and opportunity for complete oxida- 
tion, before the H^Qj is decomposed by the basic solution, 2 to 3 ml 
of 3% H 2 Oj may be used instead. However, failure to obtain a good 
test for chromium is often due to incomplete oxidation of Cr '*"*"* to 


1 A separation of the Fe+++ solution and the MnO a precipitate 
can be made here, but it is unnecessary as the tests for Fe+ ++ and 
far Mn + + do not interfere with each other. 

ml with water, and divide it into two portions, To 
one, add 6F NH^OH to neutralize excess acid, and 
then add 6 F HC1 dropwise to make the solution 
slightly acid again. Add 1 ml of 0.1 F KCNS. A 
deep red color of FeCNS++ proves iron. 4 

Test for Manganous Ion. To the other portion of 
the above solution, add a small amount (about a 
fourth of the volume of a pea) of solid sodium 
bismuthate (NaBiO*). A purple color of perman- 
ganate ion (MnOi~) proves manganese. 

Test for Aluminum Ion. Acidify Solution 33 with 
6 F HC1 (test by a drop of the mixed solution on 
litmus paper), and add about 5 drops in excess. 
Then make the solution just basic with 6F NEUOH 
(faint ammonia odor after mixing). An almost in- 
visible, flocculent precipitate indicates aluminum. 
Centrifuge, and wash the residue. Save the solution 
(S34) for the test for chromium. 

To confirm, dissolve Residue 34 in 1 ml of 6 F 
HC1, add 2 drops of aluminon reagent, and then 
make the solution just basic by dropwise addition 
of 6F NH 4 OH. A red precipitate, which flocculates 
as the mixture stands, or on gentle warming, and 
which is suspended in an otherwise colorless solu- 
tion, proves aluminum. 8 If in doubt, observe the 
test tube against a white background. If a good 
red color, which is completely absorbed in the pre- 
cipitate, is not obtained, the solution is probably 
too basic. In this case, re-acidify with HC1, and 
again make just basic with NH 4 OH. 

Test for Chromate Ion. To Solution 34, add 1 F 
HC 2 H 3 O 2 to make it slightly acid, and then add 1 
ml of 0.\ F Pb(C 2 H80 2 )2. A yellow precipitate of 
PbCrO4 proves chromium. If necessary, centrifuge 
the mixture to obtain a better observation of the 

4 Do not report iron if only a slight coloration results, which may be 
due to impurities of iron in the reagents used, or more likely to rust 
particles carelessly introduced into the sample from test tube clamps, 
wire gauze, etc. If in doubt, test a sample of the original unknown 

with 0.1 F KCNS, 

5 This precipitate is quite characteristic. Silica, often present as 
an impurity from the NaOH used, gives a white precipitate. Cr(OH)a* 
which might be present from any unoxidized Cr** + , gives a pre- 
cipitate somewhat similar to that of A1(OH) 3 , but may be dis- 
tinguished as it will redissolve on the addition of 0.5 ml of 3 F 
(NH)aCQ|. Any Fe** + which failed to be completely separated from 
Solution 32 would give a reddish color to the original A1(OH) pre- 
cipitate before the aluminon reagent is added. In this case, treat this 
precipitate with a little 6 F NaOH, centrifuge the undissolved 
Fe(OH),, and acidify the solution with 6 F HC1. Then follow the 
aluminon reagent treatment as above. 


REPORT: Exp. 42 

The Ammonium Sulfide Group 



Locker Number^ , , , , ,^ 

A, Typical Reactions of the Ammonium Sulfide Croup 

1. Behavior of AI+++, Fe+++, Cr+++, Mn++ 9 and Zn++. 

In the spaces provided, write the formulas of the precipitates formed, or new ions formed in solution, if any, 
and indicate also any characteristic colors, etc. when each ion of the group is treated with the reagent in the left 
hand column. 







6 F NH 4 OH (excess) 


6 F NaOH (excess) 

1 F Na 2 CO 3 

H2S (acid solution) 

(NH 4 ) 2 S 

2. Red Lake Formation with Aluminum Hydroxide. 

The equation for the reaction of A1+++ with 
NH 4 OHis 

Explain how the addition, first of an excess of HC1, then of a slight excess of NEUQH, insures that the solution 
will not become too basic to form a satisfactory red adsorption compound of the dye with the precipitate. 

3. Reactions of Ferrous and Ferric Ions. 

a. Tests for Fe + + and Fe+++. Indicate the characteristic colors of the precipitates or solutions formed whei 
Fe+ + and Fe+ * + are treated with the reagents listed. 

Summarizing, then, the best reagent to use in testing for: 

Fe++ would be , giving a. 

would be ~_ ,. . , giving a. 








b. Changes in the Oxidation State of Iron. Write net ionic equations and indicate colors of any precipitates or 
iolutions formed when the following reactions were carried out: 

STaOH is added to ferrous ion .... 

8202 is added to the above product . . . 

Iction of moist air on ferrous hydroxide . . _.__ 

Br2 water is added to ferrous ion ... 

tron filings are shaken with Fed* solution . 

J. Higher Oxidation States oj Chromium and Manganese. 

Write net ionic equations and indicate colors of any precipitates or solutions formed when the following sub- 
stances are mixed in solution, (indicate if there is no action) : 

Hydrogen peroxide and manganous 

litrate (basic) 

Hydrogen peroxide and chromic nitrate 

Hydrogen peroxide, heated in a basic 

Chromate ion and sulfuric acid . 

AJbove product, and sodium hydroxide . 

Manganese dioxide and hydrochloric acid 

Manganese dioxide and nitric acid (warm) 

Above mixture, and hydrogen peroxide . 

Manganous ion and sodium bismuthate 

6. A Study In Buffer Action. 

a. Ammonium Acetate as a Buffer. The volume of 1 F HC1 needed to increase the acidity from neutral (pR 7) 
tolO-Jf H+for: 

5 ml water was _ , and for 5 ml 1 F NH^CWsC^ was --- The volume of 
1 F NaOH needed to increase the basicity from neutral (pR 7) to 10~ 2 M OH~ for: 

5 ml water was _ , and for 5 ml 1 F NH 4 C 2 H 3 O2 was -- 

Suppose the same volume of 1 F HC1 which you added to the 5 ml of 1 F NH 4 C2H3O 2 to change it to 10~*M H+, 
were added to the 5 ml of H a O instead. What would the H+ concentration of this solution now be? Show calculations. 


Name _ 

Explain, by discussion and equations, how ammonium acetate is able to buffer the solution against the addition 
of both acids and bases. 

b. Sodium Hydrogen Carbonate as a Buffer. The volumes of the following reagents required for reaction witb 
5-ml samples of an NaHCOg solution were: 

1 F HC1 ml, 1 F NaOH ml. 

Explain, by discussion and equations, how NaHCO 3 is able to buffer the solution against the addition of both 
acids and bases. 

c. Control of the Precipitation of Sulfides by Buffer Action. 

The addition of H 2 S to a solution of Zn 4 +, Fe++, and HC 2 H 3 O 2 gave a 
precipitate of 

The subsequent addition of NH 4 OH to the above mixture, which was saturated 
with H 2 S, gave a precipitate of 

Explain these results, considering the relative solubilities of the precipitated salts, and the effect of any buffer 

Explain, by discussion and equation, how a mixture of 1 F Na 2 S04 and 1 F NaHSOi is able to separate a 
mixture of ZnS and FeS: 

B. The Analysis of the Ammonium Sulfide Croup 1 

Summary, Unknown No Ions founxL. 

Unknown No Ions found. 

1 Hand in, also, with each analysis, a summary of the steps in the procedure, as given for the preceding unknowns. 


The Alkaline Earth and Alkali Groups. 


Review of Fundamental Concepts 

College Chemistry, Chapter 26 

The elements comprising the alkaline earth and 
alkali groups are quite similar and have many 
properties in common. Their compounds exhibit 
only one stable oxidation state*, they do not form 
amphoteric hydroxides, being distinctly basic; and 
they do not readily form complex ions. The sepa- 
ration of the alkaline earth elements depends al- 
most entirely on differences in the solubilities of 
their salts, which show a regular gradation through 
the periodic table. The salts of the alkali metals 
are almost all readily soluble. The following table 
will be useful in interpreting the analytical pro- 

(g/100 g H 2 O, fct room temperature) 















so 4 





CrO 4 










We shall include the ions Ba++, Ca + +, Mg++, 
Na+, K+, and NH 4 + in our study of these groups. 
Strontium ion (Sr + +) will be omitted, as its in- 
clusion would introduce no new principles of sepa- 
ration and analysis. While ammonium ion is not 
an alkali ion, its salts, likewise, are nearly all 
soluble, so it is convenient to consider it at this 
time. The test for ammonium ion must always be 
carried out with a separate portion of the original 
Unknown, since ammonium hydroxide and am- 
monium salts are used as reagents in the general 
group separations* 

Precipitation with Ammonium Carbonate 

The group reagent used to precipitate the alka- 
line earth ions is ammonium carbonate. (It is 
really an approximately equimolal mixture of am- 

monium hydrogen carbonate [NHJICOJ and am- 
monium carbamate [NH 4 CO 2 NH 2 ], but we shall 
speak of it simply as ammonium carbonate, 
(NH 4 )aCO 3 ). Since this is a salt of a weak bade and 
of a weak acid, it will hydrolyze in solution to a 
marked extent* forming the hydrogen carbonate 

NEU+ + CO," + HOH :p HCOr + NH40H. 

This results In a lower concentration of the car- 
bonate ion, on which the precipitation of insoluble 
carbonates depends. To prevent this, and thereby 
increase the carbonate ion concentration, the pre- 
cipitation is carried out in a strong ammonium 
hydroxide solution. The hydrogen carbonates of 
the alkaline earth ions are quite soluble. (Recall 
that calcium and magnesium hydrogen carbonates 
are the principal substances present in temporarily 
hard water.) 

The Separation of Magnesium 

Magnesium carbonate, being fairly soluble, is 
not precipitated readily unless the solution is con- 
centrated and the carbonate ion concentration is 
very high. Furthermore, it tends to form super- 
saturated solutions and does not precipitate 
promptly. Since magnesium hydroxide is even less 
soluble than the carbonate, the magnesium ion 
might precipitate as this compound when the am- 
monium carbonate and ammonium hydroxide re- 
agent is added. This can be prevented by buffering 
the solution with a high concentration of ammon- 
ium ion so as to repress the hydroxide ion concen- 
tration. In the analysis of a general unknown, 
sufficient ammonium chloride for this purpose will 
be present from the previous treatment of the solu- 
tion in the separation of the Group 3 ions. 

While in some schemes of analysis, all excess 
ammonium ion is removed and the carbonate ion 
concentration is made as high as possible so as to 
precipitate magnesium ion with the alkaline earth 
group, we shall leave it in solution and test for it 
in the solution containing the alkali ions. 

Experimental Procedure 

Chemical* t 6 F HCH*O 2 , 3 F NH 4 CH/)i, 3 F (NH^aCO,, agent" (p-nitrobenzene-azoresorcinol), 0.1 FKBr, 1 FK&rO* 
3 F NH 4 C1, 1 F NH.C1, NH 4 C1 solid, 15 F NH^OH, 0.1 F IF KE^PO* 1 F KaCaO* 0.1 F NaCl, Na*Ct>(NO 2 )e reagent. 
BaCla, 0.1 F CaClj, 0.1 F Mg(NOi)* "Magnesium R*. 




A. Typical Reactions of the Alkaline Earth 
and Alkali Groups* 

For this experiment, instead of performing a 
number of separate, prescribed tests to learn the 
properties of the ions, study the procedure for 
Analysis of the Groups, and answer the questions 
in the report sheet. You may wish to try out va- 
rious test tube reactions on your own initiative, 
such as the flame tests for sodium and potassium 
ions, illustrated in Figure 20-5; or refer to previous 
experiments such as Experiment 35, where the 
behavior of magnesium hydroxide in ammonium 
hydroxide and ammonium chloride solutions is 

B. The Analysis of the Alkaline Earth and 
Alkali Groups (Groups 4 and 5) 

Prepare a known solution containing all or some 
of the ions in the groups, and analyze it according 
to the Procedure. Then obtain one or more un- 
known solutions from the instructor for analysis. 
Procedure for the Analysis of the Alkaline 
Earth and Alkali Groups, Ba++, Ca++, Mg++, 
Na+, K+, and NH 4 +. 

Test for Ammonium Ion. Place 3 ml of the solu- 
tion to be tested in an evaporating dish, and make 
it basic with 6 F NaOH. Cover the dish with a 
watch glass, on the under side of which is at- 
tached a moist strip of red litmus paper. Warm 
the solution gently to liberate any ammonia pres- 
ent as the gas. (Avoid boiling, which would con- 
taminate the litmus with spray droplets of the 
NaOH solution.) An even, unspotted blue color 
proves ammonium ion. The characteristic odor of 
ammonia, observed soon after the solution is 

warmed, would also serve as a positive test. (Be 
very cautious in bringing your nostrils close to a 
hot sodium hydroxide solution, however, as it 
might be superheated, and splatter in your face 
or eyes.) 

Precipitation of the Alkaline Earth Group. If the 
unknown is being analyzed for this group only, 
evaporate a 4-ml sample of it to 2 ml, add 2 ml 
of 3 F NH 4 C1, and reheat it almost to boiling. If 
the sample is Solution 31 from a general unknown 
analysis, evaporate it to 3 to 4 ml, but omit the 
addition of ammonium chloride as this is already 
present. To either of these hot samples, add 2 ml 
of 3F (NH 4 )2CO 3 reagent, and agitate the mixture 
at intervals for 5 minutes. Centrifuge. Save the 
solution (S41) for the analysis of magnesium ion 
and the alkali ions. 

Test for Barium Ion. To Residue 41, add 6 F 
HCaHsOa, a few drops at a time, and warm the 
mixture to dissolve it. (Avoid much excess, or 
partially neutralize the excess with NH 4 OH.) Buf- 
fer the solution with 0.5 ml of 3 F NH 4 C 2 H 3 O 2 , 
dilute it to 2 ml with water, and add 1 ml of 1 F 
K 2 CrO 4 . Heat the mixture, and centrifuge. Yellow 
BaCrO 4 (R42) indicates barium. The solution 
(S42) will contain any Ca+ + . To confirm barium, 
add 3 to 5 drops of 6 F HC1 to Residue 42 to dis- 
solve the yellow precipitate, dilute to 2 ml with 
H 2 O, and add 2 drops of 3 F H 2 SO 4 . A white pre- 
cipitate of BaSO 4 proves barium. (It may be neces- 
sary to centrifuge to show whether the precipitate 
is white, in the yellow solution.) 

Test for Calcium Ion. To Solution 42, add 0.5 ml 
of 1 F K 2 C 2 O 4 and then make slightly basic with 


Test the original unknown for NH 4 +. 
Sol. 31 Cations of Groups -4-5. 
Add 3 F NH 4 C1, if not present from previous treatment. Add 3 F (NH 4 )2C0 8 . Agitate. Centrifuge and wash the residue. 

Res. 41 BaCOt* CaC0 9 . 
Add 6 F HC 2 H 8 O 2 , 3 F NH 4 C 2 H 8 O 2 , 1 F K 2 CrO 4 . 
Heat. Centrifuge. 

Sol. 41 Mg++ 9 Na+, K + . Divide into two portions: 

(1) Test for Mg++. 
AddNH 4 OH,lFKHjP0 4 . 
White MgNH 4 PO 4 indi- 
cates magnesium. To con- 
firm: centrifuge, to the 
residue add HCI, "mag- 
nesium reagent," then 
NaOH. Blue lake in 
Mg(OH)s proves magne- 

(2) Test for Na+, K+. 
Evaporate, add HNO 8 , evaporate 
and ignite to remove NH 4 + salts. 
Moisten residue with HCI. 
Flame tests: Yellow proves so- 
dium, violet proves potassium. 
To confirm K+: add H 2 O, 
Na*Co(NQ 2 ). Yellow precipitate 
of KjNaCo(NOj)e proves potas- 

Res. 4% BaCrO*. 
Yellow precipitate indi- 
cates barium. To con- 
firm: add HCI, then 
H2SO 4 . White BaSO 4 
proves barium. 

Sol. 42 Ca++. 
Add 1 F K 2 C2O 4 , then 
NH 4 OH. White CaCaO* 
proves calcium. 



6 F NH 4 OH. Let stand 10 minutes if a precipitate 
does not appear before that time. White CaC 2 O4 
proves calcium. It may be necessary to centrifuge, 
to show that the precipitate is white in the yellow 
solution. Further confirmation may be made, if 
desired, by decanting the yellow solution. Then to 
the precipitate add a few drops of 6 F HC1 to dis- 
solve it, add 1 ml of water, 2 drops of 1 F K2C2O4, 
and again make basic with NH 4 OH. The white pre- 
cipitate of CaC2O 4 will reappear. 

Test for Magnesium Ion. Divide Solution 41 into 
two portions. To one portion, add 1 ml of 6 F 
NH 4 OH and then 0.5 ml of 1 F KH 2 PO 4 . Agitate 
at intervals for 5 to 10 minutes. A white pre- 
cipitate of magnesium ammonium phosphate 
(MgNH 4 PO 4 6H 2 O) indicates magnesium. 1 Let the 
solution stand at least a half hour and observe, 
before deciding that Mg 4 " 4 " is absent, as this pre- 
cipitate is often slow in forming. To confirm mag- 
nesium, centrifuge the mixture, discard the solution, 
and dissolve the precipitate by adding several drops 
of 6 F HC1. Dilute this to 1 ml with water, add 1 
drop of "Magnesium Reagent" (para-nitrobenzene- 
azoresorcinol), and then make the solution alkaline 
with 6 F NaOH. Mix, and let stand for 1 to 5 
minutes. Observe against good light for a char- 
acteristic blue "lake." This is a flocculent precipi- 
tate of Mg(OH) 2 , colored blue by the adsorbed dye. 

Test for Sodium and Potassium Ions. Evaporate 

i If Ba 4 "*" and Ca ++ were not removed completely before, they 
would give insoluble phosphate precipitates here. 

the second portion of Solution 41 almost to dry 
ness in a porcelain evaporating dish, cool it, add '. 
ml of 6 F HNO 8 , and evaporate to dryness. Con 
tinue to heat the evaporating dish intensely wit! 
the bare flame, without using the wire gauze, s< 
that all portions of the dish are hot and all ammon 
ium salts have been vaporized, i.e., until whit 
fumes cease to be evolved. (This process is calle< 
"igniting" the precipitate.) Let the dish cool. Air 
solids remaining are probably salts of sodium o 
potassium. Moisten these with about 5 drops of 6 J 
HC1. Carry out flame tests for sodium and potas 
slum ions, using a clean nichrome wire, and tw< 
pieces of blue cobalt glass for observation of th 
potassium flame if much sodium is present. 8 Re 
view Figure 20-5 for the method of cleaning th 
nichrome wire, and the proper manner in which t 
carry out these tests. 

If the potassium flame test is not definite, ad< 
1 ml of H 2 O to the sail residue, transfer this solu 
tion to a 10-cm test tube, and add 0.5 ml of sodiuE 
cobaltinitrite reagent Na 3 Co(NO2)e. Let stand an< 
observe after 5 to 10 minutes if necessary, to se 
the yellow precipitate of K2NaCo(NO2)e> whicl 
proves potassium. 3 

8 Since sodium salts are so widely distributed and occur as train 
in many salts and reagents, do not report sodium unless it gives 
flame test as pronounced as a 0.005 M Na+ solution. 

8 If NH<+ salts were not completely removed by the above ign 
tion, a very similar yellow precipitate of ammonium cobaltinitrii 
would be formed. As a comparison, add 1 ml of Nu 3 Co(NOj) fl n 
agent to 1 ml of 1 1/) which contains 2 drops of 1 F NII 4 Cl. 


REPORT: Exp. 43 


The Alkaline Earth 

and Alkali Groups Seciwn 

Locker Number 

A. Typical Reactions of the Alkaline Earth and Alkali Groups 

Refer to the Solubility Chart and to the Outline for the Analysis, given in the discussion on this experiment, 
in answering the following. Try any experimental tests necessary to be certain of the results. 

1. Write net ionic equations for any reactions occurring when dilute solutions of the following are mixed: 
MgCl 2 , CaCl 2 , and NH 4 OH 

CaCl 2 , BaCl 2 , and K 2 CrO 4 .__. 

MgCl 2 , BaCl 2 , and H 2 SO 4 

CaCl 2 , K 2 C 2 O 4 , and NH 4 OH 

MgCl 2 , NH 4 C1, K 2 HPO 4 , and NH 4 OH .... 

2. On the basis of the periodic table relationships, what would you predict as to the general solubility (indi 
cate as soluble, slightly soluble, or insoluble) of each of the following: 

Be(OH) 2 Ra(OH)2 

BeSO 4 RaSO 4 

3. Why does the addition of HC1 dissolve such insoluble salts as BaCrO 4 and CaC 2 O 4 ? 

4. Why is there a difference in the behavior of a magnesium salt solution with (a) NH 4 OH, and (b) NEUOH 
NH 4 C1? 

5. What happens when a solid ammonium salt, as NH 4 C1, is heated intensely in an evaporating dish? 

6. What happens when HNO* is added to an ammonium salt solution, and the mixture evaporated and heated! 

B. Analysis of the Alkaline Earth and Alkali Groups 1 

Summary, Unknown No Ions found 

Unknown No. Ions found 

1 Summarize the procedure &s before, 


The Qualitative Analysis off Cd++, BI+++, AsOs- 
Sb+++, NI++, and Sr++ , 


A Study Assignment 

In the qualitative analysis scheme which has 
been presented in this manual, several cations 
were omitted. In general, the separation and de- 
tection of these ions require the application of the 
same principles which are illustrated by the 
analytical scheme as given. Therefore, for students 
who have the time to study the properties of these 
ions, the following data are included. These data 
will give the information necessary for the analysis 
of these ions in the analytical scheme. 

Properties of the Ions 

In the acid hydrogen sulfide group the additional 
ions are cadmium, Cd+ + , and bismuth, Bi+++, in 
the copper subgroup, and antimonous ion, Sb+++, 
and arsenite ion, AsO 3 , in the tin subgroup. 

Cd++. Th6 nitrate, acetate, sulfate, chloride, 
bromide, and iodide salts of bivalent cadmium are 
soluble, colorless compounds. The white, slightly 
soluble hydroxide is not amphoteric, but is soluble 
in excess ammonium hydroxide because of the for- 
mation of the complex Cd(NH 8 )4 + " 1 " ion. The sulfide 
is yellow and has a Ka p of approximately 10 ~ 28 . It 
does not dissolve in alkali sulfides, but is soluble in 
fairly concentrated hydrochloric acid and in hot, 
dilute nitric acid. The cadmium ion often forms 
complex ions, such as CdCl 4 and Cd(CN)4 , 
with halide and cyanide ions. The Cd++ ion con- 
centration is so low in solutions containing high 
concentrations of chloride ions that H 2 S will give 
no precipitate when added to an acid solution. 

Bi + + + . Bismuth is a metalloid which shows oxi- 
dation states of +3 and +5. In the higher oxida- 
tion state it is usually found as the bismuthate ion, 
BiOs"", which is a very strong oxidizing agent. In 
the analytical scheme it is usually encountered as 
the simple cation, Bi+++. The chloride, bromide, 
nitrate, and sulfate of this cation are readily 
hydrolyzed to form the insoluble basic or oxy salts : 

B1+++ + Cl- + H 2 O HF: BiOCl (*) + 2 H+. 

If excess hydrogen ions are present, the hydrolysis 
reaction is reversed to give a clear solution of the 
salt. The white hydroxide is insoluble both in ex- 
cess alkali bases and in ammonium hydroxide. The 
sulfide, Bi 2 S 3 , is dark brown and has a very small 

K0 P of approximately 10~ 72 . It is not amphoteric 
but is soluble in boiling dilute nitric acid and in 
concentrated hydrochloric acid. The Bi+++ ion 
does not form complex ions readily, the yellow 
BiI 4 ~ being the only stable one. Bismuth ion may 
be reduced by an alkaline solution of sodium stan- 
nite, Na2Sn(OH) 4 , to the black metallic bismuth. 

AS+++ or AsOz . Arsenic is a metalloid, and 

from its position in Group V it might be expected 
to be less metallic than its congeners antimony and 
bismuth. The hydroxides of arsenic in both of its 
oxidation states, +3 and +5, are definitely acidic, 
HsAsO 8 and HaAsO 4 . It is only in strongly acid 
solutioms that there is present an appreciable 
amount of the cations AS+++ and As+ 5 in equilib- 
rium with the anions AsOa and AsO 4 . When 

strongly acid solutions of these anions are saturated 
with H 2 S, the yellow sulfides As 2 S 8 and As^Se are 
precipitated. Both are insoluble in concentrated 
hydrochloric acid, but will dissolve in hot concen- 
trated nitric acid. Both of the sulfides are ampho- 
teric and dissolve readily in alkali sulfides to form 

thioarsenite, AsS 3 , and thioarsenate, AsS 4 

ions, respectively. Stannous chloride will reduce 
arsenic compounds to metallic arsenic in concen- 
trated hydrochloric acid solutions. The Marsh test 
for arsenic involves the reduction of arsenic com- 
pounds by zinc in an acid solution. The reduction 
product is arsine, AsH 8 , which may be decomposed 
to form a mirror of metallic arsenic. 

S&+++. Of the two oxidation states possible, +2 
and +5, the antimonous ion, Sb+++, is the one 
usually encountered. Its hydroxide is more bask 
than that of As + ++. It is slightly soluble in watei 
but is soluble in excess strong base, gaits of Sb+ + + 
such as the halides, nitrates, and sulfates, hy 
drolyze readily to form insoluble basic or oxy salts 
such as SbOCl. In a slightly acid solution, Sb2S 
may be precipitated with H 2 S, and the orange-rec 
sulfide is soluble in alkali sulfides, forming th< 

thioantimonite ion, SbS . The sulfide is soluble 

in moderately concentrated hydrochloric acid 
Antimony compounds may be reduced to metallic 
antimony by metals such as iron or zinc in moder 
ately concentrated acid solutions, but are not re 
duced by stannous salts. 




In the ammonium sulfide group the additional 
ions to be considered are Co++ and Ni++. 

Co++. Soluble salts of the cobaltous ion include 
the chloride, bromide, iodide, nitrate, and sulfate. 
In dilute water solutions the Co 4 " 1 " is pink. The 
carbonate, chromate, cyanide, oxalate, phosphate, 
hydroxide, and sulfide are common slightly soluble 
compounds of cobalt. The black CoS, with K. p of 
approximately 10~ 21 , is precipitated almost com- 
pletely from slightly basic solutions by saturation 
with H 2 S. The freshly precipitated sulfide is 
readily soluble in dilute hydrochloric acid, but 
after standing for a while its rate of solution in a 
buffered acid solution of pH 2 is very slow. Dilute 
nitric acid will dissolve CoS readily. The hydrox- 
ide of Co 4 "*" is blue, is slightly soluble in excess 
NaOH solution, and is readily soluble in excess 
ammonium hydroxide because of the formation of 
the complex ammonia ion of Co +4> . Cobaltous ion 
is sometimes oxidized in the presence of a com- 
plexing agent and forms complexes such as 

Co(NO 2 )e , cobaltinitrite ion, or Co(NH 3 )fl+++, 

where the cobaltic ion is the central ion. All cobalt 
salts when melted with borax in an oxidizing flame 

form a blue bead on cooling. 

Ni++. The chemistry of the nickelous ion is very 
similar to that of Co 4 " 4 * in many respects. The sul- 
fide is black, has a K. p of approximately 10- tl , and 
is precipitated completely by HaS only in slightly 
basic solutions. Like ZnS and CoS, NiS is dissolved 
slowly by a dilute acid solution. The hydroxide is 
not amphoteric, but dissolves in excess ammonium 
hydroxide because of the formation of the complex 
ion Ni(NHa) 6 + + . A very specific reagent for the 
detection of Ni++ is the organic substance di- 
methylglyoxime, which forms a red precipitate 
with nickel in a solution alkaline with NH 4 OH. 

In the alkaline earth group the strontium ion, 
Sr + +, was omitted. A study of the solubility table 
in Experiment 43 will reveal that, in general, salts 
of this ion have solubilities between those of the 
Ca++ and Ba 4 * 4 * ions. Note that the carbonates 
have about the same solubility. The chromate of 
Ba++ is much less soluble than that of Ca 4 * 4 * or 
Sr* 4 *. After removal of most of the fla 4 * 4 * as 
chromate, the Sr++ may be selectively precipitated 
as the sulfate without affecting the Ca++, Stron- 
tium salts impart a bright red color to the flame. 

Optional Exercises on the Qualitative Analysis of Cd++, BI+++, 

S Co++ f ami Sr++ 

2. The Ammonium Sulfide Group Zn++, Ni++, 
Co 4 - 4 -, Mn++, Fe++, Fe+++, Cr+++, AI+++. 

(a) Write net ionic equations for each of the re- 
actions of Co 4 * 4 * and Ni+ + with the reagents given 
in the discussion of the properties of these ions. 

(b) Expand the schematic outline of the analysis 
of the ammonium sulfide group, Experiment 42, 
to include all the ions listed above. 

(c) Carry out analyses of unknown solutions of 
these ions, as directed by your instructor. 

3. The Alkaline Earth and Alkali Groups 

(a) Write net ionic equations for the reactions 
of Sr 4 ** with typical reagents used in this group, 
and distinguish between the relative solubilities of 
the various salts. 

(b) Expand the schematic outline of the analysis 
of the alkaline earth and alkali groups, Experiment 
43, to include all the ions listed above. 

(c) Carry out analyses of unknown solutions of 
these ions, as directed by your instructor. 

After careful consideration of the discussion of 
the properties of the additional cations, outline, 
on a separate report sheet, the chemistry and the 
analytical scheme for each of the groups specified, 
and include the cations listed above. As far as 
possible, use the same reagents specified in the 
group analyses. When solutions or precipitates are 
obtained which contain only one or possibly two 
ions, select an appropriate reagent to make a 
positive identification for each ion, 

1. The Hydrogen Sulfide Group Cu++, Cd++, 
JJt+++, P6++, Hg++, Sn++ 9 Sb+++, AsO, . 

(a) Write net ionic equations for each of the 

reactions of Cd++, Bi+++, Sb 4 -* 4 , and AsO, 

with the reagents given in the discussion of the 
properties of these ions. 

(b) Expand the schematic outline of the analysis 
of the hydrogen sulfide group, Experiment 41, to 
include all the ions listed above. 

(c) Carry out analyses of unknown solutions of 
these ions, as directed by your instructor. 

The Analysis of General Unknown 
Inorganic Substances. The Solution of Solids. 


Review of Fundamental Concepts 

Previous $amples which the student has analyzed 
for the several groups as studied in Experiments 
40 to 43 all have consisted of solutions of soluble 
substances, and have required no other preliminary 
treatment than a suitable adjustment of the pR. 
If one is given, for analysis, a solid which does not 
dissolve in water, he must first get it into solution 
before applying the regular analytical procedures. 
Let us review briefly several principles which fre- 
quently are involved in the solution of insoluble 

Solvent Action by Hydrogen Ions 

Dilute acids, as hydrochloric acid or nitric acid, 
will dissolve many metallic oxides and hydroxides, 
as zinc oxide and calcium hydroxide, and also the 
salts of weak acids, as magnesium carbonate and 
barium phosphate. These substances dissolve be- 
cause of the formation of slightly ionized sub- 
stances and the consequent displacement of the 
saturated solution equilibria involved. Hydrogen 
ion, which is a mild oxidizing agent, will dissolve 
many of the more active metals, as magnesium, 
and zinc. Dilute nitric acid is not so satisfactory for 
the solution of some metals, however, as aluminum, 
iron, and tin. 

Solvent Action by Oxidizing Agents 

The most commonly used oxidizing agent is 
nitric acid. When dilute, it usually is reduced to 
nitric oxide (NO), but sometimes to nitrous oxide 
(N2O), nitrogen, or even ammonia. When concen- 
trated, the reduction product is generally nitrogen 
dioxide (NOj). It is an effective solvent for most 
sulfides (see p. 299) and for most of the metals. 
Nitric acid reacts with tin, or alloys containing 
tin, but leaves the tin as an insoluble hydrated 
oxide, called beta-stannic acid. This latter sub- 
stance is difficult to dissolve either in concentrated 
hydrochloric acid or in strong sodium hydroxide 
solution, in spite of its amphoteric character. With 
the very noble metals, as gold or platinum, nitric 

acid fails to react. 

The free halogens, as chlorine water or bromine 
water, may be used to oxidize the insoluble mer- 
curous chloride (HgaCU) to a higher oxidation 
state to form soluble mercuric chloride (HgCla). 

Solvent Action by Reducing Agents 

In a few cases, reducing agents may be used to 
transform a substance into a soluble form. Thus 
manganese dioxide, which is unaffected by nitric 
acid, is readily attacked by concentrated hydro- 
chloric acid or by hydrogen peroxide, which re- 
duces it to manganous ion. (Remember this when 
you want to remove the brown manganese dioxide 
stain from a vessel which has been used with 
potassium permanganate.) 

Silver can be recovered from scrap silver chloride 
residues by first mixing such residues with 12 F 
hydrochloric acid, which favors the formation of 
silver chloride complex ion (AgCl 2 ~), and then re- 
ducing this with metallic zinc. The metallic silver 
then can be separated and dissolved with nitric 

Solvent Action by Complex Ions 

The great stability of many complex ions can 
be utilized to effect solution in a number of cases. 
The easiest way to dissolve silver chloride is to 
treat it with ammonium hydroxide solution, form- 
ing the stable ammonia complex ion, Ag(NH*)% + . 

Concentrated hydrochloric acid, in a number of 
cases, owes its effectiveness in part to the forma- 
tion of a complex chloride ion. Tin may be dis* 
solved in warm hydrochloric acid, partially due to 
the displacement of the equilibrium by the stan- 
nous chloride complex ion, SnCU , which is 
formed. (If a little oxidizing agent is present also, 
the stannic chloride ion, SnCU""', will be formed, 
and promote the solvent action still more.) Like- 
wise, a hematite ore (Pc2Os), which strangely is 
difficultly soluble in dilute nitric acid, dissolves 
readily in hydrochloric acid because of the forma- 




tion of the ferric chloride complex, FeCU"". 

Oxcdate ion (CaO 4 ) is often used to control the 
solubility of substances in qualitative and quan- 
titative separations. Oxalic acid is effective in re- 
moving rust stains and ink spots formed by the 
iron-tannin inks, because it renders ferric com- 
pounds soluble by reducing them and forming the 
complex ion, Fe(CsO 4 ) 2 

Aqua regia, the frequently used mixture of nitric 
and hydrochloric acids, owes its effectiveness to a 
combination of the oxidizing action of nitric acid, 
and the complex ion forming action of the chloride 
ion. Mercuric sulfide, gold, and platinum are at- 
tacked readily by it. (See also p. 299.) 

Cyanide ion (CN~), while not used very much 
in elementary qualitative analysis because of its 
poisonous character, forms very stable complexes 
with a number of the metal ions. Coupled with the 
oxidizing action of the air itself, it is very effective 
in the commercial extraction of gold and silver 
from their ores. The equation is 

4 Ag + 8 CN- + 2 + 2 H 2 *4 Ag(CN) 2 - + 4OH~. 

Solvent Action by Metathesis 

While insoluble salts of weak acids are generally 
easily dissolved by adding dilute nitric acid, in- 
soluble salts of strong acids, as barium sulfate, 
present a more difficult problem. Since most metal- 
lic carbonates are insoluble, it is possible, by boil- 
ing the insoluble salt with a strong sodium car- 
bonate solution, to obtain at least a partial re- 
placement of the anion of the insoluble salt by 
carbonate ion in many instances. This type of 
action is called a metathesis. The resulting car- 
bonate salt then can be dissolved by dilute acid. 

The extent of the metathesis will depend on the 
relative solubility products of the original insoluble 
salt and of the corresponding metal carbonate. For 
example, if an excess of barium sulfate solid is 
boiled with 50 ml of 1 F Na2COa solution, we can 
calculate the amount of barium sulfate which will 
dissolve, as follows. 1 The reaction is 

CO, + BaS0 4 + BaCO, + S0 4 ~~ , 
for which the equilibrium constant expression is 

Both solid BaSO 4 and solid BaCO 8 will be present 
in the equilibrium mixture, so we can obtain this 
ratio, of (SO 4 ) to (CO 3 ), by calculating the 
ratio of their solubility products, 2 

(S0 4 ) 

(C0 8 

1 X 10~ 
5 X 10- 

U ' U ' 

If the concentration of SO 4 formed by the re- 
action above is x, the concentration of COa re- 
maining at equilibrium from & I F Na2CO 8 solu- 
tion will be 1 x, so that 

0.02, or x - about 0.02 M S0 4 ~. 

1 Such calculations are only approximate, since the activity of the 
ions may be considerably affected at such high concentrations. See 
also page 279 on the concept of activity. 

(S0 4 ) _^ x 
(C0 8 ) 1 - x 

In a volume of 50 ml, this amounts to 0.001 formula 
weight or about 0.23 gram of BaSO 4 dissolved. 
Using 100 ml of 2 F Na 2 CO 3 , 0.9 gram of BaSO 4 
would dissolve. Thus, even though barium car- 
bonate is slightly more soluble than barium sulfate, 
it is possible to metathesize the latter, and then 
after filtration, dissolve the product in acid. 

Solvent Action by Fusion Processes 

Where more powerful treatment is needed to 
effect solution, the chemist resorts to a fusion, dis- 
solving the substance to be analyzed in a suitable 
molten flux. The most widely used alkaline flux is 
sodium carbonate. This is able to disintegrate the 
silicate compounds which are present in most rocks, 
minerals, and ores, and which resist the othei 
methods previously discussed. They undergo s 
fusion metathesis, so that on cooling and adding 
water, sodium silicate which is soluble, and the 
insoluble metal carbonate or oxide, are formed. Foi 
example, if the mineral, talc, is powdered, fused 
with sodium carbonate, then cooled and watei 
added, we may write as a typical equation, 

Mg,Si 4 Oio(OH), + 4 Na s CO, * 

3 MgCO, + 8 Na+ + 4 SiO,~ + H 2 + C0 a 

The addition of acid to this mixture will dissolve 
the magnesium carbonate and precipitate silicic 

MgCOs + 2 H+ >- Mg++ + H 2 + CO* 

SiO 8 + 2 H+ > H 2 SiO. 

With an acid flux, as potassium bisulf ate KHS0 4 
or better, the pyrosulfate K2S2O7, substances of * 
basic character, such as a hematite ore, Fe 2 O 8 , oftei 

1 See also the discussion of fractional crystallization. Example C 
page 281. 



containing silica also, would be attacked. 

If oxidizable substances are present in the solid 
to be fused, the addition of an oxidizing agent, as 

Experimental Procedure 

sodium peroxide (Na*Os), or potassium nitrate 
(KNOa), is an aid to solution. 

Chemicals: All reagents for Experiments 40 to 43. If the 
solution of solids is to be considered: Brj, KNOj, KiSjO? or 
KHSO 4 (fused), Na a CO 3 solid, 1 F Na s CO 8 , Na 2 0*, Na 2 B 4 O 7 , 
Zn (mossy). 

Special supplies: For fusion only, nickel crucible, Fisher or 
Meker burner, and flower pot shield. 

1. Typical Solution Reactions. 

Instead of performing a series of prescribed ex- 
periments on the solution of insoluble substances, 
the student is invited to try out, on his own initia- 
tive, such procedures as are suggested to him by 
the questions in the application of principles sec- 
tion of the report sheet. 

2. General Unknown Analyses. 

The student will obtain a general unknown solu- 
tion containing any of the ions of the limited 
scheme as outlined in Experiments 40-43. Follow 
the procedures in these experiments, step by step. 
Start with a single 4-ml sample, and use the filtrate 
from each group precipitant for the next succeed- 
ing group procedure. Report the analysis as you 
have done previously, as suggested in the report 

Analyze additional unknowns as the instructor 
directs, and as the remaining time of the laboratory 
course permits. These may be additional unknown 
solutions, or unknown solid samples. The latter 
may be metal alloys, or nonmetallic mixtures, ores, 
minerals, etc., dissolved according to the directions 
in paragraph 4, following. (They should not in- 
clude metals not provided for in the scheme of this 
manual, as these might interfere in the analysis, 
unless removed.) In the case of nonmetallic solids, 
the student may also analyze for the negative ions 
as outlined in Experiment 39, correlating these 
with the solubility of the substance. For example, 
if the solid is soluble in nitric acid and silver ion 
is found, it is not necessary to test for chloride ion. 
Also do not test such a solution for nitrate ion, 
which should be tested for in the original water 
extract of the solid, since all nitrates are soluble. 

3. Preliminary Examination of a Solid. 

A preliminary examination sometimes gives val- 
uable clues as to the nature of a solid. It is wise 
to verify these through the subsequent analysis of 

the solution before reporting them, however. Note 
whether the sample appears homogenous, or is 
made up of several substances. Observe any char* 
acteristic crystalline form, colors, or odors. Flame 
tests may be made, using a nichrome wire. Borax 
bead tests may be made. For these, touch the 
heated end of a platinum wire (the "lead" of a 
mechanical pencil is a satisfactory substitute) to 
some borax, and reheat to form the colorless bead. 
Then touch to a fragment of the unknown, and 
again fuse it. Consult one of the chemistry hand- 
books for details as to the coloration of the bead 
with various metals, and also as to the flame colora- 

Organic matter, if present, interferes with many 
qualitative tests. It probably will not be present 
in samples given for elementary analyses. It may 
be tested for by heating a bit of the solid in a 
closed glass tube, or 10-cm test tube. Charring 
indicates organic matter. A drop of concentrated 
H^SO* is sometimes added before warming, to give 
a more sensitive test. Some organic substances do 
not char by heat alone. If found, organic matter 
may be removed by oxidation with H^SOi and 
HNOa. Consult a general qualitative text for de- 

4. The Solution of a Solid Unknown. 

(a) The Sample is a Metal or Alloy. Most metals 
are attacked by warm 6 F HNOa. Some, as Fe, and 
the Al and Mg light alloys, dissolve more readily 
in warm 12 F HC1. Sometimes a drop or two of 
liquid Br2 (obtain from the instructor) added to 
the HC1 assists the solution. In a few instances, 
aqua regia may be necessary. Try a very small bit 
of your unknown to determine the best solvent to 
use, then dissolve about 0.2 to 0.5 g of the metal, 
warming it in 3-8 ml of the acid chosen. (Use less if 
concentrated.) If necessary, add more acid, but 
avoid a large excess. When the metal is completely 
disintegrated, evaporate just to dryness, add sev- 
eral drops of acid, then 10 ml of H 2 O, If a precipi- 
tate forms by hydrolysis, add a little more acid to 
dissolve it. If solution is complete, add HiO to 
make about 20 ml volume for each 0.1 g of un- 
known metal alloy taken, then use about 5 ml of 



this for the analytical procedure. 

A white residue remaining after solution in aqua 
regia, which dissolves in NH^OH or hot water, may 
be tested for Pb +**" and Ag+ by the Group 1 pro- 
cedure, and the filtrate tested beginning with 
Group 2. 

If a white residue remains from the HN0 8 treat- 
ment, which cannot be dissolved, it may be hy- 
drated Sn0 2 . Separate this by centrifuge from the 
solution, and treat it with 1 ml of 6 F NaOH and 
2 ml of 2 F NaHS, warm and agitate it for 10 
minutes. Dilute with 3 ml of H 2 and just acidify 
with 6 F HNO 8 . Centrifuge the precipitated sulfides 
and discard the filtrate. Dissolve the residue by 
wanning it with 2 ml of 12 F HC1, 1 evaporate the 

A flower pot 
helps to con- 
the heat. i 

( Broken to 
show details) 


Meker or 
Ffoher burner 

FIG. 44-1. The disintegration and solution of a silicate by 
fusion with sodium carbonate. 

solution to 1 ml and combine with the main filtrate, 
(b) The Sample is a Non-Metallic Solid. Try its 
solubility in water, and note the effect of this solu- 
tion on litmus. If insoluble, make preliminary tests 
using a very small amount of the solid with 2-ml 
portions of various solvents, to determine the best 
one to use. Try them in this order: 6 F HN0 8 , 16 F 
HNO*, 12 F HC1, and aqua regia (i. e> add ^ its 
volume of 16 F HN0 8 to the 12 F HC1 trial). Dis- 
solve about 0.5 g of the solid mixture 2 with 2-5 ml 

1 Part of this solution may be tested now for tin, by following the 
procedure as given for Sol. 24, 'Test for Tin," page 302. 

* If the sample is quite non-homogenous, a larger aftrtunt may be 
advisable to secure a -representative sample. 

of the reagent chosen (use less if concentrated). 
Warm and add more reagent if needed, but avoid a 
large excess. Finally! evaporate almost to dryness 
to remove excess acid, and dilute with water to 
give a total volume of about 10 ml for each 0.1 g 
of sample taken. Use about 5 ml of this for the 
analytical procedure. 

If a residue still remains undissolved from the 
treatment with acid by the above procedures, 8 it 
may be subjected to a sodium carbonate fusion. 
Dry the residue by warming, place it in a small 
nickel crucible, and mix with 3-4 times its bulk 
of anhydrous Na 2 C0 8 , and an equal amount of 
K 2 C<X 4 Heat with a Fisher burner at its maximum 
temperature, until all is molten. (See Fig. 44-1.) 
If undissolved solid remains, add a very small 
amount of NaNOa (or Na2C>2), and reheat. Cool, 
and place the crucible on its side in a small beaker. 
Add 15-20 ml of distilled water, and boil gently 
to disintegrate the mass. (The crucible may be 
left soaking for some time, or over night.) Decant, 
filter or centrifuge the mixture. To the residue, add 
2 ml of 6 F HNOj to dissolve it, warming if needed. 

Likewise, the solution from the carbonate melt 
is made just acid with 6 F HNOs, evaporated just 
to dryness to dehydrate any silicic acid, several 
drops of HN0 8 added, then 5 ml of H 2 O to dis- 
solve it. This would contain any negative ions from 
the insoluble substance, which could be tested for 
here, if desired. This solution probably would not 
contain metal ions, unless present as part of the 
negative ion, such as CrO* . It may be tested 
separately, or combined with the fusion residue 
which was dissolved in HN0 3 . This, in turn, may 
be analyzed separately or combined with the main 
filtrate from the acid treatment. This latter should 
first be tested for Na+ and K+, since they have 
been added by the fusion. A sodium carbonate 
fusion in a nickel crucible will probably add traces 
of this element, which will precipitate along with 
the zinc sulfide in our scheme. It should not inter- 
fere with these tests* 

1 Substances which may remain undissolved by the previous acid 
treatment include: sulfates of lead, barium, and calcium, halides of 
silver, some ignited sulfates or oxides or native ores of iron, chromium 
and tin, and most silicates. 

4 A mixture of two salts, such as this, melts at a lower temperature 
than either one singly. 

REPORT: Exp. 44 

Analysis of a General Unknown- 
Solution of Solids 



Locker Number. 

L. Application of Principles 

What reagent (s) and procedures would you choose to bring about the solution of each of the following sub- 
stances or mixtures? Indicate by number the type(s) of solvent action: (1) H+ to neutralize or form a weak acid, 
(2) H + to oxidize, (3) Oxidation in general, (4) Reduction, (5) Complex ion formation, (6) Metathesis, (7) Fusion. 
Write equations for any important reactions in each case. 


Reagent(s) and Procedures 



Dolomite rock, 
CaCO 8 and 

Brass, an 
alloy of Cu 
and Zn 

Stainless steel, 
Fe and Cr 

Apatite, a 

Ca 8 (P0 4 )2 

CaS0 4 '2H 2 O 

Pyrolusite ore, 
MnO 2 

Magnal alloy, 
Al and Mg 

H g2 CU 


Chromite ore, 

Fe(Cr0 2 ) a 

Fuse with sodium peroxide. 

Gold metal, 


2. Problems 

Would metathesis be a satisfactory method of bringing about the solution of each of the following substances? 
Calculate the weight of salt which could be dissolved by boiling an excess of the solid with 50 ml of 3 F Na 2 CO 
solution, in each case. 

(a) PbS0 4 

_g PbS0 4 

(b) AgCl (You will have to square the solubility product of AgCl in this case. Why?) 


3* Analysis of General Unknown Substances 


Type and Sample No. Ions found 

Type and Sample No Ions found__ 

Type and Sample No Ions found 

(For type indicate as solution, alloy, or non-metallic solid) 

For each unknown, on a separate sheet, describe any preliminary tests performed, the method of solution if 
a solid, and make a chart of each step of the analysis, showing sample, reagent, observations, and conclusions, as 
in previous experiments. 


Appendix I 

The Measurement of Physical Quantities 

A. The Use of Dimensions 

The measurement of any physical quantity rep- 
resents two factors, the number itself, and the unit, 
or units, in which the measurement is made. These 
units are known as the dimensions of the measure- 
ment. For example, a piece of silver is measured 
and the dimensions reported as 3.00 wide and 5.00 
long. This gives little information even the shape 
is uncertain, for we do not know that both quan- 
tities are expressed in the same units. But if the 
measurement is reported as 3.00 cm wide by 5.00 
cm long, adequate information is given to describe 
both the size and the shape. 

In the solution of all problems dealing with 
physical quantities, the units must be given. These 
may be subjected to the usual mathematical opera- 
tions of multiplication and division, the same as 
any other quantity. Thus, the surface area of one 
side of the above piece of silver is obtained by 
multiplying the units, as well as the numbers, 

3.00 cm X 5.00 era 15.0 cm 2 , or square centimeters. 

Furthermore, if the thickness is given as 2.00 mm 
and we wish to compute the volume, the dimen- 
sions first must be converted to the same units, 
i.e. 2.00 mm = 0.200 cm. The volume is then cal- 
culated by multiplying both the numbers and the 

15.0 cm 2 X 0.200 cm = 3.00 cm 8 , or cubic centimeters. 

Again, if the weight of this piece of silver is 
found to be 31.5 grams, we may calculate the 
density of silver in accordance with the defining 
equation for density 




31.5 g 
3.00 cm 3 



The units, or dimensions, of density are g/cm 8 , 
read grams per cubic centimeter. Since these are 
unlike dimensions (weight and volume), they can- 
not be reduced further, and are left simply with 
the indicated division. 

Units frequently may be cancelled in processes 
of division: Example 1. Compare the relative 
weights of equal volumes of the heaviest metal, 

osmium (density 22.48 g/cm 8 ), and of the lightest 
metal, lithium (density 0.53 g/cm 8 ). 

Density of osmium 22.48 g/cm 8 _ 
Density of lithium 0.53 g/cm 8 

The units cancel out, and the result, 42, a ratio of 
two densities, is a pure number, without dimensions. 

Example 2. If the density of silver is 10.5 g/cm* 
what will a block of silver weigh whose measure- 
ments are 4.00 cm, 5.00 cm, and 10.0 cm? 

The volume is: 4.00 cm X 5.00 cm X 10.0 cm 
200 cm 8 . From the formula for density, D M/V, 
we have, by transposing, 

M = D X V 

10.5 -&-, X 200 cm 8 
cm 8 

2100 g. 

By cancelling the cm 3 in both numerator and de- 
nominator, the dimension is given correctly in 

Students who have trouble in deciding whether 
to multiply or to divide in the solution of a given 
problem, will be helped by a consideration of di- 
mensions. This is illustrated by the following 
Example: The density of gasoline is 7.0 pounds per 

gallon (7.0 p). How many gallons would it take 


to weigh 100 pounds? 

If we multiply, the dimensions would be, Ib X 

_. _. which is obviously incorrect, 
gal gal J 

If we divide, the dimensions would be, Ib -*- 

gal . . 

X ~rr~ = gal. Since volume is required, 

= Ib 

this is obviously the correct procedure. Therefore, 
100 lb-4-7.0 -~j 100 Ib X y^ 85 14-3 gallons of gasoline. 

B. Significant Figures 

When the scientist records or publishes experi- 
mental data, he indicates the degree of certainty 
of each item of data by expressing the numerical 
value to the proper number of figures which are 
justifiable, or significant, according to the accuracy 
of the equipment, method, and observation. 

Definition: The number of significant figures in 


a quantity is the number of trustworthy figures in 
it, the last "trustworthy" figure, however, being 
somewhat doubtful. Thus, in writing the value 492 
ml, we mean that the quantity is definitely known 
to the nearest 10 ml, but not necessarily to the 
nearest 1 ml. If we wish to give a more precise 
meaning to the last significant figure, we may 
write the value as 492 2 ml, or 492 4 ml, de- 
pending on whether the true value is known to lie 
between 490 and 494 ml, or between 488 and 496 
ml. In general, the last significant figure in a value 
is to be regarded as somewhat uncertain. The num- 
ber of significant figures in 0.49 is two, in 6.96 is 
three, in 20.5 is three, in 491736 is six, and so on. 

When a zero is part of a number, it is a sig- 
nificant figure if it occurs between any two digits 
other than zero, or to the right of the number be- 
yond the decimal point. It is not a significant figure 
if it occurs on either side of the number merely as 
a means of fixing the decimal point. For example, 
the zero in 205.1 is significant, likewise in 2.50; 
but the zeros in 0.025 or in 22,400 are not sig- 
nificant figures. If we measure a length as 5.2 
meters, we might likewise write this using different 
units as 520 centimeters, 5200 millimeters, or even 
as 0.0052 kilometers. We have not changed the 
accuracy of the expression by the units we use. In 
each case, the values are expressed to two sig- 
nificant figures to the nearest 10 centimeters in 
this measurement. 

Zeros to the right of a number, which are needed 
to fix the decimal point, in general merely repre- 
sent unknown digits. Because of the ambiguity as 
to the number of significant figures in such cases, 
scientists commonly resort to a notation using ex- 
ponential powers of ten. Suppose a student meas- 
ures the volume of thirty-two grams of oxygen as 
22,400 ml. If the reliability of the measurement 
justifies only two significant figures, we write it 
2.2 X 10 4 ml; if three significant figures, 2.24 X 10 4 
ml; or if five significant figures, 2.2400 X 10 4 ml. 
Unless there is some particular reason to do other- 
wise, one digit is usually retained to the left of the 
decimal point. 

While a rigorous treatment of the number of 
figures which are significant when one performs 
various mathematical operations is a rather com- 
plicated problem, the following approximate rules 
will suffice for our purposes. 

In addition and subtraction no more figures 
should be used than can be trusted in the quantity 

having the fewest trustworthy figures. 

Examples: 401 . 1 cm 

10165.605 = 10166cm. 

307.4 g 

301.2 g. 

In the addition, the sum should be expressed 
only to the nearest centimeter, since 9763 is ex- 
pressed only to the nearest centimeter; also in the 
subtraction the difference should be expressed only 
to the nearest tenth of a gram. In the actual opera- 
tion of adding or subtracting, do not bother to add 
or subtract columns which obviously cannot affect 
the final result. In rounding off a value to the de- 
sired number of significant figures, increase the 
last figure retained by one if the first discarded 
figure is 5 or greater. 

In a subtraction, the number of significant fig- 
ures is not necessarily the same in the answer as 
in the values subtracted. 

Example: 29.744 g (wt. of dish -f silver) 

29.232 (wt. of dish) 
0.512 g (wt. of silver). 

Note that five significant figures in the weighings 
are needed to obtain a weight of silver which is 
certain to three significant figures. 

In multiplication and division, the same 
number of significant figures should be retained in 
each of the factors. It is often advantageous, how- 
ever, to carry one, but never more than one, extra 
figure in the factor where such a figure is available. 
The product or quotient should be expressed to the 
same number of significant figures that are con- 
tained in the factor having the least number of 
significant figures, except in certain cases where 
this would give a lower percent of certainty than 
in the factors involved, in which case one is justi- 
fied in carrying one additional figure. 


(1) Multiplication: What is the weight in grams 
of 2.75 gram atoms of silver, whose atomic weight 
is 107.88? 

2.75 g at X 108 ^~ - 297.00 g - 297 g (answer). 


The atomic weight of silver is rounded off to 108, 
since 2.75 has three significant figures, and since 
108 differs from 107.88 by only about 1 part in 
1000. The answer is known to only three figures, 
hence it is incorrect to write it as 297.00 g. 

(2) Division: How many moles in a liter of 


glycerine (weight 1266 grams), if the molecular 
weight of glycerine is given as 92? 

No. moles - 1255 g + 92 -&- - 13.6 moles. 

Since the molecular weight of glycerine was ex- 
pressed to two significant figures, one might expect 
to express the answer as 14 moles. To do so, how- 
ever, would express the answer to an accuracy of 
only 1 part in 14, while the molecular weight was 
expressed to an accuracy of 1 part in 92, hence one 
should carry three significant figures, 13.6 moles, 
as the answer. 

Resume: The proper use of significant figures in- 
volves nothing more than the use of good common 
sense in mathematical computations. One of the 
main purposes of stressing the subject in this 
manual is to eliminate the tendency of students to 
carry out their calculations to numerous decimal 
points, blissfully ignorant of the fact that ex- 
cessive places are utterly meaningless. 

C. Experimental Errors 

Since it is impossible, in practice, to make an 
absolutely exact measurement, the scientist meas- 
ures as closely as his method, equipment, and 
powers of observation permit. In recording his 
measurements, he expresses the data to the appro- 
priate number of significant figures to indicate 
their respective degrees of certainty. Errors of 
method are those inherent in the method, irre- 
spective of the equipment or observational powers 
employed. This is the most difficult type of error 
to detect. Errors of equipment are those which are 
inherent in the construction and calibration of the 
equipment. Errors of observation are those in- 
herent in the average experimenter's powers of 
perception, and should, of course, be distinguished 
from blunders. 

In the following chart, each value under "Pre- 
cision of Measurement" represents the sum of the 
average observational error and the error involved 
in the particular type of apparatus, as used in this 
course. For example: If the bottom of the meniscus 
of a liquid contained in a 50-ml graduated cylinder 
is somewhere between the 45.0 ml and the 46.0 
ml graduations, you should read this to the nearest 
estimated 0.1 ml, as 45.7 ml. You would recognize, 
however, that this reading is uncertain by 0.2 
ml. This uncertainty could be recorded in the 
reading as 45,7 0.2 ml. The structure of the 
apparatus, the properties of the liquid, and the 

physical abilities of the eye prevent any more pre- 
cise estimation of the volume. This chart should 
be of value in locating sources of principal errors 
in your experiments, and in indicating the degree 
of certainty of your data. 


Precision of Mearuremtnt 

Mercury barometer 


Analytical, or pulp, balance 

0.001 g 

Triple-beam balance 

0.02 g 

Platform balance 

0.5 g 

50-ml burette 

0.02 ml 

100-ml gas-measuring tube 


10-inl graduated pipette 


500-ml graduated cylinder 


50-ml graduated cylinder 


10-ml graduated cylinder 


110C thermometer 


In the experimental work performed in this 
course, the inaccuracy of the results, including 
errors of method, equipment, and observation, 
should not exceed 3 to 5 percent of the accepted 
values, and in most cases will be more accurate. 
If the deviation of the experimental results from 
the true value exceeds 5 percent, a blunder usually 
is indicated. The inaccuracy of a result usually 
is expressed as percent error and is obtained by 
dividing the difference between the accepted value 
and the experimental value by the accepted value, 
and multiplying the quotient by 100. Thus, if the 
experimental result for the molal volume of oxygen 
were 22.1 liters, and the accepted value were 22.4 
liters, the percent error of the result would be' 


Accepted value Experimental value 

* - T - 7~3 : - *- x 

Accepted value 


The percent error is always expressed as a positive 
number, whether the experimental value is larger 
or smaller than the accepted value. 

It is to be noted that the percent error is a 
measure of the accuracy of the result, not of the 
precision of the measurements. Precision is a mea- 
sure of how closely measurements which are re- 


peated check with one another, while accuracy is 
a measure of how closely the final result checks 
with the true or accepted value. You should strive 
toward both precision and accuracy, as the former 
leads to the latter, and accuracy is the aim of the 
competent experimenter. 

It is often useful to calculate the percent of un- 
certainty of a measurement, which provides a meas- 
ure of the certainty of the measurement, exclusive 
of any errors of method. This is calculated by 
dividing the uncertainty of the measurement by 
the measured value and multiplying the quotient 
by 100. Thus, if a piece of metal is weighed on the 

analytical balance and its weight recorded as 
grams, since the uncertainty of the measurement 
is 0.001 g (see preceding table), the percent of un- 
certainty is: 

% uncertainty 


X 100- 0.04%. 

This means that the experimental value is known 
to lie within the range of 0.04% greater and 0,04% 
less than the true value. To calculate the percent 
of uncertainty of a final result when several meas- 
urements are involved, the sum of the percent 
uncertainties of the several measurements is taken. 

Drill Problems 

(You may refer to the answers to some of these problems, on P. 341, after solving 

them on your own initiative.) 

1. Carry out the operations on the data given 
in each of the following cases to calculate the 
quantity called for. Show your method, including 
the dimensions of measurement. (These units will 
tell you which mathematical operation to per- 

(a) Velocity = 50 mi/hr, time 0.5 hr, distance = ? 

(b) Velocity 186,000 mi/sec, distance = 93,000,000 mi, 
time ? 

(c) Time 9.3 sec, distance 100 yd, velocity = ? 

(d) Density Al 2.70 g/cm 3 , weight - 2700 g, volume ? 
If this were shaped as a cube, length of one edge ? 

(e) Weight Hg = 272 g, volume = 20 cm 8 , density - ? 

(f) Weight apples 240 Ib, amount in each box 40 
Ib/box, number of boxes ? 

(g) Weight H 2 O - 180 g, molecular weight = 18 g/mole, 
number of moles = ? 

2. How many significant figures in each of the 
following numbers. 

(a) 3005 

(b) 3500 

(c) 0.035 

(d) 0.350 

(e) 3.050 

(f) 3.0005 

3. Carry out the following operations, recording 
the answer correctly in accordance with the rules 
of significant figures: 

(a) Subtract 5.1 from 28.347 

(b) Subtract 5.10 from 28.347 

(c) Multiply 0.020 by 1.111 

(d) Divide 36.02 by (3.0) 

4. A beaker of water is weighed as 1200 grams. 
How would you write this number so as to avoid 
ambiguity, if the weight is known to the nearest 
(a) ten grams, (b) one hundred grams, (c) gram, 
(c) tenth of a gram. 

5. The length of a table is measured as 2 meters, 
3 centimeters, and 4 millimeters. Express this 
length as (a) meters, (b) centimeters, (c) milli- 
meters, (d) kilometers. How many significant 
figures in each case ? 

6. A series of beakers are weighed as follows: 
125.2 g, 90.3 g, 56.2 g, and 20.237 g. How should 
you record the sum of these weights so as to avoid 
any incorrect conclusions as to the precision of 

7. Three determinations of the percent of chlo- 
rine in sodium chloride were: 60.1%, 60.5%, and 
60.3%, averaging 60.3%, The accepted value, 
based on the atomic weights (Na 22.997, Cl 35.457), 
is 60.658% Cl. What is the percentage error in the 
analysis, and to how many significant figures should 
it be expressed ? 

8. What is the percent of uncertainty in measur- 
ing 50 ml of water in (a) a 50-ml graduated cylin- 
der, (b) a 500-ml graduated cylinder? (Note: As- 
sume the precision of measurement given in the 
table in Appendix I, Sec. C.) 


Appendix II 

Tables of Data 

co co co to <o totoiototo 

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reococo cocococoro cococoooco 

cococoooco cocooocooo cocoeocooo cooooooooo coco rococo 




- I" 





T OO *T * ! C 

> (P CM 


-< OO < CT> CM 
LO CM O r** to 

4 r^> CM f^ i <o * 1 

> r- CM to oo en en c 



- uo LO co r-". oo o 
r" r*^ r^- r**. r^. F 

cJ CM co 53- 
oo oO oo oo oo 

oooooSoS ^ICMC 
r>* r^. r^ So oo ooooo 



m i 

r*- o> cy> c 

^ to vo c 

to <* r 

r^ oo c 


*> CD cr>cn 

o en c 

> Cf> C 

c> oo to co c 

CM en 

*-4 LO QO 1-4COOOC 

)-4CO I t 



cnoooooooo oo oo r>. r> r^. 

oocncncn cnoooooooo 

I^^H oocncncn cnoooooooo r^. r* r*. r*. r*. CD coco coco co COLO to to to to to to to 

j g^cnr>.co to ^ co CM e^ *+ -* o o 

-* o o en en cnooo 

.r^ rococo coco coco to to to to to to to toto-^- < 

co I^^S 

o oo r^. r^. r^. coco coco to toiot 

co co co co co co co oo co CM CM CM 

} oooo rococ 


- | 

r- i O CO CD C3 r^C 

to co r^> oo o * i c 

IO tO iO to CD CO C 

r^cn '~ 4 59tor^.OT r-i CM ?f to r^ oo C3 CM * tocor* 




CM P* ^- f* o CM <r <o r c 
f ** <* to to to to totototot 

10^4* i 4 LO LO co cnto <cnc 
r^S S?J$Sp 5233^1 

3 ^,-4*-H -<2\}cMCMCM CO CO CO CO C 



r-Hv-4 *-CM 


r ^r Jo to L 





<- ~< -H 

^5?sSc3 sss? s 

^- ^- tS tO tO to LO tO tO tO COOc 

Reproduced by j>rmw*i<m /rom "A LaborcUory Manual o/ CoUa C/Umwerv" by H. 0. Dewing, piMithed by John Wiley A Son*, /txu 




Fundamental Units 

Conversion Factors 


1 kilogram (kg) 1000 grams (g) 
1 gram = 1000 milligrams (mg) 


1 meter (m) 
1 centimeter 

100 centimeters (cm) 
10 millimeters (mm) 


1 liter 0) * 1000 milliliters (ml) 

1000 milliliters 1000.028 cubic centimeters (cm 8 ) 


1 pound = 454 grams 
1 kilogram - 2.20 pounds 


1 inch = 2.54 centimeters 
1 meter = 30.4 inches 
1 kilometer = 0.62 miles 


1 quart = 946 milliliters 
1 liter - 1.06 quarts 




1 Angstrom unit (A) = 10 ~~ 8 centimeter 
1 micron 10~ 8 millimeter 

1 cubic foot 28.3 liters 

= 7.48 gallons 

1 ounce (US liquid) = 29.6 milliliters 

1 gram 15.4 grains 
1 ounce (avoirdupois) 28.3 grams 
1 ounce (apothecary, troy) 31.1 grams 

1 atmosphere (atm) = pressure of a mercury column 760 mm or 29.92 inches high 
= 14.696 Ib per sq in 
- 1.0133 bars (dynes per cm 2 ) 
= 1033.3 grams per cm 2 

Absolute zero (0K) -273.18C 
Conversion formulas: K C -f 273 

F - |C + 32 

C - |(F - 32) 


1 ampere (unit of current) = a flow of one coulomb per second. 

1 volt (unit of potential) ~ the potential difference necessary to cause a current of one ampere through a resistance of 

one ohm. 
1 Faraday (electrochemical) 96|500 coulombs, the quantity of electricity which will deposit one gram equivalent of 

a substance at an electrode. 

Avogadro's number = 0.6023 X 10 84 molecules per mole 
The gas constant (R) in the perfect gas law equation (PV nRT) 82.0 ml atm per degree per mole 

1.9868 calories per degree per mole 
= 8.3145 joules per degree per mole 

1 calorie (cal) = the energy required to raise one gram of water one C (more precisely, from 14.5 to IS.S^C) 
1 British thermal unit (BTU) * tht energy required to raise one pound of water one F (more precisely, from 39 to 

1 British thermal unit 252 calories. 




Vapor Pressure 
(mm of mereury) 


Vapor Pressure 
(mm of mercury) 


Vapor Pressure 
(mm of mercury) 

-10 (ice) 

- 5 " 































































Subtract the corrections to allow for differences in the expansion of mercury and of the brass scale on 

the barometer, at various temperatures. 


Barometer Reading (mm) 














































































Percent Solute 

Acetic acid, glacial 
Acetic acid, dil 


17 F 

1.05 g/ml 


Hydrochloric acid, cone 
Hydrochloric acid, dil 






Nitric acid, cone 
Nitric acid, dil 





Sulf uric acid, cone 
Sulf uric acid, dil 





Ammonium hydroxide, cone 
Ammonium hydroxide, dil 

NH 4 OH 




Sodium hydroxide, dil 










B. P. 

Volatile acids: 

Carbonic acid 
Hydrogen sulfide 
Sulf urous acid 
Hydrochloric acid 20% 
Acetic acid 

very volatile 
very volatile 
very volatile 

Nitric acid 68% 
Hydrobromic acid 47% 
Hydroiodic acid 57% 


Practically non- volatile : 
Phosphoric acid 
Sulfuric acid 

-H 2 O213 


Note: Where percent composition is given, this is the 
composition of the constant boiling mixture. 

K ^ 

Very active with HgO or 




Oxides reduced by electro- 


lysis, but not by H 2 or CO. 



Active with acids, or with 
.steam when metal is hot. 

Zn 1 Oxides reduced by C or Al 
(^ r j (hot), but not by H a or CO. 



Less reactive with acids. < 



Oxides reduced by heating 


with H 2 or CO. 


Active only with stronger 


oxidizing acids, as HNOa, 


HtSO 4 . No H 2 formed. 



Oxides reduced to metal 

Active only with aqua regia. 


(decomposed) ,by heat alone. 





I/ 100 I H 2 

Very soluble thousands of vol- 

SO a 


umes of gas in 100 volumes of 

NH 3 


Henry' 's Law The solubility 
of a gas is proportional to the 

Fairly soluble hundreds of vol- 
umes of gas in 100 volumes of 


C1 2 
H 2 S 



partial pressure of the gas 
above the solvent. This law 
does not hold for very soluble 
gases at high concentration. 

Slightly soluble one to four vol- 

N 2 


umes of gas in 100 volumes of 

H 2 








C 2 H 3 2 ~ . . . 

S0 4 . . . . 
CO 3 , and PO 4 "" 



Na+, K+, NH 4 + 
Ag+ . . . 

All nitrates are soluble. 

All acetates are soluble, (AgC 2 H 3 O2 only moderately). 

All chlorides are soluble, except AgCl, Hg 2 Cl 2 , and PbQ 2 . (PbCl 2 is slightly soluble in cold water, 
moderately soluble in hot water.) 

All mljate* are soluble, except BaSO 4 and PbSO 4 . (OaSO 4 , Hg 2 SO 4 , and AgsSO 4 are slightly soluble; 
the corresponding bisulfates are more soluble.) 

All carbonates and phosphates are insoluble, except those of Na"*", K" 1 *, and NH 4 " t ". (Many acid phos- 
phates are soluble, as Mg(H 2 PO 4 ) 2 , and Ca(H2PO 4 ) 2 .) 

All hydroxides are insoluble, except NaOH, KOH, NH 4 OH, and Ba(OH) 2 , (Ca(OH) 2 is slightly 

All sulfides are insoluble, except those of Na+, K + , and NH 4 +, and those of the alkaline earths: 
Mg + +, Ca + +, and Ba++. (Sulfides of A1+++ and Cr+++ hydrolyze and precipitate the correspond- 
ing hydroxides.) 

All salts of sodium 9 potassium^ and ammonium are soluble, except several uncommon ones, as 
Na 4 Sb 2 7 , K 2 NaCo(N0 2 )6, CNH).NaCo(NO,)., KjPtCU, (NH 4 ) 2 PtCU. 

All silver salts are insoluble, except AgNO s and AgClO 4 . (AgC 2 H 3 O 2 and AgtSOi are only mod- 
erately soluble.) 







Strong electrolytes .... 




Almost all soluble salts 

Moderately weak electrolytes . 

H 2 C 2 4 

H,P0 4 
HN0 2 

Ca(OH) 2 (b) 

HC 2 H 8 2 

NH 4 OH 

Weak electrolytes .... 

H 2 CO, (a) 


Mg(OH) 2 (b) 

Insoluble hydroxides of 
the heavy metals: as 
Al(OH), and Fe(OH) 8 

Salts of certain 'metals, as HgCla 
and Pb(C 2 H 3 O 2 ) 2 , are poorly 
ionized, or form complex ions 
with excess of the negative ion. 

(a) A saturated solution of CO 2 at 20* C is about 0.04 F. 
(b) Many metallic hydroxides, although they may be highly ionized, are too insoluble to give a high concentration of ions. The 
solubility of Ca(OH) 9 at 20 C is about 0.02 F, and of Mg(OH) 9 is about 0.0002 F. 


Name of Indicator 

pH Interval 

Color Change 


Methyl violet . 
Thymol blue . . . 
Benzopurpurin 4B 
Methyl orange 
Bromphenol blue . 
Congo red .... 

0.2 3.0 
1.2 2.8 
1.2 4.0 
3.1 4.4 
3.0 4.6 
3.0 5.0 

Yellow, blue, violet . 
Red to yellow .... 
Violet to red 
Red to orange yellow 
Yellow to blue violet 
Blue to red ..... 

Water (+NaOH) 
20% alcohol 
Water (+NaOB[) 
70% alcohol 

Bromcresol green . 
Methyl red . . . 
Chlorphenol red . 
Bromcresol purple 

3.8 5.4 
4.4 6.2 
4.8 6.8 
5.2 6.8 
4.5 8.3 

Yellow to blue .... 
Red to yellow .... 
Yellow to red .... 
Yellow to purple ... 
Red to blue 

Water (+NaOH) 
Water (+NaOH) 
Water (+NaOH) 
Water (+NaOH) 

Bromthymol blue . 
Phenol red .... 
Thymol blue . . , 
Phenolphthalein . . 
Thymolphthalein . 
Alizarin yellow R . 
Indigo carmine 
Trinitrobenzene . 

6.0 7.6 
6.8 8.2 
8.0 9.6 
8.3 10.0 
9.3 10.5 
10.0 12.0 
11.4 13.0 
12.0 14.0 

Yellow to blue .... 
Yellow to red .... 
Yellow to blue .... 
Colorless to red .... 
Yellow to blue .... 
Yellow to red .... 
Blue to yellow .... 
Colorless to orange . 

Water (+NaOH) 
Water (+NaOH) 
Water (+NaOH) 
70% alcohol 
70% alcohol 
95% alcohol 
50% alcohol 
70% alcohol 

. 337 


Substances at unit activity (effective concentration) and at 26 e C. 





Li Li 4 . 


+ 1.66 
+ 1.18 

As HAsO 2 (H 4 ) 


K K 4 " 

Bi BiO+ (H+) 

Cs Cs+ 

Cu Cu++. 

Ba Ba++ 

H 2 O O 2 (OH~) 

g r __ Sr 4 " 4 - . 

Mn(OH) 8 MnO 2 (NH 4 +) 

Ca Ca++ 

I- I 2 

Na Na+ 
Mg Mg+ 4 
Al A1+++ 

MnO 2 MnO 4 - (OH~) .... 
H 2 O 2 O 2 (H+) 

p ++ _ p e +++ 

Mn Mn' f - + 

Hg Hg 2 ++ 

so 8 so 4 (on-) 

Ag - Ag+ 
H 2 O O 2 (10- 7 M H+) . . . . 
H g _ Hg++ . . .... 

Hj HzO (OH~) 

Zn Zn++ 

O Cr 4 -* 4 - 

H 2 H0 2 ~ (OH-) 

ci- cio- (on-) 

H 2 C 2 4 - C0 2 (H+) 
S S (OH-) 

Hg a +H- ^- Hg++ . ... 

Fe y e ++ . . 

js[O NO 3 - (H 4 -) ... 

H, HiO(10- 7 Af ii<) . . . 
CP++ Cr+++ 

Br~ Br 2 

H 2 O O 2 (H 4 ^) 

Cd Cd++ 

Mn + + MnO 2 (H + ) 

Pb __ pbSO 4 . . 

Cr+++ CrjOy (II 4 ) ... 
Cl~ C1 2 

Co Co++ 

Ni Ni++ 

Pb PbOj (H 4 ^) 

Sn Sn++ 
Cr (OH) 8 CrO 4 ~ (OIJ-) .... 
Pb Pb++ 

C1 2 C1O 8 ~ (H+) , . . 

Cl- HC1O (H+) 

A U Au 4 " 4 - 4 - 

HO 2 ~ O 2 (OH~) .... 

Mn ++ MnO 4 - (H 4 ) 

H 2 H+ 

BiO+ Bi 2 O< (H + ) 

HjS S (H+) 

PbSO 4 PbO 2 (H 4 -) 

g n ++ _ Sn 4 -^ 4 - 4 - 

MnO 2 MnO 4 - (H+) 

HjjSOs SO 4 (H + ) . . 

H 2 O H^ (H + ) 

gb SbO+ (H+) 

F- -; _ F 2 .... 

A g AgCl 

HP F 2 (H 4 ") 

The potentials listed are thoie which the several "half-cells" would give when connected with a H 2 ~~ H * half-cell, in which all 
ions present are at an effective concentration of 1 M . It is necessary to use this effective concentration, or "activity," rather than the 
ordinary inolarity, since dissolved substances do not behave as "perfect solutions" in concentrations as great as 1 J/. 
Since, in many cases, the H + or OH ~ enters into the reaction for the cell, changes in the pH will affect the potentials. It i* f there- 
fore, necessary to specify whether the data given are for 1 M H 4 of 1 M OH~. Cases in which this is not given are not greatly affected 
by this factor, since the H* 1 " or OH~ plays no part in the half-reaction. 

1 Values in this chart are taken from the data given in La timer, Oxidation Potential*, 2nd ed., Prentice-Hall, Inc., 1952. 




Water EUO - H+ + OH~ 10~ 14 

Weak Acids: 

Acetic . . . > HCJBUO* - H+ -f CJEUO,- 1.8 X 1Q~* 

Boric HJBO, H+ 4 H 2 B0 8 ~ 5.8 X lO"" 10 

Carbonic H 2 CO 8 - H+ + HCO 3 ~ K! - 4.5 X 10^ 

HCO 8 ~ - H+ + CO, K a - 6 X 10-" 

Chromic H 2 CrO 4 - H+ + HCrO 4 - K! - 5.9 X lO" 1 

HCrO 4 - - H+ 4- G0 4 K a - 6 X 10 - 

Formic HCHO 2 - H+ 4 CHO 2 - 2 X 10~ 4 

Hydrocyanic HCN - H+ + CN~ 4 X 10~ l 

Hydrofluoric HP - H+ 4 F~ 7.2 X 10~ 4 

Hydrogen peroxide .... HsOj - H + + HO 2 ~ 2.4 X 10"" 1 * 

Hydrogen sulfide H^ - H+ + HS~ K! - 1.1 X 10~* 

HS- = H + 4- S" K 2 - 1.0 X 10- 

Nitrous HNO 2 - H + + NO 2 - 4.5 X 10^ 4 

Oxalic H 2 C20 4 - H^ + HCJOr Kt = 5.9 X 10^ 

HC 2 O 4 - H^ 4- C 2 O 4 ~ K 2 6.4 X 1Q-* 

Phosphoric H,PO 4 - H+ 4 H 2 PO 4 - Ki 7.5 X 1Q-* 

HaPO,- - H^- + HP0 4 K 2 - 2 X 10-* 

HP0 4 ~ - H^ 4 P0 4 K, - 1 X 10-" 

Phosphorous HaPOs H^ 4 H 2 PO 3 - Kj - 1.6 X 10" 1 

Bisulfate ion HSO 4 - - H+ + SO 4 ~ K s - 1.2 X 1Q-* 

Sulfurous H 2 SO S = H+ 4 HSO 8 ~ K, - 1.2 X 10" 1 

HSOa~ - H + + SO," K 2 * 1 X 10- 7 

Weak Bases: 
Ammonium hydroxide . . . NH 4 OH = NH 4 + 4 OH- 1.8 X 10-* 

Complex Ions and Amphoteric Hydroxides: 

Cupric ammonia Cu(NH 8 ) 4 ++ - Cu^ + 4 4 NH 3 5 X 10" 14 

Silver ammonia Ag(NH*) 2 + - Ag+ 4- 2 NH 8 6 X lO^ 8 

Zinc ammonia Zn(NH 8 ) 4 ^"- Zn ++ 4 4 NH 8 3 X IQ- 10 

Mercuric chloride HgCl 4 ~ - HgCl 2 + 2 Cl~ 1 X lO" 2 

Silver chloride AgCl 2 - - Ag 4 ^ 4 2 Cl~ 1 X IQ- 6 

Aluminum hydroxide .... A1(OH) 4 - - A1(OH) 8 4 OH- 2.5 X 10~ a 

Chromic hydroxide .... Cr(OH) 4 - - Cr(OH) 8 4 OH" 10 2 

Lead hydroxide Pb(OH) 8 - - Pb(OH) 2 4 OH- 50 

Stannous hydroxide .... Sn(OH) 8 ~ - Sn(OH) 2 + OH- 2 X 10 8 

Zinc hydroxide Zn(OH) 4 ~ - Zn(OH) 2 4- 2 OH~ 10 





Hg 2 Cl 2 
PbCl 2 
PbI 2 

2 X 10-* 

1.6 X 10~ 10 
4 X 10 - 18 
1 X 10~ 16 
1 X 10 ~ 18 

1.7 X 10~ 6 
9 X 10- 9 


Ag 2 CO 8 8 X 10~ 12 

BaCO 8 5 X 10~ 9 

CaCO 3 4.8 X 10~ 9 

CuCO 3 IX 10~ 10 

FeCOa 2 X 10 ~ u 

MgCO 3 J X 10~ 8 

MnCO 8 9 X 10~ n 

PbCO 3 IX 10~ 13 

SrCO 3 IX 10~ 9 


Ag 2 Cr0 4 IX 10~ 12 

BaCrO 4 2 X 10~ 10 

PbCr0 4 2 X 10-" 

SrCrO 4 3.6 X 10~ 5 


A1(OH) 3 IX 10- 33 

Ca(OH) 2 . 8 X 10-' 

Cr(OH) a IX 10- 30 

Cu(OH) 2 6 X 10- 20 

Fe(OH) 4 IX 10- 1 * 

Hydroxides >, continued: 

Fe(OH) 3 ......... I 

Mg(OH) 2 ......... 6 

Mn(OH) 2 ......... I 

Pb(OH) 2 ......... I 

Sn(OH) 2 ......... I 

Zn(OH) 2 ......... I 


X 10- 12 
X 10-" 
X 10~ 16 
X 10- 26 
X 10- 17 


CaC 2 O 4 ......... 2 X 10~ 9 

MgC 2 O 4 ......... 9 X 10-* 

BaC 2 O 4 ......... IX 10~ 7 


Ag 2 S0 4 ... ....... 1.2 X 10-* 

BaSO 4 .......... IX 10~ l 

CaSO 4 .2H>O ........ 2.4 X 10" 8 

Hg 2 SO 4 ......... 6 X 10~ 7 

PbSO 4 .......... 2 X 10-* 

SrSO 4 .......... 2.8 X 10~ 7 


10~ 61 


10~ 22 






MnS (flesh colored) 10~ 16 

NiS 1Q- 21 

PbS 10- 28 

SnS 10- 28 

ZnS (j8) 10- 24 

1 See Study Assignment E for a brief discussion of the limitations which one should place on the interpretation of calculations based on 
solubility products. In the case of exceedingly insoluble substances, the values of the constants are not very precisely known. 




Positive Ions 

Negative Ions 










NH 4 + 
Li + 







Sodium . 





cio a - 




cio 4 - 






Hg 2 + + 











NO 2 - 


Ca ++ 


co t ~ 




HCO 8 ~ 




CrO 4 ~ 




Cr 2 O 7 




Cs0 4 




HC 2 O 4 ~ 




SO 4 " 




HSO 4 - 


Mn + ++ 



Iron, ferrous 

p c + + 



ferric . . 

Fe M '+ 



Tin, stannous 



BO 8 

stannic ........ 

Sn ++++ 


PO 4 ~- ' 




PO 8 ~ 





Fe(CN) 6 "- 

Ferrocyanide , 

Fe(CN) 8 

Answers to Drill Problems 

Study Assignment A, Page 11. 

1. (a) 10, (c) 5.5 X 10, (e) 10" 1 , ( 
(i) 2.006 X 10 

2. (a) 4.26 X 10 4 , (c) 3.75 X 10"*, (e) 4.65 X 10- 
(g) 3.0103 X 10 4 

3. (a) 10 6 , (c) 4.3 X 10 4 , (e) 10 1 , 
(k) 4 X 10*, (m) 1.4 X 10 4 

4. (6) KE - H mv, (e) V, - V - V (t - t p )/273 

5. (6) 2 X 10* cm/sec 

6. (o) 2.1668, (c) 7.5660 - 10, or 3.5660, (e) 23.7798 

7. (a) 304.0, (c) 3.583 X 10 4 (e) 0.0005508 

3 X lO" 11 , 

10. (i) 6.3 X 10', 

8. (a) 231 ml 

9. (a) 0.42, (c) 30.43, (e) 58.40, (?) 9.0 X 10* 
10. (a) 1.6, or -7.6, (c) 4.1 X 10~ 

Appendix /, Page 332: 

1. (6) 500 sec or 8.3 min, (d) 1000 cin 1 , 10 cm, (/) 6.0 boxes 

2. (6) 2, (d) 3, (/) 5 

3. (a) 23.2, (c) 0.022 

5. (6) 203.4 cm, (d) 0.002034 km. Four significant figures, 

regardless of the unit used. 
7. 0.6%