Full text of "Laboratory Exercises In General Chemistry"

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D. I. MENDELEYEV'S PERIOD!!

V
/.

; E L E M E N fT

>II

II!

IV j V

1 ,00797 1

Li 3 ,

6,939 2

Be 4 -,

9,0122 2

3 5 B

2 10,811

6 c

4 ^

2 12,01115

5 ? N
2 14,008

III

Na" -

22,99! 2

Mg 12 !

24,312 2

I 13 Al

2 26,9815

', !* si

2 28,086

I 15 P

2 30,9738

v 19 i

K 8

39,102 2

r* 20 2
Ca s

40,08 f

c 21 2

Sc 9

44,956 2

Ti 22 2

1 1 10

47,90 2

23 2
V n

50,942 1

l 29

2 63,54

2 30

ib Zn

2 65,37

3 31 ^

| Ga

2 69,72

4 32 '

is Ge

o 72,59

5 33

18 8 AS

2 74,9216

37 i

Rb .|

85,47 2

38 2

Sr iS

87,62 1

39 1

88,905 2

40 2

91,22 !

41 l

Nb la 2

92,906 1

VII

1 47

! i A *

2 107,870

2 48
is pH
is v>a

2 112,40

3 49

IS In

2 114,82

4 50

IS 1 ' Sn

9 118,69

I 51

i8 Sb

2 121,76

VIII

55 i

PC 18

^ 18

132,905. 2

56 2

Ba IS

13134 2

57 2.

I a* 18

La i 8

138,91 ' 2

72 , 2 o

Hf .-

178,49 1

73 2

Ta i

180,948 |

1^ 79
fa " AU

2 196,967.

, 2 8 * '

M Hg

2 200,59

.i 81

2 204,37

, 82 '

32 DK
18 U

2 207,19

? 8 2 Bi

2 208,980

87 i

18
[223] \

88 a
r l8

Ra 32

[226] |

89 2

A ** 18
AC 32

[227] 8 2

(Th)

(Pa)

*LANTHA

58 2

140,12 2

M ,

P'.' '1

.140,907 |

60 2

Nd i

144,24 2

61 2

[147]'

62 2

Sm r,

150,35 2

63 2

Eu I

151,96 *

64 2

fid 25

^-* 18

157,25 |

:}t:} y(cTl

90 2

10
Tt l8

Th . 32

18
232,038 2

91 2

Pa i

[231] |

92 2

u !

238,03 2

93 2

Np i

. 18
1^37]' *

94 2

PU 3 2 1

[242]

95 2

A 25

Am 32

18
[243]

9 V

r* ^

Cm ?2

[247] 1

Figures in square brackets are mass numbers of stablest isotopes

TABLE OF ELEMENTS

GROUPS

' VI

VII

VIII

(H)

He 2

4,0026 2

2 15,9994

2 18,9984

Ne 10 .

20,183 *

6 16

2 32,064

7 8 17 Cl

2 35,453

Ar 18 I

39,948 1

Cr 24 13

51,996 2

25 2

Mn 13

54,9381 2

26 2

Fe i

55,847 2

n 27 2
CO 15

58,9332 2

28 2

58:71 2

6 34

w Se

2 78,96

7 35

is Br

1 79,909

Kr 36 -i

83,80 2

42 i

Mo 11

95,94 2

43 1

[97] |

44 (

Ru It

101,07 2

45 is

102,9p5 2

106,4 2

6 52

18 x ^

18- 1 C

2 127,60

18 w
18 1

2 126,9044

54 8

Xe \l

131,30 2

74 ,i

W32
18

183,85 2

75 2

Re 32

IVC 18

186,2 I

76 2

fte 32

UJ> 18

190,2 I

77 2

* 5
If 32

11 18

192,2 2

78 i

Pt 32
l 18

195,09 1

6 34

18
32 I) r\
18 * U

! [210]

, 7 8 8B

I At

[2.0]

86 ,1

Rn fs

[222] . |

(u)

WIPES 58-71

65 2

66 2

67 2

68 2

69 2

70 2

71 2

-, 8

Tb 27

Dy i

Er 30

C ' 1 J8

TU ?!

Yb 3!

Lu |

158,924 9

162,50 |

164,930 1

167,26 \

168,934.1

173,04 |

174,97 2

Symbol
Atomic
number

NIDES

97 2

98 2

99 2

100 2

101 2

102 2

103 2

Bk S

^ 28
Cl 32

Es 11

FT**- 30

rm 32

Md 3 3 2

Nn 32

JNU 32

Lw g

[247] 1

[249] 8 2

[254] i

[253] f

[256] |

[255] 5

[257]

( Electron-
\ layers

Atomic

weight

OSMANIA UNIVERSITY LIBRARY

Call No. S^ Accession No. *D V Q_ l\

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Author SemisVviva , \l -

Title l_nbc>*<\t:c>Yy exeveiSeJ m ^evie^alc/
This book should be returned on or before the date last marked below.

B. H.

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V. Semishin

LABORATORY EXERCISES

IN
GENERAL CHEMISTRY

Translated from the Russian by

BO HIS BEL ITSKY

PEACE PUBLISHERS MOSCOW

CONTENTS

Page

Introduction 7

EXERCISES

1. Heating and Weighing 9

2. Solution, Filtration and Hydrometry 16

3. Determining the Molecular Weight of a Gas (Vapour) 24

4. Laboratory Techniques 35

5. Determining the Chemical Formula of a Substance 39

6. Determining Chemical Equivalents 46

7. Determining the Purity of a Substance 52

8. Thermal Effects of Reactions 62

9. Reaction Rates and Chemical Equilibrium , 69

10. Preparation of Solutions 81

11. Solubility of Substances 86

12. Solutions of Fluids 94

13. Properties of Solutions 102

14. Electrolytes 109

15. Reactions in Electrolyte Solutions 118

16. Oxidation-reduction Reactions 129

17. Hydrogen, Oxygen, and Ozone 145

18. Water and Hydrogen Peroxide 152

19. General Properties of Metals and Alloys 159

20. Alkali Metals 172

21. Copper Subgroup Elements 179

22. Complex Compounds 186

23. Beryllium, Magnesium, and the Alkaline Earth Metals 193

24. Zinc, Cadmium, and Mercury 200

25. The Elements of the Third Group of the Periodic System 207

26. Carbon, Silicon, and Their Compounds 213

27. The Elements of the Germanium and Titanium Series and Their Com-
pounds 222

28. Colloidal Solutions 229

29. The Elements of the Fifth Group of the Periodic System and Their Hydrides 235

30. Oxygen Compounds of the Elements of the Fifth Group 243

31. Oxygen Compounds of the Elements of the Fifth Group (Continued) . . 250

32. The Elements of the Sixth Group of the Periodic System and Their Hy-

drogen Compounds * 256

33. Oxygen Compounds of the Elements of the Sixth Group 263

34. Oxygen Compounds of the Elements of the Sixth Group (Continued) . . . 268

35. The Elements of the Chromium Subgroup 275

36. The Halogens 280

37. Hydrogen Compounds of the Halogens 285

38. Oxygen Compounds of the Halogens 291

Contents

39. The Elements of the Manganese Subgroup 296

40. Iron and Its Analogues 301

41. Cobalt, Nickel, and Their Analogues 306

42. Synthesis of Inorganic Substances 312

Answers to Problems 325

APPENDICES

I. Relative Densities and Concentrations of Some Salt Solutions .... 329

II. Solubilities of Some Salts 329

III. Relative Densities of Sulphuric Acid Solutions 330

IV. Relative Densities of Nitric Acid Solutions 330

V. Relative Densities of Hydrochloric Acid Solutions 331

VI. Relative Densities of Ammonia Solutions 331

VII. Relative Densities of KOH and NaOH Solutions at 15 332

VIII. Relative Densities and Degrees Batime at 17.5 (for liquids heav-
ier than water) 332

IX. Relative Densities and Degrees Baume at 17.5 (for liquids lighter

than water) 332

X. Cryoscopic Constants 333

XI. Ebullioscopic Constants 333

XII. lonisation Constants 333

XIII. Solubility Products 335

XIV. Complex Ion Instability Constants 335

XV. Normal Redox Potentials 336

XVI. Some Ionic Radii 337

XVII. Logarithms 337

Index 341

TABLES IN THE TEXT

1. Effectiveness of Dehydrating Agents in Drying Air 27

2. Aqueous Vapour Tension 28

3. Vapour Pressure over Saturated Solution of Sodium Chloride 59

4. Heat of Solution of Some Gases in Water fi6

5. Coefficients for Absorption of Several Gases by Water 97

6. Absorption Coefficient of Air 101

7. Salts Used to Prepare Cooling Mixtures 108

8. Mixtures of Salts Used to Prepare Cooling Mixtures 108

9. Degree of lonisation of Some Electrolytes 110

10. Solubility of Some Substances in Water . . 120

11. D. I. Mendeleyev's Periodic Table of Elements 131

12. Composition and Properties of Some Composite Negative Ions 136

13. Reducing and Oxidising Agents 138

14. Melting Points of Zn-Cd System 161

15. Electromotive Series of Metals 164

16. Colour of Borax Beads 212

INTRODUCTION

Laboratory exercises are an integral part of a course of chemistry.
To perform them profitably, the student should first study the subjects
listed in small print at the beginning of each exercise. The questions
included enable him to make certain that such preparation for each
exercise has been sufficiently thorough. It is especially important
to solve the problems included, which are based on the subject-matter
of the exercise. Atomic weights approximated to one decimal should
be used in solving the problems.

All laboratory work should be done by students individually.
The student should be familiar with the laboratory routine, and
he should follow it unswervingly.

From the very first exercise thought should be given to the time-
saving factor. The student should endeavour to organise his work in
such a way that during lengthy operations not requiring his undi-
vided attention (filtration, evaporation, calcination, etc.) he could
perform other experiments.

Reagents should always be taken in the quantities indicated. The
stoppers of reagent bottles should be promptly replaced, and the bot-
tles should be put back where they belong. A reagent taken in excess
should never be poured back into the bottle. Used reagents should be
poured down the drain; in the case of fuming reagents, however,
this should be a drain fitted with a ventilated hood, while spent rea-
gents of considerable value should be collected in special vessels.

No work requiring a ventilated hood should be done without it, on an
unhooded laboratory table.

The results of all the experiments conducted should be recorded in
a notebook, a margin being left for the instructor's comments. All
records should be clear and concise. The conditions in which the expe-
riment is conducted should be specified, and the phenomena observed
described. The chemical process should be expressed by proper equa-
tions, and a sketch or diagram of the apparatus used should be added
if necessary.

Exhaustive written answers should be given to all the questions
put in the exercises.

All calculations should be done by means of a table of logarithms
or slide rule. If the treatment of the experimental data involves
atomic weights, their values should be approximated to two decimals.

Introduction

The results of an experiment should be entered in the student's
notebook before a new experiment is begun.

On completing a laboratory exercise, the student should submit
his notebook with the results of all the experiments to the instructor.
Before leaving the laboratory, the student should put his workplace
in order and wash his hands.

The instructor should be informed at once of any accident in the lab-
oratory.

THE STUDENT'S LABORATORY EQUIPMENT

Each workplace should have laid on: gas, hot and cold water, elec-
tricity (a. c. and d. c.), suction, and a drain. In addition, each stu-
dent should be provided with the following objects:

A burner; B rlngstand; C test-tube ..rack; -D wire gauze with
asbestos centre; E test-tube holder; F brushes for cleaning utensils.

Exercise 1

HEATING AND WEIGHING

LABORATORY WORK

Apparatus and materials: gas burner; crucible tongs; 5-7 cm of nichrome wire
about 1 mm in diameter; porcelain crucible cover; watch glasses, weighing bottles,
and crucibles; glass tube about 20 cm in length and 5-6 cm in diameter; test tubes;
pins; matches; sheets of paper 10X 10 cm; potassium bromide; anhydrous sodium
sulphate, and an object for test weighing.

Heating

Many chemical processes can be accelerated by heating. This is
effected by electric appliances (furnaces and heaters) or burners
(spirit, petrol, and gas), depending upon the laboratory equipment
and the nature of the process studied.

1. Gas Burner. Take the burner shown in Fig. 1 apart: unscrew
the gas tube (2) from the base (1); unscrew the disk (3) from the lower,

Fig. 1. Gas burner (Teclu burner)

1 base; 2 tube; 3 disk regulating flow of air;
4 valve regulating supply ot gas.

wider part of the tube, which is called the gas-mixing chamber.
Examine each part and ascertain the purpose of the disk and of the
valve (4).

Exercise 1

Reassemble the burner* and draw a diagram cf it in your note-
book, designating the names of all the parts.

Fig. 2. Crucible tongs

Connect the burner to a gas-supply tap by means of rubber tub-
ing; screw in the disk close to the chamber; turn on the gas by turn-
ing the cock 90, and light the burner.

Now turn the valve slowly and observe the changes in the size of
the flame. What purpose does the valve serve?

By means of a pair of crucible tongs (Fig. 2) take a porcelain cru-
cible cover and introduce it into the flame. What is formed on the

surface of the cover? How is such a flame
called? Is the burning of the gas in such a
flame complete? **

Turn the disk slowly, observing the changes
in the flame, until a sharply defined inner
cone appears within it. Introduce a clean por-
celain crucible cover into the middle of the
outer cone. Is any carbon black formed now?
How is such a flame called? What does the
disk serve for?

Other types of gas burners are sometimes
used in the laboratory too (Fig. 3).

2. Flame Flash-back. By turning the valve
smoothly reduce the height of the flame to
about 2 cm, after which turn the disk until it
is fully released from the tube. The flame has
now flashed back, i. e., combustion takes place
within the burner (verify this!). This causes
the flame over the tube either to disappear
or to become elongated and luminous, combus-
tion being often accompanied by a whistling
sound. If the burner is made of a copper alloy, the flame may after
a time acquire a green tinge.

When the flame flashes back, it means that the burning of the gas
is incomplete, with the result that the air in the laboratory becomes

Fig. 3. Gas burner
(Bunsen) with ring

1 base; 2 tube; 3 ring
regulating flow of air.

* Screw the disk on so that its smooth side faces upwards.
'* Hot objects should be placed on asbestos millboard or on a ceramic plate.

Heating and Weighing

poisoned. Moreover, since combustion takes place inside the tube, the
latter becomes very hot, which may cause the rubber tubing feeding
the gas to the burner to catch fire.

When a flame flash-back occurs, turn off the gas, allow the burner
to cool (beware of burning your hands!), and then light it again in
accordance with the rules
(see p. 12).

Why is a flash-back of the
flame undesirable?

3. Gas-poisoning Hazard.
The gas used in the laboratory
(which is the same gas as that
used in industry and for house-
hold needs) is chiefly coke-
oven gas, either pure or mixed
with natural gas. Coke-oven
gas is prepared by the gasifica-
tion of coal, while natural gas
is obtained by tapping natu-
ral deposits (in the U.S.S.R.
there are big deposits at
Saratov on the Volga, Dasha-
va in the Ukraine, and else-

Fig. 4. Structure of gas burner flame

where). These
(in %):

gases have the following average composition

H 2

Crl4, CfcH0

c x H #

H 2 O

N 2

CO,

Coke-oven gas

0.5

7.5

Natural gas (Sara-

95.2

1.3

3.3

0.2

tov)

The carbon monoxide contained in coke-oven gas is highly poison-
ous. A mere 0.3% of it in the air can cause fatal poisoning in 12-15 min-
utes. For this reason always keep an eye on gas burners. Before leav-
ing the laboratory, make sure the gas is turned off.

4. Flame Structure. Quickly introduce a sheet of paper for 2-3 sec-
onds into the lower part of the small nonluminous flame (broken
line in Fig. 4 ,4). Observe the appearance of a ring of charring. Intro-
duce a glass tube into the flame, as shown in the figure, and light
the gas at its upper end. Lift the tube slowly, and note the position
of its lower end when the burning at the upper end ceases. Put out the

12 Exercise 1

burner by turning off the gas, and insert a match, attached to a pin,
into the tube of the burner (Fig. 4 B). Now turn on the gas and light
the flame. Does the match ignite? Lifting the pin by the tongs, move
the match into the upper cone of the flame. What happens now?
What conclusions can be drawn about the burning of the gas in the
inner and the outer cone on the basis of these experiments?

5. Flame Temperature. Draw a diagram of the burner flame in
your notebook. Mark all the points and corresponding temperatures
shown in Fig. 4 C. Introduce a nichrome wire, whose end has been
joined to a glass tube, into the various zones of the flame, and observe
the colour of the heated wire. Depending upon the temperature to
which it is heated, the wire acquires the following colours:

dark red at 500 orange at 1100

red at 700 white at 1500

In the diagram mark the points corresponding to the experimental-
ly observed temperatures of 700 and 1100.

Put 1 g of powdered potassium bromide (m. p. 728) into a dry test
tube and heat it for some time in the hottest zone of the flame. Does
the salt melt? Now conduct a similar experiment with 1 g of anhy-
drous sodium sulphate (m. p. 884). Does this salt melt? On the basis
of these experiments state the average temperature achieved in prac-
tice when heating objects with a gas burner.

Rules for using a gas burner

1. Drive in the disk close to the wide part of the gas tube.

2. Turn on the gas and light the burner.

3. Screw out the disk slowly until a sharply defined inner cone
appears in the flame.

Weighing

Weighing is one of the most important operations in quantitatively
estimating the results of a chemical process, as well as in determining
the numerical values of certain characteristics of a substance (its
equivalent, its atomic and molecular weights, etc.). The reliability
of the quantitative assessment of the data of a chemical experiment
depends to a considerable extent upon the accuracy of weighing.
Three types of balances are used by the student in the laboratory of
inorganic chemistry: the technical balance (for rough weighing),
the chemical balance (for weighing to 0.01 g), and the analytical bal-
ance (for weighing to 0.0001 g).

1. Chemical Balance. The chemical balance (Fig. 5) is mounted on
an uncovered base with special screws for adjusting the balance accord-
ing to the plumb weight attached to the pillar (1). After adjustment
according to the plumb weight the balance should not be shifted.

Heating and Weighing

0.5 0.2 02 0.1 0.05 002 0.02 O.Ot

Fig. 5. Chemical balance and weights

l -pillar- 2 -catch; 3 - pans; 4 pointer; 5 graduated scale; 6 plumb
weight; 7- screws for levelling the balance; 8 - beam; 9 - screws for balancing

empty pans.

Before weighing, the balance should be checked. By shifting the
catch (2) smoothly to the right, the balance is made operational: the
beam is lowered so that the edge of its central prism rests upon its
support and begins to swing together with the pans (3) suspended

Exercise 1

from side prisms of the beam. If the pointer (4) now swings back antf
forth across the graduated scale (5) an equal number of divisions away
from the middle line each way (or one or two divisions more in one
direction than in the other), the balance may be considered adjusted
correctly. After the balance has been checked, the beam should be
arrested; this is done by moving the catch back into the non-operation-
al position.

Fig. 6. Vessels for weighing
I watch glass; II and' III weighing botflos; IV crucible.

2. Rules for Weighing. The objects that are being weighed and the
weights may be placed on the balance pans or removed from them only
after the balance has been arrested.

Solids are weighed on watch glasses, in weighing bottles, crucibles,
or on sheets of clean paper, whereas liquids are weighed in weighing

bottles (Fig. 6) and beakers only.
Hot crucibles or casseroles
should, before weighing, be cooled
in desiccators to room temper-
ature (for about 15 minutes). A
desiccator (Fig. 7) is a thick-
walled glass vessel with a ground-
in lid; placed in the bottom
part is a substance capable of
absorbing water vapour (see p. 27).
The object to be weighed should
be placed on the left-hand pan
of the balance; a weight approxi-
mately equal to the weight of the
object should then be placed by means of forceps on the right pan. If
the weight is too big, it should be replaced with another of the next
lower denomination, and so on until the object is counterbalanced by
the weights (the pointer should swing an equal number of divisions
to the right and left of the middle line on the scale). The weights
should now be added up; this is done by adding up the grams, de-
cigrams, and centigrams separately. The weight of the object to 0.01 g
should then be recorded by the student in his notebook.

Fig. 7. Desiccator and porcelain plate

Heating and Weighing 15

A 100 g set of weights is decimally constituted (this system was pro-
posed by D. Mendeleyev):

Grams 1 50 20 20 10

5221
0.5 0.2 0.2 0.

0.05 0.02 0.02 0.01

Fractions of a gram ^ 0.5 0.2 0.2 0.1

The weights of such a set can be combined to make up any weight
ranging from 0.01 to 111.10 g (with an accuracy to 0.01 g).

All weighing for one and the same experiment should be done on
the same balance and with the same set of weights.

After the weighing, the weights should be replaced in their respec-
tive nests in the box with the aid of forceps. After operations the bal-
ance and the weights should always be left iru order.

3. Absolute and Relative Error. The absolute error e in weight
.determination is the difference between the true weight W and the expe-
rimentally obtained weight w:

e= W-w

In 'practice it is more customary to calculate the relative error ex-
pressed in per cent. This is the ratio of the absolute error to the true
weight multiplied by one hundred:

4. Weighing an Object. Obtain an object for test weighing from
the laboratory assistant and weigh it to an accuracy of 0.01 g. Re-
cord the result in your notebook as follows:

Name of object

Weight, g

Plastic disk No. 17

8.74

Check the weight of the object together with the laboratory assistant
and, in the event of a discrepancy, calculate the relative error.

1. An object weighed on a chemical balance has been counterbal-
anced exactly by two weights: 20 and2g. How should the exact weight
of the object be written down?

2. The weight of an object weighed on a chemical balance is writ-
ten down as 11.270 g. Is this correct?

3. Two weights 1 and 10 g were used on a chemical balance.
For which of them is the relative error of weighing larger and why?

16 Exercise 2

Exercise 2

SOLUTION, FILTRATION, AND HYDROMETRY

SUBJECTS FOR STUDY

Mechanical mixtures and chemical compounds; laws of definite proportions, mul-
tiple proportions, and weight equivalents, which govern the composition of chemical
compounds; compounds of constant and variable composition, and percentage and
molar concentration of solutions.

In conducting chemical experiments, it is often necessary to dis-
solve substances, filter liquids, wash precipitates, and determine the
concentration of solutions by means of a hydrometer.

Fig. 8. Mortars for grinding

/ glass; // porcelain; /// metal.

1. Dissolving Solids. Substances in the form of large crystals are
sometimes ground to a powder prior to dissolving. The grinding is
done in porcelain, glass, agate, or metal mortars (Fig. 8). When grind-
ing j caustic alkalis or highly poisonous substances, the face should
be protected by a mask with a respirator; the hands, by rubber gloves.
Highly poisonous substances should be ground in a ventilated hood.

Approximate volumes of solvents are determined in measuring
cylinders or glasses (Fig. 9) and expressed in millilitres (ml).

The substance to be dissolved and the solvent are placed in a beaker
(or flask) and mixed with a glass rod. In most cases it is advisable to
heat the solution (Fig. 10) in order to speed up the process. Aqueous
solutions are prepared with distilled water.

2. Filtration. Substances which have not dissolved in the solution
are separated from it by filtration.

Various porous materials, such as filter paper, cotton wool, cloth,
porous plates of glass or porcelain, finely divided charcoal and asbes-
tos, and glass wool, can serve as filters. The material most commonly
used in the laboratory is filter paper, which is employed for making
two types of filters: ordinary and folded.

Solution, Filtration and Hydrcmetry

, To make an ordinary filter (Fig. 11), a square sheet of filter paper
(A) is folded in half twice (5, C). The ends of the filter are then cut
neatly by scissors, as shown by the broken line in Fig. 11 (D).
The filter is finally opened up into a cone ().

I IT

Fig. 9. Measuring vessels

/* measuring glass; // measuring
cylinders.

Fig. 10. Dissolving sol-
ids in water

To make a folded filter, at first follow the same procedure (A-D);
then unfold the filter (F) and fold each segment in half twice towards
each other (G-7); now turn the filter over (J) and, after it has been fold-
ed twice more (/C, L), it may be opened up to assume its final shape
(M). In this case it is more convenient to trim the ends after folding.

The trimming should be done in such a way that the top of the
liter is 3-5 cm below the upper rim of the funnel.

The ordinary filter should be fitted into the funnel, held in posi-
tion with a finger, and moistened with distilled water, which is al-
lowed to drip away. Chemical funnels should form a 60 cone; the ordi-
nary filter in this case fits snugly into the funnel. If the angle is big-
ger or smaller than 60, air gaps form between the wall of the funnel
and the filter, and this slows down filtration (Fig. 12). The stem of
the funnel should be filled with liquid during filtration; the presence
sf air bubbles in it likewise delays filtration.

The speed of filtration depends upon temperature, since with a rise
in temperature there is a decrease of the internal friction of the liq-
uid in the filter pores (for example, the internal friction of water

2 -795

Exercise 2

at 100 is nearly 85% lower than at 0). If possible, liquids should
therefore always be filtered hot.

It is customary to use ordinary filters wherever the precipitate has
to be retained for further operations, since it is easier to remove the
precipitate from such a filter. If, on the other hand, the solution only

J K L

Fig. 11. Making filters

Fig. 12. Funnels for filtration

is needed for subsequent operations, a folded filter is used; filtration
in this case proceeds faster, owing to a greater filtering area.

Two filtration techniques are in common use: either the liquid is
transferred to the filter with the precipitate or the liquid is poured
off from the precipitate into the filter (decant at ion). In the former
case, the liquid is stirred with the precipitate and poured into the
filter down a glass rod (Fig. 13); the filter should be filled to such an
extent that the level of the liquid is 2-3 mm below the filter rim.

Solution, Filtration and Hydrcmetry

Fig. 13. Filtration

In the latter case, the liquid is poured
off carefully down a rod into the filter
without disturbing the precipitate.

To prevent splashing, the stem of the
funnel should rest against the side of
the receiving vessel. When the entire
liquid has been filtered, the precipitate
should be washed.

3. Washing the Precipitate. The precip-
itate transferred to the filter is washed
by means of a wash bottle (Fig. 14),
which is a flat-bottom flask with a stop-
per that has two bent tubes passing
through it. If air is blown orally into
the shorter tube, the water in the flask
will be forced by the pressure to issue
in a thin stream from the other tube.
The filter loaded with the precipitate is
filled with the washing liquid (in most
cases, cold or hot water), which is allowed
to drain through the paper. This wash-
ing is repeated two or three times, anew
portion of water not being poured into
the filter until the previous portion has
drained away. The last drops of the fil-
trate are collected in a test tube and tested
with reagents to make sure that the
soluble impurities have been washed out
from the precipitate.

When the precipitate is left in the beak-
er, it is washed by decantation. For this
purpose the washing liquid is poured
into the beaker, stirred with a rod, and,
after the precipitate has settled, trans-
ferred to the filter down a rod. After 2-3
washings, the wash water is tested for
impurities.

4. Concentration of Solutions. The
concentration of the dissolved substance
(solute) is determined by the amount
of it contained in a definite weight or
volume of the solution.

Percentage concentration is expressed by the number of weight units
of the solute in 100 weight units of the solution.

Molar concentration (M) is expressed by the number of mols of the
solute in 1 litre of the solution.

Fig. 14. Wash bottle

Exercise 2

12 -

-1124

Other units expressing the concentra-
tion of solutions are defined in Exer-
cise 10.

5. Determining the Concentration of
Solutions by the Relative Density*.
The relative density of solutions changes
with their concentration; for this reason
the concentration of a substance in solu-
tion can be determined by its relative
density, use being made of suitable ref-
erence tables. Relative density is affect-
ed by temperature and should there-
fore be determined at the temperatures
specified in the tables (or else allowance
should be made for variation from the
standard temperature indicated).

The symbol yf in reference books
means** that the relative density has been
determined at 20 and referred to the
weight of water at 4, the temperature of

~. i- ~ ... fu , . its maximum density. Relative density
Fief. 15. Position of hydrometer , , , . , , J c J

in cylinder with solution and can be determined by means of a pycno-
reading the hydrometer scale meter, by hydrostatic weighing, etc. For

technical purposes the relative density is
in most cases determined by a hydrometer.

The hydrometer (Fig. 15) is a hollow glass float with a graduated
upper part and a weighted (usually with lead shot) lower part. The
weight keeps the instrument in a vertical position when it is im-
mersed in a liquid. The hydrometer is calibrated for a definite tempera-
ture, which is indicated on the scale. The divisions of the scale corre-
spond to relative densities. By means of a set of hydrometers the
relative density of a liquid can quickly be determined.

For this purpose the test solution, first heated or cooled to the tem-
perature indicated in the tables of relative densities, is poured into a
250-300 ml dry narrow cylinder. The hydrometer (clean and dry) is
then lowered into the cylinder with the solution so that it should not
touch the walls of the vessel. The reading of the hydrometer scale
is taken according to the lowest point of the level of the liquid in the
cylinder. The scale is read downwards with an accuracy of + 0.003.

In Fig. 15 the hydrometer reading corresponds to a relative density
of 1.124. It may happen that the table of relative densities does not

* The ratio of the weight of 1 rnl of a solution to the weight of 1 ml of water at the
temperature of its maximum density (4) used to be called specific gravity.
** Some reference books use the symbol d^.

Solution, Filtration and Hydrometry 21

give the value 1 . 124, but gives close values, as in the case of a solution
of sulphuric acid:

Relative density Concentration (%)

1.120 17.01

1.130 18.31

In such cases the percentage concentration is found by the method of
interpolation, i.e., by determining an intermediate quantity accord-
ing to two known extreme quantities. The following procedure is
used:

1. The respective differences are found for the two consecutive val-
ues of relative densities and of concentrations taken from the table:

1.130 18.31%

1.120 17.01%

0.010 1.30%

2. The difference is found between the hydrometrically determined
value and the smaller of the two table values:

1.124-1.120 = 0.004
The proportion is set up:

0.010 1.30%

0.004 x

1.30-0.004 _ 0520/

x ~~ oW~ ~~ u '^ /0

3. The quantity obtained is added to the smaller of the two concen-
tration values taken from the table:

17.01 + 0.52= 17.53

QUESTIONS

1. What is meant by the analysis and synthesis of a substance?
Give three examples each of analysis and synthesis. What are these
processes used for?

2. Give examples of compounds of constant composition and of
variable composition; explain the difference between them.

3. What conclusions can be drawn from the law of multiple propor-
tions?

4. In what cases are ordinary filters used, and in what cases are
folded filters used?

5. What is decantation?

22 Exercise 2

6. Give the equations of the reactions that serve to check wash wa-
ter for sulphates and chlorides.

7. Could a hydrometer be calibrated in such a way that the divi-
sions of its scale would correspond to the percentage content of a sub-
stance in a solution instead of corresponding to relative densities?

Problems *

1. Three samples consisting of sulphur and iron were found, upon analysis, to have
the following respective percentage compositions:

i II in

Fe . . . . 63.6 60 46.6
S .... 36.4 40 53.4

Which of these correspond in composition to chemical compounds?

2. Three compounds of chromium and oxygen contain 76.5, 68.4, and 52% of
chromium respectively. Show by calculations that these proportions of the elements
in the compounds conform to one of the basic laws of chemistry.

3. A copper oxide contains 79.9% of copper and 20.1% of oxygen. How much cop-
per can be obtained from 15.9 g of the oxide?

4. When a metal is burned, 2.07 g of it combines with 2.4 g of oxygen. What
amount of hydrogen can be obtained from water if 1.38 g of this metal reacts with
the water?

5. In the formation of zinc sulphide, 32.1 g of sulphur is used up for every 65.4
g of metallic zinc. How much of which substance, will be left over if 30 g of zinc and
30 g of sulphur are taken for the reaction?

6. Determine the molar concentration of a- 17.1% solution of nitric acid which
has a relative density of 1.1.

7. A laboratory prepared 4.7 kg of a 7.6% solution of soda. How much soda and
how much water were used?

8. What amount of table salt containing 98.5% NaCl is needed to prepare 5,000
kg of an 8% solution?

9. Calculate the percentage concentration of sodium hydroxide in the solution
if 3.4 kg of caustic soda (commercial sodium hydroxide) which analyses 97.2% NaOH
was dissolved in 8.6 kg of water.

10. The relative density of a KOH solution at 15 is 1.054.With the aid of Table
VII calculate the percentage concentration of the solution.

LABORATORY WORK

Apparatus and materials: test-tube rack and test tubes; 250 ml measuring cylin-
der; 300 ml narrow measuring cylinder for determining relative density; 200 ml beak-
er; 250 ml flat-bottom flask; wash bottle; funnel 7 cm in diameter; glass rod; hy-
drometer with a 1.0-1.2 relative density scale; thermometer for 100; spatula; scissors;
sheets of paper 8 X 8cm; filter paper; mixture of table salt and sand; 0.1 N solution
of silver nitrate, and 0.1 N solution of nitric acid.

* Atomic and molecular weights approximated to one decimal should be used
in solving the problems throughout all the exercises.

Solution, Filtration and Htjdrometnj

Preparation of sand and table salt mixture. The sand for the mixture is first washed
with water and dried; the salt is ground in a mortar to a 'powder and mixed thorough-
ly with the sand. Several mixtures with different NaCl contents are needed.

Determining the Percentage Content of NaCl in a Mixture. Weigh
about 20 g of a sand and table salt mixture to an accuracy of 0.01 g*
The mixture should be weighed on a watch glass or in a weighing bottle
weighed in advance on a chemical balance. It should be placed on
the watch glass or in the weighing bottle with a spatula (Fig. 16).
The weighed mixture should be poured into a beaker, and 100 ml of
distilled water, measured in a cylinder, should be added. The beaker
with the liquid should then be heated to boiling point, the contents
being stirred with a glass rod.

Prepare a folded filter, placeitin a funnel, moisten it with distilled
water, and filter the solution, collecting the filtrate in a beaker or
a flask. During filtration heat the water in the
wash bottle to boiling point. After the whole
of the liquid has been filtered, pour 10-12 nil
of hot water into the beaker, shake it, allow
the precipitate to settle, and transfer the solu-
tion to the filter. Wash the filter thrice with
hot distilled water, collecting the last 2-3 ml
of the filtrate in a test tube (the average-
sized test tube has a volume of 15 ml). Add
2-4 drops of nitric acid solution and as many
drops of silver nitrate to the liquid in the test
tube (write the equation of the reaction). If no
precipitate is formed and the liquid does not
become cloudy, the washing may be consid-
ered completed; otherwise it should be contin-
ued.

Now add the wash water to the solution,
transfer the solution to a measuring cylinder,
and cool it to 20. By adding water, bring the

volume of the solution up to 250 ml. Then transfer the solution to
a narrow cylinder and stir it thoroughly. Lower a dry hydrometer
into the solution slowly and take a reading, first becoming familiar
with the calibration of the hydrometer and the value of a single di-
vision on its scale. Take the average of 3-4 readings *. After the rela-
tive density determination, the hydrometer should be rinsed with
clean water, dried, and put away in its case.

On the basis of the relative density of the solution, determine its
percentage concentration, making use of Table I (p. 329). If the table

Fig. 16. Spatulas

* For example, the average of three readings will be:

1.024 1.027 1.026

average: 1.026

Exercise 3

does not list the exact value of the relative density obtained, find'
the quantity sought from the two closest values given in the table
(a smaller and a bigger) by the method of interpolation.
Record the results of the experiment in your notebook as follows:

Weight of
mixture in g

Volume of
solution in ml

Relative density
according to
hydrometer at 20

Percentage concentra-
tion (from table or
found by interpola-
tion)

Quantity of salt in
solution in g

To determine the quantity of salt in the solution, it is necessary to
calculate the weight of the solution in g from its volume and relative
density, the quantity sought finally being established from the per-
centage concentration.

Knowing the weight of the salt and the weight of the mixture, cal-
culate the NaCl content in the initial mixture to 0.1%.

From the percentage concentration of the NaCl solution and its
relative density calculate the molar concentration.

Exercise 3

DETERMINING THE MOLECULAR WEIGHT
OF A GAS (VAPOUR)

SUBJECTS FOR STUDY

Equation of state of a gas; reducing the volume of a gas to N. T. P.; Avogadro's
Law and conclusions from it; gram-molecule; gas constant and Mendeleyev-Clapeyron
Equation; density of gases and the density of one gas in terms of another; re-
lation between the density of a gas (vapour) and its molecular weight; partial pressure.

1. The Molecular Weight and Density of a Gas. Molecular weight
is a number indicating how many times heavier a molecule of a given
substance is than 1/12 of a carbon atom.

A gram-molecule (mol) is the quantity of a substance equal in grams
to its molecular weight.

A gram-molecule of any substance in the gaseous or vaporous state
under normal conditions (N. T. P.*) occupies a volume of 22.4 litres.

The volume of a gas is reduced to N. T. P. by the equation of state

of a gas:

vpTo /T\

v -~^ (l)

* An abbreviation for normal temperature and pressure (i. e., 0C and 760 mrn
Hg). Another abbreviation used to express this in British and American literature
is S. T. P. (standard temperature and pressure). Tr.

Determining the Molecular Weight of } a Gas (Vapour) 25-

The conversion from the international hundred-degree scale fC)
to the absolute scale (K) is effected by the formula:

T - 273 + t

where T is the temperature in K, while t is the temperature in C
The ratio^~- from the equation of state of a gas is constant for one

gram-molecule of any gas. It is called the universal ga$ constant and is
denoted by the letter R. Its value and dimension depend upon the
system of units chosen:

R = * = - 2 ^i ^ 0.082 1-alm/deg

= 62,369 ml - mm/deg

The value and dimension of R should always be in keeping with the
dimensions adopted for pressure and volume.

By substituting the gas constant into equation (I), we obtain the
Mendeleyev-Clapeyron Equation:

(II)

If n mols of the gaseous substance are taken, the Mendeleyev-
Clapeyron Equation assumes the form:

pv^nRT (III)

But ft, the number of mols of the gas, is equal to the ratio of m, the
weight of the gas in grams, to M, the weight of a gram-molecule:

When this expression for n is substituted into equation (III), we
obtain

from which

M = ^L (IV)

pv v 7

This formula makes it possible to calculate the molecular weight
of a gas provided the weight m of a certain volume v of it is known,
as well as its pressure p and temperature T.

If we know the density of the gas (absolute density) D a , i. e., the
weight in g of 1 litre of the gas at N. T. P., the molecular weight M

26 Exercise 3

can be calculated by means of the formula:

M - 22.4D a (V)

It is more convenient, however, in determining molecular weight,
to make use of the density of one gaseous substance in terms of another.

The density D of one gas in terms of another is the ratio of the weight
of one to the weight of an equal volume of the second at the same tem-
perature and pressure.

The molecular weight of a gaseous substance is equal to its density
in terms of another gas multiplied by the latter's molecular weight:

MA^MB-D (VI)

Example. Calculate the molecular weight of a gas if equal volumes of it and of
chlorine under the same conditions weigh 0.240 and 1.065 g respectively.

First, it is necessary to determine the density of the gas in terms of chlorine:

0.240

Since the molecular weight 'of chlorine is 70.9, the molecular weight of the gas
M -70.9- 0.225 -15.95^ 16

The density of a gas is usually determined in terms of hydrogen
or air. The molecular weight of hydrogen is 2.016. Air is 14.38 times
heavier than hydrogen, and its mean molecular weight is 28.99. The
formulae for determining molecular weight are therefore:

The above formulae make it possible to calculate the molecular
weights of gaseous substances and those liquid and solid substances
that are vaporised at low temperatures without chemical decomposi-
tion.

The molecular weight is an important characteristic of a pure sub-
stance. Impurities alter the density of a gas (vapour) and lead to
errors in molecular weight determination.

Accordingly, all the necessary steps should be taken, in experimen-
tally determining the molecular weight of a gas, to remove possible
impurities from the gas. Chemically pure products should be used in
determining the molecular weights of highly volatile liquids and so-
lids.

2. Gas Purification. Gaseous impurities are removed by passing
a gas through certain substances (often in the form of solutions) that
react chemically with the impurity, but not with the principal gas.
If, for instance, it is necessary to remove CO 2 from a mixture of CO
and CO 2 , the mixture is passed through a solution of an alkali or a car-

Determining the Molecular Weight of a Gas (Vapour)

bonate of an alkali metal. The carbon dioxide in such cases reacts
according to one of the following equations:

2KOH
K 2 CO 3

CO 2
H 2 O

K 2 CO 3 + H 2 O
CO 2 - 2KHCO 3

The CO 2 admixture is in this way removed from the carbon mono-
xide.

Water vapour is removed from gases by passing it through bot-
tles containing dehydrating agents (concentrated sulphuric acid, cal-
cium chloride, zinc chloride, alkalis, phosphorus pentoxide, etc.).
Dehydrating agents are not equally effective: phosphorus pentoxide
and magnesium perchlorate are the best (Table 1).

Table 1

Effectiveness of Dehydrating Agents in Drying Air

Dehydrating agent

Residual
humidity
of gas in nig/1

Dehydrating agent

Residual
humidity
of gas
in mg/1

PoOs

2-10- 5

NaOH (molten)

Mg(ClO 4 )o

5-10~ 4

CaO

KOH (molten)

0.002

H 2 SO 4 (95.6%)

HoS0 4

003

CaClo (molten) . . .

CaSOa

004

CuSO 4 .

1 40

Gases cannot be dried by means of substances that react with them.
For instance, ammonia cannot be dried by sulphuric acid, or hydrogen
chloride by calcium oxide, to mention only two such cases. The types
of apparatus used for drying gases are shown in Fig. 17.

3. Partial Pressure. The pressure of each component of a mixture
of gases is called the partial pressure of that gas. If a gas is collected
over water, there will be a mixture of the gas and of water vapour over
its surface; the degree to which the water vapour saturates the gas will
depend upon the temperature.

(a) If the water in the cylinder in which the gas has been collected
and in the vessel into which the cylinder has been lowered is at the
same level (Fig. 18, /), the sum of the gas pressure p and the water
vapour pressure h equals the atmospheric pressure p:

To determine the gas pressure proper, i. e., its partial pressure, the
partial pressure of the water vapour has to be subtracted from the

Exercise 3

total pressure:

When the volume of a gas collected over water is'reduced to N. T. P.,
the difference;; h should be substituted for p in the equation of state,
which, accordingly, becomes:

Fig. 17. Types of apparatus for drying gases

/, //, and ///bottles with dehydrating agents; 7V drying" tower; V U-tubc.

The pressure of the water vapour, which is commonly called the
aqueous vapour tension, depends upon the temperature (Table 2).

Table 2

Aqueous Vapour Tension h (in mm Hg) at Temperature t (C)

9.2

13.6

19.8

28.3

9.8

14.5

21.1

30.0

10.5

15.5

22.4

31.8

11.2

16.5

23.8

55.3

12.0

17.5

25.2

92.5

12.8

18.6

26.7

100

760.0

(b) If the water in the cylinder and in the vessel is at different
levels (Fig. 18, //), the atmospheric pressure is balanced by the sum of

Determining the Molecular Weight of, a Gas (Vapour)

.the gas pressure //, the aqueous vapour tension /z, and the pressure
of the column of water p w (the latter, converted to pressure in mm Hg,
equals the height of the column in mm divided by the density of mer-
cury, i. e., by 13.6 g/cm 3 ):

13.6

The partial pressure of the gas can then be expressed:

13.6

ter..-

Fig. 18. Detennin ng the pressure of a gas collected over water

The equation for reducing the volume of a gas to N. T. P. then becomes:

/- M-V r.

This equation is used in experiments in which it is not possible to
bring the water in the cylinder and in the vessel to the same level.

If the gas is collected over some liquid other than water (a salt
solution), it is necessary to use the value of h for that solution.

QUESTIONS

1. How should the formula for the molecular weight of a gas be
written if its density is determined in terms of nitrogen or carbon
dioxide?

2. Which is the heavier: dry carbon dioxide or an equal volume of
the gas containing water vapour, other conditions being identical?
How can the same question be decided for methane (dry and contain-
ing water vapour)?

3. Knowing the volume of a gas at N. T. P., but not its den-
sity, how can we establish its weight?

30 Exercise 3

4. By means of the corollaries to Avogadro's Law determine
weight of an atom of nitrogen and the weight of a molecule of chlo-
rine dioxide.

5. Which of the substances listed in Table 1 can be used to remove
water vapour from carbon dioxide?

6. What is the partial pressure of carbon monoxide collected in
a cylinder over water at 17 U and 752 mrn Hg if the water in the cylin-
der and in the vessel is at the same level (use the data in Table 2)?

Problems

1. Equal volumes of chlorine and of nitric oxide weigh 3.16 and 1.34 g respec-
tively. Calculate the molecular weight of nitric oxide from its density in terms of chlo-
rine, knowing that the molecular weight of the latter is 70.9.

2. One litre of a gaseous substance at N. T. P. weighs 0. 76 g. A litre of air and
a litre of hydrogen under the same conditions weigh 1.293 and 0.09 g, respectively.
Calculate the molecular weight of the gas by three different methods.

3. A gaseous substance at 770 mm Hg and 27 occupies a volume of 3JO ml and
weighs 0.5 g. Calculate its molecular weight.

4. Given the molecular weights of oxygen and nitrogen, calculate the mean mo-
lecular weight of air on the assumption that it consists of 24% of oxygen and 76% of
nitrogen by weight.

5. Calculate the mean molecular weight of a gaseous mixture consisting of 60%
of carbon monoxide and 40% of hydrogen by weight, knowing that their molecular
weights are 28 and 2 respectively.

6. Calculate the molecular weight of a gas and its density in terms of air if 0.62
litre of the gas weighs 1.24 g at 21 and 750 mm Hg.

7. Calculate the molecular weight of a substance and the density of its vapour in
terms of nitrogen if 355 ml of the substance in the vaporous state weighs l.llg at 20
and 770 mm Hg.

8. The density of liquid chlorine is 1.5 g/cm 3 . How many litres of gaseous chlo-
rine can be obtained from 80 ml of liquid chlorine?

9. What is the weight of 410 ml of acetylene collected at a temperature of 17
and a pressure of 740 mm Hg if it is known that its molecular weight is 26?

10. Calculate the weight of the oxygen in a 5 X 2.6 X 2 m air-filled chamber con-
taining 21% by volume of oxygen if the temperature in the chamber is 17 and
the pressure is 752 mm Ilg.

LABORATORY WORK

Apparatus and materials: Kjpp gas generator for preparing carbon dioxide or
oxygen (with two wash bottles for gases); the*apparatus shown in Fig. 21; bottles with
water; 250 to 500 ml measuring cylinders; room thermometer; dry 250 ml flat bottom
flask with stopper and rubber band; barometer; glass ampoules; marble in lumps;
manganese dioxide catalyst; 1 :6 hydrochloric acid; concentrated sulphuric acid;
3% hydrogen peroxide solution; saturated solution of sodium bicarbonate; chloro-
form; carbon tetrachloride, and a metal ruler.

Preparation of manganese dioxide catalyst. Add water to a mixture of 150 g of ce-
ment and 70 g of granular manganese dioxide until it acquires a thick doughy con-
sistency. Then divide the mass into big lumps and leave them to dry overnight.

1. Kipp Gas Generator. Gases are often prepared by means of the
Kipp gas generator, an apparatus consisting of three glass bulbs*

Determining tne Molecular Weight of a Gas (Vapour)

usually of spherical shape (Fig. 19). Two of the bulbs are intercon-
nected. In the bottom one there is an outlet with a stopper, through
which the spent liquid is drained. The middle bulb also has an outlet
with a stopper, through which a glass tube with a tap has been passed.
Inserted into the top opening of this bulb is a funnel ground to fit
the opening; the funnel widens upwards into a bulb, while its lower
tip reaches the bottom of the generator.

HC1

Fig. 19. Arrangement for preparing and purifying carbon dioxide

To prepare carbon dioxide, lumps of marble are placed into the mid-
dle bulb, while hydrochloric acid is poured into the top one. When the
tap is opened, acid runs down the funnel into the bottom bulb and
rises into the middle bulb to come in contact with the marble. When
the tap is shut, the gas which is formed forces the acid out of the middle
bulb and into the upper one through the bottom one; this brings the
reaction to a halt.

The Kipp apparatus can be used to generate several other gases.
For example, it can be used to prepare oxygen from a 3% solution
of hydrogen peroxide, which decomposes to evolve oxygen under
the catalytic action of manganese dioxide.

2. Determining the Molecular Weight of Carbon Dioxide. Weigh
a dry stoppered flask with air on a chemical balance (weight Wi).
The stopper should be inserted into the neck of the flask to a definite
point marked by a rubber band. Now lower into the flask a glass tube
attached to the Kipp gas generator producing carbon dioxide (Fig. 19) .
The gas is first washed and dried by being passed through two wash
bottles for gases with a solution of sodium bicarbonate and concentrat-
ed sulphuric acid respectively. Remove the clip (Fig. 20) from the
rubber tubing. Open the tap of the Kipp gas generator and allow car-
bon dioxide to issue from it slowly for 7-8 minutes. Then stopper up
the flask to the marking and weigh it (oj 2 ). Pass carbon dioxide into
the flask again for 3-4 minutes, after which weigh it once more (w 3 ),
If the latter two weighings produce identical results or results that do
not differ by more than 0.01 g, the filling of the flask may be consid-

Exercise 3

ered completed. Otherwise repeat the filling procedure, until a constant ,
weight of the flask is achieved. This done, record the temperature and
the pressure by referring to a room thermometer and barometer.

To measure the volume of the flask, fill it to the marking with water
at room temperature (why?); then pour the water into a measuring

Fig. 20. Metal clips

/ screw clip; // spring clip.

cylinder, determine its volume, and record the results of the experi-
ment.

Results of experiment

Weight of flask
with air and
stopper

Weight of flask
with CO 2 and
stopper

Volume of
flask in ml

Temperature
t in C

Pressure
p in rum Hg

J.'o =

W 3 =

By means of the equation of state of a gas reduce the volume of
the gas in the flask to N. T. P. (V Q ).

Knowing that at N. T. P. 1 litre of air weighs 1.293 g, calculate
the weight of the air in the flask (ze/ 4 ).

The difference^ w& will be the weight of the empty flask with the
stopper (i0 B ).

The difference w 3 w$ will be the weight of the carbon dioxide in the
flask (WQ). From the weights w* and w calculate the density of carbon
dioxide in terms of air and the molecular weight of carbon dioxide
(to the second decimal). Calculate the absolute error and the relative
error of the experiment.

Make a sketch of the Kipp gas generator, explaining the principle
of its operation.

Write the equation of the reaction of carbon dioxide preparation.
Why is the CO 2 produced by the Kipp generator first washed with

Determining the Molecular Weight of a Gas (Vapour)

a sodium bicarbonate solution and then with sulphuric acid? What
does bringing something to a constant weight mean? What experi-
mental data are necessary to calculate the molecular weight of a gas?
3. Determining the Molecular Weight of Oxygen. Conduct this
experiment in the same way as the previous one. The oxygen needed
for the experiment is produced in the
Kipp generator and then dried by pas-
sage through a wash bottle for gases
tilled with concentrated sulphuric acid.
From the experimental data calcu-
late the molecular weight of oxygen (to
the second decimal) and determine the
absolute and the relative error.

How is it possible to find out whether
the air has been fully expelled from the
flask?

If the flask had been filled with un-
dried oxygen, would the figure obtained
for the molecular weight be bigger or
smaller than the true value?

4. Determining the Molecular
Weight of Volatile Substances. A
weighed quantity of a volatile substance
(chloroform, acetone, carbon tetrachlo-
ride, ether, etc.) is heated to convert
it into a vapour; the vapour expels
from the apparatus an equal volume of
air, which is measured in a eudiometer.
From the known weight of the substance
and its volume in the vaporous state,
reduced to N. T.P., it is possible to cal-
culate the molecular weight of the
volatile substance.

The apparatus employed for the de-
termination (Fig. 21) consists of a
tube (1), a bulb tube vaporiser (2), a
side arm (3), a vessel (4), and a eudio-
meter (5).

Pour 70-80 ml of water into the bulb
tube vaporiser and heat it to boiling
point. The air expanding in the tube is forced out through the side
arm and bubbles through the water in the vessel.

Weigh a glass ampoule (Fig. 22,4), heat its bulb slightly in the
flame of a burner, and immerse the open capillary into the liquid
whose molecular weight is to be determined. The cooling of the ampoule
causes the liquid to be sucked up into it (Fig. 225). When the

3795

Fig. 21. Apparatus for determin-
ing molecular weight

1 tube; 2 bulb tube vaporiser; 3
side arm; 4 vessel; 5 eudiome-
ter; 6 rod for supporting .impoule.

Exercise 3

liquid has risen 5 mm above the tapered part of the ampoule, with-
draw the ampoule from the jar and turn it over. Now weigh the ampoule
with the liquid. There should be no more than about 0.20-0.30 g
of the liquid. If there is more, turn the ampoule over and, by tapping
your finger on the capillary tube, remove the excess liquid. Then
weigh the ampoule again. Heat the capillary carefully on a small
flame of a burner until it softens (Fig. 22C), draw it out quickly,
and seal it (Fig. 22D).

B CD

Fig. 22. Filling and sealing ampoules

Lower the ampoule, capillary downwards, into the upper part of
tube 7, so that it rests on rod 6, and stopper up tube / tightly.

When no more air bubbles issue from tube 3, place it under the
eudiometer (make certain there are no more air bubbles!) and care-
fully pull rod 6. This causes the ampoule to fall to the bottom and the
thin capillary to break. The substance inside the ampoule is vaporised
and displaces its own volume of air, which ascends the eudiometer.

When the escape of air from tube 3 ceases, it is removed from the
water vessel. The volume of the displaced air and the distance from
the level of the water in the vessel to the level of the water in the cyl-
inder are measured (by means of mm ruler). These data, as well as the
temperature t and barometric pressure p at which the experiment is
conducted, are recorded. The value of /z, the aqueous vapour tension,
is found in Table 2 (p. 28).

From the experimental data calculate the molecular weight of the
substance. Compare the value found with the value derived from the
formula of the substance. Determine the relative error.

Note. If the experiment has to be repeated, air should be passed through
tube 1 to remove the remaining vapour (consult instructor).

Laboratory Techniques

Results of experiment

Ampoule weight

Weight of
substance
in g

Volume of
vapour in
ml

t In C

p in mm
Hg

h in mm
Hg

PS
in mm

with sub-
stance in g

empty in g

Exercise

LABORATORY TECHNIQUES

SUBJECTS FOR STUDY

1. Selection and Treatment of Stoppers. The materials used for
stoppers are the bark of the cork oak (these stoppers are simply
called corks), rubber, and glass.

Corks should be selected in such a way that the diameter of the ta-
pered end is slightly bigger than that of the vessel neck or tube. The
cork chosen should be carefully squeezed along
the circumference in a cork press (Fig. 23);
after this it should fit the vessel neck tightly
(not more than one-half to two-thirds of the
cork being inserted).

Corks are bored by means of special cork
borers (Fig. 24) or hand drilling machine
(Fig. 25). In selecting the bore, it is necessary

Fig. 23. Cork press

Fig. 24. Set of
cork borers

to choose one with a diameter somewhat smaller than the diameter of
the tube which is to be passed through the cork, or else the joint
will not be airtight. The boring should be done from the tapered, end

Exercise 4

of the cork, which should be held in the left hand (Fig. 26). The borer,
held in the right hand, should be centred and then turned to and fro',
a slight pressure being applied to the cork. When the borer has passed
through half the cork, it should be carefully withdrawn; the boring
should then be resumed from the wide ena of the cork. The borer should
always be perpendicular to the end of the cork. When the boring is
finished, the bits of cork left inside the hole should be pushed out by
means of a ramrod. If the borers become blunted, they should be sharp-
ened (consult the laboratory assistant).

Fig. 26. Boring corks

Fig; 25. Hand drilling machine

Fig. 27. Position of hands when
inserting tube into stopper

When boring rubber stoppers, it is advisable to put a drop of alkali
solution or glycerol on the stopper at the point of application of the
borer; this acts as a lubricant and facilitates boring.

When the boring is done by means of a hand drilling machine, the
bit should be fastened in a vertical position by a special nut and
centred on the tapered end of the stopper*. A hole is then drilled in
the stopper by turning the wheel.

When inserting a glass tube in the hole drilled, moisten the end of
the tube with water. If the tube is bent, do not hold it at the bend,
since the tube can break and cut your hand. The tube should be held
as shown in Fig. 27.

2. Processing Glass Tubes, To prepare simple laboratory apparatus,
it is enough to be proficient in cutting and drawing out glass tubes,
rounding off the ends, and making capillaries. Tubes of small diame-

* To avoid blunting the drill, place a wide cork or a thin wooden plank under-
neath the stopper.

Laboratory Techniques

ter are cut in the following manner: a file or special knife is used to
make a deep scratch at the place where it is required to cut the tube;
the tube is then taken in both hands in such a way that the thumbnails
are over the scratch; by bending and at the same time stretching the
tube, it is broken at the scratch (Fig. 28).

Tubes are best bent over a gas burner with a dovetail nozzle (Fig. 29).
This produces an extended flame which covers the required length

Fig. 28. Position of hands when breaking
scratched glass tube

Fig. 29. Dovetail nozzle

of tube. While continuously rotated, the tube should be heated until
it softens; it is then held at one end and allowed to bend under its
own weight. The use of a dovetail nozzle is essential when working

Fig. 30. Bent glass tubes

,4 correctly bent; B and C incorrectly bent.

with a nonluminous flame. The nozzle slit through which the gas
issues should be even; uneven heating may produce incorrectlv bent
tubes (Fig. 30).

Tube ends are rounded off on a nonluminous flame until they become
smooth. When the burner flame becomes yellow, this is a sure sign
that edge melting has begun.

To draw out a tube, first heat it slightly over a flame, then intro-
duce its middle part into the outer cone of the flame, and, while rotating
it between the thumb and index finger, heat it until it softens. Now
remove the tube from the flame and stretch it slowly outwards in both
directions until it is of the required diameter.

Capillaries are drawn out in the same way except that the glass
should be made softer and the stretching done somewhat faster (but
not in a jerk!).

Exercise 4

QUESTIONS

1. What advantage has a rubber stopper over a cork?

2. What conditions have to be observed in boring stoppers?

3. Why is it advisable to use a dovetail nozzle when bending glass
tubes?

4. Why is it advisable to blacken the heated part of a tube over
a smoking flame after treating it in a nonluminous flame?

5. How should a bent tube be inserted in the hole of a stopper?

LABORATORY WORK

Apparatus and materials: set of cork borers; knife for scratching tubes; round
file; burner nozzle; scissors; 750 ml wide mouth reagent bottle; test tube with a hole
in the bottom; straight funnel; refractory bulb tube; glass tube 5-6 mm in diameter;
corks and rubber stoppers; rubber tubing 5 mm in diameter, and asbestos millboard.

1. Laboratory Apparatus Components. Assemble the apparatus
shown in Fig. 31.

Fig. 31. Apparatus for reducing cupric oxide

bottle; 2 funnel; 3 test tube; 4 wash bottle for gases; 5 bulb tube;
6 gas-delivery tube.

The apparatus consists of a bottle (1) with a stopper through which
a funnel (2) and a test tube (3) with a hole in the bottom have been
passed. The test tube has a rubber stopper with a bent tube passed
through it. This is connected by short lengths of rubber tubing with
a wash bottle for gases (4) and a bulb tube (5). The latter is con-
nected to a short length of glass tube (6) bent at a right angle.

Determining the Chemical Formula o/j a Substance 39

To assemble the apparatus, it is necessary to prepare two glass
tubes bent at a right angle and to select stoppers for the bottle
(cork) and for the test tube (rubber).

2. Selection of Stoppers. Bore two holes in a cork: one for the funnel
and the other for the test tube, selecting the borers accordingly.
If the diameter of the biggest borer is smaller than that of the test
tube, widen the hole in the cork by means of a round file. The hole
should enable the test tube to be moved easily up and down. In the
centre of the rubber stopper bore a hole equal in diameter to the glass
tube chosen.

3. Preparation of Tubes. Cut two lengths of glass tube of about
15 cm each and one 30 cm length. Round off their ends and bend the
short tubes at right angles*. Do not place the heated glass tubes direct-
ly on the table: place them on asbestos millboard.

4. Assembly of Apparatus. Insert one of the bent tubes in the rub-
ber stopper so that its end should not protrude from the tapered end
of the stopper.

Make certain that the apparatus assembled according to Fig. 31
is airtight. For this purpose pour water into bottle / so that it is half
full, lower test tube 3 into it, and blow air orally into tube 6\ if the
air bubbles through the water, the apparatus is airtight.

Show the apparatus to the instructor. If it is approved, take it
apart and put the parts away for the next exercise.

5. Making of Capillaries. Make four capillaries from a glass tube. For
this purpose, cut a 15-17 cm length of tube and draw out a capillary
tube of 1.5 mm diameter from it. Mark it off carefully into 45 mm
lengths and break it at the scratches.

This done, seal the capillaries at one end and show them to the
instructor. If the capillaries are found satisfactory, put them away
for Exercise 7.

Exercise 5

DETERMINING THE CHEMICAL FORMULA
OF A SUBSTANCE

SUBJECTS FOR STUDY

Analysis of compounds; calculation of percentage compositions; atoms and atom-
ic weights; determination of atomic weights; Dulong and Petit 's rule and Canniz-
zaro's method; derivation of empirical formulae; true formulae of substances; qual-
itative and quantitative meaning of chemical symbols and formulae; equations of
reactions, and stoichiometry.

* The straight tube and the capillaries will be needed for the first experiment in
Exercise 7.

40 Exercise 5

1. Atomic Weight. This is the most important characteristic of
an element. The atomic weight is a number indicating how many times
an atom of a given element is heavier than 1/12 of a carbon atom. Approx-
imate values of atomic weights (for the heavier elements) can be de-
termined on the basis of Dulong and Pewit's rule: the product of the
specific heat of a simple substance in the solid state by the atomic weight
of the element at average temperatures equals about 6.2. The specific
heat can be determined from the thermal balance equation. If a cer-
tain mass m l of a substance with a temperature t l and specific heat c l
is placed in water whose mass is m^ and temperature t 2 (/ 2 < *i),
the temperatures level out to t 3 . If we assume the specific heat of water
to be 1, we get:

whence:

/7t2 (^3

Example. 100 g of metallic vanadium is heated to 52 and then immersed in 120
g of water (at 16.8). The water is thereby heated to 20. Calculate the atomic weight
of vanadium.

By substituting the figures given into the thermal balance equation, we obtain:

120-3.2
100-c(52 20) = 120-1(20 16.8) or c= 1Q0 . 32 " = - 12

The atomic weight (A) will then be:

6.2

= 51.7

/1 - 0.12
(the exact atomic weight of vanadium is 50.942).

To determine the atomic weight of an element according to Canniz-
zaro's method, we must take several volatile compounds of that ele-
ment and, by analysis, establish their percentage composition and
molecular weight. By then setting up a proportion, we find the weight
of the element per molecular weight of its compound. After this has
been done for several compounds, the least of the weights is assumed
to be the atomic weight,

Example. Determine the atomic weight of phosphorus from the following data con-
cerning its compounds.

Molecular Phosphorus

weight content in %

Phosphorus hydride ... 34 91.2

Phosphorus trichloride . . 137.5 22.5

Phosphorus hydride (liquid) 66 94.0

Determining the Chemical Formula of a Substance 41

A proportion is set up for phosphorus hydride:

100 parts by weight contain 91.2 parts by weight of phosphorus
34 parts by weight contain Xi parts by weight of phosphorus

34 91.2

Xl - 100 - 31

From similar proportions for the other two compounds we obtain:

137,5.22.5 66-94.0

= 31 and x 3 = = 6L

The smallest of the obtained values (i. e., 31) is assumed to be the atomic weight
of phosphorus.

Approximate atomic weights can be calculated from Mendeleyev's
Periodic Table as the arithmetic mean of the atomic weights of the
neighbouring elements (to the left and right, as well as above and
below the given element).

Example. Calculate the atomic weight of vanadium from the atomic weights of
the neighbouring elements: phosphorus, arsenic, titanium, and chromium.
We find the arithmetic mean of the four atomic weights.

30.98 + 74.91 + 47.90 + 52.01
x = - -7 - = 51.45

(the exact atomic weight of vanadium is 50.942).

Atomic weights approximated to the first decimal can be found by
the method of determining equivalents. Modern physical methods
(mass spectrography) make it possible to determine the relative masses
of atoms with an accuracy to the third and even the fourth decimal.

2. Chemical Formulae. The tiniest particle of a complex substance
which retains all its chemical properties is called a molecule. Chem-
ical formulae are conventional designations of the molecules of sub-
stances. By means of formulae we express the qualitative and quanti-
tative composition of molecules. For instance, the formula N^H^
indicates that a substance consists of nitrogen and hydrogen. This is
its qualitative composition. Without knowing the values of x and y,
we cannot say what the substance is. It might be NH 3 , N 3 H, or N 2 H 4 .
To find the formula of the substance, we must determine the number
of atoms in the molecule, or, in this particular case, determine the
values of x and y.

(a) The simplest (empirical) formulae give the relative numbers
of atoms in the molecule.

To write the empirical formula of a substance, we must know its
composition by weight and the atomic weights of its constituent
elements.

Example. Find the empirical formula of a substance consisting of 87.5% of nitro-
gen and 12.5% of hydrogen, knowing that the approximate atomic weights of these
elements are 14 and 1, respectively*

42 Exercise 5

We denote the formula as Nj^H^ and find the ratio of the number of atoms of one ,
element to the number of atoms of the other:

87 5 12 5
x: y= -jy- : -f- = 6.25 : 12.5 = 1 : 2

This means that the molecule of the substance contains twice as many hydrogen
atoms as nitrogen atoms. Its empirical formula is therefore NH2.

(b) The true, or molecular, formula gives not the relative, but the
actual number of atoms in the molecule. To arrive at the molecular
formula of a compound, we must know not only its percentage compo-
sition and the atomic weights of the constituent elements, but also
the molecular weight. In the case of some substances the empirical
formula and the molecular formula coincide.

Example. Find the molecular formula of the substance from the data of the pre-
vious example and the additional information that the density of its vapour in terms
of air D air ^ 1.1.

We first find the empirical formula of the substance, as in the previous example.
Since the ratio x : y = 1 : 2, the molecular formula can be NH2, or N2H4, or NaHi,
etc.

From the formula M =29D fl/r , we find the molecular weight M = 31.9^32.

If we assume that the empirical formula NFU is the molecular formula, the molec-
ular weight adds up to 144-2-1=16. We know, however, that the molecular weight
is double that figure. Consequently, the molecule contains twice as many atoms.
The substance must therefore have the formula N2H4 and be hydrazine.

To find the formula of a simple substance, one must know its molec-
ular weight and the atomic weights of its constituent elements.

Example. Determine the composition of a molecule of white phosphorus, knowing
that its atomic weight is 30.98 and its absolute vapour pressure is 5.53 g/1.
First, we must calculate the molecular weight:

M =22.4> fl =22.4.5.53^123.9

The number of atoms in the molecule is found from the ratio 123.9:30.98=4.
The composition of the molecule is therefore expressed be the formula P*.

From the chemical formula of a gaseous substance we can find some
of its quantitative characteristics: the percentage composition, the
molecular weight, the density in terms of any other gas, and the abso-
lute weight of the molecule.

Chemical formulae also make it possible to calculate the quanti-
ties (by weight and by volume) of the reactants and resultants of
a reaction from its equation. All such calculations involving the
use of chemical formulae and equations are called stoichiometric.
The term stoichiometric amounts means amounts corresponding to the
equation of the reaction.

3. Nomenclature of Inorganic Compounds. The following principles
form the basis of the international nomenclature of salts in chemical
literature.

Determining the Chemical Formula of a Substance 43

The names of salts are formed from the name of the metal used
adjectivally and followed by the name of the electronegative ion or
anion *. Metals which form more than one oxide and give rise to cor-
responding series of salts have the different series distinguished by
an inflecting suffix to the name of the metal. The suffix ous is used
to indicate the lower and the suffix ic to indicate the higher condition
of oxidation, as in the copper compounds cuprous chloride (CuCl)
and cupric chloride (CuCl 2 ). The name of the electronegative ion or
anion is derived from the name of the element which forms it. In the
case of salts of acids containing no oxygen, the name of the electro-
negative element is inflected by the suffix ide. In the case of salts of
oxygen-containing acids, the termination ite is used for the lower
valence of the element forming the acid, while the termination ate
is used for the higher valence. Examples: CaS calcium sulphide;
CaSO 3 calcium sulphite; CaSO 4 calcium sulphate; Cu 2 SO 4
cuprous sulphate, and CuSO 4 cupric sulphate.

For acid salts the name of the radical is inflected by the prefix
hydro (or bi). Examples: NaHS sodium hydrosulphide; NaHSO 3
sodium hydrosulphite (bisulphite), and NaHSO 4 sodium hydro-
gen sulphate (bisulphate).

For basic salts the name of the radical is inflected by the prefix
hydroxy (salts containing the hydroxyl) or oxy (salts containing oxy-
gen). Examples: Bi(OH) 2 NO 3 bismuth hydroxynitrate (subnitrate)
and BiONO 3 bismuth oxynitrate (bismuthyl nitrate).

QUESTIONS

1. Why has the oxygen unit for measuring atomic weights been
given up in favour of the carbon unit?

2. For which elements do the atomic weights found according to
Dulong and Petit 's rule differ considerably from the real values?

3. From the formula of ammonia derive its quantitative character-
istics (molecular weight, absolute density and relative density
in terms of air, and percentage composition).

4. How are absolute weights of atoms and molecules calculated?
Calculate the absolute weights of the copper atom and the phosphorus
hydride molecule.

5. What is the qualitative and quantitative meaning of the equa-
tions:

2KOH + H 2 SO 4 = 2H 2 O + K 2 SO 4 (weight units)
4NH 3 + 5O a = 4NO + 6H 2 O Mp0ttr (volume units)

* The names of some anions (acid radicals) are given on the inside back cover
of the book.

44 Exercise 5

6. Explain why Fe 2 O 3 is called ferric oxide, while FeO is called
ferrous oxide. Explain why H 3 AsO 4 is called arsenic acid, while
HgAsOg is called arsenous acid.

7. Name the following salts:

Mg 3 N 2 , Mg(NO 2 ) 2 , Mg(NO 3 ) 2 , Na 2 HAsO 8 , NH 4 VO 3 , and A1OHC1 2

Problems

1. Calculate the atomic weight of carbon from the following data:

Molecular Carbon content

weight in %

Carbon dioxide 44 27.3

Ethyl alcohol 46 52.2

Acetone 58 61.2

Benzene 78 92.3

2. Calculate the atomic weight of chlorine from the following data:

Molecular Chlorine

weight content in %

Hydrogen chloride .... 36.5 97.3

Potassium chloride ... 74.6 47.6

Silicon chloride 170.1 83.5

Chloroform 119.5 89.1

3. A piece of metal weighing 40 g was heated to 79.2 and immersed in 80 g of wa-
ter at a temperature of 17. 1, the water thereby being heated to 20. Calculate the atom-
ic weight of the metal on the basis of Dulong and Petit's rule.

4. Vanadium consists of two stable isotopes: 0.23% of V 50 and 99.77% of V 51 .
Calculate the mean atomic weight of vanadium.

5. Determine the formula of a substance containing 64.87% of C, 21.62% of O f
and 13.51% of H, if the density of its vapour in terms of air is 2.56.

6. Determine the formula of a substance containing 24.24% of C, 4.05% of H,
and 71.71% of Cl if its density in terms of air is 3.42.

7. The burning of a substance containing carbon, hydrogen, and chlorine yields
0.22 g of CO 2 and 0.09 g of water. When an identical amount of the same substance
was analysed for chlorine content, 1.44 g of silver chloride was obtained. The vapour
density of the compound in terms of hydrogen was found to be 42.5. Determine the
formula of the substance.

8. The formula of a substance is P2H 4 . Determine its percentage composition, its
density in terms of hydrogen and in terms of air, and its absolute density.

9. How much will the pressure in an oxygen-filled vessel diminish if 8.7% by vol-
ume of the oxygen is converted to ozone?

10. How much zinc will be dissolved in 1.5 kg of 35% hydrochloric acid, and
what will be the volume of hydrogen evolved (p ==770 mm, t =17)?

LABORATORY WORK

Apparatus and materials: the parts of the arrangement shown in Fig. 31; desic-
cator with sulphuric acid; test tubes and rack; copper or iron wire 1.5-2 mm in diam-
eter; absorbent cotton-wool; dilute (1 : 6) sulphuric acid; granulated zinc, and cal-
cined granulated cupric oxide.

Determining the Chemical Formula of a Substance 45

Note. The cupric oxide should be calcined thoroughly in an open crucible, cooled
in a desiccator, and put in a jar with a ground-in lid,

Determining the Formula of Cupric Oxide *. Assemble the appa-
ratus shown in Fig. 31, using the parts prepared in Exercise 4.

Weigh the bulb tube (5) and place about Ig of granular cupric
oxide in its bulb (if powdered cupric oxide clings to the inner surface
of the tube, remove it with a wad of cotton-wool attached to a wire).
Then weigh the tube again to 0.01 g and connect it up with the rest
of the apparatus, as shown in Fig. 31, making sure that the tube is
properly inclined.

Now place 15-20 lumps of zinc in the test tube and pour enough
sulphuric acid into the bottle through the funnel to immerse one-
third of the test tube into the acid. Lower the test tube into the acid.
Upon coming in contact with the acid, the zinc begins to react with
it, and the hydrogen evolved gradually forces the air out of the appara-
tus. Water vapour is removed from the hydrogen by passing the latter
through a wash bottle containing sulphuric acid (see Fig. 31).

To make certain that all the air has been forced out of the apparatus,
a test tube is placed over the end of the delivery tube (6). This test
tube should be removed every 2 minutes and, without being turned over,
held over the flame of a burner. Pure hydrogen burns quietly,
whereas hydrogen mixed with air produces a slight explosion. If the
evolution of hydrogen subsides, more acid should be added through
the funnel.

After the air has been forced out, heat the bulb of the tube careful-
ly, continuing reduction until the cupric oxide becomes red (the cc
lour of metallic copper). If drops of water should form at the end of
the bulb tube, heat that part of the tube slightly. Then remove the
burner and allow the tube to cool with hydrogen flowing through it
until room temperature is achieved (why?). Remove the test tube con-
taining the zinc from the acid, disengage the tube with the copper in
it, and, after cooling it for 10 minutes in a desiccator, weigh it (oy t ).
Attach the tube to the rest of the apparatus again, continue the reduc-
tion, and, after cooling, weigh the tube once more (o; 2 ).

If the weights w l and w% are equal or differ by not more than 0.01 g,
reduction may be considered completed. Otherwise, the process of
reduction should be continued until a constant weight is reached.

From the data obtained calculate the percentage composition of
cupric oxide and establish its empirical formula.

Write the equations for the reactions by which hydrogen was pro-
duced and the cupric oxide reduced.

* The formulae of lead oxide, bismuth oxide, etc., can also be determined. The
oxides of lead and bismuth are reduced to a drop of molten silvery metal.

46 Exercise 6

Calculate the volume of hydrogen (reduced to N. T. P.) used up
to reduce the amount of cupric oxide taken and the amount of zinc
and sulphuric acid (24%) consumed.

Why should not the bulb tube be heated from the start of the expe-
riment?

Why must the hydrogen entering the bulb tube be dried?

Why is the tube cooled with hydrogen flowing through it?

Why must reduction be repeated?

Exercise 6
DETERMINING CHEMICAL EQUIVALENTS

SUBJECTS FOR STUDY

The concept of chemical equivalents; hydrogen equivalents; relationship between
equivalent, valence, and atomic weight; derivation of atomic weight from equiva-
lent; equivalent of a compound; valence; determination of valence; formation of
molecules, and electrovalent and covalent bonds.

Chemical Equivalents. A chemical equivalent is the weight of a sub-
stance which will replace or combine with 8 parts by weight of oxygen
or 1.008 parts by weight of hydrogen. The quantity of a substance
equal in grams (milligrams) to its equivalent is called a gram-equiva-
lent (milligram-equivalent).

Substances react in equivalent amounts. If we know the equivalent
and valence of an element, we can determine its atomic weight, as
these quantities are linked by the following relationship:
atomic weight =equivalentx valence *

Experimentally the equivalent of an element can be found chemi-
cally: by estimating the amount of hydrogen, oxygen, or other ele-
ment of known chemical equivalent added or replaced by the given
element.

Chemical equivalents can also be determined elect rochemically on
the basis of Faraday's Law, whereby the passage of 96,496 coulombs
of electricity through an electrolyte solution always deposits one gram-
equivalent of a substance on the electrodes.

The guiding principle in calculating the chemical equivalents of
acids, bases, and salts from their formulae is the principle that the
chemical equivalent of an acid is equal to its molecular weight divided
by the basicity of the acid, i.e., by the number of hydrogen atoms
contained in the acid molecule and capable of being rep laced by a met-
al. Similarly, the equivalent of a base is equal to its molecular weight
divided by the valence of the metal or the number of hydroxyl groups

This relationship is inapplicable to inert gases.

Determining Chemical Equivalents 47

. in the molecule. The equivalent of a salt is equal to its molecular weight
divided by the product of the number of atoms of the metal in the
salt molecule by the valence of the metal.

Examples:

98
HsPO 4 mol. weight 98, equivalent -g- =32.7

74
Ga(OH) 2 rnol. weight 74, equivalent -y = 37

342
AU (804)3 mol. weight 342, equivalent -^=57

If we know in what amounts HC1 and AgNOa interact and the equivalent of HC1,
we can experimentally determine the equivalent of AgNOs. Thus, if a parts by weight
of HC1 react with b parts of AgNOa, we can, knowing that the equivalent of HCl =
= 36.5, calculate the equivalent of AgNOs= x from the proportion:

b __ 36.5- fr
36.5 x X ~~ a

QUESTIONS

1. Calculate the exchange equivalents of the following substances
from their formulae:

SiH 4 FeA Na 3 PO 4

2. Write the equation for the burning of SiH 4 and calculate the
substance *s equivalent. Compare this with the exchange equivalent
and explain the discrepancy.

3. Determine the valence of As, P, N, and S in the compounds:

H 4 As 2 O 5 K 2 HPO 4 Al (NO 3 ) 3 ZnSO 4

4. Write the structural formulae of the following compounds:

CIA H 2 Mn0 4 A1(NO 3 ) 3 Na 2 B 4 O 7 Ca 3 (PO 4 ) 2
Na 2 HPO, Bi (OH) 2 NO 3

5. Give diagrams for the formation of the ionic molecule K 2 S and
the atomic molecule SiH 4 . Explain the mechanism giving rise to elec-
trovalent and covalent bonds.

Problems

1. When 1.11 g of a metal was dissolved in acid, 404.2 ml of hydrogen (measured
,at 19 and 770 mm) was evolved. Calculate the equivalent of the metal and its atomic
weight, knowing that the metal is bivalent.

2. Calculate the equivalent of a metal if 3.4 g of its iodide contains 1.9 g of iodine,
whose equivalent is 126.9.

3. Calculate the equivalent of a metal if 0.347 g of it releases 180 ml of hydrogen
(at 15 and 748 mm) from water.

Exercise 6

4. Silver nitrate is used to precipitate 1.49 g of potassium chloride, which con-
tains 52.36% of- potassium. The weight of the silver chloride obtained is 2. 868 g. Cal-
culate the equivalent of silver if the equivalent of potassium is 39.

5. Find the equivalent of phosphoric acid if 1.68 g of potassium hydroxide, whose
equivalent is 56, is used up to neutralise 9.8 g of its 10% solution.

6. The heating of 4.3 g of a metal oxide yielded 580 ml of oxygen (at 17 and
850mm). Calculate the equivalent of the metal, knowing that 1 litre of oxygen
weighs 1.43 gat N. T. P.

7. Calculate the equivalent of CaCOa, knowing that 20 g of 7.3% hydrochloric
acid dissolves 2 g of it.

8. The reduction of 1.305 g of manganese dioxide by aluminium yields 0.825 g
of manganese. Calculate the equivalent of the latter.

9. The heating of 0.954 g of a metal in oxygen yields 1.194 g of its oxide. Calcu-
late the atomic weight of the metal, knowing it to be bivalent.

10. Calculate the atomic weight of a metal if 5.6 litres of oxygen (measured at N.
T. P.) was used up in burning up 30.67 g of it. The specific heat of the metal is known
to be 0.033.

LABORATORY WORK

Apparatus and materials: the arrangement shown in Fig. 33; drying cabinet;
10-25 ml measuring cylinder; 100 ml beaker; wash bottle; burette; funnel for burette;
porcelain casserole; weighing bottle; watch glass; forceps; room thermometer; desic-
cator; barometer; marble in lumps; N solution (titrated) of hydrochloric acid; 1:3

hydrochloric acid; 1:1 nitric acid; metallic sodium;
weighed amounts of the metals zinc, magnesium,
manganese, and aluminium; magnesium ribbon; ethyl
alcohol; phenolphthalein solution, and filter paper
(strips).

Note. The metals in the form of ribbon or shav-
ings should be weighed on an analytical balance to
0.001 g. The amount of zinc should range from 0.030
to 0.040 g; the amounts of magnesium and aluminium,
from 0.010 to 0.014 g; the amount of manganese should
be 0.026 g. Each weighed amount should be wrapped
in paper, and the weight written on it.

H " 1. Determining the Chemical Equivalent

/ 4- of Calcium Carbonate Weigh a clean 100

ml beaker on a chemical balance (weight
a>i). Put several lumps of clean marble in
it, and weigh it again (&y 2 ). The marble
should weigh 2.50-3.00 g.

rj[ Now measure 15 ml of a normal (N) so-

:_ lution of hydrochloric acid (a normal solu-

tion contains 1 gram-equivalent of a substance
in a litre). To do this, first pour 3-4 ml of
the acid solution into the burette through a
funnel to wash the inner walls of the burette;
pour the solution out through the stop-cock.
Repeat this procedure with the same

amount of the solution. Then fill the burette with the solution in
such a way that there are no air bubbles left in the tip of the burette

Fig. 32. Burette and how
to take a burette reading

Determining Chemical Equivalents

(consult the instructor) and that the level of the liquid is slightly above
the zero mark (remove the funnel from the burette). Burette readings
may differ depending upon the position of the observer's eye in re-
lation to the meniscus (Fig. 32). The eye should therefore always be
on a level with the lower meniscus of the liquid (Fig. 32 B). Open
the stop -cock gently to drain the extra acid, bringing the lower me-
niscus down to the zero graduation (Fig. 32). Now place the beaker
with the weighed marble under the tip of the burette and, opening
the stop-cock carefully once more, pour acid into the beaker until the
lower meniscus descends to the mark 15.

The reaction which takes place in the beaker involves the evolution
of the gas CO 2 . Write the equation of the reaction. When the evolu-
tion of the gas bubbles slows down considerably, heat the beaker to
60-70. Bubble formation will at first be speeded; when it stops, the
reaction may be considered completed. Now pour the solution formed
out of the beaker and wash the remaining lumps of marble twice with
distilled water by the method of decantation. Dry the beaker with
the remainder of the marble in a drying cabinet and, after cooling it
in a desiccator, weigh it (^ 3 ).

Record the results of the experiment as follows:

Weight of the beaker in g

Weight of the
marble in g

Amount of
marble to
react in g

Acid

empty

with marble

before
reaction

after
reaction

normality

amount
taken in ml

before exp.

jafter exp.

tt>3

W2 W\

WaWl

W2 Ws

From the experimental data calculate the chemical equivalent of
calcium carbonate.

2. Determining the Hydrogen Equivalent of a Metal. The apparatus
for the determination (Fig. 33) consists of two long tubes that are at-
tached vertically to a stand and communicate by means of rubber tub-
ing. One of the tubes is graduated (a burette); connected to it by
a stopper is a test tube with a slight bulge in the upper part. The
second tube is open at the top. The tubes are filled with water.

Before beginning the experiment, make sure that the apparatus is
airtight. This is done by the following procedure. When the appara-
tus has been assembled, raise and lower the tube that is open at the
top; the level of the water in the other tube should neither rise nor drop
much, but should only fluctuate slightly. If this is not the case,
insert the stoppers more tightly. Once the apparatus has been proved

Exercise 6

C-'-'

airtight, adjust the level of water in the tubes at the same height and
attach the tubes to the stand in this position. Record the level in the
graduated tube with an accuracy to 0.05 ml (according to the lower me-
niscus).

Now pour 3-4 ml of dilute (1 : 3)* hydrochloric acid into the test
tube by means of a funnel, taking care not to wet the bulge in the

upper part of the test tube] (if
this should be wetted, dry it with
filter paper). Now obtain a
weighed amount of metal from the
laboratory assistant, and record
its weight in ytmr notebook. Lift
the test tube into the position
shown by the broken line in
Fig. 33 and carefully transfer the
weighed metal into the bulge of
the test tube, taking care that
none of it should fall into the
acid. Stopper up the test tube
tightly again while it is still in
this inclined position, and then
shake it, allowing the metal to
fall into the acid. The pressure of
the hydrogen formed by {the in-
teraction of the metal and the
acid forces the liquid out of the
burette into the other tube.

When the reaction ends, allow
the test tube to cool and then
bring the water in the tubes to the
same level by lowering the tube

Fig. 33. Apparatus for equivalent
determination

open at the top. Record the new position of the level in the graduat-
ed tube. Record the readings of the thermometer and the barometer.

Results of experiment

Amount of
metal in g

Conditions of experiment

Position of level in burette
in ml

Volume of
hydrogen in
ml

temperature
in C

pressure In mm

before exp.

after exp.

* This designation means that the dilute acid was obtained by mixing one vol-
ume of concentrated acid with three volumes of water.

Determining Chemical Equivalents 51

From the data of the experiment, calculate the equivalent of the
metal. In the calculation, take into consideration the aqueous vapour
tension (Table 2, p. 28), as the gas was collected over water. Will
the equivalent of the metal be bigger or smaller if: (1) the metal con-
tains impurities insoluble in acid; (2) no allowance is made for the
aqueous vapour tension?

3. Determining the Oxygen Equivalent of a Metal. Weigh a dry
porcelain casserole. Then weigh 0.18-0.22 g of magnesium ribbon irr
it. Subsequent operations should be conducted in a ventilated hood.
Pour dilute (1 : 1) nitric acid in small portions (2-3 ml) into the cas-
serole until the whole of the metal dissolves. When this has taken
place, put the casserole on a wire gauze with an asbestos centre and
evaporate slowly (without splashing) until there is no water left. Then
heat the casserole with the dry salt, causing it first to melt and then
to decompose, with the evolution of a brown gas and the formation
of white magnesium oxide. Following this, cool the casserole in a des-
iccator and weigh it. To make sure that the salt has all been decom-
posed, repeat the calcination and weighing. If the results of two con-
secutive weighings are identical or differ by 0.01 g, decomposition is
complete. From the weights of the metal and the resulting oxide, cal-
culate the oxygen equivalent of the metal and the relative error of
the determination.

Submit a report on the experiment with the results obtained for
the equivalent and a brief description of the method to the instructor.

4. Determining the Chemical Equivalent of Chlorine from the Equiv-
alent of Sodium. Weigh a dry porcelain casserole, pour 15 ml
of ethyl alcohol into it, and cover it with a watch glass. Weigh a dry
weighing bottle. Obtain a lump of dry metallic sodium from the lab*
oratory assistant, place it quickly into the weighing bottle, close
the bottle tightly, and weigh it again (the sodium should weigh about
0.20 g).

Now transfer the weighed sodium with forceps to the casserole and
cover it at once with the watch glass. Write the equation for the inter-
action of the sodium with the alcohol. When the reaction ends, wash
the splashes of the solution from the surface of the watch glass back
into the casserole, using a small amount of water from a wash bottle
for this operation. Add 1-2 drops of phenolphthalein to the solution.
What is the colour of the solution now? Write the equation for the
reaction of sodium ethylate with water.

Pour M HC1 into a dry burette and add it to the solution in the cas-
serole until the crimson colouration disappears. Place the casserole
on a wire gauze with an asbestos centre and evaporate the solution
over a small flame until there is no water left in the casserole. Cool
the casserole in a desiccator for 7-10 minutes and weigh it. What is
the substance obtained? Write the equation for the reaction that pro-
duced it.

52 Exercise 7

From the weights of the sodium chloride and the sodium, knowing
the chemical equivalent of sodium, calculate the chemical equivalent
of chlorine and the relative error of the determination.

Exercise 7

DETERMINING THE PURITY OF A SUBSTANCE
SUBJECTS FOR STUDY

Pure substances and their properties; isolating individual substances from mix-
tures; purifying substances and determining the degree of their purity; grades of
purity (commercially pure, analytical reagent, and chemically pure), and standard
requirements with respect to chemical products.

Every pure substance has certain characteristic physical proper-
ties: colour, taste, odour, density, melting point and boiling point,
hardness, viscosity, etc.

When we say that water is a colourless, odourless, and tasteless
liquid that boils at 100 C, freezes at C, has a density of 1 g/cm 3
at 4 C, etc,, we are referring to pure water without any impurities,
i. e., to the individual substance.

However, despite the exceptional progress made by reagent chem-
istry, it should not be thought that what is termed a "pure substance**
does not contain minute impurities of other substances.

In the U.S.S.R. chemical products are manufactured in accord-
ance with technical standards known as GOST (which is the Russian
abbreviation for U.S.S.R. State Standard); these standards specify
exact requirements with respect to the quality of such products as
chemical reagents.

For instance, the following are the maximum permissible impurities, according
to GOST, in the different reagent grades of sodium hydroxide.

Maximum Permissible Impurities (in %)

Chemically Analytical Pure

pure reagent

Chlorides 0.005 0.01 0.025

Sulphates 0.005 0.01 0.03

Phosphates 0.003 0.005 0.01

Silica 0.01 0.02 0.1

Heavy metals 0.003 0.003 0,003

Calcium 0.012 0.024 0.06

Sodium carbonate 1 2 4

Iron 0.0005 0.001 0.002

Substances precipitated by ammonia

(A1 2 O 3 , Fe 2 O 3 ) 0.01 0.02 0.1

Total nitrogen (nitrates, nitrites, am-
monia, etc.) 0.001 0.001 0.001

Determining the Purity of, a Substance 53

Factory-made products are analysed at the factory laboratory;
on the basis of the analysis and the relevant QOST, they are graded
"chemically pure", "analytical reagent ", "pure", or "commercially
pure". The product is then packed and labelled accordingly.

The "chemically pure" substances are the purest. The amounts of
impurities in them are so negligible that they can be detected only
by special analytical techniques. Products marked "analytical re-
agent" may contain impurities in amounts that cannot prevent the use
of these substances in chemical analysis.

Products marked "pure" contain greater amounts of impurities than
products labelled "chemically pure" or "analytical reagents". The
"commercially pure" products usually contain the most impurities.
They are cheaper and are used for industrial purposes,

Of great practical importance is the development of techniques for
isolating pure substances from natural materials and from industrial
products or waste, as well as for testing the purity of substances.

Purity Testing Techniques. The purity of a substance of known com-
position can be tested either by physical methods (determination of
density, melting point, boiling point, etc.) or by methods of chemical
analysis (by treating a weighed amount of the substance with a reagent).
In the latter case (quantitative analysis) the composition of the sub-
stance being tested is determined from the quantity (by weight or
by volume) of the resultant substance or from the expenditure of the
reagent. The detection of impurities may sometimes be confined to
qualitative analysis, i.e., to employing reactions that in the case of
a certain impurity yield products easily distinguishable by visible
signs (the formation of a precipitate, a change in colour, etc.); the
quantity of the impurity is not estimated in this case.

Below we shall consider some of the methods used in assessing the
quality of a substance, namely, the determination of the melting and
boiling points of a substance, the quantitative estimate of a substance
according to the volume of a gas evolved when the product being
tested is decomposed by reagents, and the qualitative analysis of a
chemical product according to standards.

(a) The melting point of a substance is one of its most important
characteristics; by means of this characteristic we can determine the
purity of a solid crystalline * substance. The melting point is the tem-
perature at which a solid substance, in conditions of equilibrium,
passes into the liquid state.

In the case of a pure crystalline substance the temperature range
from the moment melting begins (the appearance of a liquid phase)
to its comlpetion usually does not exceed 0.5. Substances contain-
ing impurities, on the other hand, do not have a sharply defined
melting point; they usually melt over a range of several degrees.

Amorphous substances do not have a distinct melting point.

Exercise 7

Melting point determinations are carried out as follows. A finely
ground powder of the substance being tested is packed tightly in
a 2-3 mm layer into a capillary. Two capillaries are filled in this way.
A capillary is then attached to a thermometer by a rubber band
(Fig. 34,4). The thermometer, fitted into the hole of a stopper, is
then put into a dry test tube, which serves as something like an "air
sleeve" to prevent overheating. The test tube is lowered into a beaker,
flask, or special vessel (Fig. 345 and C) filled with water, paraffin
butter, or concentrated sulphuric acid, depending upon the melting
point of the substance. The beaker or flask is placed on the ring of
a stand and heated slowly.

Fig. 34. Types of apparatus for melting and boiling point determination

Before beginning to melt, the substance usually shrinks, becoming
disengaged from the walls of the capillary; after that a drop of liquid
appears at the very bottom of the capillary. That is the moment to
note the melting point on the thermometer.

The first capillary is used for a rough estimate. Let us assume that
the melting point in this case was found to lie in the 75-80 range.
The second capillary is then fixed in position and the heating con-
ducted to 65; after that the flame of the burner is reduced so that the
temperature does not rise faster than 1 a minute. The liquid in the
beaker is mixed with a stirrer. The temperature is now noted at which
the substance begins to melt.

In the case of some substances, which decompose as they melt, the
melting point cannot be regarded as a reliable indication of purity
and is used only as an auxiliary characteristic.

Melting point determinations are also carried out in a special vessel
of refractory glass, such as shown in Fig. 34C. The small flame of

Determining the Purity o/j a Substance 55

the burner is applied to that part of the vessel which is covered with
asbestos cord (between the broken lines in the figure). The heated
layer of liquid moves counterclockwise in the vessel. Thanks to ideal
conditions for the thermal movement of the liquid and for its mixing,
melting point determinations can be carried out in such a vessel with
a high degree of accuracy.

(b) The boiling point is another important characteristic of a chem-
ically pure liquid substance. This is the temperature at which the
vapour pressure of the liquid becomes equal to the atmospheric pres-
sure (i. e., 760 mm Hg).

The boiling point of a small quantity of a liquid can be determined
in the apparatus shown in Fig. 34D. Pour 3-4 ml of the liquid into
a dry test tube and throw a small lump of pumice or pieces of capil-
laries, sealed at one end, into the test tube so that the liquid should
boil without bumping. Into a stopper with a slit (Fig. 34) insert
a thermometer in such a way that its lower end is 3 cm above the lev-
el of the liquid. The liquid in the test tube is then heated over a small
flame until drops of condensing liquid begin to trickle continually
from the end of the thermometer. This temperature is noted, after
which the test tube is cooled and the determination repeated.

(c) Determining the Purity of a Substance by the Chemical Method
According to theVolumeof Gas Evolved. The substance to be tested is
treated with a reagent whose interaction with the substance causes
a gas to be evolved. The gas generated is collected over water, and its
volume measured. The purity of the initial substance is determined
from the volume of the gas, reduced to N.T.P., and the equation of
the reaction.

If the gas evolved dissolves in water, it is collected over mercury,
toluene, salt solutions, etc.

Example. Determine the degree of purity of calcite (CaCOs) if 1 g of it, when treat-
ed with acid, yields 244 ml of carbon dioxide measured at 20 and 750 mm.
First, the volume of the gas should be reduced to N.T.P.:

244*750.273

= 224 ml

= 760.293
From the equation of the reaction

CaCOa + 2HG1 = H 2 O + GO 2 + CaCl 2
100 g 22.4 lit

it follows that 1 g of calcite should yield 224 ml of C0 2 .

From a comparison of the volumes it is easy to see that the product tested is
100% purity.

(d) Determining Concentration and the Qualitative A
a Substance According to Standards. The technique of
the concentration of a solution by the relative density,
a hydrometer, has been described in detail in Exerci

56 Exercise 7

be pointed out that the relative densities of aqueous solutions are usual-
ly greater than unity, while the relative densities of solutions of
gaseous substances of low molecular weight (e. g., ammonia solution)
are less than unity.

Qualitative reactions to determine the presence of impurities are
conducted according to respective standards. To determine the pres-
ence of an impurity, a solution of the substance under test is treated
with a characteristic reagent, i. e., a substance which reacts specifical-
ly with that impurity (forming a precipitate, generating a gas, or
altering the colour of the solution).

Example. Ammonium nitrate which is chemically pure in accordance with the
proper standard should not, in a solution of definite concentration, exhibit an acid
reaction more pronounced than a certain permissible limit. To test whether a sample
conforms to standard requirements, 5 g of it, weighed to 0.01 g, is dissolved in 50 ml
of freshly boiled and cooled distilled water. A drop of 2% alcoholic solution of the
indicator methyl-red is then added, which causes a red colouration to appear. If
the colour turns to yellow upon the addition of not more than 2 ml of 0.01 M NaOH,
the chemical conforms to standard requirements.

QUESTIONS

1. In what temperature range ought a chemically pure substance
melt?

2. At what rate should heating be conducted in melting point de-
terminations?

3. The testing of a substance by the method of chemical decompo-
sition produces gaseous ammonia. Name a liquid over which the
ammonia can be collected.

4. By means of what reaction is it possible to establish the presence
of chlorides as an impurity in sulphuric acid?

Problems

1. Determine the degree of purity of calcium hydride (CaH 2 ) if the interaction of
7.6 g of it with water at 17 and 740 mm yields 8.75 1 it of hydrogen.

2. The decomposition of 1.1 g of malachite yields 111.3 ml of carbon dioxide,
measured at N.T.P. Determine the degree of purity of the malachite, which has
the formula Cu 2 (OH) 2 Cq 3 .

3. It has been established by analysis that 32.2 g of sodium hydroxide contains
an amount of sodium carbonate equivalent to 24 ml of M HC1. Determine the degree
of purity of the sodium hydroxide.

4. Determine the degree of purity of tellurium if the burning of 12.76 g of it to
tellurium dioxide requires 2.3 lit of oxygen, measured at 17 and 780 mm.

5. Determine the percentage content of impurities in pyrite if the burning of 15 1
of FeS 2 yields 5,376 cu rn of sulphur dioxide (reduced to N. T. P.). The equation for
the burning process is:

4FeSi + HO 2 = 2Fe 2 O 3 + 8SQ 2

6. Determine the degree of purity of marble if 109.4 g of it upon decomposition
yields 25 lit of carbon dioxide, measured at 15 and 780 mm..

Determining the Purity of\ a Substance

7. Determine the carbon content in coal if 205.8 g of the coal at N.T.P. yields
Leu m of a gas that contains 2% of CH 4 , 29% of CO, and 2% of CO 2 .

8. An alloy used in the manufacture of nibs contains in addition to iron, tung-
sten, chromium, and nickel the metal rhenium. The amount of rhenium in 26.5 ^
of the alloy is such that treatment with the proper reagents converts it to 0.72Jg o!
perrhenic acid HReO 4 . Determine the percentage content of rhenium in the alloy.

9. The decomposition of 2.88 g of aluminium carbide by hot water:

A1 4 C 3 + 12H 2 O = 4A1 (OH) 8 + 3CH 4

yields 1.23 lit of methane, collected at 20 and 770 mm. Determine the degree of
purity of the aluminium carbide.

10. It has been established by analysis that 50 g of metallic thallium contains
3.5 milligram-equivalents of calcium, 5.4 milligram-equivalents of magnesium,
and 6.1 milligram-equivalents of aluminium. Express the contents of the impurities
in percentages.

LABORATORY WORK

Apparatus and materials: the arrangement shown in Fig. 34 and D and in Fig.
35; test tubes and rack; 250 ml measuring cylinder; 10 ml measuring cylinder; 200 ml
beaker with stirrer; funnel; thermometer with 100 range; room thermometer;
barometer; hydrometer with 0.8-1.0 relative density range; rubber bands for at-
taching capillaries to thermometer; powdered sodium thiosulphate; set of substances

Fig. 35. Apparatus for purity determination

for melting point determinations; carbon tetrachloride; saturated solution of sodi-
um chloride; 1 : 3 sulphuric acid; 0.5 N solution of cupric sulphate; zinc dust; 25%
solution of ammonia; 10% solution of barium chloride; 2N solution of hydrochloric
acid; 10% solution of acetic acid; 0.01 N solution of iodine; starch solution; hydro-
gen sulphide solution; lime water; red and blue litmus paper; cigarette paper; ruler
with mm calibration, and weighed amounts of calcium carbide.

Note. The amounts of calcium carbide should weigh from 0.20 to 0.24 g. The
hydrogen sulphide solution should preferably be prerared for each exercise, since it
is easily oxidised by the oxygen of the air. The hydrogen sulphide prepared for
this purpose in the Kipp gas generator should first be washed in a wash bottle for
gases. Cloudy hydrogen sulphide solution should not be used.

Exercise 7

I. Determination of Melting Point, (a) Two of the capillaries pre-
pared in advance (Exercise 4) should be packed tightly with a 2-3 mm
layer of finely ground sodium thiosulphate. This is done in the fol-
lowing manner. Some sodium thiosulphate powder is placed on a
watch glass or on glossy paper. The open end of a capillary is then
thrust into the powder, so that 5-6 mm of its length is filled with
it. Following this, the capillary is dropped, sealed end first, into a
30 cm glass tube placed vertically on the table. When the capillary
strikes the table, the substance moves towards its sealed end. By
repeating this 5-6 times, the capillary is packed tightly. This done,
the capillary is attached by a rubber band to a thermometer, and the
thermometer is lowered into a beaker with water (Fig* 34).

By heating the first capillary, we now determine the approxi-
mate melting point of the substance. The second cappilary is then
heated to a temperature 10 below the found approximate melting
point; from there on heating is continued so that the tempera-
ture rises at a rate of 1 a minute until the first drop of the molten
substance forms. The temperature at which this takes place is the melt-
ing point.

Determination

i
approximate

11
exact

Melting point in C

(b) Determine the melting point of an unknown substance received
from the instructor. Record the results obtained in the above table
and check them with the instructor.

2. Determination of Boiling Point. Fill the test tube in the appara-
tus shown in Fig. 34D with carbon tetrachloride to the level illustrat-
ed and carry out three determinations of its boiling point in accord-
ance with the procedure described in the introduction to the exercise
(see Para. b).

Determination

Boiling point in C

Average value t=

Determining the Purity of, a Substance

3. Determining the Purity of Calcium Carbide. Calcium carbide
reacts readily with water:

CaC 2 + 2H 2 - Ca (OH) 2 + C 2 H 2

The acetylene evolved is collected over a saturated solution of so-
dium chloride in water (since acetylene is slightly soluble in water),
and its volume is measured. From this volume and the weight of the
calcium carbide taken, it is easy to determine the degree of purity of
the CaC 2 .

The apparatus (Fig. 35) consists of a long-necked flask with a stop-
per, through which a delivery tube has been passed, a measuring cylin-
der, and a vessel.

Pour a saturated solution of sodium chloride into the beaker until
it is half full and into the cylinder until it is quite full. Cover the
cylinder with a glass plate and place it upside down in the beaker.
Make certain that the apparatus is airtight.

Now pour 10 ml of distilled water into the flask, trying not to wet
the walls of its neck. Obtain a weighed amount of calcium carbide from
the laboratory assistant and record its weight. Place the calcium car-
bide inside the neck of the flask, which is in a horizontal position
(shown by broken line in Fig. 35), stopper up the flask quickly, place
the end of the delivery tube under the cylinder, and let the lumps of
calcium carbide slip into the water. The acetylene evolved is collected
in the cylinder.

When the reaction terminates, remove the delivery tube from the
vessel; by means of a ruler measure the distance from the level of
the liquid in the vessel to its level in the cylinder (height p w in Fig. 35);
determine the volume of acetylene. Record the temperature and the

pressure.

Table 3

Vapour Pressure h (mm Hg) over Saturated Solution of Sodium Chloride at

Temperature t (C)

7.4

10.3

14.1

19.0

7.9

11.0

15.0

20.2

8.5

11.7

15.9

21,4

9.1

12.4

16.9

22.6

9.7

13.2

17.9

24.0

From Table 3 find the vapour pressure h of the solution for the given
temperature t.

Exercise 7

Results of experiment

Weight of
CaC 2 In g

Volume of the
acetylene in ml

p >yj in mm

pfr ar in mm
Hg

h in mm Hg

Calculations:

(a) On the basis of the equation of state of a gas, reduce the volume
of the acetylene to N.T.P., bearing in mind that the pressure of the
acetylene

1 13.6

(b) From the equation of the reaction calculate the volume o
acetylene obtained from the given amount of calcium carbide (100%
when treated with water.

4. Determination of Zinc Dust Purity. This experiment, like the
previous one, is conducted in the apparatus shown in Fig. 35. The
vessel and the cylinder are filled with water, and 10 ml of 1 : 3 H 2 SC>4
and 3 drops of a dilute solution of cupric sulphate are poured into
the flask; 0.24-0.28 g of zinc dust is weighed in a packet of cigarette
paper and placed inside the neck of the flask in the paper.

Record the results as in Experiment 3. From the experimental data
determine the purity of the zinc dust.

5. Testing a Solution of Ammonia (Ammonia Spirit).
Excerpt from GOST 786-41:

"Ammonia Spirit, Medical.

"A transparent, colourless solution containing not less than 24%
of ammonia (NH 3 ).

" It should pass tests for the absence of:

a (a) sulphates, in accordance with Para. 12.

w (b) carbonates, in accordance with Para. 13.

tt (c) sulphides, in accordance with Para. 14, and

tt (d) salts of the heavy metals, in accordance with Para. 15. "

a ...Para. 12. Test for Sulphates. Dilute 5-8 ml of ammonia solution
with 20 ml of distilled water, and add hydrochloric acid until the
solution exhibits an acid reaction with respect to litmus. Heat the
solution to boiling and add 2-3 ml of 10% barium chloride solution.
Then cool the liquid and examine it in passing light. In the absence

Determining the Purity of a Substance 61

of any trace of opacity, the product is regarded as satisfying require-
ments.

"Para. 13. Test for Carbonates. Dilute 10 ml of ammonia solution
with 10 ml of water and add 40 ml of filtered lime water to the solu-
tion. Examine the liquid in transmitted light. In the absence of any
trace of opacity, the product is regarded as satisfying the require-
ments.

"Para. 14. Test for Sulphides. Add a 10% solution of acetic acid to
5 ml of an ammonia solution until it exhibits a faintly acidic reaction
with respect to litmus; then add 2 drops of an 0.01 N iodine solution.
The appearance of a blue colour in the presence of starch points to
the absence of sulphides in the product.

a Para. 15. Test for Saltsof the Heavy Metals. Dilute about 5 ml of
ammonia solution with 40 ml of distilled water and add 20-30 ml of
hydrogen sulphide solution. If no dark (salts of iron, copper, etc.) or
white (salts of zinc) precipitate is thrown down, there are no salts
of the heavy metals present...."

Testing Procedure. Pour the solution to be tested into a 250-300 ml
cylinder and make certain (by the naked eye) that it is transparent
and colourless. Lower a dry hydrometer into the cylinder and measure
the relative density to + 0.003. After the measurement dry the hydro-
meter thoroughly. On the basis of the relative density value ob-
tained, determine the percentage concent ration of ammonia in the
solution from the table on p. 331. If the experimental value of the rela-
tive density does not appear in the table, use the method of interpo-
lation (see Exercise 2).

The qualitative reactions should be carried out with the reagents
used to detect the corresponding ions (see Paras. 12-15 of the above
excerpt from GOST). Record the results in your notebook and, by com-
paring them with the standard requirements, draw a conclusion con-
cerning^ the quality of the solution tested.

RECORD OF AMMONIA SOLUTION TESTING
1. External appearance

2. Relative density shown by hydrometer

3. NHs content

4. Qualitative reactions to detect:

(a) sulphuric acid

(b) carbonic acid

(d) salts of the heavy metals
Conclusion

Date .

Name

62 Exercise 8

Exercise 8

THERMAL EFFECTS OF REACTIONS

SUBJECTS FOR STUDY

Energy transformations in chemical reactions; equivalence of different forms of
energy; chemical energy and thermal energy; exothermal and endothermal reactions;
thermal effects of reactions and thermochemical equations; Hess's Law, and the ther-
mal effects of solution and hydration.

Every substance has not only a definite composition, but also a
definite stock of chemical energy. Chemical reactions involve changes
both in the composition of substances and, at the same time, in
their stocks of energy. The difference between the chemical energy of
the initial reactants and of the final reaction products is converted
into an equivalent amount of some other form of energy: mechanical,
radiant, thermal, or electric. Typical of most chemical reactions is
the transformation of chemical energy into thermal; reactions in which
heat is liberated are called exothermal, while those in which it is ab-
sorbed are termed endothermal.

Example.

H 2 + S= H2S + 4.8 Cal. (Exdthermal reaction)
H 2 + Se = H 2 Se 18.5 Cal. (Endothermal reaction)

The quantity of heat evolved or absorbed in a chemical reaction
is called the thermal effect of the reaction. It is usually referred to one
mol of the reaction product and expressed in large calories (Cal.).
Chemical equations including the thermal effect value are called
thermochemical equations. For example, the burning of carbon mono-
xide may be represented by the following thermochemical equation:

O 2 -2CO 2 +135Cal.

This may likewise be expressed by the equation:
CO + y O 2 = CO 2 + 67.5 Cal.

It is general practice in thermochemistry to employ the latter
type of equation, i. e., to refer the thermal effect to one gram-mole-
cule of the product, using fractional coefficients if necessary.

The thermal effects of the formation and decomposition of the same
substance are equal in absolute value, but opposite in sign (first law
of thermochemistry).

For instance:

Pb + I f - PbI 2 + 41.77 Cal. PbI 2 = Pb + I a 41.77 Cal.

Thermal Effects of Reactions

This law serves as the basis for calculating the thermal effects of
various reactions important in practice (formation, decomposition,
burning, solution, hydration, and neutralisation).

One and the same end product may be obtained from the same ini-
tial substances by different reactions. Let us assume, for example,
that from the substances A and B it is necessary to prepare the product
AB 2 . The process may be conducted in different ways:

A+2B=AB 2 +QCai.

According to the second law of thermochemistry, established by
the Russian scientist G. Hess (1836), the quantity of heat evolved in
a chemical process does not depend on whether the process takes place
in a single stage or in several stages (provided volume and pressure
remain constant). Consequently, for our case Q = Qi+ Q*- Since
the thermal effect of some stages of the process may be negative, the
general principle may be formulated as follows: The algebraic sum of
the thermal effects of a chemical process consisting of several stages
is equal to the thermal effect of the same process taking place in a single
stage.

States and allotropic modifications of substances should be indicat-
ed in thermochemical equations, as the burning, say, of amorphous
carbon, graphite, and diamond yields different quantities of heat
(96.98, 94.23, and 94.45 Cal. respectively); similarly, the thermal
effects of the formation of water, water vapour, and ice from their
elements are not the same either (68.35, 57.84, and 69.95 Cal. re-
spectively).

The quantity of heat in calories generated by the burning of 1 mol
of a substance is called the heat of combustion. The burning of a com-
pound substance may follow either of the two patterns.

1. The direct interaction of the molecules of the substance with
oxygen (a single-staged process).

2. The decomposition of the molecule of the substance, with the
subsequent burning of each component in oxygen (a process taking
place in several stages: a, b, c, d...).

64 Exercise 8

In accordance with Hess's Law, the thermal effect is the same in
either of the two above cases, provided the end products of the reac-
tion are identical in composition and number of molecules.

Example. Calculate the heat of combustion of acetylene, knowing that the heat of
formation of liquid water is 68.35 Cal., the heat of formation of CO2, 94.23 CaL, and
the heat of formation of acetylene, 53.9 Cal.

It follows from the above data that acetylene C2H2 is an endothermal compound
and that its decomposition is, therefore, accompanied by the liberation of heat,

Let us denote the unknown heat of combustion of acetylene as x Cal. We can then
write:

1. C 2 H 2 + O 2 = 2CO 2 + H 2 O + x Cal.

2. (a) CaH 2 = 2C + Ha + 53.9 Cal.

(b) 2C + 2Oa = 2CO2 + 2-94.23 Cal.

From Hess's Law it follows that

x = 53.9 -|- 2-94.28 -f 68 35
x = 310.71 Cal.

The heat of formation and the heat of decomposition are calculated in the same way.
Hydration is the process whereby the molecules of a substance that
is being dissolved combine with the molecules of the solvent, specifi-
cally water. The process is accompanied by the evolution of heat,
which is called heat of hydration. For instance, from the equation

ZnS0 4 + 6H 2 - ZnSCV 6H a O + 19.3 Cal.

it follows that 19.3 Cal. is evolved when a molecule of zinc sulphate
combines with six molecules of water.

The process of solution may also be expressed as a thermochemical
equation, e. g.:

ZnSCV 7H a O + aq = ZnSO 4 -aq 4.26 Cal.

Here aq denotes the solvent, water.

It is evident from the above equation that 4.26 Cal. is absorbed when
one mol of the heptahydrate of zinc sulphate dissolves. It should be
pointed out that an anhydrous salt forming a crystal hydrate dis-
solves in two stages: (a) hydration, and (b) the solution of the hydrate.

Example. Calculate the heat of hydration of anhydrous zinc sulphate, know-
ing that its heat of solution is 18.43 Cal., while the heat of solution of the
hydrate ZnSO 4 -7H 2 O is 4.26 Cal.

Thermal Effects of> Reactions 65

Equations should be written, in which the heat of hydration is denoted as x:

1. ZnSO 4 + aq = ZnSCVaq + 18.43 Cal.

2. (a) ZnSO 4 + 7H 2 O = ZnSO 4 -7H 2 O + A: Cal.
(b) ZnSO 4 -7H 2 O + aq = ZnSO 4 -aq 4.26 Cal.

From Hess's Law it follows that

18.43 = x 4.26
x = 22.69 Cal.

QUESTIONS

1 . Where is there a greater stock of chemical energy: in 1,000 atoms
of hydrogen or in 500 molecules of it? Explain the reasons for the
difference in the amount of energy in this case.

2. The formation of a gram-molecule of water is accompanied by
the evolution of 68.35 Cal., while the formation of an equal amount
of water vapour is accompanied by the evolution of only 57.84 Cal.
Why the difference?

3. The same molar quantities of MgSO 4 and MgSO 4 -7H 2 O are
dissolved in equal amounts of water. Why are the thermal effects differ-
ent?

4. The values of the heat of formation for H 2 O, P 2 O 5 , CO 2 , and
MgO are 68.35, 360.0, 94.23, and 146.1 Cal. respectively. If we take
hydrogen, phosphorus, carbon, and magnesium in equal quantities
by weight, in which case will burning produce the most heat?

5. Explain the thermochemical equation

To what amount of PC1 3 by weight is the thermal effect referred?

6. On the basis of Hess's Law write the thermochemical equations
for the burning of the exothermal compound propane (CgHs) and for
its decomposition to its elementary constituents, with their subse-
quent separate burning. Denote the separate thermal effects Q x , Q 2 ,
Q 3 , and Q 4 respectively. Express Q 3 in terms of the other thermal
effects.

Problems

1. The values of the heat of formation for SiO 2 and MgO are 203.3 and 146.1 CaL
respectively. Calculate the heat of the reaction

2Mg + SiO 2 = 2MgO + Si.

2. The heat of the formation of K2O is 86.2 Cal. In the reaction between molecu-
lar quantities of K2O and SO 2 , 110.6 Cal. is evolved.

5-795

66 Exercise 8

Calculate the thermal effect of the formation of SO 2 , knowing that the heat of
formation of K^SOs from its elements is 267.7 Cal.
3. Calculate the heat of the reaction

It is known that the values of the heat of formation of 862 and SOs from their ele-
ments are 70.92 and 93.9 Cal. respectively.

4. Calculate the heat of the reaction

y0 2

It is known that the heat of formation of H2S from its elements is 4.8 Cal.; the heat
of formation of SO 2 , 70.9 Cal., and the heat of formation of H 2 O t;flp , 57.84 Cal.

5. Calculate the heat of formation of CaCOs from its constituent elements, knowing
that the thermal effect of the reaction CaCOa= CaO 4- CO 2 equals 43.35 Cal.
and the heat of formation of CaO and CO 2 from their constituent elements equals
151.7 and 94.23 Cal. respectively.

6. Given the thermochemical equations of the reaction:

MgSO 4 + aq = MgS0 4 -aq + 20.3 Cal.
MgSO 4 -7H 2 O + aq = MgS0 4 -aq 3 8 Cal.

Calculate the heat of hydration of anhydrous magnesium sulphate.

7. The heat of formation of AUOs is 380 Cal. In the reaction

2A1 + Fe 2 3 = A1 2 3 + 2Fe

the reduction of every 53.2 g of Fe 2 0a is accompanied by the evolution of 60.5 CaL
Calculate the heat of formation of ferric oxide.

8. The reaction of the burning of benzene obeys the equation:

15
C 6 H 6 + - O 2 = 6CO a + 3H 2 O //< + 782.3 Cal,

Calculate its heat of formation from its constituent elements, knowing the heat
of formation values for C0 2 and HaO/ty (see Question 4).

9. The heat of formation values for HzO vap and CO 2 are known (see Questions-
2 and 4); the heat of combustion of C 2 H 2 , with the formation of water vapour, is
300.2 Cal. Calculate the heat of formation of C 2 H 2 .

10. Calculate the heat of combustion of C 2 H6, knowing that its heat of formation-
is +20.24 Cal. and knowing the heat of formation values for water and CO 2 (see-
Question 4).

LABORATORY WORK

Apparatus and materials: sand bath with a thermometer up to 250; porcelain
crucible; crucible tongs; desiccator with sulphuric acid; calorimeter with cover; fun-
nel and stirrer for calorimeter; 100 ml measuring cylinder; Dewar vacuum flask (or
50 ml beaker with heat-insulating jacket); Assman thermometer; glass stirring rod;
powdered cupric sulphate; 0.5 M NaOH, and 0.5 M HC1.

Note. The Assman thermometer is used to measure the temperature of the air
in determinations of air humidity by means of a psychrometer. It is graduated from 1
20 to +50, with divisions of 0.1 (sometimes 0.2).

Making the heat-insulating jacket. A heat-insulating jacket for a beaker is made
from asbestos sheet that should first be soaked in water. The beaker is wrapped in
the wet asbestos, wound with wire, and then dried.

Thermal Effects of, Reactions

1. Determining the Water of Crystallisation in Cupric Sulphate.

The purpose of the experiment is to determine the number of mole-
cules of water in the cupric sulphate molecule CuSO 4 -nH 2 O. To do
so, it is necessary to heat a weighed amount of the hydrate, thereby
expelling all the water of crystallisation and obtaining the anhydrous
salt. The difference in weights is the weight of the expelled water,,
from which it is not difficult to derive the num-
ber of water molecules.

Weigh a clean, previously ignited crucible to
0.01 g. Put 1-1.2 g of finely ground cupric sul-
phate in it and weigh it again. The difference in
weights gives the weight of the substance.

Now place the crucible (weighed with the sul-
phate) on a sand bath, so that -^ of the crucible is

submerged in the sand. Place a thermometer into
the sand, so that its tip is level with the bottom
of the crucible (Fig. 36).

Heat the sand bath to a temperature not high-
er than 220, since at a higher temperature an
undesirable side reaction may take place, with pf 36 heat}ng
the formation of a basic salt (grey-coloured): on ' a S a nd bath

2CuSO 4 */zH 2 O = Cu 2 (OH) 2 SO 4 + SO 3 + (n 1) H 2 O

With the vaporisation of the water, the blue powder becomes paie;
when the salt in the crucible turns white, the heating should be dis-
continued and the hot crucible put in a desiccator for cooling and
then weighed (wj. Afterwards the crucible should be placed on the
sand bath again, heated for 10-15 minutes, and, after cooling, weighed 1
again (w 2 ). If the weights w 1 and w 2 coincide or differ by not more
than 0.01 g, i. e., if the crucible with the substance has been brought
to a constant weight, ignition should be stopped. Otherwise the whole-
operation should be repeated. Record the experimental data (giving
the weight after each ignition):

1. Weight of crucible with CuSO 4 .rtH 2 O
Weight of empty crucible

Weight of CuSO 4 .rzH 2 O . . .

2. Weight of crucible with CuSO 4
Weight of empty crucible . .

Weight of CuSO 4

From the experimental data, calculate the number of water mole-
cules in a molecule of cupric sulphate.

Exercise 8

2. Determining the Heat of Neutralisation. Measure 100 ml of
0.5 M HC1 in a cylinder and pour it into a beaker. Pour 100 ml of
NaOH of the same concentration into another beaker. Place the beak-
er with the acid in a calorimeter (t&ick-walled battery jar) and
cover it with a lid that has three orifices. Insert an accurate thermome-
ter (with divisions of 0.1) into the middle orifice, lowering it into
the acid solution. Pass a funnel and a stirrer through the other ori-
fices. After 10 minutes measure the temperature of the acid solution.
Then quickly pour the alkali solution through the funnel and, while
stirring the solution continuously, note the highest temperature of
the solution to 0.1.

Record the experimental data in the following manner:

HCl

NaOH

Temperature In C

concentration

volume in ml

concentration

volume in ml

initial

final

With the data obtained, carry out the following calculations:

(1) On the basis of the initial and the final temperature, as well
as the over-all volume of the solution, calculate the quantity of heat
in Gal. evolved in the course of the reaction (assume the relative den-
sity of the solution to be equal to 1, and the thermal capacity of the
solution, equal to that of water).

(2) Write the equation for the reaction of neutralisation that
takes place and calculate the thermal effect of the neutralisation of 1
gram-equivalent of the acid.

(3) Determine the relative error of the experiment, knowing that
13.7 Gal. is evolved when 1 gram-equivalent of a strong acid is neutral-
ised by an equal amount of a strong alkali.

3. Determining the Heat of Solution of an Anhydrous Salt Pour
25 ml of distilled water into a Dewar vacuum flask or into a beaker
with a heat-insulating jacket. Place an Assman thermometer into
the vessel (1 cm above its bottom), fixing the thermometer in posi-
tion by a clamp attached to a ringstand. Note the temperature of
the water to 0.1. Now quickly transfer into the water the anhydrous
cupric sulphate obtained in the first experiment and, while stirring
the water, note the maximum temperature of the solution to 0.1.

Since we know the temperature difference, the over-all mass of
the solution (equal to the weight of the salt and the water), and the
specific heat of the solution (which we assume to be equal to 1), we
can calculate the amount of heat evolved in this process (the hydra-

Reaction Rates and Chemical Equilibrium 69

tion of the anhydrous salt and its solution). This amount of heat can
then be referred to one mol of the anhydrous salt.

Next, on the basis of Hess's Law, calculate the thermal effect of
the reaction

1. CuSO 4 + aq = CuSO 4 -aq + Q

it being known that the hydration of the anhydrous salt and the so-
lution of the pentahydrate are expressed by the following thermo-
chemical equations:

2. CuSO 4 + 5H 2 O = CuSO 4 -5H 2 O + 18.6 Cal.
CuSO 4 -5H 2 O + aq = CuSO 4 *aq 2.75 CaL

Compare the experimental and theoretical thermal effects and
determine the relative error of the experiment in per cent.

Exercise 9

REACTION RATES AND CHEMICAL EQUILIBRIUM

SUBJECTS FOR STUDY

The rate of a chemical reaction and the effect of concentration and temperature
on it; irreversible and reversible reactions; concentration and active mass; Law of
Mass Action; chemical equilibrium constant; shifting of chemical equilibrium under
the influence of various factors;

Le Chatelier's Principle; the rate of heterogeneous reactions and the effect of the
solid phase surface; catalysis.

1. The Rate of Chemical Reactions. Substances interact chemically
in the gaseous or the liquid state when their molecules collide.

The number of effective collisions depends upon the active mass
of the substance, which is proportional to the total number of mole-
cules per unit volume, or the molecular (molar) concentration of the
substance, expressed in number of mols per unit volume (1 litre).
With a change in volume, there is a change in concentration. The
volume itself depends upon the pressure (in the case of gaseous sub-
stances) or upon the dilution of the solution. But even at equal con-
centrations different reactions may proceed at different rates, these
depending upon the chemical nature of the reactants.

The rate of chemical reactions depends upon the nature of the inter-
acting substances and is proportional to the product of their molecular
concentrations.

When two substances, A and B, interact, the rate of the reaction
(v) may be calculated by means of the equation

0= JC[A]-[B]

70 Exercise 9

where K is the rate constant of the reaction, depending upon the
nature of the interacting substances and upon the conditions in which
the reaction takes place (temperature and pressure), while [A] and
[B] are the molar concentrations of the substances. If [A] = [B]=l,
K is numerically equal to the rate of the reaction. The rate of a reac-
tion is measured by the change in molar concentration per unit time
(in a second or minute).

Example. One litre of 2M acetic acid is mixed with one litre of 3M ethyl alcohol.
Determine 1he rate of the reaction of the formation of the ester, ethyl acetate, at the
initial moment. How will the initial rate change, if the solutions are each diluted
by an equal volume of water before mixing?

The total volume of the mixture of the solutions is 2 lit,

The molar concentrations are: [CH S COOH] = 1 and [C 2 H 6 OH]=1.5.

The rate of the reaction is calculated according to the equation

CH 8 COOH : + C 2 H 5 OH -* CH 8 COOC 2 H5 + HO
v = /C.[CH 3 COOHHC 2 H 5 OH] = i.5K

When the solutions are diluted, the total volume of the mixture becomes 4 lit.
The molar concentrations will, accordingly, be:

[CHsCOOH] = 0.5 and [C a H 6 OH] = 0.75

a =*K. 0.5-0.75 = 0.375K

Consequently, when the solution is diluted by an equal volume of waler, the re-
action rate drops to-r-.

If two substances react in such a way that for every m molecules
of the substance A there are n molecules of the substance B (where m
and n are integers greater than 1), the reaction rate for the interaction
mA + nB will be:

In other words, the exponent of the concentration in the equation of
the reaction rate is equal to the coefficient for that substance in the
equation of the chemical reaction.

Example. How will the rate of the reaction 2H2S + C>2 change if the pressure
is increased threefold?

If we denote [H2S] at the initial pressure as Ci and [Oal as Cz, the reaction rate v
can be expressed by the equation: v = K-C^-C^

If the pressure is now trebled, the concentration of the reacting gases wil' increase
accordingly: [H2S]=3Ci and [O2l=3C 2 . The reaction rate under the new conditions
will be:

v = /C-(3Ci) 2 -3C 2 = 27.K-Cf.C2
The reaction rate, hence, increases 27 times.

5. The Law of Mass Action. All chemical reactions can be divided
into two types: irreversible and reversible reactions. Reactions pro-
ceeding in both opposite directions at the same time are called re-

Reaction Kates and Chemical Equilibrium 71

versible, while those which proceed practically to the end in one di-
rection are called irreversible. The number of irreversible reactions is
limited; most reactions are in practice reversible.
A general equation for reversible reactions may be written as follows:

mA + nB ^ pC]+ qD

The rates for the forward process (vj and the back process (i> 2 ) are

expressed by the respective equations:

where [A], [B], [C], and [D] are the molar concentrations of the sub-
stances.

When equilibrium is established, the forward and back reactions
do not cease; they continue in opposite directions at equal rates.

If v l = i> 2 the right-hand parts of the above equations are like-
wise equal, i. e.:

The Law of Mass Action may be formulated thus: When equilibri-
um is attained, the product of the active masses of the reactants by the
velocity constant of the forward reaction equals the product of the ac-
tive masses of the resultants by the velocity constant of the back reac-
tion.

From the previous equation it follows that

The ratio of two constants is a constant; hence

The chemical equilibrium constant K is the ratio of the product of
the molar concentrations of the resultants to the product of the molar
concentrations of the reactants at the moment when equilibrium is
attained.

In deriving the equilibrium constant, it is customary to have the
product of the concentrations of the resultants as the numerator.
For instance, for the reversible reaction

72 Exercise 9

the equation for the equilibrium constant should be written as fol-
lows:

~~ [HC1JMO*!

If in such a system which has attained equilibrium, we were to in-
crease the concentrationof HC1, there would have to be either an increase
in the numerator or a decrease of the second factor in the denomina-
tor, i. e., the quantity [O 2 L This could take place through a shift of
the equilibrium from left to right, leading to the formation of addi-
tional quantities of H 2 O and C1 2 . In the new steady state of equilib-
rium reached, the concentrations of all four substances will be different
from the initial values, whereas the value of /(will not have changed.

Consequently, to shift the equilibrium from left to right, it is nec-
essary either to increase the concentration of one of the substances
which appear to the left of the reversibility sign (^)in the equation
of the reaction or to reduce the concentration of one of the substances
which appear to the right of that sign.

By means of the equation for the equilibrium constant, it is possi-
ble to determine the concentrations of substances at equilibrium,
knowing the value of K and the initial concentrations, or vice versa.

If the initial concentration of substance A is denoted as [A]/ rt //,
while the number of mols of this substance that has reacted by the
time of equilibrium is denoted [A] rea c, the molar concentration at
the moment of equilibrium [A]^ may be expressed as

lA]reac

Example. For the system

CO + H 2 O ; H 2 + CO 2

in equilibrium, /C=1.0 at 830. Determine the concentrations of all the substances
at equilibrium, knowing that the initial concentrations [CO] 2M and [H2O]=3M.
The equation for the equilibrium constant may be written thus:

The following table should now be drawn up:

CO H 2 O CO 2 H

Initial concentrations ......... 2 3

Mols that have reacted ....... x x

Mols formed . . . . ^ ........ x x

Concentration at equilibrium ..... 2-x 3-# x x

In that case

==1 -

(2-*) (3-*)
From this it follows that

6
5x = 6 and x = =1.

Reaction Rates and Chemical Equilibrium 73<

Consequently, at equilibrium:

[C0] = 2 1.2 = 0.8M
[H 2 O] = 31.2= 1.8M
[H 2 ]= fC0 2 ]= 1.2M

In the above example the coefficients of all the substances involved
in the reaction are equal to 1, and determining the concentrations at
equilibrium is therefore not difficult.

On the other hand, in the system

4HC1+O 2 ;-2H 2 O + 2C1 2

which is likewise in equilibrium, the coefficients are not equal to 1,
and they must be taken into consideration in determining the molar
concentrations at equilibrium. If x mols of HC1 have reacted by the
time equilibrium is attained, the number of mols of O 2 to react will

A Y

have been -T- of that amount, i. e., -j- , while the number )}f mols of H 2 O
and C1 2 formed by that moment will be half the amount of mols of
HC1, i. e.,|-.

3. Reaction Rates in Heterogeneous Systems. The relationships
considered above apply only to homogeneous systems.

A system in chemistry is a part of space filled with a substance or
mixture of substances and isolated from the surrounding medium.
The parts of a system which have throughout the same physical prop-
erties, a uniform chemical composition, and an interface are termed
phases.

Mixtures of gases, liquid water, and solutions are all examples of
homogeneous systems (the number of phases is 1).

Heterogeneous systems are those which consist of several phases,
e. g.:

water ice water vapour (the number of phases is 3);

water oxygen hydrogen (the number of phases is 2);

magnesium carbonate magnesium oxide carbon dioxide (the
number of phases is 3).

The laws which govern reactions proceedeng in homogeneous sys-
tems do not apply in full to heterogeneous systems. For instance, in'
the heterogeneous system gas solid the molecules of the gas can
collide with the molecules of the solid only at the interface; hence,
the concentration of the solid as a whole (the mass of the solid phase)^
has no effect on the reaction rate. When ferrous oxide is reduced by
hydrogen

the rate of the forward reaction is proportional only to the con-
centration of the hydrogen, i. e., v = /C'[H 2 ].

74 Exercise 9

For a system in equilibrium such as

Fe 3 4 + 4H 2 ; 3Fe + 4H Aap

the equation for the equilibrium constant should be written thus:

[Fep.[H 2 0]* _ K
~

But since the concentrations of the solid phases [Fe 3 O 4 ] and [Fe]
are constants, we can transfer them to the right-hand side of the equa-
tion and combine them with the constant /C x . This gives rise to a new
constant, K:

[H 2 O]* _ K [Fe 3 O4] __ K
[H 2 ] 4 -~ A I' [Fe] 3 "" A

The final equation for the equilibrium constant may therefore be
written thus:

[H 2 0]* _

[Ht]

In a heterogeneous system the reaction takes place at the interface;
the reaction rate therefore increases with the surface area of contact.
This is why solids react much more quickly in the pulverised state.

4. Effect of Temperature on Reaction Rate. Any rise in temperature
speeds the reaction, as the absorption of energy increases the number
of active molecules and the velocity of molecular movement; the
number of molecules colliding per unit time therefore increases.
Every 10 temperature increase just about doubles the reaction rate.
In some cases, however, the reaction rate increases even more (e. g.,
3-4 times).

If a temperature rise of 10 increases the reaction rate twofold*
and if we denote the reaction rate at the ultimate temperature as vt s
and the reaction rate at the initial temperature as vt l9 the relationship
between the rates may be expressed thus:

Example. How much will the reaction rate increase as a result of a temperature
rise from 10 to 100, if every 10 increment doubles the rate?

100-10

The rate thus increases 512 times.

* The number indicating how many times the reaction rate increases with a 10
rise in temperature is called the temperature coefficient of the reaction.

Reaction Rates and Chemical Equilibrium 75

5. Shifting of Chemical Equilibrium. The direction in which the
equilibrium of a system shifts whenever concentration, temperature,

or pressure changes is determined by Le Chatelier's Principle. //
a reversible system in a state of mobile equilibrium is subjected to some
external influence, the equilibrium shifts in such a way that the influ-
ence is reduced.

Let us consider the following system at equilibrium:

N ,+ 3H 2 2NH 3 + 2 1 1 Gal.

In what direction will the equilibrium of the system be shifted if
the pressure is increased and the temperature lowered?

The volume of the initial gases N 2 + 3H 2 is double the volume of
the product 2NH 3 . The system will therefore react to an increase in
^pressure by an equilibrium shift towards the smaller gas volume.

The formation of ammonia is accompanied by the evolution of
heat (+2-11 Cal.), while the reverse process (ammonia decomposi-
tion) proceeds with the absorption of heat ( 2-11 Gal.). Consequent-
ly, the system will react to a drop in temperature by an equilibrium
.shift from left to right in the direction of the exothermic process,
i. e., the evolution of heat.

Exercise. How does a rise in temperature and pressure affect the equilibrium of
the following systems:

2N 2 -fO 2 ^2N 2 O 2-19.65 Cal.
H 2 +Br 3 ;2HBr +2-8.65 Cal.
2C +p 2 ^ 2CO;+ : 2 26.42 Cal .

6. Catalysis, Catalysis is a change in the speed of a chemical proc-
ess under the influence of a foreign substance present in the reaction
system, the composition and amount of the foreign substance remain-
ing unchanged at the end of the reaction. Solid, liquid, and gaseous
substances may alter the speed of a chemical process, i. e., act as
catalysts.

In homogeneous catalysis the reactants and the catalyst constitute
a one-phase system (e. g., the interaction of the gases NH 3 and HC1
in the presence of H 2 O vapours).

In heterogeneous catalysis the reactants and the catalyst form a two-
phase system (e. g., the ignition of a mixture of oxygen and hydrogen
on the surface of platinum black, a solid).

A catalyst does not initiate a chemical process; it has the same
-effect on the rate of both the forward reaction and the back reaction,
thereby hastening the establishment of equilibrium in the system
without, however, affecting that equilibrium in any way.

In heterogeneous catalysis relatively large amounts of reactants may
<react in the presence of small amounts of the catalyst. In homogeneous

Exercise 9

catalysis the rate of the reaction is practically proportional to the
amount of the catalyst.

The activity of catalysts and sometimes the specific character of
their effect depend markedly upon temperature.

QUESTIONS

1. How will the rate of a reaction change with concentration? The
concentrations of reactants A and B are given in the following table,
together with the reaction time:

Concentration of substance A in mols/1

Concentration of substance B in mols/1

0.66

0.50

0.33

Reaction time in seconds ......

Plot a chart of the reaction rate as ordinate against the concentration
as abscissa. What type of a line is the function? Should it pass through
the origin of the coordinates?
2. Write the equations for the rates of the forward reactions:

4NH 3 + 50 2

3. Give examples of homogeneous and heterogeneous systems. In
what way do they differ?

4. Given the reversible system CO 2 + H 2 5 H 2 O +CO. Will
there be an identical shift of equilibrium in two such systems under
the same conditions if 100 g of water vapour is added to one system,
while 100 g of carbon monoxide is added to the other?

5. Considering the state of the substances in the system, write equa-
tions for the equilibrium constants of the following reactions:

CuO + H 2 "; H 2 O + Cu
CaCO 3 CaO + CO 2

6. Under what conditions should the reaction

N 2 + O 2 ? 2NO - 43.06 Cal.

be conducted to obtain a maximum yield of nitrogen oxide? (Apply
Le Chatelier's Principle).

Reaction Rates and Chemical Equilibrium 77

Problems

1. The equilibrium constant for the reaction CO + C1 2 ^ COC1 2 equals 39.4. If
at equilibrium [CO]=0.2 and [COCl a ]=0.8 M, calculate the initial quantity [C1 2 ].

2. The equilibrium constant of the system CO+ IfoO j* H2+ COa at a certain
temperature is equal to 1. If the initial concentrations [CO] =1 and [H2O]=3,
what will be the concentrations of all the substances at equilibrium?

3. At 250 the equilibrium constant of the system PC1 5 ^PC13+ C1 2 equals 0.0414.
How many mols of PCU were placed in a 1 litre vessel at that temperature, if the val-
ue of [C1 2 ] turned out to be 0.1 M?

4. The concentrations at equilibrium in the system 2N 2 + O2 2N 2 O were found
to be [N 2 ]=0.72, [O 2 ]=1.12, and [N 2 O]=0.84. Determine the equilibrium constant.

5. The equilibrium constant of the system H 2 4 I 2 ^2HI equals 40. How many
mols of hydrogen have to be introduced into the system per mol of iodine to ensure
that 60% of the iodine be converted to HI (at a constant volume of the system)?

6. Calculate the equilibrium constant of the system N 2 O 4 ^ 2NO 2 if the initial
concentration of N2O 4 : =0.02 and the dissociation at equilibrium is 60%.

7. Find the equilibrium constant of the system 4HC1 + O 2 ^ 2H 2 O -f 2C1 2 if
at equilibrium [H 2 O]=[C1 2 ]=0.14, [HC1]=0.2, and [O 2 ]=0.32

8. The equilibrium constant of the system H 2 + I 2 ;2HI at 443 equals 50. At
equilibrium the concentration of HI proved equal to 0.0315 mols/1, while the con-
centration of I 2 proved equal to 0.0114 mols/1. Calculate the hydrogen concentration
at equilibrium.

9. Equilibrium in the reaction of acetic acid with ethyl alcohol is expressed by
the equation:

CH 3 COOH + C 2 H 8 OH ^ H 2 O + CHiCOOC a H 5

How many mols of the ester are there in the system at equilibrium, if the equilibrium
constant equals 4, and the initial concentrations of the alcohol and the acetic acid are
5 M and 2 M respectively.

10. Onemillimol of a substance reacts in 1 minute at 80. How many millimols
will react during the same time, if the temperature is raised first to 120 and then to

180 (the temperature coefficient of the reaction equals 2)?

LABORATORY WORK

Apparatus and materials: the set-up shown in Fig. 37; porcelain mortar; 250-400
ml beaker; 50 ml beaker; 3 burettes; 3 burette funnels; test tubes and rack; horn or
glass spatula; thermometer up to 100; metronome; stop watch; 25 ml measuring cylin-
der; splinters; 10x10 cm squared paper; filter paper; wax pencil; manganese dioxide
in powdered form; 1:9 powdered mixture of manganese dioxide and ferric oxide; pow-
dered ferric oxide; powdered silica; crystalline potassium chloride; potassium iodide;
lead nitrate; chalk in lumps; ferric chloride saturated solution; potassium thiocya-
nate saturated solution; 2N HC1; 3% and 0.2% solutions of hydrogen peroxide; so-
dium thiosulphate solution containing 75 g of Na 2 S 2 Os per litre, and sulphuric acid
solution (15 ml of H2S0 4 , relative density 1.84, per litre).

1, Reaction Rate in Homogeneous Systems. In the reaction between
sulphuric acid and sodium thiosulphate

(a) Na 2 S 2 O 3 + H 2 SO 4 = H 2 S 2 O 3 + Na 2 SO 4
(b)

the insoluble sulphur is precipitated- in the form of slime. The time
elapsing from the beginning of the reaction to the appearance of the

Exercise 9

first traces of slime depends upon the concentration of the reactants-
and the temperature.

Fill three burettes with water, a solution of sodium thiosulphate,
and a solution of sulphuric acid respectively. Fill four numbered test
tubes with the number of millilitres of thiosulphate solution and wa-
ter indicated in the table below. Pour 6 ml of sulphuric acid solutiom
from a burette into each of four other test tubes.

Now pour the measured quantity of sulphuric acid into test tube
No. 1, stir its contents quickly, and count the number of metronome
beats from the moment when the liquids were combined until any slime
appears. The reaction rate is inversely proportional to this period

of time (v = -7=-).

No. of test tube

Volume in ml

Na 2 S 2 O 8
concentration
a

Time (No. of met-
ronome beats) /

Reaction rate

-f

Na 2 S 2 O 8
solution a

H 2 b

a + b

4 .

2/3

1/2

1/3

Perform similar experiments with all the other -test tubes. Draw
a chart (this can be done conveniently on squared paper), plotting the
concentration as abscissa against the reaction rate as ordinate *.
A suitable scale is 3 cm for the minimum concentration and 8 cm for
the maximum rate.

What kind of a line is the function?

Does it run through the origin of the coordinates? If so, why?

2. Effect of Temperature on Reaction Rate. Pour 4 ml of sodium
thiosulphate solution from a burette into each of four clean numbered
test tubes (1, 2, 3, 4) and pour 4 ml of sulphuric acid solution into
each of four others (la, 2a, 3a, 4a). Place all the test tubes into a beak-
er with water; five minutes later, after measuring the temperature
of the water, combine the contents of test tubes 1 and la; count the
number of metronome beats until any slime appears (do not forget
to stir the solutions first).

Heat the beaker with the test tubes 10 above the original tempera-
ture and repeat the experiment, this time with test tubes 2 and 2a.
Carry out similar experiments with the other test tubes, each time
raising the temperature by another 10.

JOG.

* It is advisable to express v in decimals to hundredths and then multiply by

Reaction Rates and Chemical Equilibrium

79>

Record the experimental results in the form of a table:

No. of test
tube

Temperature
of experiment
in C

Time (No. of
metronome
beats) *,

Reaction rate

'-?:

Show the effect of temperature on the reaction rate by means of a'
chart, plotting the temperature as abscissa. A suitable scale is 2 cm
for 10 and 8 cm for the maximum reaction rate.

What effect has temperature on the rate of the reaction studied?

Does the rate temperature curve pass through the origin of the coor-
dinates?

3. Chemical Equilibrium. A classic reversible reaction is that be-
tween ferric chloride and potassium or ammonium thiocyanate. The-
resulting solution of ferric thiocyanate Fe(SCN) 3 exhibits a red col-
ouration whose intensity depends upon concentration. A shift of
equilibrium is therefore easily detected by the changed intensity of
the colour.

Pour 20 ml of water into a beaker and add 1-2 drops of saturated
solutions of FeCl 3 and KSCN. Pour equal portions of the resulting
red solution into four test tubes.

Write the equation of the reversible reaction and the equation of
the equilibrium constant.

Add 2-3 drops of the saturated solution of FeQ 3 to the contents of
one of the test tubes. Write up your observations and explain what
has happened on the basis of the equation for the equilibrium constant.

Add 2-3 drops of the saturated solution of KSCN to the contents
of another test tube. Explain, the change observed. Why does a change-
in the concentration of KSCN cause a more pronounced shift of equi-
librium than does a change in the concentration of FeCl 3 ?

Drop a small piece of solid potassium chloride into the third test
tube and shake it vigorously. Compare the colours of the solutions in
test tubes 3 and 4. Give an explanation.

What changes in concentration are required to shift the equilibrium
to the right or to the left?

4. Reaction Rate in Heterogeneous Systems.

(a) Pour hydrochloric acid into two test tubes, filling V^f each.
Take two lumps of chalk of equal size, grind one of them to a powder
in a mortar, and transfer the powder to a sheet of paper. Then pour
the powder into one test tube and, simultaneously, drop the lump of

Exercise 9

chalk into another. In which of the test tubes does the reaction pro-
ceed faster? Why?

(b) Mix a few crystals of lead nitrate and potassium iodide in a
mortar gently without grinding them with a pestle. Is there any change
of colour?

Then grind the crystals vigorously with a pestle and note how the
powdered mixture acquires a yellowish tinge. Write the equation of
the observed process. Add a few drops of water by means of a dropping

glass: the mixture instantly ac-
quires a bright yellow colour.
Explain the whole experiment.

5. Effect of Catalysts on Reac-
tion Rate. Pour 3 ml of a hy-
drogen peroxide solution into each
of three test tubes. Add a pinch
of manganese dioxide to one, an
equal amount of ferric oxide to
the second, and of silica to the
third, all simultaneously. Observe
the decomposition of the hy-
drogen peroxide with the evolu-
tion of oxygen (glowing
splinter test). Does the process
proceed at the same rate in each
case? Write the equation of the
reaction.

6. The Kinetics of the Cataly-
tic Decomposition of Hydrogen
Peroxide. The experiment is

conducted in the set-up shown in Fig. 37. It consists of a flask (1),
a small test tube (2), rubber tubing (3), a delivery tube (4), a vessel
(5), and a eudiometer (6).

Fill the eudiometer with water and lower it into the vessel with
water. Place the end of the delivery tube from the flask under it. Pour
40 ml of 0.2% hydrogen peroxide, measured in a measuring cylinder,
into the flask, dry the interior of its neck with rolled up filter paper,
and attach the rubber tubing to the neck.

Weigh 0.10-0.12 g of the powdered mixture of manganese dioxide
and ferric oxide in a test tube. Connect the test tube with the rubber
tubing. Then lift the test tube into a vertical position (as shown by
the broken line in the figure); start the stop-watch as soon as the first
bubble of oxygen emerges from the delivery tube. Note the volume of
oxygen displaced at the end of every minute; conduct the experiment
for 10 minutes. Then repeat it.

Plot the results of the experiment in the form of a curve represent-
ing the rate of the decomposition of the hydrogen peroxide as a func-

Fig. 37. Apparatus for hydrogen perox-
ide decomposition

1 flask with side arm; 2 test tube;

3 rubber tubing; 4 gas-delivery tube;

5 vessel; 6 eudiometer.

Reacticn Rates and Chemical Equilibrium

*tion of time. In doing so, plot the time as abscissa (1 min.= 1 cm) and
the volume of oxygen displaced every minute as ordinate (1 ml of
oxygen = 1 cm).

Describe how the process proceeded during the interval of time in
which it was observed.

Record the results of the experiment as follows:

Time

1 min

2 min

3 min

10 min

Exper. 1

Total volume of O 2
in ml

Exper. 2

Exper. 1

Volume of O 2 in ml

Exper. 2

for every minute

Average
figures

Exercise 10

PREPARATION OF SOLUTIONS

SUBJECTS FOR STUDY

The concentration of a solution and various ways of expressing it; percentage,
molar molal, and normal concentrations; titres; calculations in converting one
type of concentration to another; the preparation of a solution according to a
weighed amount and by diluting a concentrated solution.

The concentration of solutions is in most cases expressed by the
amount of the solute by weight in a definite weight or volume of the
solution or the solvent. More specifically, concentration is usually
expressed in one of the following ways: as percentage concentration
(%) as molar concentration (M), as molal concentration (m), or as
normal concentration (N). In addition to this, the concentration of
a solution may be expressed in terms of its titre (T). Percentage and
molar concentrations were defined in Exercise 2. The molal concen-
tration indicates how many mols of the solute are dissolved in l.OUUg
of the solvent. The normal concentration is the number 01 gram-
equivalents of the solute in 1 litre of the solution. Finally, the
litre is the number of grams of the solute in 1 ml of the solution.

6-795

82 Exercise 10

Since the percentage and the molal concentration involve weights
of the solution, while the molar and the normal concentration, as well
as the titre, involve volumes, a conversion of one type of concentra-
tion to another requires a knowledge of the relative density of the
solution.

Example. Determine the molar concentration, the normal concentration, and the
titre of a 40% solution of sulphuric acid with a relative density of 1.307.

First of all, find the amount of sulphuric acid contained in 1 litre of the solu-
tion (x).

One litre of the solution weighs 1,000-1.307 = 1,307 g. We can then set up the
proportion:

100 40

1,307 x

Hence.

1,307-40 rooo

* = Ioo~~ ^ 522 - 8 g

^22 S
The titre of the solution is not difficult to determine: T = , n * n = 0.5228.

1 ,UUU

Since we know that the molecular weight of sulphuric acid is 98 and its equivalent

98 522 8

is = 49, we are able to determine the molar concentration M = TTTT = 5.33
Z Uo

522.8
and the normal concentration N = -Tq=10.66.

It is evident from the above example that such a conversion in-
volves first turning the known concentration into the amount of the so-
lute and then turning that into the concentration sought.

Very common calculations are those connected with diluting so-
lutions.

Example. What volumes of a 44% solution of KOH (relative density 1.46) and
water should be mixed to prepare 80 lit of a 12% solution with a relative density o
1.10?

First, determine the amount of anhydrous-KOH that-the required solution should
contain. The relative density of the solution is 1.10; the required amount, 80 lit.
Its weight will therefore be 80-1.10 = 88kg. From the required concentration we
derive the weight of the anhydrous KOH needed to prepare the solution:

100 kg contains 12 kg of KOH
88 kg " x * " *

88-12
x = -j^j- = 10.56 kg of KOH

We then find what amount of the initial solution contains this quantity of the
alkali:

44 kg of KOH is contained in 100 kg of the solution
10 56 *' w w " y n n

10.56-100
* = 44 = 24 k S

Preparation of Solutions 83

Since the relative density of the initial solution is known, it is not difficult
determine its volume:

lit

We now determine the volume of the water: 80 16.4 =63.6 lit.
Answer. We must take 16.4 lit of 44% KOH and 63.6 lit of water.

When weighed quantities (a, b) of several solutions of the same
substance, with different percentage concentrations (A%, J3%), are
mixed, the concentration (x%) of the substance in the mixture is cal-
culated by means of the formula:

a-A + b-B = (a + b)-x

The formula is derived as follows. Let a grams of the solution of con-
centration A% be mixed with b grams of the solution of concentration
B%. The weight of the resulting mixture will be a +6, while .its
concentration will be x%. In the first solution the quantity of the

solid (100%) substance is^jrr, while in the second it is -T7^> in the mix-
ture it will be 7m- We may therefore write the sum:

I UU

a-A , b-B (a + b).x
100" ' 100 ~~ 100

When we eliminate the common denominator, this yields the for-
mula given above.

Example. Twenty kg of a 6% solution of sodium chloride is mixed with 12 kg of
a 14% solution of the same salt. Find the concentration of the salt in the mixture.
Substitute the data given into the formula:

20-6 + 12.14^ (20 -i- 12). .v

This yields

20-6 + 12- 14 288

x =

20+12 32

The concentration of the salt in the mixture is 9% .

When given volumes of solutions are mixed, these should first be
converted to weights by means of the relative densities; this makes
it possible to use the same formula for the rest of the calculations.

In engineering the concentration of solutions is sometimes expressed
in degrees Baume. In the Baume scale corresponds to the rela-
tive density of water, while 10 corresponds to the specific gravity of a
10% solution of NaCl. Tables for converting degrees Baume to rela-
tive density may be found on p. 332. The following formulae may also
be used for such conversion.

84 Exercise 10

For liquids heavier than water the formulae are:

144.3 , , , ,, 144.3

and Be = 144.3

1 "" 144.3 Be W11 " ^ ~~ x "' v 7

For liquids lighter than water the formulae are:

144.3 , D , 144.3 - AA Q

and Be =- -- 1443

144.3 + Be-

1. Define the concentration of a solution.

2. For what substances do the values of the molar and normal con-
centrations coincide?

3. Find the gram-equivalents of the following substances:

Pb (N0 3 ) 2 H 3 P0 4 Cr 2 (SOJ 8 18H 2 O

4. The relative density of a solution is 1.184. Determine the con-
centration in degrees Baume according to the formula and by means
of the table (p. 332).

5. Derive the formula for finding the molal concentration of a
solution from the molar concentration. Use the designations: m for
molal concentration, M for molar concentration, v for the volume of
the solution, Y f r the relative density, and G for gram-molecule.

Problems

1. Eighteen litres of a 48% solution of sulphuric acid with a relative density of
1.38 is mixed with 2 litres of a 20% solution of the acid with a relative density
of 1.143. Determine the percentage and the molar concentration of the resulting
solution.

2. What weights of potassium hydroxide and water should be taken to prepare
75 lit of a 12% solution with a relative density of 1.1? Determine the titre of such a
solution.

3. Determine the titre and percentage concentration of a 3 N solution of H 3 PO 4
if its relative density is 1.055.

4. The titre of a nitric acid solution is 0.122; its relative density is 1.064. Deter-
mine its percentage and its normal concentration.

5. One litre of a solution contains 112 g of lead nitrate Pb(NO 3 ) 2 ; the relative
density of the solution is 1.082. Determine the percentage, molar, and normal con-
centrations of the solution and its titre.

6. What volumes of water and of a 25% solution of ammonia (relative density
0.91) should be taken to prepare 1 litre of a 10% solution with a relative density of
0.96?

7. How many mols of ammonia are contained in 1 litre of a 12.74% solution of it
with a relative density of 0.95?

8. Determine the percentage and the molar concentration of a nitric acid solution
prepared by mixing 85 litof a 60%solution (relative density 1.373) and 25 lit of a 24%
solution (relative density 1.145).

9. Determine the concentration of a sulphuric acid solution, knowing that 4 kg
of that solution, when mixed with 16 kg of a 40% solution of the same acid, pro-
duced 20 kg of a 36% solution.

10. Determine the titre and the normal concentration of a 25% solution of sodium
nitrate with a relative density of 1.185.

Preparation of, Solutions 85

LABORATORY WORK

Apparatus and materials: a 100 and a 250 ml measuring cylinder; 300 ml narrow
cylinder for measuring relative density; 500 ml flat- bottom flask; hydrometer (for
relative densities 1-1.4); glass stirrer (30 cm long glass tube with a bulb about 3 cm
in diameter at the end); saturated solution of potassium sulphate; crystalline mag-
nesium sulphate; 22.5 and 7.5% solutions of sodium chloride.

1. Preparation of a Solution from a Concentrated Solution and
Water. Receive an individual assignment from the instructor: for
example, to prepare 250 ml of a solution of K 2 SO 4 with a relative den-
sity of 1.025 from the laboratory solution.

To carry out the assignment it is necessary to establish the per-
centage concentrations of the required solution and the available one.

In a reference book find the percentage concentration of the re-
quired solution on the basis of the given relative density (1.025). Deter-
mine the relative density of the laboratory solution by means of a
hydrometer to + 0.003 and note the temperature at which the meas-
urement was made. In the reference book find the percentage con-
centration corresponding to that relative density. If the value re-
corded by the hydrometer is not given in the book, find the required
concentration by the method of interpolation (explained in Exer-
cise 2).

Now that all the data needed for the calculation are available de-
termine the volumes of the laboratory solution and water that have
to be mixed to give 250 ml of the required solution. To do this, first
find the weight of the substance needed to obtain the required solu-
tion; then determine the weight of the laboratory solution containing
the required amount of the substance; finally, knowing the relative
density of the solution, determine its volume. The difference between
the volume required and that found for the laboratory solution gives
us the volume of the water.

When the calculation is finished, measure off the calculated amounts
of the laboratory solution and water, pour them into a high cylin-
der, mix them thoroughly with a stirrer (tube with ball), and meas-
ure the relative density with a hydrometer.

Then wash the hydrometer with clean water, wipe it dry, and put
it away. Pour the prepared solution into a special carboy. Establish
(in per cent) the discrepancy between the calculated relative density
and the experimental value. A considerable discrepancy points to an
error either in calculations or in the experiment.

2. Preparation of a Solution from Two Solutions of Different Con-
centration, Receive an individual assignment from the instructor:
for example, to prepare 250 ml of a 9 4% solution of NaCl from a 7.5%
solution and a 22.5% solution of the same salt.

In the appropriate reference tables find the relative density values
for all three solutions and calculate the volumes of the 7.5 and 22.5%
solutions necessary to prepare the required solution. Measure o2 the

86 Exercise 10

calculated amounts of the solutions, pour them together, mix them
thoroughly, and determine the relative density of the prepared solu-
tion. Establish the discrepancy between the calculated and measured
relative densities.

3. Preparation of a Solution from a Weighed Amount of a Solid
Substance and Water. Receive an individual assignment from the
instructor: for example, to prepare 250 ml of a 6.5% solution of MgSO 4 .

To carry out the assignment, first calculate how much solid salt
this is, considering that the salt crystallises with 7 molecules of water.

Weigh the calculated amount of salt, transfer it into a beaker, and
dissolve it in half the required water (stir). When the salt dissolves,
pour it into a cylinder, add water to bring the volume up to 250 ml,
and mix the solution thoroughly. Determine the relative density of
the solution by means of a hydrometer and establish the discrepancy
between this value and that in the reference book.

Exercise 11

SOLUBILITY OF SUBSTANCES

SUBJECTS FOR STUDY

Solutions; processes that take place when a substance dissolves; crystallisation,
the reverse process; saturated and supersaturated solutions; solubility and the solu-
bility coefficient; solubility curves, and thermal effects of solution.

There is a limit to the ability of a solid substance to dissolve in a
certain quantity of a solvent at certain conditions (temperature and
pressure). An excess of the solid creates a saturated solution, i. e., a
system consisting of two components in which there is mobile equi-
librium between the liquid phase (the solution) and the solid phase
(the solute).

The concentration of a saturated solution at given conditions is
constant for every substance and is called its solubility. Solubility is
usually expressed by the number of parts by weight of the anhydrous
substance saturating 100 similar parts by weight of the solvent at
given conditions. In some books the number indicating the solubil-
ity of a substance is termed the solubility coefficient.

Not infrequently solubility is expressed in percentages, i. e., by the
number of parts by weight of the solute in 100 parts by weight of the
saturated solution.

In practice the solubility of liquid and solid substances depends
upon temperature, since their volume is not affected appreciably by
pressure. This dependence can be expressed graphically in the form
of a solubility curve (Fig. 38) by plotting the temperature as ab-
scissa against the solubility as ordinate.

Solubility of Substances

A solubility curve is a graphic representation of the solubility of
a substance at different temperatures; it can be used for the calcula-
tions involved in purifying solids by the process of recrystallisation.

Example. It is required to recrystallise 1 kg of cupric sulphate (in terms of the an-
hydrous salt). Calculate the amount of water in which it should be dissolved and
the yield of recrystallised salt if the cupric sulphate is dissolved at 100 and the so-
lution then cooled to 15.

&20

10 20 30 40 50 60 70 80 90 100
Temperature (*C)

Fig. 38. Blue vitriol solubility curve

On the chart (Fig. 38) find the solubility of the salt at 100 and at 15,

The solubility in 100 g of water at 100 is given by the ordinate y% =73.6; the so-

lubility at 15, by the ordinate y\ 20.5.

Since dissolving 73.6 g of the anhydrous salt at 100 requires 100 g of water,

to dissolve 1,000 g of the salt, we shall need x g of water:

100.1,000
*~ 73.6 -MoJg

When the solution saturated at 100 is cooled to 15, the number of grams of the
.salt precipitated will be equal to the difference between the ordinates on the chart:

y 2 ^. tjl ^ 73.6 20.5 = 53.1

Since the cooling of a solution prepared from 73.6 g of the salt yields 53.1 g of
the solid salt, the cooling of -a solution prepared from 1,000 g of the salt will yield
.v grams (in terms of the anhydrous salt):

53.1-1,000

=- 722

The yield of the recrystallised salt will therefore be 72.2%.

Substances whose solubility is reduced by cooling can be crystal-
lised by lowering the temperature of a saturated solution (recrystalli-

Exercise 11

sation). Slow cooling causes the substance to form large crystals, while
rapid cooling produces small crystals. The smaller crystals are a purer
product with fewer impurities, since the growth of larger crystals is
accompanied by the inclusion of minute quantities of the mother-
liquid containing admixtures of foreign matter. A chemically pure
substance is obtained by conducting recrystallisation (sometimes
repeatedly) at a drastically reduced temperature of the solution with
vigorous stirring *. Saturation is in most cases achieved at the tem-
perature at which the solvent boils; ice, snow, or cold water is used
for cooling solutions.

Buchner
Funnel

filter

Fig. 39. Hot-water funnel

Fig. 40. Suction filter

An aqueous solution saturated at 100 often has to be filtered to
remove various mechanical impurities (dust, turbidity, etc.). This
is accomplished by means of a special hot-water funnel. If the filter-
ing were done by the ordinary procedure, the drastic cooling of the
solution in the funnel would cause crystallisation of the substance and
hence clogging up of the filter pores; this would reduce the speed of
filtration and, more important still, would result in partial loss of
the substance.

A hot-water funnel (Fig. 39) is usually made of a copper alloy. It
has double walls and a side arm (the latter is unnecessary in the event
of electric heating). The space between the walls is filled with water
through the inlet at the top. The side arm is heated by a burner. An

* Substances whose solubility rises with cooling, such as Li^COs, Ca(CH3COO)2>
and Sr(CHsCOO)2, are recrystallised by the reverse procedure: a solution is saturat-
ed with such a substance at the lowest possible temperature and is then heated to
boiling. The crystals formed are filtered off from the boiling solution.

Solubility of Substances

ordinary glass funnel with a short stem and a plain paper filter is
placed inside the hot-water funnel.

After the glass funnel has been warmed (this should take 10- 15 min-
utes), 5-6 ml of boiling distilled water is poured into the filter and
allowed to pass through it. A clean beaker is then placed under the
filter, and the hot solution is poured into the filter. The beaker with
the filtrate, in which some crystals have formed on account of cooling,
is heated until they dissolve and then placed in a cooling medium
(snow, ice, or cold water) and stirred vigorously; small crystals are
precipitated.

Fig. 41. Filtration with the application of suction
1 j flask with porcelain funnel for filtration; 2 safety flask; 3 filter pump.

The recrystallised substance is separated from the mother-liquid
by filtration with the application of suction. This is necessary because
ordinary filtration leaves a great deal of mother-liquid on the surface
of the crystals (which contaminates the product with impurities) and
they require prolonged drying.

The porcelain funnel used for this purpose and known as a Biichner
funnel (Fig. 40) has a perforated bottom. Two pieces of filter paper
are placed over the funnel, and its upper rim is traced on the paper
by the blunt end of a pair of scissors. Two circular filters are then cut
out: the first coinciding exactly with the traced circle, the second
having a diameter 3-4 mm smaller. The smaller filter is put into the
funnel first. When both filters are in the funnel, they should be moist-
ened with water and their edges pressed tight to the rim with a finger
so that there are no cracks. The stopper supporting the funnel is then
fitted into a filter flask (Fig. 40), and this, through a safety flask, is
connected to a suction pump (Fig. 41). Suction is applied, and the
solution with the crystals is transferred from the beaker to the filter.

90 Exercise 11

Suction should be continued until the dripping of the solution from
the funnel ceases *. The remaining small quantity of the solution can
be removed by pressing the material retained on the filter with a clean
glass stopper (without discontinuing suction). When filtration is
complete, the filter flask should first be disconnected from the safety
flask; only then should suction be discontinued.

QUESTIONS

1. How will the solubility of Na 2 SO 4 - 10H 2 O be affected by a change
of temperature? Why does the solubility curve of this compound have
a point of inflection and what does it designate?

2. Three flasks contain a saturated, a supersaturated, and an un-
saturated solution of the same substance respectively. How are we
to determine which flask contains which solution?

3. Why does the solubility of most solids increase with a rise in
temperature?

4. Plot a solubility curve on the basis of the solubility data for
CaSO 4 (see table on p. 329). Why is calcium sulphate listed among ano-
malously dissolving substances?

5. Plot a solubility curve for Pb(NO 3 ) 2 on the basis of the following
data:

Temperature (in C) . . . 10 20 30 50 60 70

Solubility 44.5 52.2 60.8 78.6 88.0 97.6

Determine the substance's solubility at 37.

6. Who was the first scientist to study crystallisation processes?

Problems

\. The solubility of a salt at 80 is 64.7, while at 17 it is 13.8. What quantities
of the solid salt and of water ought to be taken to obtain 1 kg of the salt, if it is
recrystallised from a solution saturated at 80 and cooled to 17?

2. A salt whose molecular weight is 168 forms a saturated solution at 18 that
has a concentration of 3.4 M. If the relative density of the solution is 1.22, what
is the solubility of the substance and the percentage concentration of the solution?

3. The solubility of K 2 SO 4 at 40 is 13.1%. Determine the molar and the normal
concentration of a solution saturated at 40, if its relative density is 1.092.

4. A solution saturated at 100 is cooled to 14, yielding 112 g of crystallised salt.
How much water and how much salt were taken for the recrystallisation, if the so-
lubility of the salt at 100 is 52.7, while at 14 it is 7.9?

5. Twenty millilitres of a saturated solution of ammonium sulphate weighs
20.94 g and contains 3.24 g of (NH 4 ) 3 SO 4 . Determine the solubility of the salt and
the percentage, molar, and normal concentrations of the solution.

6. A salt has a solubility of 48.6 at 100 and of 16.4 at 20. Recrystallisation,
with cooling in that temperature range, yields 0.5 kg of recrystallised substance.
What amounts of water and salt were taken for recrystallisation?

* If the filtrate is not required for subsequent work, there is no need for a safety
flask.

Solubility of Substances 91

7. It is known that 42.34 g of a solution contains 7.28 g of the solute. Determine
the solubility of the solute and the percentage concentration of the solution.

8. Twenty kilograms of a salt solution saturated at 60 was cooled with snow.
What amount of recrystallised salt was recovered if the solubility of the salt at
60 is 110, while at it is 13.1? Calculate the yield in per cent.

9. A 9.92% saturated solution of K 2 SO 4 has a relative density of 1 .082, Determine
the solubility of K2SO 4 and the titre of the solution.

10. The percentage solubility of K2Cr a O 7 at 100 is 50.5, while at 10 it is
7.5. What amounts of K'aChO? and water should be taken to obtain 215 g of
the pure salt recrystallised in the above temperature range?

LABORATORY WORK

Apparatus and materials: drying cabinet; hot-water funnel; Biichner funnel for
filtration with the application of suction; desiccator; test tubes and rack; porcelain
mortar; porcelain casserole 7 cm in diameter; 100 ml measuring cylinder;
500 ml beaker with stirrer; four 50 ml beakers; thermometer up to 100; glass rod;
scissors; 10 X 12 cm squared paper; filter paper; crystalline potassium bichromate;
crystalline sodium thiosulphate; crystalline sodium acetate; crystalline cupric sul-
phate; 0.5 N solution of potassium iodide; 0.5 N solution of lead nitrate, and some
salt whose solubility is to be determined (RCl, KNO 3 , K 2 SO 4 , K 2 Cr 2 7 , Ba(NO 3 ) 2 ,
etc.).

1. Determining Solubility. This exercise should be carried out by
a group of students, each determining the solubility at some one
temperature. The individual assignments should be given by the in-
structor.

Weigh a porcelain casserole to 0.01 g.

Pour 10 ml of water into a small beaker, add 1-2 g of a finely ground
salt, place the beaker into a wire ring, and lower it into a large beak-
er with water (Fig. 42). Heat or cool the large beaker to the required
temperature and then adjust the flame of the burner so that the tem-
perature of the water in the beaker remains constant.

If, after a certain time, all of the substance has dissolved, another
portion is added, and so on until part of the solid salt remains undis-
solved. All this time the solution should be stirred and the tempera-
ture watched. The saturation of the solution, at a constant tempera-
ture, should be continued for 20-25 minutes.

Then remove the thermometer and the stirrer from the beaker. Allow
the undissolved solute in the saturated solution to settle and quickly
transfer the liquid to a weighed casserole, leaving the crystals in the
beaker. Weigh the casserole with the solution.

Now place the casserole on a ring with wire gauze and heat it until
all the water has been evaporated. Towards the end, conduct the
evaporation over as small a flame as possible to avoid any splashing
of the solution and consequent loss of the substance. Then place the
casserole for 20-25 minutes into a drying cabinet (Fig. 43) heated to
115.

Cool the casserole with the dry residue in a desiccator and weigh
it. Repeat the drying of the residue and the weighing of the casserole
(do this until the results tally!).

Exercise 11

From the data of the experiment determine the weight of the solu-
tion that the casserole contained and then the weight of the dry salt

Fig. 42. Preparation of
saturated solution

Fig. 43. Drying cabinet with gas heating

in it. Calculate the solubility coefficient and the percentage concen-
tration of the solution.

Substance

Temperature

Weight of casserole (in g)

Weight (in g) of

with solution

with dry salt

empty

solution

dry salt

water

On the basis of the experiments carried out by all the students of
the group, plot a solubility curve on squared paper, as shown in Fig. 44.

The broken line in Fig. 44 is an incorrectly plotted solubility
curve; the solid line is the correctly plotted curve. Itisevident from

Solubility of Substances

the diagram that the experimental determinations of points A, B,
and C were carried out with errors.

2. Purifying a Salt by Recrystallisation. Receive an individual
assignment from the instructor: for instance, to prepare 5 g of re-
crystallised potassium bichromate in fine crystals.

Temperature
Fig. 44. Experimental solubility curve

Choose the temperature range for the proposed recrystallisation-
On the basis of the initial and final temperatures, using the solubil-
ity table at the end of the book, calculate the amounts of the salt
and water that should be taken to prepare the required amount of the
recrystallised product.

Pour the calculated amount of water into a beaker. Weigh the
required amount of the salt and pour it into the beaker with the water.
Prepare a hot-water funnel for use. Heat the beaker with the water
and salt on wire gauze with an asbestos centre until the water be-
gins boiling and then pour it quickly into the filter in the funnel.

Heat the filtered solution, in which crystals have appeared with
cooling, until the crystals dissolve; then cool the beaker, using snow
or water from the tap. The solution should be stirred. The crystals
formed should be removed by filtration with the application of suc-
tion, transferred from the funnel to filter paper, pressed in it to re-
move moisture, and then dried in the air for 20 minutes, with occa-
sional stirring by a glass rod. Drying may be considered complete
when the small crystals no longer cling to a dry glass rod.

\Veigh the crystals and calculate the yield in per cent.

3. Preparing a Supersaturated Solution, (a) Pour 1 ml of water
into a dry test tube containing 5-6 g of powdered sodium thiosulphate.
Heat the contents of the test tube carefully until the entire salt dis-
solves by lowering the test tube into a beaker with hot water. Make

94 txerctse 11

sure that there are no small crystals of the salt on the test tube walL
Close the test tube with some cotton wool and cool the solution slowly
to room temperature.

Open the test tube and introduce a tiny crystal of the same salt
(a "seed crystal") into it; this initiates crystallisation in the supersat-
urated solution. Hold the test tube in your hand and note that the
solution grows warmer during crystallisation. Why is the precipita-
tion of the solid phase from a supersaturated solution attended by
the evolution of heat? Pour the solution with the crystals into a spe-
cial bottle.

(b) Conduct a similar experiment with rodium acetate. Write up
the experiment.

List the substances you know that can be used to prepare super-
saturated solutions.

4. Effect of Surface Area on Rate of Solution. Pour 10 ml of dis-
tilled water into each of two small beakers. Weigh two 1 g amounts
of cupric sulphate in large crystals. Grind one of the portions of cup-
ric sulphate in a mortar to a fine powder. Transfer both porticns into
the beakers with water and observe the time (in minutes) which elapses
before each portion dissolves completely.

Which crystals dissolve faster and why?

5. Effect of Temperature on Crystal Growth. Add 1 ml of a potas-
sium iodide solution to an equal volume of a lead nitrate solution;
this produces a yellow precipitate. Add 10 ml of water to the preci-
pitate and heat until it boils. If the precipitate does not dissolve
fully, add more water and continue the boiling. Pour the colourless
solution into two test tubes, heat them to boiling point, and then
place one into a rack for slow crystallisation, while the other is cooled
with water. When the solutions have both cooled to room temper-
ature, compare the size of the crystals. How should a saturated solu-
tion be cooled when it is desirable to obtain small crystals?

Exercise 12

SOLUTIONS OF FLUIDS

SUBJECTS FOR STUDY

Liquid-liquid systems with unlimited and limited solubility; critical solution
temperature; solubility of gases; absorption coefficient; Henry's Law and Dai-
ton's Law; solubility of gas mixture; partial pressures; thermal effect of the solution
of gases and changes in their solubility with temperature.

Liquid-liquid systems can be divided into three groups:
(1) systems with an unlimited miscibility of the two components
at ordinary conditions (e. g., alcohol water}\

Solutions of Fluids

(2) systems with a limited miscibility of the two components, which
changes with temperature and at a definite temperature becomes un-
limited (e. g., phenol* water);

(3) systems with a limited miscibility of the two components, which
at no temperature (pressure remaining ordinary) becomes unlimited
(e.g., OCl 4 -H a O).

j k

10 20 30 40 50 60 70 60 90 100% 8
tOO 90 80 70 60 50 40 30 20 10 0% X

Fig. 45. Composition-property diagram for phenol (A) water (E)

system

Systems of the latter two types are heterogeneous: they are charac-
terised by an interface separating the phases. When a system of the
second type, with a definite content of the components, is heated, the
interface will at a certain temperature disappear. This happens when
complete miscibility is achieved. Should the proportion of the compo-
nents be altered, miscibility is observed at a different temperature.
The temperature above which the system is homogeneous whatever
the proportion of the components is called the critical solution tem-
perature.

When such a solution is cooled below a definite temperature, it
breaks into phases and becomes heterogeneous again. If we prepare
several solutions with differing contents of the two components by
weight, a definite temperature of miscibility can be established for
each solution. For the system phenol water the miscibility tempera-
tures at first, with a rise in phenol concentration, mount, but after

In the presence of small amounts of water, phenol melts at room temperature.

Exercise 12

a certain peak they begin to decline. If we plot the percentage com-
position of such systems as abscissa against the miscibility tempera-
ture as ordinate, we obtain a composition-property diagram (Fig. 45).

Let us take a solution (its temperature ,/ and 60% content of com-
ponent A and 40% content of component B are designated on the
diagram by point D') that is a heterogeneous system and start heating
it (this is represented by the movement of point D' upwards along
the broken line). Until the temperature reaches t l the system re-
mains heterogeneous (with two phases). When the temperature t l
is reached (point D) 9 complete miscibility sets in and the sys-
tem becomes homogeneous. At any point above D (such as D")
the system is homogeneous, but if it is cooled (movement from D"
downwards along the broken line), it will separate into two phases
and become heterogeneous as soon as point D is reached. Point D is
thus both the miscibility and the immiscibility temperature. Since
the curve CMG is the geometric locus of points similar to D, it may
be called the miscibility or immiscibility curve. The region above
the curve is that of homogeneous solutions, while the region below
the curve is that of the existence of the two-phase system. Thus,
the system corresponding to the point M' on the isotherm t l will be
heterogeneous. A special point on the curve is point M: at the tempera-
ture / 2 corresponding to this point the system is completely miscible
irrespective of the concentration of the components (critical solution
temperature). For the phenol water system the critical solution tem-
perature is 68.4*.

Gases dissolve in liquids with a considerable diminution of volume
and evolution of heat. The thermal effects observed when some
gases dissolve in water are given in Table 4.

Heat of Solution of Some Gases in Water

Table 4

Gas

Number of gram-
rnols of water
per 1 mol of gas

Heat of
solution
(in Cal./mol)

Gas

Number of gram-
mols of water per
1 mol of gas

Heat of
solution
(in Cal./mol)

H 2 S

900

4.6

200

11.56

Cli

1,000

4.9

NO a

300

14.1

CO 2

1,700

5.88

HC1

400

17.47

SOa

250

7.5

HBr

400

19.9

NH 3

200

8.35

500

19.2

If a gas dissolves exothermically, while its generation from a so-
lution is endothermic, then in a system such as

H 2 S + H 2 O (solvent) ; solution + 4.6 CaU

Solutions of Fluids

a rise in temperature will, in accordance with Le Chatelier's Princi-
ple, cause equilibrium to shift in the direction of the endothermic
process, i. e., in the direction of the evolution of gas from the solution.
Consequently, the solubility of a gas diminishes with a rise in temper-
ature.

The amount of gas that dissolves in a liquid also depends upon
pressure. Under Henry's Law, the amount of a gas (by weight) that dis-
solves in a liquid is directly proportional to the pressure of the gas.

The solubility of a gas is expressed by the absorption coefficient;
this is the number of volumes of a gas, reduced to N. T. P., that dis-
solve in a single volume of the solvent (Table 5).

Table 5
Coefficients for Absorption of Several Gases by Water at C

Gas

Absorption
coefficient

Gas

Absorption
coefficient

Hydrogen ....

021

Chlorine

4 61

Nitrogen ....

0.023

Hydrogen sulphide . . . .

4.70

Oxvcren

049

Sulphur dioxide . . .

79 8

Methane

0.056

Hydrogen chloride. . . .

506.5

Carbon dioxide . .

1 713

Ammonia *

1,176

When a mixture of gases is dissolved, each gas, in accordance with
Dalton's Law, dissolves in proportion to the share of the pressure it
accounts for. If we denote the pressure of the gas mixture as P m and
the pressures of the individual gases comprising the mixture as P XJ
P y , and P 2 , we may write

Pm - P x + P y + Pz

The partial pressures of the gases (see p. 27) depend upon the con-
tent by volume of the gases in the mixture.

Example. Suppose there is a mixture consisting of 40% of oxygen and 60% of
methane at a pressure of 1 atm. In what proportion (by volume) will these gases dis-
solve in water?

The partial pressures of oxygen and methane will be 0.4 and 0.6 atm respectively.
Since the absorption coefficient for O 2 is 0.049 and for methane 0.056, their solubil-
ities at their partial pressures are 0.049-0.4=0.0196 and 0.056-0.6=0.0336. The
ratio of the volumes in the solution will be 0.0196:0.0336= 1:1. 71, whereas in the ini-
tial mixture it is 40:60=1:1.5.

Henry's Law and Dalton's Law are valid for gases of low solubility.

7- 795

Exercise 12

QUESTIONS

1. How many components and phases do the following systems have:

(a) CaCO 3 ; CaO + CO 2 (b) NH 3 + water ; NH 8 solution

2. By referring to Fig. 45 describe the state of the system at points
E, F, and //.

3. In the laboratory ammonia is often prepared by heating its 25%
solution in water. Give a theoretical explanation of this method of
preparing ammonia.

4. When a mixture of gases is dissolved, does the composition of
the dissolved gases (by volume) depend upon the pressure of the mix-
ture?

5. Why do bubbles of gas appear when a bottle of mineral or soda
water is opened? When does the evolution of gas cease?

Problems

(for absorption coefficients see Table 5)

1. How many grams of carbon dioxide will dissolve in 1 cu m of water at and
6 atm?

2. How much carbon dioxide (by weight) will be evolved at 15 from 0.5 litre of
an aqueous solution if the pressure is changed from 3 to 1 atm? The absorption co-
efficient of CO 2 at 15 is 1.02.

3. How much ammonia (by volume) will be obtained if we boil an aqueous solu
tion of it prepared by saturating 800 ml of water with gaseous ammonia at N.T.P.
After boiling the solution has been found by analysis to have an 0.2 M concentration.

4. From the absorption coefficient of hydrogen determine the weight of the hydro-
gen that dissolves in 5 cu m of water at 0.

5. A mixture of gases consisting of 52% by volume of CO and 48% of CO 2 is at
a pressure of 1 atm washed in ice-cold water. Calculate the volumes (in litres) of
both gases dissolved in 1 cu m of water and the percentage composition (by volume)
of the gas mixture dissolved in the water. The absorption coefficient of CO is 0.035.

6. From the absorption coefficients of nitrogen, oxygen, and methane calculate
the amounts by weight of each of them that dissolve in 1 cu m of ice-cold water at a pres-
sure of 10 atm.

7. The absorption coefficient of acetylene in water at 20 is 1.03. while in a sat-
urated solution of sodium chloride it is 0.05. How much acetylene by weight will
dissolve in 100 lit of water and in 100 lit of the salt solution if the pressure is 12 atm?

8. Calculate the solubility of the gases in volume percentages from their mixture
consisting of 33.3% of O 2 , 42.4% of N 2 , and 24.3% of CH 4 . The total pressure of
the mixture is 1 atm.

9. Solutions of hydrogen sulphide and sulphur dioxide have been saturated at 0.
What volumes of these solutions reacted if the reaction following the equation

2H 2 S ~ S0 2 ---- 2H 2 + 3S

yielded 96 g of sulphur?

10. The reaction

MnO 2 + 4HC1 = MnCl 2 + C1 2 + 2H 2 O

has consumed 43.45 g of MnO 2 . The chlorine generated is dissolved in water to
prepare a solution saturated at 0. What amount of water will this require?

Solutions of Fluids

LABORATORY WORK

Apparatus and materials: round flask with stopper and two tubes, as shown in
Fig. 47; 1 litre beaker with stirrer; large crystalliser; 25 ml measuring cylinder;
20 ml calibrated test tube; set of test tubes with phenol solutions; thermometer up
to 100; spring or screw clips; rubber tubing 5 mm in diameter, and squared paper.

Note. The phenol used to prepare the solutions should be either colourless or
faintly pink. Dark phenol should be distilled. To prepare the solutions pour the neces-
sary amount of water into a flask and add the weighed quantity of phenol; the
amounts should be calculated on the basis of the required concentration and the
number of people working in the laboratory. Heat the flask until it contains a homo-
geneous solution; then increase the temperature by 10 and pour the solution
into test tubes. If the solution in the flask becomes cloudy during the pouring, it
should be heated again. The test tubes should be stoppered up with corks and
numbered.

1. Determining the Critical Solution Temperature for Phenol in
Water. The arrangement for determining the critical solution temper-
ature (Fig. 46) consists of a large beak-
er heated by a burner and containing
a glass stirrer and a thermometer up
to 100 attached to the ringstand and
lowered 5-6 cm below the level of the
water in the beaker. The exercise also
calls for a rack with numbered test
tubes containing solutions of phenol in
water with known proportions of the
components by weight (see table).

Place test tube No. 1 in the beaker
with water and heat it over a small
flame. From time to time shake the test
tube vigorously without removing it
from the beaker (take care not to break
the beaker or the test tube!); every now
and then also stir the water in the beak-
er with the stirrer. Note the temper-
ature at which the milk-like solution
suddenly becomes transparent (misci-
bility). Remove the burner from under
the beaker. The water in the beaker
now begins to cool, and at a certain
temperature the solution in the test
tube becomes cloudy (separates into
phases). Note this temperature. The dif-
ference between the miscibility temperature and the immiscibility
temperature should not exceed 1. If it does, the experiment should
be repeated.

Now lower test tube No. 7 into the beaker and shake the mixture
in the test tube vigorously. If the system remains turbid, heat the

Fig. 46. Apparatus for studying
phenol water system

100

Exercise 12

beaker and note the temperature at which the system becomes misci-
ble; then allow the water in the beaker to cool and note the tempera-
ture at which the system becomes immiscible. If the system becomes
transparent as soon as the test tube is lowered into the beaker and
shaken, conduct the experiment in the reverse order: first cool the
beaker, recording the immiscibility temperature, and then heat it
to find the miscibility temperature.

In the same way conduct experiments with the rest of the solutions,
taking the test tubes in the following order: No. 2 and No. 6, No. 3
and No. 5, and finally No. 4. Record the results in the form of the
following table:

No. of
test tube

Mixture contains

Temperature (in C)

phenol
(in g)

water
(in g)

phenol
(in % by
weight)

water
(in % by
weight)

of misci-
bility

of immis-
cibility

mean

0.50

4.50

0.85

4.15

1.50

3.50

2.00

3.00

2.50

3.00

2.00

3.40

1.60

On the basis of the experimental data plot a solubility diagram on
squared paper. Draw an abscissa axis and two ordinate axes: the
right-hand axis for pure water and the left-hand axis for pure phenol.
A suitable scale for the abscissa is 1 cm^ 10%; for the ordinate, 1cm
= 10*. Plot the points and then connect them by drawing a smooth
curve. Determine the coordinates of the peak.

2. Determining the Solubility of Air in Water. Fill a round flask
(volume ^ 500 ml), fitted with a cork through which two tubes have
been passed (Fig. 47), with water from the tap. Do this by connecting
one of the tubes (1) with the tap by means of rubber tubing.

When the water has filled the whole of the flask and the delivery
tube (2), close the tube (1) leading to the tap by means of a clip. No
bubbles of air should remain either in the flask or in the tube. Sub-
merge the end of the delivery tube in a vessel (3) with water *. Lower
the calibrated test tube (4), filled with water, into the vessel and
place the end of the delivery tube under it.

After measuring the temperature of the water, place a burner under
the gauze and heat the water in the flask to boiling point. Allow it
to boil for 15-20 min and then note the level of the water in test

It is better to fill the vessel and the calibrated test tube with a salt solution.

Solutions of Fluids

101

tube 4. Discontinue the heating, measure the volume occupied by air
in the test tube and the volume of the flask with the delivery tube.

Fig. 47. Apparatus for determining the solubility of air in water

1_ water-supply tube with clip; 2 gas-deli very tube; 3 vessel with water;
4 calibrated test tube.

With the data of Table 6, the volume of the flask, and the temper-
ature of the water from the tap, calculate the volume of the air that
was dissolved in the water.

Table 6

Absorption Coefficient (a) of Air at Various Temperatures

0.0292

0.0214

0.0197

0.0228

0.0210

0.0194

0.0223

0.0205

0.0190

0.0219

0.0201

0.0187

Compare the calculated volume of the air with the experimental
value and explain the discrepancy.

102 Exercise 13

Exercise 13

PROPERTIES OF SOLUTIONS

SUBJECTS FOR STUDY

Properties of solutions; osmotic pressure; vapour pressure of pure solvent and of
solution; Raoult's Law; changes of the boiling and freezing points of a solution with
concentration; cryoscopy and ebullioscopy, and the determination of a substance's
molecular weight from the boiling and freezing points of its solutions.

Osmotic Pressure. The osmotic pressure of a solution is due to the
presence of the solute particles. Quantitative measurements of the
phenomenon made it possible to establish these regularities for non-
electrolyte solutions:

(1) Osmotic pressure is independent of the chemical nature of the
solute and is directly proportional to its concentration (at a constant
temperature).

(2) Osmotic pressure is directly proportional to the absolute tem-
perature.

From the equation

in which the osmotic pressure of the solution P osm has been substi-
tuted for gas pressure, we can determine P OS m, the molecular weight,
and the other quantities involved.

If, in solving problems according to this formula, the volume is
expressed in litres and the pressure in atmospheres, the quantity R
has to be assumed equal to 0.082.

Example. Determine the molecular weight of cane sugar if 50 ml of a solu-
tion containing 2 g of sugar at 25 exhibits an osmotic pressure of 2.86 aim.
By substituting all the known quantities into the formula, we obtain:

AA 2 > 0.082 (273 + 25)

M - - 0.05.2.86 - =341.7342

The study of the physical properties of nonelectrolyte solutions (va-
pour pressure, boiling point, freezing point) revealed that the relative
lowering of the vapour pressure of a solution is independent of the chem-
ical nature of the solute and is equal to the mol fraction of the solute
(Raoult's Law).

If the vapour pressure of the solvent at one and the same temperature
is denoted as p and the vapour pressure of the solution as p (it being
stipulated that p >p), the relative lowering of the vapour pressure

will be / t^~. Raoult's Law can then be expressed mathematically thus:

m
N v '

Properties of Solutions 103

where n is the number of mols of the solute; TV is the number of mols
of the solvent, and T-T/ is the ratio of the number of mols of the solute

n-}-N

to the total number of mols of the solute and the solvent, or the mol
fraction of the solute.

The number of mols may be represented as n = -~pand N = ~,

where m l and m z are the quantities by weight of the solute and the
solvent, while M l and M 2 are their respective molecular weights ex-
pressed in grams (gram- molecules).
Substituting these values into formula (I), we obtain

Po mi m 2

Mi + Aia

(")

Formula (II) makes it possible to determine the vapour pressure
of the solvent and the solution, the amount of the solvent (m 2 ) and
of the solute (m^, and the gram-molecules (or molecular weights) of
the solute (MJ and the solvent (M 2 ).

Example L Pure water at 20 has a vapour pressure of 17.5 mm. Determine the va-
pour pressure of an aqueous solution 01 urea CO(NH 2 ) 2 at 20, if 6 g of urea is dis-
solved in 178.2 g of water.

First it is necessary to determine the number of mols of urea:

Then we determine the number of mols of water:

The known quantities are substituted into formula (I):

17.5 -P__ 0.1 QQ1

17.5 ~"0.1 + 9.9 " U1
Therefore

p - 17.5 0.175 - 17.325 mm

Example 2. Determine the molecular weight of a substance if a solution of 1 g
of it in 37.8 g of acetone at 30 has a vapour pressure of 275.2 mm. The vapour pres-
sure of pure acetone at 30 is 281 mm.

First we determine the number of mols of the solute and of the acetone:

1 37.8

n ^~M tf = -gg-=0.05

The known quantities are then substituted into formula (II):

JL
281275.2 M

Hence

4.97M = 275.2 or M - 55.37

104 Exercise 13

The following is a corollary to Raoult's Law: A depression of the
freezing point or elevation of the boiling point of a solution is inde-
pendent of the chemical nature of the solute and is directly proportional
to its molal concentration.

This relationship is expressed by the formulae:

= K c m and

where m is the molal concentration, and K c and K e are proportional-
ity coefficients.

From the above formulae it follows that at m = 1 the quantities
ktfr and A/fo,// are numerically equal to the respective constants
K c and K e .

If we take several nonelectrolytes and dissolve a gram-molecule of
each in 1,000 g of one and the same solvent, the solutions, called molal,
exhibit an identical depression of the freezing point or elevation of
the boiling point. For each solvent the molal depression of the freez-
ing point K c and elevation of the boiling point K e are constants]
they are called the cryoscopic and the ebullioscopic constant respective-
ly. The values of these constants for variuos solvents are given in
Tables X and XI (p. 333).

By means of these constants, it is possible to determine the molec-
ular weights of dissolved substances. Indeed, let a grams of a sub-
stance with the unknown molecular weight M be dissolved in b grams
of a solvent. It has been established experimentally that the solution
boils (or freezes) at a temperature A^ degrees higher (or lower) than
the pure solvent does. From the data given it is easy to determine the
molal concentration m of the solution, taking into account that the

quantity -^ denotes the number of mols of the solute in b grams of

the solvent:

a _ h

m - 1,000

a- 1,000

m =

M-b

By substituting this value of m into one of the above formulae, we
obtain:

A __ K.fl.l.OOO M _ /C.q.1,000

A/ - or M- --

Example 1. Determine the molecular weight of glucose if a solution containing
1.35 g of it in 100 g of water freezes at 0.139. The cryoscopic constant of water is
1.86.

1.86-1,000.1.35
M = 100-0.139 = 18

Properties o Solutions 105

Example 2. What will be the boiling point of a solution of 3 .46 g of urea CO(NH 2 ) 2
in 100 g of water? The ebullioscopic constant of water is 0.52.

/(.q. 1,000 0.52-346-1,000

== Wb ^ 60.100 ^ '
The boiling point of the urea solution will be 100+0.3 = 100.3.

Cryoscopic and ebullioscopic methods provide important means of
determining the molecular weights of substances that decompose
upon vaporisation (when heated). It should be pointed out, however,
that Raoult's Law applies only to dilute solutions of nonelectrolytes.

QUESTIONS

1. Do the concentrations of solutions change with temperature?
If so, which and why?

2. How do we call solutions exhibiting the same osmotic pressure?

3. From what law does it follow that the melting point or the
crystallisation point of a substance is lowered by the addition of an
admixture of another substance?

4. At what concentration does a nonelectrolyte solution have an
osmotic pressure of 1 atm?

5. The cryoscopic and ebullioscopic constants of water are 1.86
and 0.52 respectively. What do these figures express?

Problems

(for the values of K c and K e see Tables X and XI)

1. The vapour pressure of a solution of 1.8 g of glucose in 179.8 g of water at 29
is 29.97 mm and is 0.03 mm less than the vapour pressure of pure water. Cal-
culate the molecular weight of glucose.

2. What is the freezing point of a solution of 18 g of cane sugar C 12 H 22 On in
200 g of water?

3. A solution of 1.6 g of naphthalene in 40 g of benzene exhibits a boiling point
elevation of 0.8. Calculate the molecular weight of naphthalene.

4. What is the osmotic pressure exhibited at 17 by a solution containing 68. 4 g
of cane sugar dissolved in 3 litres?

5. Calculate the molecular weight of glucose if a solution of 16.2 g of it in 671 ml
at exhibits an osmotic pressure of 3 atm.

6. Aqueous vapour tension at 20 is 17.5 mm. Calculate the vapour pressure of
a solution of 18.4 g of glycerol C 3 H 6 (OH) 3 in 176.4 g of water at the same tempera-
ture.

7. Determine the formula of a substance containing 94.38% of C and 5.62% of H,
knowing that when 4. 34 g of this substance is dissolved in 100 g of ethyl alcohol, the
boiling point of the solution is 0.29 higher than that of pure alcohol.

8. How many degrees will the freezing point of benzene be depressed if 9.27 g
of naphthalene Ci H 8 is dissolved in 225 g of it?

9. A solution of glucose CeH^Og and a solution of cane sugar C^t^Ou exhibit
the same osmotic pressure. How many grams of sugar are contained in 1 litre of the
second solution if 1 litre of the first contains 9 g of glucose?

10. The boiling point of a solution of 1.28 g of sulphur in 50 g of carbon disul-
phide is 0,23 higher than the boiling point of pure carbon disulphide. Determine the
number of atoms in a sulphur molecule.

106

Exercise 13

LABORATORY WORK

Apparatus and materials: the arrangements shown in Fig. 34D and Fig. 48; beak-
er with heat-insulating jacket; 25 ml measuring cylinder; 10 ml pipette; glass rod;
Assman thermometer; magnifying lens; weighing' bottle or watch glass; crystalline
calcium nitrate; commercially pure sodium chloride; weighed amounts of urea; snow
or ice, and pumice stone.

The urea should be taken in amounts of about 0.050 g and weighed on an analyt-
ical balance to 0.001 g.

1. Determining the

sists in experimentally

Molecular Weight of Urea. The method con-
determining the freezing points of a pure solvent
and of a solution, and then calculating the
molecular weight from the depression of the
freezing point.

The depression of the freezing point of
the solution is determined by means of an
arrangement (Fig. 48) consisting of a thick-
walled beaker (1) for the cooling mixture,
a stirrer (2), a cover (3), two test tubes (4)
and (7) of different diameters, a stirrer (5)
for the solution, and a Beckmann thermo-
meter (6).

Pour 10 ml of distilled water by means of
a pipette into the inner test tube and close
it with a stopper through which a stirrer and
a Beckmann thermometer have been passed.
The Beckmann thermometer should first be
adjusted to the required temperature range*.
Insert the test tube into a wider one, which
serves as a jacket to prevent excessive
cooling.

Put snow or ice in the beaker, add so-
dium chloride, and mix the contents thor-
oughly with the stirrer. To measure the tem-
perature of the cooling mixture, lower an
ordinary thermometer into it; see that the
temperature remains at about 6.

Now lower the test tube with the Beck-
mann thermometer into the cooling mixture.
While stirring the water in the test tube
with the stirrer, note the temperature (using
a magnifying lens) at which crystals of ice
appear. It is usual in such experiments for a
slight supercooling to take place, with the
temperature dropping below the freezing

* The thermometer should be adjusted in such a way that the mercury meniscus
in melting ice is within the 1.750-2.250 range.

Fig. 48. Apparatu for
determining the freezing
point of a solution

1 thick-walled beaker; 2
stirrer for cooling mixture;
3 lid; 4 test tube with
substance under investigation;
5 stirrer for solution; 6
Beckmann thermometer; 7
test tube.

Properties of Solutions

107

point. But after a certain drop, the mercury thread will abruptly
rise and come to a stop at the freezing point of the solvent or the so-
lution.

Record the temperature (to 0.005*). Then melt the ice in the test
tube (by warming it with your hand) and repeat the experiment.
The discrepancy between repeated determinations should not ex-
ceed 0.005.

Next remove the test tube with the water and pour a weighed amount
of urea (received from the laboratory assistant) into it; stir the
contents until the urea dissolves. Then close the test tube with the
stopper through which the thermometer and stirrer have been passed,
lower the test tube into the cooling mixture, and, while stirring the
solution, determine the freezing point of the .solution. Warm the
test tube with your hand, so that the crystals of ice melt, and repeat
the determination.

Record the results of the experiment in the form of a table such as
this:

Weight of
water in g

Freezing point of water
according" to Beckmann
thermometer

Weight of
urea in g

Freezing point of solu-
tion according to Beck-
mann thermometer

A t from
experiment
in C

average

From the data obtained calculate the molecular weight of urea by
means of the formula on p. 104.

2. Cooling Mixtures. Receive an individual assignment from the
instructor: for example, to prepare a cooling mixture with a tempera-
ture of 13.6.

With the aid of Tables 7 and 8, calculate the amount of salt that
has to be mixed with 25 parts by weight of water or snow to reduce
the temperature to that required.

Weigh the calculated amount of salt on a balance and measure
off 25 ml of water in a cylinder. Pour the water into a beaker with a
heat-insulating jacket, cool it to +13 and measure the temperature
to +0.2 (Assman thermometer).

Pour the weighed amount of salt into the beaker with water, stir
the salt vigorously in the water, and note the lowest temperature
reached (corresponding to the moment when all the salt dissolves).

To prepare a cooling mixture of snow and salt, transfer the weighed
amount of salt (or mixture of salts) into the beaker with the weighed

108

Exercise 13

Salts Used to Prepare Cooling Mixtures

Table 7

Salt

Salt + water

Salt -f snow

parts by weight
of salt to 100
parts by weight
of water at 13

lowest
temperature
in C

parts by weight
of salt to 100
parts by weight
of snow

lowest
temperature
in C

KNO 3

30
30
60
75
36
250

3.0
0.6
5.1
13.6
5.3
10.1
12.4

30
25
45
50
33
143

2.9
11.1
15.8
17.3
17.8
21.2
55.0

KC1

NH 4 C1 . ...

NH4NO 3 . .

NaNO 3 . . .

NaCl .......
CaCl 2 -6H 2 O . . .

Table 8
Mixtures of Salts Used to Prepare Cooling Mixtures

Composition of mixture (parts by weight
of salt to 100 parts by weight of snow)

KN0 3 (14) + NH 4 C1 (26) .
NH 4 NO 3 (52) + NaNO 3 (55) .
KNO 3 (9) + NHiSCN (67) .
NH 4 NO 3 (32) + NH 4 SCN(59)
HNO 3 (2)+ KSGN(112) . .
NH 4 SCN(57) + NaNO 3 (57) .

Lowest temperature
in C

17.8
25.8

28.2
30.6
34.1
29.8

snow, mix thoroughly, and measure the lowest temperature of the
mixture by a thermometer.

Compare the experimentally determined drop in temperature with
the data of the table.

3. Determining the Boiling Point of a Pure Solvent and a Solution.
Into the test tube of the arrangement shown in Fig. 34 D (p. 54) place
a small lump of pumice stone and add 4 ml of distilled water. Deter-
mine the boiling point of the water by the thermometer. Does it
differ from the true boiling point of water?

Weigh 3 g of calcium nitrate crystalline hydrate Ca(NO 3 ) 2 -4H 2 O,
put it into the test tube with water, and, when all the salt has dis-
solved, measure the boiling point of the solution by the procedure
described in Para, "b" on p. 55.

How does a rise in the concentration of the solute affect the boiling
point of a solution?

Electrolytes 109

Exercise 14

ELECTROLYTES

SUBJECTS FOR STUDY

Theory of electrolytic dissociation; electrolytes and nonelectrolytes; degree of
electrolytic dissociation; strong and weak electrolytes; electrolytic dissociation as
a reversible process; equation for ionisation constant; electrolytic dissociation of wa-
ter; ion product for water; pH-value and the acidity or alkalinity of a solution;
determining [H-] and [OH'] in solutions of weak electrolytes.

1. Electrolytic Dissociation. Electrolytes are substances that, in
a medium with a high dielectric constant (water, alcohols, etc.),
break up into ions *. This process of the breaking up of a substance
into ions is called electrolytic dissociation.

The dissociation of a substance into ions is accompanied by the
salvation of the ions, i.e., their interaction with the polar molecules
of the solvent. If the solvent is water, the term hydration is used for
solvation.

The process of electrolytic dissociation ** should be expressed by
equations such as these:

NaCN -f mH 2 O ^ [Na-/iH 2 O]' + [CN^H 2 O]'
HCN + mH 2 O ; [H-H 2 O]' -f [CN-</H 2 O]'

In practice it is customary to use simplified equations:
NaCN ^ Na* + CN'
HCN ; H' + CN'

Electrolytic dissociation is a reversible process, with unionised mole-
cules as well as ions present at equilibrium. The ratio of the number
of ionised molecules to the total number of the molecules of the elec-
trolyte is called the degree of electrolytic dissociation, or, more often,
the degree of ionisation (a).

For one and the same electrolyte, a increases with dilution; when
the solution is highly dilute, the electrolyte is dissociated completely
(a = 1). Only solutions of the same normality can be compared as
far as degree of ionisation is concerned. Sometimes the degree of
ionisation is expressed as a percentage (a% =a-100). According to
their degree of ionisation, electrolytes are classified as strong, medium,
or weak (Table 9).

* Electrolytes in the molten state likewise break up into ions.
** Molecules break up into ions at the site of an ionic or markedly polar bond.

110

Exercise 14

Table 9

Degree of lonisation of Some Electrolytes in 0.1 N Solution

Strong electro-
lytes (oc>30%)

a %

Medium electro-
lytes (0 = 2-30%)

a %

Weak electrolytes
(a< 2%)

a %

HN0 3

H 3 P0 4

CH 3 COOH

1.36

HC1

H 2 S0 3

H 2 C0 3

0.17

H 2 SO 4

H 2 S

0.07

KOH

HCN

0.01

NaOH

NH 4 OH

1.4

Ba(OH) 2

Nearly all salts belong to the group of strong electrolytes. In an
0.1 N solution the various types of salts (M stands for the metal; X,
for the acid radical) have roughly the following degrees of ionisation:

+1-1
MX

80-90 %

+2-1

MX 2

+1-2

and
70-80%

+2-2

35-45%

+3-1

MX a

and
60-65%

+ 1-3

M 3 X

The equilibrium arising in the solution of a weak electrolyte, say
HCN, may be expressed by the equation:

HCN ^ H* + CN'

If we apply to this equilibrium the Law of Mass Action, we get the
following equation for the equilibrium constant:

[H'HCN'

- = Kion

[HCN]

In this case the equilibrium constant, i.e., the ratio of the product
of the ion concentrations to the concentration of the undissociated,
or unionised, molecules of the weak electrolyte at equilibrium, is
called the dissociation constant, or, more often, the ionisation constant.

Unlike the degree of ionisation, the ionisation constant does not
change when the solution is diluted.

TableXII (on p. 334) lists the ionisation constants of some substances.

By increasing the concentration of the H* ions or the CN' ions it
is possible to shift the balance in an HCN solution, reducing its degree
of ionisation.

2. Electrolytic Dissociation ol Water. Water is an exceedingly weak
electrolyte, which dissociates according to the equation: H 2 O ^
'

The product of the hydrogen and hydroxyl ion concentrations is
termed the ion product for water [H*]-[OH'] = Kw Its numerical
value at 22 is MO' 1 *.

Electrolytes 111

> In pure water [HI- [OH'] and, accordingly, [H']
-= 1-10' 7 . The value of [H'] may be used to express the acidity or
alkalinity of a solution:

in an acid solution [H']> [OH']>1- 1(T 7 ;

in a neutral solution [H*] - [OH'] - M(T 7 , and

in an alkaline solution [H']<[OH']<1, 1(T 7 .

Instead of using the actual hydrogen ion concentration [H*], it is
more customary to use the logarithm of that quantity, with its sign
reversed. This is known as the hydrogen exponent or, simply, the
pH-value: pH log [H']. The following diagram shows the pH-
value for solutions of varying acidity or alkalinity:

pH = 01 23456

acidity increases

neutral

8 9 10 11 12 13 14

alkalinity increases

Strong electrolytes break up into ions completely; for this reason
[H'] or [OH'] in solutions of strong acids or alkalis may be assumed
to be equal to the molar concentration of the acid or alkali.

Examples, (a) Determine the hydrogen ion concentration and the pH-value in an
0.01 M solution of HC1.

HC1 is a strong electrolyte, and it therefore dissociates completely in solution:

HC1 -- H' + C1'
0.01 0.01 0.01

Consequently, the gram-ion concentration [H*] is numerically equal to the molar
concentration of HC1:

[H*] = 0.01 = 1.10~ 2 ; pH = log 1 -10-2 ^ 2 *

(b) Determine [H'] and pH for an 0.01 M solution of KOH.
The concentration of OH' ions is numerically equal to the KOH concentration
(as in "a"):

[OH'] 0.01 = 1-10- 2

From the ion product for water we then find that

- " 1 ' or P = = i2

For solutions of weak monobasic acids and monovalent alkalis these
quantities are determined somewhat differently. For example, in
the case of acetic acid, in whose solution we have the equilibrium
CH 3 COOH H*-j-CH 3 COO', the equation for the ionisation con-
stant will be:

[H-].[CH 3 COQ']
[CH 3 COOH] ~ A

* To calculate the pH-value exactly for. solutions of strong acids and alkalis, it
necessary to use ion activities instead of ion concentrations.

112 Exercise 14

Since acetic acid is a weak electrolyte, there will be but a small error
involved in assuming [CH 3 COOH] to be equal to C, the total concen-
tration of the acid. The equation therefore becomes:

[H1-[CH 3 COO'] = K-C
Furthermore, since [HI - [CH 3 COO'], we get
[H'P - K-C and [H'l =

Example. Determine [H*] and pH for an 0.1 M solution of CH 3 COOH, knowing
that K

By substituting the known quantities into the formula [H'J = ]/^C-C, we obtain:

[H*] = |A. 82.10-5- MO" 1 = 1.34-10-3
pH - log [H'] = (log 1.34 + log 10- 3 ) - (0.12713) 2.87

Knowing [H'], it is not difficult to calculate the degree of ionisation from the re-

fH*l
lationship a = -p .

In this particular case

1.34-10-3
a = ifiQ - l = 1.34-10-2 or a% = 1.3496

3. Dissociation by Steps. In the case of polybasic acids and poly-
valent bases, all the hydrogen or hydroxyl ions in their molecules do not
split off at once. Electrolytic dissociation proceeds by steps.

The following are examples of stepwise dissociation equations:

I step H 2 S ^ H* + HS' Co (OH) 2 ^ CoOH* + OH'

It It

1 1 step H* + S" Of + OH'

Two equations of ionisation constants may be written for ionic equi-
librium in the H 2 S solution:

[H'HHS'] ___ K _ q i _ 8 i

"" A "" U an

[H 2 S] "" l "" [HS'] ~ 2 -

A comparison of the values of /Ci and AT 2 shows that /Ci is more than

225,000 times (9-10" 8 : 4-10" 13 ) bigger than /C 2 , which may therefore

be neglected in calculating [H'] and the pH-value. For weak polybasic

acids there will be a small degree of error involved in taking into

account only the first step of ionisation.

Some salts, such as Ca(NO 3 ) 2 , FeCl 3 , and BeF 2 , likewise dissociate
stepwise in solutions, but their stepwise dissociation is less pronounced
than that of polyvalent bases or polybasic acids.

4. Molecular Weights of Electrolytes. In determinations of the molec-
ular weight of electrolytes according to the depression of the freezing
point or the elevation of the boiling point of the solution, or according

Electrolytes 113

to the osmotic pressure, use is made of formulae in which the coefficient
M = l ' K -\'' tt and M= l y m p RT

' ostn

The theory of electrolytic dissociation has made it possible to estab-
lish the physical meaning of the coefficient /. If we denote the number
of the solute molecules as n and the degree of ionisation as a, then na
will be the number of ionised molecules, while n na will be the num-
ber of unionised molecules. The /za number of molecules that undergo
ionisation give rise to nap ions, where P is the number of ions arising
from a single molecule of the solute.

Let us denote the ratio of the sum of the number of unionised mole-
cules and the number of ions formed to the total number of molecules
taken as i. In that case:

For a binary electrolyte p = 2 and i = 1 -fa; for a ternary elec-
trolyte p = 3 and i = 1 + 2 a, etc.

In this way the quantity /, known as the van't Hoff factor, is a
measure of the increase in the number of particles in an electrolyte
solution. By means of this factor, it is possible to calculate the molec-
ular weight of electrolytes according to the above formulae, provided
their degree of ionisation is known.

Example L Calculate the molecular weight of potassium cyanide if a solution
containing 2.48 g of that salt in 500 g of water freezes at 0.27 and if cc=0.9.

We tind i =l-f 0.9=1.9. The cryoscopic constant of water K =1.86. By substi-
tuting all these values into the formula, we obtain

1.9 -1.86. 1,000. 2.48
M - 500T.T)^7 ^ 6o

Example 2. Calculate the osmotic pressure of the solution in the previous example
at 17 (assuming the relative density of the solution to be 1).

i-m-RT 1.9.2.48-0.082.290

65-0.502

3 ' 42 atm

QUESTIONS

1. What experimental data suggested the theory of electrolytic
dissociation to Arrhenius?

2. Write the equations- of electrolytic dissociation for the following
substances:

Fe 2 (S0 4 ) 3 NaHC0 3 MgOHCl KCr (SO 4 ) 2
3 Why does a solution of hydrogen chloride in water have acidic
properties, while a solution of the same substance in benzene does not
exhibit such properties?

8-795

114 Exercise 14

4. What are the factors that cause electrolytes to break up into ions
in an aqueous solution?

5. Write the equation for the ionisation constant of hypochloric
acid HOC1. What substances should be introduced to reduce the ioni-
sation of this acid?

6. From the theoretical considerations outlined in the introduction
to this Exercise derive the formula

[OH'] = V K base' C base

for the ionisation reaction NH 4 OHSNH' 4 + OH'.

7. Write the ionisation equations for H 3 PO 4 on thejt basis of the
ionisation constant values. Show which of the negative ions will^be
present in the solution in a relatively higher concentration.

Problems

(for ionisation constants see Table XII on p. 334)

1. Taking into account the first step of the ionisation of H 2 S, calculate a% for
an 0.01 M solution of it.

2. Calculate [H*] and a% for an 0.05 M solution of hydrofluoric acid HF.

3. Calculate the degree of ionisation of a binary electrolyte solution containing
0.50 g of salt in 100 g of water. The solution boils at 100.04; the molecular weight
of the salt is 126.

4. Determine which solution has a higher pH-value: an 0.01 M solution of HC!
or an 0.1 M solution of HCN.

5. Where is the pH -value greater: in an 0.0001 M solution of HNO 3 or in an 0.01
M solution of H 2 S?

6. Calculate the ot% and the pH-value of a 4% solution of boric acid, assuming
the relative density of the solution to be 1.

7. Calculate the osmotic pressure of an 0.1 M solution of BaCl 2 at 20, its apparent
degree of ionisation being 0.72.

8. Calculate the molecular weight of a binary salt (ot%~82) if 0.33 g of it is
dissolved in 100 g of water and the boiling point of the solution is 100.026.

9. What will be the freezing point of a solution of CaCl 2 containing 8.0 g of the
salt in 400 g of water if <x=70%?

10. What will be the boiling point of an 0.2-molal solution of CuS0 4 if its degree
of ionisation is 45%?

LABORATORY WORK

Apparatus and materials: the arrangement for determining conductivity shown
in Fig. 49 and that for determining the conductivity of molten salts shown in Fig. 50;
the arrangement for observing ion migration shown inFig.51;the cryoscopy arrange-
ment shown in Fig. 48; test kibes and rack; 10 ml measuring cylinder; 1 ml and 10
ml pipettes; crystalline sodium acetate; crystalline potassium nitrate; anhydrous
acetic acid; crystalline ammonium chloride; recrystallised potassium chloride in
weighed amounts; commercial sodium chloride; 0.5 N solution of potassium iodide;
5% solution of alcohol; 5% solution of sugar; 10% solution of hydrochloric acid;
5% solution of potassium nitrate; 5% solution of sodium hydroxide; 25% and 1%
solutions of ammonia; solution of phenolphthalein; solution of methyl-orange; lit-
mus solution; starch solution; distilled water; boiled distilled water; snow or ice.

Note. The potassium chloride should be weighed in amounts of about 0.050 g;
the weighing should be done on an analytical balance to 0.001 g.

Electrolytes

115

1. Conductivity of Electrolyte Solutions. The apparatus (Fig. 49)
consists of a wide mouth bottle with a rubber stopper, through which
a funnel and two carbon electrodes have been passed. The electrodes
have metal caps with terminals at the top. A line should be drawn with
a file at the bottom ends of the
electrodes to serve as a mark for
the amount of electrolyte poured
in.

Through the funnel pour dis-
tilled water into the bottle to the
mark on the electrodes; plug in
the apparatus and switch on the
electricity. Does the lamp glow?
The glowing of the lamp is a sign
that there is an electric current

Fig. 49. Apparatus for determining th
conductivity of solutions

in the circuit.

Repeat the experiment, filling
the bottle consecutively with so-
lutions of alcohol, sugar, hydro-
chloric acid, pottassim nitrate, and potassium hydroxide. Before each
experiment wash the electrodes, the funnel, and the bottle thor-
oughly; rinse them with distilled water.

Which are the substances that conduct electricity in solution?
What is an electrolyte? Write equations of electrolytic dissociation
for the electrolytes tested. What are the ions characteristic of solutions
of acids and alkalis?

2. Degree of lonisation of an Acid. Pour concentrated acetic acid
into the clean and dry bottle of the apparatus shown in Fig. 49 and
switch on the electricity. Then add distilled water through the funnel,
raising the electrodes over the stopper in such a way that the level of
the solution in the bottle should not rise above the mark (in order to
keep the effective surface of the electrodes constant). Note the changes
in the glowing of the lamp. What accounts for these changes? Check
your conclusions by calculating the values of a % for 0.1 and 0.001 N
solutions of CHgCQOH.

3. Degree of lonisation of Alkali and Salt. Pour a small amount
of a 25% solution of ammonia (1 volume) into the bottle of the conduc-
tivity apparatus and switch on the electricity. Note the brightness of the
lamp. Switch off the electricity. Carefully add glacial acetic acid

7p volume) in small portions. Stir the solution, cool it, and switch

on the electricity. Note the change in the glowing of the lamp.
Explain this, writing the necessary equations. Are salts weak or
strong electrolytes?

4. Determining the Degree of lonisation of a Salt. This experiment
is conducted with the apparatus shown in Fig. 48, employing the same

116

Exercise 14

procedure as that described in Para. 1 of Exercise 13. First determine
the freezing point of the pure solvent water. Two determinations
should be made, and the freezing point should not differ by more than
0.005. An accurately weighed amount of recrystallised potassium
chloride is then dissolved in the water. The freezing point of the solu-
tion should now be determined twice.

Since we know the weight of the salt taken for the experiment, its
molecular weight, the weight of the solvent, its cryoscopic constant,
and the depression of the freezing point of the solution, we are able

liiiiiiiiiiiiiiiiiiiiiiiiiniiiiiiiiiiiiiiuimn

Fig. 50. Apparatus for determining
the conductivity of molten salts

Fig. 51. Arrangement for observing
ion migration

o calculate the van't Hoff factor, from which we derive the degree
of ionisation. Compare the obtained value with the theoretical and
establish the relative error of the determination.

5. Conductivity of Molten Electrolytes. Assemble the apparatus
shown in Fig. 50. In the clamp attached to the ringstand fasten a stop-
per with two graphite electrodes passed through it. Lower the elec-
trodes into a porcelain crucible containing finely ground potassium
nitrate * and switch on the electricity. Does the lamp glow? Heat
the crucible by means of a burner and note the appearance of current
in the circuit when the salt melts (m. p, 336).

6. Ion Migration. Hold the glass tube of the apparatus shown i n
Fig. 51 verticalli and close one end of it with a rubber stopper that

* Potassium nitrate may be replaced by other low-melting salts, such as tin
chloride (m. p. 241) or zinc chloride (m. p. 365),

Electrolytes

117

has a graphite electrode passed through it. Pour 1 ml of a potassium
iodide solution, 1 ml of a starch solution, and 1 drop of an alcoholic
solution of phenolphthalein into the tube; fill the tube to the top with
distilled water, allowing for the closing of the tube with a second rub-
ber stopper. Turn the closed tube over repeatedly to mix the solution,

Then fasten the tube in the clamp and connect thin wires from a
storage battery or some other direct-current source to the terminals at
the end of the graphite electrodes.

After a certain period of time a blue colouration appears at the
positive electrode, while a crimson colouration appears at the nega-
tive electrode.

What accounts for these colourations at the electrodes? What ions
migrated in the solution and in what direction? What are cations and
anions?

7. Colour of Indicators. Pour 3 ml of distilled water into each oi
9 clean test tubes. Add 1 drop of phenolphthalein into each of the
first three, 1 drop of a methyl-orange solution into each of the next
three, and 1 ml of a litmus solution into each of the remaining three.
Now add 2-3 drops of a hydrochloric acid solution to 3 test tubes (one
with phenolphthalein, another with methyl-orange, and the third
with litmus). Add 2-3 drops of a sodium hydroxide solution to three
other test tubes with different indicators.

Record the colour of the indicators in different solutions as follows:

Indicator

Type of solution

acid, pll < 7

neutral, pH=7

alkaline, pH> 7

Phenolphthalein
Methyl-orange

Litmus . .

8. Equilibrium in a Solutionof a WeanAcid. Pour 4-5 ml of water
into a test tube; add 1 drop of concentrated acetic acid and 2 drops oi
a methyl-orange solution. What colour does the solution acquire
and why? Write the equation for the electrolytic dissociation of
acetic acid and the equation for the ionisation constant.

Divide the solution into two equal parts and add a few crystals of
sodium acetate to one half. Shake the test tube and compare the col-
ours of the solutions in the two test tubes. Explain this in terms oi
the equilibrium constant.

What should be introduced into a solution of a weak acid to shift
the equilibrium towards the formation of undissociated molecules?

118 Exercise 15

9. Equilibrium in a Solution of a Weak Base. Carry out the same
experiment as above, taking an ammonia solution instead of acetic
acid, phenolphthalein instead of methyl-orange, and ammonium chlo-
ride instead of sodium acetate. Write uf) the experiment.

What should be introduced into a solution of a weak base to shift
the equilibrium towards the formation of undissociated molecules?
Formulate a general conclusion about the shifting of equilibrium in
solutions of weak electrolytes.

Exercise 15

REACTIONS IN ELECTROLYTE SOLUTIONS

SUBJECTS FOR STUDY

Equilibrium in a precipitate-sctuticn system; solubility product; formation of
precipitates and their solution; Berthollet's Rule in the context of the theory of elec-
trolytic dissociation; reactions between ions in solution; neutralisation; hydrolysis
and particular cases oi it; amphoteric electrolytes; properties of hydroxides and ra-
dius of positive ion charge.

1. Solubility Product. The solubility product (SP) of a slightly sol-
uble binary electrolyte is a quantity equal to the product of the con-
centrations of its ions in a saturated solution *. For example, the solu-
bility products of AgCl, CaCO 3 , and A1PO 4 are expressed as follows:

SP A gci = [Agl -ICl'l;

The solubility product characterises the solubility of slightly soluble
substances at a given temperature', the smaller the solubility product,
the less soluble the compound.

If 5P Ag ci = 1.7- 1CT 10 and SP Asl - 8.5- 1CT 17 , it follows that AgCl
is more soluble than Agl.

The solubility product is easily converted to the molar or ionic
(gram-ions per litre) solubility. Let us consider, for instance, the
case of barium sulphate, for which SP Ba so 4 = [Ba"l- [SO/]. If
we denote the ion concentration as x, we get: SP = x-x, or SP = x 2 ,
from which it follows that x = VSP~. In this way from the solubility
product we can determine the number of mols of BaSO 4 or the number
of gram-ions of Ba"orSO 4 " in 1 litre of a saturated solution. To deter-
mine the solubility in grams per litre (B), we must multiply the molar

* In accordance with the theory of strong electrolytes, the values of concentra-
tions for the more soluble compounds ought to be replaced by activity values. These
are products of the concentration by the activity coefficient.

Reactions in Electrolyte Solutions 119

or ionic solubility (x) by the molecular weight (M) or the ionic, atomic,
or group weight (A):

4 4

Example. Calculate how many grams of Agl are contained in 500 ml of a saturat-
ed solution at 25 if SP AgI = 8.5- 10~ 17 .

[Ag"HI'J = SP ==8.5-10-"; * = YSP = V"8.5- 'to-* 7 = 9.22 !<)-

Since M AgI =234.8, the solubility of |AgI equals 9.22 -lO" 9 . 234.8 g/1. Consequently,
500 ml contains

9.22- 10-- 234.8- 0.5= 1.08-lO^g Agl

The formation and solution of precipitates can be explained in
terms of SP values. A precipitate is thrown down when the product of
the ion concentrations exceeds the solubility product; a precipitate
dissolves when the product of the ion concentrations is less than the
solubility product.

Example. How many times greater is the solubility of Agl in pure water than in
an 0.01 M solution ofNal if SP Agl = 8.5- 10~ 17 ?

From the value of 5P AgI we derive the solubility of Agl in pure water:

Sodium iodide is a strong electrolyte that breaks up into ions completely; there-
fore [T] is equal to the total concentration of the salt, i. e., to 0.01, or l-10~ a .
Hence:

8. 5 -1C- 17
[Ag']. 1-10-2- 8.5- 10-17 anc j [Ag'J = 1>1Q _ 2 =- 8.5-10~ 15 rnols/1

Since [AgI]=[Ag'H8.5-10- 15 , the ratio of the solubility of Agl in pure water

9.22-10 9

to its solubility in a Nal solution will be g 5.10~ 15 **' e " ttle so ^ u ^^^y ^
Agl in pure water will be 1,085,000 greater than in an 0.01 M solution of Nal.

2. Ionic Reactions. Chemical reactions in solutions of electrolytes
are reactions not between molecules, but between ions.

In equations of reactions the ions of an electrolyte may be separated
by a plus sign. For instance, sodium nitrate and sodium chloride may
be written thus:

Ag + NO;

and

Na + Cl'

The interaction between these substances may be expressed by the
equation:

Ag' -f NO; -f- Na' + Cl' - AgCl + Na' +

120

Exercise 15

From this equation it follows that only 2 of the 4 ions participating in
the reaction actually interact. These are the Ag' and Cl' ions; the
Na* and NO 3 ' ions do not take part in the reaction. Reactions are,
however, possible in which all 4 ions take part, e. g.:

Zn" + SOl + Ba" + S" = ZnS + BaSO 4

3. Equations of Exchange Reactions in Solutions. Chemical reactions
can be represented by equations of three types: (1) molecular, (2)
ionic-molecular, and (3) ionic. For example:

(1) BaCl 2 + H 2 SO 4 = BaSO 4 + 2HC1

(2) Ba" + 2C1' + 2H* + SOl = BaSO 4 + 2H* + 2CI'

(3) Ba" + SOl = BaSO 4

The latter (ionic) equation indicates only the ions that take a direct
part in the reaction. Henceforward, in writing the equations of the
reactions between electrolytes in solutions, we shall use equations of
the second type.

Reactions between ions may be referred to one of the following four
types.

A. Reactions Proceeding with the Formation of a Precipitate

When writing the equations of reactions of this type, it is necessary
to bear in mind the solubility of the substances involved (Table 10).

Table 10
Solubility of Some Substances in Water

(S stands for soluble; SS, slightly soluble; I, insoluble; DW, decomposed by water)

Cation

An ion

Na-

NH'

Ag'

Mg"

Ca"

Ba-

Fe"

Cu"

Zn"

Hg"

Pb"

AP'

Fe-

Cd"

3r'

)H'

:o 3 "

>0 4 "

: 2 H 3 o 2 '

Reactions in Electrolyte Solutions 121

To obtain an insoluble compound (precipitate), it is necessary to
proceed from two soluble substances containing the required ions.
In the equation of the reaction, the substance thrown down as a pre-
cipitate is represented by its molecular formula.

Below, for instance, are equations of reactions for the precipitation
of ZnS and A1PO 4 :

Zn" + SOl + 2K' + S" = ZnS + 2K' +
Al~ + 3C1' + 3Na' + PO 4 ' - A1PO 4 + 3Na' + 3d'

8. Reactions Proceeding with the Evolution of a Gas

The gases formed in ionic exchange reactions are complex gaseous
substances. This group of substances, except for some gaseous fluo-
rides, consists mainly of the hydrogen compounds of nonmetals: CH 4 ,
SiH 4 , NH 3 , PH 3 , AsH 3 , SbH 3 , H 2 S, H 2 Te, H 2 Se, HF, HC1, HBr,
HI, etc.

To obtain one of the above gases by an ionic exchange reaction, it
is necessary to proceed from a compound of that particular nonmetal
with a metal and a compound containing H'ions (acid, water), e. g.:

2Na* + S" + 2H' + SO' 4 - H 2 S + 2Na' + SOl

] f the initial substance (the compound of the nonmetal with a metal)
has a low solubility in water, it is written in the equation in mo-
lecular form:

FeS + 2H' = SOl = H a S + Fe" +

Alternatively, the soluble part of the substance in the ionised state
may be represented together with the insoluble part (precipitate)
as a system in equilibrium:

Fe" + S" + 2H* + S0 4 ' - H 2 S + Fe" + SO 4

C. Reactions Proceeding icith the Formation of a Weak Electrolyte

Weak electrolytes include water, weak acids and bases, the acid
radicals of polybasic acids containing the H'ion, and the basic radi-
cals of polyvalent bases containing OH' ions.

(a) Neutralisation. Neutralisation is a reaction in which the hy-
droxyl ions of a base (or basic salt) combine with the hydrogen ions of
an acid (or acid salt) to form water (a weak electrolyte):

Na' + OH' + H* + NO = H 2 O + Na* +

122 Exercise 15

K' + HSOi + K" + OH' - H 2 O + 2K' + SO^
A1OH" + 2C1' + H' + Cl' ^ H 2 O + AP + 3Cr

(b) Displacement of Weak Acids from Their Salts by Strong Acids.
These reactions proceed owing to the formation of a weak acid (a
weak electrolyte). The initial substances are the salt of a weak acid
and a strong acid:

2Na* + SiO^ -f- 2H* + 2C1' - H 2 SiO 3 + 2Na' + 2C1'
2K' + 2CN' + 2H* + SOl - 2HCN + 2K' + SOl

If the resulting weak acid is a compound of low stability (such as H 2 CO 8
or H 2 SO 3 ), it decomposes to form water and an anhydride:

2Na* + COg + 2H' + 2Ci' = H 2 CO 3 + 2Na* + 2C1'
H 2 CO 3 - H 2 -I- CO 2

(c) Displacement of Weak Bases from Their Salts by Strong Bases.
Weak bases include magnesium, beryllium, and aluminium hydrox-
ides, the hydroxides of the heavy metals, and ammonium hydroxide.

+ Cr + K" + OH' = NH 4 OH + K* + Cl'
Of + SOl + 2Na" + 20H' - Cu (OH) 2 + 2Na" +

(d) Converting Normal Salts to Acid Salts. Polybasic acids break
up into ions stepwise. Foi example, carbonic acid dissociates accord ing
to the equation:

H* + HCOs

co;

In this system the electrolytes are: H 2 CO 3 at the first step and the
COa ion at the second. The ionisation constants for each step are re-
spectively:

- 3.5. 10-. /C. = = B .g. IP-

23 [HGO 3 ]

From these values it is evident that the HCO 3 ion is a weaker
electrolyte than is carbonic acid (the latter is a nearly 10,000 times
stronger electrolyte).

Reactions in Electrolyte Solutions 123

The conversion of normal salts to acid salts is due to an interaction
of ions that gives rise to the acid salt ion (a weak electrolyte):

Mg" + CO; + H' + OH' + C0 2 - Mg" + 2HCO;

MgC0 3
3Ca" + 2P0 4 " -f 2H' + SO 4 - CaSO 4 + 2Ca" + 2HPO4

j7~'^

Ca 3 (P0 4 ) 2

(e) Hydrolysis. Hydrolysis is the interaction of a substance with
ions of water, a reversible piocess. In pure water dissociated according
to the equation H 2 O^H' + OH' the concentrations of the H' and OH'
ions are equal, i. e., [H*] = [OH'], owing to which water is neutral
(pH == 7). If one of the ions of a salt, in the course of hydrolysis, com-
bines with one of the ions of water, this increases the relative con-
centration of the other ion of water; as a result, the salt solution
becomes either acidic ([H']> [OH'])or alkaline (IH*]< [OH']). The
following are illustrations of both cases:

First Case Second Case

( KCN ; K' + CN' ( NH 4 NO 3 ^ NH 4 + NOa

] H 2 ^ OH' + H' j H 2 j OH' + H*

u u

I HCN I NH 4 OH

pH>7 pH<7

The equations for these reactions are written thus:

' + H' -f OH' ^ HCN + K" + OH'
; + NO,; ~h H* + OH 7 ; NH 4 OH + H' +

Salts formed by a strong base and a weak acid (/st case) exhibit
an alkaline reaction in solution; salts formed by a weak base and a
strong acid (2nd case) exhibit an acid reaction in solution.

Salts formed by a weak base and a weak acid undergo hydrolysis
most readily of all, since the ions of these salts are bound up simulta-
neously by the ions of water to produce weak electrolytes. A case in
point is the hydrolysis of ammonium acetate:

NHi + CHaCOO' -f H' + OH' ^ NH 4 OH + CH 3 COOH

124 Exercise 15

Some salts of this type undergo practically irreversible complete
hydrolysis, e. g.:

Cr a S 3 + 6H' + 6OH' - 3H 2 S + 2Cr (OH) 3
Al, (C0 3 ) 8 + 6H' + 6OH' - 2A1 (OH), + 3H 2 CO 3

3H 2 O + 3CO 2

A salt of a polybasic acid or a polyvalent base upon hydrolysis yields
an acid salt and a base or a basic salt and an acid:

2Na + CO 3 + H' + OH' ^ Na' + HOCK, + Na' + OH'
Sn" + 2C1' + H' + OH' ^ SnOH* + Cl' + H' + Cl'

SnONCl

precipitate

The salts of weak polyvalent bases undergo hydrolysis stepwise.
Ferric chloride, for instance, reacts with water thus:

Fe'" + 3C1' + H* + OH' ; FeOH" + 2C1' + H' + Cl'
FeOH" + 2Cr + H* + OH' ; Fe (OH)a + Cl' + H* + CP

Further hydrolysis is prevented by the build-up of H* ions in the
solution. Diluting the solution and increasing the temperature facil-
itates hydrolysis.

Salts formed by a strong acid and a strong base undergo practically
no hydrolysis at all.

The study of the process of hydrolysis has shown that hydrolysis
and neutralisation are reverse processes:

neutralisation

K' + OH' + H' + CN' ^_ ~* H 2 O + K' + CM'

hydrolysis

The compounds of certain nonmetals, such as PC1 5 , PI 3> and SO 2 C1 2 ,
undergo hydrolysis readily, the process being practically irreversible
and giving rise to two acids:

PI 3 + 3H" + 3OH' - H 3 PO 3 + SHI
SO 2 C1, -f 2H* + 2OH' - H 2 SO 4 + 2HC1

(/) Amphoteric Electrolytes. The hydroxides of the heavy metals,
as well as of magnesium, beryllium, and aluminium, are practically
insoluble compounds, but dissolve in acids:

Fe (OH) 2 + 2H* + 2C1' '= 2H 2 O + Fe" + 2CP

Reactions in Electrolyte Solutions 125

Some hydroxides dissolve not only in acids, but also in solutions of
alkalis *, e. g.:

Sn (OH) 2 + 2H' + 2Cr = 2H 2 O + Sn" + 2C1'
H 2 SnO a + 2K* + 20H' = 2H 2 O + 2K* + SnO a

Such hydroxides are termed amphoteric.

It follows from the above equations that amphoteric hydroxides
behave as bases in reactions with acids and as acids in reactions with
alkalis. This dual behaviour is due to the fact that in solutions they
can dissociate according to both the acid and the base pattern. The
electrolytic dissociation of Sn(OH) 2 , for example, can be summed up
as follows:

Sn" + 2OH' ; Sn (OH) 2 ^ 2H' + SnO 2

Or, considering the stepwise character of dissociation, we can put
down the process thus:

SnOH* + OH' ; Sn (OH) 2 ; H* + HSnO 2 '

II ti

SrT + OH' . --- 7-

H' SnO a

D. Reactions Proceeding with the Formation of a Complex, Ion

A saturated solution of the compound, mercuric iodide, which dis-
solves in water with difficulty, contains Hg" and I' ions, produced by
electrolytic dissociation according to the equation:

These ions can be detected by the proper reagents. But if to this solu-
tion we add such an amount of potassium iodide that there are 2 mole-
cules of it to every molecule of mercuric iodide, we will no longer be
able to detect any Hg" or I' ions. Instead there will be a new ion with
new properties: [HgI 4 ]". This is a complex ion consisting of 1 mercury
ion and 4 iodide ions. It is extremely stable and practically does not
dissociate. All other complex ions are, likewise, weak] electrolytes.
Unlike simple ions, they are usually designated by square brackets,
e. g.: [Ag(NH 3 ) a r, [Co(H 2 O) r, and [PtCl 6 ]".

The following equations illustrate the formation of a few complex
ions:

Ac;Cl + 2NH 3 - [Ag (NH ? ) 2 r + Cl'
PtCl 4 + 2H' + 2C1' = 2H* + [PtCl 6 ]"
AgCN + K* + CM' = K' + [Ag (CN) 2 ]'

* The solubility of amphoteric hydroxides in alkali solutions, with the formation
of complex hydroxysalts, is dealt with in the introduction to Exercise 22.

126 Exercise 15

QUESTIONS

1. On the basis of the SP values given in Table XIII (p. 335) explain
why ZnS dissolves in dilute hydrochloric acid, while HgS does not.

2. Why is BaSO 4 precipitated from a solution of barium chloride
by the addition of dilute sulphuric acid, whereas the precipitation of
CaSO 4 requires the addition of concentrated sulphuric acid?

3. Write ionic equations for the formation of the insoluble substances
Ag 3 PO 4 , HgCrO 4 , CaCO 3 , Cu(OH) 2 , and FeS.

4. Why does the neutralisation of a gram-equivalent of any strong
acid by a gram-equivalent of any strong alkali (in solution) produce
the same amount of heat?

5. Write the equations for the hydrolysis of the following salts:
A1C1 3 , K 2 SO 3 , Cu(NO 3 ) a , and NH 4 CN. State whether the solution in
each case is acidic, neutral, or alkaline.

6. Write ionic-molecular eqautions to show how the amphoteric
hydroxide Be(OH) 2 dissolves in an acid solution and in an alkali
solution.

7. Why do not all salts in practice undergo hydrolysis?

Problems

(The SP values needed for solving some of these problems are given \n
Table XIII on p. 335.)

1. What amount of BaCO 3 in grams is contained in 5 litres of a solution saturat-
ed at 25?

2. Calculate the SP of a binary electrolyte with a molecular weight of 140 if 200
ml of a saturated solution contains 0.00016 g of it.

3. How many times greater is the solubility of PbSO 4 in pure water than in an
0.01 M solution of MgSO 4 ?

4. A 500 ml saturated solution contains 0.79-10 3 g of AgCl. Calculate the SP
of the salt.

5. How many grams of SrSO 4 does a 400 ml saturated solution contain?

6. Will a precipitate of AgCl form when we combine equal volumes of 0.0002 M
solutions of AgNO 3 and HC1?

7 .Will SrSO 4 be precipitated when 0.05 litre of an 0.002 M solution of SrS is
combined with an equal volume of an 0.002 M solution of MgSO 4 ?

8. What amount of water should be taken to dissolve 1 g of CaSO 4 -2H 2 O and ob-
tain a saturated solution?

9. A saturated solution of AgCN is combined with an equal volume of an 0.02
M solution of KI. Will Agl be precipitated?

10. Will MgCO 3 be precipitated when 0.6 litre of an 0.01 M solution of MgSO,
is combined with 0.4 litre of an 0.002 M solution of Na 2 CO 3 ?

LABORATORY WORK

Apparatus and materials: test tubes and rack; 10 ml measuring cylinder; casse-
role; glass rod; ammonium acetate; potassium carbonate; iron sulphide; cupric sul-
phide,-zinc sulphide; manganese chloride; sodium chloride; sodium phosphate; concen-
trated nitric acid; concentrated sulphuric acid; saturated solution of strontium chlo-
ride; saturated solution of calcium sulphate; saturated solution of calcium chloride;
saturated solution of silver acetate; saturated solution of strontium sulphate; 20%
solution of silver nitrate; 20% solution of sodium acetate; 10% solution of ammonia;

Reactions in Electrolyte Solutions 127

1 :6 sulphuric acid; 2N solution of sodium hydroxide; N solution of sodium phosphate;
2 N solution of hydrochloric acid; 0.5 N solution of ferric chloride; 0.5 N solution of
sodiumsulphide; N solution of sodium sulphate; 0. 5 N solution of zinc sulphate; N so-
lution of magnesium sulphate; 0.5 N solution of cupric sulphate; N solution of barium
chloride; 0.5 N solution of mercuric nitrate; 0.5 N solution of potassium iodide; 0.5 N
solution of aluminium sulphate; 0.5 N solution of bismuth nitrate; phenolphthalein
solution; neutral litmus solution; litmus paper, and boiled distilled water.

Preparation of Saturated Solution of Silver Acetate. Add a solution of 3.30 g of
anhydrous soda salt in 75 ml of water to a solution of 10. 6g of silver nitrate in 50 ml
of water. Filter off the precipitate formed, wash it on the filter three or four times
with cold water, transfer it to a 1 litre beaker, add 8 ml of 80% acetic acid in 92
nil of water, and heat the contents of the beaker to a temperature not exceeding 50
(until the evolution of carbon dioxide in the form of bubbles ceases). Now pour 900 ml
of water into the beaker, heat the solution to boiling point, and, after it has cooled,,
filter off the undissolved precipitate.

1. Conditions at Which Precipitates Form, (a) Mix 2 ml of a satu-
rated solution of strontium chloride in a test tube with an equal vol-
ume of a saturated solution of calcium sulphate. What happens? Re-
peat the experiment with 2 ml amounts of saturated solutions of
calcium chloride and strontium sulphate. Does a precipitate form?
Account for the results of these experiments in terms of SP values
(Table XIII).

(b) Pour 2 ml of a saturated solution of silver acetate into each of 2
test tubes. Add 2 ml of a 20% solution of silver nitrate to one and an
equal volume of a 20% solution of sodium acetate to the other. What
happens? Give an explanation of the results.

What are the conditions at which a substance is precipitated from
a solution?

After the experiment pour the solutions with precipitates of silver
salts into special bottles.

2. Dissolving Precipitates. Into each of 2 test tubes put a pinch
of zinc sulphide and a pinch of cupric sulphide, taken separately on
the end of a knife. Add 2 ml of water to each of the test tubes and shake
them. Write the SP equations for both substances. Then add 3 ml of
a hydrochloric acid solution to each of the test tubes. What happens
to the precipitates? Give an explanation of this.

3. Ionic Reactions, (a) Choose from the available reagents those
whose interaction can produce precipitates of magnesium phosphate,
ferric hydroxide, and cupric sulphide. Bring about the necessary
reactions and record the colours of the precipitates formed. Write
ionic-molecular equations of the reactions.

(b) Pour 2 ml of a sodium sulphate, zinc sulphate, and aluminium
sulphate solution respectively into 3 test tubes; add 2 ml of a barium
chloride solution to each of them. Write three ionic-molecular equa-
tions and one over-all ionic equation. Which ion is a reagent for detect-
ing the sulphate ion?

(c) Reactions proceeding with the evolution of a gas (conduct the
experiment in a ventilated hood!). Treat a lump of ferric sulphide (in

128

Exercise 15

a test tube) with 3 ml of a sulphuric acid solution, heating the test
tube. Identify the gas evolved by its odour. Write the equation of
the reaction.

(d) Treat about 0.5g of table salt in a test tube with 2-3 ml of con-
centrated sulphuric acid. Observe the evolution of hydrogen chloride.
Write the equation of the reaction. Why is not dilute sulphuric acid
used for this reaction?

(e) Pour 5 ml of a 2 N solution of hydrochloric acid (measured in
a cylinder) into a casserole and add 5 ml of a 2 N solution of sodium
hydroxide first in small portions and then in drops. Stir the solution
with a glass rod and test it with litmus paper (this should be done by
putting a drop of the solution on small strips of red and blue litmus
paper and observing the change in colour). Add alkali until the solu-
tion ceases changing the colour of either litmus paper. Evaporate the
obtained neutral solution in the casserole. What is the residue? Taste
it. Write the equation of the neutralisation reaction. What other two
reactions of complex substances can produce sodium chloride? Write
the equations of these reactions.

4. Hydrolysis. Pour 3 ml of a neutral litmus solution into each of
six numbered test tubes. Introduce a few crystals of the following
substances into five of the test tubes respectively: potassium carbo-
nate, sodium phosphate, sodium chloride, aluminium chloride, and
ammonium acetate. Shake the solutions and compare their colours with
the colour of the litmus solution in the sixth test tube, examining
them against the background of a sheet of white paper.

Record the results of the experiments as follows:

No. of test
tube

Substance

Colour of
litmus

Is the solution acidic, neutral, or alkaline?

What type of salts underwent hydrolysis? Write the equations of
the hydrolysis reactions.

5. Shifting the Equilibrium of Hydrolysis. Pour several drops of
a solution of bismuth nitrate into a test tube and dilute the solution
gradually with distilled water. A precipitate forms. Write the equa-
tion of the hydrolysis reaction. Add a few drops of concentrated nitric
acid to the test tube with the precipitate. What happens to the pre-
cipitate? Explain this on the basis of the equation for the equilibrium
constant.

6. Effect of Temperature on Hydrolysis. Pour 4-5 ml of a sodium
acetate solution and 1-2 drops of a phenolphthalein solution into a

Oxidation-reduction Reactions 129

test tube. Heat the solution until it boils and observe the appearance
of a pink colour. Write the equation of the hydrolysis reaction and
explain what happened. What factors influence equilibrium in hydrol-
ysis and why?

7. Amphoteric Hydroxides. Prepare a precipitate of aluminium
hydroxide and pour the solution with the precipitate into two test
tubes. Add an excess of alkali to one and of acid to the other. Watch
the precipitates dissolve. Write the equations of the three reactions
carried out. Why do amphoteric hydroxides dissolve both in acids
and in alkalis?

. 8. Reactions Proceeding with the Formation of a Complex Ion. (a)
Prepare a cupric hydroxide precipitate by means'of an exchange reac-
tion between solutions of a cupric salt and sodium hydroxide. Write
the equation of the reaction.

Pour the solution off from the precipitate and add an ammonia
solution to the precipitate in the test tube. What happens? Write the
equation of the reaction, which produces the complex ion [Cu(NH 3 ) 4 ]'\

(b) Add a few drops of a potassium iodide solution to a solution of
a mercuric salt. What is formed? Write the equation of the reaction.

Write the equation for the solubility product of HgI 2 . What should
happen if the concentration of iodine ions is increased? Add an excess
of a KI solution to the precipitate. Why does the precipitate dissolve?

Exercise 16

OXIDATION-REDUCTION REACTIONS

SUBJECTS FOR STUDY

Oxidation-reduction reactions from the electronic standpoint; oxidation-reduction
characteristics of neutral atoms and ions; writing equations for oxidation-reduc-
tion reactions; effect of the pH-value.

Unlike ion-exchange reactions, oxidation-reduction reactions are
marked by a transfer of electrons from one reactant to another, this
altering the charge or valence of the interacting substances.

From the electronic standpoint a process involving the loss of
electrons is termed oxidation, while a process involving the gain of
electrons is termed reduction. Accordingly, substances that lose elec-
trons in oxidation-reduction reactions are called reducing agents, while
substances that gain, or receive, electrons are called oxidising agents.

Neutral atoms (molecules), elementary ions, and complex ions
can take part in oxidation-reduction reactions.

The oxidation-reduction properties of neutral atoms and of ions
can easily be explained on the basis of the Mendeleyev Periodic Sys-
tem of Elements (Table 11).

9-795

130 Exercise 16

1. D. Mendeleyev's Periodic System of Elements. In 1869 the great
Russian scientist Dmitry Mendeleyev discovered one of the most
fundamental laws of nature: the Periodic Law of Chemical Elements.
This Law he formulated thus: "The properties of the elements (and,
hence, of the simple and compound substhnces they form) are periodic
functions of their atomic weights" .

The Law was embodied in the Periodic System of Elements, which
has had a most profound influence on the progress of chemistry.

The most common method of representing the Periodic System is
in the form of an eight-group table, such as is given on the inside cover
of the book.

The elements are arranged in the Periodic System according to their
atomic numbers (from 1 to 103). These numbers indicate the positive
charge of the atomic nucleus (the number of protons) or the number
of electrons in the electron layers of the atom.

Each period in the Periodic Table begins with an element whose
atom contains but one electron in its outer layer; the periods end with
elements whose atoms have 8 electrons in the outer layer (atoms of
inert gases). The first period ends with the inert gas helium, which
has two electrons in its outer layer. In all, there are 7 periods in the
Table, the number of each period indicating the number of electron
layers in the atom. The only exception is the fifth-period element
palladium, whose atom has 4 electron layers.

According to the manner in which their electron]sublevels are filled,
the elements may be divided into four groups:

(a) The elements in whose atoms electrons are added to the s sublayer
of the outer layer, or shell, are called s-elements. There are 14 such
elements in the Table: the first two in each period v

(b) The elements in whose atoms electrons are added to the p sub-
layer of the outer shell are called p-elements. The last 6 elements in
each of five periods (2-6) are p-elements.

(c) The elements in whose atoms electrons are added to the d sublayer
of the next to last layer are called d-elements. Since the d sublayer
contains a maximum of 10 electrons, there are 10 such elements (atom-
ic numbers 21-30, 39-48, 57, and 72-80) in each of periods 4, 5, and
6 and one element (number 89) in period 7, which makes a total of
31 elements. In their outermost shells these atoms usually have two
electrons, seldom one (this is true of 9 elements, which are underlined
by a single line in Table 11), while palladium has none because both
of the s-electrons are in the d sublayer of the neighbouring layer (in
Table 11 it is underlined with a double line).

(d) The rare-earth elements, or lanthanides, in whose atoms elec-
trons are added to the / sublayer of the fourth shell. Since the / sublayer
can hold a maximum of 14 electrons, there are 14 lanthanides in all
(atomic numbers 58-71). They are situated in period 6, their atoms
having 6 electron shells each. The outermost layer in these atoms has 2

132 Exercise 16

-electrons (6s 2 ), while the one next to last (layer V) has 8 electrons.
The / sublayer begins to be built up by two electrons at number 58
(Ce). The 64th element (gadolinium) and the 71st (lutetium), in ad-
dition to the /-electrons, add one electron each to the d sublayer of
<the fifth shell.

The actinides are elements whose atoms, like those of the lanthanides,
'build up the / sublayer, but of the fifth shell instead of the fourth.
In elements 93-102 the distribution of electrons is similar to that in
the above-situated (61-70) lanthanides. The six valence-electrons of
uranium are distributed among the layers thus: 5/ 3 6d l 7s 2 . For the 5
valence-electrons of protactinium (91) and the 4 valence-electrons of
thorium (90) there are two possible combinations of electron distri-
bution: for protactinium 5/ 2 Sd 1 7s 2 or 5/ 1 6d 2 7s 2 and for thorium 5/ 1
fid 1 7s 2 or 6d 2 7s 2 . Thirteen actinides are known at present (90-102).

The chemical properties of an element are related to the distribution
of electrons in the atom. A number of elements with similar chemical
properties due to an identical number of valence-electrons are called a
homologous series, or a series of analogues. The elements of one and
the same homologous series, or series of analogues, are connected in
Table 11 by either a solid or a broken line.

The solid line is used to connect the analogues in the principal sub-
groups; the broken line, to connect the analogues in the subordinate
subgroups, the lanthanides, and the actinides.

The similarity in the properties of the elements in the principal
subgroups may be traced to the same number of electrons in the outer
shells of the atoms of these elements. In the subordinate subgroups,
however, the similarity of the elements depends not only on the outer
electrons, but also on the electrons that are added (above eight) to
the next layer. Thus, the element rhenium (Z=75), with the distri-
bution of electrons in the atom )2)8) 18)32)8 + 5)2, is an analogue of
the element manganese (Z=25), which has the electron distribution
)2)8)8 +5)2. The actinides are chemically analogous to the lanth-
anides.

The study of atomic structure threw light on the underlying reason
lor periodicity. The periodic change in the properties of the elements
is related to a periodic recurrence of similar electron structure.

2. Oxidation-reduction Properties of Neutral Atoms. All the ele-
ments can, according to their chemical properties, be divided into three
groups: reducing agents, oxidising-reducing agents, and elements
that are neither one nor the other.

(a) The reducing agents are elements whose atoms have from 1 to
3 electrons in the outer shell. In chemical reactions they do not receive
-electrons, but only give them up:

R ne- - R+"
In Table 11 these elements are situated to the left of the dotted line*

Oxidation-reduction Reactions 133

This group of elements consists primarily of metals, the exceptions
being H, He, and B.

(b) The oxidising- reducing agents are elements whose atoms have
4, 5, 6, and 7 electrons in the outermost shell. They can either take
up electrons (to 8), i. e., act as oxidising agents, or furnish electrons,
i. e. t act as reducing agents *. Thus, an atom with 6 electrons in its
outer shell can react according to either of the following two equations:

R + 2e- = R- 2 (as an oxidising agent)
R 6e~ = R +6 'as a reducing agent)

In Table 1 1 the elements of this group are situated to the right of
the dotted line (the inert gases are an exception). Hydrogen and boron
are likewise oxidising-reducing agents.

(c) The inert gases are elements whose atoms have 8 electrons in the
outer shell (helium has 2). They neither receive nor give up electrons
in chemical reactions and, consequently, exhibit neither oxidising
nor reducing properties. Six elements make up this group.

The oxidising-reducing properties of neutral atoms are expressed in
terms of ionisation potential and electron affinity.

The ionisation potential (unit: electron-volt, eV) is the work required
to split off an electron from a neutral atom and can serve as a measure
of an atom's reducing activity.

The electron affinity (usually expressed in Calories) is the energy
of the addition of one electron to a neutral atom. It can serve as a
measure of the relative oxidising activity of neutral atoms, which is
inversely proportional to the reducing activity. Ionisation potential
and electron affinity data can be used to characterise the changes in
oxidation-reduction activity in the periods and homologous series of
elements in the Periodic Table.

In the periods the ionisation potential, as a rule, increases from left
to right, while the reducing activity declines and the oxidising activity
mounts, owing to rising electron affinity. In the homologous series of
the s-, /-, and p-elements the ionisation potential diminishes as we
move downwards; the reducing activity increases and the oxidising
activity falls off, since the electron affinity decreases. In the case of
the d-elements, however, the change in ionisation potential is more
complex. The strongest oxidising agents are in the upper right-hand
part of the Table (fluorine, oxygen, chlorine, sulphur, etc.), while the
reducing agents are concentrated in the lower left-hand part (francium,
caesium, radium, barium, lanthanum, etc.).

3. Oxidation-reduction Properties of Elementary Ions**. Elementary
ions can be charged either negatively or positively.

* Oxygen and fluorine are oxidising agents only; their atoms do not give up
electrons in reactions.

** The radii of some ions are given in Table XVI (see Appendix).

134 Exercise 16

(a) Negative Ions. The ions of this group are formed by the addition
of one or several electrons to a neutral atom of a nonmetal.

For instance, a neutral atom of sulphur S, whose atomic number is
16, has the electron distribution )2)8)6.,It can add another two elec-
trons, becoming a doubly charged negative ion S" 2 with the electron
distribution )2)8)8.

In their outer shells elementary negative ions have 8 electrons;
accordingly, they can only give up electrons and are reducing agents.

Examples: The reducing ionsS" 2 , I" 1 , Br' 1 , Se~ 2 , Te~ 2 , etc., in such compounds as
H 2 S, HI, HBr, H 2 Se, and H 2 Te, or in corresponding salts: Na 2 S, KI, KBr, Na 2 Se,
and I^Te.

(b) Positive Ions. The ions of this group can be divided into 2 sub-
groups according to their charge; there can be ions of maximum charge
and ions of lower charge.

Positive elementary ions can have charges of 1 to 4 (free quadriposi-
tive ions are seldom encountered). Ions of maximum charge are formed
when neutral atoms give up all their outer, or valence, electrons. For
instance, the Al atom with the electron arrangement )2)8)3 or the
Sn atom with the arrangement )2)8)18)18)4 can lose their respective
3 and 4 outer electrons to become Al +3 and Sn+ 4 with the electron ar-
rangements )2)8 and)2)8)18)18. In their outer shells ions of maximum
positive charge have either 8 or 18 electrons, which means that such
ions can no longer relinquish electrons and can only recapture earlier
lost electrons, acting as oxidising agents.

Examples: The oxidising ions Sn+ 4 , Pb+ 4 , Hg+ 2 , Ag +1 , etc., in various com-
pounds.

Ions of lower charge are formed when neutral atoms give up only
part of their outer electrons. The Sn atom, with the electron arrange-
ment shown above, may lose only 2 electrons. In that case it turns into
the bipositive ion Sn^ 2 , which has the electron distribution )2)8) 18) 18)2
and can therefore give up another 2 electrons, exhibiting reducing
properties (or can recapture the earlier lost 2 electrons, in this case
exhibiting oxidising properties). Consequently, ions of lower positive
charge can act as reducing (as well as oxidising) agents. The reducing
properties of the ions of this subgroup are more pronounced than
their oxidising properties.

Reducing ions: Sn 42 , Ti +3 , Ce+ 3 , Fe+ 2 , etc.

4. Oxidising-reducing Properties of Composite Ions. Most composite
ions are negatively charged (e. g., NO^ CrO^f MnO*, ClOa, AsOl",
NO 2 , SO 3 , and SnO 2 ); there are, however, a few positive composite
ions as exceptions (H 3 O*, NH 4 ', PH 4 ').

These ions are known to form in aqueous solutions as a result of
the electrolytic dissociation of acids and salts. Nitric acid and potas-
sium permanganate, for instance, are ionised thus:

HN0 3 ^t H" +

Oxidation-reduction Reactions 135

KMnO 4 ^ K" ^

The chemical bond between H" and NO 3 , as well as between K'
andMnO 4 , isionogenic; it is for this reason that the HNO 3 and
KMnO 4 molecules break up into such ions. Can the composite ions
NO 3 and MnO 4 in turn break up (dissociate) into components? Exper-
imental findings lead to the conclusion that composite negative
ions practically do not dissociate, since the chemical bond between

their component atoms is covalent (nonionogenic). The NO 3 and

MnOl ions and, hence, all other composite negative ions act
as monolithic groups.

As pointed out above, free positive ions of charges above +3 do not
exist in aqueous solutions; indeed, even tripositive ions, if they have
a small radius, turn into composite ions. This is due to the fact that
if multipositive ions did form in an aqueous solution, they would at
once, owing to the high specific density of the electric charge, com-
bine with the oxygen ions of the water to form composite negative ions.
This is what would happen to the heptapositive manganese a ion w :

H H

0-2 0-2

'/ \

H H

Mn+ 7
H -H

\ /;

The heptapositive Mn "ion" polarises the oxygen ions to such a de-
gree that the ionic bonds between them are converted entirely to 'co-
valent bonds, which produces a monolithic composite negative ion. This
is corroborated by the hydrolysis of halogen anhydrides, such as PC1 5 :

PC1 5 + 4H'OH' - H 3 P0 4 + 5HC1

I _ j

POl"

In other words, pentavalent phosphorus in an aqueous medium is
at once converted to the composite negative ion PO 4 . Hence, there
are no free multipositive ions; as for Mn +7 inMnO 4 and P +5 in PO7,
they can be regarded as having a positive valence, but not charge.
Elements with a high positive valence form composite negative ions
of the types RO~ 4 m > ROT, and ROT 7 .

For instance, MnO 4 is an oxidising ion and in certain conditions
may be reduced to the ion MnO 3 , adding three electrons. With respect

136

Exercise 16

Table 12
Composition and Properties of Some Composite Negative Ions

Positive valence of element R

R 4S

R+4

R+S

R-H5

R +7

NO2

NO3 *

AsOs

AsCY" .

S0 8 "

SCY'

SeO s "

SeO"

CKV

CrO 4 "

C1CV

C10 3 '

C1CV

I0 2 '

I0 3 '

10*'

BrOa'

BrCV

MnO s "

Mn0 4 "

Mn0 4 '

rise in positive valence oxidation

a shift of 1 box to the right loss of \e~
decrease in positive valence reduction

a shift of 1 box to the left gain of I er

Oxidising ions are underlined; the rest are reducing ions.

to the MnO 4 ion the NO^ ion is a reducing agent: it can give
up 2 electrons, undergoing oxidation to the NO 3 ion.

5. Writing Equations of Oxidation-reduction Reactions of the Sim-
plest Type. These are reactions between two substances only: a reducing
agent (A) and an oxidising agent (B). The equation is therefore written:

mA + nB =

where m and n are coefficients. The problem of writing the full equa-
tion is a problem of finding these coefficients. They are found by
applying the electron balance rule: the total number of electrons lost
by the reducing agent must equal the total number of electrons gained
by the oxidising agent.

Example. Write the equation of the reaction Al + O a =
This is done in 3 stages:

Oxidation-reduction Reactions 137

(1) First of all, it is necessary to identify the reducing and the oxidising agent.
Aluminium is a metal of group III; its atom gives up 3 electrons, i.e., acts as a reduc-
ing agent. Oxygen, on the other hand, is a nonmetal and with respect to the reducing
agent Al it acts as an oxidising agent: the O 2 molecule takes on 4 electrons.

(2) We can now write the skeleton electron balance:

O 2

where 4 and 3 are additional factors that have to be introduced into the equation
of the reaction to balance the number of electrons lost and gained.
(3) The equation is now written in its final form:

4A1 + 30 2 = 2A1 2 8
Exercise. Write the equations of the following reactions:

(1) P 4- O 2 = (2) S0 2 + O 2 = (3) Na + S =

(4) K + Br 2 = (5) H 2 S + I 2 = (6) Mg + N 2 =

(7) FeCl 2 + C1 2 = (8) Al + CuSO 4 = (9) Fe (NO 3 ) 2 + AgNO 3 =

6. Writing Equations of Oxidation-reduction Reactions That
Take Place in Aqueous Solutions. These equations are of the gen-
eral type:

mA + nB + pC =

where A is the reducing agent, B is the oxidising agent, and C is the
molecule of the medium. The coefficients m and n are found as ex-
plained above from the skeleton electron balance. The coefficient p
is in most cases determined when writing the equation, after the coef-
ficients m and n have been found.

Reactions in Acid Solutions

Example 1. Find the coefficients for the following reaction:
KN0 2 + KMn0 4 + H 2 SO 4 - KNO 3 + K 2 SO 4 + MnSO 4 + H 2 O

To trace the whole process of the writing of the equation, we must
break it up into steps:

(a)JJnderline the reducing ion and its oxidised form once, the oxi-
dising ion and 'its reduced form twice. The NOa ion is oxidised to the
NOs ion, while theMnO^ ion is reduced to theMn" ion.

(b) An additional skeleton electron balance is drawn up*, and the
additional factors (the coefficients m and n) are found in this way:

Least common Additional
multiple factors

NO; -26-= NO; io 5

* Tables 12 and 13 may be used to determine the number of electrons lost by the
reducing agent and the number gained by the oxidising agent.

-i

"5
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S^ * *" to o

c ^^ ^T3 3

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*CJ ., CL)

c *""*

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*2^ *3D"3

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<L> C 'co

5 -5

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Q <

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III

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LO cr> -S

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JQ CO Q* M_

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*-, to < 3

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s:i "g

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QJ ' *-* "- ( 1

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C w ^ 03

c3 g %

? x. x

+ oS c ^^ ) cS GG oo

4, CM ^ CM CM

j-,
!S

1 1 1 1 1 1 1 II II

CMCMCMVf^CMCM CM-*H COLO

* #
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0**^,*.^^ bb **

<5 s -^^ GcQ GC c?c5
2. J GO

G <D <L) G C

*~*

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c/3 03 03 3 to to

QJ K/I O <l> <U
C 2? C C G G

r: = .0 ^ .0 r: :=:

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03 l O O ^3 03

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li?l

?|G~ 8 0*G 99

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ilif-

1 1 1 1 t 1 II III

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^ * w * * /~\ # *

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gbm^a G g= fi &

* : * *

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140 Exercise 16

(c) Now substitute the coefficients found into the equation:
5KNO 2 + 2KMnO 4 + H 2 SO 4 = KNO 3 + K 2 SO 4 + MnSO 4 + H 2 O

(d) From the coefficients found determine the number of molecules
of the products, except the water:

5KNO 2 + 2KMn0 4 + H 2 SO 4 = 5KNO 3 + K 2 SO 4 +
+ 2MnSO 4 + H 2 O

Rule. In an acid solution all the ions of metals with charges of I
to 3 combine with acid radicals to form salts.

(e) Count the number of sulphuric acid (medium) radicals in the
right-hand side of the equation (in this case it is 3), in this way find-
ing the coefficient for the acid molecules; insert it into the equation:

5KN0 2 + 2KMn0 4 + 3H 2 SO 4 = 5KNO 3 +:K 2 SO 4 +:2MnSO 4 +H 2 O

(f) From the number of hydrogen ions in the amount of acid taken
determine the number of water molecules formed (in this particular
case it is 3) and insert the last coefficient:

5KNO 2 + 2KMn0 4 + 3H 2 SO 4 - 5KNO 3 + K 2 SO 4 +
+ 2MnSO 4 + 3H 2 O

Note. The equation written is checked by counting the number of oxygen atoms
in the left-hand and the right-hand side.

Steps a c", a d", u ", and a /" are performed mentally, and the equation
need not be rewritten several times.
Example 2. Find the coefficients in the following equation:

FeSO 4 + KMnO 4 + H 2 SO 4 = Fe 2 (SO 4 ) 3 + K 2 SO 4 + MnSO 4 + H 2 O

After carrying out steps* *a" ,*b n 9 *c" , and *d* as in Example!, we
obtain:

10FeSO 4 + 2KMnO 4 + H 2 SO 4 '= 5Fe 2 (SO,) 3 + K 2 SO 4 +
+ 2MnSO 4 + H 2 O

Then (step a e") count the number of sulphuric acid radicals in the
right-hand side of the equation (it equals 18) and deduct from it the
number of the same radicals contained in the reductant (it equals 10);
the difference 18 10=8 yields the coefficient for the acid molecules.
This element in writing equations should be borne in mind.

* The additional factors found have to be doubled because the molecule of the
resulting ferric sulphate contains an even number of ferric ions.

Oxidation-reduction Reactions 141

Determine (step a / w ) the number of water molecules formed. The
equation now assumes the final form:

10FeS0 4 + 2KMnO 4 + 8H 2 SO 4 - 5Fe 2 (SO 4 ) 3 + K 2 SO 4 +
+ 2MnSO 4 + 8H 2 O

Example 3. In this example the acid is the medium and plays
the part of an oxidising or reducing agent. This being so, the formula
of the acid should be written in the equation twice. Complete the
equation:

Cu + HNO 3 + HNO 3 = Cu (NO 3 ) 2 + NO + H 2 O
After carrying out all the steps, we obtain the final equation:

3Cu + 2HN0 3 + 6HN0 3 - 3Cu (NO 3 ) 2 + 2NO + 4H 2 O

From this equation it follows that out of the 8 molecules of nitric
acid needed for the reaction, 2 are used for oxidation (the hydrogen
ions of the oxidant and the medium are used to form water).

Exercise. Insert the coefficients In the following skeleton reactions:

(1) K 2 S + KMnO 4 + H 2 SO 4 -> S + K 2 SO 4 + MnSO 4 + H 2 O

(2) KNO 2 + K 2 Cr 2 O 7 + HNO 3 -* KNO 3 + Cr (NO 3 ) 3 + H 2 O

(3) FeSO 4 + KC1O 3 + H 2 SO 4 -> Fe 2 (SO 4 ) 3 + KCi + H 2 O

(4) S0 2 + K 2 Cr 2 O 7 + H 2 SO 4 - K 2 SO 4 + Cr 2 (SO 4 ) 3 + H 2 O

(5) Nal + Pb0 2 + H 2 S0 4 -> Na 2 SO 4 + PbS0 4 + I 2 +H 2 O

(6) SnCl 2 + K 2 Cr 2 7 + HC1 -* H 2 [SnCU] + KCI + CrCl 3 + H 2 O

(7) KI + KN0 2 + H 2 S0 4 -+ I. + NO + K 2 SO 4 + H 2 O

(8) H 2 S + K 2 Cr 2 O 7 + H 2 SO 4 -+ S + K 2 SO 4 + Cr 2 (SO 4 ) 8 + H a O

(9) FeSO* + HNO 3 + H 2 SO 4 -> Fe 2 (SO 4 ) 3 + NO + H 2 O

(10) Cu + H 2 S0 4 + H 2 SO 4 -* CuS0 4 + S0 2 + H 2 O

(11) Ag + HNO 3 + HNO 3 -> AgNO 3 + NO 2 + H 2 O

(12) HC1 + MnO 2 + HCl -* C1 2 + MnCl 2 + H 2 O

(13) S + HNOs -* H 2 SO 4 + NO 2 + H 2 O

Reactions in Alkaline Solutions
Example 1. Complete the equation:

NaCrO 2 -f Br 2 + NaOH =Na a CrO 4 + NaBr + H 2 O

After carrying outsteps a a", tf 6 n , a c", and a d"' as above, we ob-
tain the following equation:

2NaCrO 2 + 3Br 2 + NaOH = 2Na 2 CrO 4 + GNaBr + H 2 O

Rule. In an alkaline solution all the acid radicals formed (nega-
tive ions) combine with the positive ions of metals to form salts.

142 Exercise 16

Count (step *e n ) the number of metal ions in the right-hand side
of the equation that were part of the alkali molecule (in this case Na*
ions) and deduct from them the number, if any, of the same ions that
were part of the reductant or oxidant molecules (in this case 10-2).
The difference yields the coefficient for the alkali:

2NaCrO 2 + 3Br 2 + SNaOH = 2Na 2 CrO 4 + GNaBr + H 2 O

From the number of hydrogen atoms in the left-hand side of the
equation (step "/") we now find the coefficient for the water (provided
no other hydrogen-containing compounds are formed in the reaction):

2NaCrO 2 + 3Br 2 + SNaOH = 2Na 2 CrO 4 + GNaBr + 4H 2 O
Exercise. Insert the coefficients in the following skeleton reactions:

(1) NaaCrOs + C1 2 + NaOH -> Na 2 CrO 4 + NaCl + H 2 O

(2) MnO 2 + Br 2 + KOH -> K 2 MnO 4 + KBr + H 2 O

(3) NaCrO 2 + H 2 O 2 + NaOH -+ Na 2 CrO 4 + H 2 O

(4) MnO 2 + KC1O 3 + KOH -> K 2 MnO 4 + KC1 + H 2 O

(5) Bi 2 O 3 + Br 2 + KOH -+ KBiO 3 + KBr + H 2 O

(6) Al + KNO 3 + KOH -> KaAlO 3 + NH 3 + H 2 O

(7) Zn + KNO 2 + KOH -* K 2 ZnO 2 + NH 3 + II 2 O

(8) Fe 2 O 3 + NaNO 3 + NaOH -> Na 2 FeO 4 + NaNO 2 + H 2 O

Reactions in Neutral Solutions

Equations of reactions of this type are written according to
the same procedure, with molecules of water introduced instead of the
acid or alkali.

Exercise. Insert the coefficients in the following skeleton reactions:

(1) Fe (OH) 2 + 2 + H 2 O -> Fe (OH) 3

(2) Ni (OH) 2 + NaOCl + H 2 O -+ Ni (OH) 3 + NaCl

(3) MnSO 4 + KMnO 4 + H 2 O -+ MnO 2 + K 2 SO 4 + H 2 SO 4

(4) SOa + KMnO 4 + H 2 O -> MnO 2 + K 2 SO 4 + H 2 SO 4

7. Oxidation-reduction Equivalents. One and the same substance
can take part both in exchange and oxidation-reduction (redox)
reactions. Potassium chromate is a case in point. It can take part in
the following two reactions:

BaCl 2 + K 2 CrO 4 = BaCrO 4 + 2KC1
(exchange reaction)

-1 -f6 +3

6KI + 2K 2 CrO 4 + 8H 2 SO 4 -3 l2 + 5K 2 SO 4 +Cr 2 (SO 4 ) 3 + 8H 2 O

(redox reaction)

Oxidation-reduction Reactions

143

The equivalents of potassium chromate in these reactions are dif-
ferent. To find its equivalent from the exchange reaction, we have to
divide the molecular weight of potassium chromate by 2 (the valence
of the salt's acid radical). The quantity obtained in this way is termed
the normal, or exchange, equivalent. To find the redox equivalent,
we have to divide the molecular weight of the substance by the num-
ber of electrons that are lost or gained by the reductant or oxidant
molecule in the reaction. In the second reaction, for instance, the
potassium chromate molecule (or, to be more exact, the CrO^' ion)
gains 3 electrons; hence, the equivalent equals the molecular weight
divided by 3. In the case of some substances the exchange and redox
equivalents may be equal.

QUESTIONS

1. Write two equations of redox reactions in each of the different
types of solutions from the exercises above.

2. Calculate the values of the exchange and the redox equivalent
of KMnO 4 in an acid solution.

3. The substances in the upper boxes of the table below are convert-
ed by reactions to the substances in the lower boxes:

H 2 SO 4

SnCl 2

Fe 2 (S0 4 ) 3

F 2

NiS

PH 3

NaCIO

K,HAO t

SO 2

SnCl 4

FeSO 4

NiSO 4

H 3 PO 4

NaCl

KAsOa

Copy the table in your notebook and indicate in the vacant boxes
in which cases the substance was oxidised and in which it was reduced.

4. With an eye to the fact that HI is both a reductant and an oxi-
dant, write the equation for the oxidation of HI by potassium dichro-
mate K 2 Cr 2 O 7 .

5. On the basis of negative ion radii (Table XVI on p. 337) explain
why hydrogen telluride is a stronger reducing agent than hydrogen
sulphide is.

LABORATORY WORK

Apparatus and materials: test tubes and rack; copper shavings; iron nails; crys-
talline potassium permanganate; concentrated nitric acid; concentrated hydrochloric
acid; 0.5 N solution of cupric sulphate; 0.5 N solution of potassium iodide; 2 N solu-
tion of sulphuric acid; N solution of potassium nitrite; 2 N solution of sodium hydro-
xide; 0.5 N solution of manganese chloride; 0.05 N solution of potassium permanga-

144 Exercise 16

nate; 0.1 N solution of sodium metachromite: 0.3 N solution of potassium dichromate;
0.2 N solution of potassium iodate; 0.5 N solution of potassium sulphite; starch
solution; chlorine water; bromine water; hydrogen sulphide water, and sandpaper,

Preparation oJ5 Starch Solution. In a mortar, grind 0.5 g of soluble starch with
5 ml of water; pour the mixture into 95 ml of boiling water. To protect ttoe solution
from decomposition by microorganisms, add a lew grains of HgI 2 or salicylic acid.
The starch solution should with iodine produce a blue colouration; a violet or reddish-
brown colour of the starch, upon contact with iodine, indicates that the starch solu-
tion is unsatisfactory.

Preparation of t Chromite Solution. Dissolve 8. 9 g of chromic chloride or 16.5 g of
potassium dichromate in 50 ml of water and add 10 ml of a 25% ammonia solution
to the boiling solution. Filter off the precipitate*and wash it with hot water until
the Cl' and SO"* ions have been removed. Then dissolve the precipitate in 10 ml of a
40% solution 01 sodium hydroxide and dilute to 1 litre. Prepare the solution when
needed.

Oxidation-reduction Reactions *. (a) Into a solution of cupric
sulphate lower an iron nail whose surface has been cleaned with sand-
paper. Keep the nail in the solution for 3-5 minutes. What happens?

(b) To 2-3 ml of a potassium iodide solution add an equal volume
of chlorine water. Why does the colour of the solution change? Write
the equation for the reaction that takes place. What is the substance
that imparts the yellow colour to 4 the solution?

(c) To 2-3 ml of hydrogen sulphide water add bromine water by
drops. Observe how the solution loses colour and becomes turbid.
Write the equation of the reaction that takes place.

(d) To 2-3 drops of a potassium iodide solution add 2 ml of an H 2 SO4
solution and 2-3 drops of a potassium iodate solution. The solution
acquires a yellow colour, owing to the formation of free iodine. Write
the equation of the reaction that takes place. Prove the presence of
iodine in the solution, bearing in mind that the best reagent for free
iodine is starch.

(e) To 2 ml of a potassium sulphite solution add an equal volume
of dilute sulphuric acid and 1 ml of a potassium permanganate solution.
The violet colour of the latter disappears. Write the equation of the
reaction that takes place.

(f) Establish experimentally that potassium nitrite in an acid so-
lution is a reducing agent with respect to potassium dichromate (the
solution should {be warmed).

(g) Add 2 ml of concentrated nitric acid (in a ventilated hood!)
to a test tube containing some copper shavings. Observe the evolution
of a gas (what is its colour?). Write the equation of the reaction that
takes place.

(h) Pour 1 ml of concentrated hydrochloric acid on a few crystals
of "potassium permanganate in a test tube. After a short time observe
the evolution of a gas in the test tube (what colour is the gas?). Write
the equation of the reaction that takes place.

For every equation draw up a skeleton electron balance.

Hydrogen, Oxygen, and Ozone 145

(i) To 1 ml of a sodium chromite solution add an equal volume of
an alkali and 3 ml of bromine water. Boil the solution and note the
change in its colour. Write the equation of the reaction that takes place.

(j) Add an alkali solution and then some bromine water to a MnCl. 2
solution. Note the changes that occur and write the equations of the
reactions (exchange and red ox).

Exercise 17

HYDROGEN, OXYGEN, AND OZONE

SUBJECTS FOR STUDY

Hydrogen. Atomic structure and chemical properties; monoatoniiL hydrogen;
molecular hydrogen; burning of hydrogen; hydrogen as a reducing agent, and methods
of preparing hydrogen.

Oxygen. Atomic structure and chemical properties; reactions of oxidation and
combustion; oxidation by oxygen in neutral and alkaline media; methods of pre-
paring oxygen.

Ozone. Molecular structure and chemical properties; methods of preparation.

1. Hydrogen. Hydrogen is the lightest of the chemical elements;
1 litre of this gas at N. T. P. weighs 0.0899 g.

In chemical reactions hydrogen can be either a reducing or an
oxidising agent:

H l e - -> H + or H + \e- -> H~

Its oxidant properties are manifested very seldom, only in reactions
with certain metals.

Two hydrogen atoms combine to form a molecule with the evolution
of a considerable amount of heat:

H + H ; H 2 4 104 Cal.

Two hydrogen atoms have a greater store of internal energy than
a hydrogen molecule. For this reason monoatomic hydrogen is much
more active than molecular hydrogen. In chemical processes monoa-
tomic hydrogen is often used for reactions of reduction (nascent hydro-
gen). Chemical reactions involving hydrogen are in most cases conduct-
ed at an increased temperature.

In the laboratory, hydrogen is prepared by the action of metals on
water, dilute solutions of acids, or solutions of alkalis (amphoteric
metals are used in the latter case). Only metals standing above hy-
drogen in the electromotive series (Table 15 on p. 164) can be used for
this purpose.

10-795

146 Exercise 17

The processes of hydrogen preparation may be expressed
by equations such as these:

1. Ca + 2H* + 2OH' - H 2 + Ca" + 2OH'
2. Zn + 2H* + 2C1' = H 2 + Zn" + 2C1'

3. (a) Zn + 2H* + 2OH' - H 2 + Zn (OH) 2 (or H 2 ZnO 2 )
(b) H 2 Zn0 2 + 2Na' + 2OH' = 2H 2 O + 2Na*

Zn + 2H* + 2OH' + 2Na' + 2OH' - H 2 + 2H 2 O+ 2Na' -f ZnOj

Commercial zinc contains impurities (arsenic, phosphorus, sulphur,
etc.). Accordingly, the hydrogen prepared by treating an acid with
zinc contains gaseous impurities (arsenic hydride or phosphorus
hydride, hydrogen sulphide, etc.). To free hydrogen from these gase-
ous impurities, it is passed through wash bottles containing solutions
that oxidise these impurities.

A mixture of two volumes of hydrogen and one volume of oxygen is
called detonating gas. The components of this mixture, when ignited,
react with an explosion. It should be borne in mind that mixtures of
these gases in other proportions (from 6 to 67% of hydrogen by volume)
likewise explode when ignited.

2. Oxygen. Oxygen is the chemical element most widespread in
nature. The electron arrangement of an oxygen atom is )2)6. In chem-
ical reactions this atom builds up its outer shell to 8 electrons: O -f
+2 e" =O~*. In doing so, it displays pronounced oxidant properties.

Like hydrogen, monoatomic oxygen is much more active than mole
cular oxygen, since two atoms combine to form a molecule with the
evolution of a large amount of heat: O+O = O 2 +117.0Cal.

The atoms in the oxygen molecule are connected by two shared elec-
tron pairs:

At an ordinary temperature oxygen has a low reactivity, but upon
heating this increases markedly. The oxygen molecule in oxidation
processes can react in one of the following manners:

1. O, + 4e- e= 2O~ 2 (the most general type of oxidising action

of oxygen)

2. O 2 + 2e~ = [O 2 ]~ 2 (when alkaline metals, except lithium,

are burned; partly when hydrogen is
burned with sharp cooling of the flame)

3. O 2 -f \e~ = [O 2 r

Hydrogen, Oxygen* and Ozone 141

The following are examples of oxidation:

S + O 2 = SO 2 + 70.92 Cal.
CH 4 + 2O 2 - CO 2 + 2H 2 O + 210. 8 Cal.

2Na + O 2 - Na 2 [6*2] + 119.2 Cal.

K + O 2 = K[0 2 ] +67.5 Cal.

Compounds such as Na 2 [O 2 l and KIO 2 ], containing the ion [O a r*
or [O 2 ]~, are called metal peroxides.

In the laboratory, oxygen is in most cases prepared by decomposing
certain oxides or salts of oxygen-containing acids (HgO, KMnO 4 .
KC1O 3 , KNO 3 , etc.) by heating, sometimes in the presence of catalysts
Such reactions are called intramolecular redox reactions and are charac-
terised by the transfer of electrons inside the molecule from one compo-
nent to another. The following process, for example, takes place in thf
'lecomposition of potassium chlorate:

2KCK>3 = 2KC1 + 3O 2

Oxygen can also be prepared from solutions of hydrogen peroxidt
and the peroxides of alkaline metals.

Since oxygen is somewhat heavier than air, it can be collected ir
upright cylinders.

Oxygen can be detected qualitatively in the laboratory by mean?
of a glowing splint, which bursts into flame in oxygen.

3. Ozone. Ozone is an allotropic modification of oxygen; its mole
cule consists of three atoms. The question of the linkages between the
oxygen atoms in the ozone molecule is still somewhat obscure, but
on the strength of the fact that ozone is broken down in reactions to
molecular and monoatomi coxy gen (O 3 =.O 2 +O), the formula of ozone
may be written thus: [O 2 -O], Since monoatomic oxygen is much more
active than molecular oxygen, ozone too is a more active oxidant that
oxygen is. In reactions ozone acts thus:

X+[O 2 -O1 = X +6+O 3 -

reducing
agent

The ozone molecule thus gains two electrons, e. g.:

2Na+ KVO1 + H 2 O = I 2 + 2NaOH + O a

Ozone is an endothermic compound, which decomposes according to
the equation:

2O 3 = 3O 2 + 2 34.5 Cal.

148 Exercise 17

Ozone is produced in ozonisersby the action of a silent electric dis-
charge on dry oxygen.

QUESTIONS

1. Write the equations for the reactions whereby oxygen is pre-
pared from KMnO 4 and from Ag a O.

2. One of the industrial methods of preparing hydrogen consists in
passing steam over red-hot iron. Write tne equation of the reaction
and the equation of the equilibrium constant.

3. What are hydrides? How can the charge of their hydrogen ion
be confirmed experimentally?

4. Draw a diagram of the apparatus used to prepare oxygen from
air, adding a brief explanation of each stage of the process.

5. What is the significance, in the liquefaction of gases, of the
critical temperature?

6. By means of what reactions can oxygen be distinguished from
ozone?

7. How was it proved that an ozone molecule consists of three
oxygen atoms?

Problems

1. The daily output of an oxygen plant is 20,000 cu m of the gas reduced to
N. T. P. How many steel bombs will the plant require daily for its output if the bombs
have a 40 lit capacity, withstand a pressure of 150 atm, and are kept in premises where
the temperature is 15? What volume of air has to be passed through the compres-
sor daily (assuming the performance of the compressor to be ideal?)

2. In the steam-iron method a single operation yields 80 cu m of hydrogen. De-
lermine the daily expenditure of iron and the amount of Fe 3 O 4 produced if 3 opera-
lions daily are conducted at the plant.

3. Calculate the volume of the resulting gaseous mixture and the number of
grams of metallic silver that it can oxidise if 500 lit of air are passed through an ozo-
niser, 7% of the oxygen (by volume) being converted to ozone.

4. Two hundred grams of barium peroxide is decomposed by heating. What will
be the yield of oxygen (by volume) at p =700 mm and /o =27?

5. At 1200 one volume of iron dissolves 0.65 volume of hydrogen (at N. T. P.).
Determine the weight of the hydrogen that can dissolve in 1 t of iron with a density
of 7.9 g/cu cm.

6. An industrial plant has a daily output of 186 kg of sodium peroxide. What are
Its requirements in metallic sodium (by weight) and in 251itoxygen bombs, in which
the oxygen is at 150 atm and 13?

7. In a weakly alkaline medium 10 lit of ozonised oxygen is absorbed by a solution
of potassium iodide. The potassium iodide combines quantitatively with the ozone
to form KIO 8 . Determine the percentage composition of the gaseous 'mixture (by
volume) f the yield being 10.7 g of potassium iodate.

8. A mixture consisting of 1 litre of hydrogen and 0.5 litre of oxygen at N. T. P
Is ignited. Determine the pressure of the water vapour formed, the temperature
during the explosion reaching 2300.

9. How much calcium hydride must react with water for the hydrogen evolved
to reduce 32 g of ferric oxide?

10. What volume of air, taken at N. T. P. and containing 12% of ozone, is
needed to oxidise 280 ml of an 0.1 N solution of sodium iodide in an acid medium?

Hydrogen, Oxygen, and Ozone 149

LABORATORY WORK

Apparatus and "materials: the apparatus shown in Fig. 52, the test tube having a
hole in the bottom; the apparatus shown in Fig. 54; gas-holder with oxygen; Kipp gas
generator; barometer; room thermometer; metal ruler; wash bottle for gases; tesl
tubes and rack; two deflagrating spoons; two glass cylinders; 250 ml measuring
cylinder; stopper with gas delivery tube; two glass covers for the cylinders; funnel;
glass vessel; splints; potassium chlorate; manganese dioxide; potassium permanga-
nate; ammonium persulphate; granulated zinc; coal in lumps; sulphur in lumps;
diethyl ether; concentrated nitric acid; 1:6 dilute sulphuric acid; 0.1 N solution of
potassium permanganate; 0.5 N solution of potassium iodide; 0.5 N solution of
lead acetate; 2 N solution of sodium hydroxide; N solution of sodium sulphide; 0.5
N solution of manganese chloride; solution of indigo or indigo carmine, and cottonwool.

Note. It is best to use an 0.5% solution of indigo carmine, which dissolves readi-
ly in water. In the absence of indigo carmine, prepare a 1% solution of indigo in con-
centrated sulphuric acid. At first the solution will be green, but with the passage of
time it will turn blue.

The apparatus shown in Fig. 54 may be replaced by an ozoniser.

1. Preparation and Burning of Hydrogen, (a) Place 5-6 granules
of ziric in the test tube of the apparatus shown in Fig. 52; close the
test tube with a stopper through which a straight delivery tube with
a drawn tip has been passed. Lower the test tube into the sulphuric
acid solution in the flask. When all the air has been displaced (check
this as in Exercise 5), light the hydrogen that is being evolved and
hold a dry funnel over the flame for a time. Note what forms on the
glass of the funnel. Remove the test tube from the acid solution.
Write the equations for the reactions of the preparation of hydrogen
and its burning. Could any metal be used instead of zinc to prepare
hydrogen in this way?

(b) Pour 3-4 ml of a 30% solution of sodium hydroxide into a
test tube, put 2-3 aluminium shavings in it, and close the test tube
with a stopper through which a straight gas delivery tube has been
passed. Heat the test tube slightly. Collect the gas evolved in another
test tube, held upside down, and ignite it. Write the equation of the
reaction that has taken place. Name 4-5 metals that could be used
instead of aluminium in this reaction.

2. Explosion of Detonating Gas (conduct this experiment with
caution!). Fill with water a 100 ml cylinder* for collecting gases,
cover the cylinder with a flat piece of glass, and immerse it upside
down in a vessel with water. To the straight gas delivery tube of the
apparatus used in the previous experiment attach a bent tube. Place
the end of this bent tube in the vessel with water so that it is right
under the cylinder. Lower the test tube with the zinc into the acid
solution and fill two-thirds of the cylinder with the hydrogen evolved.
Then fill the rest of the cylinder with oxygen from a gas-holder (Fig,
53). Cover the cylinder with the gaseous mixture with the glass, lift
it out of the vessel, wrap it tightly in a towel, bring it up to the flame
of a burner, and, removing the glass, light the gas. What happens 7

* Do not substitute a flask or bottle for the cylinder in this experiment!

150

Exercise 17

What is called detonating gas? Draw a diagram of a gas-holder and
give a brief explanation of the principle on which it operates.

3. Activity of Monoatomic and Molecular Hydrogen. Pour 7-8 ml
of a sulphuric acid solution into a test j tube and add 2-8 drops of a
dilute solution of potassium permanganate. Shake the solution and
pour it, in equal portions, into two test tubes. Put a small lump of

Fig. 52. Set-up for preparing hydrogen Fig. 53. Filling a cylinder with oxygen

from a gas-holder

dnc into one of the test tubes, and pass a stream of hydrogen slowly
through the other from the Kipp gas generator (to free the hydrogen
from possible reductant impurities, pass it through a wash bottle
for gases that contains an oxidising mixture). Why does the solution in
the first test tube lose its colour? Write the equation for the reaction
of the reduction of potassium permanganate by monoatomic hydrogen
in an acid solution.

4. Preparation of Oxygen, (a) Fasten a dry test tube, which con-
tains 0.5 g of crystalline potassium permanganate, vertically in a
damp and heat it over a burner. By means of a glowing splint deter-
mine whether the gaseous product of decomposition is oxygen. Write
the equation for the reaction of decomposition of potassium permanga-
nate. What type of a reaction is it?

(b) Conduct the same experiment, substituting potassium chlorate
for potassium permanganate. When the salt melts, carry out the
glowing splint test. Does the process of the decomposition of potassium
chlorate proceed intensely? Remove the burner and add a pinch of
manganese dioxide to the test tube. Observe the abundant evolution

Hydrogen, Oxygen t and Ozone 151

of gas; test it with a glowing splint. Write the equation for the reaction
of the decomposition of potassium chlorate. What role does manganese
dioxide play in this reaction?

5. Oxidising Properties of Oxygen, (a) Fill two small cylinders
with oxygen from a gas-holder, having the oxygen displace water from
the cylinders *. In one of them burn a small lump of sulphur in a
spoon, lighting the sulphur in advance**. In the second cylinder burn
a small lump of coal heated red-hot on a burner. Observe the intensity
with which these substances burn in pure oxygen and in the air. Why
is combustion more intense in oxygen? Write the equations of the
reactions that have taken place.

(b) Precipitate some manganous hydroxide in a test tube by means
of an exchange reaction. Observe its colour. Then pass a stream of
oxygen slowly from a gas-holder through the solution and note the
change in the colour of the precipitate. Write the relevant equations.
What is the difference between oxidation and combustion?

6. Determining the Absolute Density of Oxygen. Put 1.6-1.7 g of
potassium permanganate into a dry test tube, stopper it up with a thin
wad of cotton wool, and weigh it. Fasten the test tube, in a horizontal
position, in a clamp attached to aringstand. Close the test tube with
a rubbei stopper with a delivery tube (the wad of cotton wool should
be next to the end of the stopper); lower the delivery tube into a vessel
with water. Fill a 250 ml cylinder with water, cover it with a piece
of glass, and turn it over into the vessel. Fasten the cylinder by
means of a large clamp attached to the ringstand.

. By means of a small flame heat the test tube, but not the part of
it containing the potassium permanganate. When air ceases bubbling
through the vessel, place the end of the delivery tube under the cylin-
der and start heating the potassium permanganate carefully. When
about 160 ml of oxygen has collected in the cylinder, stop heating
the test tube and immediately remove the stopper with the gas de-
livery tube from it.

Measure the volume of the oxygen generated and the height of the
column of water in the cylinder above its level in the vessel. Note the
room temperature and the barometric pressure. When the test tube
has been cooled to room temperature, weigh it.

From the data obtained determine the absolute density of the oxy-
gen and calculate the relative error.

7. Preparation of Ozone and Its Properties, (a) Pour 5 drops
of diethyl ether into a large dry test tube and close it with a stopper.

* The oxygen can also be prepared in the Kipp gas generator from a 3% solu-
tion of hydrogen peroxide, which decomposes under the influence of a catalyst made
from manganese dioxide (see Exercise 3). The yield is almost theoretical: from 1.7
litres of 3% hydrogen peroxide it is possible to obtain about 17 litres of oxygen.

** The spoon with the remainder of the sulphur should be ignited in a ventilated
hood.

152

Exercise 17

After the ether has evaporated, heat an iron wire-until it is red-hot (the
handle of a deflagrating spoon can serve as such a wire); insert the
wire into the test tube with the ether vapour. Remove the wire after
5-10 seconds. Notice the odour of the ozone formed. Pour a few drops

of a potassium iodide solution into the
test tube. What happens? Write the equa-
tion for the reaction of the oxidation of
potassium iodide by ozone in ^a neutral
solution.

. (b) Conduct this experiment in a -ven-
tilated hoodl Pour 3-4 ml of an indigo* so-
lution into one test tube and conduct an
exchange reaction between lead acetate and
sodium sulphide in another. The reaction
in the second test tube produces a black
precipitate of lead sulphide (pour oft the
excess solution from the precipitate).
Now put 2 g of crystalline ammonium per-
sulphate in the flask of the apparatus
shown in Fig. 54 and add 10ml of concen-
trated nitric acid to it. Close the neck
of the flask with a ground-in tube and
lower its other end into the indigo solu-
tion.

Heat the flask on a small flame and
note the disappearance of the blue colour
after a time.

Pass ozone into the test tube containing the PbS precipitate.
What is the black precipitate converted to? Write the equation for
the reaction of the oxidation of PbS by ozone. What effect has ozone on
dyes?

llillllHiiiliiiiiiiimiiiiiiiniiimii

Fig. 54. Apparatus for prepar-
ing ozone

Exercise 18

WATER AND HYDROGEN PEROXIDE

SUBJECTS FOR STUDY

Water. Separation of mechanical admixtures from water; distillation of water;
structure of water molecule and its polarity; chemical reactions involving water;
hydration, and the catalytic action of water.

Hydrogen peroxide. Its molecular structure; chemical properties; methods of
preparation; peroxides of metals; some idea of peracids.

1 Water. Water in nature contains mechanical admixtures, and
these may be separated by filtration. Soluble impurities aie separated
by distillation, which is carried out by means of a distillation

Water and Hydrogen Peroxide

15?

apparatus consisting of a flask for boiling the water, a condenser for
condensing the steam to water, and a receiver for the distilled water.

Fig. 55 shows the arrangement used in the laboratory, consisting of
a round flask (1), a condenser (2), an adapter (3), and a receiver (4).

The distilled water obtained by distillation in such an apparatus
contains dissolved gases and a negligible amount of silicates (resulting

Fig. 55. Distillation of water

round-bottom flask; 2 condenser; 3 adapter; 4 receiver.

from the leaching of the glass by steam). This is the water usually
employed in the laboratory for preparing solutions, but the dis-
solved gases can hamper certain experiments.

They can be removed by prolonged boiling. For this purpose a
1 litre flask is three-quarters filled with distilled water, several pieces
of capillary tubing are placed in it (what for?), and the water is boiled
for 30-40 minutes. After that the flask is removed from the wire gauze
with the asbestos centre and is stoppered up tightly, the stopper
having a tube containing soda lime (to absorb CO 2 ).

All chemical reactions involving water may be divided into two
groups:

(1) oxidation reactions, and

(2) addition and exchange reactions.

Examples.

(1) Oxidising action of water:

3Fe + 4H 2 O

4H 2 + Fe 3 O 4

154 Exercise 18

(2) Interaction with oxides:

CaO + H0 - Ca (OH) 2
Formation of aquo-complexes:

CoCl 2 + 6H 2 O = [Co (HaO)e] C1 2

Hydrolysis:

K2CO 3 + HOH ; KHCO 3 + KOH

Exceptionally important is the catalytic action of water. Many
reactions are greatly accelerated by the presence of traces of water
and do not proceed at all without it.

2. Hydrogen Peroxide (H 2 2 ). A study of hydrogen combustion
with flame cooling, as well as of the properties and structure of
the metal peroxides, which are salts of hydrogen peroxide, justifies

r +1 -2

the assumption that hydrogen peroxide has the formula H 2 [O ? ].

Hydrogen peroxide is a weak electrolyte (weak acid), which dissoci-
ates into ions:

HA ^ H" + HO;

H I + io 2 r

Hydrogen peroxide has both oxidant and reductant properties.
The [O 2 ]~ 2 ion receives 2 electrons (as an oxidising agent), turning
-into negative oxygen ions: [O 2 ]~ 2 +2e~= 2CT 2 . For example:

8-
I - =^-71

PbS + 4H 2 Io a ] - PbSO 4 + 4H 2 O

In other reactions the [O 2 ]~ 2 ion can display reductant properties
giving up 2 electrons and forming a neutral oxygen molecule, e. g.,

10*-

\ * -I 7 + 2

5H 2 [0 2 ] + 2KMn0 4 + 3H 2 SO 4 = 5O 2 + 8H 2 O + K 2 SO 4 + 2MnSO 4

Since peroxides can be oxidants and reductants, electrons can be
transferred from one molecule to another:

[OJ + H 2 [0 2 ] =

H 2 [0 2 ] + H 2 [0 2 ] = 2 + 2H 2

Hydrogen peroxide is an unstable compound, and in due course
'(especially rapidly in the presence of catalysts, such as silicon com-
pounds, metal oxides and dioxides, and colloidal metals) it decomposes
lo oxygen and water. The process of decomposition is hastened by
illumination. Oxidation-reduction reactions in which molecules,

Water and Hydrogen Peroxide 155

atoms, or ions of one and the same substance act both as a reductant
and as an oxidant are termed autooxidation-autoreduction, or dis-
proportionation, reactions.

In the laboratory hydrogen peroxide is usually prepared by
treating barium peroxide with sulphuric acid.

In practice it is customary to use either a 3% aqueous solution ot
hydrogen peroxide (employed medicinally as a disinfectant) or a 30%
aqueous solution, or perhydrol.

QUESTIONS

1. Write the structural formula of water and explain why a mole-
cule of water has a dipole moment.

2. Do steam, water, and ice differ in their percentage or molecular
composition?

3. How can it be proved that there are 8 parts by weight of oxygen
to 1 part by weight of hydrogen in water?

4. Why is perhydrol stored in bottles whose inner walls have been
coated with paraffin wax?

5. Give examples of chemical reactions that can serve to distin-
guish ozone from hydrogen peroxide.

, 6. How can the oxide and peroxide of sodium be prepared from
the metal?

7. Write the equation of the reaction between potassium iodide
and hydrogen peroxide in an acid solution.

8. What distinguishes metal dioxides from peroxides?

9. What substances serve to stabilise hydrogen peroxide?

Problems

1. Determine the empirical and the molecular formula of a substance that has the
composition: 5.92%of H and 94.08% of O. A solution containing 2.55 g of this sub-
stance in 500 g of water freezes at 0.279.

2. Calculate the molecular weight of steam, knowing that at the b. p. of water
it contains 3.5% of double molecules.

3. What volume of oxygen will be evolved in the complete decomposition of 400 g
of a 3% solution of hydrogen peroxide?

4. Determine the percentage concentration of the resulting solution if 8.75 kg of
a 3% solution of hydrogen peroxide is mixed with 0. 75 kg of perhydrol.

5. What will be the b. p. of a solution of 82 g of glycerol in 660 g of water? The for-
mula of glycerol is C 3 H 5 (OH)3.

6. What will be the osmotic pressure exhibited by a solution containing 100 g of
perhydrol in 2 litres? The temperature of the solution is 17 (the electrolytic proper-
ties of the dissolved substance may be disregarded).

7. Calculate the pH-value and the a% of a solution of 6 g of acetic acid in 0.5
litre.

8. What amounts of 3% hydrogen peroxide (its relative density being assumed
equal to 1) and water have to be mixed to yield 750 ml of an 0.1 M solution? Calcu-
late the [H'l and pHof the solution.

156 Exercise 18

9. How much of a 3% solution of H 2 O 2 and crystalline KMn0 4 will have reacted
in an acid solution to produce 1.12 litres of oxygen at N. T. P.?

10. What amounts of barium peroxide (by weight) and carbon dioxide (by volume)
will be used up to produce 1 kg of a 3% solution of H 2 O 2 ?

LABORATORY WORK

Apparatus and materials: Kipp gas generator for preparing hydrogen sulphide
and drying bottles (with CaCl 2 ); the apparatus shown in Fig. 55; stopper with gas de-
livery tube bent at a right angle; test tubes and rack; 100 ml beaker; two cylinders
with glass covers; 50 ml measuring cylinder; 10 ml pipette; large crystalliser; fun-
nel; glass spatula; two glass rods; deflagrating spoon; manganese dioxide; cupric chlo-
ride; potassium bromide; mercuric oxide; sodium peroxide; barium peroxide; iron
filings; cobaltous chloride; sulphur in lumps; 2 N solution of sulphuric acid;
N solution of potassium dichromate; 0.5 N solution of potassium iodide; N solution of
sodium sulphide; 0.5 N solution of sodium sulphate; 0.5 N solution of sodium chlo-
ride; 1% solution of silver nitrate; 0.5 N solution of barium chloride; 1% solution
of fuchsin; 0.5 N solution of lead nitrate; 0.1 N solution of sodium chromite; 2 N so-
lution of sodium hydroxide; 0.05 N and 2 M solutions of potassium permanganate;
5% solution of ammonia; solutions of litmus, phenolphthalein, and methyl -orange;
3% solution of hydrogen peroxide; acetone; ether; snow or ice; ethyl alcohol; pieces
of coloured cloth; filter paper; splints, and sand.

1. Reaction for Detecting Hydrogen Peroxide. Mix 2 ml of a hy-
droger? peroxide solution and an equal volume of a sulphuric acid so-
lution in a test tube; add a layer of ether 0.5 cm high. Take a clean
glass rod, dip it into a bottle with a potassium dichromate solution,
and then lower it into the test tube. Chromium peroxide is formed at
once (note its colour!); careful shaking of the test tube causes it
to dissolve in the ether.

2. Preparation of Hydrogen Peroxide. Hydrogen peroxide is
prepared by treating barium peroxide with dilute sulphuric acid.
Calculate the amounts of barium peroxide and of 2 N sulphuric acid
needed to obtain 0.5 g of H 2 O 2 . Measure off the calculated amount
of sulphuric acid, pour it into a beaker, and place the beaker into a
crystalliser with snow to be cooled. Weigh the required amount of
barium peroxide and pour it in small portions into the beaker with
the cooled solution of sulphuric acid, stirring it with a glass rod.
Leave the mixture for 30-40 minutes in the snow, stirring the contents
of the beaker from time to time. Filter the liquid through a small
filter and test the filtrate for the presence of hydrogen peroxide.

3. Oxidant Properties of H 2 a . (a) Add a solution of hydrogen
peroxide to a solution of potassium iodide that has been acidified
by the addition of an equal volume of dilute sulphuric acid. What is
the substance formed? Write the equation of the reaction that takes
place.

(b) Add a solution of an alkali and hydrogen peroxide to a sodium
chromite solution and heat it. How does the colour of the solution
change? Write the equation of the reaction that takes place.

(c) Add an equal volume of sodium sulphide solution to 2 ml of a
lead nitrate solution; heat the solution to boiling point and, when the

Water and Hydrogen Peroxide 157

black precipitate has settled, pour o2 the solution. Add 3 ml of 3%
hydrogen peroxide to the precipitate and heat the solution slightly.
How does the colour of the precipitate change? Write the equation of
the reaction that takes place.

4. Reductant Properties of H 2 2 . (a) Pour 1 ml of a concentrated
solution of potassium permanganate, 2 ml of a sulphuric acid solution,
and 1-2 ml of a hydrogen peroxide solution into a test tube. Introduce
a glowing splint into the test bube. What happens? Write the equation
of the reaction that takes place.

(b) Add 5-6 drops of an ammonia solution and a pinch (on the tip
of a knife) of powdered mercuric oxide to 3-4 ml of 3% hydrogen pero-
xide. What happens? Write the equation of the reaction that takes
place.

5. Catalytic Decomposition of Hydrogen Peroxide. Pour 2-3 ml
of a hydrogen peroxide solution into a test tube and add a pinch of
manganese dioxide (on the tip of a penknife). Introduce a glowing
splint into the test tube. What happens? Write the equation of the
reaction that takes place.

6. Bleaching by Means of Hydrogen Peroxide. Place a small bit
of coloured cloth into a test tube. Add 1 ml of a 5% solution of ammo-
nia (what for?) and 2 ml of a hydrogen peroxide solution. Observe
the change in the colour of the cloth after a time.

7. Interaction of Sodium Peroxide and Water. Pour 3 ml of water
into a test tube and add a pinch of sodium peroxide to it. Establish
experimentally what gas issues from the test tube and what has been
formed in the solution. Write the equations of the reactions of
hydrolysis and decomposition that take place.

8. Decomposition of Sodium Peroxide. Heat 0.5 g of sodium pero-
xide in a dry test tube. Continue this until a glowing splint ceases
to flare up inside the test tube. What is the substance remaining in
the test tube? Use it for experiment 9.

9. Interaction of Oxides and Water. Pour 4-5 ml of water into a cooled
test tube with the substance obtained in the previous experiment.
Test the solution with various indicators. What is the substance
formed? Write the equation of the reaction that takes place.

10. Oxidant Properties of Water. Place a small quantity of wet
sand at the bottom of a refractory test tube (1); fasten the test tube
in a horizontal position, as shown in Fig. 56. By means of a small
glass spatula, put some iron filings on the wall of the test tube (2).
Close the test tube with a stopper through which a gas delivery tube
bent at a right angle has been passed. Another test tube (3) is placed
over the end of the delivery tube. Warm the reaction test tube where
the iron filings are and then heat it until they are red-hot. From time
to time heat the sand to remove part of the water. After a time show
that there is hydrogen in the test tube used as a gas collector. Write
the equation for the interaction of iron with steam.

158

Exercise 18

11. Hydration of Cations and Formation of Aquo-complexes.

(a) Dissolve a few crystals of cobaltous chloride hexahydratein2-3 ml
of water and note the colour of the solution. Dissolve the same amount
of crystals in 2-3 ml of ethyl alcohol, Beating the solution. Account
for the difference in the colours of the alcohol and the aqueous solu-
tion. Dilute the alcohol solution with water, and observe the change
in the colour of the solution. Sprinkle a few drops of the aqueous solu-
tion on a piece of filter paper, and dry
the filter paper over a burner. Does the
pink colour remain? What colour are
theCo ++ and [Co(H 2 O) 6 ]" ions? Write
the equation of the reactions that take
place.

(b) Pour 3 ml of acetone into each
of two dry test tubes. Dissolve a few
crystals of potassium bromide in one
of them and about 1 g of cupric chlo-
ride in the other; the colour of the lat-
ter solution should be a grassy green.
Now pour half of the cupric chloride
solution into the potassium bromide
solution; a brownish solution of cupric
bromide is formed. Add water by drops
to both solutions and observe the
appearance of the blue colouration
typical of the hydrated cupric ionsfCu
(H 2 O) 4 ]". Write the equation of the
hydration reactions that take place.

Can the colour of Me + " ions in a
solid salt serve as any indication of the
colour of solutions of that salt?

12. Catalytic Action of Water (experiment to be conducted in a
ventilated hood!). Prepare two identical cylinders with glass covers.
In one of them burn a small quantity of sulphur. Fill the other one
with hydrogen sulphide from the Kipp gas generator (to dry the gas
pass it through drying bottles filled with granulated calcium chloride).
Place one cylinder on top of the other orifice to orifice, remove the
glass lids, and mix the gases. Observe whether any changes take place.
Move the upper cylinder slightly, pour a few drops of water from a
pipette into the bottom cylinder,andmix the gases again thoroughly.
Observe the formation of a light yellow deposit of sulphur on the
walls of the cylinders. Write the equation of the reaction that takes
place. What is the role of the water in this reaction?

13. Distillation of Water. Pour 100 ml of water from the tap into
a beaker and add 2-3 ml of a sodium chloride solution and the same
amount of a sodium sulphate solution. By means of the proper rea-

Fig. 56. Set-up for preparing hy-
drogen by the action of steam on
iron

i _ refractory test tube; 2 iron
filings; 3 test tube for collect! ng gas.

General Properties of Metals and Alloys

gents prove the presence of chloride and sulphate ions in the solution.
Assemble the apparatus shown in Fig. 55. Pour the water from the
beaker into the flask and add 10 ml of a fuchsin solution. Boil the
water in the flask, collecting the distillate in the receiver. Pour the
first 10 ml of the distillate down the drain. Discontinue distillation
when a second portion (10-20 ml) of the distillate has collected in the
receiver.

Does the distillate have a colour? Prove the absence of chloride OF
sulphate ions in the distillate.

Exercise 19

GENERAL PROPERTIES OF METALS AND ALLOYS

SUBJECTS FOR STUDY

General properties of metals; alloys; physico-chemical analysis; cooling curves;
fusibility diagrams; eutectic mixtures; solid solutions; intermetallic compounds.

Electric current generated by chemical reactions; normal electrode potentials
of metals; electromotive series; displacement of hydrogen by metals and displacement
of some metals by others; galvanic cell.

Electrolysis and Faraday's Laws; electrolysis ^.different types of compounds;
principal methods of preparing metals.

1. Physico-chemical Analysis. The methods of physico-chemical
analysis, which determine changes in the physical properties of a
system with changes in the concentration of the components, serve
to establish that the two or more components have formed a new
chemical compound.

The physical property studied most frequently is the melting
(crystallisation) point of a system; the technique of studying its change
with changes in the concentration of the components of the system
is known as thermal analysis, or the fusibility method.

(a) The melting point is one of the most important constants of a
chemically pure substance. When chemically pure cadmium is heated,
it melts at 321; if the molten metal is heated, say, to 340 and then
cooled, at 321 it will begin to crystallise. This means that for one
and the same substance the melting point and the crystallisation
point coincide. The transition from the liquid to the solid state is
accompanied by the release of the latent heat of fusion, and for this
reason the temperature of the system remains constant until the en-
tire liquid phase has been turned into the solid phase. If the process
of crystallisation or melting is studied as a function of time, the
variation of the temperature with time may be expressed graphically
by means of cooling or heating curves. Fig. 57 shows the cooling and
heating curves for Cd. Both curves, itwill beobserved,havetemperature

160

Exercise 19

halts, or breaks, at 321 (lines parallel to the abscissa), which
indicate phase conversion (solid phased liquid phase) *.

If a second component, say Zn (m. p. 419), is added to Cd, the
crystallisation point of Cd drops. The cooling curve (1) in Fig. 58
has a far less pronounced temperature halt at point B (2.5% Zn), since
{here is only partial crystallisation of the principal component (Cd)

Heating

Time

Fig. 57. Cadmium cooling and heating
curves

321

Time

Fig. 58. Cooling curves for Cd with
impurities: 12.5% Zn; 115% Zn.

here; a second and more pronounced temperature halt(C D) corre-
sponds to the simultaneous crystallisation of both components (Cd-f
4-Zn). The AB section of the curve refers to the liquid phase; the BC
section, to a system consisting of a solid and a liquid phase in equi-
librium, and theD section, to a system consisting of the two solid
phases Cd and Zn.

If the Zn concentration is increased to 5%, the first temperature
halt (Fig. 58, II) shifts even further downward, whereas the position
of the second is unchanged.

The picture will be the same in systems where Zn is the principal
component. If to that metal we add mounting quantities of Cd, the
crystallisation point of Zn will gradually be lowered (Table 14).

The variation of the crystallisation (melting) point with the com-
position of the system may be represented graphically: the values
characterising the composition (weight or molecular ** percentages for
compounds and atomic percentages for simple substances) are plot-
ted as abscissae against the temperatures as ordinates. Such a graph
is called a fusibility diagram (Fig. 59).

The fusibility diagram of the Zn-Cd system, based on the data of
Table 14, has the form of a V-curve (Fig. 59) and points to the absence
of chemical interaction between the components.

* Supercooling is disregarded.
** Molecular percentages are expressed in the molar fraction multiplied by 100.

General Properties of Metals and Alloys

161

Table 14

Melting Points of Zn-Cd System

Weight composition in %

M. p. in C

100

419

305

402

211

381

16.5

83.5

263

362

282

343

300

324

100

321

(J
500

(

300
D

200
Zn!

321
F

--+A'

S-^_

i *

A ^*<

^^N^iA'

P, \

t ly'

- 1

^r^

4^*5

i^ 1

1 '
i l!

00% 80 60 40 20
20 40 60 80 100% Ct

(b) Fusibility Diagram Analysis. At point A' in the diagram the
system has the constitution: 70% Zn and 30% Cd; at the given tem-
perature (470) it is in a liquid state (melt). If the system is cooled
(movement from A' downward),
there is no change in it
until the temperature corre-
sponding to the point /(' is
reached. At this point the Zn
undergoes partial crystallisa-
tion, the system is converted
from a homogeneous to a heter-
ogeneous one, a solid phase
appears, and the Zn concentra-
tion in the remaining liquid
phase diminishes. With
further cooling, changes in the
composition of the liquid phase
follow the curve A'E, which
is part of the AE branch, un-
til the point E (263 C ) is reached,
crystallises.

If we take the system at any point above the BE branch (say, at
5'), cooling will produce the same change, but the primary crystal-
lisation product wall be Cd. Finally, if a system consisting of 83.5%
of Cd and 16.5% of Zn is heated, for example, to 375 (E') and then
cooled, a mixture of both components will crystallise simultaneously
at 263 (D).

This most readily melting mixture of the two components is
termed a eutectic, while its melting point is termed iheeutectic tempera-
ture (denoted by the letter E). The line DF parallel to the abscissa
and passing through E is called the eutectic horizontal.

11-705

Fig. 59. Fusibility diagram for Zn-Cd
system

where the Zn and Cd mixture

162

Exercise 19

r so,"

-Ti

--^

-U-

"+ S

o/."

+ 5

The part of the diagram above the AEB curve is called the liquidus,
i. e., the area of liquid melts or solutions. The areas AED and BEF
correspond to systems in which the solid and the liquid phase are
in equilibrium: Zn so /+ (Zn +Cd) m o/r are in equilibrium at P lf while
Cd so / + (Cd + Zri) mo it are in equilibrium at P 2 .

The part of the diagram below the eutectic horizontal is called the
solidus, i. e., the area of solid phases.

The Zn-Cd system is not the only one with a eutectic point (E)
this is also characteristic of the systems Pb-Sn, Pb-Sb, Cd-Bi, Al-Si,

and others. It should be added
that such systems are formed not
only by metals; they may also be
formed by other components, e. g.:
metal metal oxide (Cu-CuO),
salt salt (KCl-LiCl), salt water
(KC1-H 2 O), and organic com-
pounds (naphthalene phenol). If
one of the components is water,
the most fusible mixture of crys-
tals of ice and crystals of the
second component is called
a cryohydric mixture. Since ice

t melts at 0, all cryohydric

Fig. 60. Galvanic cell mixtures melt below 0.

When the components of a system form chemical compounds or
solid solutions, the fusibility diagrams differ from those for systems
with a eutectic mixture.

2. Galvanic Cell. In a galvanic cell the energy of a chemical
process is transformed into the energy of an electric current. What type
of chemical reactions arising in galvanic cells may be ascertained by
considering the operation of a cell consisting of two galvanic couples,
for instance, Zn/Zn"and Cu/Cu" (Fig. 60).

This type of galvanic cell was first proposed by the Russian phys-
icist B. Yakobi.

The neutral atoms of zinc give up electrons, which travel along the
outer circuit (metal wire) to the Cu/Cu" couple, while SO 4 ions
move along the inner circuit (tube with electrolyte solution) in the
opposite direction. In this process the zinc plate gradually dissolves,
whereas the copper one grows thicker at the expense of copper from
the solution. The overall chemical process can be expressed by the
equation:

\~^~\

Zn + Cu" + SO 4 = Cu + Zn" + SO 4

From this equation it is evident that the reaction involves the trans-
fer of electrons, i. e., it is a redox reaction. In this galvanic cell the

General Properties of. Metals and Alloys 163

Zn /Zrf couple is the reducing agent, while the CuVCu" is the oxi-
dising agent. At the metal solution of metal salt interface in every
couple there arises a definite potential, which depends (at a given con-
centration and temperature) upon the nature of the metal. By means
of a voltmeter inserted in the outer circuit it is possible to measure the
difference of potentials, or electromotive force (e. m. f.), of a cell. The
electromotive force is a measure of the impulse of the redox reaction
and is expressed by the oxidant-reductant potential difference. The
e. m. f. of a reaction is equal to the E Q of the oxidant minus the
of the reductant.

The symbol denotes the electrode potential of a particular couple
with respect to the couple 2H'/H 2 , whose potential is assumed to be
equal to zero.

Example. In the case of the galvanic cell considered above E Q for the Zn/Zn"
couple is equal to 0.76 V; for the Cu/Cu" couple, to 0.34 V. Accordingly, the
e. m. f.=0.34 (0.76) + 1.10 V.

A galvanic cell operates if the potential difference is a positive
value. This means that a redox reaction proceeds in a chosen direction
only provided the potential difference is positive.

The value of is usually determined at a temperature of 25 for
unimolar concentrations * of the salt solutions.

If the electrode potentials are measured for various couples Me/Me+"
and arranged in a series of rising values, this will be the electromotive
series of metals (Table 15).

A study of the electromotive series leads us to several conclusions:

(1) An element situated in the series before another is a reducing
agent with respect to it.

(2) All metals with negative potentials dissolve (oxidise) in acids
with the evolution of hydrogen, while the metals that follow hydrogen
do not displace it from acids.

(3) Any metal of the series displaces (reduces) from a salt solution
a metal of a higher potential, but does not displace a metal of a lower
potential.

The reducing activity of metals mounts and the oxidising ac-
tivity of their ions diminishes in the series from the positive poten-
tial value to the negative.

3. Electrolysis. The passage of an electric current through an electro-
lyte solution causes the positively charged ions to move towards
the negatively charged electrode (cathode) and the negatively charged
ions towards the positive pole (anode). The positive ions acquire
electrons (reduction) from the cathode, while the negative ions give
up electrons (oxidation) to the anode. For example, the process of

* Or, to be more exact, when the activity of the ions of that particular metal in
the solution is equal to 1.

11*

164

Exercise 19

Table 15

Electromotive Series of Metals

(C = 1M; t = 25)

MC/M6+"

E in V

Me/Me + /l

in V

Cs/Cs' . . .

3.02
--3.02

> OQ

i- . c/t7

2 . 92
2.90
2.89
2.87
2.71
2.34
1.70
1.67
1.05
0.76
0.71
0.52
0.44
-0.40
0.34

Co/Co" . .

0.28
0.25
0.14
0.13
0.04
+ 0.00

Li/Li' . . .

Ni/Ni"

Rb/Rb*

Sn/Sn" . .

K/K* - - -

Pb/Pb" . .

Ba/Ba" . .

Fe/Fe

Sr/Sr"

H 2 /2H' . .

Ca/Ca" . .

Na/Na*

Q n / < sn""

0.01
0.21
0.23
0.30
0.34
0.43
0.52
0.80
0.83
0.85
1.42
1.68

Mg/Mg"

Sb/Sh'"

Be/Be"

Ri /Ri'"

A1/A1'"

As/As'"

Mn/Mn" . . .

Cu/Cu"
Co/Co'"

Zn/Zn"

Cr/Cr'"

Cu/Cu'

Ga/Ga* "

A u / A ft'

Fe/Fe"

^s/^s
PH/PH"

Cd/Cd"

Her/Her"

Tl/TT

n s/ n s
Au /Au"*

Au/Au'

electrolysis for hydrochloric acid (Fig. 61) may be written down as
follows:

cathode process 2H* -[ 2e~ = H 2 (reduction)

anode process 2C1' 2e~ = C1 2 (oxidation)

Ions will be neutralised at electrodes only provided there is a cur-
rent of sufficient voltage. Under the action of the electric current

,, n ,..,,. there takes place a chemical process of

"'" "" ^3, decomposition:

2MCI

2HC1 - H

C1 2

Electrolysis is a process of chemical
decomposition of substances by an electric
current; the process comprises reduc-
tion at the cathode and oxidation at the
anode.

The direct decomposition of a substance at the electrodes by means
of an electric current is called the primary electrolysis process. But

Fig. 61. Electrolysis of HC1

General Properties of Metals and Alloys 165

the electrolysis products can also interact with one another (to prevent
them from coming into contact it is customary to use an electrolytic
cell * with a diaphragm), with the molecules of the solvent (this can
be avoided by conducting electrolysis of the molten substance), and
with the material of the electrodes (this can be avoided by the use
of inert electrodes made of graphite or platinum). Such processes are
called secondary electrolysis processes. Some secondary products of
electrolysis (NaOH, KC1O 3 , etc.) are manufactured on an industrial
scale.

The quantitative study of electrolysis led to the establishment of
the following laws:

1 . The amount of substance deposited on the electrodes in electrolysis
is directly proportional to the quantity of electricity passed through the
electrolyte.

2. The passage of equal quantities of electricity through different
electrolytes causes amounts of substances proportional to their chemical
equivalents to be deposited on the electrodes (Faraday's Laws),

Let us denote as Q the weight in grams of the substance deposited
in electrolysis, as A the atomic weight of the element, as n the charge
of the element in the compound subjected to electrolysis, and as I-t
the number of coulombs (coulomb ampere X second) of electricity
passed through the electrolyte solution. In that case:

_ A*I*t
^ "n- 96, 496

Theoretically, 96,496 (or, approximately, 96,500) coulombs of
electricity have to be expended to deposit 1 gram-equivalent of a
substance. But actually, owing to a number of reasons, the amount of
electricity expended is larger. The ratio of the theoretical amount to
that actually used up is equal to the ratio of the amount of actually
deposited substance to the theoretical amount and is called the cur-
rent efficiency.

Electrolysis is used extensively in industry to prepare hydrogen,
oxygen, chlorine, fluorine, and alkalis, to prepare and refine metals,
to effect the anodic coating of metals with oxides, and of some metals
with others.

4. General Methods of Preparing Metals. Since metals occur natu-
rally either in the form of nuggets (gold, silver, etc.) or in the form of
compounds (oxides, sulphide ores, salts), their preparation consists
basically in either separating them from the gangue or in reducing
them from their compounds.

(a) The separation of noble metals from gangue involves two proces-
ses: the oxidation of the metal by oxygen in the presence of potassium

The bath in which electrolysis is conducted.

166 Exercise 19

cyanide and the subsequent reduction of the cyan complex:

4Au + O 2 + 8KCN + 2H 2 O - 4K[Au (CN) 2 1 + 4KOH

2Zn +4KfAu (CN) 2 ] - 4Au + 2K 2 fzn (CN) 4 1

(b) Cathodic reduction is the electrolysis of molten hydroxides or
salts and the electrolysis of salt solutions with a mercury cathode and
subsequent vacuum distillation of the amalgams obtained. The method
is employed to prepare alkali metals, alkaline earth metals, and
metals of the third group of the Periodic Table (including the rare
earths). Cathodic reduction is also used to refine metals prepared by
other techniques.

carbon

C + SnO 2 - Sn -f CO 2

hydrogen

_ |

7H 2 + 2KRe6 4 - 2Re + 2KOH -[- 6H 2 O
aluminium

2A1 + Cr 2 O 3 = 2Cr + A1 2 O 3

Reduction by means of aluminium has come to be known as alu-
minothermics. Other reducing agents used are sodium, magnesium,
calcium carbide.

(d) Intramolecular reduction is employed to prepare noble metals
from their oxides or nitrates, pyrophorous metals by the decomposi-
tion of oxalates and formiates (salts of oxalic acid and formic acid),
and several other metals (Cr, Fe, Co, and Ni) by decomposing carbo-
nyls by heating them. The following are examples of the preparation
of metals from:

an oxide

2HgO = 2Hg + 2
a nitrate

2AgNO 3 - 2Ag + 2NO 2 + O 2
an oxalate

FeC 2 O 4 = Fe + 2CO 2

General Properties of Metals and Alleys 167

a formiate

Co(HCOO) 2 - Co + CO + CO 2 + H 2 O

a carbonyl

Ni(CO) 4 = Ni -t- 4CO

(e) Preparation of metals from sulphides. Most of the heavy metals
are prepared from sulphide ores (pyrites, glances, blendes, etc.).
The process is conducted in two stages: (1) ore roasting and (2) reduc-
tion.

Ore roasting involves the oxidation of the ore by oxygen (air) at
high temperatures:

2Bi 2 S 3 -|- 9O 2 = 2Bi 2 O 3 + 6SO 2

In this reaction it is only S~ 2 (which gives up 6 electrons) that is oxi-
dised; in some cases the metal ion toojandergoes oxidation. The result-
ing oxide is reduced by carbon in the second stage of the process.

QUESTIONS

1. Pure tin melts at 232; pure lead, at 327. The eutectic point
of their alloy is 181 at a composition of 64% by weight of tin and
36% of lead. On the basis of these data draw a fusibility diagram (de-
picting the branches of the curve as straight lines). In the diagram
draw an isotherm * for 216 and describe the system at the points of
intersection of the isotherm with the branches of the curve. Through
the point with the coordinates 300 and 50% of Pb draw a line paral-
lel to the ordinate and describe all the changes in the system from the
given temperature to the eutectic.

2. Give examples of the use of eutectic mixtures in industry.

3. What special point characterises the fusibility diagrams of systems
whose components form chemical compounds?

4. What properties has the crystal lattice of solid solutions?

5. What galvanic couples of metals should be chosen (Table 15 on
p. 164) for a galvanic cell to produce a maximum e. m. f.?

6. Draw schemes illustrating the electrolysis of CuSO4 and NaBr
solutions and indicate all the processes that take place at inert
electrodes.

7. What chemical reactions can be used to obtain free antimony from
antimony glance Sb 2 S 3 ? Write the equations of the reactions that take
place.

* An isotherm is a line of equal temperatures.

168 Exercise 19

Problems

1. Draw a scheme of a cell consisting of the couples Ni/Ni" and Cu/Cu". Indi-
cate the direction of the current in the inner and the outer circuit, determine the
e. m. f. of the cell, and write the equation of the reaction on the basis of which the
cell operates (solutions of M concentration are used; the temperature is 25).

2. During the operation of a galvanic cell consisting of the couples Zn/Zn" and
Cu/Cu" the weight of the cathode diminished by 0.1634 g. Calculate the amount of
electricity produced.

3. The passage of a current through a dilute solution of H 2 SO 4 for 10 minutes pro-
duces 652.2 ml of detonating gas (at a temperature of 18 and pressure of 746 mm).
Calculate the current intensity.

4. A solution prepared by dissolving 0.45 g of brass (an alloy of zinc and copper)
is subjected to electrolysis. The amount of electricity required to deposit all the cop-
per from the solution is 482.5 coulombs. Calculate the percentage composition of cop-
per in the brass.

5. Prove by adducing figures that nickel displaces copper, but does not displace
aluminium, from solutions of their salts.

6. How many grams of a molal solution of HC1 are needed to dissolve the amount
of metallic zinc deposited in electrolysis on the cathode as a result of the expenditure
of 4,825 coulombs of electricity?

7. Make up a galvanic cell of the couples Mg/Mg" and Hg/Hg", indicate the di-
rection of the current in the circuit, write the equation of the reaction that takes
place in the cell, and determine the e. m. f. if the concentration is M and the tempera-
ture is 25.

8. How much iron will be deposited in two galvanic baths connected in series,
one containing a FeSO 4 solution and the other a FeCl 3 solution, if "a 2 ampere current
is passed through them for 1 hour?

9. The electrolysis of molten LiCl for 2 1 /2 hours yeilds 1.2 g of lithium at the
cathode. Determine the current intensity and the volume of chlorine (reduced to
N. T. P.) evolved at the anode.

10. The weight of a nickel plate immersed in a solution of silver nitrate increases
by 0.73 g. How much silver is deposited on the plate?

LABORATORY WORK

Apparatus and materials: the apparatus shown in Figs. 62 and 63 (or 60 and 64);
5 ampere ammeter; 6-8 volt storage battery; 5 volt voltmeter; slide rheostat; electric
key; copper electrode; zinc electrode; porous vessel; sand bath; iron crucible; thermo-
meter for 100; porcelain mortar; test tubes and rack; 300-400 ml beaker; 200 ml
beaker; two 100 ml beakers; weighing bottle or watch glass; slides; glass tubing 5 rnm
in diameter; knife for cutting glass tubing; dovetail nozzle; electrolytic bridges with
agar; rubber bands for attaching capillaries; steel nibs; bismuth; tin; lead; cadmium;
zinc plate; copper plate; magnesium turnings; granulated zinc; copper turnings;
powdered antimony; powdered iron; ferric oxalate; cupric oxide; dry crystalline phe-
nol; crystalline naphthalene; M solution of cupric sulphate; M solution of zinc sul-
phate; 0.5 N solution of potassium ferricyanide; N solution of magnesium chloride;
N solution of cupric chloride; 2 N solution of sodium chloride containing phenol-
phthalein; 0.5 N solution of mercuric nitrate; N solution of hydrochloric acid; 2N
solution of sulphuric acid; 1:1 nitric acid; platinum or silver wire; iron wire; pa-
raffin wax; charcoal; squared paper; thick paper, and clean rags.

Note. The phenol should be colourless and dry. If the product is coloured, it ought
to be distilled and dried over potassium hydroxide for 10 days,

1. Fusibility Diagram of Naphthalene-phenol System. This is a
collective job for the whole group of students, each determining one

General Properties of Metals and Alleys

point on the fusibility curve of the system. Ten points in all should be
determined for the system (see table below). The students who receive
the assignments 1 and 10 should determine the melting points of pure
naphthalene and phenol, conducting the experiment
as described in Exercise 7. The students receiving the
assignments 2-9 determine the melting point and the
crystallisation point of naphthalene-phenol mixtures.

By means of a chemical balance weigh to 0.01 g
the amounts of the initial substances indicated in the
table (do the weighing in weighing bottles or watch
glasses). The work with phenol requires caution, as it is
poisonous; besides that, upon coming into contact
with the skin, it produces burns.

Place the weighed amounts into a special test tube
with a glass jacket to prevent overcooling or overheat-
ing. Close the test tube with a stopper through which
a thermometer and a stirrer (careful!) have been
passed (Fig. 62). Place the apparatus into a beaker
with water, fastening it in a clamp.

Heat the water in the beaker slowly. When the
crystals begin to melt, reduce the flame of the burner
so that the rise in temperature continues at a rate of
0.5 e) a minute.

As soon as the solid phase begins to melt, it is
necessary to start stirring the contents of the test tube
with the stirrer (be careful not to break the stirrer!).
Note the temperature at which the last crystals melt.
Then discontinue the heating and, whi le stirring the melt
with the stirrer, begin cooling the beaker with water.
Note the temperature at which the first crystals appear.

Make three determinations of the melting point and
the crystallisation point, and record the results.

On the basis of the data in the table, plot a fusibility diagram
(a convenient scale for the abscissa is 10% by weight = 1 cm; for the
ordinate, 10=- 1 cm). Find the point of intersection of the branches
in the diagram and determine the coordinates of the eutectic point.

2. Preparation of Wood's Metal. This experiment should be con-
ducted in a ventilated hood. Weigh 5,3 g of Bi (m. p. 271), 2gof Sn
(m. p. 232), 1.5 g of Pb (m. p. 327), and 1.2 g of Cd (in. p. 321).
Melt about 10 g of paraffin wax in an iron crucible and place the tin
in it. Heat the crucible, mixing the contents with an iron wire, until the
melting point is reached; then add the lead, the cadmium, and the
bismuth in that order. When all the metals have melted, cool the
alloy, first pouring off the molten paraffin wax. Wipe the alloy with
a rag, melt it again, and pour it into a cigarette-wrapper, which
serves as a mould. When the rod of alloy has cooled, lower it into a

Fig. 62. Set-
up for melt-
ing point and
crystallisation
point measu-
rements

170

Exercise 19

No. of
assign-
ment

Composition of mixture

Determination of
melting point (a)
and crystallisa-
tion point (b)

Mean
tempera-
ture in C

Name of
student
perform-
ing ex-
periment

naph-
thalene
In g

phenol
In g

naphtha-
lene in %
by
weight

phenol
In % by
weight

100

)

1.29

0.11

1.02

0.38

0.81

0.59

0.65

0.75

0.52

0.88

0.36

1.04

0.29

1.11

0.24

1.16

100

glass of water heated to 55; slowly, at the rate of 1 a minute, con-
tinue heating the water and note the melting point. Record the result
in your notebook and hand the alloy over to the laboratory assistant.
Why does the alloy melt at a temperature lower than the m. p. of
any one of the four constituent metals?

Fig. 63. Galvanic cell

I thick-walled.? jar; 2 cylindrical copper electrode; 3 vessel of clay;
4 zinc electrode; 5 voltmeter; 6 key.

3. Galvanic Cell. Assemble the cell shown in Fig. 63 (or in Fig. 60
on p. 162, in which case an electrolytic bridge filled with a solution
of KG with agar serves as the "key"). Into a thick-walled jar (1) con-
taining an M solution of cupric sulphate lower a cylindrical electrode
(2) made of sheet copper. Inside the electrode place a porous vessel
of clay (3) with an M solution of zinc sulphate. Into the clay vessel
lower a zinc electrode (4). Connect up the voltmeter (5) and key (6).
Close the circuit by pressing the key and determine the e. m. L by

General Properties of Metals and Alloys 171

means of the voltmeter. Disconnect the wires from the voltmeter and
apply them to a slide on which is a strip of filter paper moistened
with a solution of NaCl with phenolphthalein. Note which of the
wires produces a crimson colouration (negative pole!). Make a draw-
ing of the apparatus and a scheme of the cell, indicating the direc-
tion of the current in the inner and the outer circuit. Calculate the
e. m. f. of the cell. Is there a discrepancy between the calculated and
the experimental value? Write the equation of the chemical reaction
that produces the electric current in the circuit.

4. Electromotive Series, (a) Pour 4 ml of an N solution of hydrochlo-
ric acid into each of 4 test tubes; add a piece of magnesium, zinc,
copper, and antimony respectively. Do all the metals displace hydro-
gen from acids? Give an explanation.

(b) Select two lumps of zinc with surface areas as near equal as
possible. Place one of them for 3-4 minutes into a test tube containing
1 ml of a cupric sulphate solution. Pour the solution from the test
tube and wash the copper-plated zinc with water several times. Pour
3-4 ml of a hydrochloric acid solution into each of 2 test tubes; put
the lump of ordinary zinc into one of them; the lump of copper-
plated zinc, into the other.

In which of the test tubes is the evolution of hydrogen more vi-
gorous?

On the surface of which metal is hydrogen evolved? Give an expla-
nation,

(c) Into the slit of a steel nib insert a small plate of zinc; into the
slit of another nib, a plate of copper of the same size. Pour 5-6 ml of
a sulphuric acid solution into each of 2 test tubes and add 3-4 drops
of a solution of potassium ferricyanide to each (the salt is a reagent
for the ion of bivalent iron, with which it forms TurnbulTs Blue).
Immerse the nibs with the plates in the test tubes. Observe the appear-
ance of a blue colouration in one of them (which?) in a few minutes.
Give an explanation of this.

(d) Pour 2 N H 2 SO 4 into a small beaker and lower a plate of zinc
into it. Is there evolution of hydrogen on the surface of the zinc?

Touch the plate with the bent end of a platinum or silver wire.
On which of the metals is hydrogen evolved and why?

(e) Pour 3-4 ml of a MgCl 2 solution into one test tube and an equal
amount of a HgCl 2 solution into another; lower copper plates into
the test tubes. After a period of time examine the surface of the
plates, explain the phenomenon observed, and write an equation of the
reaction.

5. Electrolysis. Pour a solution of CuCl 2 into a U-tube (Fig. 64)
in such a way that 3 cm lengths of the carbon electrodes are immersed
in the solution.Connect wires from a storage battery to the carbon elec-
trodes. After 5-10 minutes observe what has been deposited on the
cathode. By its odour identify the gas evolved at the anode. Remove

172

Exercise 20

BHF

the copper from the cathode by dissolving it in nitric acid and-wash
the apparatus with water. Draw a scheme of the electrolysis process.

,6. Preparation of Metals, (a) Mix 1 g of mercuric sulphide (cinna-
bar) in a small mortar with 1 g of powdered iron and put the mixture
in a dry test tube. Heat the test tube over the flame of a burner in a

ventilated hood and, after a time, observe the
formation of a "mercury" belt on the coldwalls
of the test tube. Write the equation of the reac-
tion that takes place.

(b) Mix 1 g of cupric oxide with 1 g of coal
in a mortar. Place the mixture in a dry test
tube and, after warming it over the flame of a
burner, heat it for 7- 10 minutes. After cooling,
empty the test tube on a sheet of paper; ascer-
tain (from the colour) that metallic copper has
been formed. Write the equation of the reac-
tion that has taken place.

(c) Place about 1 g of powdered ferric oxa-
late into a dry test tube. Fasten the test tube in
a wooden clamp and heat it over the flame of a
burner, holding the test tube so that the upper
part is slightly below the bottom. The yellow
powder is decomposed by heating, and grey-
ish-black metallic iron is formed. After the oxalate has been com-
pletely decomposed, remove the powdered iron from the test tube, expo-
sing it to the air. What happens? Note the colour of the oxide formed!
What is the term used for metals that in the powdered state ignite
spontaneously in the air? Write the equations for the reactions of
the decomposition of the oxalate and the burning of the iron in the

Fig. 64. Apparatus for
electrolysis

air.

Exercise 20

ALKALI METALS

SUBJECTS FOR STUDY

AlKali metals; atomic structure of these elements; their reductant properties;
atomic radii and relative activity of alkali metals; attitude to oxygen, hydrogen,
water, and acids; oxides and hydroxides of akali metals and their properties; salts
of alkali metals, and distinctive features of lithium among the alkali metals.

The alkali metals are: Li, Na, K, Rb, Cs, and Fr. The atoms of
all these elements have only one electron each in their outermost
shells. In chemical reactions they part with this electron readily
according to the equation Me l<T=Me + , thereby exhibiting highly

Alkali Metals 173

pronounced reductant properties. The atomic radii increase with the
rise in atomic numbers, and reductant activity mounts in the same
order:

Li Na K Rb Cs

Atomic number 3 11 19 37 55

Radius (A) 1.56 1.86 2.23 2.43 2.62

The alkali metals are kept under a layer of kerosene, since exposure
to the air causes them to undergo rapid oxidation:

4Me + O a - 2Me 2 O

All the alkali metals come before hydrogen in the electromotive
series; they are readily oxidised by water and by acids, interacting,
for instance, as follows:

2Na + 2HOH = H 9 + 2NaOH

2Li + 2HC1 - H 2 + 2LiCl

The oxides of these metals can be prepared by reducing their per-
oxides by an excess of the metal:

- J

2Na + Na 2 tO 2 l - 2Na 2 O

The oxides of the alkali metals are solid, highly hygroscopic sub-
stances that combine readily with water to form hydroxides, solid
substances that dissolve well in water.

The aqueous solutions of the hydroxides exhibit pronounced alkaline
properties, due to dissociation into ions:

MeOH ; Me* + OH'

The most important of the hydroxides are sodium hydroxide NaOH
and potassium hydroxide KOH, which are prepared primarily by the
electrolysis of aqueous solutions of NaCl and KC1 (with subsequent
evaporation).

The salts of the alkali metals are, with a few exceptions, soluble and
belong to the strong electrolyte group. The salts of weak acids under-
go hydrolysis in aqueous solutions, and their solutions are for this
reason alkaline. The volatile salts impart characteristic colours to
the colourless flame of a burner: in the case of sodium compounds the
colour is yellow; lithium, carmine; potassium, violet; rubidium, red-
dish-violet, and caesium, violet.

174 Exercise 20

QUESTIONS

1. Which metal, sodium or potassium, is oxidised more readily?
Why?

2. Write the equations for the reactions of the oxidation of potas-
sium by oxygen at room temperature and in combustion.

3. How can sodium hydroxide be prepared from Na 2 CO 3 and from
NaCl? Write the equations for the respective reactions.

4. Given solutions of the salts:

K 2 C0 3 NaN0 3 K 2 SO 4 K 2 S KHCO 3

Write the equations for their interaction with water and explain why
some of these solutions are alkaline.

5. Explain why CsOH is a stronger alkali than KOH.

6. Draw diagrams for the electrolysis of molten KC1 and of a KC1
solution, indicating the processes that take place at inert electrodes.
Can the K' ion be reduced without resorting to electrolysis?

Problems

1. Demonstrate whether it is possible for potassium to form compounds with oxy-
gen in which there would be 17% of oxygen and 45.1% of oxygen.

2. It is known that 0,43 g of a certain metal reacts with water to displace
123.2 ml of a gas (measured at N. T. P.). It is known too that 1,56 g of the same
metal reacts with 1,415 g of chlorine. What is the metal?

3. Raising the temperature of 1 g of metallic sodium by 12 requires 3.48 Cal.
Calculate the atomic weight of sodium and the relative error of the determination,
the exact atomic weight being 22.997.

4. On the assumption that rubidium consists of the isotopes Rb 85 (72.8%) and
Rb 87 (27.2%) determine the atomic weight of the element.

5. Calculate the titre and the normality of 8% KOH, the relative density of the
solution being 1.065.

6. What volumes of 52% NaOH (relative density 1,56) and water should be mixed
to prepare 1 ton of 16% NaOH?

7. The interaction of 1 g of sodium amalgam with water produces an alkali solu-
tion whose neutralisation requires 50 ml of a hydrochloric acid solution with a titre
of 0.00365. Calculate the sodium content of the amalgam.

8. What will be the boiling point of a solution of 25 g of KOH in 800 g of water
if the apparent degree of ionisation is 86%?

9. Calculate the osmotic pressure of a 6% solution of KOH (relative density 1,05
at 20) if the apparent degree of ionisation is 84%?

10. What will be the yield of the products in the electrolysis of molten NaOH
by an 0.2 a current for 5 hours?

LABORATORY WORK

Apparatus and materials: the apparatus shown in Fig. 33; 100 ml Erlenmeyer
flask; 10ml measuring cylinder; 25ml burette; funnel for burette; 10ml pipette; two
porcelain casseroles; two watch glasses 7-9 cm in diameter; 100 ml measuring flask;
test tubes and rack; forceps; scalpel; two glass rods; platinum wire; room thermometer;
barometer; indigo or cobalt prism; sodium; potassium; 0.04-0.05 g weighed amounts of
sodium; potassium chloride; sodium sulphide; sodium nitrate; potassium carbonate;.

Alkali Metals 175

titrated solution of 0.02 N HC1; 0.5 N NaCl; 0.5 N KC1; saturated solution of potassium
antimonate; 0.4 N solution of sodium bitartrate; solution of sodium cobaltinitrate;
ethyl alcohol; solutions of phenolphthalein, methyl-orange, and neutral litmus; set
of test tubes with saturated solutions of lithium, potassium, and sodium chlorides
and concentrated hydrochloric acid; boiled distilled water, and filter paper.
Note. The preparation of sodium cobaltinitrate is described in Exercise 22.

1. Interaction of Alkali Metals with Air and with Water, (a) By

means of forceps take a lump of metallic sodium from a jar in which
it is kept in kerosene, place the sodium on filter paper, and make a
cut in the lump with a knife or scalpel. Note how the freshly exposed
lustrous metal surface tarnishes. Cut o't a tiny bit of the metal the
size of a match-head and throw it into a porcelain casserole with water.
Cover the casserole with a piece of glass. Observe the vigorous reac-
tion that takes place.

(b) Carry out a similar experiment with metallic potassium. Add
a drop of phenolphthalein to the solutions obtained in each of the
casseroles. What happens? Why?

Write the equations for the interaction of potassium and sodium
with water. What accounts for the greater activity of potassium?

2. Experimental Verification of the Equivalent of Sodium. The equiv-
alent of sodium is determined by the procedure described in Exercise
6 (experiment 2). Before conducting the experiment make certain that
the apparatus is airtight.

For the acid substitute alcohol (why not water?), with which alkali
metals react thus:

2Na + 2C 2 H 5 OH = H 2 + 2C 2 H 5 ONa

Take 5-6 ml of ethyl (methyl) alcohol for the experiment. After
pouring the alcohol into a test tube, carefully dry the inner walls of
the test tube with strips of filter paper.

By means of forceps take a lump of metallic sodium weighing about
0.04-0.05 g from a jar with kerosene, dry it quickly on a piece of
filter paper, and place it in the bulb (broken line in Fig. 33) of an
inclined test tube.

Close the test tube tightly with the stopper of the apparatus and
lower the test tube in a vertical position, so that the sodium slips
into the alcohol and reacts with it.

After the reaction has ended and the apparatus has been allowed
to cool, measure the volume of the hydrogen evolved (don't forget
to equalise the levels of the liquid in the burettes!) and record the
barometric pressure and the temperature.

From the weight of the metal and the volume of the hydrogen
evolved calculate the equivalent of the sodium. To determine the
exact weight of the metal sodium taken, pour the solution of sodium
ethyl ate formed in the test tube into a measuring flask. Rinse the

176

Exercise 20

Level in burette

Hydrogen
volume in ml

Barometric
pressure p
in mm Hg

Temperature
in C

Vapour
tension h
in mm Hg

before
reaction

after reaction

test tube 3-4 times with small portions (4-5 ml) of distilled water,
pouring the solution each time into the measuring flask.
The hydrolysis reaction produces sodium hydroxide:

C 2 H 5 ONa + HOH - C 2 H 5 OH + NaOH

From the amount of the sodium hydroxide it is possible to calculate

the amount of metal originally taken.
Bring the level of the solution in the measuring flask up to the mark

so that the bottom meniscus rests on the level of the mark on the

neck of the flask (Fig. 65). Close
the flask with a stopper and mix
the solution in it thoroughly.

Lower a dry 10 ml pipette into
the flask, suck the solution into
the pipette above the mark, and
cover the upper end of the pipet-
te with the index finger (Fig. 66).
By reducing the pressure of the
finger bring the level of the liquid
down to the mark (according to
the bottom meniscus) and then
close the upper end of the pipet-
te tigthly with the index finger
again. Now lower the tip of the
pipette into a clean Erlenmeyer
flask (Fig. 66) and release the
measured amount of alkali solu-
tion into it (the tip of the pip-
ette should touch the inner wall of

Fig. 65. Measuring Fig. 66. Pipette and
flask Erlenmeyer flask

the neck of the flask at an angle of
45-50). Add 2-3 drops of methyl-
orange indicator to the solu-
tion; the solution turns yellow.

Pour a titrated solution of hydrochloric acid into a burette (Fig. 67),
making sure that the tip of the burette is filled and the bottom meniscus
touches the zero line. The glass ball (3) fitted snugly inside the rubber

Alkali Metals

ill

tubing (2) prevents any acid from issuing from the burette. But by
pressing the rubber tubing next to the ball it is possible to form a
channel (C), through which acid passes. If any bubbles of air remain
inside the burette tip, these should be removed as shown in Fig. 67D.

Place an Erlenmeyer flask with an alkali solution and an indicator
underneath the burette and pour 1 ml of acid from the burette at a
time, shaking the flask after each
addition of the acid, until the yel-
low colour of the solution changes
to orange * (normal titration)
or to a pinkish-red (overtitrated
solution!). Let us assume that
the addition of 9 ml of acid
leaves the solution yellow, while
10 ml makes it pinkish-red. In
that case the exact volume of the
acid is greater than 9, but less
than 10 ml.

Empty the Erlenmeyer flask,
rinse it thoroughly with water,
measure another 10 ml of the
alkali solution with a pipette,
and add the same amount of
indicator. Titrate the solution
again, at first adding several
millilitres at a time (up to 9 ml in the above example), then adding
the solution by drops until the colour changes to orange. Record
the amount of acid solution used (with an accuracy of up to
0.05 ml). Repeat the titration two more times. Take the average of
the three titrations.

Fig. 67. Burette with pinch-cock clamp

A burette; 1 calibrated part; 2 rubber
tubing; 3 pinch-cock clamp; 4 drawn-
out tip; B position of ball when burette
is closed; C position of ball when liquid
issues from burette; D filling the burette
tip.

Volume in ml

Amount of HCI solution used for titration in ml

Concentra-
tion of
HCI so-
lution (N)

flask

pipette

approx

exact

average of
3 exact
titrations

From the data obtained calculate the equivalent of the metallic
sodium that reacted with the alcohol.

* To be able to observe the change of colour clearly, place a sheet of white paper
under the Erlenmeyer flask

178

Exercise 20

To do this, determine the concentration of the alkali from the con-
centration and volume of the acid used for the titration and the volume
of the alkali. From the concentration of the alkali calculate the amount
of it (in g) contained in the flask. From the formula (NaOH) calculate
the amount of the metal corresponding to the found amount of the
alkali. Finally, calculate the equivalent of the metal.

3. Detecting Alkali Metals by Flame Colourations. Receive from
the laboratory assistant a set consisting of 4 test tubes, a rack, and a
glass rod with a platinum wire protruding from it. One of the test tubes
contains concentrated hydrochloric acid, while the
other three contain solutions of lithium, sodium, and
potassium chlorides. Introduce the platinum wire into
the outer cone of the flame. The clean wire glows with-
out altering the colour of the flame. Should the flame
become coloured, immerse the wire in hydrochloric
acid and then heat it again. Next, immerse the wire
in the lithium salt solution and heat it, observing the
characteristic colouration of the flame. Before testing
the next salt solution in the flame, keep the wire in
the flame for a time. The flame colouration produced
by potassium salts should be observed through an
indigo or cobalt prism (Fig. 68).

4. Hydrolysis of Salts. Place a few crystals of
NaNO 3 , Na 2 S, KC1, and K 2 CO 3 into four test tubes re-
spectively and add 2-3 ml of boiled distilled water to
each. Then add 1 ml of neutral litmus solution to
each, observing the change in colour against a back-
ground of white paper. Write the equations of the hy-
drolysis reactions that have taken place. Which of the
four salts undergo hydrolysis?

5. Preparation of Slightly Soluble Salts, (a) To a neutral solution
of a sodium salt add an equal volume of a solution of potassium hydro-
xoantimonate K[Sb(OH) 6 l; to the potassium salt solution, a solu-
tion of sodium hydrogen tartrate NaHC 4 H 4 O 6 . If no precipitates appear
at once (owing to the formation of supersaturated solutions), rub the
test tubes inside with a glass rod. To speed the formation of the potas-
sium hydrogen tartrate crystals, it is advisable to add 2-3 ml of ethyl
alcohol. How do the precipitates differ by appearance? Write the
equations of the reactions that have taken place.

(b) To a neutral potassium salt solution or one acidified with acetic
acid add an equal volume of a solution of sodium cobaltinitrate
Na 3 [Co(NO 2 )6]. A yellow crystalline precipitate of K 2 Na[Co(NO 2 ) 6 l
is formed. Write the equation of the reaction that has taken
place.

6. Analysis of Solid Salt. Receive from the instructor a salt sample
for analysis (NaCl, KC1, Na 2 SO 4 , or K 2 SO 4 ). Write a brief plan of

Fig. 68. Indi-
go or cobalt
prism

Copper Subgroup Elements 179

how to detect the cation and anion of the salt and submit the plan to
the instructor. Carry out the analysis, write it up, and submit the
report to the instructor.

Exercise 21

COPPER SUBGROUP ELEMENTS

SUBJECTS FOR STUDY

The copper subgroup: copper, silver, and gold; their atomic structure and a com
parison of the electron shells of their atoms and those of the alkali metals; points of
similarity and difference in the properties of these metals; the position of copper,
silver, and gold in the electromotive series; the attitude of these metals to oxygen,
water, and acids; gold dissolved in aqua regia\ oxides and hydroxides; major salts;
oxidant properties of the ions of the noble metals, and complex compounds.

The atoms of copper, silver, and gold have one electron in their
outermost shell and are in this respect structurally similar to the
atoms of the alkali metals. The atoms of the alkali metals, however,
have 8 electrons in the next to last shell, whereas in the case of cop-
per, silver, and gold this shell has 18 electrons. This accounts for the
marked difference between the properties of these elements and the
properties of the alkali metals.

For example, let us compare the properties of potassium and copper:

K Cu

Atomic number 19 29

Erectron arrangement ) 2) 8) 8) 1 )2) 8) 18) 1

Atomic radius (A units) 2.23 1.28

lonisation potential (ev) 4.30 7.69

Melting point (C) 63.5 1083

Boiling point (C) 776 2360

Density 0.86 8.95

Hardness 0.5 3.0

Conductivity 14 57

Electrode potential (V) 2.92 0.52

With the change in electron structure from potassium (with its
8-electron structure) to copper (with its 18-electron structure), the
relative density, melting point, and boiling point rise, while the
atomic radius and the reductant activity diminish.

Silver and gold are noble metals and are difficult to oxidise; their
positive ions, unlike the ions of the alkali metals, can act as oxidants
and have a marked polarising effect, which accounts for the coloura-
tion of most of their compounds, the low thermal stability of the
oxides and hydroxides, and the ability to form complexes.

180 Exercise 21

The 18-electron shell of the Cu, Ag, and Au atoms is not particular-
ly stable: in chemical reactions the atoms of these elements give up
not only their outer electrons, but also electrons from the I8e~ shell,
turning into uni- to tripositive ions.

Copper, silver, and gold rank below hydrogen in the electromotive
series; since their normal potentials are positive, they are not oxi-
dised by hydrogen ions, but can be oxidised (except gold) by acids
whose radicals have a high oxidising ability (concentrated and dilute
nitric acid, concentrated sulphuric acid, etc.). In these reactions the
acids act both as oxidant and as the medium.

As we know from earlier work (Exercise 19, paragraph 2), the
normal electrode potential values can be used to find the e. m. f . of redox
reactions and to predict their direction. By using a hydrogen electrode,
it is possible to determine the E not only of metal couples (Table
15), but also of couples consisting of more complicated oxidising
agents and their reduced forms or of complicated reducing agents and
their oxidised forms. For instance, the nitric acid ion can by reduc-
tion be converted to nitrogen peroxide; the potential of the NOa +
-f-2H7NO 2 +H 2 O couple, measured by means of a platinum elec-
trode, proved equal to 0.81V. The normal redox potentials of other
couples (Table XV on p. 336) can be measured in the same way.

Example. Can metallic copper be oxidised by concentrated nitric acid that is
reduced to nitrogen peroxide?

(a) Write the equation of the reaction (see Exercise 16, experiment 6):

21 _

I -> |+5 +2 +4

Cu + 2HNO 3 + 2HNO 3 Q^(N0 3 ) 2 + 2NOH- 2H 2 O

(b) In the equation underline the reductant and its oxidised form once; the oxidant
and its reduced form, twice. The galvanic cell corresponding to this reaction is the
cell:

Cu/Cu'y/NOg + 2H'/N0 2 + H 2 O

Consequently, the Cu/OT couple is the reductant, while the NO3+2H7NO a +H 2 O
couple is the oxidant.

==0 - 81 v ( Table xv > P- 336 ) determine the e. m. f.

, o u . ^ u ^

-|-2H /NO g -f-H 2 O

f ' - E N03 + 2HVNO, + H,0 ~ ^" = ' 81 " QM = ' 47

The e. m. f. being positive, the reaction between these substances is possible.

Metallic gold, which is insoluble in hydrochloric, sulphuric, or
nitric acid, dissolves in aqua regia (a mixture of three volumes of
concentrated hydrochloric acid to one volume of concentrated nitric
acid):

Au + 4HC1 + HN0 3 - H [AuClJ + NO + 2H 2 O

Copper Subgroup Elements 181

The reaction can be explained as consisting of the following stages;
(1) the interaction of the two acids to form monoatomic chlorine
and nitrosyl chloride NOC1:

I I

2HC1 + HNO a + HC1 = 2CH- NOC1 -|- 2H 2 O

(2) the oxidation of the gold by the monoatomic chlorine and nit-
rosy 1 chloride:

26*

I | -> | +3 +3-1 +2

Au + 2C1 + NOC1 = AuCl 3 + NO

(3) the combination of the auric chloride with excess hydrochloric
acid to form chloroauric acid:

AuCl 3 + HC1 - H [AuCl 4 ]

At ordinary temperatures the oxygen of the air has no effect on
copper, silver, or gold. But when copper is heated, it is oxidised to
cupric oxide or, when the access of air is limited, to cuprous oxide.
Dissolved oxygen, in the presence of substances that form complex
compounds with Cu", Ag', and Au*, oxidises these metals, e. g,:

2Cu + O 2 -|- 8NH 4 OH - 2[Cu (NH 3 ) 4 1 (OH) 2 + 6H 2 O (1)

4Au -|- 2 + 8KCN -I- 2H 2 O - 4K [Au (CN) 2 ] + 4KOH (2)

Reaction (2) is used to extract gold from quartz sand.
The treatment of copper, silver, and gold salt solutions with alkalis
produces hydroxides, e. g.:

Cu" -J- 2OH' = Cu (OH) 2
2Ag* + 20H' ; 2AgOH ; Ag 2 O |- H 2 O

Silver hydroxide at ordinary temperatures decomposes reversibly to
water and the oxide; cupric hydroxide decomposes irreversibly, more-
over only when heated. The oxides of these metals do not dissolve in
water, except silver oxide, whose saturated solution exhibits alkaline
properties perceptibly. Upon being heated, the oxides of silver and
gold decompose to the metal and oxygen.

182 Exercise 21

The salts of the univalent cations are mostly colourless and in-
soluble. The compounds of bivalent copper are blue-green or green,
while the salts of trivalent gold are golden yellow. The salts of cop-
per, silver, and gold undergo hydrolysis,

The positive ions Cu", Ag', Au', Cu", and Au"' may, in reactions,
exhibit oxidant properties, e. g.:

2 +

SSnCl., -f 2A + uCl 3 + 6HC1 = 3H 2 [SnClJ + 2Au

2FeS0 4 + Ag 2 S0 4 - Fe 2 (SO 4 ) 3 -|- 2Ag

Characteristic of these ions are complex compounds,the most important
being ammoniates, such as [Cu(NH 3 ) 4 ]SO4 and [Ag(NH 3 ) 2 ]Cl, and
cyanides, such as K[Ag(CN) 2 ] and K[Au(CN) 2 ].

QUESTIONS

1. What are the points of similarity and difference between the
metals of the copper series and the alkali metals? Explain this.

2. Copper objects are often covered with a green coating of basic
copper carbonate Cu 2 (OH) 2 CO 3 . Write the equation for the reaction
of the formation of this salt, bearing in mind that it is formed in the
presence of copper, water, oxygen, and carbon dioxide.

3. Write the equations for the reactions that take place when metal-
lic copper is dissolved (with heating) in concentrated sulphuric acid
and when silver is dissolved in concentrated nitric acid.

4. Why is silver oxide decomposed by heating, while potassium
oxide is not?

5. By writing the equations of relevant reactions, demonstrate
that Au(OH) 3 is an amphoteric compound.

6. What is the effect of aqueous solutions of CuSO 4 and AuCl 3
on litmus, and why is this so?

7. Write the equations of two reactions illustrating how the Ag*
ion is precipitated and reduced.

Problems

1. The heating of 20 g of metallic copper from 97 to 100requires5.71 cal. Calcu-
late the atomic weight of copper and express in percentages the discrepancy between
ithe value found and the exact value of 63.54.

2, What amount of 20% HC1 is needed for the complete precipitation of the silver
from a solution obtained by dissolving 50 g of a silver alloy of the 875 standard in
nitric acid?

Copper Subgroup Elements 183

3. How many litres of nitric oxide (at 18 and 750 mm Hg) will be produced when
I kg of copper is dissolved in an excess of dilute nitric acid?

4. Twenty grams of 8.5% ammonia solution is used to dissolve 7.17 g of silver
chloride. Determine the composition of the complex compound formed.

5. Calculate the amounts of substances needed to extract 1 kg of gold from quartz
sand by the cyanide method.

6. Calculate the percentage and molar concentrations of a solution prepared by
mixing 18 lit of a 4% solution of CuCl 2 (relative density 1.036) with 4 lit of a 22%
solution of CuCl 2 (relative density 1.23).

7. The heat of formation of cuprous oxide, from its elements is 42.5 Cal.; of cupric
oxide, 38.5 Cal. Calculate the heat of the reaction of the oxidation of cuprous oxide
to cupric oxide.

8. The surface of a cylinder 16 cm high and 6 cm in diameter is to be coated gal-
vanically by a layer of gold 0.005 mm thick. How much electricity and K[Au(CN) 2 ]
salt will this require, if the density of the gold is 19.3 g/cu cm?

9. Draw a diagram of a galvanic cell consisting of the couples A1/A1'" and
Au/Au'". Explain the operation of the cell in terms of values.

10. Calculate the solubility product for AgCl, knowing that 5.15-10-* g of AgCl
dissolves in 100 ml of the solution at 50.

LABORATORY WORK

Apparatus and materials-. 6-8 V storage battery; 5 A ammeter; 5 V* voltmeter; stop
watch; slide rheostat; electric key; two 2 mm copper plates; electric wires; test tubes
and rack; funnel; 300 ml beaker; crucible tongs; copper turnings; copper wire; copper
gauze turned into a cylinder; litmus paper; starch paper; sandpaper; filter paper;
Fehling's solution (in two solutions); concentrated nitric acid; 2 N H 2 SO 4 ; 1 : 1 and
15% nitric acid; concentrated, 2 N, and 6% HC1; 1% and 0.1 N AgNO 3 ; 10% ammo-
nia solution; 1% and 2 N NaOH; 0.5 N KI; 0.5 N KBr; 0.5 N KC1; 0.5 N CuSO 4 ; N
Na 2 CO 3 ; N solution of sodium thiosulphate; 2% solution of glucose; ethyl alcohol;
hydrogen sulphide water, and copper plating solution.

Preparation of Fehling's Solution. Two solutions are prepared: I- 34.64 g of
cupric sulphate in 500 ml of water, and II 52 g of sodium hydroxide and 173 g of
potassium sodium tartrate NaKC 4 H 4 O 6 in 500 ml of water.

Preparation of Electrolysis Solution. Dissolve 150 g of cupric sulphate in 1 litre
of water containing 50 g of sulphuric acid (relative density 1.84) and 50 ml of ethyl
alcohol.

Precaution! All the spent solutions of silver salts should be collected in special bot-
tles!

1. Reductant Properties of Copper, (a) This experiment should be
conducted in a ventilated hood.Tesi the action of dilute and concentrat-
ed H 2 SC>4 on metallic copper with and without heating.

(b) This experiment too should be conducted in a ventilated hood.
Test the action of dilute and concentrated nitric acid on metallic
copper with heating. Note the colour of the gases evolved. Determine
the e.m.f. of the first reaction from redox potential values (Table
XV on p. 336).

(c) Take some bits of copper wire with forceps and heat the wire in
the outer cone of a nonluminous flame of a burner. What changes are
observed? Write the equation of the reaction that takes place.

184 Exercise 21

2. Action of Alkalis on Cupric and Silver Salts. Study the action
of an alkali solution on solutions of cupric and silver salts. Record
the external changes that take place and write the relevant equations.
Shake the solution with the cupric hydroxide precipitate and pour it
into four test tubes. Heat one of the test tubes and describethechanges
observed. Treat the precipitate in another test tube with an acid
solution. Write the equations of the reactions that take place.

Pour 2-3 ml of concentrated sodium hydroxide solution into the
third test tube and shake it vigorously. Let the precipitate settle.
What is now the colour of the solution? Write the equation of the
reaction between cupric hydroxide and sodium hydroxide. What
are the properties of cupric hydroxide? What is cuprite? Leave the
fourth test tube with the cupric hydroxide precipitate for experiment
10.

3. Preparation of Cuprous Oxide. Mix 1.5 nil each of the two so-
lutions used to make Fehling's solution in a test tube with an equal
volume of a 1% solution of glucose and boil the contents of the test
tube. A yellow precipitate of CuOH is formed at first; further boiling
converts this to red Cu 2 O.

The reaction may be set down as follows:

/ +2

CH 2 OH - (CHOH) 4 - C + 2Cu (OH) 2 -

glucose \

= CH 2 OH - (CHOH) 4 C + 2CuOH + H 2 O

fflyconic acid \

2CuOH = H 2 O + Cu 2 O

4. Preparation of Cupric Oxide and Its Oxidant Properties. Heat a

cylinder of copper gauze in the oxidising flame of a burner. Observe
the formation of black cupric oxide. Immerse the heated "sausage"
in a test tube with 3-4 ml of ethyl alcohol. What happens? Write
the equation of the reaction that takes place on the assumption that
the alcohol is oxidised to acetaldehyde.

5. Hydrolysis of Salts, (a) Test solutions of some cupric and silver
salts with litmus paper. Observe the changes in its colour and write
the equations of the hydrolysis reactions involved.

(b) Add a saturated soda solution to a concentrated solution of cup-
ric sulphate. Observe the formation of the precipitate [Cu 2 (OH) 2 CO 3 )

Copper Subgroup Elements

185

and the evolution of a gas. Write the equation of the reaction
that takes place.

6. Oxidant Properties of Cu" Ion. Add a solution of potassium iod-
ide to a solution of cupric sulphate. A white precipitate is formed
(Cu 2 I 2 ) while the solution turns yellow, owing to the formation of
free iodine (test with starch paper). Write the equation of the reaction
that takes place. What is the role of the iodine ion?

7. Silver Plating of Glass. In a test tube mix 1 ml of 10% AgNO^
with 1 ml of 1 % NaOH and dissolve the brown precipitate formed in
2 ml of a 10% ammonia solution.

Add 2 ml of a 2% solution of glu-
cose and place the test tube in a
beaker with water heated to 60.
After a time a silver mirror forms on
the walls of the test tube. What prop-
erties of the silver ion does this
reaction illustrate?

8. SilverHalides. By means of
exchange reactions prepare precip-
itates of silver chloride, bromide,
and iodide. Note the colours of the
precipitates and write the equations
for the reactions by which they
were prepared. Filter off the brom-
ide and iodide precipitates, unfold
the filter papers with the precipi-
tates, and expose them to the light.
After a time note how they have
darkened (photochemical reactions).
Leave the silver chloride precipitate
for experiment 10.

9. Cupric and Silver Sulphides.
By means of exchange reactions

prepare cupric and silver sulphides. Note the colours ot the precipi-
tates. Test the action of concentrated HC1 and HNO 3 solutions on
them. Write the equations of the reactions that take place. Why do not
these sulphides dissolve in hydrochloric acid?

10. Complex Compounds of Copper and Silver, (a) Add ammonia
solution to a cupric hydroxide precipitate. What happens to the pre-
cipitate? Note the colour of the resulting solution and write the equa-
tion of the reaction that takes place, the formula of the product being
[Cu(NH 3 ) 4 ](OH) 2 .

(b) Shake the silver chloride precipitate with the solution and pour
it into two test tubes. Add ammonia solution to one of them. What
happens to the precipitate? Write the equation of the reaction that
takes place, the formula of the product being [Ag(NH 8 ) 2 ]Cl. Add sodium

Fig. 69. Diagram of arrangement
for electrolysis of copper

1 storage battery; 2 ammeter;
3 -rheostat; 4 -key; 5 - voltmeter;
6 copper anode; 7 iron cathode.

186 Exercise 22

thiosulphate solution to the other test tube. Note the changes that
occur and write the equation of the reaction. What is the practical
significance of this last reaction?

11. Detection of the Cupric and the Silver Ion. Receive from the
instructor a test solution containing one of these ions. By means of
the maximum number of the reactions you have studied, prove the
presence of one of these ions in the solution.

12. Electrolysis. Clean the surface of a copper plate with sandpaper
until it shines, immerse it for 3-4 minutes in a 15% solution of nit-
ric acid, rinsing it afterwards thoroughly with water under the tap.
Assemble the apparatus shown in Fig. 69. Into the electrolytic cell
pour an amount of the solution sufficient to cover three-quarters of
the plates. Lower the electrodes into the solution and adjust the rheo-
stat to maximum resistance. Close the circuit by means of the key,
note the time of the beginning of the experiment (start the stop watch),
and reduce the resistance so as to bring the voltage up to 1.2-1.4 V.
JRead the current intensity on the ammeter. Exactly 15 minutes later
break the circuit and rinse the iron plate with water. The copper coat-
ing is a compact reddish layer without lustre. Measure the surface
area of the plate. Knowing the current intensity and the duration of
electrolysis, calculate the thickness of the coating (the density of
copper is 8.9 g/cu cm).

Exercise 22

COMPLEX COMPOUNDS

SUBJECTS FOR STUDY

Basic principles of the Coordination Theory; central atom; coordination number;
tigands; the inner and the outer coordination sphere.

Electrolytic properties of complex compounds and their conductivity; dissoci-
ation of a complex ion and the instability constant of complex compounds; complex
ions in exchange reactions, and complex ions in redox reactions.

The so-called complex compounds contain more elaborate groups of
ions, with properties differing from those of elementary ions, e. g.:

Na 3 [Co (N0 2 ) 6 1 KlCo (NH 3 ) 2 C1 4 ] En (NH 8 ) 6 1 (OH) 2
[Pt (NH 3 ) 4 C1 2 1C1 2

A complex compound has a central atom, or complex-forming ion,
with a characteristic valence (charge) and a coordination number,
which indicates how many groups, atoms, ions, or molecules are
linked to the complex former in the inner sphere. With these ligands*

* This term has superseded the earlier term addend.

Complex Compounds 187

the complex former constitutes what is known as a complex ion, whose
charge is the algebraic sum of the charges of the complex former and
the ligands.

If the complex ion has a negative charge, the outer sphere consists
of positive ions, and vice versa. For example:

K* [Au (CN) 2 ]' [Cu (NH 3 ) 4 ]"Ci;

complex . complex

an ion cation

To write the formula of a complex compound, it is necessary to
know the valence (charge) of the complex former and the ligands, as
well as the coordination number. The following are the coordination
numbers of the most important complex formers:

24 6

-fl -fl +2 +2 +2 +3 +2 -f2

Ag, Au Cu, Hg Co, Co, Ni, Fe

4-2 +3 4-3 4-4 43 42

Pt, Au Fe, Pt, Cr, Mn

+2 +3 42 +2 +4 43

Pb, B Zn, Cd, Sn, Al

These are the most widespread coordination numbers of central
atoms. Less frequent coordination numbers are 3, 8, and 12.

One and the same complex former may have several different coordi-
nation numbers: for instance, A1+ 3 , Co+ 2 , and Zn" 1 " 2 are known to
have the coordination numbers 4 and 6.

Both electrically neutral molecules (H 2 O, NH 3 , NO, etc.) and
negative ions (O~, OH", NO 2 ~, CN", Cl", Br, I~, CO 3 "" f etc.) can
be ligands.

Example in Writing the Formula of a Complex Compound. Write the formula of
a complex compound consisting of the ions Fe +3 , CN^and K + . Since the positive ions
K+ cannot be part of the inner sphere, the CN'ions form the inner coordination sphere
around the Fe+ 3 ion. The latter ion has the coordination number 6; the formula is
therefore K 3 [Fe(CN) 6 ].

The most important complex compounds with neutral ligands are
the ammoniates, e. g., [Co(NH 3 ) 6 ]Cl 3 , and aquocomplexes, e. g.,
[Cr(H 2 O) 6 ]Cl 3 ; the latter group includes the so-called crystalline
hydrates. Complex compounds in which negatively charged acid radi-
cals are grouped around the central atom are known as acid complexes,
e. g., K 3 [Co(N0 2 ) 6 ].

In an aqueous solution complex compounds dissociate into a com-
plex ion and the ions that make up the outer sphere. For example:

[Cu (NHa)J S0 4 ; [Cu (NH 3 ) 4 ]"
Na 3 [Co (N0 2 ) 6 ] j> 3Na' + [Co (NO 2 ) 6 ]

188 Exercise 22

In exchange reactions complex ions pass from one compound to another
without any change in their composition:

2Cu" + 2SOI + 4K* + fFe (CN) e r = Cu 2 [Fe (CN) 6 1 + 4K* + 2SOl
In solution complex ions dissociate to a certain degree:

[Cu (NH 3 ) 4 r^OT + 4NH 8
For this case of equilibrium we may write:

[[Cu (NH,),n ~~

The quantity /C (the ratio of the product of the concentrations of the
particles into which the complex ion dissociates to the concentration
of the complex ion) is called the instability constant of the complex ion.
The greater this constant, the less stable the complex.

2 r^Ag +2NH 3
# = 6.8 -10- 8
tAg(CN) 2 ]';Ag'+2CN'
K = 1 10- 21

It is evident from these figures that the ammoniate of silver is less
stable a complex than the cyanide.

The instability constants of some complex ions are listed in Table
XIV (p. 335).

A decrease in the concentration of one of the components of a com-
plex can cause its destruction. Let us take solutions of two nickel salts:
NiSO 4 and [Ni(NH 3 ) 6 ]SO 4 . Both solutions are in a state of equilibri-
um:

NiS0 4 ; Ni" + SOl
[Ni (NH 3 ) 6 ]S0 4 1 [Ni (NH 3 ) 6 r + SOl

Ni" + 6NH.3

The addition of a solution of NaOH causes the precipitation of
Ni(OH) 2 in the first solution, but no precipitation in the second.

For a precipitate to be formed, it is necessary that the product of
the concentrations [Mi"]- [OH 7 ] 2 be greater than the solubility prod-
uct SPNi(OH) 2 - Although the complex ion [Ni(NH 3 ) 6 ]" does dissociate
into Ni" ions and NH 3 molecules, the extent to which it does this is
so small that the concentration [Ni"] is insufficient for the SP value
to be reached and no precipitate is therefore formed.

The addition of ammonium sulphide to the NiSO 4 and[Ni(NH 3 ) 6 ]SO 4
solutions causes a precipitate of NiS to be formed in both cases. Nickel
sulphide is much less soluble (SPnis = 1.4-10" 24 ) than is nickel hyd-

Complex Compounds 189

roxide (SPNi(OH)* = 1-6- 10 14 ). The concentration of Ni" ions even in
the complex salt solution is sufficient for the S^NIS value to be reached.

As the S" ions bind the Ni" ions, there is a shift in the equilibrium
in the system [Ni(NH 3 ) 6 ]":Ni" +6NH 3 towards the dissociation
of the complex ion, which leads to the complete destruction
of the complex ion.

Redox reactions of complex compounds involve either a change in
the charge of the complex former without any substantial disturbance
in the composition of the complex (example l)or the complete destruc-
tion of the complex with the formation of oxidation products of sim-
pler composition (example 2):

+3 -1

(1) 2K 4 IFe (CN).l + C1 2 = 2K 3 [Fe (CM),] + 2KC1

le-

+2 -1

(2) 2K 2 [Ni (CN)J + 9Br a + 6KOH - 2Ni (OH) 3 -j 8CNBr + lOKBr.

If a complex contains positive ions, which are oxidising agents, as
complex formers, they can be extracted as neutral atoms by the
method of reduction:

26-

Zn + 2K [Au (CN) 2 1 = 2Au + K 2 [Zn (CN) 4 1

The positive ions can be reduced by means of an electric current (in
galvanisation, for gold- and silver-plating, etc.).

The process whereby amphoteric hydroxides dissolve in solutions
of alkalis is now regarded as a process of the formation of specific com-
plex compounds called hydroxy salts, in which the complex ion consists
of a central atom and hydroxyl ions in its coordination sphere. For
example, when stannous hydroxide is dissolved, the process may be
represented by the following equation:

Sn (OH) 2 + 2K' + 2OH' - K 2 [Sn (OH) 4 1
This compound is called potassium hydroxystannite.

QUESTIONS

1. Is there a difference between double and complex salts?

2. 'Write the coordinate formulae for the ammoniates CoCl 3 -6NH 8
and CoCl 8 -5NH 3 . Why does the precipitation of the chloride ions from
1 mol of the former compound require 3 mols of silver nitrate, whereas
the precipitation of the chloridfe ions from 1 mol of the latter requires
only 2 mols?

190 Exercise 22

3. For the system in equilibrium

[Ag (NH 3 ) 2 1 Cl ^[Ag (NH 3 ) 2 r + Cl'

write the equations of the ionisation constant and the instability con-
stant. Why is no AgCl precipitate formed despite the presence of Ag^
and Cl' ions in the solution? Why does the addition of nitric acid to
the solution cause the precipitation of AgCl?

4. Why does the treatment of a CdCl 2 solution with an alkali pro-
duce a Cd(OH) 2 precipitate, while the treatment of a [Cd(NH- 8 ) e ]Cl ft
solution does not?

5. Write the equations for the oxidation of yellow prussiate of
potash KjFe(CN) 6 ] by hydrogen peroxide in an acid solution and for
the oxidation of the complex compound K 4 [Co(CN) 6 l by the oxygen
of the air in a neutral solution.

6. Write the equation for the reaction of the formation of potassium
hydroxyaluminate from aluminium hydroxide.

Problems

1. A certain amount of metallic zinc reacts with a solution of K[Au(CN) 2 ] to
yield 7.89 g of metallic gold. The amount of 10%HC1 needed to dissolve the same
amount of metallic zinc is 14.6 g. Calculate the equivalent of the gold.

2. A complex compound contains Co'", NH 3 , and Cl'. The precipitation of the
Cr from 11.67 g of this salt requires 8.5 g of silver nitrate. When the same amount
of the complex salt was broken down, the yield of ammonia was 4.48 litres (reduced
to N. T. P.). The molecular weight of the salt is 233.3. Write the coordinate formula
of the complex compound.

3. What amount of a 5% solution of (NH 4 ) 2 S is needed for the complete precipi-
tation of the copper as .sulphide from 120 ml oi an 0.1 M solution of [Cu(NH 3 ) 4 ] SO 4 ?

4. The oxidation of a 3% solution of H 2 O 2 in an alkaline medium by potassium
ferricyanide K 3 [Fe(CN) 6 ] yields 560 ml of 2 , measured at N. T. P. Calculate the
amounts of 3% H 2 O 2 , solid K 3 [Fe(CN) 6 ], and 20% KOH that were used up.

5. It is necessary to prepare 80 litres of an 8% solution (relative density 1.043)
of K 3 [Fe(CN) 6 ] by diluting an 18% solution (relative density 1.10). Calculate the
amounts of water and of the initial solution that have to be mixed.

6. Determine the molarity of a solution of [Ag(NH 3 ) 2 ]Cl if 4.4 ml of HNO 3 (titre
0.063) is used to bind the ammonia contained in 20 ml of the solution.

7. What will be the boiling point of 0.1 M K 3 [Fe(CN) 6 ], its relative density being
1.02 and its degree of ionisation 65%?

8. Calculate the osmotic pressure at of 8% K 3 [Fe(CN) 6 ], its relative density
being 1.043 and a being 62%.

9. For how long is it necessary to pass a 2-ampere electric current through a
K[Ag(CN) 2 ] solution to deposit 0.24 g of metallic silver?

10. A 5-ampere current is passed for 20 minutes through a [Ni(NH 3 ) 6 ]SO 4 so-
lution. Calculate by how many grams the weight of the cathode increases.

LABORATORY WORK

Apparatus and materials: test tubes and rack; funnel; 2 watch glasses; filter
paper; wash bottle; litmus paper; splints; 2 N KOH; N BaCL; N FeCl 3 ; 2 N (NH 4 ) 2 S;
0.5 N KSCN. 2N HNO 3 ; 0.5 N solution of aluminium potassium sulphate; 2 NKOH;

Complex Compounds 19f

0.4 N solution of sodium bitartrate; solution of sodium cobaltinitrate; saturated and-
0.1 N solutions of silver nitrate; saturated solution of CoCl 2 ; saturated solution of
ammonium rhodanate; saturated solution of potassium iodide; 0.5 N ZnCl 2 ; 0.5 N
KC1; 0.5 N CuSO 4 ; 10% ammonia solution; 3% H 2 O 2 ; N solutions of potassium ferro-
and ferricyanide; N FeSO 4 , and chlorine water.

Preparation of Sodium Cobaltinitrate. Dissolve 50 g of CoCl 2 -6H 2 0, with gentle
heating, in 50 ml of water, cool the solution, and mix it with a solution of 150 g of
sodium nitrite in 500 ml of water. Add 50 ml of 5% CH 3 COOH, while stirring.
When the precipitate has settled, pour off the liquid into reagent bottles.

1. Ions of a Double Salt. Pour 2 ml of a KA1(SO 4 ) 2 solution into*
each of three test tubes; add a sodium bitartrate solution (see Exercise
20, experiment 5) to the first, a few drops of an NaOH solution to
the second, and a barium salt solution to the third. Write ionic equa-
tions for all three reactions. What ions may be assumed to be present
in the initial solution on the basis of these reactions? Write the equa-
tion for the electrolytic dissociation of KA1(SO 4 ) 2 .

2. Ions of a Complex Salt, (a) Pour 2 ml of a FeCl 3 solution into each
of two test tubes; add a solution of (NH 4 ) 2 S to one and a KSCN so-
lution to the other. What is formed? Write ionic equations of the re-
actions. What ion may be assumed to be present in the FeCl 3 solution
on the basis of the reactions with the S" and SCN" ions?

(b) Carry out analogous reactions, substituting a K3[Fe(CN) 6 T
solution for the FeCl 3 solution. Is a black precipitate * formed? Does
a red colouration appear? Are there Fe'" ions in the K 3 tFe(CN) 6 l
solution?

Write the equation for the electrolytic dissociation of K 3 [Fe(CN) 6 ].
Prove experimentally that the solution of this salt contains potassium
ions (see Exercise 20, experiment 5). What are complex salts? Can a
sharp dividing line be drawn between double and complex salts? Write
equations of electrolytic dissociation for the salts:

NH 4 Cr(S0 4 ) 2 [Cu(NH 3 ) 4 lCl 2 K 2 [PtCl 6 ]

3. Compounds with a Complex Negative Ion. (a) Place a drop of &
saturated solution of silver nitrate on a watch glass and add a saturated
solution of potassium iodide io it by drops until the precipitate formed
dissolves. Write the equations of the reactions that have taken place.
Keep the solution for experiment 6.

(b) Place 3-4 drops of concentrated solutions of CoCl 2 and NH 4 SCN
on a watch glass. The solution acquires the blue colour typical of
[Co(SCN) 4 ]" ions. Write the equation of the reaction. ? : Keep the so-
lution for experiment 6.

(c) Add an alkali solution by drops to a zinc salt solution until a-
white precipitate is formed. Write the equation of the reaction. Add

* The yellow precipitate formed is sulphur produced by a side reaction of oxi-
dation.

192 Exercise 22

excess alkali to the precipitate. The precipitate dissolves owing to
the formation of hydroxyzincate. Write the equation of the reaction.

4. Compounds with a Complex Positive Ion. Prepare the precipi-
tates of AgCl and Cu(OH) 2 by exchange reactions, pour off the liquid
from the precipitates when they have settled, and add excess ammonia
.solution to the precipitates. Write the equations of the reactions
whereby the precipitates are dissolved. Note the colour of the solutions
of the complex compounds. Keep the ammonia solutions for experi-
ment 6.

5. Complex Compounds in Exchange Reactions, (a) Add a solution
of yellow prussiate of potash to an equal volume of a dilute, acidified
solution of FeCl 3 . What is formed? Write the equation of the reaction
-that takes place.

(b) Add a Na 3 [Co(N0 2 ) 6 ] solution to a KCl solution. A yellow
crystalline precipitate of K 2 Na [Co(N0 2 ) 6 ] is formed. Write the equation
of the reaction.

6. Destruction of Complex Ions, (a) Dilute with water (by drops)
the complex compound solutions on watch glasses left over from ex-
periment 3. Observe the changes that take place. Write the equations
for the electrolytic dissociation of the complex ions. How is the balance
shifted by the addition of water? How does the concentration of the
-solution affect the stability of complex compounds?

(b) Add nitric acid to the solution of silver diamminochloride
[Ag(NH 3 ) 2 ]Cl from experiment 4 until the solution changes the colour

of litmus paper. A white precipitate of silver chloride is formed.
Write the equation of the electrolytic dissociation of the complex
compound and explain the role of nitric acid.

(c) Divide the blue solution of cupric tetramminohydroxide left
over from experiment 4 into two parts. Add a sodium hydroxide
solution to one part and an ammonium sulphide solution to the other.
Explain why a precipitate is formed in one case only.

7. Complex Compounds in Redox Reactions, (a) Mix 2 ml of a
.hydrogen peroxide solution in a test tube with an equal volume of a
potassium hydroxide solution; add 2 ml of a K 3 [Fe(CN) 6 ] solution.
Confirm by a glowing splint that oxygen is evolved. Write the equa-
tion of the reaction that has taken place.

(b) Test the effect of a newly prepared FeS0 4 solution on potassium
ferrocyanide and ferricyanide. Which of these salts produces a blue
colouration? Add a few drops of potassium ferrocyanide to 3-4 ml of
chlorine water and boil the solution thoroughly (to remove the excess
chlorine). Add FeSO 4 to the resulting solution. What is observed?
Write the equation for the oxidation of potassium ferrocyanide by
chlorine.

Beryllium, Magnesium, and Alkaline Earth Metals 193

Exercise 23

BERYLLIUM, MAGNESIUM, AND THE ALKALINE EARTH

METALS

SUBJECTS FOR STUDY

Atomic structure of beryllium, magnesium, and the alkaline earth metals; posi-
tion in the electromotive series; attitude to water, acids, oxygen, and oxidants;
oxides and hydroxides, their preparation and chemical properties; principal soluble
and insoluble salts; magnesium and calcium salts in natural waters; the hardness of
water and the methods of its removal.

In their compounds Be, Mg, Ca, Sr, Ba, and Ra are bipositive.
This is because the outer electron shells of their atoms contain two
electrons each, which are easily given up under the action of oxidising
agents:

Me 2e~ = Me + *

The atomic radii increase from Be to Ra; reductant activity rises
in the same direction:

Be Mg Ca Sr Ba Ra

Atomic number ... 4 12 20 38 56 88

Radius in Angstrom
units 1.05 1.62 1.97 2.13 2.17

Barium and strontium are usually kept under a layer of kerosene
or, like radium, in sealed glass tubes, while the other metals of this
series, which are weaker reducing agents, may be kept simply in
closed jars.

Heated in oxygen or air, the metals burn with a bright flame. In the
electromotive series all of them rank above hydrogen and are, accord-
ingly, easily oxidised by water with (Be and Mg) or without (Ca-Ra)
heating:

Me + 2HOH = H 2 + Me(OH) 2
Oxidation proceeds even better in acid solutions:

^ I
e + 2HC1 = H 2 + MeCl 2

Beryllium, because it is amphoteric, also dissolves in alkali solu-
tions, forming hydroxyberyllates: Me 2 [Be(OH) 4 ].

1 3-795

194 Exercise 23

The oxides of the metals are for practical purposes prepared by the
thermal decomposition of the carbonates (Mg and Ca) or the nitrates
(Sr and Ba):

CaCO 3 ^CaO-f>CO 2

2Ba(NO 3 ) 2 = 2BaO + 4NO 2 + O 2

The hydroxides of the group may be prepared by treating the oxides
with water; the hydroxides of magnesium and beryllium may also be
prepared by exchange reactions of salts with alkalis, thanks to the
poor solubility of these hydroxides in water.

The oxides and hydroxides become increasingly alkaline with the
rise in the radius of the Me* 2 ion; the hydroxide corresponding to Be* 2 ,
the ion with the smallest radius, is amphoteric.

In the laboratory it is customary in most cases to use a solution of
Ca(OH) 2 , known as lime water, or a suspension of solid calcium hydro-
xide in a saturated solution, known as lime milk, and a solution of
Ba(OH) 2 , known as baryta water.

The volatile salts of calcium, strontium, and barium impart a brick
red, carmine, and yellowish green colour respectively to a nonlumino-
us flame of the Bunsen burner.

Among the insoluble salts, mention should be made of the carbo-
nates MeCO 3 and the ortho-phosphates Me 3 (PO 4 ) 2 . The solubility of the
sulphates MeSO 4 decreases from Be to Ra: the sulphates of Sr, Ba,
and Ra are practically insoluble, CaSO 4 is somewhat soluble, and the
sulphates of Be and Mg are highly soluble.

Barium chromate BaCrO 4 , unlike strontium chromate, which dis-
solves in acetic acid, dissolves in hydrochloric acid. The chromates
have a yellow colour.

Treatment of the carbonates with an aqueous solution of carbon
dioxide causes them to dissolve with the formation of bicarbonates:

CaCO 3 + H 2 O + CO 2 = Ca(HCO 3 ) 2

Natural water with large amounts of calcium and magnesium salts
in solution is known as hard water. Soft water contains relatively small
amounts of dissolved salts. According to the Standards in operation
in the U.S.S.R., the hardness of water is expressed in the number of
milligram-equivalents of Ca" or Mg" ions in 1 litre of water. Water
with a hardness of less than 4 mg-equiv./lit is called soft; water with
a hardness of from 4 to 8 mg-equiv./lit is called medium; from 8 to 12,
hard, and above 12, very hard.

One mg-equiv. of Ca" corresponds to 20.04 mg; 1 mg-equiv. of Mg",
to 12.16 mg.

The presence of Ca(HCO 3 ) 2 and Mg(HCO 3 ) 2 in natural water is re-
sponsible for what is known as temporary (or removable) hardness. The
chlorides and sulphides of calcium and magnesium are the cause of
what is termed the permanent hardness of water. Together, the tempo-

Beryllium, Magnesium, and Alkaline Earth Metals 195

rary and the permanent hardness of water add up to the total hardness.
This is easily calculated from the number of mg of Ca" and Mg" in
1 litre of the water:

total hardness 2Q -\- -^JQ~ mg-equiv./lit

Temporary hardness can be determined by titrating a definite volume
of water with a hydrochloric acid solution of known concentration
in the presence of the indicator methyl-orange. The reaction that takes
place is expressed by the equation:

Ca(HC0 8 ) a + 2HC1 - 2H 2 O + 2CO 2 + CaCl 2

In this reaction two mols of HC1 are used up to one mol of Ca(HCO 3 ) 2 .
Since a mol of Ca(HCO 3 ) 2 is equivalent to a gram-ion ofCa",we can,
knowing the amount of acid used up, calculate the number of mg-
equiv. of Ca** in 1 litre of water and in this way determine the tempora-
ry hardness.

Example. The titration of 100 ml of tap water requires the expenditure of 2.65
ml of an HC1 solution whose titre is equal to 0.0032. Determine the temporary hard
ness of the water.

First, it is necessary to determine the number of mg of HC1 used up for the
titration:

2.65-0.0032 = 0.00848 g = 8.48 mg

From the ratio Ca" : 2HC1, bearing in mind ionic and molecular weights, find the
number of mg of Ca" corresponding to the quantity of the acid used up:

40.08 2-36.46
x 8.48

Accordingly,

Hence, 100 ml of water contains 4.66 mg of Ca"; 1 litre, consequently, contains
46.6 mg. To establish the hardness of the water, this number should be divided by
20.04, i. e., by the equivalent of calcium. The hardness of the water will therefore
be equal to

/ 1 f*

2Lj~ = 2.3 mg-equiv./litre

The total hardness can be determined by precipitating the calcium
and magnesium salts by an alkaline mixture consisting of equal
volumes of 0.1 N NaOH and 0.1 N Na 2 CO 3 . Part of the mixture is used
up for the precipitation, e. g.:

CaSO 4 + Na 2 CO 3 = CaCO 3 + Na 2 SO 4
Ca(HC0 3 ) 2 + 2NaOH - CaCO 3 + Na 2 CO 3 + 2H 2 O

13*

196 Exercise 23

The unreacted part of the mixture is titrated (after the precipitate has
been filtered off) by 0.1 N HC1 with methyl-orange. The difference
between the titration of the initial mixture and the titration of the
mixture with water gives the number qf ml of 0.1 N NaOH used up
to precipitate the hardness-causing salts.

Examplf. Fifteen ml of alkaline mixture is added to 100 ml of tap water. After
the precipitate has been filtered off, the titration of the surplus mixture requires
9.90 ml of 0.1 N HC1. Determine the total hardness.

Equinormai solutions react in equal volumes. Accordingly, 9.90 ml of 0.1 N HC1
is equivalent to 9.90 ml of 0.1 N NaOH, which is the remainder of the alkali after
precipitation. The amount used up for the precipitation of the salts from 10 ml of
water is therefore:

15.00 9.90 =,5.10 ml of 0.1 N NaON

Precipitating all the salts from 1 litre of water would require 51 ml of 0.1 N NaOH.

This volume is equivalent to the same volume of an 0.1 N solution of the Ca"'
ion.

Now determine the amount of the Ca" ion contained in 51 ml of its 0.1 N soiu-
tion, knowing that the equivalent of Ca" is 20.04:

inm- 20 - 04

51 x
Therefore

* = = 0.1022 g = 102.2 mg

The number of mg-equiv. of Ca" in 1 litre of water (or its hardness) equals

102 ' 2 .^5.1

20.04

QUESTIONS

1. Why are the alkaline earth metals less potent reducing agents
than the alkali metals?

2. Draw a diagram of a galvanic cell consisting of the couples
Mg/Mg" and H 2 /2H"; determine its e. m. f. Will metallic magnesium
dissolve in M HC1?

3. How would you distinguish lime water from baryta water?

4. By means of what chemical reactions is it possible to effect the
following processes:

(a) BaSO 4 -> BaS -> BaCO 3 -+ BaCl 2

(b) Ca(N0 3 ) 2 -> CaO -> Ca 3 (PO 4 ) 2

Write the equations for these reactions. Which of the reactions that
you have proposed are redox reactions?

5. Write the equations for the electrolytic dissociation of beryllium
hydroxide and for dissolving it in acid and alkali solutions.

Beryllium, Magnesium, and Alkaline Earth Metals 197

6. A potassium chromate solution is added to solutions of CaCl 2
and BaCl 2 having the same concentration. Why does a precipitate form
only in the barium salt solution?

7. How is the temporary and the permanent hardness of water
removed?

Problems

1. How many grams of a unimolal solution of sodium hydroxide are needed to
convert to beryllate the beryllium chloride contained in 600 g of a 6% solution
of it?

2. What volume of 35.38% HC1 (relative density 1.18) will be needed to dissolve
1 kg of an alloy consisting of 24% of magnesium and 76% of aluminium?

3. What will be the yield of CaSO 4 from 1 cu m of its saturated solution if the
solution is heated from 40 to 100? The solubility of CaSO 4 is given in Table II at
the end of the book.

4. What is the hardness of water if 7 g of calcium hydroxide per 100 litres is re-
quired to soften it?

5. Fifty litres of water contain 4.5 g of magnesium bicarbonate. Express the
temporary hardness of the water in mg-equiv./lit.

6. It is required to reduce the temporary hardness of water from 14 to 6 mg-
equiv./lit. How much calcium hydroxide will be needed to soften 1,000 cu m of the
water?

7. One mol of anhydrous calcium chloride dissolves with the evolution of 17.41
Cal. The hexahydrate dissolves with a thermal effect of 4.31 Cal. What is the
thermal effect of the hydration of the salt?

8. The thermal effects of the formation of barium peroxide and aluminium oxide
are 144.2 and 378 Cal., respectively. Determine the thermal effect of the reaction of
the aluminothermic reduction of barium peroxide.

9. How many faradays of electricity should be passed through molten carnallite
(KCl-MgCl 2 ) to obtain 1 kg of metallic magnesium?

10. Draw a diagram of a galvanic cell consisting of the couples Mg/Mg" and
Cd/Cd", indicate the direction of the current in the inner and the outer circuit,
determine the e. m. f. and write the equation of the reaction by virtue of which
hte cell operates.

LABORATORY WORK

Apparatus and materials: Kjpp gas generator for preparing carbon dioxide; blow-
pipes; test tubes and rack; crucible tongs; file; four 200 ml Erlenmeyer flasks; 25 ml
burette; funnel for the burette;' 100 ml pipette; 20-25 ml pipette; 50 ml beaker; fun-
nel; sandpaper; filter paper; platinum wire; chalk or marble; magnesium shavings;
calcium shavings; 10% ammonia solution; 5% calcium chloride solution; 5%
strontium chloride solution; 5% barium chloride solution; 2 N and 0.1 N titrated
HC1; 2 N and concentrated H 2 SO 4 ; 2 N CH 3 COOH; 2 N NaOH;* N NH 4 Cl; saturated
solution of ammonium sulphate; N K 2 CrO 4 ; N MgCl 2 ; N Na 2 HPO 4 ; 0.5 N BeCl 2 ; N
Na 2 SO 4 ; saturated solution of strontium sulphate; alkaline mixture; methyl -orange
and phenolphthalein solutions; lime water.

Preparation of Alkaline Mixture. Mix equal volumes of 0.1 N NaOH and 0.1 N
Na 2 CO 3 .

1. Reductant Properties of Magnesium and Calcium, (a) Test the
action of small lumps of these metals on water (if necessary, heat the

198 Exercise 23

water) and on dilute hydrochloric acid. Record the changes observed
and write the equations of the reactions that take place.

(b) Carefully examine the surface of metallic magnesium and cal-
cium, then by means of a file or sandpaper remove the top layer of the
metal . How has the surface of the metal changed? Take the cleaned lumps
of the metals by means of tongs and heat each, in turn, in the flame of a
burner. Write up the experiment, giving the equations of the reactions.
Collect the product of the oxidation of magnesium for the next ex-
periment.

2. Oxides of the Metals, (a) Take a small lump of chalk or marble
by means of tongs and heat it in the flame of a blowpipe for 3-5 min.
After the lump has cooled, wet it with such an amount of water (adding
it by drops from a dropper or pipette) that all the water is absorbed by
the calcium oxide and none remains. What do you observe? Write the
equations of the reactions that take place. What is the name used in
industry for the reaction of calcium oxide with water?

Mix the powder obtained with a small quantity of water in a porce-
lain casserole and add a drop of an alcoholic solution of phenolphthalein.
Explain the appearance of the colouration.

(b) Treat with water (in a test tube) the white powder of magnesium
oxide obtained in experiment 1. Test the solution with phenolphtha-
lein. Note the shade of colour and heat the solution. Does the colour
become deeper? Explain the experiment.

3. Beryllium and Magnesium Hydroxides. By what methods can
the hydroxides of these metals be prepared? Perform the necessary
experiments and write the equations of the reactions that take place.
Prove experimentally that beryllium hydroxide is amphoteric.

4. Temporary Hardness of Water. Pour 100 ml of tap water into an
Erlenmeyer flask by means of a pipette and add 5-6 drops of a methyl-
orange solution; titrate from a burette with 0.1 N HC1 until a single
drop of the HC1 changes the colour of the water from yellow to orange.
Record the volume of the acid used up with an accuracy of up to
0.05 ml. Repeat the titration with another portion of water. The second
result should not differ from the first by more than + 0.05 ml of HC1.
Calculate the temporary hardness of the water.

5. Permanent and Total Hardness of Water. Pour 100 ml of tap
water by means of a pipette into each of two Erlenmeyer flasks, add
20 ml of alkaline mixture (also by means of a pipette) to each flask,
and boil each for 5 min. Then filter the solutions through folded
filters into two other Erlenmeyer flasks. After filtration wash the
precipitates twice with 10-15 ml of distilled water, adding the wash
water to the filtrate. Add 5-6 drops of methyl-orange solution to each
of the flasks and titrate the filtered solutions with 0.1 N HC1 until the
colour of the solution changes from yellow to orange.

From the titration results calculate the total hardness of tap water.
From the data of experiments 4 and 5 calculate the permanent hardness.

Beryllium, Magnesium, and Alkaline Earth Metals 199

6. Calcium Carbonate and Bicarbonate. Pour 3 ml of lime water and
7ml of distilled water into a beaker and pass a current of CO 2 from the
Kipp gas generator through it. The carbon dioxide should be passed
through the water at a rate making it possible to count the bubbles
of the gas. A precipitate is formed, which dissolves when an excess of
CO 2 has been passed through the water (write the equations). How can
calcium carbonate be precipitated by two different methods from the
calcium bicarbonate formed in the solution? Perform these experiments,
dividing the solution into two portions. Write the equations of the
reactions that take place.

7. Sulphates of the Metals, (a) Test the solutions of beryllium,
magnesium, calcium, strontium, and barium chlorides with a dilute
solution of sulphuric acid. In which cases are precipitates formed?

(b) To a calcium salt solution add an equal volume, of concentrated
sulphuric acid. Why does a precipitate form in this case?

(c) Pour 1 ml of SrCl 2 solution into one test tube, and 1 ml of CaCl 2
solution, into another. Add first a small amount and then an excess
of a saturated solution of ammonium sulphate to each of them. The
CaSO 4 precipitate dissolves because of the formation of the soluble
complex compound (NH 4 ) 2 [Ca(SO 4 ) 2 ]. Write the equations of the
reactions that take place.

(d) To a solution of barium salt add a saturated solution of stron-
tium sulphate. What happens? Is a precipitate formed if a saturated
solution of barium sulphate is added to a strontium salt solution?

8. Chromates of the Metals. Pour 1 ml of a calcium salt solution into
one test tube; 1 ml of a strontium salt solution, into a second, and 1 ml
of a barium salt solution, into a third test tube. Add 1 ml of a po-
tassium chromate solution to each of them. In which cases do precipi-
tates form and what are their colours? Add 3-4 ml of an acetic acid
solution to each of the precipitates. Do all the precipitates dissolve?
Can strontium ions be precipitated by the chromate ion in the presence
of acetic acid?

9. Precipitation of Magnesium Ammonium Phosphate. To 1 ml of
a magnesium salt solution add an equal volume of NH 4 OH and
3-4 ml of an NH 4 C1 solution (the magnesium hydroxide precipitated
at first should dissolve).

Add 2-3 ml of Na 2 HPO 4 solution to the solution obtained; large
crystals of MgNH 4 PO 4 are precipitated. Why do hydroxyl ions have
to be introduced into this reaction?

10. a Dry Reactions". By means of a platinum wire introduce vola-
tile salts of beryllium, magnesium, calcium, strontium, and barium
into the flame of a burner (as described in experiment 3, Exercise 20),
observing the changes in the colour of the flame. Which are the metals
whose salts change the colour of the flame? What colours do they impart
to the flame?

200 Exercise 24

11. Cation Detection. Receive from the instructor a test solution
that may contain one of the following catiotis: Ba", Sr", Ca", Mg'\
or Be".

Take part of the solution and add to it half as much (by volume) of
a sulphuric acid solution. The formation oi a white precipitate points
to the presence of the Ba" or Sr" ion in the solution; if no white pre-
cipitate is formed, the solution may contain one of the three other
cations. By means of the reactions studied in this Exercise
(3, 7, 8, 9, and 10) establish the presence of a definite cation. Write up
the analysis briefly and submit your report to the instructor.

Exercise 24

ZINC, CADMIUM, AND MERCURY

SUBJECTS FOR STUDY

The metals of the zinc subgroup; their atomic structure, reductant properties,
and position in the electromotive series; features distinguishing mercury from zin;
and cadmium; attitude of zinc, cadmium, and mercury to various oxidising agentsc
oxides and hydroxides of these metals; their simple and complex salts; amalgams.

The neutral atoms of zinc, cadmium, and mercury have two elec-
trons each in their outermost shells; in chemical reactions they give
them up

Me 2e~ = Me* 2

forming the bipositive ions Zn^ 2 , Cd* 2 , and

Apart from the compounds of bivalent mercury, called mercuric
compounds, there is a series of compounds containing the Hg
Hg group and called mercurous.

The similarity between the elements of the zinc subgroup and the
alkaline earth metals is due to their having the same number of elec-
trons in the outermost shell. But in the next to last shell the atoms of
the zinc subgroup elements have 18 electrons, which gives rise to
certain distinct properties: a weaker reductant activity, a more pro-
nounced tendency to form complex compounds, etc.

Zinc and cadmium, at ordinary conditions, are oxidised by oxygen
and the air only at the surface (their tarnishing is caused by the for-
mation of an oxide film), but, when heated, they burn up, e. g.:

2Zn + O 2 = 2ZnO

Mercury, at ordinary conditions, is not oxidised either by oxygen
or by the air.

Zinc, Cadmium, and Mercury 201

The reductant activities of the elements in the row zinc-cadmium-
mercury are reflected in the position of these metals in the electro-
" motive series:

Zn/Zn" Cd/Cd" H 2 /2H* Hg/Hg" . . .

-0.76V 0.4V +O.OOV + 0.85 V

Zinc and cadmium dissolve in dilute hydrochloric acid and sulphur-
ic acid with the evolution of hydrogen, but scarcely react at all
with water, since when they are immersed in water the reaction com-
mences on the surface according to the equation

Zn + 2HOH - H 2 + Zn(OH) a

but is at once halted by the formation of a hydroxide film. Unlike
cadmium and mercury, metallic zinc also dissolves in alkali solutions
(it is amphoteric).

Mercury can be oxidised by concentrated nitric and sulphuric acids.
It dissolves many other metals, forming alloys that are known as
amalgams; with some metals mercury forms chemical (intermetallic)
compounds.

The following oxides are known: ZnO, CdO, HgO, and Hg 2 O. The
oxides of mercury are prepared by exchange reactions of its salts with
alkalis. The corresponding hydroxides Hg(OH) 2 and Hg 2 (OH) 2 are
unstable; no sooner are they formed than they break up, e. g.:

Hg(OH) 2 = HgO + H 2

The red modification of mercuric oxide is prepared by heating the
nitrates of mercury. ZnO and CdO can be prepared either by oxidising
the metals or, best of all, by the thermal decomposition of the hydro-
xides, carbonates, nitrates, etc. The oxides of the metals of the zinc
subgroup do not react with water.

The hydroxides Zn(OH) 2 and Cd(OH) 2 are prepared by the inter-
action of salts and alkalis; the former is an amphoteric hydroxide, the
latter, a base.

Some of the mercurous salts are unstable compounds and decompose
the moment they are formed, yielding mercury and mercuric com-
pounds, e. g.:

Hg 2 S = HgS+Hg

The salts of mercury that have practical applications are: mercurous
chloride Hg 2 Cl 2 , known as calomel, which is insoluble in water;
mercurous nitrate Hg 2 (N0 3 ) 2 ; mercuric nitrate Hg(NO 3 ) 2 , and mercuric
chloride HgCl 2 , known as corrosive sublimate. All the soluble com-
pounds of mercury are very poisonous.

202 Exercise 24

The nitrates and sulphates of zinc, cadmium, and mercury are soluble
in water and undergo hydrolysis. Among the insoluble salts, mention
should be made of the sulphides. These have characteristic colours:
ZnS is white, CdS is yellow or orange, and HgS is black or red.

Some salts, such as mercuric iodide Hg'I 2 , are polymorphic, i. e.,
they can exist in several forms, each of which has a distinct colour
of its own. For instance, the ot-form of mercuric iodide is bright red
and stable to a temperature of 127, while the p-form is yellow and
unstable at ordinary conditions. At 127 one modification is converted
to the other. This temperature is called the transition point.

Mercuric salts possess oxidant properties, the reduction of the Hg"
proceeding either to mercurous salts or to free mercury, e. g.:

SnCl a + 2HgCl 2 + 2HC1 - H 2 [SnCl 6 l + Hg a Cl a *

S + nCl 2 + Hg 2 Q 2 + 2HC1 = H 2 [SnCl 6 l + 2Hg

In complex compounds the ions of this series usually have the
following coordination numbers: Zn 42 , 4 and 6; Cd 42 , 4 and 6; Hg + ,
2 and 4.

Typical of the Hg 42 ion is the formation of what are called ammono
compounds of mercury (such as HgNH 2 Cl or Hg 2 NH 2 ONO 3 ), which
should be regarded as products of the substitution of Hg" for hydrogen
ions in the ammonia group.

Among the complex compounds with halogens, mention should be
made of the soluble colourless compound K 2 [HgI 4 ]; its alkaline so-
lution (Nessler's solution) is a sensitive reagent for free ammonia, with
which it forms the reddish brown iodide precipitate [OHg 2 -NH 2 ]I.

QUESTIONS

1. What are the points of similarity and difference between the
elements of the zinc subgroup and the alkaline earth metals?

2. Write the equations for the treatment of metallic mercury (upon
heating) with concentrated nitric and sulphuric acids.

3. Why does metallic zinc dissolve in an ammonium chloride -so-
lution? Write the relevant equations.

4. How can potassium hydroxyzincate be prepared from zinc sul-
phate?

* To prevent hydrolysis, hydrochloric acid is added to the stannous chloride so-
lution; the resulting stannic chloride combines with the acid to form a complex
compound, chlorostannic acid.

Zinc, Cadmium, and Mercury 203

5. Write the equations of the reactions in the consecutive prepara-
tion of the following substances:

Cd(N0 3 ) 2 -> CdO ~> Cd - CdCl 2 - [Cd(NH 3 ) 6 ](OH) 2

6. How can metallic zinc be prepared from zinc blende and cadmium
sulphate from cadmium nitrate? Write the equations of the reactions
that take place.

7. By means of what reactions is it possible to distinguish a solution
containing the Ba" ion from a solution containing the Cd" ion?

8. Why does the Ca" ion have a less pronounced complex-forming
tendency than the Zn" ion? Why is the coordination number of Ca"
greater than of Zn"?

Problems

1. A compound of mercury and oxygen consists of 96.23% of Hg and 3.77% of
O. When 43.32 g of another compound of the same two elements is heated, the yield
of oxygen is 2,345 ml, measured at 1.05 atm and 27. Are these figures in keeping
with the Law of Multiple Proportions?

2. The heating of 10 g of metallic mercury from 77.2 to 41.8 requires 11.65
cal. Determine the atomic weight of mercury from these data and the discrepancy
between the found value and the exact value equal to 200.61.

3. How many atoms make up a molecule of cadmium (atomic weight 112.4) va-
pour, if 1.686 g of the vapour, obtained at boiling point and reduced to N. T. P.,
occupies a volume of 336 ml?

4. At boiling point 6.02 g of a metal produces a vapour whose volume, after re-
duction to N. T. P., equals 672 ml. Determine the number of atoms in the molecule
of the metal, knowing that its specific heat is 0.0319.

5. A zinc ore contains 30% of ZnS. What is the volume of sulphur dioxide that
can be obtained by burning a ton of the ore?

6. Determine the titre, molarity, and normality of a 25% solution of CdSO 4 ,
which has a relative density of 1.294.

7. What are the volumes of a 45% solution of ZnCl 2 (relative density 1.489) and
of water that have to be mixed to prepare 1 cu m of an 8% solution with a relative
density of 1.084?

8. The solubility of corrosive sublimate at 100 is 35.1 g, whereas at it is 4.1
g. How much corrosive sublimate and water should be taken to obtain 1 kg of the
salt by recrystallisation in that temperature range?

9. From the value of the solubility product determine the number of grams of
CdS contained in 500 ml of a saturated solution.

10. How many 200 ml jars can be- filled with the metallic mercury (density 13.55
g/cu cm) prepared by reducing 314.4 kg of pure cinnabar?

LABORATORY WORK

Apparatus and materials: conical test tube with stopper and thermometer; re-
fractory test tube; test tubes and rack; crucible tongs; tray or pans; porcelain crucible
with lid; pipestem triangle; 200 ml and 1 litre beakers; asbestos cone; glass stir-
rer; glass rod; indigo or cobalt prism; splints; filter paper; platinum wire; mercury
(in a special dropper); zinc; sodium amalgam; zinc oxide; cadmium oxide; mercuric
oxide; mercuric iodide; concentrated nitric acid; concentrated and 2 N H 2 SO 4 ;
0.05 N KMnO 4 ; 2 N HC1; 30% and 2 N NaOH; 10% ammonia solution; 20% KOH;
saturated solution of strontium sulphate; 2 N Na 2 CO 3 ; 0.5 N ZnCl 2 ; 0.5 N[CdCl 2 ;

204 Exercise 24

0.2 N Hg(NO 3 ) 2 ; N HgNO 8 ; N solution of sodium dihydrogen phosphate; N SnCl;
2N (NH 4 ) 2 S; 2 N NH 4 C1; 2 N NH 4 NO 8 ; saturated solution of ammonium sulphate;
0.5 N KI; 0.4 N solution of sodium bitartrate; saturated solution of potassium an-
timonate; Nessler's solution, and vaseline oil.

Preparation of Nessler's Solution. Dissolve 32 g of HgI 2 and 20 g of KI in 50 ml
of water and dilute the solution with distilled water to 200 ml. Mix the solution
with 300 ml of 20% KOH.

Special bottles with ground-in stoppers should be on the laboratory tables for
left-over solutions containing metallic mercury that has not reacted.

1. Granulating of Zinc. Place several lumps of zinc in a porcelain
crucible, put it on a pipestem triangle, and heat it in a burner flame
until the metal melts. Fill a 1 litre beaker with water almost to the
brim, insert an asbestos cone with a 2 mm orifice in a ringstand ring
over the beaker, and pour the molten metal from the crucible (picking
it up with tongs) into the cone. Store away the zinc granules for
future work.

2. Reductant Properties of Metals, (a) By means of three separate
tests, study the action of water, hydrochloric acid, and a concentrated
alkali solution on metallic zinc. Write the equations of the reactions
that have taken place. How will cadmium react to the same solvents?

(b) From a special dropper pour* 1-2 drops of metallic mercury into
each of three test tubes (mercury should be poured over a tray without
spilling any of it on the table or the floor!). Test the action on it of
water, hydrochloric acid, and concentrated nitric acid (with and with-
out heating). Write the equations of the reactions that have taken
place. Precaution! Pour the remains of the metallic mercury into the
special bottles! Conduct the experiment in a ventilated hood!

3. Reductant Properties of Amalgam. Pour 4 ml of a dilute solution
of sulphuric acid and 2 drops of a potassium permanganate solution
into a test tube, shake the contents, and pour the solution into two
test tubes. Immerse a lump of sodium amalgam into one of the test
tubes and, whe nthe reaction has ended, compare the colour of the solution
with that of the solution in the other test tube. Write the equation of
the reaction that has taken place.

4. Preparation of Hydroxides. Add a few drops of an alkali solution
to solutions of the salts ZnCl 2 , CdCl 2 , Hg(NO ? ) 2 , and Hg 2 (NO 3 ) 2 .
What is formed? Write the equations of the reactions that have taken
place. Add excess alkali to all the precipitates. What happens to the
zinc hydroxide precipitate? Write the equation of the reaction that
has taken place. In what way do the hydroxides of zinc, cadmium,
and mercury differ?

5. Calcination of Oxides, (a) Place a pinch of zinc oxide on the lid
of a porcelain crucible; handling the lid with tongs, heat it in the
flame of a burner until it is red-hot. Does the colour of the zinc oxide
change when it is heated and after cooling?

* This experiment should be conducted only in those laboratories that are
equipped for work with mercury!

Zinc, Cadmium, and Mercury 205

(b) Carry out a similar experiment with cadmium oxide. Do these
oxides undergo chemical decomposition upon heating?

(c) Carry out a similar experiment with mercuric oxide in a refracto-
ry test tube held in the clamp of a ringstand. Establish the presence
of oxygen in the test tube and of drops of mercury on its walls. Write
the equation of the reaction that has taken place.

Why is it that mercuric oxide, unlike zinc oxide or cadmium oxide,
decomposes upon heating?

6. Preparation and Properties of Sulphides* By treating solutions
of the salts of the zinc subgroup metals with an ammonium sulphide
solution prepare sulphide precipitates. Note their colours and the ac-
tion of hydrochloric acid upon them. By referring to the SP values,
explain why the sulphides of these metals react differently with hydro-
chloric acid.

7. Determining the Transition Point of Mercuric Iodide. Put 2-3 g
of mercuric iodide into a conical test tube and close it with a stopper
through which a thermometer has been passed. The bulb of the thermo-
meter should be submerged in the powdered salt. Fasten ihe test
tube in a ringstand clamp and lower it into a beaker filled with vaseline
oil and fitted with a stirrer (an oil bath). Heat the beaker over the
flame of a burner, stirring the oil in the beaker gently. When the tem-
perature reaches 105-110, reduce the flame so that the temperature
does not rise by more than 1 per minute (continue the stirring). Note
the temperature at which the colour of the salt changes from red to
yellow.

Remove the test tube from the oil and allow it to cool. Does the
colour of the salt revert to red? Open the test tube and rub the yellow
powder with a glass rod against the wall of the test tube. What hap-
pens? Give a definition of the transition point. Why is the recon-
version of the yellow modification to the red delayed?

8. Oxidant Properties of Mercury Salts, (a) Place a drop of a mercu-
ry salt solution on a copper or bronze coin. Two or three minutes
later wash the solution off and rub the grey stain on the coin with a
piece of filter paper. Write the equation of the reaction that has taken
place.

(b) Test the action of stannous chloride on mercuric salts (for both
an excess and a shortage of SnCl 2 ). Write the relevant equations.

9. Zinc, Cadmium, and Mercury Complexes, (a) Test the action of
an ammonia solution (shortage and excess) on solutions of salts of
these metals. Write the equations of the reactions that take place.

(b) Add a KI solution by drops to 1 ml of 0.2 N Hg(NO 3 ) 2 until
the red precipitate formed is fully dissolved. Write the equation of the
reaction that has taken place. Add to the resulting solution an equal
volume of 20% KOH. The solution prepared in this way, which is
very close to what is known as Nessler's solution, can be used for the
qualitative detection of ammonia and ammonium salts. Carry out

206 Exercise 24

such a reaction and write its equation. What is the reddish brown
precipitate formed?

10. Analysis of a Solution. A test solution should contain only one
of the following cations:

Na* K* Mg" Ca" Sr" Ba" Zn" Cd" Cu" Ag* Hg"

Detect this cation by qualitative reactions.

One cation can be detected in a solution by the combined use of
group precipitants and identifying reagents. The group precipitants
separate several cations from the solution simultaneously. The given
mixture of cations can be tentatively divided into three groups:

(1) Zn", Cd", Cu", Ag", and Hg", which are precipitated by a solution of
(NH 4 ) 2 S;

(2) Mg", Ca", Sr", and Ba", which are precipitated by a solution of Na2CO 3 ^
and

(3) Na* and K", which have no group precipitant.

Receive your test solution from the instructor.

(a) Add 3 ml of an ammonium nitrate solution and 1-2 ml of a so-
lution of (NH 4 ) 2 S to 1 ml of the test solution. If a precipitate is formed,,
the solution contains one of the cations of the first group (note the
colour of the precipitate). In this case by testing the solution separate-
ly by specific reactions establish the presence of a definite cation:

Ag" is detected as in experiment 8, Exercise 21;

Cu" is detected as in experiment lOa, Exercise 21;

Zn" and Cd" are detected as in experiments 4 and 6 of the present Exercise,.

and
Hg" is detected as in experiment 8 of the present Exercise.

If no precipitate is formed with (NH 4 ) 2 S, the solution is analysed
by the procedure described in "b".

(b) Add 1-2 ml of a Na 2 CO 3 solution to the rest of the test solution.
If a precipitate is formed, the solution contains one of the cations of
the second group. In this case by testing the solution separately by
specific reactions establish the presence of a definite cation:

Ca", Sr", and Ba" are detected as in experiments 7, 8, and 10, Exercise 23,

and
Mg" is detected as in experiment 9, Exercise 23.

If the addition of soda produces no precipitate, the solution contains
no cations of the second group and can contain only the cation Na'
or K' (see below).

(c) Since the cations K and Na' have no group precipitant, they are
detected separately by specific reactions (experiments 3 and 5,
Exercise 20).

Write up the analysis carried out, giving the equations of the re-
actions by which the cation in the solution was detected. Submit
your report to the instructor.

Elements of the Third Group 207

Exercise 25

THE ELEMENTS OF THE THIRD GROUP OF THE
PERIODIC SYSTEM

SUBJECTS FOR STUDY

The elements boron, aluminium, gallium, indium, and thallium; the structure
of their atoms and outermost electron shells; features distinguishing boron from the
rest of the group.

Boron; boric anhydride; boric acids and their salts; borax; reductant activity in
the A1-T1 series; attitude of aluminium to oxygen, water, acids, and alkalis; alu-
minium oxide and hydroxide; comparison of hydroxides; aluminium salts and the
alums.

The atoms of the boron group elements have three electrons each in
their outermost shells. All of them, with the exception of boron,
exhibit the chemical properties of metals and, upon oxidation, give
up three electrons: Me 3e~ = Me* 3 . The valence +1 occurs in
the compounds of aluminium and its analogues, but is most common
in the case of thallium.

Amorphous boron is best dissolved by concentrated nitric acid,
which is reduced to nitrogen dioxide:

B + 3HN0 3 - H* 8 BO 8 + 3NO 2

The electrode potential of the A1/A1'" couple is 1.67 V. Alumini-
um is oxidised by H" ions, but the reaction comes to a standstill quick-
ly because of the formation of a protective film of hydroxide, which is
insoluble in water:

2A1 + 6HOH = 3H 2 + 2A1 (OH) 3

In hydrochloric acid and dilute sulphuric acid, aluminium dissolves
with the evolution of hydrogen.

Because of its amphoteric properties, aluminium dissolves in alka-
lis. Concentrated nitric acid without heating renders the surface of
metallic aluminium passive, owing to the formation of a film of
oxide *.

* The termination of the reaction between a metal and a medium (acid, alkali,
etc.), owing to the formation of a surface oxide film, constitutes the essence of the
"film" theory of metal corrosion formulated by the late Academician V. A. Kistya-
kovsky.

208 Exercise 25

In oxygen, boron and aluminium burn up to form B 2 O 3 and A1 2 O 3
respectively.

Elementary boron is a nonmetal; it possesses oxidant, as well as
reductant, properties. Upon heating, it interacts vigorously with some
metals to form borides, e. g.:

ee-

I -* I +2-3

3Mg + 2B - Mg 3 B 2

The chemical character of the hydroxides corresponding to the
tripositive ions changes with their radius:

Ion radius 0.20 0.50 0.62 0.81 0.95

(in Angstrom units)
Hydroxide H 3 BO 3 Al (OH) 3 Ga (OH) 3 In (OH) 3 Tl (OH) 3

Its character acidic amphoteric basic

With a decrease in radius, the acidic properties of the hydroxide
become more pronounced, while with an increase in radius, basic
properties come to the fore.

Boric anhydride forms several acids: orthoboric acid H 3 BO 3 , meta-
boric acid HBO 2 , and the polyboric acids that have the general formu-
la (B 2 O3)jc-(H a O)0. The simplest of the polyboric acids (x = 2 and
y = 1) is tetraboric acid H 2 B 4 O 7 . All the boric acids are weak electro-
lytes. The stablest in aqueous solution is H 3 BO 3 ; the others combine
with water to form orthoboric acid:

HB0 2 + H 2 - H 3 B0 3
H 2 B 4 O 7 + 5H 2 O = 4H 3 BO 3

The stablest boric acid salt is sodium tetraborate, which is common-
ly known as borax. It is prepared by crystallisation from an aqueous
solution at a temperature not higher than 50-60 in the form of the hy-
drate Na 2 B 4 O 7 -10H 2 O.

Borax interacts with strong acids to form orthoboric, or simply
boric, acid:

Na 2 B 4 O 7 + 2HC1 + 5H 2 O - 4H 3 BO 3 + 2NaCl

When borax is heated, the crystals swell and give up the water of
crystallisation:

Na 2 B 4 O 7 - 10H a O 10H 2 O + Na 2 B 4 O 7

Molten borax dissolves metal oxides to form metaborates, salts
that rank among the most stable thermally, e. g.:

CoO + Na 2 B 4 O ? = 2NaBO 2 -f Co(BO 2 ) a

Elements of the Third Group 209

Boric anhydride, upon being fused with metal oxides, likewise
dissolves them readily to form metaborates, e. g.:

CoO + B. 2 O 3 - Co(BO 2 ) 2

Many of the metaborates have distinctive colours, a property that
is used in preparing coloured glasses or the borax beads employed in
analytical chemistry.

As the salt of a weak acid, borax undergoes hydrolysis:

Na 2 B 4 O 7 + 2HOH + 5H 2 O ^ 4H 3 BO 3 + 2NaOH

Borax can therefore be used as an alkali for titrating acids.

All the volatile compounds of boron impart a bright green colour
to the flame of a burner. A drop of boric acid solution produces a red
stain on a yellow curcuma test paper; moistened with an alkali solu-
tion, the stain becomes greenish black.

Aluminium hydroxide is amphoteric.

Two aluminium acids are known: orthoaluminic acid H 3 A1O 3 and
metaaluminic acid HA1O 2 . The process that takes place when A1(OH) 3
dissolves in alkalis may be represented by the equations:

H 3 A1O 3 + 3NaOH - 3H 2 O + Na 3 AlO 3
H 3 Ai0 3 + NaOH - 2H 2 O + NaAlO a

Aluminium hydroxide possesses highly pronounced adsorption
properties and is used in the purification of water and as a mordant in
dyeing.

Among the soluble salts of aluminium uses have been found for the
sulphate A1 2 (SO 4 ) 3 , the chloride A1C1 3 , the nitrate A1(NO 3 ) 3 , and potas-
sium, or potash, alum KA1(SO 4 ) 2 - 12H 2 O. All of these salts experience
hydrolysis in aqueous solutions and give acid reactions. Some alumi-
nium salts, such as A1 2 S 3 or A1 2 (CO 3 ) 3 , are completely hydrolysed in
aqueous solutions:

A1 2 S 3 + 6HOH - 3H 2 S -|- 2A1(OH),

These salts cannot therefore be prepared by exchange reactions in solu-
tions.

QUESTIONS

1. In what ways does boron differ in chemical properties from the
other elements of its group?

2. Why cannot the reaction of reduction by charcoal be used to
prepare boron from B 2 O 3 ? When B 2 O 3 is reduced by magnesium,
why is it inadvisable to have the magnesium in excess?

3. Write the equation for the reaction between Al and an alkali
solution.

14-795

210 Exercise 25

4. How can orthoboric acid be converted to metaboric and to tetra-
boric acid, and vice versa? Write the structural formulae of those three
acids.

5. Write the equation for the neutralisation of H 3 BO 3 by a sodium
hydroxide solution.

6. The boiling of an aqueous solution of ammonium aiuminate
precipitates aluminium hydroxide, whereas the boiling of a sodium
aiuminate solution causes no precipitate to be formed. Explain why
this is so.

7. Which base is the stronger: T1(OH) 3 or T1OH? Why?

Problems

1. The heating of 8 g of aluminium from 273 to 283 requires 19.13cal. Determine
the atomic weight of aluminium and the discrepancy (in per cent) between this
value and the exact value (26.98).

2. What amount of boric acid can be prepared from amorphous boron upon its
oxidation by 65.3% nitric acid if the amount of acid used is 1 litre (relative density
1.4)?

3. What is the titre and normality of a borax solution if 20 ml of that solution
is neutralised by 16 ml of 0.1 N HC1?

4. What volume of M NaOH is required to neutralise 200 g of 3% H 3 BO 3 ?

5. The heat of decomposition of MnO 2 is 123 Cal., while the heat of formation
of A1 2 O 3 is 399 Cal. Determine the thermal effect of the reaction of the alumi-
nothermic reduction of manganese dioxide.

6. Determine the [H'] f pH, and a% of 0.001 M H 3 BO 3 , taking into considera-
tion only the first step in its dissociation.

7. For how long should a 20,000 A current be passed through an electrolyte to
deposit 10 kg of metallic aluminium on the cathode?

8. A galvanic cell consisting of the coupies A1/A1"* and Cu/Cu" is operated for 1
hour, with the weight of the cathode diminishing by 1.12 g. Determine the intensity
of the current in the circuit.

9. Draw the diagram of galvanic cell consisting of the couples A1/A1'" and Bi/
Bi'" ? indicate the direction of the current, aim determine the e. m. f. (the salt solu-
tions are of molar concentration).

10. What amounts of aluminium tritluonue and 40% HF are needed to prepare
39.6 g of hexafluoroaluminic acid H 3 [AlF 6 j.3H 2 O?

LABORATORY WORK

Apparatus and ^materials: hot-water funnel; suction filter; 50 ml beaker; 100 ml
beaker; 50 ml measuring cylinder; funnel; poicelain casserole; porcelain mortar;
test tubes and rack; glass bath; watch glass; two glass rods; platinum wire; scissors;
aluminium plate; sandpaper; litmus paper; curcuma paper; lead paper; filter paper;
pieces of white fabric; asbestos millboard; aluminium turnings; aluminium dust;
magnesium ribbon; powdered sulphur; anhydrous borax; concentrated nitric acid;
concentrated sulphuric acid; concentrated hydrochloric acid; 30% and 2 N NaOH;
10% ammonia solution; 0.2 N Hg(NO 8 ) 2 ;0.5 N A1 2 (SO 4 ) 3 ; 10%. solution of potas-
sium alum; crystalline copper nitrate or sulphate; saturated solution of ammonium
chloride; 0.05% solution of methyl-violet; ethyl alcohol; alkaline solution of aliza-
rin; snow or ice.

Elements of the Third Group 2H

1. Preparation of Boric Acid. Weigh about 4 g of anhydrous borax
to 0.01 g. Transfer the weighed amount of borax to a beaker, add
20 ml of distilled water to it, and heat it over a small flame until
all of the borax dissolves (do not allow the solution to boil). If the so-
lution is turbid after the borax has been dissolved, it should be filtered
through an ordinary filter, using a hot-water funnel. Test a clear
borax solution with litmus paper; what change does the litmus paper
undergo and why? Write the equation for the reaction of the hydroly-
sis of borax.

Heat the solution to 80-90 and add to it a hydrochloric acid solution
prepared by mixing 6 ml of concentrated HC1 with an equal volume of
water. Make sure that the amount of acid taken is sufficient (how?).

Cool the beaker with the solution to room temperature and then
place it into a casserole with ice or snow. Filter off thecrystals formed,
using a Buchner filter, and dry them between filter paper until they
are as dry as possible.

By means of a glass rod spread the crystals on a piece of filter paper
in a thin layer and dry them in the air for 20-30 min., stirring them
from time to time. Weigh the dried crystals and calculate the percent-
age yield of the product. Submit the crystals to the instructor. Keep
the filtrate for experiment 2.

2. Reaction for Detecting Boric Acid, (a) By means of a glass rod
place a drop of the boric acid solution prepared in the previous experi-
ment and a drop of an HC1 solution on a piece of curcuma paper and
dry it over the flame of a burner. What is the colour of the stain?
Moisten the stain with a drop of sodium hydroxide solution. How
does the colour change?

(b) Evaporate 3-4 ml of the boric acid solution in a porcelain
casserole, cool the casserole, moisten the crystals with 3-4 drops of
concentrated sulphuric acid, add 2-3 ml of methyl or ethyl alcohol,
and stir the mixture well. Put the casserole in a ventilating hood,
ignite the alcoholic solution, and observe the colour of the flame.
Write the equations for the reactions of the formation of methyl-borate
(ethyl-borate) and its combustion.

3. Preparation of Borax Beads. Heat the end of a platinum wire in
an oxidising flame until it is red-hot and touch some borax crystals
with it. Carry the clinging crystals into the flame. The borax first swells
and then fuses to become a clear glass; this is known as a colourless
bead. The bead should best be 1 .5 mm in diameter. Wet the bead on the
wire loop with water; then touch some copper nitrate or sulphate crys-
tals with the moist bead and heat it in an oxidising flame until it
becomes transparent. Note the colour of the bead when it cools. Then
place the bead into the central zone of a reducing flame and heat it for
5-7 min. Cool the bead first in the lower part of the flame, near the end
of the nozzle, and then in the air; determine its colour. Salts of the
other metals listed in Table 16 may be used instead of the copper salt.

212

Exercise 25

Table 16

Colour of Borax Beads

Metal

Oxidising flame

Reducing flame

Vanadium

Greyish yellow

Pale green

Tungsten

Colourless

Yellow

Iron

Yellow

Pale green

Cobalt

Blue

Manganese

Pale violet

Colourless

Copper

Blue

Reddish brown

Molybdenum

Light yellow

Dark brown

Nickel

Reddish brown

Grey

Titanium

Colourless

Yellow

Uranium

Yellow

Green

Chromium

Green

4. Attitude of Aluminium to Air and to Water, (a) Clean an 80 mm
length of aluminium wire with sandpaper, bend it to form an angle
of 30, and immerse one end in a beaker with water. Is there any evolu-
tion of hydrogen from the water?

(b) Immerse the aluminium wire from the previous experiment in
a mercuric nitrate solution for 2 min, wipe it with a piece of filter
paper, and immerse one end of the wire in a beaker with water. Observe
the evolution of hydrogen from the water and the oxidation of the
aluminium surface exposed to the air. Write the equation of the reac-
tion that takes place.

5. Passivation of Aluminium. Immerse an aluminium shaving for
3-4 min in a test tube containing 2-3 ml of concentrated nitric acid.
Are any changes noticeable? Pour the solution off and, carefully,
without shaking (1), wash the metal with water.

6. Attitude of Aluminium to Acids and to Alkalis.

(a) Pour 2-3 ml of concentrated nitric acid into each of two test tubes.
Place an ordinary aluminium shaving into one of them, and a shaving
with a passivated surface (from the previous experiment) into the
other.

What happens? Give an explanation.

(b) Test the action of concentrated nitric acid upon aluminium
with boiling (ventilated hood!). Note the colour of the gas evolved.
Write the equation of the reaction that takes place.

(c) Test the action of a 30% solution of alkali upon aluminium with
slight (!) heating. Demonstrate that hydrogen is evolved (be careful!)
in the reaction. Write the equation of the reaction.

Carbon, Silicon, and Their Compounds 213

7. Amphoteric Character of Aluminium Hydroxide. Prepare alumi-
nium hydroxide and test its effect on an acid and on an excess of al-
kali. Write the equations of all the reactions that have taken place.

8. Hydrolysis of Aluminium Salts, (a) To 1-2 ml of an alkaline
solution of an aluminate add an equal volume of a saturated solution
of ammonium chloride and a triple volume of water. Boil the solution,
observing the formation of a precipitate. Write the equations of the
reactions that have taken place. Why does the precipitation of A1(OH) 3
require the addition of NH 4 C1 and water?

(b) Mix 0.54 g of aluminium dust and 0.96 g of sulphur in a mortar
and then transfer the mixture to a sheet of asbestos to form a small
heap. Insert a small strip of magnesium ribbon into the mixture.
In a ventilated hood ignite the magnesium by the flame of a burner.
After the reaction has taken place, cool the product, transfer it to a
test tube, add 3-4 ml of water to it, and cover it with a moist lead
paper. What colour does the lead paper acquire? Write the equations
of the reactions of the formation of aluminium sulphide, of its hydrol-
ysis by water, and of the interaction of hydrogen sulphide with a
lead salt.

9. Adsorption of Dyes by Aluminium Hydroxide.

(a) Prepare a precipitate of aluminium hydroxide and filter it off.
Wash the precipitate on the filter once with water. Pour a faintly
coloured solution of methyl-violet on the filter and observe that the
filtrate is a colourless solution.

(b) Moisten a strip of white cotton fabric on a watch glass with a
10% solution of potassium alum. Pour off the excess alum solution
and pour a few drops of a 10% ammonia solution on the fabric.

What is deposited in the fibres of the fabric? Five minutes later
wash the treated fabric with water and place it in a beaker with an
alkaline solution of alizarin together with a strip of fabric not treated
with an alum solution. After 5-7 min wash both strips thoroughly
in water and compare their colours.

Exercise 26

CARBON, SILICON, AND THEIR COMPOUNDS

SUBJECTS FOR STUDY

The elements of the fourth group of the Periodic System; atomic structure; car-
bon and silicon; attitude of these elements to oxygen and to acids; hydrogen com-
pounds of carbon and silicon; carbides and silicides.

Carbon monoxide and silicon oxide; carbon dioxide and silicon dioxide; car-
bonic acid and silicic acids; carbonates, bicarbonates, and silicates; hydrolysis of
salts; glass.

214 Exercise 26

The atoms of the elements of the carbon lead series have 4 elec-
trons each in their outermost shells. In chemical reactions they exhibit
reductant (in reactions with oxidising agents) and oxidant (in reactions
with reducing agents) properties:

+4 +2 -4

R 4e~ = R R _ 2e- = R R + 4e- = R

Owing to this equally pronounced tendency to give up and to gain
electrons, the atoms of these elements combine with the atoms of other
elements primarily by means of covalent bonds, forming nonpolar
compounds.

Carbon, silicon, and their compounds are dealt with in this Exercise,
while the elements of the germanium and titanium series and their
compounds are examined in the next Exercise.

At ordinary temperatures oxygen has practically no effect on either
carbon or silicon. At higher temperatures the elements undergo more
or less vigorous oxidation with the formation of the oxides RO 2 and
RO.

Upon heating with concentrated nitric and sulphuric acids, carbon
is oxidised to carbon dioxide. Acids, except hydrofluoric, have no
effect on silicon.

In the presence of alkalis, silicon displaces hydrogen from water:

Si + 2NaOH + H 2 O - 2H 2 + Na 2 SiO 3

The compounds of carbon with metals are called carbides (e. g.,
calcium carbide CaC 2 ), while the compounds of silicon with metals
are called silicides (e. g., magnesium silicide Mg 2 Si).

Some of these compounds may be prepared synthetically, for
instance:

2Mg + Si = Mg 2 Si

The carbides and silicides of the alkaline and alkaline earth metals
and of aluminium are easily decomposed by water (hydrolysis) and
by acids to form gaseous hydrogen compounds:

CaC 2 + 2HOH = C 2 H 2 + Ca (OH) 2
Mg 2 Si + 4HC1 - SiH 4 + 2MgCl 2

The simplest hydrogen compounds, in which the number of hydrogen
atcms corresponds to the maximum valency of the element, may be
represented by the general formula RH 4 . These compounds (CH 4 and
SiH 4 ), with covalent bonds between the atoms, are gaseous and
nonpolar; they exhibit reductant properties.
Methane CH 4 , when ignited, burns according to the equation:

Mixtures of methane with air or oxygen are explosive, like detonating
gas.

Carbon, Silicon, and Their Compounds 215

The hydrogen compounds of silicon, called silanes (SiH 4 , Si 2 H 6 ,
etc.), are spontaneously inflammable in air.

Carbon monoxide CO and silicon oxide SiO are indifferent oxides,
since they do not combine with water and exhibit neither acidic
nor basic properties. Carbon monoxide may be prepared by heating
a mixture of formic acid and concentrated sulphuric acid. Carbon
monoxide is poisonous and exhibits reductant properties.

Carbon dioxide is prepared by burning coal or by decomposing car-
bonates or bicarbonates by heating them or by treating them with
acids.

At N. T. P. one volume of water dissolves 1.7 volumes of CO 2 .
The resulting carbonic acid is a very weak acid. The solution is, in
fact, a system in equilibrium:

CO + HO ; HoCO ^ H"

H-+CO;

Heating tilts the balance towards the evolution of CO 2 .

Among the silicic acids mention should be made of metasilicic acid
H 2 SiO 3 and orthosilicic acid H 4 SiO 4 , which are formed as gels in the
exchange reactions between silicic acid salts and mineral acids.
Many so-called polysilicic acids are known, their general formula
being (SiO^-^O^.The simplest of these (x = 2; y = 1) is disilicic
acid H 2 Si 2 O 5 . The silicic acids are very weak acids.

Carbonic acid gives rise to two series of salts: carbonates (with the
CO 3 " ion) and bicarbonates (with the HCO 3 ' ion). The carbonates of
the alkaline metals (with the exception of lithium) and of ammonium
are soluble in water; the most important of these are sodium carbonate,
or soda ash, Na 2 CO 3 , hydrated sodium carbonate, or washing soda,
Na 2 CO 3 - 10H 2 O, anhydrous and hydrated potash, K 2 CO 3 and
K 2 GO 3 -2H 2 O, and ammonium carbonate (NH 4 ) 2 CO 3 . The most impor-
tant of the bicarbonates is sodium bicarbonate, or baking soda, NaHCO 3 .

The silicic acid salts are called silicates, and the only water-soluble
ones among them are Na 2 SiO 3 and K 2 SiO 3 (soluble glass).

The average composition of ordinary window glass is expressed by
the formula Na 2 O-CaO-6SiO 2 ; water partly dissolves glass.

The soluble salts of the above acids undergo hydrolysis; their solu-
tions are alkaline. The hydrolysis of soda ash, for instance, follows
the equation:

2Na* + CO 3 + HOH s Na' + HCC4 + Na' + OH'

The other salts behave similarly, except for Na 2 SiO 3 and K 2 SiO 3>
which are converted by hydrolysis to salts of disilicic acid:

4Na + 2SiO; + HOH ^2Na* + Si./^ + 2Na* + 2OH'

216 Exercise 26

Strong reducing agents (such as Mg and P) are oxidised at combus-
tion temperature by carbon dioxide or silicon dioxide:

2Mg + CO 2 = 2MgO + C

QUESTIONS

1. What is activated charcoal made from and how?

2. Briefly explain the uses of the radioisotope C 14 in science and
engineering.

3. How can pure carbon monoxide be prepared from oxalic acid?

4. List the main constituents of water gas, producer gas, and illu-
minating gas. Which of these gases has the highest calorific value?
Why is this so?

5. How can pure carbon dioxide be prepared from exhaust gases?

6. What are the processes that have to be carried out to obtain
silicic acid gel from silica?

7. By means of what chemical reactions can a carbonate be turned
into a bicarbonate, and vice versa?

8. Why is it that in one of the stages of the ammonia-soda process

NH 4 HCO 3 + NaCl NaHCO 3 + NH 4 C1
the balance is shifted to the right?

Problems

1. Calculate the "mean molecular weight" of a mixture consisting of 42% of car-
bon dioxide and 58% of nitrogen.

2. A compound consists of 42.86% of carbon and 57.14% of oxygen. The densi-
ty of this gas in terms of chlorine is 0.396. Determine the formula of the substance
and its density in terms of air.

3. A mixture of 100 ml of methane and 100 ml of oxygen is ignited. Determine
fhe volume of the gas that has not reacted.

4. Determine the atomic weight of silicon from the following data:

Mol. weight Content of Si (%)

Silicon tetrafluoride . . . 104.06 26.96

Disilane 62.17 90.27

Silicochloroform 135.45 20.72

5. What are the amounts (by weight) of each of the initial products needed to
prepare 1 ton of soda ash by the Le Blanc process, the yield of soda ash being 35%?

6. One hundred and eighty grams of washing soda is dissolved in 1,300 g of water.
Determine the titre and the percentage, molar, and normal concentrations of the
solution in terms of the anhydrous salt, the relative density of the solution being
1.16.

7. According to the data of O. Alekin, 100 ml of the water of the Belaya River
contains 1 1.4 mg of Ca" and 2.5 mg of Mg", while 2 litres of the water of the Ishim
River contains 163 mg of Ca" and 154.6 mg of Mg". Which of the rivers has the
harder water?

Carbon, Silicon, and Their Compounds 217

8. The gas from the oil wells of Soviet Azerbaijan contains 85% (by volume) of
CH 4 , 2.8% of C 2 H 6 , and 1.2% of C 4 Hi , the rest of the gases being noncombustible.
The combustion heats of these gases are 212.8, 372.8, and 687,9 Cal. respectively.
Determine the calorific value of 1 cu m of the gas. Express in percentage the dis-
crepancy between this result and the practical figure of the calorific value, which
is equal to 8,014 Cal./cu m.

9. The heat of formation of SiH 4 is 8.7 Cal. Determine the combustion heat
of SiH 4 , knowing that the heat of formation of water is 68.35 Cal. and the heat of
formation of SiO 2 is 203.34 Cal.

10. Determine the pH and the a % of 0.001 M H 2 CO 3 , taking into account only
the first step in its electrolytic dissociation.

LABORATORY WORK

Apparatus and materials: apparatus shown in Figs. 35 and 70; Kipp gas gener-
ator; barometer; room thermometer; metal ruler; two porcelain crucibles; pipestem
triangle; porcelain mortar; jar; burette and funnel for it; two funnels; gas-delivery
tube with stopper; refractory test tube; test tubes; test tube with a hole in the bot-
tom and a gas-delivery tube; two Erlenmeyer flasks; 100 ml flask with stopper and
cooling pipe; glass rod; two cylinders with glass covers; 250 ml measuring cylinder;
cane sugar; activated charcoal; 4 : 5 and 8 : 5 mixtures of magnesium and quartz
sand; ferrosilicon; fused sodium acetate; soda lime; calcium carbide; bismuth oxide;
powdered glass; powdered (0.25-0.5 mm) silica gel; sodium carbonate; sodium
carbonate in tablets; potassium carbonate; calcium carbonate; sodium bicarbonate;
concentrated sulphuric acid; concentrated, 2 N, 3 M, and 0.01 N titrated hydro-
chloric acid; formic acid;2NNaOH; 25%Na 2 Siu 3 ; 0.1 N AgNO 3 ; saturated solution
of ammonium chloride; 10% solution of sodium carbonate; 10% solution of commer-
cial sodium chloride; 10% ammonia solution; O.b N A1 2 (SO 4 ) 3 ; 10% CuSO 4 ; hydrogen
sulphide water; lime water; indigo solution, neutral litmus solution; phenolphtha-
lein solution; palladium paper, and filter paper.

Preparation of Palladium Paper. Impregnate a piece of filter paper with an
0.01% solution of PdCl 2 , 100 ml of which contains 15 drops of N HC1. Dry the paper
in the absence of CO. Keep the paper in a jar with a ground-in lid. It is advisable
to prepare it every time it is needed.

Preparation of Powdered Glass. Bottle, window, or chemical-ware glass should
be ground thoroughly in a mortar. Approximately 1 g quantities of the powder
should then be weighed on a technical balance, wrapped in paper, and marked,
indicating the type of glass.

1. Preparation of Charcoal from Sugar. Put about 3 g of pure cane
sugar (powdered) in a porcelain crucible, place the crucible on a pipe-
stem triangle, and heat it. The sugar at first melts and is charred, a
froth rising on account of the burning of the volatile products. Stir
the mass with a glass rod and heat it for several minutes over a
blowpipe until a large porous mass is obtained. Write the equation of
the reaction that has taken place. Keep the product for experiment 4.

2. The Adsorption Properties of Charcoal. Pour some hydrogen
sulphide water into one Erlenmeyer flask and some indigo solution
into another. Add 1 g of finely ground charcoal to each and shake the
contents of the flasks vigorously. Ten or 15 minutes later filter off
the charcoal and establish the absence (by odour and colour) of hydro-
gen sulphide or indigo in the solutions.

What distinctive property does activated charcoal exhibit?

218 Exercise 26

3. Preparation of Silicon. Conduct this experiment in a ventilated
hood. Place 2 g of a mixture of metallic magnesium and quartz sand
(4 : 5 by weight) in a dry crucible and heat the crucible carefully
over the flame of a burner until the mixture catches fire. If the quartz
sand is very fine, the reaction may even be accompanied by an explo-
sion. When the mixture has cooled, transfer it to a beaker with a
hydrochloric acid solution. The magnesium oxide and silicide dissolve,
leaving a dark brown precipitate of amorphous silicon in the beaker
(the decomposition of the magnesium silicide by the acid produces
si lanes, which upon contact with the air ignite!).

Filter oft the precipitate and keep it for experiment 4. Write the
equations of the reactions whereby Si is prepared and Mg 2 Si is formed.

4. Reductant Properties of Charcoal and of Silicon, (a) Test the ac-
tion of concentrated sulphuric acid, when heated, upon the charcoal
prepared in experiment 1. Identify (with caution!) the gas evolved by
its odour. Write the equation of the reaction that has taken place.

(b) Test the action of a concentrated alkali solution on the amorphous
silicon prepared in experiment 3 or on a lump of ferrosilicon. Write
the equation of the reaction that takes place.

5. Preparation and Properties of Methane. Mix 3 g of fused sodium
acetate and the same amount of soda lime in a mortar, and transfer
the mixture to a dry test tube. Close the test tube with a stopper
through which a gas-delivery tube has been passed; fasten the test
tube horizontally in the clamp of a ringstand, and immerse the end of
the gas-delivery tube in a large jar with water (crystalliser tank).
Place two cylinders in the beaker, one of them full of water and the
other half-full. Warm the test tube with the flame of a burner and
then heat it from the bottom up. Fill both cylinders with the gas
that is evolved, cover them under water with lids, remove them from
the jar, and bring them up to the flame of the burner. The pure methane
burns quietly, whereas the mixture with air explodes (careful!).
Write the equations for the reactions of the preparation and burning of
methane. Does the mixture explode whatever the proportion of the
volumes in which CH 4 and O 2 are mixed?

6. Preparation and Properties of Acetylene. Conduct this experi-
ment in a ventilated hood. Place 2-3 small lumps of calcium carbide
in a dry test tube with a hole in its bottom, close the test tube with
a stopper through which a straight gas-delivery tube has been passed,
and lower it into a beaker with water. After a minute or two ignite the
gas and note the type of flame. Blow at the acetylene flame carefully
and observe whether it changes. Write the equations for the reactions
<of the preparation and burning of acetylene. Why, unlike methane,
does acetylene burn in the ; ir with a smoking flame?

7. Preparation of Hydrogen Compounds of Silicon. Conduct this
experiment in a ventilated hood. Place 1.5-2 g of a mixture of metallic
magnesium and quartz sand (8 : 5 by weight) in a dry test tube and

Carbcn, Silicon, and Their Compounds

219

fasten the test tube vertically in the clamp of a ringstand. Warm the
test tube and then heat it until the mixture ignites (careful!). When
the test tube has cooled, break it, transfer the mixture into a hydrochlor-
acid solution, and observe the spontaneous combustion of the

hydrogen compounds of silicon in the air. Write the equations of the
reactions that have taken place.

8. Carbon Monoxide and Its Reductant Properties. Conduct the
experiment in a ventilated hood. Prepare the apparatus shown in Fig.
70. Pour 10-15 ml of concentrated sulphuric acid into the flnsk and an

Fig. 70. Apparatus for preparing carbon monoxide

equal volume of formic acid into the dropping funnel. Place about
1 g of bismuth oxide into the bulb of the refractory tube. Add the acid
from the funnel to the flask and heat the flask gently. When the carbon
monoxide begins to flow evenly, ignite the gas at the outlet of the
gas-delivery tube. Heat the bulb with a burner until a drop of molten
metal forms in it. Then stop heating the bulb and hold a palladium
paper, moistened with a drop of water, over the outlet of the tube.
What happens to the paper? Turn the tip of the gas-delivery tube down-
wards and immerse it in a test tube with a heated solution of silver
diamminochloride. What happens? Write the equations for the reac-
tions of the preparation of CO, its burning, and its action uponBi 2 O 3 ,
PdCl 2 and [Ag(NH 3 ) 2 ]Cl.

9. Dissolving Co 2 in Water. Pour 5 ml of a neutral litmus solution
into a test tube and pass a stream of carbon dioxide from the Kipp gas

220 Exercise 26

generator through it slowly, washing the CO 2 with water in a wash
bottle for gases. How does the colour of the litmus change? Why is
this? Write the equation of the reaction that has taken place. Pour
half of the solution into another test tube, and heat it carefully for
some time. How does the colour change? Why?

10. Preparation of Silicic Acid Gel. Add 4 ml of concentrated hydro-
chloric acid to an equal volume of a 25-30% solution of sodium silicate.
Observe the formation of silicic acid gel 20 min later. The contents
of the test tube, it should be noted, remain in it even when the test
tube is turned over. Write the equation of the reaction that has taken
place.

11. The Adsorption Properties of Silica Gel. Add an ammonia
solution by drops to 15-20 drops of a cupric sulphate solution until
the solution becomes dark blue owing to the formation of a cuprammo-
niurti complex. Now add 2 g of ground silica gel (0.25-0.5 mm grain
diameter) to the solution and shake the mixture for some time. The
colourless powder of silica gel acquires a dark blue colour, whereas the
solution becomes pale. Pour off the solution, wash the silica gel 3-4
times by decantation, add 2 ml of hydrochloric acid, and shake the
solution. The silica gel loses its blue colour because the cuprammonium
complex is washed away by the acid.

12. Calcination of Carbonates and Their Treatment with Acids.
(a) Place 0.5 g of the salts CaCO 3 , NaHCO 3 , Na 2 CO 3 , and K 2 CO 3 into
four dry test tubes respectively. Close each test tube in turn with a
stopper through which a gas-delivery tube has been passed, the tube
being bent at a right angle and immersed in a test tube with lime
water. Heat the test tube with the substance over the flame of a burner,
noting that the lime water becomes cloudy. Which carbonic acid salts
are not decomposed by heating? Write the equations for all the reac-
tions carried out.

(b) Treat 0.5 g of the same salts in test tubes with 2 N HC1. Do all
the salts dissolve in the acid? What characteristic phenomenon is
observed when carbonates dissolve in acids? Write the equations for
the reactions carried out. Explain why sodium bicarbonate, rather
than carbonate, is used in fire extinguishers.

13. Hydrolysis of Salts, (a) Test a solution of sodium carbonate and
a solution of sodium silicate with red litmus paper. Write the equations
for the reactions of hydrolysis that take place. Which of the salts
undergoes hydrolysis more readily? Why is this so?

(b) Add a sodium carbonate solution to an aluminium sulphate solu-
tion. A voluminous white precipitate of aluminium hydroxide is
formed, with the simultaneous evolution of bubbles of CO 2 . Write
the relevant equations.

(c) Mix a solution of sodium silicate (1-2 ml) with a double volume
of ammonium chloride solution. Heat the mixture slightly, observing
the formation of a precipitate. Identify the gas evolved (by its odour).

Carbon, Silicon, and Their Compounds

221

Write the equations of the reactions that take place. In which case is
hydrolysis plactically irreversible?

14. Determining the Percentage Content of Bound Carbon Dioxide
in Sodium Carbonate. The experiment is conducted in the apparatus
shown in Fig. 35. Fill one-third of the beaker with a 10% solution of
sodium chloride. Fill a 250 ml measuring cylinder with the same solu-
tion, cover it with a glass lid, and turn it over into the beaker. Fasten
the cylinder with a clamp attached to a ringstand.

Pour 30-40 ml of 3 M HC1 into the flask and dry the inner wall of the
neck of the flask with some rolled-up filter paper. Weigh about
0.8-0.85 g of sodium carbonate pressed into tablets. Place the weighed
sodium carbonate into the neck of the flask, which is in a horizontal
position (shown by broken line in Fig. 35); close the flask tightly with
the stopper through which a delivery tube has been passed. Place the
end of the delivery tube under the cylinder, and let the sodium carbo-
nate tablets slip into the acid. The carbon dioxide collects in the cylin-
der. When all the sodium carbonate has dissolved, measure the volume
of the gas evolved and the height of the column of the liquid in the
cylinder over the level of the liquid in the beaker; note the room temper-
ature and the barometric pressure.

Experimental Results

Weight of
Na 2 CO in g

Volume of gas
in ml

Temperature
In C

Barometric
pressure in
mm Hg

Height P w

in mm

Vapour
tension h in
mm Hg

From the data obtained calculate the percentage content of bound
CO 2 in the sodium carbonate. Calculate the percentage content of
CO 2 from the formula of sodium carbonate; compare this figure with
the experimental result.

15. Determining the Resistance of Glass to the Action of Water.
Receive from the instructor a glass sample for testing (weighing 1 g).
Place the weighed sample in a flask, add 25 ml of distilled water, and
close the flask with a stopper through which a long tube has been
passed to act as a cooling pipe *. Attach the flask to a ringstand and
heat it on a wire gauze with an asbestos centre to boiling point. Boil
the water for 20 min in such a way that all the vapour should be con-
densed in the cooling pipe. After cooling the flask, add 2-3 drops of a

* The flask and the tube should first be treated with live steam.

222 Exercise 27

phenolphthalein solution: the solution will acquire a pink colour. Fill
a burette with 0.01 N HC1 and titrate the extract carefully until the
addition of one drop of extra acid causes the solution to become col-
ourless. From the data of the following table determine the grade of the
^glass:

Grade of glass

Unaffected

Resistant

Hard for chemical ware

Soft for chemical ware

Ordinary (window glass, bottle glass)

Number of ml of 0.01 N HC1

used up to titrate 25 ml

of the extract

0-0.32

0.32-0.65

0.65-2.8

2.8-6.5

over 6.5

Write the equation for the hydrolysis of the soluble part of the glass.

Exercise 27

THE ELEMENTS OF THE GERMANIUM AND TITANIUM
SERIES AND THEIR COMPOUNDS

SUBJECTS FOR STUDY

The elements of the germanium series: germanium, tin and lead; simple sub-
stances, their preparation and properties; valence in compounds; oxides and hydro-
xides of the bivalent elements, and their amphoteric nature; the most important
salts and their chemical properties; stannic oxide and lead dioxide.

The elements of the fourth subgroup: titanium, zirconium, and hafnium; their
atomic structure and chemical properties; the nature of their oxides; titanium dio-
xide and its hydroxide; pertitanic acid.

The elements germanium, tin, and lead belong to the fourth main
group and are chemical analogues of carbon and silicon. In compounds
they are bi- and tetrapositive, and they form gaseous compounds with
hydrogen.

As to physical properties, tin and lead are metals with low melting
points.

Water at ordinary temperatures has practically no effect on either
tin of lead: it oxidises them, a protective coating appearing on their
surface and preventing a continuation of the reaction.

Hydrochloric acid acts only on Sn and Pb, turning them into biva-
lent ions:

Sn + 2HC1 = H 2 + SnCl 2

Elements of the Germanium and Titanium Series 223

With lead the reaction requires heating, since PbCl 2 is almost in-
soluble in cold water.

Nitric acid oxidises lead to Pb". Tin is oxidised by concentrated
nitric acid to SnO 2 or p -stannic acid, which probably has the composi-
tion

4*-

.n + 4HNO 3 - SnO 2 + 4NO 2 + 2H 2 O

Dilute nitric acid oxidises tin to form Sn(NO 3 ) 2 , while at the same
time undergoing reduction to NO; very dilute acid is reduced to ammo-
nium nitrate:

4Sn + 10HNO 3 - 4Sn(NO 3 ) 2 + NH 4 NO 3 + 3H 2 O

Dilute sulphuric acid has no effect on lead (owing to the formation
of a protective PbSO 4 film, which is insoluble in acid); concentrated
sulphuric acid (80% and higher concentration) dissolves the protective
film:

PbSO 4 + H 2 SO 4 - Pb (HSO 4 ) 2

The oxides of lead and tin dissolve in acids and alkalis.

The hydroxides Sn(OH) 2 and Pb(OH) 2 are prepared by exchange
reactions between salts and alkalis. Water scarcely dissolves them at
all, but they do dissolve in acids and alkalis, in the latter case forming
hydroxystannites Me 2 [Sn(OH) 4 ] or hydroxyplumbites Me 2 IPb(OH) 4 l.

The most soluble salt of bivalent tin is stannous chloride SnCl 2 .

The soluble salts of lead are the nitrate Pb(NO 3 ) 2 and the acetate
Pb(CH 3 COO) 2 ; the chloride PbCl 2 and the iodide Pbl 2 are soluble
in hot water. The salts of bivalent tin are used as good reducing
agents in diverse media, e. g.:

SnCl 2 + 2FeCl 3 + 2HC1 - H 2 [SnCl 6 ] + 2FeCl 2

The salts of bivalent lead can also undergo oxidation to tetravalent
lead compounds

Pb (CH 3 COO) 2 + NaOCl + H 2 O = PbO 2 + NaCl + 2CH 3 COOH

but they are very weak reducing agents and are therefore not used for
reduction reactions.

224 Exercise 27

The sulphides SnS (brown) and PbS (black) are insoluble in water.
The formation of the black precipitate of PbS serves as a reaction for
detecting the S" ion; the precipitate dissolves in nitric acid.

Stannous sulphide dissolves in a solution of ammonium polysul-
phide, undergoing oxidation from the bivalent to the tetravalent state:

2 ^

SnS + (NH 4 ) 2 [S 2 ] - (NH 4 ) 2 SnS 3

The resulting salt, ammonium thiostannate, is a salt of thiostannic
acid H 2 SnS 3 , which is extremely unstable and decomposes the moment
it is formed:

H 2 SnS 3 = H 2 S -f SnS 2

The insoluble salts of lead that deserve mention are the sulphate
PbSO 4> the basic carbonate Pb 3 (OH) 2 (CO 3 ) 2 , also known as white
lead, and the chromate PbCrO 4 .

Stannic oxide SnO 2 and lead dioxfde PbO 2 are insoluble in water
and do not react with it chemically; they are amphoteric.

Two acids correspond to stannic oxide: a-stannic acid H 2 SnO 3 and
p-stannic acid H 4 SnO 4 ; similar acids correspond to lead dioxide. In the
H 2 C,O 3 H 2 PbO 3 series acidity declines from carbonic to metaplumbic
acid. The hydroxides of tetravalent germanium, tin, and lead exhibit
amphoteric properties.

The salts of the stannic acids are called a- andp-stannates; the salts
of the plumbic acids, meta- and orthoplumbates. The plumbates include
minium (or red lead) Pb 3 O 4 and the trioxide Pb 2 O 3 , which are lead
salts of orthoplumbic and metaplumbic acid respectively:

+2 +4 +2 +4

Pb 2 PbO 4 and PbPbO 3

The salts of tetravalent lead are extremely unstable in aqueous
solution and experience complete hydrolysation to the more stable
compound PbO 2 .

The compounds PbO 2 and Pb 3 O 4 are often used for oxidation reac-
tions, e. g.:

-1 +4

2KI + PbO 2 + 2H 2 SO 4 - I 2 + PbSO 4 + K 2 SO 4 + 2H 2 O

2KI + Pb 2 PbO 4 + 4H 2 SO 4 - I 2 -f 3PbSO 4 + K 2 SO 4 + 4H 2 O

The elements of the titanium series are titanium, zirconium, and
hafnium. The atoms of these elements have two electrons in their
outermost shells, with 8+2 electrons in the next shell.

Elements of the Germanium and Titanium Series 225

In chemical reactions the atoms of these elements can give up 4
electrons (2 from the outermost shell and 2 from the next). The maxi-
mum valence of these elements in compounds is 4 (a point of analogy
with the elements of the carbon series). However, these elements, as
reducing agents, do not, unlike the carbon elements, form negative
ions and do not produce gaseous hydrides with hydrogen.

Although titanium, zirconium, and hafnium may be bivalent,
trivalent, and tetravalent in compounds, the latter valence is the
most characteristic and the one typical of their stable compounds.
Some of the compounds of titanium TiO 2 , TiCl 4 , etc. are used in
the laboratory.

The colourless compounds of Ti +4 , upon interacting with strong
reducing agents, turn into pale violet compounds of Ti 43 . Exposure to
air causes the violet colour to disappear because of the oxidation of
Ti+ 3 to Ti+ 4 . When Ti+ 4 salts react with an alkali solution, a white
gelatinous precipitate of hydroxide is formed, which dissolves only in
acids. With hydrogen peroxide in water, Ti^ 4 compounds form the
yellow-orange pertitanic acid. The reaction follows the equation

OH H

- [0 2 ] - H HO [O 2 ] - H

Ti + 2H 2 O

- [0 2 ] - H HO [0 2 ] - H

and is used for the qualitative detection of titanium.

QUESTIONS

1. Write the equations for the reactions whereby tin is dissolved in
dilute and lead, in concentrated nitric acid.

2. Why is hydrochloric acid added to water in preparing an aqueous
solution of SnCl 2 ? Write the equations of the reactions that take
place.

3. Write the equations for the reactions of the decomposition of
ammonium thiostannate by hydrochloric acid.

4. Write the equations for the reactions of the preparation of sodium
zirconate and its hydrolysis.

5. What can be used to dissolve metallic zirconium? Write the
equations for the reactions whereby it is dissolved.

6. Write the equations for the reactions whereby the following
consecutive transformations are effected:

TiCl 4 -> Ti -* TiO 2 - TiCl 4 - Ti (OH) 4

7. Write the structural formulae for Na 2 SnS 3 and Pb 3 O 4 .

226 Exercise 27

8. Draw the diagram of a lead storage battery. What chemical reac-
tions take place at the cathode and at the anode when the battery is
charged and when it is discharged?

Problems

1. The interaction of 23.7gof metallic tin with excess hydrochloric acid produces

an amount of hydrogen sufficient to obtain 12.7 g of metallic copper by reducing
cupric oxide. Determine the equivalent of tin.

2. What volume of nitrogen dioxide will be produced when 1 kg of lead is dis-
solved in concentrated nitric acid? The gas is collected at 20 and 750 mm Hg.

3. What will be the volume (at N. T. P.) of nitrogen dioxide produced if 50 g
of an alloy containing 70% of copper and 30% of tin is treated with concentrated
nitric acid in excess?

4. Assuming lead to consist of the isotopes Pb 2J4 (1.37%), Pb 206 (25.15%), Pb 2J7
(21.11%), and Pb 208 (52.38%), determine the atomic weight of lead.

5. Determine the percentage concentration and the normality of a SnCl 2 solution
prepared by mixing 2.5 litres of a 22% solution (relative density 1.19) and 1.5
litres of a 4% solution (relative density 1.03).

6. What volume of 2 N KOH should be added to 200 g of a 5% solution of SnCl 2
to convert all of the latter to stannite?

7. One hundred grams of minium was treated with excess nitric acid; after the
reaction the solution was filtered and evaporated; the dry residue was then dissolved,
the solution filtered and diluted to 2 litres. Determine the molar concentration
and normality of the solution prepared.

8. One gram of a chemically resistant zirconium alloy yielded 0.88 g of zirconium
dioxide, 0.37 g of ferric oxide, and 0.145 g of aluminium oxide. Determine the pro-
portion (%) in which these metals are contained in the alloy.

9. The burning of a gram-molecule of titanium carbide generates 202.45 Cal. of
heat. The heats of formation of titanium dioxide and carbon dioxide are 218 and
94.45 Cal. respectively. Determine the heat of formation of titanium carbide.

10. The reduction of the complex compound K^IHfFg] by metallic potassium
involves the expenditure of 3.13 g of the metal. What is the yield of hafnium?

LABORATORY WORK

Apparatus and materials: apparatus shown in Fig. 69; felt or cloth polishing
wheel; blowpipe; funnel; test tubes and rack; coal in lumps; lead dioxide; 1 : 1 mix-
ture of coal and lead oxide; 1 : 1 mixture of coai and stannic oxide; zinc in granules;
stannic chloride; concentrated and 2 N HNO 3 ; 2 N H 2 SO 4 ; concentrated ^and 2 N
HC1; 2 N NaOH; 2 N KOH; 0.5 N SnCl 2 ; 0.5 N Pb(NO 3 ) 2 ; 0.5 N Pb(CH 8 COO) a ; 0.5
N Bi(NO 3 ) 2 ; 0.5 N KI; N solution of ammonium polysulphide; 0.5 N K 2 CrO 4 ; 3
N Na 2 CO 3 ; 2% solution of titanium sulphate; 0.3 N SnCl 4 ; 3% H 2 O 2 ; saturated
solution of lime chloride; electrolysis solution; hydrogen sulphide water; filter
paper, Htmus paper, and sandpaper.

Preparation of Titanium Sulphate Solution. Boil titanium dioxide slowly with
concentrated sulphuric acid; then either dissolve it in water or fuse the titanium
dioxide with potassium pyrosulphate (1 : 7); dissolve the melt in 0.2 N H 2 SO 4 .

Preparation of Solution for Electrolysis. Dissolve 6 g of crystalline SnCl 2 and 2g
of NH 4 C1 in 300 ml of water containing 2 ml of concentrated hydrochloric acid.

1. Preparation of Lead and Tin. (a) Make a hollow 1 cm in diameter
in a lump of coal (or coke) and fill it with a 1 : 1 mixture of lead oxide
and powdered coal. Moisten the mixture with 2-3 drops of water. Hold
a blowpipe at the fringe of the reducing flame of a burner and, blowing

Elements of the Germanium and Titanium Series

227

Fig. 71. Reduction on a lump of
coal

air into it carefully by mouth, direct the elongated flame at the mix-
ture (Fig. 71). Carry on reduction in this way until a drop of molten
metal is formed. Write the equation of the reaction that has taken
place.

(b) Conduct a similar experiment with stannic oxide, mixing it
with an equal amount of coal (or coke). Write the equation of the
reaction. Keep the resulting beads of
lead and tin for the next experiment.

2. Dissolving Lead and Tin in Acids.
(a) Test the action of hot concentrated
nitric acid on the bead of lead. What
is the gas evolved? Write the rele-
vant equations.

(b) Dissolve the bead of tin in con-
centrated hydrochloric acid, heating it
gently. Write the equation of the
reaction that takes place. How is stan-
nic chloride prepared?

3. Electrolytic Tinning. Assemble
the apparatus shown in Fig. 69. Clean
40x80 mm copper electrodes thorough-
ly with fine sandpaper, wash them
with water, and wipe them with filter

paper. Pour 300 ml of the electrolysis solution into the jar. Immerse
55 mm of the electrodes in the solution and fasten them at a distance
of 50 mm from each other. Switch on the current (the voltage should
be about 2.2 V, the current intensity 0.2-0.25 A) and conduct
electrolysis for 10 min. Then switch off the current, remove the
cathode from the solution, wash it in water, and wipe it dry with
filter paper.

Polish the dry plate on a felt or cloth polishing wheel. Note the
outward appearance of the tin coating. How can this layer be re-
moved from the copper electrode? Dismantle the apparatus.

4. Amphoteric Nature of Stannous and Plumbous Hydroxides.
Prepare precipitates of hydroxides from salt solutions containing the
Sn" and Pb" ions and prove by experiment that they are amphoteric.
Write the equations of all the reactions carried out. Why is plumbous
hydroxide usually dissolved in nitric rather than hydrochloric or
sulphuric acid? Keep thehydroxystannite solution for the next experi-
ment.

5. Reductant Properties of Sn" 2 and Pb+ 2 . (a) Prepare a precipitate
of bismuth hydroxide by an exchange reaction and add to it the alka-
line solution of hydroxystannite prepared in the previous experiment.
The white precipitate turns black, owing to the reduction of Bi +s to
metallic bismuth. Write the equations of the reactions that have taken
place.

228 Exercise 27

(b) Pour 2 ml of a Pb(CH 3 COO) 2 solution and 3-4 ml of a saturated
solution of lime chloride* (CaClOCl) into a test tube and boil the so-
lution. A brown precipitate of lead dioxide is thrown down. Write the
equation of the reaction that has taken place. Filter off the precipitate,
wash it with hot water several times, and keep it for experiment 10.
Why are the salts of bivalent lead not used in practice as reducing
agents?

6. Insoluble Compounds of Pb+ 2 and Sn 42 . (a) Test solutions of
SnCl 2 and Pb(NO 3 ) 2 in separate test tubes with the following reagents:
sulphuric acid, hydrogen sulphide, and potassium iodide. Do precipi-
tates form in all the test tubes? Note the colours of the precipitates
formed. Write the equations of the reactions that have taken place.

(b) Divide the solution with the PbI 2 precipitate into two parts;
let the precipitate settle in each test tube and pour off the solution.
Add an excess of KI solution to one of them. Observe that the precipi-
tate dissolves. Explain this. Write the equation of the reaction that
has taken place. Add 5-10 ml of 2 N CH 3 COOH to the precipitate in
the other test tube and heat it to boiling point. When it cools, lustrous
golden crystals of PbI 2 are precipitated.

(c) Pour off the solution from the SnS precipitate (colour?) and add
ammonium polysulphide solution to the precipitate. What happens to
the precipitate? Write the equation of the reaction that has taken
place. Add 2NHC1 to the solution until litmus paper dipped in the
solution exhibits an acid reaction and heat the solution slightly.
Observe the formation of a precipitate (colour?). Write the equation of
the reaction that has taken place.

(d) Add a solution of potassium chromate and sodium carbonate to
the lead salt solution in two test tubes. Note the colours of the preci-
pitates formed in each of them. Write the equations of the reactions
that have taken place, bearing in mind that the reaction with sodium
carbonate produces a basic lead salt. What are the commercial uses of
these products?

7. Preparation and Burning of Stannic Hydride. (This experiment
should be conducted in a ventilated hood!) Pour 5-6 ml of concenrtat-
ed HC1 andlO-15 drops of a stannic chloride solution into a crucible;
add 2-3 granules of metallic zinc. Stir the contents of the crucible
with a. test tube half full of cold water. Quickly place the test tube in
the flame of a burner: the stannic hydride ignites, producing bright
blue scintillations. Write the equations of the reactions that have
taken place.

8. Preparation and Properties of Stannic Acid. Add a NaOH
solution by drops to a stannic salt solution. Divide the white amor-

* In lime chloride, which has the structural formula Ca_i the uniposi-

\C1,
tive chlorine, as a strong oxidising agent, can receive two electrons.

Colloidal Solutions 229

phous precipitate formed, together with the solution, into two parts.
Add excess alkali to the part in one test tube, excess acid, to the part in
the other. What happens? Write the equations of the reactions that
have taken place.

9. Hydrolysis of Stannic Chloride. Inside a ventilated hood open a
jar containing liquid stannic chloride. What do you observe? Write
the equation of the reaction that takes place. What practical appli-
cation has the reaction?

10. Oxidant Properties of Pb02. Pour 5-6 drops of a KI solution
and 4 ml of H 2 SO 4 into a test tube; add some of the lead dioxide pre-
pared in experiment 5b, introducing it on a glass spatula. Heat the test
tube, allow the precipitate to settle, and determine the colour of the
solution.

11. Titanium Hydroxide and Its Properties. Add 2 ml of an alkali
solution to an equal volume of titanium sulphate solution. What is
formed? Write the equation of the reaction that has taken place.
Shake the solution with the precipitate and divide it into two parts*
Add excess alkali to the part in one test tube; a sulphuric acid solution,
to the part in the other. What conclusions may be drawn from these
experiments?

12. Reduction of Tetravalent Titanium. Dilute 2 ml of a titanium
salt solution with an equal volume of 2 N HC1 and lower a bit of zinc
into the solution. After a time the solution acquires a violet colour.
Write the equation of the reaction that has taken place. Pour off the
violet solution into another test tube and note that it loses its colour.
Explain this.

13. Preparation of Pertitanic Acid. Add a hydrogen peroxide so-
lution to a slightly acidified solution of a titanium salt and observe
the appearance of a yellow-orange colour. Write the equation of the
reaction that has taken place.

14. Analysis of a Test Solution. Receive from the instructor a test
solution that may contain one of the following ions: Pb", Sn*", or Ti"".
Establish by means of the reactions studied which of these ions is
present in the solution. Write up the results of the experiment and
submit the report to the instructor.

Exercise 28
COLLOIDAL SOLUTIONS

SUBJECTS FOR STUDY

Dispersed systems; colloidal solutions; preparation of colloidal solutions and
their distinctive properties; degrees of dispersion; micelles; sols; lyophilic and ly-
ophobic colloids; coagulation, sedimentation, and the causes of the formation of

Exercise 28

a precipitate in colloidal systems; gels; mutual coagulation of colloids; reversible
and irreversible colloids.

Colloidal solutions occupy an intermediate position among the
dispersed systems somewhere between the suspensions and the true
solutions: the diameter of the dispersed particles in the liquid phase
of a colloidal solution ranges from 1 to 100 m^*. These solutions can
be prepared by two different methods: the dispersion method (reducing
the size of the particles of coarser dispersed systems, suspensions) and
the condensation method (increasing the size of the particles of true so-
lutions, in which there is molecular or ionic dispersion of the sub-
stance). Colloidal solutions are also called sols. Unlike true solutions,
colloidal solutions are optically inhomogeneous systems, since a beam of
light experiences scattering in them. This accounts for the opalescence
of colloidal solutions (their different colours in reflected and transmit-
ted light), which is a distinguishing feature of such systems. The
particle size of colloidal solutions of one and the same substance rang-
ing within broad limits, their colour too may differ. Owing to their
extremely high degree of dispersion, colloidal solutions exhibit all
the phenomena that occur at the interface of two phases, especially
the surface absorption of various substances (adsorption). One of the
products of adsorption from solutions may be molecules of the solvent,
specifically water. Colloidal systems in which the particles are sub-
jected to surface hydration by a thin layer of water molecules are
called hydrophob ic (e. g., colloidal metals, sulphides, etc.). Hydrophilic
colloids are distinguished by the fact that, in addition to surface
hydration, their particles bind a large number of water molecules by
a "loose" internal structure (e. g., silicic acid, glue, etc.). The particles
of a colloidal solution can, in addition to the molecules of water,
adsorb ions on their surface. They adsorb from a solution the ions
that are part of their composition and that are present in the solution
in excess.

Colloids that adsorb positive ions are called positive (e. g., the
hydroxides of metals), while those that adsorb negative ions are called
negative (e. g., sulphides and colloidal metals). The colloidal particles
with the adsorbed ions are termed granules, while with the ions of
opposite charge ("counter-ions") linked to the granules they are called
micelles. The qualitative composition of micelles is illustrated by
the following two formulae:

micelle of arsenic trisulphide

{m (As 2 S 3 ) nHS'-(n x) //'}"
nucleus

granule

mjJi is the designation for millimicron, which is a millionth of a millimetre.

Colloidal Solutions 231

micelle of ferric hydroxide

{m [Fe (OH) 3 1 nFe&-(n x) Cl'}+ + xCV
nucleus

granule

An electric current sends the granules moving towards one electrode
and the counter-ions towards the other. This migration of colloidal
particles under the influence of an electric current is Called catapho-
rests. Through cataphoresis it is possible to determine whether the
granules in micelles are positively or negatively charged.

Colloidal solutions are rather stable systems and can be kept for
some time without undergoing any change. The relative stability
of colloidal systems is governed by the relationship between the
forces of attraction, which tend to cause particle growth, and the forces
of repulsion preventing this. The repulsion effect is caused by the elec-
trostatic forces arising between the grains, since they carry like
charges. In lyophilic colloids, aggregation to larger particles is also
prevented by the solvate envelope around the solvent molecules.

When the electric charges of the granules are neutralised, this pro-
duces particle aggregation, and the process is known as coagulation. The
larger particles are precipitated (sedimentation). The precipitates
formed in the coagu