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PROFESSOR: OK, so we're going
to leave off where we were last

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time talking about buffers.
 

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And I'm a big fan of buffers,
actually, and buffers are

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extremely useful in my research
because they keep the p

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h of things constant.
 

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So if you're going to do any
kind of biology research or

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biochemistry research, you
need to know about buffers.

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And as I'll mentioned later,
also your body needs to be

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buffered appropriately, you
don't want a large changes in

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p h in your body or things
won't function properly.

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So buffering is very important.
 

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So let's go over buffers.
 

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So if we talk about an acid
buffer, that's something that's

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buffering on the acidic
side of the p h range.

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And in that, you want to
have play on both sides.

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So if you have a strong acid,
it can be neutralized,

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the p h stays the same.
 

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00:01:22 --> 00:01:24
If you had a strong base,
that would be neutralized

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so the p h stays the same.
 

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So what happens, you have to
have an acid and its conjugate

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base in the mixture to
have a good buffer.

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00:01:32 --> 00:01:36
So what the weak acid is going
to do is it will transfer

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protons or hydrogen ions to
hydroxide ions that are

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supplied by the strong base to
neutralize that strong

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base that was added.
 

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So you need to have an acid in
there that's capable of giving

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up a proton to neutralize the
hydroxide ion that was added.

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00:01:54 --> 00:01:57
You also need to have
a conjugate base.

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The conjugate base is going to
accept protons if an acid is

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added, thereby neutralizing the
effect of that added acid.

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00:02:06 --> 00:02:09
So again, you need a weak acid
and its conjugate base, you

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need both to have
a good buffer.

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So I've been talking about weak
acids and weak conjugate bases,

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so why would a strong acid and
the salt of its conjugate

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base not make a good buffer?
 

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What do strong acids do?
 

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They dissociate almost
completely, and that's not

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going to do you any good.
 

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You want to be able to shift
that equation both directions.

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00:02:39 --> 00:02:42
So a strong acid goes to
completion is going to drive it

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to the right, and for a good
buffer you want to be able to

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switch things around that if
you add a strong acid, then you

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neutralize that, you add a
strong base, you neutralize it.

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So you need the conjugate
to be effective as a base,

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and a strong acid has an
ineffective conjugate base.

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It's not good as a base at
all, so that won't work.

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So buffers are made up of weak
acids and their conjugate

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bases, or weak bases and
their conjugate acids.

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They need to be in the
weak range or there's

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no buffering capacity.
 

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So let's look at a
base buffer example.

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The only difference here is
that a basic buffer is going

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to buffer on the basic
end of the p h range.

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So here's an equation.
 

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We have a base in water forming
a conjugate of that weak

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base and hydroxide ions.
 

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So let's think about what
happens if a strong

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acid is added.
 

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Well if a strong acid is added,
then this base, n h 3, can

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accept protons from the
incoming acid and make more n h

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4 plus, thereby removing the
strong acid from the solution

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and neutralizing the p h.
 

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If on the other hand a strong
base is added, n h 4 plus or

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ammonium ions can donate the
proton forming its conjugate

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base in water, and again, the
p h remains about the same.

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So the base and its conjugate
acid can both react,

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neutralizing the p h.
 

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So in the base buffer acid, a
weak base, b, will accept

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protons supplied by the strong
acid, removing a strong

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acid from the solution.
 

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The conjugate acid of that weak
base, which again, would be a

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weak acid, can transfer protons
to the hydroxide ions added

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by the strong base, again,
neutralizing the effect there.

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So a buffer is a mixture of a
weak acid base conjugate pair

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that will stabilize the p h
serving as a source or a

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sink for the added protons.
 

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So that's how a buffer works.
 

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So, I mentioned buffering
is very important.

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It's very important
in your blood.

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Your blood is buffered
in the range of p h 7 .

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35 to 7 .
 

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45, and you have a buffering
pair and a conjugate acid

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00:05:22 --> 00:05:25
base pair that's in your
blood that does to work.

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So nature has its own buffering
system design to keep the p

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h of your blood constant.
 

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Well what happens if the p
h of your blood changes?

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00:05:37 --> 00:05:43
So let me just tell you about
one disease that can change

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00:05:43 --> 00:05:45
the p h of your blood.
 

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00:05:45 --> 00:05:47
This is from a vitamin B12
dependent enzyme called

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00:05:47 --> 00:05:54
methylmalonic coa mutase, and
this enzyme is your friend.

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00:05:54 --> 00:05:57
I think that most of you are
excited about the idea of

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having all your fat go to
energy, so you should be

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00:06:00 --> 00:06:02
all fan of this enzyme.
 

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00:06:02 --> 00:06:07
So it breaks down from fatty
acids -- you get methylmalonyl

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00:06:07 --> 00:06:11
coa, which is converted to
succinyl coa and goes into

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00:06:11 --> 00:06:12
the citric acid cycle.
 

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00:06:12 --> 00:06:16
But if this step is blocked, if
there's some kind of genetic

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00:06:16 --> 00:06:19
disease associated with that
particular enzyme, you

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get a condition called
methylmalonic aciduria.

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00:06:22 --> 00:06:25
So this is excreted in the
body, it can't be converted to

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00:06:25 --> 00:06:29
the succinyl coa, so it has to
be excreted from the body,

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00:06:29 --> 00:06:31
and it has an acidic p h.
 

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And so when it's excreted, it
will actually, there's a large

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00:06:35 --> 00:06:38
amount of it, and it can
change the p h of blood.

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00:06:38 --> 00:06:41
So even though you have this
buffering system, it can

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00:06:41 --> 00:06:45
overwhelm the buffering system
and cause a change in the p h,

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00:06:45 --> 00:06:51
which can lead to neurological
disorders and sometimes death.

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00:06:51 --> 00:06:55
So they can look for this in
newborns, they can do a test,

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and I was -- one of the tests
is -- it's not in every state,

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but it is in Massachusetts, so
my daughter, Samantha,

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00:07:03 --> 00:07:06
had this test.
 

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But they can simply look at the
urine of a newborn and look at

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the p h and whether any of
this is being excreted.

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00:07:13 --> 00:07:16
And some infants that have this
condition can excrete like

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a gram of this acid a day.
 

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So it's really large quantities
of acid that are excreted

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and it goes beyond the
buffering capacity.

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So, I'm going to tell you a
little story and it sort of

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00:07:29 --> 00:07:32
emphasizes some of the
importance of genetic research.

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00:07:32 --> 00:07:36
So before the human genome
project, scientists were trying

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00:07:36 --> 00:07:38
to find the gene to figure out
what was wrong with these

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00:07:38 --> 00:07:40
patients that had
this condition.

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00:07:40 --> 00:07:45
And here are all the amino
acids translated from the gene.

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00:07:45 --> 00:07:50
This is not a small protein,
that's a lot of amino acids.

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00:07:50 --> 00:07:53
And they found that the problem
was right here, or one of the

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00:07:53 --> 00:07:54
main problems is right here.
 

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There's a glycine residue,
which glycine is the smallest

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00:07:57 --> 00:08:00
amino acid and that had been
changed to something else and

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00:08:00 --> 00:08:02
that led to the disease.
 

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00:08:02 --> 00:08:06
Now I want you to imagine if
you were parents of a small

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00:08:06 --> 00:08:10
child and you found out your
child had this condition and

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00:08:10 --> 00:08:12
then you had heard that there
were advances, they now know

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what caused it, and so, they
went to the doctor and the

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doctor said, yes, your child,
the DNA is wrong, and so let's

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00:08:21 --> 00:08:27
substitute a different amino
acid for glycine here.

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OK.
 

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What does that do for you?
 

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00:08:30 --> 00:08:32
Like why is that a problem?
 

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00:08:32 --> 00:08:35
Well, they didn't know why it
was a problem, they just knew

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00:08:35 --> 00:08:38
that it had the wrong
amino acid in there.

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00:08:38 --> 00:08:43
And so, the next step was
to try to understand what

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00:08:43 --> 00:08:45
was actually going on.
 

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00:08:45 --> 00:08:49
So what happened in this case
was that the vitamin B12

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00:08:49 --> 00:08:53
cofactor had changed its
conformation when it

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00:08:53 --> 00:08:54
bound to the protein.
 

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And instead of having this
sort of loop structure here,

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it had more of a tail.
 

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00:09:00 --> 00:09:05
So this base here had moved
down, and so there was a

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00:09:05 --> 00:09:10
different shape to the vitamin
than had been expected.

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So when the first structure was
solved of one of these enzymes,

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it found that here's kind of
what the vitamin B12 looked

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like, it had this extended
region which no one was

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expecting, and so that had
to fit into the protein.

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So this is the domain of the
protein that binds the vitamin,

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and it had a large hole, and
you put these together and

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00:09:31 --> 00:09:35
that's great, it fits and
you have happy enzyme.

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Well, how do you create a hole
-- nature supposedly abhors

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a vacuum, so how do
you have this hole?

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Well, what nature did was it
put the smallest amino acids

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there are, glycines, right
along here to make room

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00:09:51 --> 00:09:53
for this to fit together.
 

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So, what was happening in this
genetic case is that the switch

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of a glycine to an argenine
did this to the hole.

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00:10:02 --> 00:10:07
So, what people were trying to
do is treat these patients with

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00:10:07 --> 00:10:10
extra B12, you give them
injections of it, you have a

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00:10:10 --> 00:10:14
lot more B12, and you want to
the B12 to bind and get the

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00:10:14 --> 00:10:16
activity back going again.
 

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But really, no matter how much
vitamin B12 you added to it, it

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wasn't really going to do much
good, it just wasn't

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going to fit.
 

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00:10:24 --> 00:10:27
So instead, there was a
suggestion that instead of

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giving the whole vitamin,
perhaps you could just give a

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00:10:30 --> 00:10:33
truncated version of the
vitamin, which was commercially

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available, and that that might
bind and restore activity

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00:10:39 --> 00:10:41
of the protein.
 

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So this is what was suggested
once you knew what the

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00:10:44 --> 00:10:47
problem actually was.
 

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00:10:47 --> 00:10:51
So this is just an example of a
genetic disease, and it's not

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that common, but as I said, it
is tested for here

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in Massachusetts.
 

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00:10:55 --> 00:11:00
And how knowing something about
what the problem is can lead

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00:11:00 --> 00:11:02
to a potential solution.
 

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00:11:02 --> 00:11:06
But for most of you, your blood
is doing just fine, you don't

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00:11:06 --> 00:11:09
have a genetic condition which
is pumping too much acid into

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00:11:09 --> 00:11:13
your bloodstream, so that
nature's buffering capacity

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00:11:13 --> 00:11:18
is working quite well.
 

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00:11:18 --> 00:11:23
So buffers are important.
 

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00:11:23 --> 00:11:29
It's great to have my important
statements emphasized up there.

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We've been rehearsing at home.
 

198
00:11:32 --> 00:11:33
OK.
 

199
00:11:33 --> 00:11:36
So let's do a sample
buffer problem.

200
00:11:36 --> 00:11:41
All right, suppose we have an
acid, 0.1 moles of an acid, and

201
00:11:41 --> 00:11:46
0.5 moles of its conjugate base
added in the form of a salt.

202
00:11:46 --> 00:11:50
And so they're put into water
and diluted to 1.0 liter.

203
00:11:50 --> 00:11:54
And you're given information
about the k a of the acid, 1 .

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00:11:54 --> 00:11:58
77 times 10 to the minus
4, and you're asked

205
00:11:58 --> 00:12:00
to calculate the p h.
 

206
00:12:00 --> 00:12:02
So what do you do?
 

207
00:12:02 --> 00:12:05
All right, first, it's always
good to write the equation

208
00:12:05 --> 00:12:06
that you're talking about.
 

209
00:12:06 --> 00:12:11
And so you have acid in water
forming hydronium ions

210
00:12:11 --> 00:12:14
and the conjugate base.
 

211
00:12:14 --> 00:12:19
So here, the acid is giving up
a proton to the water, forming

212
00:12:19 --> 00:12:24
h 3 o plus and it's
conjugate base.

213
00:12:24 --> 00:12:28
Now we want to think about
what happens at equilibrium.

214
00:12:28 --> 00:12:30
So we know how many moles
we have and we know what

215
00:12:30 --> 00:12:32
the total volume is.
 

216
00:12:32 --> 00:12:35
And in this table we should be
using molarity, but the math

217
00:12:35 --> 00:12:40
here is pretty easy because we
have 1 mole and 1 liter or 1,

218
00:12:40 --> 00:12:46
and 0.5 moles in 1 liter
or 0.5 for molarity.

219
00:12:46 --> 00:12:50
And now at equilibrium what's
going to happen, well, when

220
00:12:50 --> 00:12:54
this equilibrates some of this
will go away and more of these

221
00:12:54 --> 00:12:56
are going to be formed.
 

222
00:12:56 --> 00:13:01
So we have minus x over here,
plus x and plus x over there,

223
00:13:01 --> 00:13:05
and we add it all together you
get 1 minus x plus

224
00:13:05 --> 00:13:08
x and 0.5 plus x.
 

225
00:13:08 --> 00:13:10
So the important thing to
remember is in buffering

226
00:13:10 --> 00:13:14
problems, you have the
acid and the conjugate.

227
00:13:14 --> 00:13:17
So we're used to seeing these
tables where we only have

228
00:13:17 --> 00:13:21
something over here and it's
all zero on the other side.

229
00:13:21 --> 00:13:24
But in a buffering problem,
you have things on both sides

230
00:13:24 --> 00:13:25
from the very beginning.
 

231
00:13:25 --> 00:13:28
And that's a really important
point, and that alone, if you

232
00:13:28 --> 00:13:31
can remember that, will get you
a long way through this unit.

233
00:13:31 --> 00:13:35
So you remember that in a
buffer you got to have both an

234
00:13:35 --> 00:13:39
acid in its conjugate base or a
base in its conjugate acid.

235
00:13:39 --> 00:13:42
So here we're talking about an
acid in water, it's an acidic

236
00:13:42 --> 00:13:47
buffer, and so we can use our k
a value -- we want to know what

237
00:13:47 --> 00:13:49
the concentrations are
at equilibrium to

238
00:13:49 --> 00:13:51
calculate our p h.
 

239
00:13:51 --> 00:13:57
So we can use our k a and we
can set it up, so we have

240
00:13:57 --> 00:14:01
on the top products over
reactants, and we we're not

241
00:14:01 --> 00:14:05
including water in this because
it's dilute in solution and so

242
00:14:05 --> 00:14:06
it's not changing very much.
 

243
00:14:06 --> 00:14:10
And now we can plug in our
values, so we have 0.5

244
00:14:10 --> 00:14:15
plus x, x over 1 minus x.
 

245
00:14:15 --> 00:14:19
All right, now this is just
written again from the last

246
00:14:19 --> 00:14:24
slide, we can try an
approximation, which is it

247
00:14:24 --> 00:14:28
that x is small compared to
1 molar or 0.5 molar.

248
00:14:28 --> 00:14:32
And so we can get rid of the
plus x and the minus x down

249
00:14:32 --> 00:14:37
here and just have one x term,
which makes it simpler to

250
00:14:37 --> 00:14:39
solve, but we'll have to
go back and check that

251
00:14:39 --> 00:14:42
approximation in a few minutes.
 

252
00:14:42 --> 00:14:46
So making that approximation,
we can calculate x as 3 .

253
00:14:46 --> 00:14:51
54 times 10 to the minus 4
molar, and now we need to

254
00:14:51 --> 00:14:52
check that assumption.
 

255
00:14:52 --> 00:14:56
Well, you can probably guess
it's going to be OK, because

256
00:14:56 --> 00:15:01
something times 10 to the minus
4 compared to 0.5 and 1, that's

257
00:15:01 --> 00:15:03
probably going to be
OK that it's small.

258
00:15:03 --> 00:15:07
But our assumption here is
that it needs to be less

259
00:15:07 --> 00:15:11
than 5%, and here it's 0.1%.
 

260
00:15:11 --> 00:15:14
So you could just take this
value of x, divide it by the

261
00:15:14 --> 00:15:20
smaller of the two, 0.5 ,
times 100, and calculate

262
00:15:20 --> 00:15:21
the percent ionization.
 

263
00:15:21 --> 00:15:26
If it's less than 5% you're OK,
if it's more you need to use

264
00:15:26 --> 00:15:30
the quadratic equation
to solve the problem.

265
00:15:30 --> 00:15:35
So x here is our concentration
of hydronium ion, which is

266
00:15:35 --> 00:15:39
great, because then we can
easily calculate the p h.

267
00:15:39 --> 00:15:42
So p h again is minus log
of the hydronium ion

268
00:15:42 --> 00:15:47
concentration, and so
the p h here is 3 .

269
00:15:47 --> 00:15:48
45.
 

270
00:15:48 --> 00:15:51
And again, significant figures,
we'll be talking about

271
00:15:51 --> 00:15:53
this as we go through.
 

272
00:15:53 --> 00:15:55
The volume had two, it was 1 .
 

273
00:15:55 --> 00:16:00
0 liters, and so we're going to
have two significant figures

274
00:16:00 --> 00:16:03
after the decimal point,
because the volume was limiting

275
00:16:03 --> 00:16:08
insignificant figures.
 

276
00:16:08 --> 00:16:14
All right, now let's consider
what happens if some strong

277
00:16:14 --> 00:16:17
acid had been added
in the solution.

278
00:16:17 --> 00:16:20
So the volume is still going to
be 1 liter, because the acid

279
00:16:20 --> 00:16:28
was added in before we went
up to the 1 liter mark.

280
00:16:28 --> 00:16:32
So strong acids, they
go to completion.

281
00:16:32 --> 00:16:37
So, if we added 0.1 moles of
that strong acid, it will react

282
00:16:37 --> 00:16:41
with equal number of moles
of the conjugate base to

283
00:16:41 --> 00:16:44
form the conjugate acid.
 

284
00:16:44 --> 00:16:47
So we can just do
subtractions here.

285
00:16:47 --> 00:16:49
We don't have to worry about
setting up any kind of

286
00:16:49 --> 00:16:52
equilibrium, we assume
it goes completely.

287
00:16:52 --> 00:16:58
So for the conjugate base we
had 0.5 moles before, and it's

288
00:16:58 --> 00:17:03
going to be reacting then with
a strong acid, so 0.1 of those

289
00:17:03 --> 00:17:08
will react, leaving us with 0.4
moles of the conjugate,

290
00:17:08 --> 00:17:10
and that's in 1 liter.
 

291
00:17:10 --> 00:17:13
So now we have 0.4 molar.
 

292
00:17:13 --> 00:17:17
For the acid, we had 1 mole to
begin with, but now we formed

293
00:17:17 --> 00:17:24
more as they reaction has taken
place, and so we have 0.1 more.

294
00:17:24 --> 00:17:25
Now we have 1 .
 

295
00:17:25 --> 00:17:29
1 moles again in 1 liter.
 

296
00:17:29 --> 00:17:33
So now we can do the
same thing again.

297
00:17:33 --> 00:17:36
So after this has happened,
after we have added the strong

298
00:17:36 --> 00:17:40
acid, then a new equilibrium
is going to be reached.

299
00:17:40 --> 00:17:46
And so we can plug in
this table as well.

300
00:17:46 --> 00:17:47
So we have 1 .
 

301
00:17:47 --> 00:17:55
1 now over here, and
0.4 on the other side.

302
00:17:55 --> 00:17:59
As equilibrium is approached,
you lose some of the acid as it

303
00:17:59 --> 00:18:05
reacts with water and ionizes
and you form more of the h 3 o

304
00:18:05 --> 00:18:08
plus and more of the conjugate.
 

305
00:18:08 --> 00:18:09
So we have 1 .
 

306
00:18:09 --> 00:18:14
1 minus x, x, and 0.4 plus x.
 

307
00:18:14 --> 00:18:18
So again, the trick here is
just to remember that there is

308
00:18:18 --> 00:18:21
a reaction with a strong acid
that has been added, and you

309
00:18:21 --> 00:18:25
need to figure out the new
molarities, and then go back

310
00:18:25 --> 00:18:33
to your equilibrium table
with those new molarities.

311
00:18:33 --> 00:18:37
So here we can set up with a
k a again, and do the same

312
00:18:37 --> 00:18:42
problem that we did before.
 

313
00:18:42 --> 00:18:45
And so, we can make the
assumption again that x

314
00:18:45 --> 00:18:49
is small, try it out
and see if it works.

315
00:18:49 --> 00:18:52
X turns out to be 4 .
 

316
00:18:52 --> 00:18:55
87 times 10 to the minus
4 molars -- so again,

317
00:18:55 --> 00:18:57
it's a small number.
 

318
00:18:57 --> 00:19:03
And so, it turns out to be 1
. -- less than 1% of 0.4 .

319
00:19:03 --> 00:19:06
If it's less than 5% of the
smaller number, you don't have

320
00:19:06 --> 00:19:10
to worry about the bigger
number, and so your assumption

321
00:19:10 --> 00:19:13
is OK, you don't need to use
the quadratic equation.

322
00:19:13 --> 00:19:17
And so, then we can
calculate at the end

323
00:19:17 --> 00:19:19
that the p h is now 3 .
 

324
00:19:19 --> 00:19:22
31, again, two significant
figures after the decimal

325
00:19:22 --> 00:19:24
place because of the volume.
 

326
00:19:24 --> 00:19:27
So addition of 0.1 moles
of a strong acid changed

327
00:19:27 --> 00:19:29
our p h from 3 .
 

328
00:19:29 --> 00:19:32
45 to 3 .
 

329
00:19:32 --> 00:19:33
31.
 

330
00:19:33 --> 00:19:34
So we buffered pretty well.
 

331
00:19:34 --> 00:19:38
There was a change in p h, we
added acid and the p h did go

332
00:19:38 --> 00:19:41
down, but not by an
enormous amount.

333
00:19:41 --> 00:19:45
And so, that's because this
turned out to be a pretty

334
00:19:45 --> 00:19:48
decent buffer here, so we
didn't have a really

335
00:19:48 --> 00:19:54
huge effect.
 

336
00:19:54 --> 00:20:07
So, those are -- that's an
example of a buffer problem.

337
00:20:07 --> 00:20:10
So, let's think for a minute
about designing a buffer.

338
00:20:10 --> 00:20:17
Suppose you wanted to create a
buffer at a particular p h,

339
00:20:17 --> 00:20:19
what do you need
to think about?

340
00:20:19 --> 00:20:22
Well, you need to think about
the ratio of your acid to the

341
00:20:22 --> 00:20:26
conjugate, and you need to
think about the p k a of the

342
00:20:26 --> 00:20:32
acid, and the p h
that's desired.

343
00:20:32 --> 00:20:38
So here is a generic equation
for an acid in water, so we

344
00:20:38 --> 00:20:41
have our acid, h a, in water
forming hydronium ions

345
00:20:41 --> 00:20:44
and our conjugate base.
 

346
00:20:44 --> 00:20:47
And now we're going to do a
little derivation to come up

347
00:20:47 --> 00:20:50
with an equation that would be
useful to you in thinking about

348
00:20:50 --> 00:20:53
buffers and designing buffers.
 

349
00:20:53 --> 00:20:57
So, we can write a
generic term for k a.

350
00:20:57 --> 00:21:01
K a equals products over
reactants, and so we have our

351
00:21:01 --> 00:21:05
hydronium ions are conjugate
base over our acid.

352
00:21:05 --> 00:21:09
And now we can take this
term and rearrange it.

353
00:21:09 --> 00:21:14
So we can pull a hydronium ion
concentration over to one side.

354
00:21:14 --> 00:21:19
And now we're going to take the
logs of both sides, so we get

355
00:21:19 --> 00:21:23
the log of hydronium ion
concentration, the log of k

356
00:21:23 --> 00:21:27
a, and the log of h a
concentration over a minus

357
00:21:27 --> 00:21:31
concentration, and now we're
going to multiply everything by

358
00:21:31 --> 00:21:36
negative sign, so we have
minus logs of things.

359
00:21:36 --> 00:21:40
And I'm just going to move this
now up to the top of the

360
00:21:40 --> 00:21:43
screen, and we have
this equation.

361
00:21:43 --> 00:21:49
So what is minus log of
hydronium ion concentration?

362
00:21:49 --> 00:21:49
P h.
 

363
00:21:49 --> 00:21:53
What's minus log of k a?
 

364
00:21:53 --> 00:21:55
P k a.
 

365
00:21:55 --> 00:22:00
So, we have p h equals p
k a minus the log of a

366
00:22:00 --> 00:22:03
concentration of your acid
over the concentration

367
00:22:03 --> 00:22:05
of your conjugate base.
 

368
00:22:05 --> 00:22:08
And these are equilibrium
concentrations for that,

369
00:22:08 --> 00:22:10
because remember, we derived
it from our equilibrium

370
00:22:10 --> 00:22:13
expression for k a.
 

371
00:22:13 --> 00:22:16
But a lot of times when you're
working buffer problems, you

372
00:22:16 --> 00:22:19
know the concentrations that
you added of these things, not

373
00:22:19 --> 00:22:23
necessarily the concentrations
of equilibrium.

374
00:22:23 --> 00:22:27
And if we rewrite this
expression in terms of the

375
00:22:27 --> 00:22:31
initial concentrations or
original or o concentrations,

376
00:22:31 --> 00:22:36
then we have p h is
approximately equal to p k a

377
00:22:36 --> 00:22:40
minus log of the initial
concentrations of your acid

378
00:22:40 --> 00:22:41
over your conjugate base.
 

379
00:22:41 --> 00:22:45
And this is known as a
Henderson Hasselbalch equation.

380
00:22:45 --> 00:22:48
And people love the Henderson
Hasselbalch equation, and it

381
00:22:48 --> 00:22:51
can be incredibly useful to you
in solving these problems.

382
00:22:51 --> 00:22:55
But I'm going to emphasize when
it's OK to use it and when it's

383
00:22:55 --> 00:22:58
not OK to use it, because
people try to apply it to

384
00:22:58 --> 00:23:02
everything, and it's for buffer
problems, and people apply it

385
00:23:02 --> 00:23:05
to all sorts of things that
are not buffer problems.

386
00:23:05 --> 00:23:07
So I want to make sure that
it's clear when you can use it

387
00:23:07 --> 00:23:10
and when you can't use it.
 

388
00:23:10 --> 00:23:13
And it's fine to use it and
you should use it when

389
00:23:13 --> 00:23:17
it's acceptable, but
not at other times.

390
00:23:17 --> 00:23:19
It's one of the few equations
in acid base, so people

391
00:23:19 --> 00:23:22
get very excited by it.
 

392
00:23:22 --> 00:23:26
All right, so remember that
it's really equal when they're

393
00:23:26 --> 00:23:29
at equilibrium concentrations,
and this is just an

394
00:23:29 --> 00:23:32
approximation that we're saying
that those are initial

395
00:23:32 --> 00:23:34
concentrations.
 

396
00:23:34 --> 00:23:39
So when is it OK to say well,
the equilibrium concentration

397
00:23:39 --> 00:23:43
is more or less the same as
the initial concentration.

398
00:23:43 --> 00:23:47
And that's true when x, which
is in this particular example

399
00:23:47 --> 00:23:50
your hydronium ion
concentration, is small

400
00:23:50 --> 00:23:54
compared to your initial
concentration of the acid and

401
00:23:54 --> 00:23:57
the conjugate base that you
added to the buffer.

402
00:23:57 --> 00:24:01
So, and when x is small, then
pretty much, the equilibrium

403
00:24:01 --> 00:24:04
concentration equals the
initial concentration.

404
00:24:04 --> 00:24:07
So there's not really very
much change if x is a

405
00:24:07 --> 00:24:10
really small number.
 

406
00:24:10 --> 00:24:13
So, a lot of the
time this is true.

407
00:24:13 --> 00:24:15
Remember, we're making buffers
from weak acids and weak

408
00:24:15 --> 00:24:19
conjugate bases, and the
definition of a weak acid is

409
00:24:19 --> 00:24:22
that it only loses a tiny
fraction of its protons.

410
00:24:22 --> 00:24:26
It only ionizes
slightly in water.

411
00:24:26 --> 00:24:29
And a weak base typically only
accepts a fraction of the

412
00:24:29 --> 00:24:31
protons it can accept.
 

413
00:24:31 --> 00:24:36
It's a weak base, so it's
not doing a whole lot

414
00:24:36 --> 00:24:37
of chemistry there.
 

415
00:24:37 --> 00:24:41
So this approximation is
good a lot of the time.

416
00:24:41 --> 00:24:44
So just you know when you can
and can not use it, we're going

417
00:24:44 --> 00:24:47
to give a rule and say it's
the same rule as before.

418
00:24:47 --> 00:24:52
When we say when x is small
or your hydronium ion

419
00:24:52 --> 00:24:55
concentration is small compared
to the acid or the conjugate

420
00:24:55 --> 00:24:58
base, when it's less than 5%,
that's what we're going to call

421
00:24:58 --> 00:25:03
small, so the same rule
we've been using all along.

422
00:25:03 --> 00:25:06
When x is small,
this approximation

423
00:25:06 --> 00:25:08
works pretty well.
 

424
00:25:08 --> 00:25:11
And you can plug in your
concentrations right there --

425
00:25:11 --> 00:25:14
your initial concentrations
not your equilibrium

426
00:25:14 --> 00:25:19
concentrations.
 

427
00:25:19 --> 00:25:22
All right, so let's use
Henderson Hasselbalch and

428
00:25:22 --> 00:25:26
design a buffer system
if we want a p h of 4 .

429
00:25:26 --> 00:25:27
6.
 

430
00:25:27 --> 00:25:32
And a good rule to remember is
that a buffering solution is

431
00:25:32 --> 00:25:38
most effective when it's plus
or minus 1 away from the p k a.

432
00:25:38 --> 00:25:40
So that's a good thing
too I keep in mind

433
00:25:40 --> 00:25:43
for research as well.
 

434
00:25:43 --> 00:25:46
If you're really far away from
the p k a, it's not going to be

435
00:25:46 --> 00:25:49
a good buffering system, and
this is a mistake the people

436
00:25:49 --> 00:25:52
make in the lab sometimes, so
you may run into that in some

437
00:25:52 --> 00:25:55
of your laboratory courses.
 

438
00:25:55 --> 00:25:59
So we want a buffer
about p h 4 .

439
00:25:59 --> 00:26:00
6.
 

440
00:26:00 --> 00:26:03
So we can look at our
ionization constants of acid,

441
00:26:03 --> 00:26:05
and of course, we're always
doing everything at room

442
00:26:05 --> 00:26:09
temperature in this unit, and
so we can look at what p k a's

443
00:26:09 --> 00:26:14
are of some weak acids, and
acetate, acetate is an easy

444
00:26:14 --> 00:26:19
buffer to get your hands on,
acetate is quite available, and

445
00:26:19 --> 00:26:21
it has a good p k a 4 .
 

446
00:26:21 --> 00:26:22
75.
 

447
00:26:22 --> 00:26:27
So that's great, we can use
that in designing our buffer.

448
00:26:27 --> 00:26:32
So, acetic acid has the right
p k a, and we're all set,

449
00:26:32 --> 00:26:34
we can prepare a buffer.
 

450
00:26:34 --> 00:26:38
But we'll need to add the acid
and the conjugate, and we need

451
00:26:38 --> 00:26:42
to know how much acid and how
much of the conjugate to add.

452
00:26:42 --> 00:26:46
So we can use Henderson
Hasselbalch equation for this.

453
00:26:46 --> 00:26:49
We're creating a buffer,
and so this is an equation

454
00:26:49 --> 00:26:51
we can use for buffers.
 

455
00:26:51 --> 00:26:55
We know what p h we want, we
know the p k a of the thing

456
00:26:55 --> 00:26:58
we're going to use, and we want
to figure out how much of the

457
00:26:58 --> 00:27:01
acid to use and how much
of the conjugate to use.

458
00:27:01 --> 00:27:05
We need to know the ratio
of one to the other to set

459
00:27:05 --> 00:27:07
up our buffer ideally.
 

460
00:27:07 --> 00:27:12
So we can plug our numbers
in and calculate that.

461
00:27:12 --> 00:27:15
So, we can rearrange this if
you want so that the unknown is

462
00:27:15 --> 00:27:18
on one side and we have to p k
a and the p h on

463
00:27:18 --> 00:27:21
the other side.
 

464
00:27:21 --> 00:27:26
And so, we have here our p k a
minus our p h and we get 0 .

465
00:27:26 --> 00:27:27
15.
 

466
00:27:27 --> 00:27:32
Now we need to inverse log and
come up with the ratio, and

467
00:27:32 --> 00:27:34
the ratio that we want is 1 .
 

468
00:27:34 --> 00:27:35
4.
 

469
00:27:35 --> 00:27:38
So we want to have a ratio
from our acid to the

470
00:27:38 --> 00:27:40
conjugate of 1 .
 

471
00:27:40 --> 00:27:40
4.
 

472
00:27:40 --> 00:27:43
Well how much exactly are we
going to add them, we know

473
00:27:43 --> 00:27:46
the ratio now, how much
are we going to add?

474
00:27:46 --> 00:27:48
Well, you could use 1 .
 

475
00:27:48 --> 00:27:53
4 molar and 1 molar, for
example, that would have

476
00:27:53 --> 00:27:54
the correct ratio.
 

477
00:27:54 --> 00:27:59
But how do you know if
that's going to be good?

478
00:27:59 --> 00:28:02
Well, the ratio is more
important than the exact

479
00:28:02 --> 00:28:07
amounts, however, the amounts
get to this issue of sort of

480
00:28:07 --> 00:28:10
the capacity of the buffer --
how resistant it is to

481
00:28:10 --> 00:28:12
changes in the p h.
 

482
00:28:12 --> 00:28:15
So if you use too low
concentrations of both, it

483
00:28:15 --> 00:28:17
won't be all that resistant.
 

484
00:28:17 --> 00:28:22
So if you need to have a better
buffering capacity, the higher

485
00:28:22 --> 00:28:24
concentrations that
you should use.

486
00:28:24 --> 00:28:27
And you can use Henderson
Hasselbalch then to calculate

487
00:28:27 --> 00:28:31
what sort of the minimum amount
you use, the minimum amount

488
00:28:31 --> 00:28:35
to have that 5% rule work.
 

489
00:28:35 --> 00:28:39
So in the example I told you
about this condition, the

490
00:28:39 --> 00:28:43
buffer in the blood was pretty
good, but the acid with so much

491
00:28:43 --> 00:28:46
that it overwhelmed
buffering capacity.

492
00:28:46 --> 00:28:48
So if there had been larger
concentrations of the buffer,

493
00:28:48 --> 00:28:52
that would have certainly
helped in that case.

494
00:28:52 --> 00:28:55
So the higher the
concentrations you use, the

495
00:28:55 --> 00:28:57
more resistant to change.
 

496
00:28:57 --> 00:29:02
And that's the idea of
buffering capacity.

497
00:29:02 --> 00:29:04
So, and if you use too low
concentrations, then your

498
00:29:04 --> 00:29:08
Henderson Hasselbalch equation
won't even be valid.

499
00:29:08 --> 00:29:12
So we can go back and calculate
what is sort of the minimum

500
00:29:12 --> 00:29:16
concentrations we need to use
for Henderson Hasselbalch to be

501
00:29:16 --> 00:29:18
valid to meet that 5% rule.
 

502
00:29:18 --> 00:29:20
So for a p h of 4 .
 

503
00:29:20 --> 00:29:24
6, a hydronium ion
concentration is a 2 .

504
00:29:24 --> 00:29:26
5 times 10 to the minus 5.
 

505
00:29:26 --> 00:29:31
And so then if we work our
equation backwards, we want to

506
00:29:31 --> 00:29:35
be less than 5%, so we have
this over the concentration of

507
00:29:35 --> 00:29:39
either one of those times
100% gives you 5%.

508
00:29:39 --> 00:29:42
So the concentrations need to
be greater than 5 times 10 to

509
00:29:42 --> 00:29:46
the minus 4 moles or you
won't meet this 5% rule.

510
00:29:46 --> 00:29:48
So at least to be greater
than that and they need

511
00:29:48 --> 00:29:50
to be in the right ratio.
 

512
00:29:50 --> 00:29:55
Other than that you're somewhat
free to decide what you're

513
00:29:55 --> 00:29:56
going to do with that,
and there may be other

514
00:29:56 --> 00:30:01
considerations involved if
you're working with a certain

515
00:30:01 --> 00:30:04
concentration of protein, you
might not want your buffer

516
00:30:04 --> 00:30:07
concentration to be too high,
it might start interfering with

517
00:30:07 --> 00:30:10
enzyme assay, for example.
 

518
00:30:10 --> 00:30:14
And often buffers are somewhere
around 100 millimolar, that's a

519
00:30:14 --> 00:30:19
pretty common concentration
for buffers.

520
00:30:19 --> 00:30:24
All right, so this slide
we're doing really well.

521
00:30:24 --> 00:30:28
So we talked about weak acids
in water, weak bases in water,

522
00:30:28 --> 00:30:31
and I'm trying to convince you
that these are the same as

523
00:30:31 --> 00:30:33
the salt and water problems.
 

524
00:30:33 --> 00:30:35
So we're going to talk
much more about salt and

525
00:30:35 --> 00:30:36
water in a few minutes.
 

526
00:30:36 --> 00:30:39
And I told you about
buffers, so we only have

527
00:30:39 --> 00:30:40
2 more things to go.
 

528
00:30:40 --> 00:30:43
We need to talk about strong
acids in water, and strong

529
00:30:43 --> 00:30:47
bases in water, and then you'll
have all of the 5 types of

530
00:30:47 --> 00:30:50
problems to do the problem-set.
 

531
00:30:50 --> 00:30:55
And as I mentioned last time,
it looks deceivingly short,

532
00:30:55 --> 00:30:58
the number of questions on
them, but the titration

533
00:30:58 --> 00:31:00
problems are long.
 

534
00:31:00 --> 00:31:04
So don't leave those to the
last minute or there will

535
00:31:04 --> 00:31:06
not be a whole lot
of sleep involved.

536
00:31:06 --> 00:31:10
So just a word of
warning from the past.

537
00:31:10 --> 00:31:12
Everyone looks and go, "Oh,
this is an easy problem-set,

538
00:31:12 --> 00:31:19
there aren't many problems."
So, titration problems

539
00:31:19 --> 00:31:20
do take a lot of time.
 

540
00:31:20 --> 00:31:24
So let's talk about titration,
so that as soon as class is

541
00:31:24 --> 00:31:27
over today, you can go work on
those questions on

542
00:31:27 --> 00:31:28
the problem-set.
 

543
00:31:28 --> 00:31:29
All right.
 

544
00:31:29 --> 00:31:31
So acid base titrations.
 

545
00:31:31 --> 00:31:37
How many of you in high school
titrated an acid with a base?

546
00:31:37 --> 00:31:42
Large fraction of you, OK.
 

547
00:31:42 --> 00:31:47
So usually what titration
problems are meant to do, that

548
00:31:47 --> 00:31:52
you have something known about
an acid or a base, base say of

549
00:31:52 --> 00:31:58
known concentration, and and
acid of unknown concentration

550
00:31:58 --> 00:32:01
or maybe unknown molecular
weight, and you're titrating

551
00:32:01 --> 00:32:04
them out, titrating
them together to find

552
00:32:04 --> 00:32:06
missing information.
 

553
00:32:06 --> 00:32:08
So you can determine
concentration, sometimes

554
00:32:08 --> 00:32:11
you can determine a
molecular weight.

555
00:32:11 --> 00:32:14
All right, so here's what a
plot looks like and some

556
00:32:14 --> 00:32:19
of the key terms involved
in acid base titrations.

557
00:32:19 --> 00:32:24
So you have a p h on one side,
and then volume of either base

558
00:32:24 --> 00:32:26
or acid added on
the other side.

559
00:32:26 --> 00:32:28
So p h versus a volume.
 

560
00:32:28 --> 00:32:33
And the actual experiment here,
you have a base that you're

561
00:32:33 --> 00:32:37
dripping, one drip at a
time supposedly into

562
00:32:37 --> 00:32:40
your strong acid.
 

563
00:32:40 --> 00:32:44
And then you're looking to
figure out when you're going to

564
00:32:44 --> 00:32:48
reach the either equivalence
point or the end point, these

565
00:32:48 --> 00:32:49
are basically the same terms.
 

566
00:32:49 --> 00:32:53
It's also called the
stoichiometric point, and often

567
00:32:53 --> 00:32:58
equivalence point is used as a
more theoretical amount of

568
00:32:58 --> 00:33:01
volume is added, where
as end point is the

569
00:33:01 --> 00:33:03
experimentally measured.
 

570
00:33:03 --> 00:33:06
So we're going to be talking a
lot about equivalence points,

571
00:33:06 --> 00:33:09
because we have no lab
associated with this course, so

572
00:33:09 --> 00:33:12
everything will be theorectical
in terms of these, but

573
00:33:12 --> 00:33:15
they should be the same.
 

574
00:33:15 --> 00:33:19
So, many times you will measure
p h with the p h meter --

575
00:33:19 --> 00:33:21
this just shows a p h meter.
 

576
00:33:21 --> 00:33:25
And for those of you who have
done these experiments,

577
00:33:25 --> 00:33:26
this should look familiar.
 

578
00:33:26 --> 00:33:30
And for those of you who
haven't, often what you're

579
00:33:30 --> 00:33:32
doing when you're dripping
something in, you're waiting

580
00:33:32 --> 00:33:36
for an indicator dye to change
color as an indication that you

581
00:33:36 --> 00:33:39
have reached the end point, and
often you're adding, and it's

582
00:33:39 --> 00:33:44
clear, clear, clear for what
seems to be forever, and

583
00:33:44 --> 00:33:46
just as you get incredibly
impatient, you start adding it

584
00:33:46 --> 00:33:49
faster and you go all the
way to this dark color.

585
00:33:49 --> 00:33:55
And what you want is this very,
very, very light changed color

586
00:33:55 --> 00:33:57
that indicates the end point.
 

587
00:33:57 --> 00:34:02
So you usually go too slow and
then go too fast and then have

588
00:34:02 --> 00:34:03
to do the experiment
over again.

589
00:34:03 --> 00:34:06
Those of you who know how to
calculate theoretical values

590
00:34:06 --> 00:34:09
could sit down and do a
calculation first and then go

591
00:34:09 --> 00:34:13
really fast right until you get
around this point and add it

592
00:34:13 --> 00:34:16
really slowly, and be
done with lab sooner.

593
00:34:16 --> 00:34:20
So, we'll talk about how you do
the theoretical value, which

594
00:34:20 --> 00:34:24
could save you a lot of time if
you ever run across these

595
00:34:24 --> 00:34:30
labs in any future class.
 

596
00:34:30 --> 00:34:34
So, here are the two curves
that you would see for either a

597
00:34:34 --> 00:34:36
strong acid with a strong base,
or strong base with

598
00:34:36 --> 00:34:38
a strong acid.
 

599
00:34:38 --> 00:34:41
So if you're titrating a strong
acid with a strong base, you're

600
00:34:41 --> 00:34:45
going to start down being
very acidic, because all you

601
00:34:45 --> 00:34:47
have is your strong acid.
 

602
00:34:47 --> 00:34:51
And then as you add in
base, the p h will go up.

603
00:34:51 --> 00:34:55
You'll reach this point -- s is
for the stoichiometric point,

604
00:34:55 --> 00:34:58
we also call this, again, the
equivalence point, and then

605
00:34:58 --> 00:35:01
it'll continue to go up and
then start to level off

606
00:35:01 --> 00:35:03
a bit more up here.
 

607
00:35:03 --> 00:35:06
If you're going the other
direction, you're starting with

608
00:35:06 --> 00:35:10
strong base, your p h is going
to be high, and the p h will

609
00:35:10 --> 00:35:12
decrease as you add the acid.
 

610
00:35:12 --> 00:35:14
And then you get to the
stoichiometric or equivalence

611
00:35:14 --> 00:35:20
point in the middle
of this curve here.

612
00:35:20 --> 00:35:23
And then as you go down
farther, more acidic, it

613
00:35:23 --> 00:35:24
starts to level off.
 

614
00:35:24 --> 00:35:29
So those are what the
titration curves look like.

615
00:35:29 --> 00:35:32
So let's do an example.
 

616
00:35:32 --> 00:35:35
We're going to have a strong
base being titrated with

617
00:35:35 --> 00:35:38
a strong acid first.
 

618
00:35:38 --> 00:35:47
And our strong base is n a o h,
and our strong acid is h c l.

619
00:35:47 --> 00:35:56
So let's calculate the p h
at the equivalence point.

620
00:35:56 --> 00:36:01
Well, let's just calculate the
p h at 5 mils -- I don't know

621
00:36:01 --> 00:36:03
whether it's equivalence
point, actually.

622
00:36:03 --> 00:36:08
5 mils of this have been
added to the amount of base.

623
00:36:08 --> 00:36:12
So we have 25 mils of our
strong base at 0.15 molar

624
00:36:12 --> 00:36:21
concentration, and we're adding
5 mils of our 0.34 molar acid.

625
00:36:21 --> 00:36:25
So what we want to do is figure
out how many moles of base

626
00:36:25 --> 00:36:28
we had to start with.
 

627
00:36:28 --> 00:36:33
So how many moles of o h minus
-- again, n a o h is a strong

628
00:36:33 --> 00:36:37
base, so however much n a o h
you add, is the amount of o h

629
00:36:37 --> 00:36:42
minus you get, and so we just
put in the number of liters

630
00:36:42 --> 00:36:44
times the concentration
to get moles.

631
00:36:44 --> 00:36:48
We don't need any
equilibrium table here.

632
00:36:48 --> 00:36:52
So then we want to figure out
the amount of acid added when

633
00:36:52 --> 00:36:55
you're adding 5 mils of it.
 

634
00:36:55 --> 00:36:59
So again, h c l is a strong
acid, so the amount of h c l

635
00:36:59 --> 00:37:05
added equals the amount of h 3
o plus, that's formed, goes to

636
00:37:05 --> 00:37:09
completion, and so we added 5
mils, the concentration was

637
00:37:09 --> 00:37:12
0.34 molar, so we have 1 .
 

638
00:37:12 --> 00:37:17
7 times 10 to the
minus 3 moles.

639
00:37:17 --> 00:37:21
So now, the strong acid is
going to react with the strong

640
00:37:21 --> 00:37:25
base, and we want to figure out
how much of the base is left,

641
00:37:25 --> 00:37:28
how much hydroxide ion is left
after it reacts with

642
00:37:28 --> 00:37:31
the hydronium ions.
 

643
00:37:31 --> 00:37:35
So it'll react 1:1,
so we had 6 .

644
00:37:35 --> 00:37:41
25 times 10 to the minus 3
moles, and we added 1 .

645
00:37:41 --> 00:37:45
7 times 10 to the minus
3 moles of the acid, so

646
00:37:45 --> 00:37:47
we're going to have 4 .
 

647
00:37:47 --> 00:37:53
55 times 10 the
minus 3 moles left.

648
00:37:53 --> 00:37:59
And now we can calculate the
molarity here, so we have the

649
00:37:59 --> 00:38:03
number of moles and the new
volume, so we had 25 mils

650
00:38:03 --> 00:38:04
to the base to begin with.
 

651
00:38:04 --> 00:38:09
We've added 5 mils to the acid,
so our new volume is 30, and so

652
00:38:09 --> 00:38:12
we have our new concentration.
 

653
00:38:12 --> 00:38:15
And now we can calculate p h.
 

654
00:38:15 --> 00:38:19
First we'll calculate p o
h, and then use 14 minus

655
00:38:19 --> 00:38:23
to get the new p h.
 

656
00:38:23 --> 00:38:29
So if you were making your own
little titration curve, we

657
00:38:29 --> 00:38:34
could have volume of acid on
one side and p h on the other

658
00:38:34 --> 00:38:39
side, and you have
a point of 13 .

659
00:38:39 --> 00:38:40
18.
 

660
00:38:40 --> 00:38:49
That's how much we have after
you have 5 mils added.

661
00:38:49 --> 00:38:52
So we didn't, in this curve, we
have not measured the zero

662
00:38:52 --> 00:39:00
point, but we know what one
point after 5 mils of the

663
00:39:00 --> 00:39:02
acid has been added.
 

664
00:39:02 --> 00:39:06
So now let's go right to the
equivalence point -- so the way

665
00:39:06 --> 00:39:09
I draw it should indicate that
was not an equivalence point.

666
00:39:09 --> 00:39:13
So we want to figure out how
much we need to add of the acid

667
00:39:13 --> 00:39:15
to reach the equivalence point.
 

668
00:39:15 --> 00:39:18
So the equivalence point or the
stoichiometric point means that

669
00:39:18 --> 00:39:23
you add the same amount of
moles of your titrate as you

670
00:39:23 --> 00:39:26
had, and so they're going
to be equal to each other.

671
00:39:26 --> 00:39:30
So if you're adding acid to a
strong base, you've added

672
00:39:30 --> 00:39:33
enough acid that you now have
equal number of moles to the

673
00:39:33 --> 00:39:36
number of moles of base that
you had to begin with.

674
00:39:36 --> 00:39:37
So we know we have 6 .
 

675
00:39:37 --> 00:39:42
25 times 10 to the minus 3
moles of the base were present.

676
00:39:42 --> 00:39:46
So at the equivalence point,
you need to have that exact

677
00:39:46 --> 00:39:49
same number of moles of acid.
 

678
00:39:49 --> 00:39:53
So we know the concentration of
the acid we have and the number

679
00:39:53 --> 00:39:57
of moles we need, and so we can
calculate the volume,

680
00:39:57 --> 00:39:59
which is 18 .
 

681
00:39:59 --> 00:40:04
4 millimeters, or
0.0184 liters.

682
00:40:04 --> 00:40:12
So what would the p h then be
of the equivalence point?

683
00:40:12 --> 00:40:15
We're titrating a strong
acid with a strong base,

684
00:40:15 --> 00:40:19
what should the p h be?
 

685
00:40:19 --> 00:40:22
The p h should be 7.
 

686
00:40:22 --> 00:40:27
So somewhere around here we're
going to be at 18, so you can

687
00:40:27 --> 00:40:36
just try to draw a curve, and
here we have a p h of 7 when

688
00:40:36 --> 00:40:39
we've gone to about 18 .
 

689
00:40:39 --> 00:40:44
4 milliliters added.
 

690
00:40:44 --> 00:40:47
So when you titrate a strong
acid with a strong base, you

691
00:40:47 --> 00:40:49
form a salt that's neutral.
 

692
00:40:49 --> 00:40:52
Because the conjugates of a
strong acid and a strong base

693
00:40:52 --> 00:40:53
are not basic or acidic.
 

694
00:40:53 --> 00:40:56
They're ineffective as acids
or bases, so you're going

695
00:40:56 --> 00:40:58
to get a neutral salt.
 

696
00:40:58 --> 00:41:01
So if you're doing a problem
with a strong acid and strong

697
00:41:01 --> 00:41:04
base, it's not a lot of
calculations that need to be

698
00:41:04 --> 00:41:06
done at this point, you just
need to recognize that your

699
00:41:06 --> 00:41:09
salt should be neutral.
 

700
00:41:09 --> 00:41:12
So now, we can talk about what
happens if you add an extra

701
00:41:12 --> 00:41:17
milliliter of your acid, after
you've reached that equivalence

702
00:41:17 --> 00:41:20
point, which is something that
people probably have done in

703
00:41:20 --> 00:41:24
doing these titrations not
meaning to, gone well beyond

704
00:41:24 --> 00:41:27
the equivalence point.
 

705
00:41:27 --> 00:41:31
So first we can find out the
number of moles of the strong

706
00:41:31 --> 00:41:36
acid that we've added extra, so
we added one mil of the strong

707
00:41:36 --> 00:41:39
acid, and so again, it
goes to completion.

708
00:41:39 --> 00:41:43
The concentration, the acid,
is 0.34 molar times our 1

709
00:41:43 --> 00:41:45
mil, so we've added 3 .
 

710
00:41:45 --> 00:41:52
4 times 10 to the minus 4 extra
moles of our acid, and so let's

711
00:41:52 --> 00:41:55
calculate then what the
molarity is that we have

712
00:41:55 --> 00:42:03
added that was extra.
 

713
00:42:03 --> 00:42:05
And even if you don't
have a calculator,

714
00:42:05 --> 00:42:41
we've helped you out.
 

715
00:42:41 --> 00:43:05
Let's just take
10 more seconds.

716
00:43:05 --> 00:43:10
I think this is a
first, isn't it?

717
00:43:10 --> 00:43:15
Did I mention it's a good idea
to start the problem-set early.

718
00:43:15 --> 00:43:20
So the trick here was
about the volume.

719
00:43:20 --> 00:43:25
So we had 25 mils to begin
with, then we added 18 .

720
00:43:25 --> 00:43:30
4 to get to the equivalence
point, and then we're 1 mil

721
00:43:30 --> 00:43:32
beyond the equivalence point.
 

722
00:43:32 --> 00:43:36
And the tricks to these parts
of the problems are to remember

723
00:43:36 --> 00:43:39
all the things that have been
added to get to the volume.

724
00:43:39 --> 00:43:43
So the problem itself is not
very tricky of having just an

725
00:43:43 --> 00:43:47
acid in water, but the one
trick is in calculating the

726
00:43:47 --> 00:43:50
molarity, remembering all of
the things that were added

727
00:43:50 --> 00:43:52
to get to this point.
 

728
00:43:52 --> 00:43:56
And so by the end of the
problem-set this should be

729
00:43:56 --> 00:43:59
pretty familiar to you, but
it's also something that you

730
00:43:59 --> 00:44:02
want to make sure that you
check on an exam that you're

731
00:44:02 --> 00:44:06
not making any volume mistakes.
 

732
00:44:06 --> 00:44:10
So again, here are the things
that you want to remember to

733
00:44:10 --> 00:44:13
add in in calculating this.
 

734
00:44:13 --> 00:44:17
And then, if you calculate
the p h of that, you

735
00:44:17 --> 00:44:19
get a p h of 2 .
 

736
00:44:19 --> 00:44:24
1106, so that's down
somewhere in here.

737
00:44:24 --> 00:44:33
We've added 1 more mill,
and we're at a p h of 2 .

738
00:44:33 --> 00:44:38
1106, and this can
be a check as well.

739
00:44:38 --> 00:44:40
If you forgot to add some
of your volume, you might

740
00:44:40 --> 00:44:43
get a p h that doesn't
make a lot of sense.

741
00:44:43 --> 00:44:45
So that can be a double check.
 

742
00:44:45 --> 00:44:49
Always ask yourself when you
have a strong base and you're

743
00:44:49 --> 00:44:51
adding acid, before you've
added very much, your

744
00:44:51 --> 00:44:53
p h should be basic.
 

745
00:44:53 --> 00:44:56
At the equivalence point of
a strong acid, strong base

746
00:44:56 --> 00:44:58
titration, it'll be 7.
 

747
00:44:58 --> 00:45:01
And if you continue to add
acid, then you should have

748
00:45:01 --> 00:45:03
a pretty low p h, you'll
notice the curve drops

749
00:45:03 --> 00:45:05
off pretty fast.
 

750
00:45:05 --> 00:45:07
So it should be a dramatic
change in p h by adding

751
00:45:07 --> 00:45:09
extra of your acid.
 

752
00:45:09 --> 00:45:11
So always check your work.
 

753
00:45:11 --> 00:45:14
OK, so let's stop there, and
we're going to weak acid, weak

754
00:45:14 --> 00:45:17
base titrations next time.
 

755
00:45:17 --> 00:45:18