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

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00:00:26,180 --> 00:00:28,780
last time talking
about buffers.

10
00:00:28,780 --> 00:00:33,570
And I'm a big fan of buffers,
actually, and buffers are

11
00:00:33,570 --> 00:00:37,890
extremely useful in my research
because they keep the

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00:00:37,890 --> 00:00:39,910
p h of things constant.

13
00:00:39,910 --> 00:00:42,670
So if you're going to do any
kind of biology research or

14
00:00:42,670 --> 00:00:46,510
biochemistry research, you need
to know about buffers.

15
00:00:46,510 --> 00:00:51,270
And as I'll mentioned later,
also your body needs to be

16
00:00:51,270 --> 00:00:55,450
buffered appropriately, you
don't want a large changes in

17
00:00:55,450 --> 00:00:59,880
p h in your body or things
won't function properly.

18
00:00:59,880 --> 00:01:02,580
So buffering is very
important.

19
00:01:02,580 --> 00:01:06,290
So let's go over buffers.

20
00:01:06,290 --> 00:01:09,160
So if we talk about an acid
buffer, that's something

21
00:01:09,160 --> 00:01:13,460
that's buffering on the acidic
side of the p h range.

22
00:01:13,460 --> 00:01:18,070
And in that, you want to have
play on both sides.

23
00:01:18,070 --> 00:01:21,720
So if you have a strong acid, it
can be neutralized, the p h

24
00:01:21,720 --> 00:01:22,370
stays the same.

25
00:01:22,370 --> 00:01:25,110
If you had a strong base, that
would be neutralized so the p

26
00:01:25,110 --> 00:01:26,570
h stays the same.

27
00:01:26,570 --> 00:01:29,750
So what happens, you have to
have an acid and its conjugate

28
00:01:29,750 --> 00:01:32,840
base in the mixture to
have a good buffer.

29
00:01:32,840 --> 00:01:36,260
So what the weak acid is going
to do is it will transfer

30
00:01:36,260 --> 00:01:41,130
protons or hydrogen ions to
hydroxide ions that are

31
00:01:41,130 --> 00:01:45,540
supplied by the strong base to
neutralize that strong base

32
00:01:45,540 --> 00:01:46,740
that was added.

33
00:01:46,740 --> 00:01:49,450
So you need to have an acid in
there that's capable of giving

34
00:01:49,450 --> 00:01:54,190
up a proton to neutralize the
hydroxide ion that was added.

35
00:01:54,190 --> 00:01:57,230
You also need to have
a conjugate base.

36
00:01:57,230 --> 00:02:00,600
The conjugate base is going to
accept protons if an acid is

37
00:02:00,600 --> 00:02:06,050
added, thereby neutralizing the
effect of that added acid.

38
00:02:06,050 --> 00:02:09,260
So again, you need a weak acid
and its conjugate base, you

39
00:02:09,260 --> 00:02:12,000
need both to have
a good buffer.

40
00:02:12,000 --> 00:02:14,750
So I've been talking about weak
acids and weak conjugate

41
00:02:14,750 --> 00:02:21,740
bases, so why would a strong
acid and the salt of its

42
00:02:21,740 --> 00:02:27,050
conjugate base not make
a good buffer?

43
00:02:27,050 --> 00:02:29,330
What do strong acids do?

44
00:02:29,330 --> 00:02:34,620
They dissociate almost
completely, and that's not

45
00:02:34,620 --> 00:02:35,960
going to do you any good.

46
00:02:35,960 --> 00:02:39,490
You want to be able to shift
that equation both directions.

47
00:02:39,490 --> 00:02:42,600
So a strong acid goes to
completion is going to drive

48
00:02:42,600 --> 00:02:45,700
it to the right, and for a good
buffer you want to be

49
00:02:45,700 --> 00:02:49,380
able to switch things around
that if you add a strong acid,

50
00:02:49,380 --> 00:02:51,630
then you neutralize that, you
add a strong base, you

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00:02:51,630 --> 00:02:52,720
neutralize it.

52
00:02:52,720 --> 00:02:56,620
So you need the conjugate to be
effective as a base, and a

53
00:02:56,620 --> 00:02:59,960
strong acid has an ineffective
conjugate base.

54
00:02:59,960 --> 00:03:04,190
It's not good as a base at
all, so that won't work.

55
00:03:04,190 --> 00:03:07,770
So buffers are made up of weak
acids and their conjugate

56
00:03:07,770 --> 00:03:11,210
bases, or weak bases and
their conjugate acids.

57
00:03:11,210 --> 00:03:13,380
They need to be in the
weak range or there's

58
00:03:13,380 --> 00:03:17,800
no buffering capacity.

59
00:03:17,800 --> 00:03:21,420
So let's look at a base
buffer example.

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00:03:21,420 --> 00:03:24,590
The only difference here is that
a basic buffer is going

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00:03:24,590 --> 00:03:28,340
to buffer on the basic
end of the p h range.

62
00:03:28,340 --> 00:03:29,970
So here's an equation.

63
00:03:29,970 --> 00:03:33,490
We have a base in water forming
a conjugate of that

64
00:03:33,490 --> 00:03:35,150
weak base and hydroxide ions.

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00:03:35,150 --> 00:03:39,150
So let's think about
what happens if a

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00:03:39,150 --> 00:03:41,150
strong acid is added.

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00:03:41,150 --> 00:03:46,000
Well if a strong acid is added,
then this base, n h 3,

68
00:03:46,000 --> 00:03:50,080
can accept protons from the
incoming acid and make more n

69
00:03:50,080 --> 00:03:55,390
h 4 plus, thereby removing the
strong acid from the solution

70
00:03:55,390 --> 00:03:58,060
and neutralizing the p h.

71
00:03:58,060 --> 00:04:02,540
If on the other hand a strong
base is added, n h 4 plus or

72
00:04:02,540 --> 00:04:06,960
ammonium ions can donate the
proton forming its conjugate

73
00:04:06,960 --> 00:04:11,760
base in water, and again, the
p h remains about the same.

74
00:04:11,760 --> 00:04:16,590
So the base and its conjugate
acid can both react,

75
00:04:16,590 --> 00:04:21,050
neutralizing the p h.

76
00:04:21,050 --> 00:04:25,160
So in the base buffer acid, a
weak base, b, will accept

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00:04:25,160 --> 00:04:29,480
protons supplied by the strong
acid, removing a strong acid

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00:04:29,480 --> 00:04:30,760
from the solution.

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00:04:30,760 --> 00:04:34,650
The conjugate acid of that weak
base, which again, would

80
00:04:34,650 --> 00:04:39,640
be a weak acid, can transfer
protons to the hydroxide ions

81
00:04:39,640 --> 00:04:41,840
added by the strong
base, again,

82
00:04:41,840 --> 00:04:45,380
neutralizing the effect there.

83
00:04:45,380 --> 00:04:50,630
So a buffer is a mixture of a
weak acid base conjugate pair

84
00:04:50,630 --> 00:04:55,370
that will stabilize the p h
serving as a source or a sink

85
00:04:55,370 --> 00:04:59,350
for the added protons.

86
00:04:59,350 --> 00:05:03,880
So that's how a buffer works.

87
00:05:03,880 --> 00:05:06,910
So, I mentioned buffering
is very important.

88
00:05:06,910 --> 00:05:09,410
It's very important
in your blood.

89
00:05:09,410 --> 00:05:13,510
Your blood is buffered in
the range of p h 7 .

90
00:05:13,510 --> 00:05:15,080
35 to 7 .

91
00:05:15,080 --> 00:05:22,710
45, and you have a buffering
pair and a conjugate acid base

92
00:05:22,710 --> 00:05:25,820
pair that's in your blood
that does to work.

93
00:05:25,820 --> 00:05:30,020
So nature has its own buffering
system design to

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00:05:30,020 --> 00:05:33,220
keep the p h of your
blood constant.

95
00:05:33,220 --> 00:05:37,550
Well what happens if the p
h of your blood changes?

96
00:05:37,550 --> 00:05:43,390
So let me just tell you about
one disease that can change

97
00:05:43,390 --> 00:05:45,330
the p h of your blood.

98
00:05:45,330 --> 00:05:47,960
This is from a vitamin B12
dependent enzyme called

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00:05:47,960 --> 00:05:54,070
methylmalonic coa mutase, and
this enzyme is your friend.

100
00:05:54,070 --> 00:05:57,070
I think that most of you are
excited about the idea of

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00:05:57,070 --> 00:06:00,980
having all your fat go to
energy, so you should be all

102
00:06:00,980 --> 00:06:02,790
fan of this enzyme.

103
00:06:02,790 --> 00:06:07,180
So it breaks down from fatty
acids -- you get methylmalonyl

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00:06:07,180 --> 00:06:11,270
coa, which is converted to
succinyl coa and goes into the

105
00:06:11,270 --> 00:06:12,820
citric acid cycle.

106
00:06:12,820 --> 00:06:15,530
But if this step is blocked,
if there's some kind of

107
00:06:15,530 --> 00:06:19,650
genetic disease associated with
that particular enzyme,

108
00:06:19,650 --> 00:06:22,890
you get a condition called
methylmalonic aciduria.

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00:06:22,890 --> 00:06:25,910
So this is excreted in the body,
it can't be converted to

110
00:06:25,910 --> 00:06:29,260
the succinyl coa, so it has to
be excreted from the body, and

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00:06:29,260 --> 00:06:31,010
it has an acidic p h.

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00:06:31,010 --> 00:06:35,320
And so when it's excreted, it
will actually, there's a large

113
00:06:35,320 --> 00:06:38,460
amount of it, and it can change
the p h of blood.

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00:06:38,460 --> 00:06:41,080
So even though you have this
buffering system, it can

115
00:06:41,080 --> 00:06:45,360
overwhelm the buffering system
and cause a change in the p h,

116
00:06:45,360 --> 00:06:51,340
which can lead to neurological
disorders and sometimes death.

117
00:06:51,340 --> 00:06:55,880
So they can look for this in
newborns, they can do a test,

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00:06:55,880 --> 00:07:00,550
and I was -- one of the tests is
-- it's not in every state,

119
00:07:00,550 --> 00:07:04,370
but it is in Massachusetts, so
my daughter, Samantha, had

120
00:07:04,370 --> 00:07:08,940
this test. But they can simply
look at the urine of a newborn

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00:07:08,940 --> 00:07:11,290
and look at the p h
and whether any of

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00:07:11,290 --> 00:07:13,680
this is being excreted.

123
00:07:13,680 --> 00:07:16,390
And some infants that have this
condition can excrete

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00:07:16,390 --> 00:07:18,460
like a gram of this
acid a day.

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00:07:18,460 --> 00:07:21,930
So it's really large quantities
of acid that are

126
00:07:21,930 --> 00:07:26,280
excreted and it goes beyond
the buffering capacity.

127
00:07:26,280 --> 00:07:29,560
So, I'm going to tell you a
little story and it sort of

128
00:07:29,560 --> 00:07:30,950
emphasizes some of the

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00:07:30,950 --> 00:07:32,830
importance of genetic research.

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00:07:32,830 --> 00:07:35,850
So before the human genome
project, scientists were

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00:07:35,850 --> 00:07:38,500
trying to find the gene to
figure out what was wrong with

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00:07:38,500 --> 00:07:40,710
these patients that had
this condition.

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00:07:40,710 --> 00:07:41,960
And here are all
the amino acids

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00:07:41,960 --> 00:07:45,670
translated from the gene.

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00:07:45,670 --> 00:07:50,090
This is not a small protein,
that's a lot of amino acids.

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00:07:50,090 --> 00:07:53,040
And they found that the problem
was right here, or one

137
00:07:53,040 --> 00:07:54,230
of the main problems
is right here.

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00:07:54,230 --> 00:07:57,250
There's a glycine residue, which
glycine is the smallest

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00:07:57,250 --> 00:08:00,450
amino acid and that had been
changed to something else and

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00:08:00,450 --> 00:08:02,240
that led to the disease.

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00:08:02,240 --> 00:08:06,070
Now I want you to imagine if
you were parents of a small

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00:08:06,070 --> 00:08:10,080
child and you found out your
child had this condition and

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00:08:10,080 --> 00:08:12,600
then you had heard that there
were advances, they now know

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00:08:12,600 --> 00:08:15,920
what caused it, and so, they
went to the doctor and the

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00:08:15,920 --> 00:08:21,240
doctor said, yes, your child,
the DNA is wrong, and so let's

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00:08:21,240 --> 00:08:27,570
substitute a different amino
acid for glycine here.

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00:08:27,570 --> 00:08:29,040
OK.

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00:08:29,040 --> 00:08:30,980
What does that do for you?

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00:08:30,980 --> 00:08:32,760
Like why is that a problem?

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00:08:32,760 --> 00:08:35,310
Well, they didn't know why it
was a problem, they just knew

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00:08:35,310 --> 00:08:38,930
that it had the wrong
amino acid in there.

152
00:08:38,930 --> 00:08:43,340
And so, the next step was to
try to understand what was

153
00:08:43,340 --> 00:08:45,210
actually going on.

154
00:08:45,210 --> 00:08:49,820
So what happened in this case
was that the vitamin B12

155
00:08:49,820 --> 00:08:53,690
cofactor had changed its
conformation when it bound to

156
00:08:53,690 --> 00:08:54,680
the protein.

157
00:08:54,680 --> 00:08:58,640
And instead of having this sort
of loop structure here,

158
00:08:58,640 --> 00:09:00,400
it had more of a tail.

159
00:09:00,400 --> 00:09:05,350
So this base here had moved
down, and so there was a

160
00:09:05,350 --> 00:09:10,520
different shape to the vitamin
than had been expected.

161
00:09:10,520 --> 00:09:13,270
So when the first structure
was solved of one of these

162
00:09:13,270 --> 00:09:17,890
enzymes, it found that here's
kind of what the vitamin B12

163
00:09:17,890 --> 00:09:20,460
looked like, it had this
extended region which no one

164
00:09:20,460 --> 00:09:24,100
was expecting, and so that had
to fit into the protein.

165
00:09:24,100 --> 00:09:27,190
So this is the domain of the
protein that binds the

166
00:09:27,190 --> 00:09:30,540
vitamin, and it had a large
hole, and you put these

167
00:09:30,540 --> 00:09:33,970
together and that's
great, it fits and

168
00:09:33,970 --> 00:09:35,710
you have happy enzyme.

169
00:09:35,710 --> 00:09:40,780
Well, how do you create a hole
-- nature supposedly abhors a

170
00:09:40,780 --> 00:09:43,110
vacuum, so how do you
have this hole?

171
00:09:43,110 --> 00:09:47,250
Well, what nature did was it put
the smallest amino acids

172
00:09:47,250 --> 00:09:51,350
there are, glycines, right along
here to make room for

173
00:09:51,350 --> 00:09:53,360
this to fit together.

174
00:09:53,360 --> 00:09:57,470
So, what was happening in this
genetic case is that the

175
00:09:57,470 --> 00:10:02,060
switch of a glycine to an
argenine did this to the hole.

176
00:10:02,060 --> 00:10:06,750
So, what people were trying to
do is treat these patients

177
00:10:06,750 --> 00:10:10,570
with extra B12, you give them
injections of it, you have a

178
00:10:10,570 --> 00:10:14,250
lot more B12, and you want to
the B12 to bind and get the

179
00:10:14,250 --> 00:10:16,140
activity back going again.

180
00:10:16,140 --> 00:10:20,410
But really, no matter how much
vitamin B12 you added to it,

181
00:10:20,410 --> 00:10:22,890
it wasn't really going to do
much good, it just wasn't

182
00:10:22,890 --> 00:10:24,070
going to fit.

183
00:10:24,070 --> 00:10:27,410
So instead, there was a
suggestion that instead of

184
00:10:27,410 --> 00:10:30,850
giving the whole vitamin,
perhaps you could just give a

185
00:10:30,850 --> 00:10:32,920
truncated version of the
vitamin, which was

186
00:10:32,920 --> 00:10:37,530
commercially available, and
that that might bind and

187
00:10:37,530 --> 00:10:41,020
restore activity
of the protein.

188
00:10:41,020 --> 00:10:45,500
So this is what was suggested
once you knew what the problem

189
00:10:45,500 --> 00:10:47,140
actually was.

190
00:10:47,140 --> 00:10:51,380
So this is just an example of
a genetic disease, and it's

191
00:10:51,380 --> 00:10:54,770
not that common, but as I said,
it is tested for here in

192
00:10:54,770 --> 00:10:55,900
Massachusetts.

193
00:10:55,900 --> 00:11:00,150
And how knowing something about
what the problem is can

194
00:11:00,150 --> 00:11:02,610
lead to a potential solution.

195
00:11:02,610 --> 00:11:05,920
But for most of you, your blood
is doing just fine, you

196
00:11:05,920 --> 00:11:09,350
don't have a genetic condition
which is pumping too much acid

197
00:11:09,350 --> 00:11:13,820
into your bloodstream, so that
nature's buffering capacity is

198
00:11:13,820 --> 00:11:18,040
working quite well.

199
00:11:18,040 --> 00:11:23,530
So buffers are important.

200
00:11:23,530 --> 00:11:26,980
It's great to have my
important statements

201
00:11:26,980 --> 00:11:29,560
emphasized up there.

202
00:11:29,560 --> 00:11:32,570
We've been rehearsing at home.

203
00:11:32,570 --> 00:11:33,220
OK.

204
00:11:33,220 --> 00:11:36,780
So let's do a sample
buffer problem.

205
00:11:36,780 --> 00:11:40,670
All right, suppose we have an
acid, 0.1 moles of an acid,

206
00:11:40,670 --> 00:11:44,460
and 0.5 moles of its conjugate
base added in

207
00:11:44,460 --> 00:11:46,090
the form of a salt.

208
00:11:46,090 --> 00:11:50,390
And so they're put into water
and diluted to 1.0 liter.

209
00:11:50,390 --> 00:11:54,480
And you're given information
about the k a of the acid, 1 .

210
00:11:54,480 --> 00:11:58,380
77 times 10 to the minus
4, and you're asked to

211
00:11:58,380 --> 00:12:00,840
calculate the p h.

212
00:12:00,840 --> 00:12:02,450
So what do you do?

213
00:12:02,450 --> 00:12:05,060
All right, first, it's always
good to write the equation

214
00:12:05,060 --> 00:12:06,380
that you're talking about.

215
00:12:06,380 --> 00:12:12,810
And so you have acid in water
forming hydronium ions and the

216
00:12:12,810 --> 00:12:14,390
conjugate base.

217
00:12:14,390 --> 00:12:19,050
So here, the acid is giving up a
proton to the water, forming

218
00:12:19,050 --> 00:12:24,480
h 3 o plus and it's
conjugate base.

219
00:12:24,480 --> 00:12:28,570
Now we want to think about what
happens at equilibrium.

220
00:12:28,570 --> 00:12:31,020
So we know how many moles we
have and we know what the

221
00:12:31,020 --> 00:12:32,300
total volume is.

222
00:12:32,300 --> 00:12:35,650
And in this table we should be
using molarity, but the math

223
00:12:35,650 --> 00:12:40,910
here is pretty easy because we
have 1 mole and 1 liter or 1,

224
00:12:40,910 --> 00:12:46,030
and 0.5 moles in 1 liter
or 0.5 for molarity.

225
00:12:46,030 --> 00:12:50,900
And now at equilibrium what's
going to happen, well, when

226
00:12:50,900 --> 00:12:54,480
this equilibrates some of this
will go away and more of these

227
00:12:54,480 --> 00:12:56,430
are going to be formed.

228
00:12:56,430 --> 00:13:01,520
So we have minus x over here,
plus x and plus x over there,

229
00:13:01,520 --> 00:13:05,480
and we add it all together
you get 1 minus x plus x

230
00:13:05,480 --> 00:13:08,290
and 0.5 plus x.

231
00:13:08,290 --> 00:13:10,710
So the important thing to
remember is in buffering

232
00:13:10,710 --> 00:13:14,440
problems, you have the acid
and the conjugate.

233
00:13:14,440 --> 00:13:17,170
So we're used to seeing these
tables where we only have

234
00:13:17,170 --> 00:13:21,130
something over here and it's
all zero on the other side.

235
00:13:21,130 --> 00:13:24,390
But in a buffering problem, you
have things on both sides

236
00:13:24,390 --> 00:13:25,710
from the very beginning.

237
00:13:25,710 --> 00:13:28,320
And that's a really important
point, and that alone, if you

238
00:13:28,320 --> 00:13:30,730
can remember that, will
get you a long way

239
00:13:30,730 --> 00:13:31,510
through this unit.

240
00:13:31,510 --> 00:13:35,090
So you remember that in a buffer
you got to have both an

241
00:13:35,090 --> 00:13:39,600
acid in its conjugate base or a
base in its conjugate acid.

242
00:13:39,600 --> 00:13:42,630
So here we're talking about an
acid in water, it's an acidic

243
00:13:42,630 --> 00:13:46,800
buffer, and so we can use our
k a value -- we want to know

244
00:13:46,800 --> 00:13:49,720
what the concentrations
are at equilibrium to

245
00:13:49,720 --> 00:13:51,530
calculate our p h.

246
00:13:51,530 --> 00:13:57,500
So we can use our k a and we can
set it up, so we have on

247
00:13:57,500 --> 00:14:01,510
the top products over reactants,
and we we're not

248
00:14:01,510 --> 00:14:04,360
including water in this
because it's dilute in

249
00:14:04,360 --> 00:14:06,840
solution and so it's not
changing very much.

250
00:14:06,840 --> 00:14:10,710
And now we can plug in our
values, so we have 0.5 plus x,

251
00:14:10,710 --> 00:14:15,780
x over 1 minus x.

252
00:14:15,780 --> 00:14:19,550
All right, now this is just
written again from the last

253
00:14:19,550 --> 00:14:24,180
slide, we can try an
approximation, which is it

254
00:14:24,180 --> 00:14:28,190
that x is small compared to
1 molar or 0.5 molar.

255
00:14:28,190 --> 00:14:32,400
And so we can get rid of the
plus x and the minus x down

256
00:14:32,400 --> 00:14:37,100
here and just have one x term,
which makes it simpler to

257
00:14:37,100 --> 00:14:39,400
solve, but we'll have to
go back and check that

258
00:14:39,400 --> 00:14:42,010
approximation in
a few minutes.

259
00:14:42,010 --> 00:14:46,160
So making that approximation,
we can calculate x as 3 .

260
00:14:46,160 --> 00:14:51,040
54 times 10 to the minus 4
molar, and now we need to

261
00:14:51,040 --> 00:14:52,720
check that assumption.

262
00:14:52,720 --> 00:14:56,740
Well, you can probably guess
it's going to be OK, because

263
00:14:56,740 --> 00:15:00,740
something times 10 to the minus
4 compared to 0.5 and 1,

264
00:15:00,740 --> 00:15:03,860
that's probably going to
be OK that it's small.

265
00:15:03,860 --> 00:15:08,060
But our assumption here is that
it needs to be less than

266
00:15:08,060 --> 00:15:11,320
5%, and here it's 0.1%.

267
00:15:11,320 --> 00:15:14,930
So you could just take this
value of x, divide it by the

268
00:15:14,930 --> 00:15:20,840
smaller of the two, 0.5 , times
100, and calculate the

269
00:15:20,840 --> 00:15:21,500
percent ionization.

270
00:15:21,500 --> 00:15:26,310
If it's less than 5% you're OK,
if it's more you need to

271
00:15:26,310 --> 00:15:30,740
use the quadratic equation
to solve the problem.

272
00:15:30,740 --> 00:15:35,410
So x here is our concentration
of hydronium ion, which is

273
00:15:35,410 --> 00:15:39,720
great, because then we can
easily calculate the p h.

274
00:15:39,720 --> 00:15:42,920
So p h again is minus log
of the hydronium ion

275
00:15:42,920 --> 00:15:47,230
concentration, and so
the p h here is 3 .

276
00:15:47,230 --> 00:15:48,700
45.

277
00:15:48,700 --> 00:15:51,380
And again, significant figures,
we'll be talking

278
00:15:51,380 --> 00:15:53,580
about this as we go through.

279
00:15:53,580 --> 00:15:55,830
The volume had two, it was 1 .

280
00:15:55,830 --> 00:15:59,620
0 liters, and so we're going
to have two significant

281
00:15:59,620 --> 00:16:02,760
figures after the decimal point,
because the volume was

282
00:16:02,760 --> 00:16:08,110
limiting insignificant
figures.

283
00:16:08,110 --> 00:16:14,250
All right, now let's consider
what happens if some strong

284
00:16:14,250 --> 00:16:17,730
acid had been added
in the solution.

285
00:16:17,730 --> 00:16:20,520
So the volume is still going
to be 1 liter, because the

286
00:16:20,520 --> 00:16:28,690
acid was added in before we went
up to the 1 liter mark.

287
00:16:28,690 --> 00:16:32,810
So strong acids, they
go to completion.

288
00:16:32,810 --> 00:16:37,150
So, if we added 0.1 moles of
that strong acid, it will

289
00:16:37,150 --> 00:16:41,480
react with equal number of moles
of the conjugate base to

290
00:16:41,480 --> 00:16:44,870
form the conjugate acid.

291
00:16:44,870 --> 00:16:47,070
So we can just do subtractions
here.

292
00:16:47,070 --> 00:16:49,100
We don't have to worry about
setting up any kind of

293
00:16:49,100 --> 00:16:52,920
equilibrium, we assume
it goes completely.

294
00:16:52,920 --> 00:16:58,420
So for the conjugate base we had
0.5 moles before, and it's

295
00:16:58,420 --> 00:17:03,120
going to be reacting then with a
strong acid, so 0.1 of those

296
00:17:03,120 --> 00:17:08,110
will react, leaving us with 0.4
moles of the conjugate,

297
00:17:08,110 --> 00:17:10,710
and that's in 1 liter.

298
00:17:10,710 --> 00:17:13,530
So now we have 0.4 molar.

299
00:17:13,530 --> 00:17:17,550
For the acid, we had 1 mole to
begin with, but now we formed

300
00:17:17,550 --> 00:17:22,010
more as they reaction
has taken place, and

301
00:17:22,010 --> 00:17:24,130
so we have 0.1 more.

302
00:17:24,130 --> 00:17:25,220
Now we have 1 .

303
00:17:25,220 --> 00:17:29,270
1 moles again in 1 liter.

304
00:17:29,270 --> 00:17:33,190
So now we can do the
same thing again.

305
00:17:33,190 --> 00:17:36,390
So after this has happened,
after we have added the strong

306
00:17:36,390 --> 00:17:40,920
acid, then a new equilibrium
is going to be reached.

307
00:17:40,920 --> 00:17:46,340
And so we can plug in
this table as well.

308
00:17:46,340 --> 00:17:47,970
So we have 1 .

309
00:17:47,970 --> 00:17:55,250
1 now over here, and 0.4
on the other side.

310
00:17:55,250 --> 00:17:59,880
As equilibrium is approached,
you lose some of the acid as

311
00:17:59,880 --> 00:18:04,510
it reacts with water and ionizes
and you form more of

312
00:18:04,510 --> 00:18:08,040
the h 3 o plus and more
of the conjugate.

313
00:18:08,040 --> 00:18:09,290
So we have 1 .

314
00:18:09,290 --> 00:18:14,810
1 minus x, x, and 0.4 plus x.

315
00:18:14,810 --> 00:18:18,030
So again, the trick here is just
to remember that there is

316
00:18:18,030 --> 00:18:21,810
a reaction with a strong acid
that has been added, and you

317
00:18:21,810 --> 00:18:25,410
need to figure out the new
molarities, and then go back

318
00:18:25,410 --> 00:18:33,720
to your equilibrium table with
those new molarities.

319
00:18:33,720 --> 00:18:37,790
So here we can set up with a
k a again, and do the same

320
00:18:37,790 --> 00:18:42,030
problem that we did before.

321
00:18:42,030 --> 00:18:45,390
And so, we can make the
assumption again that x is

322
00:18:45,390 --> 00:18:49,410
small, try it out and
see if it works.

323
00:18:49,410 --> 00:18:52,060
X turns out to be 4 .

324
00:18:52,060 --> 00:18:55,820
87 times 10 to the minus 4
molars -- so again, it's a

325
00:18:55,820 --> 00:18:57,440
small number.

326
00:18:57,440 --> 00:19:03,300
And so, it turns out to be 1
. -- less than 1% of 0.4 .

327
00:19:03,300 --> 00:19:06,840
If it's less than 5% of the
smaller number, you don't have

328
00:19:06,840 --> 00:19:10,100
to worry about the bigger
number, and so your assumption

329
00:19:10,100 --> 00:19:13,630
is OK, you don't need to use
the quadratic equation.

330
00:19:13,630 --> 00:19:17,580
And so, then we can calculate
at the end that the

331
00:19:17,580 --> 00:19:19,260
p h is now 3 .

332
00:19:19,260 --> 00:19:22,260
31, again, two significant
figures after the decimal

333
00:19:22,260 --> 00:19:24,300
place because of the volume.

334
00:19:24,300 --> 00:19:27,990
So addition of 0.1 moles of
a strong acid changed

335
00:19:27,990 --> 00:19:29,420
our p h from 3 .

336
00:19:29,420 --> 00:19:32,200
45 to 3 .

337
00:19:32,200 --> 00:19:33,290
31.

338
00:19:33,290 --> 00:19:34,750
So we buffered pretty well.

339
00:19:34,750 --> 00:19:38,540
There was a change in p h, we
added acid and the p h did go

340
00:19:38,540 --> 00:19:41,630
down, but not by an
enormous amount.

341
00:19:41,630 --> 00:19:45,040
And so, that's because this
turned out to be a pretty

342
00:19:45,040 --> 00:19:46,890
decent buffer here,
so we didn't have

343
00:19:46,890 --> 00:19:54,740
a really huge effect.

344
00:19:54,740 --> 00:20:07,530
So, those are -- that's an
example of a buffer problem.

345
00:20:07,530 --> 00:20:10,720
So, let's think for a minute
about designing a buffer.

346
00:20:10,720 --> 00:20:17,690
Suppose you wanted to create a
buffer at a particular p h,

347
00:20:17,690 --> 00:20:19,680
what do you need
to think about?

348
00:20:19,680 --> 00:20:22,870
Well, you need to think about
the ratio of your acid to the

349
00:20:22,870 --> 00:20:26,570
conjugate, and you need to think
about the p k a of the

350
00:20:26,570 --> 00:20:32,660
acid, and the p h
that's desired.

351
00:20:32,660 --> 00:20:38,940
So here is a generic equation
for an acid in water, so we

352
00:20:38,940 --> 00:20:42,060
have our acid, h a, in water
forming hydronium ions and our

353
00:20:42,060 --> 00:20:44,340
conjugate base.

354
00:20:44,340 --> 00:20:47,300
And now we're going to do a
little derivation to come up

355
00:20:47,300 --> 00:20:50,010
with an equation that would be
useful to you in thinking

356
00:20:50,010 --> 00:20:53,420
about buffers and designing
buffers.

357
00:20:53,420 --> 00:20:57,910
So, we can write a generic
term for k a.

358
00:20:57,910 --> 00:21:01,420
K a equals products over
reactants, and so we have our

359
00:21:01,420 --> 00:21:05,220
hydronium ions are conjugate
base over our acid.

360
00:21:05,220 --> 00:21:09,400
And now we can take this
term and rearrange it.

361
00:21:09,400 --> 00:21:11,330
So we can pull a hydronium ion

362
00:21:11,330 --> 00:21:14,380
concentration over to one side.

363
00:21:14,380 --> 00:21:18,860
And now we're going to take the
logs of both sides, so we

364
00:21:18,860 --> 00:21:24,090
get the log of hydronium ion
concentration, the log of k a,

365
00:21:24,090 --> 00:21:27,560
and the log of h a concentration
over a minus

366
00:21:27,560 --> 00:21:31,680
concentration, and now we're
going to multiply everything

367
00:21:31,680 --> 00:21:36,950
by negative sign, so we have
minus logs of things.

368
00:21:36,950 --> 00:21:40,070
And I'm just going to move this
now up to the top of the

369
00:21:40,070 --> 00:21:43,950
screen, and we have
this equation.

370
00:21:43,950 --> 00:21:49,160
So what is minus log of
hydronium ion concentration?

371
00:21:49,160 --> 00:21:49,850
P h.

372
00:21:49,850 --> 00:21:53,310
What's minus log of k a?

373
00:21:53,310 --> 00:21:55,430
P k a.

374
00:21:55,430 --> 00:22:00,110
So, we have p h equals p
k a minus the log of a

375
00:22:00,110 --> 00:22:03,970
concentration of your acid over
the concentration of your

376
00:22:03,970 --> 00:22:05,120
conjugate base.

377
00:22:05,120 --> 00:22:08,030
And these are equilibrium
concentrations for that,

378
00:22:08,030 --> 00:22:10,570
because remember, we derived
it from our equilibrium

379
00:22:10,570 --> 00:22:13,230
expression for k a.

380
00:22:13,230 --> 00:22:16,400
But a lot of times when you're
working buffer problems, you

381
00:22:16,400 --> 00:22:19,710
know the concentrations that you
added of these things, not

382
00:22:19,710 --> 00:22:23,780
necessarily the concentrations
of equilibrium.

383
00:22:23,780 --> 00:22:27,550
And if we rewrite this
expression in terms of the

384
00:22:27,550 --> 00:22:31,990
initial concentrations or
original or o concentrations,

385
00:22:31,990 --> 00:22:36,550
then we have p h is
approximately equal to p k a

386
00:22:36,550 --> 00:22:40,310
minus log of the initial
concentrations of your acid

387
00:22:40,310 --> 00:22:41,810
over your conjugate base.

388
00:22:41,810 --> 00:22:42,900
And this is known as a

389
00:22:42,900 --> 00:22:45,140
Henderson Hasselbalch equation.

390
00:22:45,140 --> 00:22:48,490
And people love the Henderson
Hasselbalch equation, and it

391
00:22:48,490 --> 00:22:51,870
can be incredibly useful to you
in solving these problems.

392
00:22:51,870 --> 00:22:55,090
But I'm going to emphasize when
it's OK to use it and

393
00:22:55,090 --> 00:22:58,560
when it's not OK to use it,
because people try to apply it

394
00:22:58,560 --> 00:23:02,300
to everything, and it's for
buffer problems, and people

395
00:23:02,300 --> 00:23:04,340
apply it to all sorts of things
that are not buffer

396
00:23:04,340 --> 00:23:07,160
problems. So I want to make sure
that it's clear when you

397
00:23:07,160 --> 00:23:10,060
can use it and when
you can't use it.

398
00:23:10,060 --> 00:23:14,110
And it's fine to use it and you
should use it when it's

399
00:23:14,110 --> 00:23:17,150
acceptable, but not
at other times.

400
00:23:17,150 --> 00:23:19,800
It's one of the few equations
in acid base, so people get

401
00:23:19,800 --> 00:23:22,400
very excited by it.

402
00:23:22,400 --> 00:23:26,680
All right, so remember that it's
really equal when they're

403
00:23:26,680 --> 00:23:29,330
at equilibrium concentrations,
and this is just an

404
00:23:29,330 --> 00:23:32,520
approximation that we're saying
that those are initial

405
00:23:32,520 --> 00:23:34,180
concentrations.

406
00:23:34,180 --> 00:23:39,080
So when is it OK to say well,
the equilibrium concentration

407
00:23:39,080 --> 00:23:43,240
is more or less the same as
the initial concentration.

408
00:23:43,240 --> 00:23:47,400
And that's true when x, which is
in this particular example

409
00:23:47,400 --> 00:23:50,500
your hydronium ion
concentration, is small

410
00:23:50,500 --> 00:23:54,110
compared to your initial
concentration of the acid and

411
00:23:54,110 --> 00:23:57,000
the conjugate base that you
added to the buffer.

412
00:23:57,000 --> 00:24:01,030
So, and when x is small, then
pretty much, the equilibrium

413
00:24:01,030 --> 00:24:04,110
concentration equals the
initial concentration.

414
00:24:04,110 --> 00:24:08,110
So there's not really very much
change if x is a really

415
00:24:08,110 --> 00:24:10,050
small number.

416
00:24:10,050 --> 00:24:13,510
So, a lot of the time
this is true.

417
00:24:13,510 --> 00:24:15,740
Remember, we're making buffers
from weak acids and weak

418
00:24:15,740 --> 00:24:19,120
conjugate bases, and the
definition of a weak acid is

419
00:24:19,120 --> 00:24:22,390
that it only loses a tiny
fraction of its protons.

420
00:24:22,390 --> 00:24:26,070
It only ionizes slightly
in water.

421
00:24:26,070 --> 00:24:29,960
And a weak base typically only
accepts a fraction of the

422
00:24:29,960 --> 00:24:31,740
protons it can accept.

423
00:24:31,740 --> 00:24:36,330
It's a weak base, so it's
not doing a whole lot

424
00:24:36,330 --> 00:24:37,650
of chemistry there.

425
00:24:37,650 --> 00:24:41,440
So this approximation is
good a lot of the time.

426
00:24:41,440 --> 00:24:44,590
So just you know when you can
and can not use it, we're

427
00:24:44,590 --> 00:24:47,870
going to give a rule and say
it's the same rule as before.

428
00:24:47,870 --> 00:24:52,040
When we say when x is small
or your hydronium ion

429
00:24:52,040 --> 00:24:55,200
concentration is small compared
to the acid or the

430
00:24:55,200 --> 00:24:58,360
conjugate base, when it's less
than 5%, that's what we're

431
00:24:58,360 --> 00:25:02,380
going to call small, so
the same rule we've

432
00:25:02,380 --> 00:25:03,800
been using all along.

433
00:25:03,800 --> 00:25:06,880
When x is small, this
approximation

434
00:25:06,880 --> 00:25:08,170
works pretty well.

435
00:25:08,170 --> 00:25:11,800
And you can plug in your
concentrations right there --

436
00:25:11,800 --> 00:25:14,380
your initial concentrations
not your equilibrium

437
00:25:14,380 --> 00:25:19,060
concentrations.

438
00:25:19,060 --> 00:25:22,060
All right, so let's use
Henderson Hasselbalch and

439
00:25:22,060 --> 00:25:26,490
design a buffer system if
we want a p h of 4 .

440
00:25:26,490 --> 00:25:27,480
6.

441
00:25:27,480 --> 00:25:32,060
And a good rule to remember is
that a buffering solution is

442
00:25:32,060 --> 00:25:36,420
most effective when it's
plus or minus 1 away

443
00:25:36,420 --> 00:25:38,300
from the p k a.

444
00:25:38,300 --> 00:25:40,540
So that's a good thing
too I keep in mind

445
00:25:40,540 --> 00:25:43,050
for research as well.

446
00:25:43,050 --> 00:25:46,490
If you're really far away from
the p k a, it's not going to

447
00:25:46,490 --> 00:25:49,200
be a good buffering system,
and this is a mistake the

448
00:25:49,200 --> 00:25:52,300
people make in the lab
sometimes, so you may run into

449
00:25:52,300 --> 00:25:55,580
that in some of your
laboratory courses.

450
00:25:55,580 --> 00:25:59,270
So we want a buffer
about p h 4 .

451
00:25:59,270 --> 00:26:00,130
6.

452
00:26:00,130 --> 00:26:03,870
So we can look at our ionization
constants of acid,

453
00:26:03,870 --> 00:26:05,830
and of course, we're always
doing everything at room

454
00:26:05,830 --> 00:26:09,840
temperature in this unit, and so
we can look at what p k a's

455
00:26:09,840 --> 00:26:14,680
are of some weak acids, and
acetate, acetate is an easy

456
00:26:14,680 --> 00:26:19,650
buffer to get your hands on,
acetate is quite available,

457
00:26:19,650 --> 00:26:21,860
and it has a good p k a 4 .

458
00:26:21,860 --> 00:26:22,980
75.

459
00:26:22,980 --> 00:26:27,340
So that's great, we can use that
in designing our buffer.

460
00:26:27,340 --> 00:26:32,780
So, acetic acid has the right
p k a, and we're all set, we

461
00:26:32,780 --> 00:26:34,490
can prepare a buffer.

462
00:26:34,490 --> 00:26:38,010
But we'll need to add the acid
and the conjugate, and we need

463
00:26:38,010 --> 00:26:42,890
to know how much acid and how
much of the conjugate to add.

464
00:26:42,890 --> 00:26:46,410
So we can use Henderson
Hasselbalch equation for this.

465
00:26:46,410 --> 00:26:50,140
We're creating a buffer, and so
this is an equation we can

466
00:26:50,140 --> 00:26:51,510
use for buffers.

467
00:26:51,510 --> 00:26:55,840
We know what p h we want, we
know the p k a of the thing

468
00:26:55,840 --> 00:26:58,730
we're going to use, and we want
to figure out how much of

469
00:26:58,730 --> 00:27:01,700
the acid to use and how much
of the conjugate to use.

470
00:27:01,700 --> 00:27:05,660
We need to know the ratio of one
to the other to set up our

471
00:27:05,660 --> 00:27:07,820
buffer ideally.

472
00:27:07,820 --> 00:27:12,200
So we can plug our numbers
in and calculate that.

473
00:27:12,200 --> 00:27:15,660
So, we can rearrange this if you
want so that the unknown

474
00:27:15,660 --> 00:27:18,560
is on one side and we have
to p k a and the p h

475
00:27:18,560 --> 00:27:21,300
on the other side.

476
00:27:21,300 --> 00:27:26,560
And so, we have here our p k a
minus our p h and we get 0 .

477
00:27:26,560 --> 00:27:27,710
15.

478
00:27:27,710 --> 00:27:32,500
Now we need to inverse log and
come up with the ratio, and

479
00:27:32,500 --> 00:27:34,640
the ratio that we want is 1 .

480
00:27:34,640 --> 00:27:35,510
4.

481
00:27:35,510 --> 00:27:38,530
So we want to have a ratio
from our acid to the

482
00:27:38,530 --> 00:27:40,260
conjugate of 1 .

483
00:27:40,260 --> 00:27:40,910
4.

484
00:27:40,910 --> 00:27:43,480
Well how much exactly are we
going to add them, we know the

485
00:27:43,480 --> 00:27:46,460
ratio now, how much are
we going to add?

486
00:27:46,460 --> 00:27:48,690
Well, you could use 1 .

487
00:27:48,690 --> 00:27:53,130
4 molar and 1 molar, for
example, that would have the

488
00:27:53,130 --> 00:27:54,860
correct ratio.

489
00:27:54,860 --> 00:27:59,010
But how do you know if that's
going to be good?

490
00:27:59,010 --> 00:28:02,310
Well, the ratio is more
important than the exact

491
00:28:02,310 --> 00:28:07,150
amounts, however, the amounts
get to this issue of sort of

492
00:28:07,150 --> 00:28:11,190
the capacity of the buffer --
how resistant it is to changes

493
00:28:11,190 --> 00:28:12,170
in the p h.

494
00:28:12,170 --> 00:28:15,440
So if you use too low
concentrations of both, it

495
00:28:15,440 --> 00:28:17,490
won't be all that resistant.

496
00:28:17,490 --> 00:28:21,490
So if you need to have a better
buffering capacity, the

497
00:28:21,490 --> 00:28:24,880
higher concentrations
that you should use.

498
00:28:24,880 --> 00:28:27,630
And you can use Henderson
Hasselbalch then to calculate

499
00:28:27,630 --> 00:28:31,350
what sort of the minimum amount
you use, the minimum

500
00:28:31,350 --> 00:28:35,050
amount to have that
5% rule work.

501
00:28:35,050 --> 00:28:39,520
So in the example I told you
about this condition, the

502
00:28:39,520 --> 00:28:42,730
buffer in the blood was pretty
good, but the acid with so

503
00:28:42,730 --> 00:28:46,120
much that it overwhelmed
buffering capacity.

504
00:28:46,120 --> 00:28:48,790
So if there had been larger
concentrations of the buffer,

505
00:28:48,790 --> 00:28:52,760
that would have certainly
helped in that case.

506
00:28:52,760 --> 00:28:55,240
So the higher the concentrations
you use, the

507
00:28:55,240 --> 00:28:57,760
more resistant to change.

508
00:28:57,760 --> 00:29:02,010
And that's the idea of
buffering capacity.

509
00:29:02,010 --> 00:29:04,870
So, and if you use too low
concentrations, then your

510
00:29:04,870 --> 00:29:08,770
Henderson Hasselbalch equation
won't even be valid.

511
00:29:08,770 --> 00:29:11,520
So we can go back and calculate
what is sort of the

512
00:29:11,520 --> 00:29:15,500
minimum concentrations we need
to use for Henderson

513
00:29:15,500 --> 00:29:18,830
Hasselbalch to be valid
to meet that 5% rule.

514
00:29:18,830 --> 00:29:20,630
So for a p h of 4 .

515
00:29:20,630 --> 00:29:24,830
6, a hydronium ion concentration
is a 2 .

516
00:29:24,830 --> 00:29:26,880
5 times 10 to the minus 5.

517
00:29:26,880 --> 00:29:31,040
And so then if we work our
equation backwards, we want to

518
00:29:31,040 --> 00:29:35,010
be less than 5%, so we have this
over the concentration of

519
00:29:35,010 --> 00:29:39,310
either one of those times
100% gives you 5%.

520
00:29:39,310 --> 00:29:42,890
So the concentrations need to be
greater than 5 times 10 to

521
00:29:42,890 --> 00:29:46,690
the minus 4 moles or you won't
meet this 5% rule.

522
00:29:46,690 --> 00:29:49,710
So at least to be greater than
that and they need to be in

523
00:29:49,710 --> 00:29:50,880
the right ratio.

524
00:29:50,880 --> 00:29:55,210
Other than that you're somewhat
free to decide what

525
00:29:55,210 --> 00:29:56,910
you're going to do with that,
and there may be other

526
00:29:56,910 --> 00:30:01,930
considerations involved if
you're working with a certain

527
00:30:01,930 --> 00:30:04,930
concentration of protein, you
might not want your buffer

528
00:30:04,930 --> 00:30:07,810
concentration to be too high,
it might start interfering

529
00:30:07,810 --> 00:30:10,700
with enzyme assay,
for example.

530
00:30:10,700 --> 00:30:12,540
And often buffers are
somewhere around 100

531
00:30:12,540 --> 00:30:15,440
millimolar, that's
a pretty common

532
00:30:15,440 --> 00:30:19,170
concentration for buffers.

533
00:30:19,170 --> 00:30:24,810
All right, so this slide we're
doing really well.

534
00:30:24,810 --> 00:30:28,790
So we talked about weak acids in
water, weak bases in water,

535
00:30:28,790 --> 00:30:31,410
and I'm trying to convince you
that these are the same as the

536
00:30:31,410 --> 00:30:34,320
salt and water problems. So
we're going to talk much more

537
00:30:34,320 --> 00:30:36,510
about salt and water
in a few minutes.

538
00:30:36,510 --> 00:30:39,560
And I told you about buffers,
so we only have 2

539
00:30:39,560 --> 00:30:40,950
more things to go.

540
00:30:40,950 --> 00:30:43,890
We need to talk about strong
acids in water, and strong

541
00:30:43,890 --> 00:30:47,340
bases in water, and then you'll
have all of the 5 types

542
00:30:47,340 --> 00:30:50,900
of problems to do
the problem-set.

543
00:30:50,900 --> 00:30:55,690
And as I mentioned last time,
it looks deceivingly short,

544
00:30:55,690 --> 00:30:58,840
the number of questions on
them, but the titration

545
00:30:58,840 --> 00:31:00,700
problems are long.

546
00:31:00,700 --> 00:31:04,400
So don't leave those to the last
minute or there will not

547
00:31:04,400 --> 00:31:06,860
be a whole lot of
sleep involved.

548
00:31:06,860 --> 00:31:11,460
So just a word of warning from
the past. Everyone looks and

549
00:31:11,460 --> 00:31:13,300
go, "Oh, this is an easy
problem-set, there aren't many

550
00:31:13,300 --> 00:31:19,500
problems." So, titration
problems do

551
00:31:19,500 --> 00:31:20,950
take a lot of time.

552
00:31:20,950 --> 00:31:24,790
So let's talk about titration,
so that as soon as class is

553
00:31:24,790 --> 00:31:27,650
over today, you can go work
on those questions on the

554
00:31:27,650 --> 00:31:28,140
problem-set.

555
00:31:28,140 --> 00:31:29,530
All right.

556
00:31:29,530 --> 00:31:31,410
So acid base titrations.

557
00:31:31,410 --> 00:31:37,300
How many of you in high school
titrated an acid with a base?

558
00:31:37,300 --> 00:31:42,630
Large fraction of you, OK.

559
00:31:42,630 --> 00:31:47,740
So usually what titration
problems are meant to do, that

560
00:31:47,740 --> 00:31:52,810
you have something known about
an acid or a base, base say of

561
00:31:52,810 --> 00:31:58,390
known concentration, and and
acid of unknown concentration

562
00:31:58,390 --> 00:32:01,660
or maybe unknown molecular
weight, and you're titrating

563
00:32:01,660 --> 00:32:04,790
them out, titrating them
together to find missing

564
00:32:04,790 --> 00:32:06,030
information.

565
00:32:06,030 --> 00:32:08,750
So you can determine
concentration, sometimes you

566
00:32:08,750 --> 00:32:11,580
can determine a molecular
weight.

567
00:32:11,580 --> 00:32:14,690
All right, so here's what a plot
looks like and some of

568
00:32:14,690 --> 00:32:19,090
the key terms involved in
acid base titrations.

569
00:32:19,090 --> 00:32:24,350
So you have a p h on one side,
and then volume of either base

570
00:32:24,350 --> 00:32:26,690
or acid added on
the other side.

571
00:32:26,690 --> 00:32:28,940
So p h versus a volume.

572
00:32:28,940 --> 00:32:33,220
And the actual experiment here,
you have a base that

573
00:32:33,220 --> 00:32:38,440
you're dripping, one drip at
a time supposedly into your

574
00:32:38,440 --> 00:32:40,340
strong acid.

575
00:32:40,340 --> 00:32:44,600
And then you're looking to
figure out when you're going

576
00:32:44,600 --> 00:32:47,050
to reach the either equivalence
point or the end

577
00:32:47,050 --> 00:32:50,620
point, these are basically the
same terms. It's also called

578
00:32:50,620 --> 00:32:54,650
the stoichiometric point, and
often equivalence point is

579
00:32:54,650 --> 00:33:00,010
used as a more theoretical
amount of volume is added,

580
00:33:00,010 --> 00:33:03,860
where as end point is the
experimentally measured.

581
00:33:03,860 --> 00:33:06,500
So we're going to be talking a
lot about equivalence points,

582
00:33:06,500 --> 00:33:09,110
because we have no lab
associated with this course,

583
00:33:09,110 --> 00:33:12,360
so everything will be
theorectical in terms of

584
00:33:12,360 --> 00:33:15,970
these, but they should
be the same.

585
00:33:15,970 --> 00:33:19,190
So, many times you will measure
p h with the p h meter

586
00:33:19,190 --> 00:33:21,400
-- this just shows
a p h meter.

587
00:33:21,400 --> 00:33:25,540
And for those of you who have
done these experiments, this

588
00:33:25,540 --> 00:33:26,890
should look familiar.

589
00:33:26,890 --> 00:33:30,090
And for those of you who
haven't, often what you're

590
00:33:30,090 --> 00:33:32,330
doing when you're dripping
something in, you're waiting

591
00:33:32,330 --> 00:33:35,970
for an indicator dye to change
color as an indication that

592
00:33:35,970 --> 00:33:38,710
you have reached the end
point, and often you're

593
00:33:38,710 --> 00:33:41,870
adding, and it's clear, clear,
clear for what seems to be

594
00:33:41,870 --> 00:33:45,920
forever, and just as you get
incredibly impatient, you

595
00:33:45,920 --> 00:33:47,960
start adding it faster and
you go all the way

596
00:33:47,960 --> 00:33:49,250
to this dark color.

597
00:33:49,250 --> 00:33:54,800
And what you want is this very,
very, very light changed

598
00:33:54,800 --> 00:33:57,000
color that indicates
the end point.

599
00:33:57,000 --> 00:34:02,030
So you usually go too slow and
then go too fast and then have

600
00:34:02,030 --> 00:34:03,950
to do the experiment
over again.

601
00:34:03,950 --> 00:34:06,880
Those of you who know how to
calculate theoretical values

602
00:34:06,880 --> 00:34:09,770
could sit down and do a
calculation first and then go

603
00:34:09,770 --> 00:34:12,970
really fast right until you get
around this point and add

604
00:34:12,970 --> 00:34:16,500
it really slowly, and be
done with lab sooner.

605
00:34:16,500 --> 00:34:20,110
So, we'll talk about how you
do the theoretical value,

606
00:34:20,110 --> 00:34:23,810
which could save you a lot of
time if you ever run across

607
00:34:23,810 --> 00:34:30,280
these labs in any
future class.

608
00:34:30,280 --> 00:34:34,140
So, here are the two curves that
you would see for either

609
00:34:34,140 --> 00:34:36,920
a strong acid with a strong
base, or strong base with a

610
00:34:36,920 --> 00:34:38,210
strong acid.

611
00:34:38,210 --> 00:34:41,410
So if you're titrating a strong
acid with a strong

612
00:34:41,410 --> 00:34:44,630
base, you're going to start
down being very acidic,

613
00:34:44,630 --> 00:34:47,450
because all you have is
your strong acid.

614
00:34:47,450 --> 00:34:51,640
And then as you add in base,
the p h will go up.

615
00:34:51,640 --> 00:34:55,260
You'll reach this point -- s
is for the stoichiometric

616
00:34:55,260 --> 00:34:58,060
point, we also call this, again,
the equivalence point,

617
00:34:58,060 --> 00:35:01,840
and then it'll continue to go up
and then start to level off

618
00:35:01,840 --> 00:35:03,880
a bit more up here.

619
00:35:03,880 --> 00:35:05,960
If you're going the other
direction, you're starting

620
00:35:05,960 --> 00:35:09,790
with strong base, your p h is
going to be high, and the p h

621
00:35:09,790 --> 00:35:12,120
will decrease as you
add the acid.

622
00:35:12,120 --> 00:35:14,750
And then you get to the
stoichiometric or equivalence

623
00:35:14,750 --> 00:35:20,090
point in the middle of
this curve here.

624
00:35:20,090 --> 00:35:23,350
And then as you go down farther,
more acidic, it

625
00:35:23,350 --> 00:35:24,680
starts to level off.

626
00:35:24,680 --> 00:35:29,880
So those are what the titration
curves look like.

627
00:35:29,880 --> 00:35:32,190
So let's do an example.

628
00:35:32,190 --> 00:35:36,040
We're going to have a strong
base being titrated with a

629
00:35:36,040 --> 00:35:42,095
strong acid first. And our
strong base is n a o h, and

630
00:35:42,095 --> 00:35:47,040
our strong acid is h c l.

631
00:35:47,040 --> 00:35:56,560
So let's calculate the p h
at the equivalence point.

632
00:35:56,560 --> 00:36:00,730
Well, let's just calculate
the p h at 5 mils --

633
00:36:00,730 --> 00:36:03,450
I don't know whether it's
equivalence point, actually.

634
00:36:03,450 --> 00:36:08,200
5 mils of this have been added
to the amount of base.

635
00:36:08,200 --> 00:36:12,920
So we have 25 mils of our strong
base at 0.15 molar

636
00:36:12,920 --> 00:36:17,710
concentration, and we're adding
5 mils of our 0.34

637
00:36:17,710 --> 00:36:21,520
molar acid.

638
00:36:21,520 --> 00:36:24,580
So what we want to do is figure
out how many moles of

639
00:36:24,580 --> 00:36:28,280
base we had to start with.

640
00:36:28,280 --> 00:36:33,050
So how many moles of o h minus
-- again, n a o h is a strong

641
00:36:33,050 --> 00:36:37,300
base, so however much n a o h
you add, is the amount of o h

642
00:36:37,300 --> 00:36:42,390
minus you get, and so we just
put in the number of liters

643
00:36:42,390 --> 00:36:44,780
times the concentration
to get moles.

644
00:36:44,780 --> 00:36:48,760
We don't need any equilibrium
table here.

645
00:36:48,760 --> 00:36:52,290
So then we want to figure out
the amount of acid added when

646
00:36:52,290 --> 00:36:55,660
you're adding 5 mils of it.

647
00:36:55,660 --> 00:36:59,950
So again, h c l is a strong
acid, so the amount of h c l

648
00:36:59,950 --> 00:37:05,040
added equals the amount of h 3 o
plus, that's formed, goes to

649
00:37:05,040 --> 00:37:09,220
completion, and so we added 5
mils, the concentration was

650
00:37:09,220 --> 00:37:12,480
0.34 molar, so we have 1 .

651
00:37:12,480 --> 00:37:17,480
7 times 10 to the
minus 3 moles.

652
00:37:17,480 --> 00:37:21,100
So now, the strong acid is going
to react with the strong

653
00:37:21,100 --> 00:37:24,850
base, and we want to figure out
how much of the base is

654
00:37:24,850 --> 00:37:28,980
left, how much hydroxide ion is
left after it reacts with

655
00:37:28,980 --> 00:37:31,210
the hydronium ions.

656
00:37:31,210 --> 00:37:35,650
So it'll react 1:1,
so we had 6 .

657
00:37:35,650 --> 00:37:41,900
25 times 10 to the minus 3
moles, and we added 1 .

658
00:37:41,900 --> 00:37:45,260
7 times 10 to the minus 3 moles
of the acid, so we're

659
00:37:45,260 --> 00:37:47,350
going to have 4 .

660
00:37:47,350 --> 00:37:53,430
55 times 10 the minus
3 moles left.

661
00:37:53,430 --> 00:37:59,270
And now we can calculate the
molarity here, so we have the

662
00:37:59,270 --> 00:38:03,900
number of moles and the new
volume, so we had 25 mils to

663
00:38:03,900 --> 00:38:04,990
the base to begin with.

664
00:38:04,990 --> 00:38:08,600
We've added 5 mils to the acid,
so our new volume is 30,

665
00:38:08,600 --> 00:38:12,980
and so we have our new
concentration.

666
00:38:12,980 --> 00:38:15,280
And now we can calculate p h.

667
00:38:15,280 --> 00:38:20,380
First we'll calculate p o h, and
then use 14 minus to get

668
00:38:20,380 --> 00:38:23,180
the new p h.

669
00:38:23,180 --> 00:38:29,630
So if you were making your own
little titration curve, we

670
00:38:29,630 --> 00:38:34,480
could have volume of acid on one
side and p h on the other

671
00:38:34,480 --> 00:38:39,080
side, and you have
a point of 13 .

672
00:38:39,080 --> 00:38:40,430
18.

673
00:38:40,430 --> 00:38:49,360
That's how much we have after
you have 5 mils added.

674
00:38:49,360 --> 00:38:52,560
So we didn't, in this curve, we
have not measured the zero

675
00:38:52,560 --> 00:39:00,420
point, but we know what one
point after 5 mils of the acid

676
00:39:00,420 --> 00:39:02,760
has been added.

677
00:39:02,760 --> 00:39:06,250
So now let's go right to the
equivalence point -- so the

678
00:39:06,250 --> 00:39:08,180
way I draw it should indicate
that was not

679
00:39:08,180 --> 00:39:09,510
an equivalence point.

680
00:39:09,510 --> 00:39:12,850
So we want to figure out how
much we need to add of the

681
00:39:12,850 --> 00:39:15,150
acid to reach the equivalence
point.

682
00:39:15,150 --> 00:39:18,290
So the equivalence point or the
stoichiometric point means

683
00:39:18,290 --> 00:39:22,900
that you add the same amount
of moles of your titrate as

684
00:39:22,900 --> 00:39:26,740
you had, and so they're going
to be equal to each other.

685
00:39:26,740 --> 00:39:30,250
So if you're adding acid to a
strong base, you've added

686
00:39:30,250 --> 00:39:33,370
enough acid that you now have
equal number of moles to the

687
00:39:33,370 --> 00:39:36,380
number of moles of base that
you had to begin with.

688
00:39:36,380 --> 00:39:37,640
So we know we have 6 .

689
00:39:37,640 --> 00:39:41,030
25 times 10 to the
minus 3 moles of

690
00:39:41,030 --> 00:39:42,590
the base were present.

691
00:39:42,590 --> 00:39:46,400
So at the equivalence point,
you need to have that exact

692
00:39:46,400 --> 00:39:49,560
same number of moles of acid.

693
00:39:49,560 --> 00:39:53,700
So we know the concentration
of the acid we have and the

694
00:39:53,700 --> 00:39:56,590
number of moles we need, and
so we can calculate the

695
00:39:56,590 --> 00:39:59,160
volume, which is 18 .

696
00:39:59,160 --> 00:40:04,480
4 millimeters, or
0.0184 liters.

697
00:40:04,480 --> 00:40:12,860
So what would the p h then be
of the equivalence point?

698
00:40:12,860 --> 00:40:15,980
We're titrating a strong acid
with a strong base, what

699
00:40:15,980 --> 00:40:19,830
should the p h be?

700
00:40:19,830 --> 00:40:22,270
The p h should be 7.

701
00:40:22,270 --> 00:40:27,120
So somewhere around here we're
going to be at 18, so you can

702
00:40:27,120 --> 00:40:36,650
just try to draw a curve, and
here we have a p h of 7 when

703
00:40:36,650 --> 00:40:39,770
we've gone to about 18 .

704
00:40:39,770 --> 00:40:44,300
4 milliliters added.

705
00:40:44,300 --> 00:40:47,620
So when you titrate a strong
acid with a strong base, you

706
00:40:47,620 --> 00:40:49,610
form a salt that's neutral.

707
00:40:49,610 --> 00:40:52,320
Because the conjugates of a
strong acid and a strong base

708
00:40:52,320 --> 00:40:53,380
are not basic or acidic.

709
00:40:53,380 --> 00:40:56,740
They're ineffective as acids or
bases, so you're going to

710
00:40:56,740 --> 00:40:58,640
get a neutral salt.

711
00:40:58,640 --> 00:41:01,470
So if you're doing a problem
with a strong acid and strong

712
00:41:01,470 --> 00:41:04,130
base, it's not a lot of
calculations that need to be

713
00:41:04,130 --> 00:41:06,910
done at this point, you just
need to recognize that your

714
00:41:06,910 --> 00:41:09,420
salt should be neutral.

715
00:41:09,420 --> 00:41:12,920
So now, we can talk about what
happens if you add an extra

716
00:41:12,920 --> 00:41:17,170
milliliter of your acid, after
you've reached that

717
00:41:17,170 --> 00:41:19,820
equivalence point, which is
something that people probably

718
00:41:19,820 --> 00:41:22,820
have done in doing these
titrations not meaning to,

719
00:41:22,820 --> 00:41:27,550
gone well beyond the
equivalence point.

720
00:41:27,550 --> 00:41:31,140
So first we can find out the
number of moles of the strong

721
00:41:31,140 --> 00:41:36,250
acid that we've added extra,
so we added one mil of the

722
00:41:36,250 --> 00:41:39,660
strong acid, and so again,
it goes to completion.

723
00:41:39,660 --> 00:41:44,480
The concentration, the acid, is
0.34 molar times our 1 mil,

724
00:41:44,480 --> 00:41:45,940
so we've added 3 .

725
00:41:45,940 --> 00:41:52,100
4 times 10 to the minus 4 extra
moles of our acid, and

726
00:41:52,100 --> 00:41:55,650
so let's calculate then what the
molarity is that we have

727
00:41:55,650 --> 00:42:03,660
added that was extra.

728
00:42:03,660 --> 00:42:05,330
And even if you don't
have a calculator,

729
00:42:05,330 --> 00:42:41,980
we've helped you out.

730
00:42:41,980 --> 00:43:05,530
Let's just take 10
more seconds.

731
00:43:05,530 --> 00:43:10,650
I think this is a
first, isn't it?

732
00:43:10,650 --> 00:43:12,300
Did I mention it's a
good idea to start

733
00:43:12,300 --> 00:43:15,660
the problem-set early.

734
00:43:15,660 --> 00:43:20,060
So the trick here was
about the volume.

735
00:43:20,060 --> 00:43:25,960
So we had 25 mils to begin
with, then we added 18 .

736
00:43:25,960 --> 00:43:30,110
4 to get to the equivalence
point, and then we're 1 mil

737
00:43:30,110 --> 00:43:32,310
beyond the equivalence point.

738
00:43:32,310 --> 00:43:35,360
And the tricks to these parts
of the problems are to

739
00:43:35,360 --> 00:43:38,160
remember all the things
that have been added

740
00:43:38,160 --> 00:43:39,630
to get to the volume.

741
00:43:39,630 --> 00:43:43,150
So the problem itself is not
very tricky of having just an

742
00:43:43,150 --> 00:43:47,640
acid in water, but the one trick
is in calculating the

743
00:43:47,640 --> 00:43:50,530
molarity, remembering all of the
things that were added to

744
00:43:50,530 --> 00:43:52,490
get to this point.

745
00:43:52,490 --> 00:43:56,060
And so by the end of the
problem-set this should be

746
00:43:56,060 --> 00:43:59,230
pretty familiar to you, but it's
also something that you

747
00:43:59,230 --> 00:44:02,670
want to make sure that you check
on an exam that you're

748
00:44:02,670 --> 00:44:06,530
not making any volume
mistakes.

749
00:44:06,530 --> 00:44:10,620
So again, here are the things
that you want to remember to

750
00:44:10,620 --> 00:44:13,640
add in in calculating this.

751
00:44:13,640 --> 00:44:17,120
And then, if you calculate
the p h of that, you

752
00:44:17,120 --> 00:44:19,620
get a p h of 2 .

753
00:44:19,620 --> 00:44:24,450
1106, so that's down
somewhere in here.

754
00:44:24,450 --> 00:44:33,080
We've added 1 more mill, and
we're at a p h of 2 .

755
00:44:33,080 --> 00:44:38,140
1106, and this can be
a check as well.

756
00:44:38,140 --> 00:44:41,100
If you forgot to add some of
your volume, you might get a p

757
00:44:41,100 --> 00:44:43,230
h that doesn't make
a lot of sense.

758
00:44:43,230 --> 00:44:45,150
So that can be a double check.

759
00:44:45,150 --> 00:44:49,290
Always ask yourself when you
have a strong base and you're

760
00:44:49,290 --> 00:44:52,160
adding acid, before you've
added very much, your p h

761
00:44:52,160 --> 00:44:53,830
should be basic.

762
00:44:53,830 --> 00:44:56,800
At the equivalence point of
a strong acid, strong base

763
00:44:56,800 --> 00:44:58,710
titration, it'll be 7.

764
00:44:58,710 --> 00:45:01,660
And if you continue to add acid,
then you should have a

765
00:45:01,660 --> 00:45:04,590
pretty low p h, you'll notice
the curve drops off pretty

766
00:45:04,590 --> 00:45:07,580
fast. So it should be a dramatic
change in p h by

767
00:45:07,580 --> 00:45:09,750
adding extra of your acid.

768
00:45:09,750 --> 00:45:11,680
So always check your work.

769
00:45:11,680 --> 00:45:14,980
OK, so let's stop there, and
we're going to weak acid, weak

770
00:45:14,980 --> 00:45:17,760
base titrations next time.