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JOCELYN: Hi.

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00:00:22,380 --> 00:00:23,370
Jocelyn here.

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00:00:23,370 --> 00:00:27,000
Today, we're going to go over
fall 2009 final exam problem

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00:00:27,000 --> 00:00:28,680
number five.

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00:00:28,680 --> 00:00:31,110
As always, we're going to start
by reading the question.

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00:00:31,110 --> 00:00:35,010
The skeletal structure of the
amino acid alanine is given

14
00:00:35,010 --> 00:00:38,790
below as it exists in its
neutral zwitterion.

15
00:00:38,790 --> 00:00:40,430
To the right is shown
its titration

16
00:00:40,430 --> 00:00:42,520
curve in aqueous solution.

17
00:00:42,520 --> 00:00:45,270
The abscissa expresses
concentration in terms of

18
00:00:45,270 --> 00:00:49,460
degree of protonation so
that a value of 0.5--

19
00:00:49,460 --> 00:00:52,390
that the neutral ion is the
only species present.

20
00:00:52,390 --> 00:00:56,150
At a value of zero, alanine is
totally deprotonated and at a

21
00:00:56,150 --> 00:01:00,220
value of one, alanine is
totally protonated.

22
00:01:00,220 --> 00:01:02,240
So let's move to part A.

23
00:01:02,240 --> 00:01:04,630
What is the hybridization
of each of the

24
00:01:04,630 --> 00:01:08,720
three carbons in alanine?

25
00:01:08,720 --> 00:01:14,180
So to begin with this one, I
would start by rewriting the

26
00:01:14,180 --> 00:01:18,350
structure of alanine because as
it's written now, it's not

27
00:01:18,350 --> 00:01:22,130
as clear what the hybridization
of the alpha,

28
00:01:22,130 --> 00:01:24,800
beta and gamma carbons are.

29
00:01:24,800 --> 00:01:30,320
So starting with the
alpha carbon,

30
00:01:30,320 --> 00:01:31,570
connecting the beta carbon--

31
00:01:37,110 --> 00:01:45,204
and here is the gamma carbon--

32
00:01:52,960 --> 00:01:55,020
and of course, we don't want
to forget the amine group.

33
00:02:00,280 --> 00:02:02,520
So this looks fairly similar
to what is on the page, but

34
00:02:02,520 --> 00:02:05,920
I've kind of expanded the
structure so that it's more

35
00:02:05,920 --> 00:02:09,420
straightforward for us to
determine the hybridization.

36
00:02:09,420 --> 00:02:13,870
The alpha carbon is the central
carbon here and we see

37
00:02:13,870 --> 00:02:18,370
that it has four bonds and
four electron domains.

38
00:02:18,370 --> 00:02:20,780
If we go back to the beginning
of the class, we learned that

39
00:02:20,780 --> 00:02:23,280
when you have four electron
domains around a central

40
00:02:23,280 --> 00:02:29,550
carbon, we have Sp3
hybridization.

41
00:02:29,550 --> 00:02:32,510
Going to the beta carbon here,
we see that it also has four

42
00:02:32,510 --> 00:02:36,700
bonds, but one of them is
a double bond to oxygen.

43
00:02:36,700 --> 00:02:40,920
So we need to remember from our
earlier lessons that the

44
00:02:40,920 --> 00:02:43,830
double bond only counts as
one electron domain.

45
00:02:43,830 --> 00:02:49,606
Therefore, this carbon has
three electron domains.

46
00:02:54,180 --> 00:02:56,520
And going back to our
hybridization rules, we know

47
00:02:56,520 --> 00:03:01,660
that this gives us an
Sp2 hybridization.

48
00:03:01,660 --> 00:03:06,500
Lastly, we have our gamma carbon
and it gives us one,

49
00:03:06,500 --> 00:03:10,790
two, three, four bonds and
four electron domains.

50
00:03:10,790 --> 00:03:18,800
Therefore, our gamma is
also Sp3 hybridized.

51
00:03:18,800 --> 00:03:23,630
This problem on the final was
dealing with first beginning

52
00:03:23,630 --> 00:03:26,940
material, but then in part B and
C, we're going to move on

53
00:03:26,940 --> 00:03:30,330
to the material we've learned
more recently.

54
00:03:30,330 --> 00:03:34,050
So moving to part B--

55
00:03:34,050 --> 00:03:36,780
we are now are dealing with the
titration curve that is

56
00:03:36,780 --> 00:03:38,080
given to us.

57
00:03:38,080 --> 00:03:40,990
So this looks hopefully fairly
similar to what is on your

58
00:03:40,990 --> 00:03:45,690
page and the question-- it's
asking us to indicate on the

59
00:03:45,690 --> 00:03:47,030
titration curve--

60
00:03:47,030 --> 00:03:51,470
one, the PKA for protonation
of the zwitterion.

61
00:03:51,470 --> 00:03:55,650
Two, the PKA for deprotonation
of the zwitterion.

62
00:03:55,650 --> 00:03:59,360
And three, the isoelectric
point.

63
00:03:59,360 --> 00:04:02,180
So before we start that, let's
talk a little bit about this

64
00:04:02,180 --> 00:04:03,770
titration curve here.

65
00:04:03,770 --> 00:04:04,800
What is it showing us?

66
00:04:04,800 --> 00:04:06,660
What is it depicting?

67
00:04:06,660 --> 00:04:08,600
What I like to do is--
we see that we have

68
00:04:08,600 --> 00:04:11,640
two buffered regions.

69
00:04:11,640 --> 00:04:13,720
That right off the bat should
tell you that there are two

70
00:04:13,720 --> 00:04:15,900
different equilibriums
going on.

71
00:04:15,900 --> 00:04:17,980
There are two different
protonation/

72
00:04:17,980 --> 00:04:19,880
deprotonation reactions.

73
00:04:19,880 --> 00:04:25,690
So I'm going to split this right
down the middle here.

74
00:04:31,010 --> 00:04:34,770
And that way we can think about
the two protonations/

75
00:04:34,770 --> 00:04:37,980
deprotonations separately.

76
00:04:37,980 --> 00:04:43,930
So on the left side here, we
have the more acidic portion

77
00:04:43,930 --> 00:04:48,360
of our titration curve and
even if you didn't quite

78
00:04:48,360 --> 00:04:51,810
understand his description of
the abscissa, we know that

79
00:04:51,810 --> 00:04:54,050
because this side--

80
00:04:54,050 --> 00:04:57,210
the solution has a lower Ph,
it's going to be dealing with

81
00:04:57,210 --> 00:05:00,790
the more protonated species
of the alanine.

82
00:05:00,790 --> 00:05:01,300
This--

83
00:05:01,300 --> 00:05:04,490
on the right side, we have the
higher Ph, so we're going to

84
00:05:04,490 --> 00:05:09,100
have a more deprotonated
alanine on this side.

85
00:05:09,100 --> 00:05:11,980
And he already told
us that we had the

86
00:05:11,980 --> 00:05:13,960
zwitterion in the middle.

87
00:05:13,960 --> 00:05:22,240
So next to 0.5, we're going to
write a generic term for a

88
00:05:22,240 --> 00:05:24,130
neutral acid.

89
00:05:24,130 --> 00:05:25,810
If we look back to
our structure,

90
00:05:25,810 --> 00:05:29,010
we see that we have--

91
00:05:29,010 --> 00:05:36,770
the neutral zwitterion has one
acidic proton here on the

92
00:05:36,770 --> 00:05:40,220
amine group and it has
a neutral charge.

93
00:05:40,220 --> 00:05:45,030
So that's why I'm going to
depict that zwitterion as HA

94
00:05:45,030 --> 00:05:48,320
Moving to the left, more
acidic portion of the

95
00:05:48,320 --> 00:05:50,620
titration curve, we have--

96
00:05:50,620 --> 00:05:52,970
we know that there's a
protonation going on.

97
00:05:52,970 --> 00:05:59,920
So I'm going to write that as
H2A plus because whenever we

98
00:05:59,920 --> 00:06:03,820
add a proton, we're adding
a positive charge.

99
00:06:03,820 --> 00:06:10,240
And then, at the more basic
side, we have doubly

100
00:06:10,240 --> 00:06:15,310
deprotonated alanine and so we
have just our backbone, which

101
00:06:15,310 --> 00:06:19,580
I'm going to call a and
a negative charge.

102
00:06:19,580 --> 00:06:22,420
So from the question,
we should know that

103
00:06:22,420 --> 00:06:24,500
this is true, correct?

104
00:06:24,500 --> 00:06:28,780
Because he told us that we have
the totally protonated

105
00:06:28,780 --> 00:06:31,310
alanine on the left side.

106
00:06:31,310 --> 00:06:35,160
We have the zwitterion in the
middle at 0.5 and we have the

107
00:06:35,160 --> 00:06:39,320
totally deprotonated
on the right side.

108
00:06:39,320 --> 00:06:39,470
OK.

109
00:06:39,470 --> 00:06:42,050
What else can we know from this
titration curve or what

110
00:06:42,050 --> 00:06:43,930
can we add to it so that
we can understand it

111
00:06:43,930 --> 00:06:45,600
a little bit better?

112
00:06:45,600 --> 00:06:49,580
Well, let's write down the
equilibrium reactions that are

113
00:06:49,580 --> 00:06:54,350
happening that make this
buffer region here.

114
00:06:54,350 --> 00:07:03,830
On left side, we have the doubly
protonated alanine

115
00:07:03,830 --> 00:07:17,500
going to proton plus
the zwitterion.

116
00:07:17,500 --> 00:07:21,780
And that's what's creating
this buffered Ph region.

117
00:07:21,780 --> 00:07:29,460
On the right side here, we
have the zwitterion in

118
00:07:29,460 --> 00:07:31,750
equilibrium with--

119
00:07:31,750 --> 00:07:38,310
again, a proton and the doubly
deprotonated alanine.

120
00:07:42,230 --> 00:07:44,390
These are things that
are true for any

121
00:07:44,390 --> 00:07:45,830
titration curve you see.

122
00:07:45,830 --> 00:07:48,350
Depending upon the number
of protonations--

123
00:07:48,350 --> 00:07:51,600
the number of times a buffer
region might change, but for

124
00:07:51,600 --> 00:07:55,750
each buffered region, you
can always determine an

125
00:07:55,750 --> 00:08:02,400
equilibrium constant that is
causing that stability in Ph.

126
00:08:02,400 --> 00:08:02,760
OK.

127
00:08:02,760 --> 00:08:04,460
Now let's go to the question.

128
00:08:07,910 --> 00:08:15,970
We need to indicate on the curve
the PKA of both this

129
00:08:15,970 --> 00:08:20,170
equilibrium equation and this
equilibrium reaction.

130
00:08:20,170 --> 00:08:22,790
So to do that, we hopefully
will remember the

131
00:08:22,790 --> 00:08:25,790
Henderson-Hasselbalch equation
and that gives us a

132
00:08:25,790 --> 00:08:32,470
relationship between the Ph of
the solution and the PKA of

133
00:08:32,470 --> 00:08:36,430
the acid that's in solution.

134
00:08:36,430 --> 00:08:37,500
So writing down the

135
00:08:37,500 --> 00:08:38,875
Henderson-Hasselbalch equation--

136
00:08:45,090 --> 00:08:47,450
we have--

137
00:08:47,450 --> 00:08:49,150
not the Ph--

138
00:08:49,150 --> 00:08:53,270
and this is the same equation
that Professor Sadoway derived

139
00:08:53,270 --> 00:08:54,520
in lecture.

140
00:09:11,220 --> 00:09:16,740
And here, I've written it in the
most generic form, but no

141
00:09:16,740 --> 00:09:20,560
matter what our equilibrium
species are, we just know that

142
00:09:20,560 --> 00:09:26,390
we put the more basic species
on top and the acid that got

143
00:09:26,390 --> 00:09:29,360
deprotonated on the bottom.

144
00:09:29,360 --> 00:09:34,550
So for the first point of this,
we are looking for the

145
00:09:34,550 --> 00:09:38,210
PKA of the protonation
of the zwitterion.

146
00:09:38,210 --> 00:09:46,420
So we know that it's an
equilibrium between the doubly

147
00:09:46,420 --> 00:09:57,340
deprotonated and
the zwitterion.

148
00:09:57,340 --> 00:10:00,770
That is the equation
we're looking at.

149
00:10:00,770 --> 00:10:05,300
So let's write down our
Henderson-Hasselbalch for that

150
00:10:05,300 --> 00:10:08,295
and I'm going to call this PK1
just to keep them separate.

151
00:10:19,080 --> 00:10:24,020
Remember here, my Ha is on
top because we have--

152
00:10:24,020 --> 00:10:29,510
our doubly protonated species
is the acid in this case.

153
00:10:29,510 --> 00:10:34,060
Now in order to put on the
titration curve where the PKA

154
00:10:34,060 --> 00:10:38,060
is, we look at the
Henderson-Hasselbalch equation

155
00:10:38,060 --> 00:10:45,390
and see that where the Ph equals
the PKA is when the

156
00:10:45,390 --> 00:10:49,190
concentration of the acid and
its conjugate base are

157
00:10:49,190 --> 00:10:51,190
equivalent.

158
00:10:51,190 --> 00:11:04,240
because if Ha equals H2a plus,
the log term goes to zero and

159
00:11:04,240 --> 00:11:08,300
we have our Ph equals PK1.

160
00:11:08,300 --> 00:11:13,740
Going back to the titration
curve, we see that the

161
00:11:13,740 --> 00:11:18,590
concentration between H2a
plus and Ha is equal

162
00:11:18,590 --> 00:11:20,050
right in the middle.

163
00:11:20,050 --> 00:11:26,040
We can assume that this is
a to-scale concentration

164
00:11:26,040 --> 00:11:27,290
abscissa here.

165
00:11:27,290 --> 00:11:31,750
And so there we go up and we can
say, OK, this point on the

166
00:11:31,750 --> 00:11:38,880
titration curve here corresponds
to our PK1.

167
00:11:46,460 --> 00:11:48,070
So from the
Henderson-Hasselbalch

168
00:11:48,070 --> 00:11:52,080
equation, we can see that given
the titration curve,

169
00:11:52,080 --> 00:11:56,360
this is where our PK1 is
and it's about two.

170
00:11:56,360 --> 00:11:58,710
For our purposes, that's--

171
00:11:58,710 --> 00:12:02,090
we weren't asked to estimate it,
but if I had to, I'd say

172
00:12:02,090 --> 00:12:04,160
it's around two.

173
00:12:04,160 --> 00:12:06,940
Now we have to do the same thing
for the other side of

174
00:12:06,940 --> 00:12:10,620
the titration curve.

175
00:12:10,620 --> 00:12:12,800
So if you haven't done that
already, maybe you can stop

176
00:12:12,800 --> 00:12:19,970
now and try that on your own
because it's the same process.

177
00:12:19,970 --> 00:12:20,830
OK.

178
00:12:20,830 --> 00:12:23,390
So it's do of the second one--

179
00:12:23,390 --> 00:12:30,630
which again, is just the Ph is
equal to the PKA when the two

180
00:12:30,630 --> 00:12:33,676
species are equal, except for
now we have different species.

181
00:12:33,676 --> 00:12:36,670
So basically we're
looking for--

182
00:12:36,670 --> 00:12:40,130
our new Henderson-Hasselbalch
equation is--

183
00:12:45,460 --> 00:12:52,530
with using the new equilibrium
reaction, our doubly

184
00:12:52,530 --> 00:12:59,650
deprotonated alanine is now on
top and our zwitterion is on

185
00:12:59,650 --> 00:13:01,200
the bottom.

186
00:13:01,200 --> 00:13:03,550
So this might be a little
confusing because we have the

187
00:13:03,550 --> 00:13:06,120
same species going from the
top to bottom in our

188
00:13:06,120 --> 00:13:08,950
Henderson-Hasselbalch equation,
but remember that

189
00:13:08,950 --> 00:13:13,350
we're always talking about the
acid is in the denominator and

190
00:13:13,350 --> 00:13:18,440
the conjugate base is in
the numerator here.

191
00:13:18,440 --> 00:13:21,740
But either way, we again want
to look for where the Ph

192
00:13:21,740 --> 00:13:26,230
equals the PK2 and so our
zwitterion concentration will

193
00:13:26,230 --> 00:13:30,760
equal our doubly deprotonated
alanine.

194
00:13:30,760 --> 00:13:34,675
And that occurs right here.

195
00:13:43,620 --> 00:13:49,040
So the last part of this
question is to indicate where

196
00:13:49,040 --> 00:13:54,810
the isoelectric point is and the
important thing for this

197
00:13:54,810 --> 00:13:56,840
part of the problem, of course,
is to know what the

198
00:13:56,840 --> 00:13:58,610
isoelectric point meets--

199
00:13:58,610 --> 00:14:04,010
and it's where the dominant
species in the solution is the

200
00:14:04,010 --> 00:14:05,590
neutral zwitterion.

201
00:14:05,590 --> 00:14:09,035
And Professor Sadoway already
told us where that is and it

202
00:14:09,035 --> 00:14:12,830
is where we're at 0.5 here,
which we've already indicated

203
00:14:12,830 --> 00:14:16,670
as where the zwitterion
is dominant.

204
00:14:16,670 --> 00:14:21,040
And so we just can go up here
and circle that and we can

205
00:14:21,040 --> 00:14:23,510
call that number three.

206
00:14:26,730 --> 00:14:29,980
So you didn't have to go through
all of the work I did

207
00:14:29,980 --> 00:14:33,047
and if you were familiar with
the titration curve, you could

208
00:14:33,047 --> 00:14:37,500
have simply put these down, but
understanding where these

209
00:14:37,500 --> 00:14:40,260
points come from is quite
important for understanding

210
00:14:40,260 --> 00:14:44,670
acid base chemistry and
how amino acids work.

211
00:14:44,670 --> 00:14:45,130
All right.

212
00:14:45,130 --> 00:14:46,380
So let's move to part C.

213
00:14:52,920 --> 00:14:56,360
In part C, we're asked to draw
the skeletal structure of

214
00:14:56,360 --> 00:14:59,330
alanine when it is solvated
in an aqueous

215
00:14:59,330 --> 00:15:03,420
solution at extreme acidity--

216
00:15:03,420 --> 00:15:07,390
ie, Ph less than one.

217
00:15:07,390 --> 00:15:12,010
So I'm going to write that down
and then we can go back

218
00:15:12,010 --> 00:15:15,720
to our titration curve and see
where that is because we've

219
00:15:15,720 --> 00:15:17,850
already spend a lot of time
understanding our titration

220
00:15:17,850 --> 00:15:22,480
curve and that tells us a lot
about the acid base chemistry

221
00:15:22,480 --> 00:15:24,790
of alanine.

222
00:15:24,790 --> 00:15:28,780
So at the titration curve, we
see that at a Ph less than

223
00:15:28,780 --> 00:15:32,990
one, we're going to have mostly
our doubly protonated

224
00:15:32,990 --> 00:15:35,440
zwitterion.

225
00:15:35,440 --> 00:15:38,090
That said, now we need to go
back and look at the structure

226
00:15:38,090 --> 00:15:42,780
of alanine and we can put
in the protonation.

227
00:15:42,780 --> 00:16:00,980
So we're given the zwitterion
form of alanine, but at very

228
00:16:00,980 --> 00:16:03,660
low acidity--

229
00:16:03,660 --> 00:16:06,270
sorry-- high acidity, low Ph.

230
00:16:06,270 --> 00:16:10,680
This carboxy group is going
to be protonated.

231
00:16:10,680 --> 00:16:14,450
And so we have both of our
acidic protons are on the

232
00:16:14,450 --> 00:16:19,580
alanine and this would be a
structure of alanine at a Ph

233
00:16:19,580 --> 00:16:20,830
less than one.

234
00:16:23,020 --> 00:16:24,473
Now moving to part D--

235
00:16:28,480 --> 00:16:32,250
we are asked for an aqueous
solution of alanine--

236
00:16:32,250 --> 00:16:35,700
calculate the ratio of the
concentration of neutral

237
00:16:35,700 --> 00:16:39,750
alanine zwitterion to the
concentration of deprotonated

238
00:16:39,750 --> 00:16:45,100
anion when the Ph is 8.091.

239
00:16:45,100 --> 00:16:48,540
So a couple things I
would write down.

240
00:16:48,540 --> 00:16:58,510
Ph equals 8.091 and we are asked
for the ratio of the

241
00:16:58,510 --> 00:17:05,710
concentration of the zwitterion
to the doubly

242
00:17:05,710 --> 00:17:07,040
deprotonated anion.

243
00:17:11,340 --> 00:17:14,960
When we're asked for this,
hopefully you think first,

244
00:17:14,960 --> 00:17:16,110
Henderson-Hasselbalch--

245
00:17:16,110 --> 00:17:17,730
I can use that.

246
00:17:17,730 --> 00:17:19,840
We're given a Ph.

247
00:17:19,840 --> 00:17:23,780
We can find the PKA and we're
asked for the ratio and

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00:17:23,780 --> 00:17:26,070
concentrations.

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00:17:26,070 --> 00:17:28,050
So let's write down the

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00:17:28,050 --> 00:17:30,580
Henderson-Hasselbalch equation again.

251
00:17:43,310 --> 00:17:45,590
And because we're asked for the
ratio, we don't need any

252
00:17:45,590 --> 00:17:49,030
more information except
for the PKA.

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00:17:49,030 --> 00:17:54,580
Luckily, from part B, we know
what the PKA is of this

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00:17:54,580 --> 00:17:57,590
equilibrium between the
zwitterion and the doubly

255
00:17:57,590 --> 00:18:00,151
deprotonated anion.

256
00:18:00,151 --> 00:18:04,850
So moving back to your titration
curve, we see that

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00:18:04,850 --> 00:18:08,000
the PKA is roughly 10--

258
00:18:08,000 --> 00:18:09,850
doesn't need to be our--

259
00:18:09,850 --> 00:18:12,690
our estimation doesn't need to
be exact, but if it's in the

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00:18:12,690 --> 00:18:16,760
ballpark range, then you
have the right idea.

261
00:18:16,760 --> 00:18:20,370
So I'm going to write that.

262
00:18:20,370 --> 00:18:22,607
We can plug that in
just down here.

263
00:18:32,060 --> 00:18:33,310
Log of our concentration.

264
00:18:35,980 --> 00:18:38,740
That's a log term.

265
00:18:38,740 --> 00:18:41,430
And from here, it's
just algebra.

266
00:18:41,430 --> 00:18:44,460
So simplify a little bit.

267
00:18:44,460 --> 00:18:50,460
If we bring the 10 over to
this side, we get -1.909

268
00:18:50,460 --> 00:18:57,131
equals the log of
a minus over--

269
00:18:59,780 --> 00:19:00,210
I'm sorry.

270
00:19:00,210 --> 00:19:02,750
I'm getting a little
bit crooked here.

271
00:19:02,750 --> 00:19:06,270
And then if you take both sides
of the power of 10,

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00:19:06,270 --> 00:19:07,520
we'll get--

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00:19:23,330 --> 00:19:25,250
the ratio equal to this.

274
00:19:25,250 --> 00:19:29,840
And if we take the reciprocal
of this,

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00:19:29,840 --> 00:19:32,020
we'll get a nicer answer.

276
00:19:32,020 --> 00:19:36,800
We'll get HA over A minus
and it won't be a

277
00:19:36,800 --> 00:19:37,650
fraction over here.

278
00:19:37,650 --> 00:19:41,030
So I suggest that you do that
and you'll see that our final

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00:19:41,030 --> 00:19:42,280
answer is--

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00:19:44,580 --> 00:19:45,830
let's put it right here.

281
00:19:48,380 --> 00:19:51,200
81.1:1--

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00:19:51,200 --> 00:19:56,820
the ratio of our zwitterion to
the doubly deprotonated anion.

283
00:20:00,540 --> 00:20:03,290
And then you've answered all
parts of the problem.