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JOCELYN: Hi.

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00:00:21,990 --> 00:00:22,750
Jocelyn here.

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00:00:22,750 --> 00:00:25,930
And today we're going to go over
fall 2009's final exam,

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00:00:25,930 --> 00:00:27,900
problem 10.

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00:00:27,900 --> 00:00:30,420
As always we're going to start
by reading the question.

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00:00:30,420 --> 00:00:34,450
You have 333 milliliters of
alkaline solution at a pH

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00:00:34,450 --> 00:00:36,330
equal to 9.9.

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00:00:36,330 --> 00:00:39,490
You wish to neutralize this
by reacting it with 222

16
00:00:39,490 --> 00:00:41,380
milliliters of acid.

17
00:00:41,380 --> 00:00:47,500
What must be the value of
the pH of the acid?

18
00:00:47,500 --> 00:00:52,940
So first off let's write down
the information that was given

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00:00:52,940 --> 00:00:55,670
and the information that
he's asking for.

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00:00:55,670 --> 00:01:04,990
So we have 333 milliliters
of a solution

21
00:01:04,990 --> 00:01:09,360
with pH equal to 9.9.

22
00:01:09,360 --> 00:01:15,590
And he's asking you what must
the pH be if you want to put

23
00:01:15,590 --> 00:01:21,330
in 222 milliliters in order to
completely neutralize that

24
00:01:21,330 --> 00:01:24,030
basic solution.

25
00:01:24,030 --> 00:01:26,040
And the key here
is to recognize

26
00:01:26,040 --> 00:01:28,150
what neutralize means.

27
00:01:28,150 --> 00:01:35,350
And a neutralization is when you
take acid, or protons, and

28
00:01:35,350 --> 00:01:41,960
base and it reacts to
form water, thus

29
00:01:41,960 --> 00:01:44,950
creating a neutral solution.

30
00:01:44,950 --> 00:01:50,780
So the next step would probably
be to figure out how

31
00:01:50,780 --> 00:01:55,590
much excess hydroxide we
have in this solution.

32
00:01:55,590 --> 00:01:59,950
And then we can figure out how
much hydrogen we need to react

33
00:01:59,950 --> 00:02:05,340
with that hydroxide and to
form water in order to

34
00:02:05,340 --> 00:02:08,610
completely neutralize
this solution.

35
00:02:08,610 --> 00:02:13,630
So first step is to
find out what our

36
00:02:13,630 --> 00:02:17,840
concentration of OH minus is.

37
00:02:17,840 --> 00:02:19,920
And we do that by remembering
a few things

38
00:02:19,920 --> 00:02:22,900
about acid/base solutions.

39
00:02:22,900 --> 00:02:24,330
Right?

40
00:02:24,330 --> 00:02:31,550
We know that if you have the pH
plus the pOH it equals 14.

41
00:02:31,550 --> 00:02:35,370
And this is from the equilibrium
constant of this

42
00:02:35,370 --> 00:02:38,810
neutralization reaction.

43
00:02:38,810 --> 00:02:45,530
And then we can figure out what
the pOH is because we're

44
00:02:45,530 --> 00:02:50,100
given the pH of our
initial solution--

45
00:02:50,100 --> 00:02:51,920
9.9--

46
00:02:51,920 --> 00:02:55,660
and that equals 4.01.

47
00:02:55,660 --> 00:03:02,720
Now using the definition of
pOH, which is that the pOH

48
00:03:02,720 --> 00:03:10,300
equals the negative log of the
concentration of the OH minus,

49
00:03:10,300 --> 00:03:17,840
we can rearrange that and
get that our OH minus

50
00:03:17,840 --> 00:03:23,350
concentration equals 10
to the negative pOH.

51
00:03:23,350 --> 00:03:23,630
Right?

52
00:03:23,630 --> 00:03:26,740
Just using our log
rules there.

53
00:03:26,740 --> 00:03:31,930
So this gives us that our
current concentration of the

54
00:03:31,930 --> 00:03:38,700
hydroxide ion is 10 to
the negative 4.1.

55
00:03:38,700 --> 00:03:40,220
That gives us the concentration
of the

56
00:03:40,220 --> 00:03:43,110
hydroxide, but what we really
want to know, if we're going

57
00:03:43,110 --> 00:03:46,720
to use our neutralization
reaction here, is we want to

58
00:03:46,720 --> 00:03:48,960
know how many moles
of hydroxide we

59
00:03:48,960 --> 00:03:50,840
have in that solution.

60
00:03:50,840 --> 00:03:53,390
And we can just do that
with stoichiometry.

61
00:03:53,390 --> 00:03:54,220
Right?

62
00:03:54,220 --> 00:04:00,290
The concentration is
in moles per liter.

63
00:04:00,290 --> 00:04:01,810
And we know--

64
00:04:01,810 --> 00:04:11,370
so our moles hydroxide equals
our concentration, which is 10

65
00:04:11,370 --> 00:04:18,630
to the negative 4.1
moles per liter.

66
00:04:18,630 --> 00:04:20,080
Times--

67
00:04:20,080 --> 00:04:26,640
if we want to get rid of that
liter unit we multiply it by

68
00:04:26,640 --> 00:04:37,890
our volume, which is 333
milliliters, converting it

69
00:04:37,890 --> 00:04:40,970
from milliliters to liters.

70
00:04:40,970 --> 00:04:58,720
And this gives us 2.65 times 10
to the negative 5 moles of

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00:04:58,720 --> 00:05:02,950
OH minus in solution.

72
00:05:02,950 --> 00:05:04,990
So that's just the first
step of this problem.

73
00:05:04,990 --> 00:05:08,600
Now that we have the number of
moles of OH minus we need to

74
00:05:08,600 --> 00:05:12,350
find the number of moles of H
plus necessary to react with

75
00:05:12,350 --> 00:05:17,110
this excess hydroxide, and then
we will be able to find

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00:05:17,110 --> 00:05:20,920
the concentration of hydrogen,
or protons,

77
00:05:20,920 --> 00:05:24,650
in that second solution.

78
00:05:24,650 --> 00:05:27,730
So first off we're going to
look at our neutralization

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00:05:27,730 --> 00:05:31,470
reaction over here and realize
that it's a one-to-one

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00:05:31,470 --> 00:05:33,340
stoichiometric ratio.

81
00:05:33,340 --> 00:05:36,990
That means in order to react
with 2.65 times 10 to the

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00:05:36,990 --> 00:05:40,760
negative 5 moles of hydroxide,
we need the same number of

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00:05:40,760 --> 00:05:43,300
moles of hydrogen.

84
00:05:43,300 --> 00:05:48,420
So over here I'm going to say
that this also equals the

85
00:05:48,420 --> 00:05:51,960
moles of H plus needed.

86
00:05:56,100 --> 00:05:57,690
All right.

87
00:05:57,690 --> 00:06:00,690
Now that we have our moles H
plus needed, we need to do the

88
00:06:00,690 --> 00:06:04,630
same process here
but in reverse.

89
00:06:04,630 --> 00:06:09,700
So first we need to find the
concentration of protons that

90
00:06:09,700 --> 00:06:18,270
we have. So given that we want
to only have 222 milliliters

91
00:06:18,270 --> 00:06:20,190
of this solution, right?

92
00:06:20,190 --> 00:06:21,440
So we have--

93
00:06:29,550 --> 00:06:38,140
and we just divide that by
our volume, 0.222 liters.

94
00:06:38,140 --> 00:06:38,430
Right?

95
00:06:38,430 --> 00:06:40,740
Because we divide by 1,000 to
convert from milliliters.

96
00:06:43,240 --> 00:06:52,500
And that equals 1.19
times 10 to the

97
00:06:52,500 --> 00:06:56,716
negative 4 moles per liter.

98
00:06:56,716 --> 00:06:59,670
So this is another problem where
you could get the answer

99
00:06:59,670 --> 00:07:03,170
but not actually be answering
the question.

100
00:07:03,170 --> 00:07:03,490
Right?

101
00:07:03,490 --> 00:07:07,370
Because this is the
concentration of protons that

102
00:07:07,370 --> 00:07:11,670
we need, but now we need
to convert that to pH.

103
00:07:11,670 --> 00:07:15,390
So moving over here we'll just
write the final steps.

104
00:07:15,390 --> 00:07:21,100
We know the definition of pH
is the negative log of the

105
00:07:21,100 --> 00:07:23,750
concentration.

106
00:07:23,750 --> 00:07:28,740
And so because we know the
concentration of protons we

107
00:07:28,740 --> 00:07:30,430
know that the--

108
00:07:34,340 --> 00:07:42,450
we just need to plug this into
our calculator and we get our

109
00:07:42,450 --> 00:07:51,180
answer, which is a pH of 3.92.

110
00:07:51,180 --> 00:07:54,110
Put a box around that one
because it's our correct final

111
00:07:54,110 --> 00:07:56,200
answer, and you'll
be good to go for

112
00:07:56,200 --> 00:07:59,230
this part of the problem.

113
00:07:59,230 --> 00:08:04,740
Now moving on to part b we are
asked, name the conjugate base

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00:08:04,740 --> 00:08:06,790
of each of the following.

115
00:08:06,790 --> 00:08:13,030
So we'll just start
part b right here.

116
00:08:13,030 --> 00:08:15,870
And the first thing to recognize
is what does a

117
00:08:15,870 --> 00:08:18,170
conjugate base mean.

118
00:08:18,170 --> 00:08:23,520
And in an acid/base pair, or
when you have a kind of

119
00:08:23,520 --> 00:08:28,470
dissociation of an acid, in the
most simple terms we can

120
00:08:28,470 --> 00:08:31,430
think about it just
losing a proton.

121
00:08:33,940 --> 00:08:37,570
In the leftover molecule, that's
the conjugate base

122
00:08:37,570 --> 00:08:38,900
right there.

123
00:08:38,900 --> 00:08:45,095
So in this problem we're asked
if the molecule loses a

124
00:08:45,095 --> 00:08:48,760
proton, what is the resulting
molecule?

125
00:08:48,760 --> 00:08:51,050
Ion, basically.

126
00:08:51,050 --> 00:09:02,800
So part i we have HPO4, and
that will dissociate to

127
00:09:02,800 --> 00:09:07,820
hydrogen phosphate ion.

128
00:09:10,460 --> 00:09:11,260
Simple enough.

129
00:09:11,260 --> 00:09:14,910
This is your conjugate base.

130
00:09:14,910 --> 00:09:16,930
So that's your answer
for part i.

131
00:09:20,970 --> 00:09:22,280
Moving to part ii.

132
00:09:22,280 --> 00:09:32,010
We have CH3NH3 plus.

133
00:09:32,010 --> 00:09:36,810
So it's kind of like an ammonia
ion but not quite--

134
00:09:36,810 --> 00:09:38,390
sorry, ammonium.

135
00:09:38,390 --> 00:09:42,310
And we know that this
dissociates by losing a

136
00:09:42,310 --> 00:09:46,440
hydrogen off of the nitrogen,
because that's where the

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00:09:46,440 --> 00:09:47,576
positive charge is.

138
00:09:47,576 --> 00:09:50,490
It makes more sense there.

139
00:09:50,490 --> 00:09:52,990
And we get--

140
00:09:52,990 --> 00:09:55,930
sorry, keeping the same
order as above--

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00:09:55,930 --> 00:09:57,180
a proton plus.

142
00:10:03,040 --> 00:10:07,850
And this would be your
conjugate base.

143
00:10:07,850 --> 00:10:10,100
If you were a little bit
confused on this one it might

144
00:10:10,100 --> 00:10:12,130
help to draw the Lewis
structure, and you would see

145
00:10:12,130 --> 00:10:17,360
that the formal charge does
actually reside on nitrogen

146
00:10:17,360 --> 00:10:20,000
and, therefore, it makes
sense that the nitrogen

147
00:10:20,000 --> 00:10:21,380
would lose a proton.

148
00:10:21,380 --> 00:10:26,560
In addition, if you think about
the ammonia/ammonium ion

149
00:10:26,560 --> 00:10:31,600
relationship you know that you
have NH3 going to NH4.

150
00:10:31,600 --> 00:10:34,760
And in this case we just have
one of the hydrogens replaced

151
00:10:34,760 --> 00:10:39,050
with a methyl group, and so you
can make the analogy there

152
00:10:39,050 --> 00:10:41,850
that, oh, the hydrogen
would come from the

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00:10:41,850 --> 00:10:44,190
nitrogen in that case.

154
00:10:44,190 --> 00:10:44,640
All right.

155
00:10:44,640 --> 00:10:47,120
So hopefully this was pretty
straightforward.

156
00:10:47,120 --> 00:10:51,340
As long as you're comfortable
with what a conjugate base is

157
00:10:51,340 --> 00:10:56,120
and what the dissociation
reactions look like for acids.

158
00:10:56,120 --> 00:11:00,150
So now on part c we are asked
to classify each of the

159
00:11:00,150 --> 00:11:02,760
following as a Lewis acid
or a Lewis base.

160
00:11:06,050 --> 00:11:09,062
Now first we need to remember
what a Lewis acid and

161
00:11:09,062 --> 00:11:10,890
a Lewis base are.

162
00:11:10,890 --> 00:11:20,450
So if you remember from lecture
or in your reading, we

163
00:11:20,450 --> 00:11:31,960
know that a Lewis acid is an
electron-pair acceptor and a

164
00:11:31,960 --> 00:11:36,230
Lewis base is an electron-pair
donor.

165
00:11:40,900 --> 00:11:43,780
These definitions are just a
little more general than the

166
00:11:43,780 --> 00:11:46,420
classic proton acceptor
or proton donor.

167
00:11:50,410 --> 00:11:55,220
The first part is asking us
about the cyanide ion.

168
00:11:55,220 --> 00:12:01,380
And to kind of clarify it for
yourself and because we're

169
00:12:01,380 --> 00:12:05,350
asked about Lewis acids and
bases, which are to deal with

170
00:12:05,350 --> 00:12:08,440
electron pairs, it might
be helpful to

171
00:12:08,440 --> 00:12:10,480
draw the Lewis structure.

172
00:12:10,480 --> 00:12:14,900
So for this molecule a good
Lewis structure looks like

173
00:12:14,900 --> 00:12:22,310
this, and it has a net
negative charge.

174
00:12:22,310 --> 00:12:27,700
And you can see that we have a
couple of lone electron pairs

175
00:12:27,700 --> 00:12:30,130
there that aren't involved
in bonding.

176
00:12:30,130 --> 00:12:34,260
And because it has a net
negative charge you can

177
00:12:34,260 --> 00:12:38,080
imagine that these electron
pairs would want to kind of

178
00:12:38,080 --> 00:12:40,170
grab onto something else that's
a little bit more

179
00:12:40,170 --> 00:12:43,110
electron positive.

180
00:12:43,110 --> 00:12:46,390
And because of that we
know that this is

181
00:12:46,390 --> 00:12:49,540
actually a Lewis base.

182
00:12:49,540 --> 00:12:53,230
It acts as a base and would grab
something like a proton

183
00:12:53,230 --> 00:13:01,790
and thus become more, I don't
know, stable in some cases.

184
00:13:01,790 --> 00:13:02,190
All right.

185
00:13:02,190 --> 00:13:04,540
Moving to part 2.

186
00:13:04,540 --> 00:13:08,120
We are asked about water, which
is a molecule that we

187
00:13:08,120 --> 00:13:12,200
deal a lot with in acid/base
chemistry.

188
00:13:12,200 --> 00:13:14,880
Again, you might be able to
answer this right off the bat,

189
00:13:14,880 --> 00:13:19,910
but we're going to look at
the Lewis structure here.

190
00:13:19,910 --> 00:13:22,580
Water has two lone pairs.

191
00:13:22,580 --> 00:13:28,560
So that means it could very
well act as a Lewis base.

192
00:13:28,560 --> 00:13:29,850
Hopefully that makes a
little bit of sense.

193
00:13:29,850 --> 00:13:30,130
Right?

194
00:13:30,130 --> 00:13:34,200
Because these electrons can go
and grab something like a

195
00:13:34,200 --> 00:13:42,140
proton and form the hydronium
ion with a positive charge,

196
00:13:42,140 --> 00:13:44,950
thus acting as a Lewis base.

197
00:13:51,750 --> 00:13:54,260
It's important to remember
that even though we're

198
00:13:54,260 --> 00:13:58,210
creating an acid in the
hydronium ion, this water

199
00:13:58,210 --> 00:14:04,400
molecule in converting to the
acid acts as a Lewis base.

200
00:14:04,400 --> 00:14:09,630
However, we also know that water
can lose a hydrogen and

201
00:14:09,630 --> 00:14:22,700
accepts those electrons to form
the hydroxide ion, and

202
00:14:22,700 --> 00:14:27,040
thus acts as a Lewis acid.

203
00:14:27,040 --> 00:14:32,780
So in the case of water either
Lewis acid or Lewis base was a

204
00:14:32,780 --> 00:14:35,580
correct answer in this case.

205
00:14:35,580 --> 00:14:39,300
Now moving on to part d.

206
00:14:39,300 --> 00:14:42,700
We will go over here, I think.

207
00:14:42,700 --> 00:14:43,300
I'm sorry.

208
00:14:43,300 --> 00:14:44,550
d, not b.

209
00:14:47,430 --> 00:14:51,660
Again, it's going to be
about acids and bases.

210
00:14:51,660 --> 00:14:55,320
Consider the effect each of the
following substances has

211
00:14:55,320 --> 00:14:58,510
on the ionization of the
weak base ammonia.

212
00:14:58,510 --> 00:15:02,620
For each state whether the
substance, one, suppresses

213
00:15:02,620 --> 00:15:07,110
ionization, two, enhances
ionization, or, three, has no

214
00:15:07,110 --> 00:15:09,550
effect on the ionization
of ammonia.

215
00:15:09,550 --> 00:15:14,040
In each instance give a reason
for your choice.

216
00:15:14,040 --> 00:15:17,290
Going back to the first part of
this question we see that

217
00:15:17,290 --> 00:15:21,170
it's asking us about the
ionization of ammonia.

218
00:15:21,170 --> 00:15:23,370
A good place to start from here
would be to write down

219
00:15:23,370 --> 00:15:26,870
the reaction, just so we have
that on our paper, in our

220
00:15:26,870 --> 00:15:29,580
mind, ready to think about it.

221
00:15:29,580 --> 00:15:33,460
So we have ammonia--

222
00:15:33,460 --> 00:15:39,910
and aqueous means that
it's in water, right?

223
00:15:39,910 --> 00:15:43,260
And that is reacting
with water to form

224
00:15:43,260 --> 00:15:45,260
the ammonium ion--

225
00:15:45,260 --> 00:15:46,740
sorry--

226
00:15:46,740 --> 00:15:54,530
by adding a proton and
a hydroxide ion.

227
00:15:54,530 --> 00:15:54,710
OK.

228
00:15:54,710 --> 00:15:57,440
So this is the reaction
we're asked about.

229
00:15:57,440 --> 00:16:00,360
Now we need to look at each of
the three substances listed

230
00:16:00,360 --> 00:16:05,510
and decide how it affects
this ionization.

231
00:16:05,510 --> 00:16:10,020
So number i, we have KOH.

232
00:16:13,100 --> 00:16:16,860
Again, a good place to start
is to write what happens to

233
00:16:16,860 --> 00:16:21,050
potassium hydroxide when
it goes into water.

234
00:16:21,050 --> 00:16:24,600
And we know that KOH is
a strong base and will

235
00:16:24,600 --> 00:16:28,700
dissociate into potassium
plus hydroxide.

236
00:16:31,630 --> 00:16:37,110
So we see that it produces
hydroxide.

237
00:16:37,110 --> 00:16:41,430
And this should indicate to
you that, well, it has the

238
00:16:41,430 --> 00:16:43,640
same product as one of
the products of the

239
00:16:43,640 --> 00:16:45,320
ionization of ammonia.

240
00:16:45,320 --> 00:16:50,240
And if we remember back a
little bit before this

241
00:16:50,240 --> 00:16:54,120
section, we know that if you
have the same ion in solution

242
00:16:54,120 --> 00:16:59,330
it will definitely affect the
solubility, or ionization.

243
00:16:59,330 --> 00:17:03,330
So that's one way to think about
it, is that I now have a

244
00:17:03,330 --> 00:17:16,510
common ion effect that will,
because there's more hydroxide

245
00:17:16,510 --> 00:17:21,050
in solution now, it will push
this reaction, this

246
00:17:21,050 --> 00:17:23,850
equilibrium, back to the left.

247
00:17:23,850 --> 00:17:25,790
Another way to think
about that is

248
00:17:25,790 --> 00:17:27,040
Le Chatelier's Principle.

249
00:17:37,840 --> 00:17:39,830
Kind of the same thing, although
it's a little bit

250
00:17:39,830 --> 00:17:43,280
more versatile, I think, so
pretty easy to think about.

251
00:17:43,280 --> 00:17:43,650
Right?

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00:17:43,650 --> 00:17:47,060
If you add a product
your reaction will

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00:17:47,060 --> 00:17:47,940
shift to the left.

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00:17:47,940 --> 00:17:52,050
If you add a reactant
your reaction will

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00:17:52,050 --> 00:17:53,750
shift to the right.

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00:17:53,750 --> 00:17:56,360
So here we're adding a product,
our reaction will

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00:17:56,360 --> 00:17:57,710
shift to the left.

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00:17:57,710 --> 00:18:03,550
Same as a common ion effect,
these two combine to decrease,

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00:18:03,550 --> 00:18:08,780
or both suggest that we will
decrease, ionization.

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00:18:16,050 --> 00:18:19,510
Moving to the second part
of this problem.

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00:18:19,510 --> 00:18:25,430
We're asked about hydrogen
chloride.

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00:18:25,430 --> 00:18:28,970
Again, let's look at what
that does in water.

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We know that it's a strong acid,
so it will dissociate

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00:18:34,090 --> 00:18:40,850
into hydrogen, or protons,
and chloride.

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00:18:40,850 --> 00:18:43,070
Now this one's a little bit
trickier because we don't have

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either of these ions in our
initial ionization reaction,

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00:18:48,390 --> 00:18:52,070
but if we think about the fact
that ammonia is a weak base we

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00:18:52,070 --> 00:18:55,940
can write an additional
ionization reaction.

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00:18:55,940 --> 00:19:07,390
So that additional would be
ammonia reacting with a proton

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00:19:07,390 --> 00:19:09,400
to form ammonium.

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00:19:12,090 --> 00:19:12,460
Right?

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00:19:12,460 --> 00:19:18,110
So this is another way
that the ammonia

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00:19:18,110 --> 00:19:20,390
molecule would be ionized.

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00:19:20,390 --> 00:19:27,160
And we can see that if we add
protons from this acidic

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00:19:27,160 --> 00:19:31,800
molecule there we will increase
the ionization,

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00:19:31,800 --> 00:19:37,040
because it will allow the
ammonia to work more as a base

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00:19:37,040 --> 00:19:40,900
because of this reaction here.

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00:19:40,900 --> 00:19:46,505
So we are increasing
ionization.

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00:19:51,460 --> 00:19:55,020
Another way to think about this
is completely valid and

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00:19:55,020 --> 00:20:01,350
kind of the same thing, but if
we look at this ionization

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00:20:01,350 --> 00:20:04,010
reaction here where the
hydrogen chloride is

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00:20:04,010 --> 00:20:07,320
dissociating into protons and
chlorides we can recognize

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00:20:07,320 --> 00:20:13,960
that these protons will then
react with the hydroxide that

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00:20:13,960 --> 00:20:17,980
is produced in the initial
ionization reaction.

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00:20:17,980 --> 00:20:21,500
This will result in a decrease
of hydroxide

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00:20:21,500 --> 00:20:23,280
in solution, right?

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00:20:23,280 --> 00:20:26,040
And so by Le Chatelier's
Principle we know that a

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00:20:26,040 --> 00:20:30,560
decrease in a product causes the
reaction to shift towards

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00:20:30,560 --> 00:20:32,645
the product, to make
more hydroxide.

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00:20:32,645 --> 00:20:34,750
So that's just another way of
thinking about the same

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00:20:34,750 --> 00:20:39,510
process, and it results in the
same conclusion that addition

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00:20:39,510 --> 00:20:45,300
of hydrogen chloride increases
the ionization of ammonia.

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00:20:45,300 --> 00:20:45,600
OK.

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00:20:45,600 --> 00:20:46,850
Moving to part iii.

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00:20:50,080 --> 00:21:04,560
We have ammonium chloride, and
this will dissociate into

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00:21:04,560 --> 00:21:08,210
ammonium and chloride.

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00:21:08,210 --> 00:21:11,360
And even if you're not sure how
much it will dissociate,

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00:21:11,360 --> 00:21:13,160
it will still dissociate
to some amount.

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00:21:13,160 --> 00:21:20,210
So it will have at least a
little bit of an effect on our

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00:21:20,210 --> 00:21:22,890
initial ionization reaction.

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00:21:22,890 --> 00:21:28,200
So here we see the dissociation
produces ammonium

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00:21:28,200 --> 00:21:31,690
ions, and just like the first
case that is one of the

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00:21:31,690 --> 00:21:34,840
products in our ionization
reaction.

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00:21:34,840 --> 00:21:38,590
And so Le Chatelier's Principle,
common ion effect,

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00:21:38,590 --> 00:21:42,575
whatever you want to call it,
tells us that it will shift

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00:21:42,575 --> 00:21:57,030
the equilibrium to the left
and decrease ionization.

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00:21:57,030 --> 00:22:00,720
So with that we're done with
problem 10 from the final.

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00:22:00,720 --> 00:22:02,980
It had a lot to do with
acid/bases, so hopefully you

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00:22:02,980 --> 00:22:05,900
feel more comfortable with
them after this.