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JOCELYN: Hi.

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00:00:22,590 --> 00:00:25,960
Jocelyn here and we're going
to go over fall 2009 exam

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00:00:25,960 --> 00:00:28,060
three problem number three.

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00:00:28,060 --> 00:00:31,280
As always, let's read the
question first. Calcium

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00:00:31,280 --> 00:00:35,110
ammonium phosphate dissolves
in water according to the

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00:00:35,110 --> 00:00:38,070
dissolution equation, for
which the value of the

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solubility product, ksp, has
been determined to be 4.4

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00:00:42,630 --> 00:00:45,050
times 10 to the -14.

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00:00:45,050 --> 00:00:47,880
Calculate the solubility of
the compound in water.

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00:00:47,880 --> 00:00:51,730
Express your answers in units
of molartiy, ie moles of the

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00:00:51,730 --> 00:00:53,910
compound per liter
of solution.

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00:00:56,620 --> 00:01:00,080
So the first thing we want to
do is write down what the

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00:01:00,080 --> 00:01:02,610
question's asking.

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00:01:02,610 --> 00:01:05,150
Part A--

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00:01:05,150 --> 00:01:17,760
how much calcium ammonium
phosphate

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00:01:17,760 --> 00:01:21,790
can dissolve in water?

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00:01:33,570 --> 00:01:38,930
And we're going to
call this cs--

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00:01:38,930 --> 00:01:44,020
or the saturation
concentration.

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00:01:44,020 --> 00:01:48,330
Next we need to figure out how
to find this out, right?

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00:01:48,330 --> 00:01:50,720
So it gives us that the ksp--

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00:01:53,730 --> 00:01:58,630
4.4 times 10 to the -14.

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00:01:58,630 --> 00:02:02,580
So we need to know something
about the ksp and although

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00:02:02,580 --> 00:02:06,410
it's called the solubility
product, it's the same as

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00:02:06,410 --> 00:02:09,270
other equilibrium constants.

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00:02:09,270 --> 00:02:14,590
So equilibrium constants tells
us about the concentrations at

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00:02:14,590 --> 00:02:19,210
equilibrium and therefore
the maximum solubility.

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00:02:19,210 --> 00:02:28,080
Looking at the equation, we
have calcium ammonium

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00:02:28,080 --> 00:02:30,760
phosphate dissolving--

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00:02:30,760 --> 00:02:33,680
which is a solid--

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00:02:33,680 --> 00:02:45,490
dissolving into calcium
ions, ammonium ions

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00:02:45,490 --> 00:02:49,190
and phosphate ions.

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00:02:49,190 --> 00:02:52,290
As with every equilibrium
constant, we can write down an

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00:02:52,290 --> 00:02:55,725
equation relating to the
concentration of the species.

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00:02:58,230 --> 00:03:11,983
So our ksp is the product
of the concentration.

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00:03:17,960 --> 00:03:20,240
And a common mistake made
on this problem

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00:03:20,240 --> 00:03:22,250
was that people included--

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00:03:22,250 --> 00:03:25,870
students included the calcium
ammonium phosphate solid on

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00:03:25,870 --> 00:03:28,400
the bottom as you
normally would--

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00:03:28,400 --> 00:03:30,170
where you would normally
put the reactant.

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00:03:30,170 --> 00:03:33,030
However, remember that for
equilibrium constants, we

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00:03:33,030 --> 00:03:37,130
don't put solids in
there because the

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00:03:37,130 --> 00:03:41,440
activities don't change--

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00:03:41,440 --> 00:03:42,900
or the concentration
doesn't change.

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00:03:42,900 --> 00:03:45,680
It doesn't really make
sense to add it in.

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00:03:45,680 --> 00:03:51,810
So here we just look at the
solvated ions and we see that

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00:03:51,810 --> 00:03:55,510
we have a 1:1:1 molar
ratio here.

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00:03:55,510 --> 00:03:57,590
That's going to make our life
a little bit easier.

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00:04:00,160 --> 00:04:03,170
So now we need to figure out
what the solubility of the

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00:04:03,170 --> 00:04:06,380
compound is and to do that,
we want to look at

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00:04:06,380 --> 00:04:08,010
this chemical equation.

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00:04:08,010 --> 00:04:11,380
We see that for one mole--

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00:04:11,380 --> 00:04:22,590
so one mole of calcium ammonium
phosphate dissolved

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00:04:22,590 --> 00:04:38,790
gives you one mole of the
calcium ion, the sodium ion

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00:04:38,790 --> 00:04:40,490
and the phosphate ion.

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00:04:43,070 --> 00:04:47,900
Thus, we can say the amount of
calcium ammonium phosphate

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00:04:47,900 --> 00:04:52,910
dissolved, which we called Cs,
is going to be equivalent to

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00:04:52,910 --> 00:04:54,330
each of these concentrations.

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00:05:05,120 --> 00:05:09,900
Because we're asked for how
much of the compound is

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00:05:09,900 --> 00:05:14,410
dissolved, but we're given the
ksp, which is in terms of the

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00:05:14,410 --> 00:05:17,600
concentrations of these
solvated ions.

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00:05:17,600 --> 00:05:18,000
I'm sorry.

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00:05:18,000 --> 00:05:20,270
This doesn't equal
the product.

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00:05:20,270 --> 00:05:21,520
These are all equal.

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00:05:24,320 --> 00:05:30,010
Now we can plug the
Cs into our ksp

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00:05:30,010 --> 00:05:33,380
equation that we had before.

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00:05:33,380 --> 00:05:35,880
So instead of the
concentration of

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00:05:35,880 --> 00:05:39,650
calcium ions, we have--

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00:05:39,650 --> 00:05:41,550
it's equal to the
concentration of

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00:05:41,550 --> 00:05:43,840
the compound dissolved.

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00:05:43,840 --> 00:05:49,470
The same goes for the ammonium
and the phosphate, because

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00:05:49,470 --> 00:05:56,050
again, we have 1:1:1:1
stoichiometric coefficients.

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00:05:56,050 --> 00:06:02,790
And we can be a little
more concise.

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00:06:02,790 --> 00:06:07,030
Now it's just a matter of
solving for the saturated

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00:06:07,030 --> 00:06:08,560
concentration.

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00:06:08,560 --> 00:06:18,650
So doing some algebra here.

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00:06:18,650 --> 00:06:19,970
Then we plug in the numbers.

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00:06:26,510 --> 00:06:36,330
And we get that the solubility
of calcium ammonium phosphate

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00:06:36,330 --> 00:06:42,210
is 3.53 times 10 to
the -5 moles--

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00:06:42,210 --> 00:06:43,290
molar, sorry.

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00:06:43,290 --> 00:06:49,440
So that means 3.53 times -5
moles per liter of water.

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00:06:49,440 --> 00:06:50,450
That's not very much.

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00:06:50,450 --> 00:06:56,030
So we can say generally or
relatively this calcium

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00:06:56,030 --> 00:07:00,120
ammonium phosphate is not
that soluble in water.

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00:07:00,120 --> 00:07:01,490
So let's move on to part B.

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00:07:08,590 --> 00:07:11,890
Part B says, calculate the
solubility of calcium ammonium

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00:07:11,890 --> 00:07:16,060
phosphate in 2.2 molar
calcium bromide.

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00:07:16,060 --> 00:07:18,890
Express your answer
in units molarity.

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00:07:18,890 --> 00:07:21,310
Assume that in water, calcium
bromide completely

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00:07:21,310 --> 00:07:23,440
disassociates into
calcium 2 plus

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00:07:23,440 --> 00:07:24,740
cations and bromide anions.

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00:07:29,510 --> 00:07:33,790
This is basically asking
us for the same value.

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00:07:33,790 --> 00:07:36,280
It's asking us for the
solubility of calcium ammonium

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00:07:36,280 --> 00:07:39,240
phosphate, but under slightly
different conditions.

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00:07:39,240 --> 00:07:40,920
So we want to write those
conditions down.

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00:07:51,920 --> 00:07:55,050
Now we have to ask ourselves
why would the fact that we

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00:07:55,050 --> 00:07:58,920
have a concentration of calcium
bromide affect the

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00:07:58,920 --> 00:08:01,440
solubility of calcium
ammonium phosphate?

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00:08:01,440 --> 00:08:03,640
And the thing to remember
here is the

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00:08:03,640 --> 00:08:06,330
common ion effect, right?

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00:08:06,330 --> 00:08:10,590
2 molar calcium bromide will--

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00:08:10,590 --> 00:08:21,130
when dissolved in water, the
concentration of calcium from

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00:08:21,130 --> 00:08:26,870
just the calcium bromide will
be 2.2 molar, right?

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00:08:26,870 --> 00:08:29,610
Because we're told that it
completely disassociates.

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00:08:29,610 --> 00:08:36,140
So if we look at the
disassociation reaction, we

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00:08:36,140 --> 00:08:40,190
see that for every mole of
calcium bromide, we get one

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00:08:40,190 --> 00:08:44,480
mole of calcium and two
moles of bromide.

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00:08:53,370 --> 00:08:58,330
So having calcium bromide will
alter our answer from the

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00:08:58,330 --> 00:09:04,300
previous problem because the ksp
will always be the same.

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00:09:04,300 --> 00:09:07,200
It's an equilibrium
constant, right?

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00:09:07,200 --> 00:09:10,960
So even though it has to do
with calcium ammonium

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00:09:10,960 --> 00:09:13,570
phosphate, we have to take into
account that we already

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00:09:13,570 --> 00:09:17,340
have calcium in the system.

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00:09:17,340 --> 00:09:20,590
So before we start plugging in
any numbers, we should think

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00:09:20,590 --> 00:09:22,190
about this problem.

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00:09:22,190 --> 00:09:24,780
Do we think that already having
calcium in the system

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00:09:24,780 --> 00:09:28,350
will increase or decrease the
solubility of the calcium

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00:09:28,350 --> 00:09:29,600
ammonium phosphate?

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00:09:32,430 --> 00:09:36,560
Hopefully we can agree that it
would probably decrease the

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00:09:36,560 --> 00:09:39,690
solubility because you already
have those calcium ions.

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00:09:39,690 --> 00:09:43,800
So putting more calcium ions
into the water is going to be

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00:09:43,800 --> 00:09:46,820
harder and therefore less
calcium ammonium phosphate

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00:09:46,820 --> 00:09:48,180
will dissolve.

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00:09:48,180 --> 00:09:51,420
Now that we know what kind of
number we're looking for, we

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00:09:51,420 --> 00:09:56,080
can start doing the actual
calculation.

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00:09:56,080 --> 00:09:59,520
So again, we'll start with our
equation for the ksp, which

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00:09:59,520 --> 00:10:03,010
for calcium ammonium phosphate
has not changed.

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00:10:03,010 --> 00:10:05,240
I'm just going to rewrite
it on this board here.

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00:10:18,560 --> 00:10:22,650
But now instead of having
each of these be equal

136
00:10:22,650 --> 00:10:24,920
concentrations, we
already have some

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00:10:24,920 --> 00:10:26,000
calcium in the system.

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00:10:26,000 --> 00:10:27,450
So we need to take that
into account.

139
00:10:37,310 --> 00:10:41,900
And I'm going to call the
saturation concentration Cs

140
00:10:41,900 --> 00:10:44,280
star because we're under
different conditions, right?

141
00:10:49,220 --> 00:10:55,400
The calcium bromide does not
contribute any ammonium or

142
00:10:55,400 --> 00:11:00,420
phosphate ions so those
concentrations will just be

143
00:11:00,420 --> 00:11:02,470
determined by how much
of the calcium

144
00:11:02,470 --> 00:11:05,710
ammonium phosphate dissolves.

145
00:11:05,710 --> 00:11:12,600
From before, we can use what we
found in part A to simplify

146
00:11:12,600 --> 00:11:14,630
this a little bit.

147
00:11:14,630 --> 00:11:18,600
So in part A, we know that
without anything else, without

148
00:11:18,600 --> 00:11:21,620
a common ion effect, that we
have decided will decrease the

149
00:11:21,620 --> 00:11:22,940
solubility.

150
00:11:22,940 --> 00:11:31,240
We have 3.35 times 10 to the
-5th molar solubility.

151
00:11:31,240 --> 00:11:41,650
Therefore, we can say with
good certainty that our

152
00:11:41,650 --> 00:11:45,710
saturation concentration with
the common ion effect will be

153
00:11:45,710 --> 00:11:52,090
much, much less than 2.2 molar
because 2.2 molar is much

154
00:11:52,090 --> 00:11:54,610
greater than our pure
solubility.

155
00:11:57,260 --> 00:12:00,070
Going back to our equation
over here, that

156
00:12:00,070 --> 00:12:09,308
means we can have--

157
00:12:15,720 --> 00:12:20,360
and now we have a fairly simple
algebraic problem.

158
00:12:23,650 --> 00:12:29,300
Dividing by 2.2 and then taking
the square root, we get

159
00:12:29,300 --> 00:12:34,460
that the saturation
concentration under these

160
00:12:34,460 --> 00:12:35,030
conditions--

161
00:12:35,030 --> 00:12:37,320
2.2 molar calcium bromide--

162
00:12:37,320 --> 00:12:45,360
will be 1.41 times 10
to the -7 molar.

163
00:12:45,360 --> 00:12:49,750
And going back to our answer
from part A, we see that this

164
00:12:49,750 --> 00:12:52,600
is indeed a lower solubility.

165
00:12:52,600 --> 00:12:59,720
There's less calcium ammonium
phosphate that can be

166
00:12:59,720 --> 00:13:03,170
dissolved and that makes sense
from our previous thought

167
00:13:03,170 --> 00:13:05,360
about this problem.

168
00:13:05,360 --> 00:13:09,888
And so now that you've
found the answer, box

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00:13:09,888 --> 00:13:12,530
it and we're done.