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JOCELYN: Hi.

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Jocelyn here and we're going
to go over fall 2009 exam

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three problem number three.

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As always, let's read the
question first. Calcium

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ammonium phosphate dissolves
in water according to the

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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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times 10 to the -14.

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Calculate the solubility of
the compound in water.

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Express your answers in units
of molartiy, ie moles of the

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compound per liter
of solution.

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So the first thing we want to
do is write down what the

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question's asking.

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Part A--

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how much calcium ammonium
phosphate

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can dissolve in water?

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And we're going to
call this cs--

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or the saturation
concentration.

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Next we need to figure out how
to find this out, right?

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So it gives us that the ksp--

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4.4 times 10 to the -14.

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So we need to know something
about the ksp and although

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it's called the solubility
product, it's the same as

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other equilibrium constants.

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So equilibrium constants tells
us about the concentrations at

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equilibrium and therefore
the maximum solubility.

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Looking at the equation, we
have calcium ammonium

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phosphate dissolving--

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which is a solid--

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dissolving into calcium
ions, ammonium ions

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and phosphate ions.

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As with every equilibrium
constant, we can write down an

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equation relating to the
concentration of the species.

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So our ksp is the product
of the concentration.

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And a common mistake made
on this problem

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was that people included--

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students included the calcium
ammonium phosphate solid on

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the bottom as you
normally would--

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where you would normally
put the reactant.

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However, remember that for
equilibrium constants, we

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don't put solids in
there because the

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activities don't change--

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or the concentration
doesn't change.

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It doesn't really make
sense to add it in.

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So here we just look at the
solvated ions and we see that

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we have a 1:1:1 molar
ratio here.

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That's going to make our life
a little bit easier.

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So now we need to figure out
what the solubility of the

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compound is and to do that,
we want to look at

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this chemical equation.

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We see that for one mole--

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so one mole of calcium ammonium
phosphate dissolved

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gives you one mole of the
calcium ion, the sodium ion

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and the phosphate ion.

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Thus, we can say the amount of
calcium ammonium phosphate

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dissolved, which we called Cs,
is going to be equivalent to

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each of these concentrations.

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Because we're asked for how
much of the compound is

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dissolved, but we're given the
ksp, which is in terms of the

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concentrations of these
solvated ions.

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I'm sorry.

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This doesn't equal
the product.

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These are all equal.

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Now we can plug the
Cs into our ksp

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equation that we had before.

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So instead of the
concentration of

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calcium ions, we have--

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it's equal to the
concentration of

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the compound dissolved.

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The same goes for the ammonium
and the phosphate, because

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again, we have 1:1:1:1
stoichiometric coefficients.

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And we can be a little
more concise.

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Now it's just a matter of
solving for the saturated

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concentration.

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So doing some algebra here.

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Then we plug in the numbers.

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And we get that the solubility
of calcium ammonium phosphate

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is 3.53 times 10 to
the -5 moles--

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molar, sorry.

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So that means 3.53 times -5
moles per liter of water.

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That's not very much.

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So we can say generally or
relatively this calcium

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ammonium phosphate is not
that soluble in water.

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So let's move on to part B.

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Part B says, calculate the
solubility of calcium ammonium

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phosphate in 2.2 molar
calcium bromide.

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Express your answer
in units molarity.

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Assume that in water, calcium
bromide completely

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disassociates into
calcium 2 plus

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cations and bromide anions.

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This is basically asking
us for the same value.

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It's asking us for the
solubility of calcium ammonium

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phosphate, but under slightly
different conditions.

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So we want to write those
conditions down.

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Now we have to ask ourselves
why would the fact that we

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have a concentration of calcium
bromide affect the

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solubility of calcium
ammonium phosphate?

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And the thing to remember
here is the

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common ion effect, right?

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2 molar calcium bromide will--

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when dissolved in water, the
concentration of calcium from

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just the calcium bromide will
be 2.2 molar, right?

00:08:26.870 --> 00:08:29.610
Because we're told that it
completely disassociates.

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So if we look at the
disassociation reaction, we

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see that for every mole of
calcium bromide, we get one

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mole of calcium and two
moles of bromide.

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So having calcium bromide will
alter our answer from the

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previous problem because the ksp
will always be the same.

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It's an equilibrium
constant, right?

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So even though it has to do
with calcium ammonium

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phosphate, we have to take into
account that we already

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have calcium in the system.

00:09:17.340 --> 00:09:20.590
So before we start plugging in
any numbers, we should think

00:09:20.590 --> 00:09:22.190
about this problem.

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Do we think that already having
calcium in the system

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will increase or decrease the
solubility of the calcium

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ammonium phosphate?

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Hopefully we can agree that it
would probably decrease the

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solubility because you already
have those calcium ions.

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So putting more calcium ions
into the water is going to be

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harder and therefore less
calcium ammonium phosphate

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will dissolve.

00:09:48.180 --> 00:09:51.420
Now that we know what kind of
number we're looking for, we

00:09:51.420 --> 00:09:56.080
can start doing the actual
calculation.

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So again, we'll start with our
equation for the ksp, which

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for calcium ammonium phosphate
has not changed.

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I'm just going to rewrite
it on this board here.

00:10:18.560 --> 00:10:22.650
But now instead of having
each of these be equal

00:10:22.650 --> 00:10:24.920
concentrations, we
already have some

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calcium in the system.

00:10:26.000 --> 00:10:27.450
So we need to take that
into account.

00:10:37.310 --> 00:10:41.900
And I'm going to call the
saturation concentration Cs

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star because we're under
different conditions, right?

00:10:49.220 --> 00:10:55.400
The calcium bromide does not
contribute any ammonium or

00:10:55.400 --> 00:11:00.420
phosphate ions so those
concentrations will just be

00:11:00.420 --> 00:11:02.470
determined by how much
of the calcium

00:11:02.470 --> 00:11:05.710
ammonium phosphate dissolves.

00:11:05.710 --> 00:11:12.600
From before, we can use what we
found in part A to simplify

00:11:12.600 --> 00:11:14.630
this a little bit.

00:11:14.630 --> 00:11:18.600
So in part A, we know that
without anything else, without

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a common ion effect, that we
have decided will decrease the

00:11:21.620 --> 00:11:22.940
solubility.

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We have 3.35 times 10 to the
-5th molar solubility.

00:11:31.240 --> 00:11:41.650
Therefore, we can say with
good certainty that our

00:11:41.650 --> 00:11:45.710
saturation concentration with
the common ion effect will be

00:11:45.710 --> 00:11:52.090
much, much less than 2.2 molar
because 2.2 molar is much

00:11:52.090 --> 00:11:54.610
greater than our pure
solubility.

00:11:57.260 --> 00:12:00.070
Going back to our equation
over here, that

00:12:00.070 --> 00:12:09.308
means we can have--

00:12:15.720 --> 00:12:20.360
and now we have a fairly simple
algebraic problem.

00:12:23.650 --> 00:12:29.300
Dividing by 2.2 and then taking
the square root, we get

00:12:29.300 --> 00:12:34.460
that the saturation
concentration under these

00:12:34.460 --> 00:12:35.030
conditions--

00:12:35.030 --> 00:12:37.320
2.2 molar calcium bromide--

00:12:37.320 --> 00:12:45.360
will be 1.41 times 10
to the -7 molar.

00:12:45.360 --> 00:12:49.750
And going back to our answer
from part A, we see that this

00:12:49.750 --> 00:12:52.600
is indeed a lower solubility.

00:12:52.600 --> 00:12:59.720
There's less calcium ammonium
phosphate that can be

00:12:59.720 --> 00:13:03.170
dissolved and that makes sense
from our previous thought

00:13:03.170 --> 00:13:05.360
about this problem.

00:13:05.360 --> 00:13:09.888
And so now that you've
found the answer, box

00:13:09.888 --> 00:13:12.530
it and we're done.