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JOCELYN: Hi.
Jocelyn here and we're going to go over fall 2009 exam
three problem number three.
As always, let's read the question first. Calcium
ammonium phosphate dissolves in water according to the
dissolution equation, for which the value of the
solubility product, ksp, has been determined to be 4.4
times 10 to the -14.
Calculate the solubility of the compound in water.
Express your answers in units of molartiy, ie moles of the
compound per liter of solution.
So the first thing we want to do is write down what the
question's asking.
Part A--
how much calcium ammonium phosphate
can dissolve in water?
And we're going to call this cs--
or the saturation concentration.
Next we need to figure out how to find this out, right?
So it gives us that the ksp--
4.4 times 10 to the -14.
So we need to know something about the ksp and although
it's called the solubility product, it's the same as
other equilibrium constants.
So equilibrium constants tells us about the concentrations at
equilibrium and therefore the maximum solubility.
Looking at the equation, we have calcium ammonium
phosphate dissolving--
which is a solid--
dissolving into calcium ions, ammonium ions
and phosphate ions.
As with every equilibrium constant, we can write down an
equation relating to the concentration of the species.
So our ksp is the product of the concentration.
And a common mistake made on this problem
was that people included--
students included the calcium ammonium phosphate solid on
the bottom as you normally would--
where you would normally put the reactant.
However, remember that for equilibrium constants, we
don't put solids in there because the
activities don't change--
or the concentration doesn't change.
It doesn't really make sense to add it in.
So here we just look at the solvated ions and we see that
we have a 1:1:1 molar ratio here.
That's going to make our life a little bit easier.
So now we need to figure out what the solubility of the
compound is and to do that, we want to look at
this chemical equation.
We see that for one mole--
so one mole of calcium ammonium phosphate dissolved
gives you one mole of the calcium ion, the sodium ion
and the phosphate ion.
Thus, we can say the amount of calcium ammonium phosphate
dissolved, which we called Cs, is going to be equivalent to
each of these concentrations.
Because we're asked for how much of the compound is
dissolved, but we're given the ksp, which is in terms of the
concentrations of these solvated ions.
I'm sorry.
This doesn't equal the product.
These are all equal.
Now we can plug the Cs into our ksp
equation that we had before.
So instead of the concentration of
calcium ions, we have--
it's equal to the concentration of
the compound dissolved.
The same goes for the ammonium and the phosphate, because
again, we have 1:1:1:1 stoichiometric coefficients.
And we can be a little more concise.
Now it's just a matter of solving for the saturated
concentration.
So doing some algebra here.
Then we plug in the numbers.
And we get that the solubility of calcium ammonium phosphate
is 3.53 times 10 to the -5 moles--
molar, sorry.
So that means 3.53 times -5 moles per liter of water.
That's not very much.
So we can say generally or relatively this calcium
ammonium phosphate is not that soluble in water.
So let's move on to part B.
Part B says, calculate the solubility of calcium ammonium
phosphate in 2.2 molar calcium bromide.
Express your answer in units molarity.
Assume that in water, calcium bromide completely
disassociates into calcium 2 plus
cations and bromide anions.
This is basically asking us for the same value.
It's asking us for the solubility of calcium ammonium
phosphate, but under slightly different conditions.
So we want to write those conditions down.
Now we have to ask ourselves why would the fact that we
have a concentration of calcium bromide affect the
solubility of calcium ammonium phosphate?
And the thing to remember here is the
common ion effect, right?
2 molar calcium bromide will--
when dissolved in water, the concentration of calcium from
just the calcium bromide will be 2.2 molar, right?
Because we're told that it completely disassociates.
So if we look at the disassociation reaction, we
see that for every mole of calcium bromide, we get one
mole of calcium and two moles of bromide.
So having calcium bromide will alter our answer from the
previous problem because the ksp will always be the same.
It's an equilibrium constant, right?
So even though it has to do with calcium ammonium
phosphate, we have to take into account that we already
have calcium in the system.
So before we start plugging in any numbers, we should think
about this problem.
Do we think that already having calcium in the system
will increase or decrease the solubility of the calcium
ammonium phosphate?
Hopefully we can agree that it would probably decrease the
solubility because you already have those calcium ions.
So putting more calcium ions into the water is going to be
harder and therefore less calcium ammonium phosphate
will dissolve.
Now that we know what kind of number we're looking for, we
can start doing the actual calculation.
So again, we'll start with our equation for the ksp, which
for calcium ammonium phosphate has not changed.
I'm just going to rewrite it on this board here.
But now instead of having each of these be equal
concentrations, we already have some
calcium in the system.
So we need to take that into account.
And I'm going to call the saturation concentration Cs
star because we're under different conditions, right?
The calcium bromide does not contribute any ammonium or
phosphate ions so those concentrations will just be
determined by how much of the calcium
ammonium phosphate dissolves.
From before, we can use what we found in part A to simplify
this a little bit.
So in part A, we know that without anything else, without
a common ion effect, that we have decided will decrease the
solubility.
We have 3.35 times 10 to the -5th molar solubility.
Therefore, we can say with good certainty that our
saturation concentration with the common ion effect will be
much, much less than 2.2 molar because 2.2 molar is much
greater than our pure solubility.
Going back to our equation over here, that
means we can have--
and now we have a fairly simple algebraic problem.
Dividing by 2.2 and then taking the square root, we get
that the saturation concentration under these
conditions--
2.2 molar calcium bromide--
will be 1.41 times 10 to the -7 molar.
And going back to our answer from part A, we see that this
is indeed a lower solubility.
There's less calcium ammonium phosphate that can be
dissolved and that makes sense from our previous thought
about this problem.
And so now that you've found the answer, box
it and we're done.