Finding solubility of salts of polyprotic acids

Question

Problem

Given, for $\ce{H3PO4}$,
$\mathrm{pK_\mathrm{a_1}=2.2}$
$\mathrm{pK_\mathrm{a_2}=7.2}$
$\mathrm{pK_\mathrm{a_3}=12.4}$
For the salt $\ce{Mg(NH4)PO4}$, $\mathrm{pK_\mathrm{sp}=12.6}$ and the equilibrium concentration of $\ce{NH4^+}$ is $\mathrm{0.1 M}$. If the solubility of $\ce{Mg(NH4)PO4}$ in the solution, with $\mathrm{pH=10}$ is S, find the value of $\mathrm{-log(S)}$.

Answer

$\mathrm{4.6}$

Question

I assumed that the contribution of $\ce{NH4^+}$ ions from the salt is to be neglected, so $\ce{[NH4^+]=10^\mathrm{-1}}$. The solubility S is then equal to the concentration of $\ce{Mg^2+}$ ions. Also, the initial amount of $\ce{PO4^3-}$ will also be S, out of which, I took some $x$ amount to have hydrolyzed to $\ce{H3PO4}$, $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$.

Then, $$\ce{K_\mathrm{sp}=\ce{[Mg^2+][NH4^+][PO4^3-]}}$$ $$\ce{K_\mathrm{sp}=\mathrm{(S)(0.1)(S-x)}}$$

And using the values of $\mathrm{pK_a}$, I can get three more equations, with two more variables for the amounts of $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. Four equations, four variables - that's fine in general, if one has a computer. But this particular exam doesn't even allow a calculator, so I don't think that is what we are supposed to do here.

Another similar, but much simpler question - one is asked to find the solubility of $\ce{AgCl}$ in an $\ce{NH3}$ solution of given molarity, with the values of $\mathrm{pK_\mathrm{sp}}$ and formation constant of complex $\ce{[Ag(NH3)2]^+}$ given. In that case we assume that all of the $\ce{Ag^+}$ initially released, turns to $\ce{[Ag(NH3)2]^+}$ and solve accordingly.

But in this case there are three different ions and products possible by hydrolysis, so I am unable to find out which of them should be considered as the dominant species.

EDIT - M.L.'s answer clarified the calculations involved. My main confusion now is regarding why $\ce{HPO4^2-}$ (and not $\ce{H2PO4^-}$ or $\ce{H3PO4}$), will be the major ionic species in the solution. Is there a way to predict this based on values of $\mathrm{pK_a}$ and the $\mathrm{pH}$ of the solution?

Answer 1

Okay, so I managed to get a solution without too many equations by making the right approximations.

First with your expressions:

$\ce{K_{sp} = [Mg^2+][NH_4^+][PO_4^3-]}$

$\ce{K_{sp} = (S)(0.1)(S-x})$

Here we want to find S-x which is $\ce{[PO_4^3-]_{eq}}$. To do this, I did the following:

$\ce{\frac{[PO_4^3-][H+]}{[HPO_4^2-]}=K_{a3} = 10^{-12.4}}$

Assuming $\ce{[H^+] = 10^{-10}}$:

$\ce{\frac{[PO_4^3-] 10^{-10}}{[HPO_4^2-] 10^{-12.4}}=K_{a3} = 1}$

$\ce{\frac{[PO_4^3-] 10^{2.4}}{[HPO_4^2-]} = 1}$

Now we could continue on and find $\ce{[H_2PO_4]}$. But it turns out the value is quite small and can be essentially ignored. Here is what we'll get if we solve for it:

$\ce{\frac{[HPO_4^2-][H^+]}{[H_2PO_4^-]} = K_{a2} = 10^{-7.2}}$

$\ce{\frac{[HPO_4^2-] 10^{-2.8}}{[H_2PO_4^-]} = 1}$

Replacing $\ce{[HPO_4^2-]}$ with $\ce{[PO_4^3-]}$ in our earlier expression, we get

$\ce{\frac{[PO_4^3-] 10^{-0.4}}{[H_2PO_4^-]} = 1}$

Now, we could set the balance for $\ce{S = [PO_4^3-]_{eq} + [HPO_4^2-] + [H_2PO_4^-] + [H_3PO_4]}$ (Since S should equal the total concentration of any phosphate species) and then replace each concentration with the ratios we solved above. If we ignore $\ce{[H_3PO4]}$ (which makes because if we solve for it's relative concentration, $\ce{[H_3PO4] = 10^{-8.2}[PO_4^3-]}$), we get:

$\ce{S = [PO_4^3-]_{eq} + [HPO_4^2-] + [H_2PO_4^-] = [PO_4^3-]_{eq} + 10^{2.4}[PO_4^3-]_{eq} + 10^{-0.4}[PO_4^3-]_{eq}}$

Solving, we get $\ce{\frac{S}{1+ 10^{-0.4} + 10^{2.4}} = [PO_4^3-]_{eq}}$ and we can plug this into the original expression:

$\ce{K_{sp} = (S)(0.1)(S-x})$

$\ce{10^{-12.6} = (S)(0.1)(\frac{S}{1+10^{-0.4} 10^{2.4}})}$

Here, just because it is non-calculator, I will approximate the denominator of the fraction to be simply $10^{2.4}$ as the $1 + 10^{-0.4}$ term is very small compared to $10^{2.4}$ and is just complicating the calculations

$\ce{\frac{10^{-12.6} 10^{2.4}}{0.1} = 10^{-9.2} = S^2}$

So we can find that $S = 10^{-4.6}$ and $-\log(S) = 4.6$

If we didn't make the approximation we did for $1+10^{-0.4} + 10^{2.4}$ term, the value we would get is $-\log(S) = 4.59879$ which obviously shows that the approximation we made didn't alter the resulting value by too much.