# Finding solubility of salts of polyprotic acids

### 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 $$\mathrm{-log(S)}$$.

$$\mathrm{4.6}$$

### 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?

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

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

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

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.

• How did you realize that the concentrations of $\ce{H2PO4^-}$ (and subsequently $\ce{H3PO4}$) will be small enough to be negligible? Is there a logic/way to predict the major ionic species in the solution depending on conditions such as pH? Thank you :)
– TRC
Jul 19 at 7:40
• Hi. I made a few edits to my answer to clarify that up. Essentially, you can try to solve for the concentration of the next phosphate species if you'd realize that it is so small that it could be ignored. For me, I understood what I was doing and my brain just spontaneously decided that it would be suitable to make the approximations I did so I didn't actually solve for $\ce{H_2PO_4^-}$ in terms of $\ce{PO_4^3-}$. But you could as it shouldn't take too long and you'd realize the value is so small that you can essentially ignore it.
– M.L
Jul 19 at 15:58