Determine the $\ce{[H+]}$ needed to just prevent precipitation by $\ce{H2PO4-}$ in a $\pu{0.25 M}$ $\ce{H3PO4}$ solution that has $\ce{[Ca^2+]} = \pu{0.15 M}$.
The $K_\mathrm{a1}$ of $\ce{H3PO4}$ is $7.1 \times 10^{–3}$. The $K_\mathrm{sp}$ of $\ce{Ca(H2PO4)2}$ is $1.0 \times 10^{-3}$.

For this problem, I first found the $\ce{[H2PO4-]}$ present in the solution when it becomes saturated, which is $\pu{8.2 \times 10^-2 M}$. So I interpreted this as at equilibrium, there is at most $\pu{0.082 M}$ $\ce{H2PO4-}$ present in the solution. When I went to set up my ICE chart for $$\ce{H3PO4 <=> H2PO4- + H+},$$ I put $\pu{0.082 M}$ as the equilibrium value of $\ce{[H2PO4-]}$. However, the answer key wrote

$ \displaystyle K_\mathrm{a1} = \frac{\ce{[H+][H2PO4-]}}{\ce{[H3PO4]}} = \frac{x (0.082 + x )}{(0.25 - x )}$

meaning that $\pu{0.0082 M}$ is the initial concentration and the equilibrium concentration of $\ce{H2PO4-}$ is $0.082 + x$. But if this was the case, then wouldn't $\ce{Ca(H2PO4)2}$ precipitate since there's more than $\pu{0.082 M}$ at equilibrium?

I think I'm probably misinterpreting the question, so any clarification on what's happening here would be appreciated.

edited - MaxW Added image (cut and paste job for needed parts...) of problem from 2002 U. S. NATIONAL CHEMISTRY OLYMPIAD NATIONAL EXAM—PART II enter image description here

  • $\begingroup$ You're correct. $$\ce{[H2PO4^{-}] =} \text{ 0.082}$$ which means that $$\ce{[H3PO4] =} \text{0.25 - 0.082 = 0.168}$$. $\endgroup$
    – MaxW
    Commented Apr 12, 2016 at 22:40
  • $\begingroup$ @MaxW So is what the answer key wrote wrong then? In this case, the question is kind of pointless since having 0.25 M $\ce{H3PO4}$ and 0.15 M $\ce{Ca^{2+}}$ won't precipitate anyway, so what does "the $\ce{H^+}$ needed to just prevent precipitation" mean? $\endgroup$
    – carbenoid
    Commented Apr 13, 2016 at 1:44
  • $\begingroup$ Think of starting with a phosphoric acid solution but increasing the pH by adding NaOH. At what pH will $\ce{Ca(H2PO4)2}$ start to ppt? // Since $\ce{[H2PO4^{-}]}$ and $\ce{[H3PO4]}$ have been calculated, you can us Ka1 to calculate $\ce{H+}$. $\endgroup$
    – MaxW
    Commented Apr 13, 2016 at 2:45
  • $\begingroup$ @MaxW But how does this relate to what the answer wrote? $\endgroup$
    – carbenoid
    Commented Apr 13, 2016 at 23:10
  • $\begingroup$ It shows that the answer key is wrong, or there is some interpretation of the problem where $\ce{[H3PO4] + [H2PO4^{-}]} \ne 0.25$ // The gist is that I can't explain an answer that doesn't make any sense... $\endgroup$
    – MaxW
    Commented Apr 13, 2016 at 23:52

1 Answer 1


I was looking for something else and stumbled across this old problem. I noticed Martin's comment about the source, and I looked for one. Noticing that this was sponsored by ACS made me wonder if my initial analysis had been correct.

The answer key, the OP, and I agree that the $\ce{H2PO4^-}$ concentration must be $8.2\times10^{-2}$ molar to cause $\ce{Ca(H2PO4)_2}$ to start to precipitate.

Problem 1 - The problem statement lists $K_{\alpha1} = 7.1\times 10^{-3}$ but the answer key uses $7.3\times10^{-3}$

My faith in the infallibility of the ACS is shaken...

Problem 2 - There are really two ways to solve the problem.

A. Neutralizing the acid

The solution starts off as 0.25 molar $\ce{H3PO4}$ and 0.15 molar $\ce{Ca^{2+}}$. You add $\ce{NaOH}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?

Mass balance for all phosphate species is:

$0.25 = \ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$

The solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so

$0.25 = \ce{[H3PO4] + [H2PO4^-]}$

$\ce{[H3PO4] = 0.25 - [H2PO4^-] = 0.25 - 0.082 = 0.168 }$

Now using the equilibrium expression for the first ionization of phosphoric acid

$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO3^-]}}{\ce{[H3PO4]}}$

Rearranging and using $7.3\times10^{-3}$ from the answer key

$\ce{[H^+]} = \dfrac{K_{\alpha1}\ce{[H3PO4]}}{\ce{[H2PO4^-]}} = \dfrac{(7.3\times10^{-3})(0.168)}{0.082} = 0.014956 \ce{->[Rounding]} 1.5\times10^{-2}$

Fortuitously using 7.1 from the problem statement yields the same rounded value.

$\dfrac{(7.1\times10^{-3})(0.168)}{0.082} = 0.014546 \ce{->[Rounding]} 1.5\times10^{-2}$

B. Making a buffer

To the 0.25 molar solution of $\ce{H3PO4}$ you add $\ce{NaH2PO4}$ until $\ce{Ca(H2PO4)_2}$ starts to precipitate. What is the pH?

Let $x$ be the nominal molarity of the $\ce{NaH2PO4}$.

Again the solution is very acidic so we can assume $0 = \ce{[HPO4^{2-}] = [PO4^{3-}]}$ so

$0.25 +x = \ce{[H3PO4] + [H2PO4^-]}$

But $\ce{[H2PO4^-] = 0.082}$, so

$\ce{[H3PO4]} = 0.25 + x - \ce{[H2PO4^-]} = 0.25 + x - 0.082 = 0.168 + x$

The charge balance is given by

$\ce{[Na^+] + [H^+] = [H2PO4^-] + 2[HPO4^{2-}] + 3[HPO4^{3-}] + [OH^-]}$

Since the solution is very acidic we can neglect $\ce{[HPO4^{2-}]}$, $\ce{[HPO4^{3-}]}$, and $\ce{[OH^-]}$. So

$\ce{[Na^+] + [H^+] = [H2PO4^-]}$

$\ce{[H^+] = [H2PO4^-] - [Na^+]} = 0.082 - x$

Now substituting into the equilibrium expression for the first ionization

$K_{\alpha1} = \dfrac{\ce{[H^+][H2PO4^-]}}{\ce{[H3PO4]}} = \dfrac{(0.082-x)(0.082)}{0.168+x}$

$0.0073(0.168+x) = 0.082^2 - 0.082x$

$(0.0073 + 0.082)x = 0.0893x = 0.082^2 - 0.0071(0.168)$

$x= 0.06194$

So the buffer solution would be 0.25 molar $\ce{H3PO4}$ and 0.062 molar $\ce{NaH2PO4}$.

$\ce{[H^+] = 0.082 - 0.6194 = 0.02006 ->[Rounding] 2.0\times 10^{-2}}$


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