6
$\begingroup$

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

Improve this question
$\endgroup$
7
  • $\begingroup$ You're correct. $$\ce{[H2PO4^{-}] =} \text{ 0.082}$$ which means that $$\ce{[H3PO4] =} \text{0.25 - 0.082 = 0.168}$$. $\endgroup$
    – MaxW
    Apr 12 '16 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
    Apr 13 '16 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
    Apr 13 '16 at 2:45
  • $\begingroup$ @MaxW But how does this relate to what the answer wrote? $\endgroup$
    – carbenoid
    Apr 13 '16 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
    Apr 13 '16 at 23:52
1
$\begingroup$

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 $\ce{NaH2PO4}$.

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.