What is the $\ce{pH}$ of the solution created when $\ce{50ml}$ of $\ce{0.2M}$ $\ce{H3PO4}$ is mixed with $\ce{50ml}$ of $\ce{0.4M}$ $\ce{NH3}$?

This problem can be approached rigorously. That's what I did, but I got this equation which can only be solved analytically:

$$[H^+] + \frac{0.2\cdot[H^+]}{K_4+{[H^+]}} = \frac{K_w}{[H^+]} + \frac{0.3\cdot K_1\cdot K_2\cdot K_3 + 0.2\cdot [H^+]\cdot K_1\cdot K_2+0.1\cdot[H^+]\cdot K_1}{[H^+]^3 + [H^+]^2 \cdot K_1 + [H^+]\cdot K_1\cdot K_2 + K_1 \cdot K_2 \cdot K_3 } $$ $K_1$ to $K_3$ are the $\ce{H3PO4}$ dissociation constants. $K_4$ is the $\ce{NH4+}$ dissociation constant, and to derive this, I used material balance and charge balance.

My question is, how can I solve this problem without the use of a computer, using only a scientific calculator? Can anybody suggest other methods to do this problem?

  • $\begingroup$ Also they said we can suppose that 7< pH<8 , so in proton balance : [H+] + [H2PO4-] + 2[H3PO4] = [NH3] + [OH-] + [PO4 3-] , we can exclude [H+] , [OH-] , [H3PO4] and [PO4 3-] , So we get [H2PO4-] = [NH3], how they arrived to this conclusion ? $\endgroup$ – Saba Tavdgiridze Jun 2 '17 at 4:19
  • $\begingroup$ We'd love to help you, but I can't seem to digest what your question is basically, with all that stuff around. Can you format your question to make it look neat and readable? $\endgroup$ – Pritt Balagopal Jun 2 '17 at 9:32
  • $\begingroup$ What is $K_{4}$? $\endgroup$ – Zhe Jun 2 '17 at 21:20
  • 1
    $\begingroup$ As a first estimation figure out what are the major phosphate components you deal with , ie which protonation step is the most important. Most cases you can forget the others. $\endgroup$ – Greg Jun 2 '17 at 21:35

The $\mathrm{p}K_{\mathrm{a}}$ values for phosphoric acid are: $$\begin{array}{|c|c|} \hline \mathrm{p}K_{\mathrm{a}1} & 2.148 \\ \hline \mathrm{p}K_{\mathrm{a}2} & 7.198 \\ \hline \mathrm{p}K_{\mathrm{a}3} & 12.319 \\ \hline \end{array}$$

The $\mathrm{p}K_{\mathrm{a}}$ for ammonium is $9.25$.

The initial conditions led me to partially neutralize the phosphoric acid, giving $0.01\ \mathrm{mol}\ \ce{H2PO4-}$, $0.01\ \mathrm{mol}\ \ce{NH3}$, and $0.01\ \mathrm{mol}\ \ce{NH4+}$.

Note that dihydrogen phosphate is a stronger acid than ammonium, so the equilibrium will shift towards making hydrogen phosphate by neutralizing more ammonia:

$$\ce{NH3 + H2PO4- -> NH4+ + HPO4^{2-}}$$

We note also that the equilibrium $\mathrm{pH}$ satisfies the Henderson-Hasselbalch equation for each equilibrium:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log\frac{\ce{[A-]}}{\ce{[HA]}}$$

That is: $$\mathrm{pH} = \mathrm{p}K_{\mathrm{a}\ \ce{NH4+}} + \log\frac{\ce{[NH3]}}{\ce{[NH4+]}} = \mathrm{p}K_{\mathrm{a}\ \ce{H2PO4-}} + \log\frac{\ce{[HPO4^{2-}]}}{\ce{[H2PO4-]}}$$

From the initial conditions and stoichiometry, I determine: $$\begin{array}{|c|c|} \hline \text{species} & \text{moles} \\ \hline \hline \ce{NH4+} & 0.01 + x \\ \hline \ce{NH3} & 0.01 - x \\ \hline \ce{H2PO4-} & 0.01 - x \\ \hline \ce{HPO4^{2-}} & x \\ \hline \end{array}$$

Solving this yields: $x=0.00878\ \mathrm{mol}$, which plugging back in gives a $\mathrm{pH}$ of $8.08$.

That seems to be about the right ballpark. The problem is equivalent to adding a weak acid to an ammonia/ammonium buffer at pH 9.25.


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.