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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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