Calculating pH of solution of phosphoric acid and ammonia


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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 $\ce{0.4M}$ $\ce{NH3}$?
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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?




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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