Accurate method to calculate the pH of a salt from a weak acid and weak base Can someone help me with this?

The problem is as follows:


  Find the $\mathrm{pH}$ of $\pu{1M}$ solution of $\ce{NH4CN}$. Their respective
  hydrolysis constants are:
  
  $K_h = 5.6\times 10^{-10}$
  
  $K_h = 2.5 \times 10^{-5}$


The alternatives given are as follows:

$\begin{array}{ll}
1.&amp;2.8\\
2.&amp;3.5\\
3.&amp;6.4\\
4.&amp;11.7\\
5.&amp;12.2\\
\end{array}$ 

What I've attempted to do was to establish the following reactions:

$\begin{array}{lllll}
\ce{NH4+ + H2O &lt;=&gt; NH3 + H3O+} &amp;k_{h}=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\
\ce{CN- + H2O &lt;=&gt; HCN + OH-} &amp;k_{h}=\frac{K_\mathrm{w}}{K_\mathrm{a}}\\
\ce{H3O+ + OH- &lt;=&gt; 2H2O}&amp;k_{h}=\frac{1}{K_\mathrm{w}}\\
\end{array}$ 

This meant that:

$K=\frac{K_\mathrm{w}}{K_\mathrm{a}\times K_\mathrm{b}}$

For the given information it can be calculated:

$K=\frac{10^{-14}}{\frac{10^{-14}}{5.6\times 10^{-10}}\times\frac{10^{-14}}{2.5 \times 10^{-5}}}=1.4$

All of this is translated into:

$\ce{NH4+ + CN- &lt;=&gt; NH3 + HCN}$

From this equation it can be established the equilibrium equation:

$K=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$

Returning to the previous equation I'm getting:

$\begin{array}{cccccc}
&amp;\ce{NH4+ + CN-  &amp; &lt;=&gt; &amp;  NH3 + HCN }\\ 
i&amp;1&amp;1 &amp; &amp; &amp;\\ 
c&amp;-x&amp;-x &amp;  &amp;+x &amp;+x\\ \hline
e&amp;(1-x)&amp;(1-x) &amp;  &amp;x &amp;x\\
\end{array}$

Using this relation with what was established earlier it becomes into:

$K=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$

$1.4=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$

$1.4=\frac{x^2}{(1-x)^{2}}$

Solving this for $x$ results into:

$x=0.54196$

and

$x=6.45804$

The thing here comes on which to choose for the concentration. I decided that the first seems more logical as it will not produce negative concentration in the denominator.

Therefore the concentration for $x=0.54196$

This means:

$\ce{[NH3]=[HCN]}=0.54196$

$\ce{[CN-]=[NH4+]}=1-0.54196= 0.45804$

Then to get the $\mathrm{pH}$ can be established from the hydrolyzation of $\ce{HCN}$:

$\ce{HCN + H2O &lt;=&gt; CN- + H3O+}$

Then:

$K=\frac{10^{-14}}{2.5\times 10^{-5}}$

$K=\frac{\ce{[CN-][H3O+]}}{\ce{[HCN]}}$

$4.0\times 10^{-10}=\frac{\ce{[H3O+]}(0.45804)}{(0.54196)}$

$[\ce{H3O+}]=4.73286\times 10^{-10}$

$\mathrm{pH}=-\log(4.73286\times 10^{-10})=9.3248$

Which seems kind of logical that the $\mathrm{pH}$ is within the range of an alkaline solution as $K_\mathrm{b}&gt;K_\mathrm{a}$.

But since this option does not appear within the alternatives. I'm not convinced whether my solution is right or not.

I've found this equation which relates the $\mathrm{pH}$ with the $\mathrm{p}K_\mathrm{w}$ and $\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ and the concentration:

It states:

$\mathrm{pH}=\frac{1}{2}[\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} + \log c]$

where $c = \text{concentration}$ of the salt.

Inserting the given values in the above equation yields:

$\mathrm{pH}=(0.5)\left[-\log(10^{-14})+\log\left(\frac{10^{-14}}{5.6\times 10^{-10}}\right)-\log\left(\frac{10^{-14}}{2.5\times 10^{-5}}\right)+\log(1)\right]$

$\mathrm{pH}=9.3248$

which also checks with what was obtained previously.

But again none of these seem to check with any of the alternatives given. Could it be that my procedure was wrong?

The thing which it bothers me here is if this could be accepted as a general procedure to calculate the $\mathrm{pH}$ of a salt which has a weak acid and weak base ions. Can somebody help me here? And more importantly, how to prove that equation for $\mathrm{pH}$? I'm not sure if the hydrolyzation constants given in the problem are right or accurate. Does somebody has access to cross reference and check if those values are correct?

Can somebody help me with that part too?
Ignoring activities, activity coefficients, ionic strength effects, and so on, I get a pH of around 9.3. Given the 1 M concentration, I would not bet a lot of money on this, though. My solution, assuming a more typical solution concentration, is given in the two figures below. Sorry these are figures rather than proper formatting: I have been away for 3 months and am beyond rusty at this point. If the mods wish to delete this answer, they should feel totally free to do so.





There are obviously other ways to do this problem and what I have posted is simply an approximation: any pH problem solution, of this type, that does not explicitly involve the formal concentration of the salt, i.e., "F", always gives me pause. But maybe this helps a bit.