[Blueberries116]

This question has gave me headaches so far due the problematic nature of how to get to the equation for finding the pH of the resulting solution from such salt.

Can someone help me with this?

The problem is as follows:

Find the pH of a solution of [itex]NH_{4}CN[/itex] [itex]1\,M[/itex]. Their respective hydrolisis constants are:

[itex]K_h=5.6\times 10^{-10}[/itex]

[itex]K_h=2.5 \times 2.5 \times 10^{-5}[/itex]

The alternatives given are as follows:

[itex]\begin{array}{ll}
1.&2.8\\
2.&3.5\\
3.&6.4\\
4.&11.7\\
5.&12.2\\
\end{array}[/itex]

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

[itex]\begin{array}{lllll}
NH_4^+&+H_2O\rightleftharpoons&NH_3&+H_3O^+&k_{h}=\frac{K_w}{K_b}\\
CN^-&+H_2O\rightleftharpoons&HCN&+OH^-&k_{h}=\frac{K_w}{K_a}\\
H_3O^+&+OH^-\rightleftharpoons&2H_2O&&k_{h}=\frac{1}{K_w}\\
\end{array}[/itex]

This meant that:

[itex]K=\frac{K_w}{K_a\times K_b}[/itex]

For the given information it can be calculated:

[itex]K=\frac{10^{-14}}{\frac{10^{-14}}{5.6\times 10^{-10}}\times\frac{10^{-14}}{2.5 \times 10^{-5}}}=1.4[/itex]

All of this is translated into:

[itex]NH_4^{+}+CN^{-}\rightleftharpoons NH_{3}+HCN[/itex]

From this equation it can be established the equilibrium equation:

[itex]K=\frac{[NH_3][HCN]}{[NH_{4}^{+}][CN^{-}]}[/itex]

Returning to the previous equation I'm getting:

[itex]\begin{array}{cccccc}
&NH_{4}^{+}&+CN^{-} & \rightleftharpoons & NH_{3} & +HCN \\
i&1&1 & & &\\
c&-x&-x &  &+x &+x\\ \hline
e&(1-x)&(1-x) &  &x &x\\
\end{array}[/itex]

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

[itex]K=\frac{[NH_3][HCN]}{[NH_{4}^{+}][CN^{-}]}[/itex]

[itex]1.4=\frac{[NH_3][HCN]}{[NH_{4}^{+}][CN^{-}]}[/itex]

[itex]1.4=\frac{x^2}{(1-x)^{2}}[/itex]

Solving this for $x$ results into:

[itex]x=0.54196[/itex]

and

[itex]x=6.45804[/itex]

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 [itex]x=0.54196[/itex]

This means:

[itex][NH_3]=[HCN]=0.54196[/itex]

[itex][CN^{-}]=[NH_4^{+}]=1-0.54196= 0.45804[/itex]

Then to get the [itex]pH[/itex] can be established from the hydrolization of [itex]HCN[/itex]:

[itex]HCN+H_2O\rightleftharpoons CN^{-}+H_3O^{+}[/itex]

Then:

[itex]K=\frac{10^{-14}}{2.5\times 10^{-5}}[/itex]

[itex]K=\frac{[CN^-][H_3O^+]}{[HCN]}[/itex]

[itex]4.0\times 10^{-10}=\frac{[H_3O^{+}](0.45804)}{(0.54196)}[/itex]

[itex][H_3O^{+}]=4.73286\times 10^{-10}[/itex]

[itex]pH=-log(4.73286\times 10^{-10})=9.3248[/itex]

Which seems kind of logical that the pH is within the range of an alkaline solution as $Kb>Ka$.

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 pH with the pKw and pKa and pKb and the concentration:

It states:

[itex]pH=\frac{1}{2}[pKw-pKb+pKa+\log C][/itex]

where c=concentration of the salt.

Inserting the given values in the above equation yields:

[itex]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][/itex]

[itex]pH=9.3248[/itex]

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 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 pH? I'm not sure if the hydrolization constants given in the problem are right or accurate. Does somebody has access to cross reference and check if those values are correct?.

[AWK]

There is no good answer between the alternatives given. Your result (after rounding) seems to be reliable.
https://www.chembuddy.com/?left=pH-calculation&right=pH-salt-simplified

[Borek]

Hydrolysis constants are equivalent to acid/base dissociation constants, just reversed (Kb for HCN, Ka for NH₃).

9.32 is a correct answer.

compare https://youtu.be/Slxxj7T4X5s

[Blueberries116]

Hydrolysis constants are equivalent to acid/base dissociation constants, just reversed (Kb for HCN, Ka for NH₃).

9.32 is a correct answer.

compare https://youtu.be/Slxxj7T4X5s

It seems that I overlooked that fact. But I'm glad that my procedure was validated but the software which you used. By the way, how can it be obtained?.

One particular aspect here if there exist a way to obtain a good approximation when there's a competition between the ions from a weak acid and weak base?. In my question I posted an equation which found online but I was looking for a simplified version which does get more or less accurate results.

[AWK]

Did you read the link I was given?
Such a simple formula exists there.

[Borek]

It seems that I overlooked that fact. But I'm glad that my procedure was validated but the software which you used. By the way, how can it be obtained?

https://www.chembuddy.com/?left=BATE&right=pH-calculator

And follow the link to teh same site posted earlier by AWK, it contains the derivation of the simplified formula similar to the one you used.