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Q. The $_{p} K_{a}$ of $HCN$ is $9.30$. The pH of a solution prepared by mixing $2.5$ moles of $KCN$ and $2.5$ moles of $HCN$ in water and making total volume up to $500 \,ml$, is:

Haryana PMTHaryana PMT 2005

Solution:

Conc. of $KCN =$ conc. of $HCN$
$=\frac{2.5}{500 / 1000}=5 \,M$
$ \therefore \log \frac{[K C N]}{[H C N]}=\log \frac{5}{5}=0$
Now, from Hendersons equation,
$p H={ }_{p} K_{a}+\log \frac{[\text { salt }]}{[\text { acid }]}$
$=9.30+0=9.30$