Question

The solubility of $AgCN$ in a buffer solution is $1.58\times 10^{-5} \,mol$ litre $^{-1}$. Given $K _{ sp }$ of $AgCN =1.2 \times 10^{-16}$ and $K _{ a }$ for $HCN$ is $4.8 \times 10^{-10}$. The $pH$ of the buffer is ___

Answer 1

Since $HCN$ is weak acid and has low degree of dissociation. Also its dissociation is suppressed in the presence of $H ^{+}$ions. Thus
$HCN \rightleftharpoons H ^{+}+ CN ^{-}$
$K _{ a }=\frac{\left[ CN ^{-}\right]\left[ H ^{+}\right]}{[ HCN ]}$
$\left[ CN ^{-}\right]=\frac{ K _{ a }[ HCN ]}{\left[ H ^{+}\right]}=\frac{ a \times 4.8 \times 10^{-10}}{\left[ H ^{+}\right]}$
$=\frac{1.58 \times 10^{-5} \times 4.8 \times 10^{-10}}{\left[ H ^{+}\right]}$
$ K _{ sp } =\left[ Ag ^{+}\right]\left[ CN ^{-}\right]$
$ 1.2 \times 10^{-16} =\frac{1.58 \times 10^{-5} \times 1.58 \times 10^{-5} \times 4.8 \times 10^{-10}}{\left[ H ^{+}\right]} $
$\left[ H ^{+}\right] =\frac{1.58 \times 10^{-5} \times 1.58 \times 10^{-5} \times 4.8 \times 10^{-10}}{1.2 \times 10^{-16}} $
$=10^{-3} $
$pH =3 $