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

Answer: 3

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