Question

At $25{}^{\circ }\text{C}$, $k_{sp}$ of $AgCN$ is $4\times {10}^{-16}$ while $K_a$ of $HCN$ is $4\times {10}^{-10}$. The value of molar solubility of $AgCN$ in $0.01MHN{O}_3$ is ${10}^{-y}$. The value of $y$ is:

Answer 1

$HCN\left(aq\right)\leftrightharpoons H^{+}\left(aq\right)+C{N}^{-}\left(aq\right)$
$AgCN\left(s\right)\leftrightharpoons A{g}^{+}\left(aq\right)+C{N}^{-}\left(aq\right)$

When $AgCN$ is added to nitric acid the following equilibrium is estabished.

$\begin{array}{cccccccc}AgCN& +& H^{+}& \rightleftharpoons & A{g}^{+}& +& HCN;& k_{\text{eq}}=\frac{k_{sp}}{k_a}\\ & & 0.01-x& & x& & x& \end{array}$

$k_{\text{eq}}=\frac{k_{sp}}{k_a}\Rightarrow \frac{x^2}{0.01-x}={10}^{-6}...\left(1\right)$
$\because x$ is sufficiently small, $0.01-x\approx 0.01$
Now from the equation (1), we get
$x^2={10}^{-8}M\Rightarrow x={10}^{-4}M$
Hence,
The value of $y$ is $4$.