Question

The solubility of $AgCN$ in a buffer solution of $pH=3$ is:
[Given $K_{sp}$ of $AgCN=1.2\times {10}^{-6}$ and $K_a$ for $HCN=4.8\times {10}^{-10}$. Assume that no cyano complex is formed.]

• A. $a=1.67\times {10}^{-5}$ $M$
• B. $a=1.58\times {10}^{-5}$ $M$
• C. $a=1.51\times {10}^{-5}$ $M$
• D. None of these

Answer 1

Answer: (B)

$a=1.58\times {10}^{-5}$ $M$

Assume solubility of $AgCN$ be $aM$ in presence of a buffer of $pH=3$

$AgCN\rightleftharpoons \underset{a}{{Ag}^{+}}+\underset{a}{{CN}^{-}}$

$\left[{Ag}^{+}\right]=a;\left[HCN\right]=a$

For weak $HCN$, $K_a=\frac{\left[H^{+}\right]\left[{CN}^{-}\right]}{\left[HCN\right]}$

or $4.8\times {10}^{-10}=\frac{{10}^{-3}\left[{CN}^{-}\right]}{a}$

$\therefore \left[{CN}^{-}\right]=4.8\times {10}^{-7}\times a$

For $AgCN$: $K_{sp}=\left[{Ag}^{+}\right]\left[{CN}^{-}\right]$

$1.2\times {10}^{-16}=a\times 4.8\times {10}^{-7}\times a$

$a^2=\frac{1.2\times {10}^{-16}}{4.8\times {10}^{-7}}$

$\therefore$ $a=1.58\times {10}^{-5}$ $M$