Question

Calculate the solubility of AgCN in a buffer solution of $pH=3$. Given $K_{sp}$ of $AgCN=1.2\times {10}^{-16}$ and $K_a$ for $HCN=4.8\times {10}^{-10}$.

• A. $4.4\times {10}^{-4}g{L}^{-1}$
• B. $3.2\times {10}^{-4}g{L}^{-1}$
• C. $1.5\times {10}^{-5}g{L}^{-1}$
• D. none of these

Answer 1

The correct option is C $1.5\times {10}^{-5}g{L}^{-1}$

$HCN$ is a weak acid in Buffer of $pH=3$

We use the formula

$={\left[k_{sp}(1+\frac{\left[H^{+}\right]}{k_a})\right]}^{1/2}$

$PH=3$

$\left[H^{+}\right]={10}^{-3}$

$={[1.2\times {10}^{-16}(1+\frac{{10}^{-3}}{4.8\times {10}^{-10}})]}^{1/2}$

$={[1\cdot 2\times {10}^{-16}(1+2.08\times {10}^6)]}^{1/2}$

$={(1.2\times {10}^{-16}+2.08\times {10}^6\times 1.2\times {10}^{-16})}^{1/2}$

Solubility $=1.5\times {10}^{-5}M$