Question

Calculate the solubility of $AgCN$ in a buffer solution of $pH=3,K_{sp}$ of $AgCN=1.2\times {10}^{-15}$ and $Ka$ of $HCN=4.8\times {10}^{-10}$.

• A. $4\times {10}^{-5}M$
• B. $4.5\times {10}^{-6}M$
• C. $5\times {10}^{-5}M$
• D. $5.5\times {10}^{-3}M$

Answer 1

Answer: (C)

$5\times {10}^{-5}M$

$AgCN\leftrightharpoons A{g}^{+}+C{N}^{-}................Ksp$

$C{N}^{-}+H^{+}\leftrightharpoons HCN...................\frac{1}{K_a}$

$AgCN+H^{+}\leftrightharpoons HCN+A{g}^{+}...................\frac{Ksp}{k_a}=K_a$

$K_0=\frac{1.2\times {10}^{-15}}{4.8\times {10}^{-10}}=25\times {10}^{-7}$

At Equilibrium

$AgCN+H^{+}\leftrightharpoons HCN+A{g}^{+}$

$AgCN=xmoles$

$\left[H^{+}\right]=0.001$ as $pH=3$

$\left[A{g}^{+}\right]=xmoles$

$HCN=xmoles$

$k_0=\frac{\left[{Ag}^{+}\right]\left[HCN\right]}{\left[H^{+}\right]}$

$2.5\times {10}^{-7}=\frac{x^2}{0.001}$

$x^2=5^5\times {10}^{-10}$

$x^2=5\times {10}^{-5}M$

Molar Solubility of $AgCN$ in buffer solution is $5\times {10}^{-5}M$