Chemistry - Ionic Equilibrium Question with Solution | TestHub

Let, the solubility be $\mathrm{x}$

Given that, 

 ${\mathrm{K}}_{\mathrm{sp}}\left(\mathrm{AgCN}\right)=2.2\times {10}^{-16}$ and ${\mathrm{K}}_{\mathrm{a}}\left(\mathrm{HCN}\right)=6.2\times {10}^{-10}$

$\mathrm{AgCN}\rightleftharpoons \underset{\mathrm{x}}{{\mathrm{Ag}}^{+}}+\underset{\mathrm{x}}{{\mathrm{CN}}^{-}} {\mathrm{K}}_{\mathrm{sp}}=2.2\times {10}^{-16}$

${\mathrm{H}}^{+}+{\mathrm{CN}}^{-}\rightleftharpoons \mathrm{HCN} \mathrm{K}=\frac{1}{ {\mathrm{K}}_{\mathrm{a}}}=\frac{1}{6.2\times {10}^{-10}}$

${\mathrm{K}}_{\mathrm{sp}}\times \frac{1}{ {\mathrm{K}}_{\mathrm{a}}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{CN}}^{-}\right]\times \frac{\left[\mathrm{HCN}\right]}{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{CN}}^{-}\right]}$

$2.2\times {10}^{-16}\times \frac{1}{6.2\times {10}^{-10}}=\frac{\left[\mathrm{S}\right]\left[\mathrm{S}\right]}{{10}^{-3}}$

${\mathrm{S}}^2=\frac{2.2}{6.2}\times {10}^{-9}$

${\mathrm{S}}^2=3.55\times {10}^{-10}$

$\mathrm{S}=\sqrt{3.55\times {10}^{-10}}$

$\mathrm{S}=1.88\times {10}^{-5}\Rightarrow 1.9\times {10}^{-5}$