Chemistry - Ionic Equilibrium Question with Solution | TestHub

${\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}^{\text{+}}\left]\right[{\text{CN}}^{-}\right]}{\left[\text{HCN}\right]}$
$6.2\times 10^{-10}=\frac{\left[{\mathrm{H}}^{+}\right]\mathrm{ }\left[0.02\right]}{\left[0.01\right]}$
$\left[\text{NaCN}\right]\text{=}\left[{\text{CN}}^{-}\right]$due to common ion effect
$\left[{\mathrm{H}}^{+}\right]=\frac{6.2\times 10^{-10}\times 0.01}{0.02}=3.1\times 10^{-10}$