Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\mathrm{HCl}$ gives $\left[{\mathrm{H}}^{+}\right]=0.05\mathrm{M}$, which suppresses ionisation of HA, so $[\mathrm{HA}]=0.2\mathrm{M}$.

$\left[{\mathrm{A}}^{-}\right]=\frac{K_a[\mathrm{HA}]}{\left[{\mathrm{H}}^{+}\right]}=\frac{10^{-7}\times 0.2}{0.05}=4\times 10^{-7}\mathrm{M}$.

$\alpha =\frac{4\times 10^{-7}}{0.2}=2\times 10^{-6}$.

Answer: (C).