Calculate the degree of ionization of $0.05 \ M$ acetic acid if its $p K_{ a }$ value is $4.74$. How is the degree of dissociation affected when its solution also contains $(a)$ $0.01 \ M$ $HCl$ and $(b)$ $0.1 \ M$ $HCl$?

$c = 0.05 \ M$
$p K_{ a } = 4.74$
$p K_{ a } = -\log(K_{ a }) \implies K_{ a } = 10^{-4.74} = 1.82 \times 10^{-5}$
For weak acid,$\alpha = \sqrt{\frac{ K_{ a } }{ c }} = \sqrt{\frac{ 1.82 \times 10^{-5} }{ 0.05 }} = 1.908 \times 10^{-2}$
When $HCl$ is added,the concentration of $H^{+}$ ions increases,shifting the equilibrium backward (common ion effect),thus decreasing the degree of dissociation.
Case $(a)$: $0.01 \ M$ $HCl$ added.
$K_{ a } = \frac{ [CH_{3}COO^{-}][H^{+}] }{ [CH_{3}COOH] } \approx \frac{ \alpha c \times 0.01 }{ c }$
$1.82 \times 10^{-5} = \alpha \times 0.01 \implies \alpha = 1.82 \times 10^{-3}$
Case $(b)$: $0.1 \ M$ $HCl$ added.
$1.82 \times 10^{-5} = \alpha \times 0.1 \implies \alpha = 1.82 \times 10^{-4}$