The ionization constant of propanoic acid is $1.32 \times 10^{-5}$. Calculate the degree of ionization of the acid in its $0.05 \, M$ solution and also its $pH$. What will be its degree of ionization if the solution is $0.01 \, M$ in $HCl$ also?

Let the degree of ionization of propanoic acid be $\alpha$.
For the $0.05 \, M$ solution:
$HA \leftrightarrow H^{+} + A^{-}$
$K_a = \frac{C\alpha^2}{1-\alpha} \approx C\alpha^2$
$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = 1.625 \times 10^{-2} \approx 0.0163$
$[H^{+}] = C\alpha = 0.05 \times 0.0163 = 8.15 \times 10^{-4} \, M$
$pH = -\log(8.15 \times 10^{-4}) = 3.089 \approx 3.09$
In the presence of $0.01 \, M \, HCl$ (a strong acid,$[H^{+}] \approx 0.01 \, M$):
$K_a = \frac{[H^{+}][A^{-}]}{[HA]} = \frac{(0.01)(0.05\alpha')}{0.05} = 0.01 \times \alpha'$
$\alpha' = \frac{1.32 \times 10^{-5}}{0.01} = 1.32 \times 10^{-3}$