Let the degree of ionization of propanoic acid be $a$.

Then, representing propionic acid as $HA$, we have:

               $HA\quad  + \quad {H_2}O\quad  \leftrightarrow \quad {H_3}{O^ + }\quad  + \quad {A^ – }$

$(.05-0.0 \alpha) \approx .05$                             $.05 \alpha$            $.05 \alpha$

$K_{a}=\frac{\left[ H _{3} O ^{+}\right]\left[ A ^{-}\right]}{[ HA ]}$

$=\frac{(.05 \alpha)(.05 \alpha)}{0.05}=.05 \alpha^{2}$

$\alpha=\sqrt{\frac{K_{d}}{.05}}=1.63 \times 10^{-2}$

Then, $\left[ H _{3} O ^{+}\right]=.05 \alpha=.05 \times 1.63 \times 10^{-2}=K_{b} .15 \times 10^{-4} \,M$

$\therefore pH =3.09$

In the presence of $0.1 \,M$ of $HCl$, let $a'$ be the degree of ionization.

Then, $\left[ H _{3} O ^{+}\right]=0.01$

$\left[ A ^{-}\right]=005 \alpha^{\prime}$

$[ HA ]=.05$

$K_{a}=\frac{0.01 \times .05 \alpha^{\prime}}{.05}$

$1.32 \times 10^{-5}=.01 \times \alpha^{\prime}$

$\alpha^{\prime}=1.32 \times 10^{-3}$