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?
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 $a$.
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, $\left[ H _{3} O ^{+}\right]=0.01$
$\left[ A ^{-}\right]=005 \alpha^{\prime}$
$[ HA ]=.05$
$K_{a}=\frac{0.01 \times .05 \alpha^{\prime}}{.05}$
$1.32 \times 10^{-5}=.01 \times \alpha^{\prime}$
$\alpha^{\prime}=1.32 \times 10^{-3}$
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