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, $\left[ H _{3} O ^{+}\right]=0.01$
$\left[ A ^{-}\right]=005 \alpha^{\prime}$
$[ HA ]=.05$
$K_{a}=\frac{0.01 \times .05 \alpha^{\prime}}{.05}$
$1.32 \times 10^{-5}=.01 \times \alpha^{\prime}$
$\alpha^{\prime}=1.32 \times 10^{-3}$
When $CO_2$ dissolves in water, the following equilibrium is established
$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $
for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-
Determine the degree of ionization and $pH$ of a $0.05 \,M$ of ammonia solution. The ionization constant of ammonia can be taken from Table $7.7 .$ Also, calculate the ionization constant of the conjugate acid of ammonia.
At $25\,^oC$, the dissociation constant of $CH_3COOH$ and $NH_4OH$ in aqueous solution are almost the same. The $pH$ of a solution $0.01\, N\, CH_3COOH$ is $4.0$ at $25\,^oC$. The $pH$ of $0.01\, N\, NH_4OH$ solution at the same temperature would be
${K_b}$ of $N{H_4}OH = 1.8 \times {10^{ - 5}}$ calculate $pH$ of $0.15$ $mol$ $N{H_4}OH$ and $0.25$ $mol$ $N{H_4}OH$ containing solution.
A compound whose aqueous solution will have the highest $pH$