Question

At ${25}^{0}C$ the ionization constants of $C{H}_{3}COOH$ and $N{H}_3$ are almost equal. The $pH$ of a solution $0.01$ M $C{H}_{3}COOH$ is $4.0$ at ${25}^{0}C$.

The $pOH$ of 0.01 M $N{H}_{4}OH$ at the same temperature will be:

• A. $3.0$
• B. $4.0$
• C. $10.0$
• D. $10.5$

Answer 1

The correct option is B $4.0$

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

$C$ - -

$C-C\alpha$ $C\alpha$ $C\alpha$

$K_a=\frac{C^2{\alpha }^2}{C(1-\alpha )}$ as $\alpha <<1$ (for weak acid); $1-\alpha \simeq 1$

$\Rightarrow K_a=C{\alpha }^2$

$\Rightarrow \alpha =\sqrt{\frac{K_a}{C}}$

$\left[H^{+}\right]={10}^{-4}=C\alpha$ as $pH=4$

Thus, ${10}^{-4}=C\sqrt{\frac{K_a}{C}}=\sqrt{K_{a}C}$

$\Rightarrow K_a=\frac{{10}^{-8}}{C}=\frac{{10}^{-8}}{0.01M}={10}^{-6}$

as $K_a=K_b$ (given)

$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}{+}^{-}OH$

$C$ - -

$C-C\alpha$ $C\alpha$ $C\alpha$

Similarly $K_b=C{\alpha }^2$ as for weak base $\alpha <<1$

$\Rightarrow \alpha =\sqrt{\frac{K_b}{C}}$

as ${[}^{-}OH]=C\alpha =C\sqrt{\frac{K_b}{C}}=\sqrt{K_{b}C}$

$\Rightarrow {[}^{-}OH]=\sqrt{{10}^{-6}\left(0.01\right)}={10}^{-4}M$

$\Rightarrow pOH=4$