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In which of the following mixed aqueous solutions, $\mathrm{pH}=\mathrm{p} K_{a}$ at equilibrium?
(1) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{MCH}_{3} \mathrm{COONa}$
(2) $100 \mathrm{mL}$ of $0.1 \mathrm{MCH}_{3} \mathrm{COOH}+50 \mathrm{mL}$ of 0.1 $\mathrm{M} \mathrm{NaOH}$
(3) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$
(4) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NH}_{3}$
Options:
(1) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{MCH}_{3} \mathrm{COONa}$
(2) $100 \mathrm{mL}$ of $0.1 \mathrm{MCH}_{3} \mathrm{COOH}+50 \mathrm{mL}$ of 0.1 $\mathrm{M} \mathrm{NaOH}$
(3) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$
(4) $100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}+100 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NH}_{3}$
Solution:
2432 Upvotes
Verified Answer
The correct answers are:
(1) is correct, (2) is correct, Both (1) and (2) are correct
pH of buffer solutions is given by
$$
\mathrm{pH}=\mathrm{pK}_{a}+\mathrm{log} \frac{[\mathrm{salt}]}{[\mathrm{acid}]}
$$
(1) $\mathrm{CH}_{3} \mathrm{COOH}$ and $\mathrm{CH}_{3} \mathrm{COONa}$
$\quad \quad$ $100 \times 0.1$ $\quad \quad$$100 \times 0.1$
No. of milimoles $=10 \quad=10$
$\because[$ salt $]=\mid a c i d]=10, p H=p K_{a}+\log \frac{10}{10}$
$\therefore \mathrm{pH}=\mathrm{pK_{a}}$
$\because \quad$ (salt) $=[\mathrm{acid}]=5, \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{5}{5}$
$\therefore \quad \mathrm{pH}=\mathrm{p} K_{\mathrm{a}}$
$[$ salt $]=10$ and $[a c i d]=0$
$p H=p K_{a}+\log \frac{10}{0}$
$\therefore \quad \mathrm{pH} \neq \mathrm{pK}_{a}$
(4) $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{COONH}_{4}$
initial $10 \quad \quad \quad \quad 10 \rightarrow$ No buffer action
Final Due to weak $\mathrm{CH}_{3} \mathrm{COONH}_{4}$ salt
$\therefore \quad p H \neq p K_{a}$
$$
\mathrm{pH}=\mathrm{pK}_{a}+\mathrm{log} \frac{[\mathrm{salt}]}{[\mathrm{acid}]}
$$
(1) $\mathrm{CH}_{3} \mathrm{COOH}$ and $\mathrm{CH}_{3} \mathrm{COONa}$
$\quad \quad$ $100 \times 0.1$ $\quad \quad$$100 \times 0.1$
No. of milimoles $=10 \quad=10$
$\because[$ salt $]=\mid a c i d]=10, p H=p K_{a}+\log \frac{10}{10}$
$\therefore \mathrm{pH}=\mathrm{pK_{a}}$
$\because \quad$ (salt) $=[\mathrm{acid}]=5, \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{5}{5}$
$\therefore \quad \mathrm{pH}=\mathrm{p} K_{\mathrm{a}}$
$[$ salt $]=10$ and $[a c i d]=0$
$p H=p K_{a}+\log \frac{10}{0}$
$\therefore \quad \mathrm{pH} \neq \mathrm{pK}_{a}$
(4) $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NH}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{COONH}_{4}$
initial $10 \quad \quad \quad \quad 10 \rightarrow$ No buffer action
Final Due to weak $\mathrm{CH}_{3} \mathrm{COONH}_{4}$ salt
$\therefore \quad p H \neq p K_{a}$
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