Q45 / 150 · MHT CET 2021 (21 Sep Shift 1) · Chemistry
Dissociation constant of propionic acid is $1.32 \times 10^{-5}$. Calculate the degree of dissociation of acid in $0.05 \mathrm{M}$ solution.
- A$2.6 \times 10^{-4}$
- Correct answerB$1.61 \times 10^{-2}$
- C$1.90 \times 10^{-2}$
- D$3.5 \times 10^{-5}$
Solution
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}$ (Propionic Acid)
$$
\begin{aligned}
& \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{(\mathrm{aq})}^{+} \\
& \mathrm{t}=0_1 \quad \mathrm{C} \\
& \mathrm{t}=\mathrm{t}_{\mathrm{aq}}, \quad \mathrm{C}-\mathrm{C} \propto \quad \mathrm{C} \propto \quad \mathrm{C} \propto \\
& \mathrm{k}_{\mathrm{a}}=\frac{\mathrm{c} \propto^2}{1-\propto} \quad(1-\propto \simeq 1) \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \propto=\sqrt{\frac{\mathrm{k}_{\mathrm{a}}}{\mathrm{C}}}=\sqrt{\frac{1.32 \times 10^{-5}}{0.05}}=\sqrt{\frac{132 \times 10^{-1}}{5}} \times 10^{-2} \quad \mathrm{k}_{\mathrm{a}}=\mathrm{c}^2 \\
& =\sqrt{2.64} \times 10^{-2}=1.61 \times 10^{-2}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.