Question bank/JEE Main/Chemistry

The dissociation constant of acetic acid at a given temperature
is 1.69 × 10^-5. The degree of dissociation of 0.01 M acetic acid in
the presence of 0.01 M HCl is

A1.69 × 10^-7B1.69 × 10^-5C1.69 × 10^-3D2.9 × 10^-2

Show answer and solution▾

${CH}_{3}COOH(aq) \rightleftharpoons H^{+}(aq) + {CH}_{3}{COO}^{-}(aq)$
t = 0 0.01
t = eq 0.01 - x x x
[H^+] = x + 0.01[IMAGE] 0.01 M
$\therefore$
$K_{a} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack CH_{3}COO^{-} \right\rbrack}{\left\lbrack CH_{3}COOH \right.\ }$
$\Rightarrow$ 1.69 ×10^-5 =
$\frac{0.01 \times \left\lbrack {CH}_{3}COO^{-} \right\rbrack}{0.01}$
$\therefore$ CH_3COO-] = 1.69 × 10^-5 M
So, degree of dissociation of CH_3COOH
$= \frac{1.69 \times 10^{- 5}}{0.01} = 1.69 \times 10^{- 3}$

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx