The molar conductivity of… - NEET UG 2021 Chemistry PYQ

NEET UG 2021 — Chemistry Physical Chemistry

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The molar conductivity of $0.007M$ acetic acid is $20S{\mathrm{cm}}^{2}{\mathrm{mol}}^{-1}.$ What is the dissociation constant of acetic acid? Choose the correct option.

${\lambda }_{{H}^{+}}^{\circ }=350S{\mathrm{cm}}^{2}{\mathrm{mol}}^{-1} {\lambda }_{{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}}^{\circ }=50S{\mathrm{cm}}^{2}{\mathrm{mol}}^{-1}$

Official previous-year question

Held on 30 Apr 2021 · Verified 9 Jul 2026.

Options

A
$1.75\times {10}^{-4}\mathrm{mol}{L}^{-1}$
B
$2.50\times {10}^{-4}\mathrm{mol}{L}^{-1}$
C
$1.75\times {10}^{-5}\mathrm{mol}{L}^{-1}$
D
$2.50\times {10}^{-5}\mathrm{mol}{L}^{-1}$

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Solution

${\lambda }_{{\mathrm{CH}}_{3}\mathrm{COOH}}^{\circ }={\lambda }_{{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}}^{\circ }+{\lambda }_{{H}^{+}}^{\circ }$

${\lambda }_{{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}}^{\circ }=350+50=400S{\mathrm{cm}}^{2}{\mathrm{mol}}^{-1}$

${\lambda }_{{\mathrm{CH}}_{3}\mathrm{COOH}}^{}=20{\mathrm{Scm}}^{2}{\mathrm{mol}}^{-1}$ given

$\alpha =\frac{{\lambda }_{{\mathrm{CH}}_{3}\mathrm{COOH}}^{}}{{\lambda }_{{\mathrm{CH}}_{3}\mathrm{COOH}}^{\circ }}=\frac{20}{400}=5\times {10}^{-2}$

${K}_{a}={c\alpha }^{2}=0.007\times {(5\times {10}^{-2})}^{2}=1.75\times {10}^{-5}{\mathrm{molL}}^{-1}$

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Options

$1.75\times {10}^{-4}\mathrm{mol}{L}^{-1}$

$2.50\times {10}^{-4}\mathrm{mol}{L}^{-1}$

$1.75\times {10}^{-5}\mathrm{mol}{L}^{-1}$

$2.50\times {10}^{-5}\mathrm{mol}{L}^{-1}$

Solution

${\lambda }_{{\mathrm{CH}}_{3}\mathrm{COOH}}^{\circ }={\lambda }_{{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}}^{\circ }+{\lambda }_{{H}^{+}}^{\circ }$

${\lambda }_{{\mathrm{CH}}_{3}{\mathrm{COO}}^{-}}^{\circ }=350+50=400S{\mathrm{cm}}^{2}{\mathrm{mol}}^{-1}$

${\lambda }_{{\mathrm{CH}}_{3}\mathrm{COOH}}^{}=20{\mathrm{Scm}}^{2}{\mathrm{mol}}^{-1}$ given

$\alpha =\frac{{\lambda }_{{\mathrm{CH}}_{3}\mathrm{COOH}}^{}}{{\lambda }_{{\mathrm{CH}}_{3}\mathrm{COOH}}^{\circ }}=\frac{20}{400}=5\times {10}^{-2}$

${K}_{a}={c\alpha }^{2}=0.007\times {(5\times {10}^{-2})}^{2}=1.75\times {10}^{-5}{\mathrm{molL}}^{-1}$

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