Question

The specific conductance $(k )$ of $0.02 \,M$ aqueous acetic acid solution a t $298 \,K$ is $1.65 \times 10^{-4} \,S\, cm^{-1}$.The degree of dissociation of acetic acid is [Given equivalent conductance at infinite dilution of $H^{-} = 349.1\, S\, cm^{2} mol^{-1} \,\,$ and $\,\ CH_{3}COO^{-} =40.9\, S\, cm_{2} \,mol_{-1}$]

• A. 0.021
• B. 0.21
• C. 0.012
• D. 0.12

Answer 1

Answer: (A)

Given, specific conductance of acetic acid $=1.65 \times 10^{-4} S\, cm ^{-2}$
Molarity of acetic acid $=0.02\, M$
As we know, molar conductance can be calculated using formula
$\lambda_{ m } =\frac{1000 \times k}{M} $
$\lambda_{ m _{ CH _{3} COOH }} =\frac{1000 \times 1.65 \times 10^{-4}}{0.02}=8.25 $
$\lambda_{ m _{ CH _{3} COOH }}^{\infty} =\lambda_{ m \left( H ^{+}\right)}^{\infty}+\lambda_{ m \left( CH _{3} COO ^{-}\right)}^{\infty} $
$=(349.1+40.9) S\,cm ^{2} mol ^{-1} $
$=390$
Degree of dissociation, $\alpha = \frac{\lambda_{m CH_3COOH}}{\lambda^{\infty}_{m CH_{3COOH}}}$
$\alpha=\frac{8.25}{390}=0.0211$