Specific conductance of $0.1 M \mathrm{HA}$ is $3.75 \times 10^{-4}$ $\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$.…

Specific conductance of $0.1 M \mathrm{HA}$ is $3.75 \times 10^{-4}$ $\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$. If $\lambda^{\infty}(\mathrm{HA})=250 \mathrm{ohm}^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$,
the dissociation constant $K_{a}$ of HA is:
  1. $1.0 \times 10^{-5}$
  2. $2.25 \times 10^{-4}$
  3. $2.25 \times 10^{-5}$
  4. $2.25 \times 10^{-13}$

Solution

$\lambda_{m}=\frac{1000 \kappa}{0.1}=\frac{1000 \times 3.75 \times 10^{-4}}{0.1}=3.75$
$\alpha=\frac{\lambda_{\mathrm{m}}}{\lambda_{\mathrm{m}}^{\infty}}=\frac{3.75}{250}=1.5 \times 10^{-2}$
$K_{\mathrm{a}}=\mathrm{C} \alpha^{2}=0.1 \times\left(1.5 \times 10^{-2}ight)^{2}=2.25 \times 10^{-5}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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