At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 \times 102 S cm2 mol - 1. At 298 K, for an aqueous solution of the acid the degree of dissociation is \alpha and the molar conductivity is y \times 102 S cm2 mol - 1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y \times 102 S cm2 mol - 1. The value of \alpha is \underline\hspace2cm .

Electrochemistry PYQ — JEE Advanced 2021 · Shift 2 · Q7 with Solution

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Correct answer: 0.22

For a weak monobasic acid, the degree of dissociation is related to molar conductivity by

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

where $\Lambda_m^\circ$ is the limiting molar conductivity.

Given:

\Lambda_m^\circ = 4 \times 10^2\; \text{S cm}^2\text{ mol}^{-1}

Initial molar conductivity:

\Lambda_m = y \times 10^2\; \text{S cm}^2\text{ mol}^{-1}

So initially,

\alpha = \frac{y \times 10^2}{4 \times 10^2} = \frac{y}{4}

Hence,

y = 4\alpha

After $20$ times dilution, concentration becomes $c/20$ . For a weak acid, using Ostwald's dilution law,

K_a = \frac{c\alpha^2}{1-\alpha}

After dilution, let the new degree of dissociation be $\alpha'$ . Then

K_a = \frac{(c/20)(\alpha')^2}{1-\alpha'}

Given that the molar conductivity becomes $3y \times 10^2$ , so

\alpha' = \frac{3y \times 10^2}{4 \times 10^2} = \frac{3y}{4}

But since $y=4\alpha$ ,

\alpha' = 3\alpha

\frac{c\alpha^2}{1-\alpha} = \frac{(c/20)(3\alpha)^2}{1-3\alpha}

Cancel $c$ :

\frac{\alpha^2}{1-\alpha} = \frac{9\alpha^2}{20(1-3\alpha)}

For $\alpha \neq 0$ , cancel $\alpha^2$ :

\frac{1}{1-\alpha} = \frac{9}{20(1-3\alpha)}

Cross-multiply:

20(1-3\alpha) = 9(1-\alpha)

20 - 60\alpha = 9 - 9\alpha

11 = 51\alpha

\alpha = \frac{11}{51} \approx 0.2157

\alpha \approx 0.22

So the required value is $0.22$ .