At 298 K, the limiting molar conductivity of a weak monobasic acid is \(4 \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). At 298 K, for an aqueous solution of the acid the degree of dissociation of \(\boldsymbol{\alpha}\) and the molar conductivity is \(\mathbf{y} \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes \(3 \mathbf{y} \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). The value of \(\boldsymbol{\alpha}\) is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Redox Reactions and Electrochemistry

At 298 K, the limiting molar conductivity of a weak monobasic acid is

4 \times 1 0^{2} S cm^{2} mol^{- 1}

. At 298 K, for an aqueous solution of the acid the degree of dissociation of

α

and the molar conductivity is

y \times 1 0^{2} S cm^{2} mol^{- 1}

. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes

3 y \times 1 0^{2} S cm^{2} mol^{- 1}

.

The value of

α

is ____.

Answer: 0.21

Using the relation for weak electrolytes:

K_{a} = \frac{Λ _{m}^{2} C}{Λ _{m}^{\circ} ( Λ _{m}^{\circ} - Λ _{m} )}

. For the original solution:

K_{a} = \frac{( y \times 1 0 ^{2} ) ^{2} C}{4 \times 1 0 ^{2} ( 4 \times 1 0 ^{2} - y \times 1 0 ^{2} )}

. For the diluted solution (concentration C/20):

K_{a} = \frac{( 3 y \times 1 0 ^{2} ) ^{2} ( C /20 )}{4 \times 1 0 ^{2} ( 4 \times 1 0 ^{2} - 3 y \times 1 0 ^{2} )}

. Equating the two expressions and simplifying:

\frac{y ^{2}}{4 - y} = \frac{9 y ^{2}}{20 ( 4 - 3 y )}

, which gives

\frac{1}{4 - y} = \frac{9}{20 ( 4 - 3 y )}

. Solving:

20 (4 - 3 y) = 9 (4 - y)

80 - 60 y = 36 - 9 y

51 y = 44

, so

y = 44/51 \approx 0.863

. The degree of dissociation

α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{y \times 1 0 ^{2}}{4 \times 1 0 ^{2}} = \frac{y}{4} = \frac{44/51}{4} = \frac{44}{204} = \frac{11}{51} \approx 0.2156

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Numerical answer