The conductivity of 0.001 mol L^-1 solution of a weak monobasic acid…

The correct answer is Option (4) → \(1.1 × 10^{-5}\, mol\, L^{-1}\)

For a weak electrolyte, the degree of dissociation is the ratio of its molar conductivity at the given concentration to its limiting molar conductivity, \( \alpha = \Lambda_m / \Lambda_m^\circ \). The dissociation constant then follows from the equilibrium \( \ce{HA <=> H+ + A-} \):

    \( \displaystyle K_a = \frac{c\,\alpha^{2}}{1 - \alpha} \)

First the molar conductivity (the factor 1000 converts litres to cm³):

    \( \displaystyle \Lambda_m = \frac{\kappa \times 1000}{c} = \frac{4.0 \times 10^{-5}\ \mathrm{S\,cm^{-1}} \times 1000\ \mathrm{cm^{3}\,L^{-1}}}{0.001\ \mathrm{mol\,L^{-1}}} = 40\ \mathrm{S\,cm^{2}\,mol^{-1}} \)

    \( \displaystyle \alpha = \frac{40}{400} = 0.1 \)

    \( \displaystyle K_a = \frac{0.001 \times (0.1)^{2}}{1 - 0.1} = \frac{1.0 \times 10^{-5}}{0.9} = 1.1 \times 10^{-5}\ \mathrm{mol\,L^{-1}} \)

So the dissociation constant is \( 1.1 \times 10^{-5} \) mol L⁻¹.

Shortcut: \( c\alpha^{2} = 1.0 \times 10^{-5} \), and dividing by \( 1 - \alpha \) raises it slightly, so the answer is the option just above \( 1.0 \times 10^{-5} \).

Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Applications of Kohlrausch Law – “The conductivity of 0.001028 mol L^–1 acetic acid is 4.95 × 10^–5 S cm^–1. Calculate its dissociation constant if Λ⁰_m for acetic acid is 390.5 S cm² mol^–1.”