The conductivity of a weak acid HA of concentration 0.001 mol L^(-1) is 2.0×10^(-5) S cm^(-1). If Λm0(HA)=190 S cm^2 mol^(-1), the ionization constant K_a of HA is equal to ____ ×10^(-6) (Round off to the Nearest Integer)

Correct answer: 12. Λ_m=1000×K/M =1000×(2×10^(-5))/0.001=20 S cm^2 mol^(-1) ⇒α=Λ_mΛm∞=20/190=2/19 =12.3×10^(-6)

Q23 / 90 · JEE Main 2021 (27 Jul Shift 1) · Chemistry

The conductivity of a weak acid $HA$ of concentration $0.001 mol L^{- 1}$ is $2.0 \times 10^{- 5} S {cm}^{- 1}$ . If $Λ_{m}^{0} (HA) = 190 S {cm}^{2} {mol}^{- 1},$ the ionization constant $(K_{a})$ of HA is equal to $____$ $\times 10^{- 6}$
(Round off to the Nearest Integer)

Correct answer: 12

$Λ_{m} = 1000 \times \frac{K}{M}$
$= 1000 \times \frac{2 \times 10^{- 5}}{0.001} = 20 S {cm}^{2} {mol}^{- 1}$

$\Rightarrow α = \frac{Λ_{m}}{Λ_{m}^{\infty}} = \frac{20}{190} = (\frac{2}{19})$

$= 12.3 \times 10^{- 6}$

Hence, the correct answer is 12.