The conductivity of a weak acid HA of concentration \(0.001 \mathrm{~mol} \mathrm{~L}^{-1}\) is \(2.0 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}\). If \(\wedge_{\mathrm{m}}^{\circ}(\mathrm{HA})=190 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\), the ionization constant (\(\mathrm{K}_{\mathrm{a}}\)) of HA is equal to ______ \(\times 10^{-6}\) (Round off to the nearest integer)

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Redox Reactions and Electrochemistry

The conductivity of a weak acid HA of concentration

0.001 mol L^{- 1}

2.0 \times 1 0^{- 5} S cm^{- 1}

. If

\land_{m}^{\circ} (HA) = 190 S cm^{2} mol^{- 1}

, the ionization constant (

K_{a}

) of HA is equal to ______

\times 1 0^{- 6}

(Round off to the nearest integer)

Answer: 12

The molar conductivity at this concentration is

Λ_{m} = \frac{κ \times 1000}{C} = \frac{2.0 \times 1 0 ^{- 5} \times 1000}{0.001} = 20 S c m^{2} mo l^{- 1}

For a weak electrolyte the degree of dissociation is the ratio of this to the value at infinite dilution:

α = \frac{Λ _{m}}{Λ _{m}^{\circ}} = \frac{20}{190} = 0.1053

For

HA ⇌ H^{+} + A^{-}

K_{a} = \frac{C α ^{2}}{1 - α} = \frac{0.001 \times ( 0.1053 ) ^{2}}{1 - 0.1053} = \frac{1.108 \times 1 0 ^{- 5}}{0.8947} = 1.24 \times 1 0^{- 5}

Rounded to the nearest integer, the ionization constant is

12 \times 1 0^{- 6}

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer