A conductivity cell filled with 0.1 M KCl (kappa = 1.29 S m^-1) shows a resistance of 100 Omega. Filled with 0.02 M acetic acid it shows 6450 Omega. Given Lambda deg_m(CH3COOH) = 390.5 S cm^2 mol^-1, the acid dissociation constant K_a x 10^5 is (two decimals). Use K_a = (calpha^2)/(1-alpha).

JEE Advanced levelNumerical answerDifficulty 3 of 3ec07

A conductivity cell filled with 0.1 M KCl ( $κ = 1.29 S m^{- 1}$ ) shows a resistance of $100 Ω$ . Filled with 0.02 M acetic acid it shows $6450 Ω$ . Given $Λ_{m}^{\circ} (CH X_{3} COOH) = 390.5 S c m^{2} mo l^{- 1}$ , the acid dissociation constant $K_{a} \times 1 0^{5}$ is (two decimals). Use $K_{a} = \frac{c α ^{2}}{1 - α}$ .

Numerical answer type, tolerance 0.02.

Cell constant

G^{*} = κ R = 1.29 \times 100 = 129 m^{- 1}

. For the acid:

κ = 129/6450 = 0.0200 S m^{- 1}

c = 20 mol m^{- 3}

, so

Λ_{m} = 0.0200/20 = 1.00 \times 1 0^{- 3} S m^{2} mo l^{- 1} = 10.0 S c m^{2} mo l^{- 1}

α = 10.0/390.5 = 0.0256

K_{a} = \frac{c α ^{2}}{1 - α} = \frac{0.02 \times ( 0.0256 ) ^{2}}{0.9744} = 1.35 \times 1 0^{- 5}