Question

The molar conductivity of 0.007 M acetic acid is 20 S cm² mol^–1. What is the dissociation constant of acetic acid ? Choose the correct option.

\(\begin{bmatrix} ∧_{H^+}^\circ = 350\, S\, cm^2 mol^{-1}\\ ∧_{CH_3COO^-}^\circ = 50\, S\, cm^2 mol^{-1} \end{bmatrix}\)

(1) 1.75 × 10^–4 mol L^–1

(2) 2.50 × 10^–4 mol L^–1

(3) 1.75 × 10^–5 mol L^–1

(4) 2.50 × 10^–5 mol L^–1

Answer 1

Correct option is (3) 1.75 x 10^-5 mol L^-1

\(\Lambda^\circ_m\) = 20 S cm² mol⁻¹

\(\Lambda^\circ_{m\,CH_3COOH} = \Lambda^\circ_{m\,CH_3COO^-} + \Lambda^\circ_{m\, H^+}\) 

= 50 + 350

= 400 S cm² mol⁻¹

\(\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}\)

\(= \frac{20}{400}\)

\(= \frac 1{20}\)

\(K_a = \frac{C\alpha^2}{1 - \alpha}\)

\(= C\alpha^2\)

\(= 7 \times 10^3 \times (\frac 1{20})^2\)

\(= 7 \times 10^3 \times (\frac 1{4})^2 \times 10^{-2}\)

\(= 1.75 \times 10^{-5} \) mol L⁻¹

Answer 2

Correct answer is (3).