The specific conductance of 0 . 0025M acetic acid is 5 × 10 ^ - 5 S...

The relation between Molar conductance ${\lambda}_{m}$ and specific conductance $κ$ is

${\lambda}_{m} = \dfrac{k \times 1000}{M}$

M = molarity

${\lambda}_{m} = \dfrac{\left(5 \times {10}^{- 5}\right) \times \left({10}^{3}\right)}{\left(2 . 5 \times {10}^{- 3}\right)} = 20S{cm}^{2}{mol}^{- 1}$

Now, the degree of dissociation, $\alpha = \dfrac{{\lambda}_{m}}{{\lambda}^{\infty}}$

$\alpha = \dfrac{20}{400} = \dfrac{1}{20}$

The acid dissociation constant, ${K}_{a} = \dfrac{{C \alpha}^{2}}{( 1 - \alpha )} = \dfrac{\left(2 . 5 \times {10}^{- 3}\right)\left(\dfrac{1}{20} \times \dfrac{1}{20}\right)}{\left(\dfrac{19}{20}\right)}$

$= 65 . 789 \times {10}^{– 7}$

$\approx 66 \times {10}^{– 7}$