The conductivity of 0.001 M acetic acid is 5 × 10^–5 S cm^–1 and ${\wedge }^0$ is 390.5 S cm² mol^–1 then the calculated value of dissociation constant
of acetic acid would be

Given concentration (C) = 0.001 mol.L^-1. 

Specific conductance (k) = 4 X 10^-5.

$Molar conductivity =\frac{k \times 1000}{C}= \frac{4\times {10}^{-5}\times 1000}{0.001}=40S.c{m}^2.mo{l}^{-1}.$

$Degree of ionization \left(\alpha \right)=\hspace{0.17em}\frac{{\wedge }_m^c}{{\wedge }_m^{\infty }}=\frac{40}{390}0.1025$

$Acid dissociation cons\tan t \left(K_a\right)= \frac{C{\alpha }^2}{1-\alpha }= \frac{0.001\times {(0.1025)}^2}{1-0.1025}=1.18\times {10}^{-5}$