The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Given ?0(H+) = 3
.6 S cm2 mol–1 and ?0 (HCOO–) = 54.6 S cm2 mol–1.

Given that
λ0(H+)= 349.6 S cm2 mol–1
λ0(HCOO–) = 54.6 S cm2 mol–1
Concentration ,C = 0.025 mol L−1
λ(HCOOH) = 46.1 S cm2 mol−1
use formula
λo(HCOOH) = λ0(H+) + λ0(HCOO–)
plug the values we get
λo(HCOOH) = 0.349.6 + 54.6
=404.2 S cm2 mol−1
Formula of degree of dissociation:
ά = λo(HCOOH)/ λo(HCOOH)
ά = 46.1 / 404.2
ά = 0.114

Formula of dissociation constant:
K = (c ά2)/(1 – ά)
Plug the values we get
K=3.67*10^-4 mol/lt