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 λ0(H+)= 349.6 S cm2 mol–1 and λ0(HCOO–) = 54.6 S cm2 mol–1

Q3.9:  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 λ⁰(H⁺)= 349.6 S cm² mol^–1 and λ⁰(HCOO^–) = 54.6 S cm² mol^–1

Solution:

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

λ^o(HCOOH)   = λ⁰(H⁺)  +   λ⁰(HCOO^–)

plug the values we get

λ^o(HCOOH)   = 0.349.6 + 54.6

                        =404.2 S cm² mol⁻¹

Formula of degree of dissociation:

ά = λ^o(HCOOH)/ λ^o(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 per liter