The molar conductivity of 0.025 mol/L methanoic acid is 46.1 Scm²mol^-1. Calculate the degree of dissociation and dissociation constant. Given (H⁺)= 349.6 Scm²mol^-1 and (HCOO^-)= 54.6 Scm²mol^-1.

Given, K, conductivity of HCOOH = 5.25 10^-5 Scm^-1,

Molarity of HCOOH, M = 0.025 M,

(H⁺) = 349.6 Scm²mol^-1 and

⋀°(HCOO^-) =54.6 Scm²mol^-1

Molar conductivity, ⋀_m = 46.1 Scm²mol^-1

According to Kohlrauch’s Law, limiting molar conductivity of an electrolyte can be represented as sum of individual contribution of cation and anion of electrolyte.

Therefore, limiting molar conductivity of HCOOH can be written as,

⋀°_m (HCOOH) = ⋀°_m (HCOO^-) + ⋀°_m (H⁺)

= 54.6 Scm²mol^-1 + 349.6 Scm²mol^-1

= 404.2 Scm²mol^-1

Degree of dissociation,

= 0.114

Dissociation constant, K =

= = 3.67

Degree of dissociation and dissociation constant of methanoic acid is 0.114 and 3.67respectively.