Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

Q.11: Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

Solution:

Given that ,          κ = 7.896 × 10⁻⁵ S m⁻¹

C=M= 0.00241 mol ^L−1

The formula of molar conductivity,

Λ_m = (k  × 1000)/M

Plug the value we get

Λ_m = (7.896 × 10⁻⁵  ×  1000)/ 0.00241

      = 32.76S cm² mol⁻¹

The formula of degree of dissociation

α    = Λ_m/ Λ^o_m

Plug the value we get

α    = 32.76S/390.5

                         = 0.084

The formula of dissociation constant

K = Cα/(1 – α)

Plug the values we get

K = 0.00241 × 0.084/(1– 0.084)

                      = 1.86 × 10⁻⁵ mol L⁻¹