The conductivity of 0.001028 mol l-1 acetic acid is 4.94*10-5 S cm-1 .can you calculate its dissociation constant if Λ 0 m foe acetic acid is 390.5 S cm2 mol-1.
The degree of dissociation is calculated by the following relation
α = Λ/ Λ0
Where Λ is the molar conductivity and Λ0 is the limiting molar conductivity of the solution.
Now Λ = к/c = [к (Scm-1) X 1000cm3L-1]/ molarity (molL-1)
where к = conductivity of solution and c is the concentration of solution
Given к = 4.94*10-5 S cm-1 and c = 0.001028 molL-1
Λ = [4.94 X 10-5 X 1000] / 0.001028 Scm-2mol-1
= 48.054 Scm-2mol-1
Given Λ0 = 390.5 Scm-2mol-1
therefore α = Λ/ Λ0 = 48.054/390.5 = 0.123