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

Therefore
 
 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

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