Question

Calculate the degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 × 10-5 S cm-1.

Given λ⁰H₊ = 349.6 S cm² mol⁻¹; λ⁰CH₃COO⁻ = 40.9 S cm² mol⁻¹

Answer 1

K =8.0×10^-5 s cm^-1

M = 0.0024

Molar conductance = (k× 1000)/m

= (8.0×10^-5×1000)/0.0024

= 33.33 s cm^2 mol^-1

Now degree of dissociation = molar conductance/sum of conductance of H+ and CH3COO-

= 33.33/390.5

= 0.0853

Answer 2

The degree of dissociation of 0.0024 M acetic acid if conductivity of this solution is 8.0 × 10-5 S cm-1 is 0.085

• ∧°(CH3COOH) = λ∘(H+) +  λ°(CH3COO−)

$349.6+ 40.9 = 390.5 S. {cm}^2 {mol}^(-1)$

∧$m=\frac{k*1000}{c}$

∧$m =\frac{8.0*{10}^5 S {cm}^1{L}^1}{0.0024 mol {L}^1} =33.33 S {cm}^2 {mol}^1$

α=(∧m)/(∧∘m)

Therefore,

           α =$\frac{33.33 S cm^2 {mol}^(-1)}{390.5 S cm^2 mol^(-1)}$ =0.085