Question

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?

Answer 1

Given, κ = 7.896 × 10⁻⁵ S m⁻¹

c = 0.00241 mol L⁻¹

Then, molar conductivity,

= 32.76S cm² mol⁻¹

Again, = 390.5 S cm² mol⁻¹

Now,

= 0.084

Dissociation constant,

= 1.86 × 10⁻⁵ mol L⁻¹