Question

The conductivity of 0.001M acetic acid is 7.8 × 10⁻⁵ S cm⁻¹. Calculate its degree of dissociation if `∧_m^0` for acetic acid is 390 S cm² mol⁻¹.

Answer 1

Given:

1. Molar conductivity (∧) = `(k xx 1000)/C`

k = 8 × 10⁻⁵ S cm⁻¹

Concentration (C) = 0.001M

2. Limiting molar conductivity `∧_m^0` = 390 S cm² mol⁻¹

Formula:

∧ = `(k xx 1000)/C`

Substitute the given values

∧ = `(7.8 xx 10^-5 xx 1000)/0.001`

∧ = 78 S cm² mol⁻¹

Calculate the degree of dissociation (α):

The degree of dissociation is given by: `alpha = ∧/∧_m^0`

Substitute the values

`alpha = 78/390`

α = 0.2

∴ The degree of dissociation (α) of acetic acid is 0.2 or 20%.