Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If ∧0m for acetic acid is 390.5 S cm2mol–1, what is its dissociation constant?
Given -
Molarity, C = 0.00241 M
Conductivity, κ = 7.896 × 10–5 S cm–1
Molar conductivity, ∧m = ?
for acetic acid = 390.5 S cm2mol–1
Molar conductivity, ∧m = S cm2 mol-1
∴ ∧m = 32.76 S cm2 mol-1
To calculate the dissociation constant, Ka, we use
Ka = → Equation 1
Here, we need to find the value of α (degree of dissociation), by the formula,
∴ α = 8.4×10-2 ⇒ Equation 2
→ Thus, substituting Equation 2 in Equation 1, we get,
Ka =
=
∴ Ka = 1.86×10 - 5
The molar conductivity, ∧m is 32.76 S cm2 mol-1 and the dissociation constant, Ka is 1.86×10-5