Conductivity of 0.00241 M acetic acid is 7.896 × 10^–5 S cm^–1. Calculate its molar conductivity. If ∧⁰_m 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, K_a, we use

K_a = → 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,

K_a =

=

∴ K_a = 1.86×10 ^{- 5}

The molar conductivity, ∧_m is 32.76 S cm² mol^-1 and the dissociation constant, K_a is 1.86×10^-5