The specific conductance of 0 . 0025 M acetic acid is 5 × 10 - 5 S cm - 1 at a certain temperature. The dissociation constant of acetic acid is ___________ × 10 - 7 .(Nearest integer) Consider limiting molar conductivity of CH 3 COOH as 400 S cm 2 mol - 1

Step-by-step solution

The relation between Molar conductance λ m and specific conductance κ is λ m = k × 1000 M M = molarity λ m = 5 × 10 - 5 × 10 3 2 . 5 × 10 - 3 = 20     S   cm 2   mol - 1 Now, the degree of dissociation, α = λ m λ ∞ α = 20 400 = 1 20 The acid dissociation constant, K a = Cα 2 ( 1 - α ) = 2 . 5 × 10 - 3 1 20 × 1 20 19 20 =   65 . 789   ×   10 – 7 ≈   66   ×   10