The molar conductivity of 0.025 mol L –1 methanoic acid is 46.1 S cm 2 mol –1 . Calculate its degree of dissociation and dissociation constant. Given λ 0 (H+) = 349.6 S cm 2 mol –1 and λ 0 (HCOO – ) = 54.6 S cm 2 mol –1 .

Question

Philoid · Accepted Answer

C = 0.025 mol L^-1

A_m = 46.1 Scm² mol^-1

λ⁰ (H^⁺) = 349.6 Scm² mol^-1

λ⁰ (HCOO^-) = 54.6 Scm² mol^-1

Λ⁰_m (HCOOH) = Λ ⁰(H^⁺) + Λ⁰(HCOO^-)

= 349.6 + 54.6

= 404.2 S cm² mol^-1

Now, the degree of dissociation:

∴ α = 0.114(approximately)

Thus, dissociation constant:

⇒ K = 3.67 × 10^-4 mol L^-1

The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Calculate its degree of dissociation and dissociation constant. Given λ0(H+) = 349.6 S cm2 mol–1 and λ0 (HCOO–) = 54.6 S cm2 mol–1.

The molar conductivity of 0.025 mol L^–1 methanoic acid is 46.1 S cm² mol^–1. Calculate its degree of dissociation and dissociation constant. Given λ⁰_(H+) = 349.6 S cm² mol^–1 and λ⁰ (HCOO^–) = 54.6 S cm² mol^–1.