Question

The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate its degree of dissoiation and dissociation constant. Given ∧°_m(H⁺) = 349.6 S cm² mol^-1 and ∧°_m(HCOO^–) = 546 S cm² mol^-1?

Answer 1

∧°_m (HCOOH^–)= ∧°_m (H⁺) + ∧°_m (HCOO^–)
= 349.6 + 54.6
= 404.2 S cm² mol^-1