Q. 94.3( 41 Votes )

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.

Answer :

C = 0.025 mol L-1


Am = 46.1 Scm2 mol-1


λ0 (H+) = 349.6 Scm2 mol-1


λ0 (HCOO-) = 54.6 Scm2 mol-1


Λ0m (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


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