# 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.

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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Electrochemistry - 0763 mins
Nernst Equation - Learn the Concept33 mins
Difference between electrochemical and electrolysis45 mins
Electrochemistry - 0464 mins
Electrochemistry - 0621 mins
Practice Questions on faraday's law44 mins
Qualitative Aspect of Electrolysis58 mins
Master Concept of Oxidising and Reducing agent42 mins
Quiz | Nernst Equation - 0341 mins
Everything to know about Electrolytic Cell55 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses