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Question

The molar conductance of 0.1 M CH3COOH at pH=4 is Scm2mol1.
(Given the limiting molar conductance of acetic acid is 390 Scm2mol1).


Solution

pH=log[H+]log[H+]=4[H+]=104
[H+]=cα, for acetic acid where, 'c' is the concentration and α is the degree of dissociation.
cα=104
Given c=0.1M
α=103
Also, α=λmλ0m=103
Given, λ0m=390 Scm2mol1
Thus λm=390×103=0.390Scm2mol1
 

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