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Electrochemistry

Question
CBSEENCH12006114

The conductivity of 0.00241 M acetic acid is 7.896 x 10–5 s cm–1. Calculate its molar conductivity and if Λ°m for acetic acid is 390.5 s cm2 mol–1, what is its dissociation constant?

Solution

                                      M = 0.00241 M
                                      K = 7.896 x 10-5 s cm-1
                                     m = 103KM = 103×7.896×10-50.00241m = 32.76 s cm2 mol-1
                                       α = m°m = 32.76390.5 = 0.0839K = 21-α = 0.00241×(0.0839)21-0.0839K = 1.85 × 10-5