The molar conductance of 0.1 M CH3COOH at pH=4 is Scm2mol−1. (Given the limiting molar conductance of acetic acid is 390 Scm2mol−1).
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pH=−log[H+]log[H+]=−4⇒[H+]=10−4 [H+]=cα, for acetic acid where, 'c' is the concentration and α is the degree of dissociation. ⇒cα=10−4 Given c=0.1M ⇒α=10−3 Also, α=λmλ0m=10−3 Given, λ0m=390Scm2mol−1 Thus λm=390×10−3=0.390Scm2mol−1