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Calculate the degree of dissociation (alpha) of acetic acid if its molar conductivity (Lambda_(m) ) is 39.05 S cm^(2) mol^(-1). Given: lambda^(@)(H^(+)) = 349.6 S cm^(2) mol^(-1) and lambda^(@)(CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1) |
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Answer» SOLUTION :`Lambda_(m)^(@) = lambda_(CH_(3)COOH)^(@) = 349.6 + 40.9 = 390.5 S cm^(2) mol^(-1)` Degree of dissociation `(alpha) = (Lambda_(m))/(Lambda_(m)^(@)) = (39.05)/(390.5) = 0.1` |
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