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The conductivity of `0.00241 M` acetic acid is `7.896xx10^(-5)Scm^(-1)`. Calculate its molar conductivity. If `wedge_(m)^(@)` for acetic acid is `390.5Scm^(2)mol^(-1)`, what is its dissociation constant ? |
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Answer» Correct Answer - `1.86xx10^(-5)` `wedge_(m)^(c)=(kxx1000)/(Molarity)` `=((7.896xx10^(-5)Scm^(-1))xx1000cm^(3)L^(-1))/(0.00241mol L^(-1))` `=32.76 S cm^(2) mol^(-1)` `alpha=(wedge_(m)^(c))/(wedge_(m)^(@))=(32.76)/(390.5)=8.4xx10^(-2)` `K_(a)=(calpha^(2))/(1-alpha)=(0.00241xx(8.4xx10^(-2))^(2))/(1-0.084)=1.86xx10^(-5)` |
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