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 ?

Answer

32.76 S `cm^(2)mol^(-1).1.85xx10^(-5)`

Explanation

`wedge_(m)^(c)=(kappaxx1000)/("Molarity")=(7.896xx10^(-5)" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.00241" mol "L^(-1))=32.76" S "cm^(2)mol^(-1)`
`alpha=(32.76)/(390.5)=0.084,K=(calpha^(2))/(1-alpha)=(0.00241xx(0.084)^(2))/(1-0.084)=1.85xx10^(-5)`.