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

`1.86xx10^(-5)`

Explanation

`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)`