Question

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

`^^_(m)^(@)=(kxx1000)/("Molarity")`
`=(7.896xx10^(-5)" S "cm^(-1))xx1000cm^(3)L^(-1))/(0.00241" mol "L^(-1))`
`=32.76" S "cm^(2)mol^(-1)`
`alpha=(^^_(m)^(c))=(32.76)/(390.5)=8.4xx10^(-2)`
`K_(a)=(Calpha^(2))/(1-alpha)=(0.0024xx(8.4xx10^(-4))^(2))/(1-0.084)=1.86xx10^(-5)`.