Conductivity of `0.00241M` acetic acid solution is `7.896xx10^(-5)S.cm^(-1)`. Calculate its molar conductivity. If `Lambda_(m)^(@)` for acetic acid is `390.5S.cm^(2).mol^(-1)`, what is its dissociation constant ?

Explanation

`Lambda_(m)=(k xx 1000)/(C )=(7.896xx10^(-5)xx1000)/(0.00241)`
`=32.76S.cm^(2).mol^(-1)`
The degree of dissociation, `alpha=(Lambda_(m))/(Lambda_(m)^(@))=(32.76)/(390.5)=8.4xx10^(-2)`
Dissociation constant, `K_(a)=(C alpha^(2))/(1-alpha)=(0.00241xx(8.4xx10^(-2))^(2))/(1-0.084)`
`=1.856xx10^(-5)`