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 ?

Explanation

`Lambda_(m)^(c) = (kappa xx 1000)/(M)`
`= (7.896 xx 10^(-5) S cm^(-1) xx 1000 cm^(3) L^(-1))/(0.00241 mol L^(-1))`
`= 32.76 S cm^(2) mol^(-1)`
`alpha = (Lambda_(m)^(c))/(Lambda_(m)^(@))`
= `(32.76)/(390.5) = 8.39 xx 10^(-2)`
`K_(a) = (C alpha^(2))/(1 - alpha) = (0.00241 xx (8.39 xx 10^(-2))^(2))/((1 - 8.39 xx 10^(-2)))`
`= 1.86 xx 10^(-5)`