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

Molar conductivity ,
`Lambda = (kappa xx 1000)/(M)`
`kappa = 7.896 xx 10^(-5) S cm^(-1) , M = 0.00241 M `
`therefore Lambda = (7.896 xx 10^(-5) xx 1000)/(0.00241)`
`=32.76 S cm^(2) mol^(-1)`
Now , `alpha = (Lambda^(e))/(Lambda^(o)) = (32.76)/(390.5) = 0.0839`
`K = ( C alpha^(2))/(( 1- alpha)) = (0.00241 xx (0.0839)^(2))/((1-0.0839))`
`= 1.85 xx 10^(-5)`