Conductivity of 0.00241 M acetic acid is `7.896 xx 10^(-5) S cm^(-1)`. Calculate its molar conductivity. If `Lambda_(m)^(@)` for acetic acid is 390.5 S `cm^2 mol^(-1)` , what is its dissociation constant?

Explanation

`k = 7.896 xx 10^(-5) S cm^(-1)`
Molar conductivity can be obtained as under :
`Lambda_(m) = (1000 k)/M = (1000 xx 7.896 xx 10^(-5))/0.00241 = (7.896 xx 10^(3))/241 = 7896/241 = = 32.76 S cm^(2) mol^(-1)`
`alpha = (Lambda_(m))/(Lambda_(m)^(@)) =32.76/390.5 = 0.0832 = 8.38 %`
Use the following relation to calculate dissociation constant, `K_(a)`
`K_(a) = (Calpha^(2))/(1- alpha) =(0.00241 xx (0.084)^(2))/(1-0.084) = 1.86 xx 10^(-5)`