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

Using the following relation and substituting the values, we get `alpha = Lambda_(m)^( C)/Lambda_(m)^(@) = (32.76)/(390.5) = 8.4 xx 10^(-2)`
Degree of dissociation may be obtained as under:
`K_(a) -(Calpha^(2))/(1- alpha) = (0.00241 xx (8.4 xx 10^(-2))^(2))/(1-0.084) = 1.86 xx 10^(-5)`