Equivalent conductivity of a weak acid HA at infinite dilution is `390 S cm^(2) eq^(-1)`. Conductivity of `1xx10^(-3)` N HA solution is `4.9xx10^(-5) S cm^(-1)`. Calculate the extent of dissociation and dissociation constant of the acid.

Explanation

`Lambda_(c) = (4.9 xx 10^(-5) xx 1000)/(1xx10^(-3))=49 S cm^(2) eq^(-1)`.
Equivalent conductance at infinite dilution `(Lambda_(0))=390 S cm^(2) eq^(-1)`.
Extent of dissociation `= alpha = (Lambda_(c))/(Lambda_(0))=(49)/(390)=0.126`.
Dissociation constant of acid `= C alpha^(2) = 1xx10^(-3) (0.126)^(2)=1.5 x 10^(-5) "mol" L^(-1)`.