The molar conductivity of 0.025 mol `L^(-1)` methanoic acid is 46.1 S `cm^(2) mol L^(-1)` . Calculate its degree of dissociation and dissociation constant. Given `lambda^(@)(H^(+)) = 349.6 S cm^(2) mol^(-1)` and `lambda^(@) (HCOO^(-)) = 54.6 S cm^(2) mol^(-1)`.

Explanation

`Lambda_(m)^(@)(HCOOH) = lambda^(@)(H^(+)) + lambda^(@)(HCOO^(-)) = (349.6 + 54.6) S cm^(2) mol^(-1) = 404.2 S cm^(2) mol^(-1)`
`Lambda_(m)^( c) = 46.1 cm^(2) mol^(-1)`
`therefore alpha = (Lambda_(m)^( c))/(Lambda_(m)^(@)) = 46.1/404.2 = 0.114`
To find out dissociation constant, proceed as follows:
`HCOOH `therefore K_(a) =(Calpha-Calpha)/(C(1-alpha)) =(Calpha^(2))/(1-alpha) =(0.025 xx (0.114)^(2))/(1- 0.114) = 3.67 xx 10^(-4)`
Thus, the dissociation constant `=3.67 xx 10^(-4)`.