The molar conductivity of `0.025 mol L^(-1)` methanoic acid (HCOOH) is `46.1 Scm^2 mol^(-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 = 404.2 S cm^2 mol^-1`
Degree of dissociation of HCOOH = `Lambda_m/Lambda_m^@= 46.1/404.2=0.114`
for weak electrolytes, `K_a = (Calpha^2)/(1-alpha) therefore K_a=(0.025xx(0.114)^2)/(1-0.0114)=3.67xx10^-4`