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

Answer

3.435

Explanation

`Lambda^(@) (HCOOH) = lambda^(@) (H^(+)) + lambda^(@) (HCOO^(-))`
`= 404. 2 S cm^(2) mol^(-1)`
`alpha = (46.1)/(404.2) = 0.114`
`K_(a) = (C alpha^(2))/(1 - alpha) = (0.025 xx (0.114)^(2))/((1- 0.114)) = 3.67 xx 10^(-4)`
`= 3.435`