The molar conductivity of 0.025 mol `L^(-1)` methanoic acid is 46.1 S `cm^(2)` `mol^(-1)`. Calculate its degree of dissociation and dissociation constant. Given `v^(0)(H^(+))=349.6S" "cm^(2)mol^(-1) and lamda^(0)(HCOO^(-))=54.6S" "cm^(2)mol^(-1)`

Explanation

`lambda_(m) (HCOOH) =46.1 Scm^(2) "mol"^(-1)`
`lambda_(m)^(0)=lambda_((H^(+)))^(0) +lambda_(("HCOO"^(-)))^(0)`
`=(349.5 +54.6) S cm^(2) "mol"^(-1) = 404.1 Scm^(2) "mol"^(-1)`
Degree of dissociation, `alpha = (wedge_(m))/(wedge_(m)^(@)) =(46.1)/(404.1) =0.114`
`=0.114xx100 =11.4%`
At eq. `underset(C(1-alpha))(HCOOH) hArr underset(C alpha)(H^(+))+underset(C alpha)(HCOO^(-))`
`K_(a) = ([H^(+)][HCOO^(-)])/([HCOOH])= (c alpha xx c alpha)/(c(1-alpha))=(c alpha^(2))/((1-alpha))`
`=(0.025 xx (0.114)^(2))/( (1-0.114)) =(3.249xx10^(-4))/(0.886) = 3.67 xx 10^(-4)"mol L"^(-1)`