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, `lambda^(@) ( H^(+)) = 349.6S cm^(2) "mol"^(-1)`
and `lambda^(@) ( HCOO^(-)) = 54.6 S cm^(2) "mol"^(-1)`

Explanation

Step I Calculation of degree of dissociation ( `alpha)` of HCOOH
`Lambda _(m) = 461. S cm^(2) "mol"^(-1)`
`Lambda_(m)^(@) ( HCOOH) = lambda_(m(HCOO^(-)))^(@) + lambda_(m(H^(+)))^(@)`
`= ( 54.6 + 349.6) S cm^(2) "mol"^(-1)`
`= 404 .2 S cm^(2) "mol"^(-1)`
`alpha = ( Lambda_(m))/( Lambda_(m)^(@)) = ((46.1S cm^(2))cm^(2) "mol"^(-1))/(( 404.2) Scm^(2) "mol"^(-1))`
= 0.1140
Step II Calculate of dissociation constant
`HCOOH (aq) overset("Water")(hArr) HCOO^(-) ( aq) + H^(+) ( aq)`
`{:("Initial conc.",c,0,0),("Equilibrium conc.",C(1-alpha),C alpha,Calpha):}`
`K_(a) = ( [ HCOO^(-) ][H^(+)])/([ HCOOH]) = ( C alpha xx C alpha)/( C ( 1- alpha)) = ( C alpha^(2))/( ( 1-alpha))`
Putting values,
`K_(a) = (( 0.025 "mol" L^(-1)) xx ( 0.114)^(2))/( ( 1- 0.114))`
`= (( 3.249xx 10^(-4) "mol" L^(-1)))/( ( 0.886))`
`= 3.67 xx 10^(-4) ` mol `L^(-1)`