Molar conductivity of `"0.025 mol.L"^(-1)` methanoic acid is `"46 S.cm"^(2)."mol"^(-1)`. Calculate its degree of dissociation and dissociation constant.
Given : `lambda^(@)(HCOO^(-))="54.6 S.cm"^(2)."mol"^(-1), lambda^(@)(H^(+))="349.6 S.cm"^(2)."mol"^(-1)`.

Explanation

From Kohlrausch law,
`Lambda_(m)^(@)(HCOOH)=lambda^(@)(H^(+))+lambda^(@)(HCOO^(-))`
`="349.6+54.6=404.2.S.cm"^(2)."mol"^(-1)`
Given, `Lambda_(m)=46.1" Thus, "alpha=(Lambda_(m))/(Lambda_(m)^(@))=(46.1)/(404.2)=0.114`
`{:(,HCOOH,hArr,HCOO^(-),+,H^(+)),("Initial concn. ",C,,0,,0),("Concn. at equilibrium", C(1-alpha),,Calpha,,Calpha):}`
Therefore, `K_(alpha)=(Calpha.Calpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)=(0.025xx(0.114)^(2))/(1-0.114)`
`=3.67xx10^(-4)`