The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Calculate the degree of dissociation constant.
Given `: lambda_((H^(o+)))^(@)=349.6S cm^(2)mol^(-1) ` and
`lambda_((CHM_(3)COO^(c-)))^(@)=54.6Scm^(2)mol^(-1)`

Explanation

We know,
`HCOOHhArrH^(+)+HCOO^(-)`
`Lambda_(m)^(0)HCOOH=lambda_(H+)^(0)+lamda_(HCOO^(-))`
`=349.6+54.6`
`="404.2 S cm"^(2)mol^(-1)`
Degree of dissociation ` alpha ` may be calculated as
`alpha=(Lambda_(m)HCOOH)/(Lambda_(m)^(0)HCOOH)=(46.1)/(404.2)=0.114`
Let us consider the ionisation of HCOOH.
`HCOOHhArrHCOO^(-)+H^(+)`
`{:(t=0,C,0,0),(teq.,C-Calpha,Calpha,Calpha):}`
Dissociation constant of formic acid may be calculated as,
`K=([HCOO^(-)][H^(+)])/([HCOOH])`