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_(HCOO^(-))^(@)=54.6Scm^(2)mol^(-1)`

Explanation

`^^_(m)^(@)(HCOOH)=lamda^(@)(H^(+))+lamda^(@)(HCOO)^(-)`
`=349.6+54.6`
`=404.2S" "cm^(2)" "mol^(-1)`
`^^_(m)^(C)=46.1S" "cm^(2)" "mol^(-1)`
`thereforealpha=(^^_(m)^(@))/(^^_(m)^(@))=(46.1)/(404.2)=0.114`
`{:(,HCOO,hArr,HCOO^(-),+,H^(+)),("Initial conc".,c,,c,,c),("At equi.,c(1-alpha),,calpha,,calpha):}`
`thereforeK_(a)=(c alpha.c alpha)/(c(1-alpha))=(c alpha^(2))/(1-alpha)`
`=(0.025xx(0.114)^(2))/(1-0.114)=3.67xx10^(-4)`