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)`

Answer

`3.67xx10^(-4)`

Explanation

`wedge_(m(HCOOH))^(@)=wedge_((H^(o+)))^(@)+wedge_((HCOO^(c-)))^(@)`
`=349.6+54.6S cm^(2)mol^(-1)`
`=404.2S cm^(2)mol^(-1)`
`wedge_(m)^(@)=46.1S cm^(2)mol^(-1)(" Given ")`
`:. alpha=(wedge_(m)^(@))/(wedge_(m)^(@))=(46.1)/(404.2)=0.114`
`{:(,HCOOH,hArr,HCOO^(c-),+,H^(o+)),("Initial conc","c mol" L^(-1),,-,,-),("Conc at Eq",c(1-alpha),,"calpha",,"calpha"):}`
`K_(a)=("calpha.calpha")/(c(1-alpha))=(calpha^(2))/(1-alpha)=(0.025xx(0.114)^(2))/(1-0.114)`
`=3.67xx10^(-4)`