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 `lamda^(@)H^(+)=349.6"S "cm^(2)" mole"^(-1) and lamda^(@)(HCO O^(-))=54.6" S "cm^(2)mol^(-1)`.

Explanation

(i) Calculation of `lamda-(m)^(@)(HCOOH)`:
`underset("methanoic acid")(HCOOH) to HCOO_((aq))^(-)+H_((aq))^(+)`
`therefore lamda_(m)^(@)(HCOOH)=lamda_(m)^(@)(H^(+))+lamda_(m)^(@)(HCOO^(-))`
`=(349.6+54.6)" S "cm^(2)mol^(-1)`
`=404.2" S "cm^(2)mol^(-1)`
(ii) Calculation for degree of dissociation `(alpha)`:
`alpha=("Molar conductivity")/("Limited molar conductivity")|"Where, "lamda_(m)(HCOOH)" "=46.1" S "cm^(2)mol^(-1)`
`=(Lamda_(m)(HCOOH))/(Lamda_(m)^(@)(HCOOH))`
`=(46.1" S "cm^(2)mol^(-7))/(404.2" S "cm^(2)mol^(-1))`
`=0.1141`
Percentage of dissociation=`100alpha`
`=100xx0.1141`
`=11.41%`
(iii) Calculation for dissociation constant `K_(a)` :
`therefore K_(a)=(cxxalpha^(2))/((a-alpha))" Where, "c=0.25mol" "L^(-1)`
`=(0.025xx(0.1141)^(2))/((1-0.1141))" "alpha=0.1141`
`=(0.025xx(0.1141)^(2))/(0.8859)`
`=0.0003673`
`=3.673xx10^(-4)`.