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).

The molar conductivity of 0.025 mol L^(-1) methanoic acid is 46.1 S cm

1.	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).
Answer» Solution :(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)`.