The molar conductivity of 0.025 mol L^(-1)methanoic acid is 46.1 S cm^(2) mol L^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda^(@)(H^(+)) = 349.6 S cm^(2) mol^(-1) and lambda^(@) (HCOO^(-)) = 54.6S 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 L^(-1) . Calculate its degree of dissociation and dissociation constant. Given lambda^(@)(H^(+)) = 349.6 S cm^(2) mol^(-1) and lambda^(@) (HCOO^(-)) = 54.6S cm^(2) mol^(-1).
Answer» Solution :`Lambda_(m)^(@)(HCOOH) = lambda^(@)(H^(+)) + lambda^(@)(HCOO^(-)) = (349.6 + 54.6) S CM^(2) mol^(-1) = 404.2 S cm^(2) mol^(-1)` `Lambda_(m)^( C) = 46.1 cm^(2) mol^(-1)` `therefore alpha = (Lambda_(m)^( c))/(Lambda_(m)^(@)) = 46.1/404.2 = 0.114` To find out dissociation CONSTANT, proceed as follows: `HCOOH `therefore K_(a) =(Calpha-Calpha)/(C(1-alpha)) =(Calpha^(2))/(1-alpha) =(0.025 xx (0.114)^(2))/(1- 0.114) = 3.67 xx 10^(-4)` Thus, the dissociation constant `=3.67 xx 10^(-4)`.