The conductivity of a 0.01 M solution of a 1:1 weak electrolyte at 298 K is 1.5xx10^(-4)Scm^(-1). (i) molar conductivity of the solution (ii) degree of dissociation and the dissociation constant of the weak electrolyte Given that lambda_("cation")^(@)=248.2" S cm"^(2)" mol"^(-1) lambda_("anion")^(@)="51.8 S cm"^(2)" mol"^(-1)

The conductivity of a 0.01 M solution of a 1:1 weak electrolyte at 298

1.	The conductivity of a 0.01 M solution of a 1:1 weak electrolyte at 298 K is 1.5xx10^(-4)Scm^(-1). (i) molar conductivity of the solution (ii) degree of dissociation and the dissociation constant of the weak electrolyte Given that lambda_("cation")^(@)=248.2" S cm"^(2)" mol"^(-1) lambda_("anion")^(@)="51.8 S cm"^(2)" mol"^(-1)
Answer» Solution :(i) Molar conductivity `C=0.01M` `kappa=1.5xx10^(-4)"S cm"^(-1)(or)kappa=1.5xx10^(-2)"S m"^(-1)` `Lambda_(m)=(kappaxx10^(-3))/(C )"Sm"^(-1)"mol"^(-1)m^(3)=(1.5xx10^(-2)xx10^(-3))/(0.01)"Sm"^(2)"mol"^(-1)` `Lambda_(m)=1.5xx10^(-3)"Sm"^(2)"mol"^(-1)` (ii) Degree of dissociation `ALPHA=(Lambda_(m))/(Lambda_(oo))(or)alpha=(Lambda_(m))/(Lambda_(m)^(@))` `Lambda_(oo)^(@)=lambda_("cation")^(@)+lambda_("anion")^(@)` `=(248.2+51.8)"S cm"^(2)"mol"^(-1)` `="300 S cm"^(2)"mol"^(-1)` `=300xx10^(-4)"S m"^(2)"mol"^(-1)` `alpha=(1.5xx10^(-3))/(300xx10^(-4))=0.05` `K_(a)=(alpha^(2)C)/(1-alpha)` `=((0.05)^(2)(0.01))/(1-0.05)` `K_(a)=2.63xx10^(-5)`