Conductivity of 0.00241 M acetic acid solution is 7.896 xx 10^(-5) S cm^(-1). Calculate its molar conductivity in this solution. If wedge_(M)^(@) for acetic acid be 390.5 S cm^(2) mol^(-1), what would be its dissociation constant?

Conductivity of 0.00241 M acetic acid solution is 7.896 xx 10^(-5) S c

1.	Conductivity of 0.00241 M acetic acid solution is 7.896 xx 10^(-5) S cm^(-1). Calculate its molar conductivity in this solution. If wedge_(M)^(@) for acetic acid be 390.5 S cm^(2) mol^(-1), what would be its dissociation constant?
Answer» Solution :Conductivity of ACETIC acid, `K = 7.896 XX 10^(-5) "S cm"^(-1), wedge_(m)^(@)` for acetic acid = `390.5 " S cm"^(2)mol^(-1)` Molar conductivity,`wedge_(m)^(C) = (K xx 1000)/("Molarity")` `= (7.896 xx 10^(-5)xx 1000)/(0.00241)= (789600xx1000xx10^(-5))/(241)` `=32.76 " S cm"^(2) mol^(-1)` Degree of dissociation, `alpha = (wedge_(m)^(c))/(wedge_(m)^(@)) = (32.76)/(390.5) = 8.4xx10^(-2)` Dissociation constant of acetic acid, `Ka = (CALPHA^(2))/(1-alpha)=((0.00241)xx(8.4xx10^(-2))^(2))/(1-0.084) = 1.86 xx 10^(-5)`