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?

Explanation

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)`