Question

The conductivity of `0.001028 M` acetic acid is `4.95xx10^(-5) S cm^(-1)`. Calculate dissociation constant if `wedge_(m)^(@)` for acetic acid is `390.5 S cm^(2) mol ^(-1)`.

Answer 1

`wedge_(m)=(kxx1000)/(M)=(4.95xx10^(-5)S cm^(-1))/(0.001028 mol L^(-1))xx(1000cm^(3))/(L)`
`=48.15 S cm^(2)mol ^(-1)`
`alpha=(wedge_(m)^(c))/(wedge_(m)^(@))=(48.15S cm^(2) mol^(-1))/(390.5 S cm^(2)mol^(-1))=0.1233`
`K_(a)=(calpha^(2))/((1-alpha))=(0.001028mol L^(-1)xx(0.1233)^(2))/((1-0.1233))`
`=1.78xx10^(-5)mol L`