The conductive of 0.001028 mol `L^(-1)` acetic acid si `4.95xx10^(-5)S" "cm^(-1)` calculate its dissociation constant if `^^_(m)^(@)` for acetic acid is

Explanation

`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`