The conductivity of `0.001028 mol L ^(-1)` acetic acid is `4.95 xx10^(-5)S cm ^(-1).` Calculate its dissociation constant if `^^(m)^(0)` for acetic acid id `390.5 S cm ^(2)mol ^(-1).`

Explanation

`^^_(m) = (k)/(C) = (4.95 xx 10^(-5) S cm^(-1))/(0.001028 mol L^(-1)) xx (1000cm^(3))/(L) = 48.15 S cm^(2) mol^(-1)` ,
`alpha = (^^_(m))/(^^_(m)) = (48.15 S cm^(2) mol^(-1))/(390.5 S cm^(2) mol^(-1))= 0.1233`
`k = (c alpha^(2))/((1-alpha)) = (0.001028 mol L^(-1) xx (0.1233)^(2))/(1-0.1233) = 1.78 xx 10^(3) mol L^(-1)`