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).`

Answer

`1.78 xx 10^(-5)`

Explanation

`Lambda = (4.95 xx 10^(-5) xx 1000)/(1.028 xx 10^(-3)) = 48.15 S cm^(2) mol^(-1)`
`alpha = (48.15)/(390.5) = 0.1233`
`K_(a) = (C alpha^(2))/(1 - alpha)`
`= (1.028 xx 10^(-3) xx (0.1233)^(2))/(1- 0.123) = 1.78 xx 10^(-5)`.