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)`.

Options

A. `1.78 xx 10^(-5)`
B. `1.87 xx 10^(-6)`
C. `2.05xx 10^(-5)`
D. `1.78 xx 10^(-6)`

Answer

A

Explanation

`:' Lambda = kappa xx (1000)/(C) = (4.95 xx 10^(-5) xx 1000)/(0.001028)`
`= 48.15 S cm^(2) mol^(-1)`
`:. alpha = (Lambda)/(overset(@)(Lambda)) = (48.15)/(390.5) = 0.1233`
Now, `K = (C.alpha)^(2)/(1-alpha) = (0.001028 xx (0.1233)^(2))/((1-0.1233))`
`= 1.78 xx 10^(-5)`