Equivalent conductivity of dilute monovalent acidic solution `(M)/(32)` is 8.0 mol `cm^(2)` and at infinite dilution equivalent conductivity is 400 mol `cm^(2)` then the dissociation constant for acid is

Options

A. `1.25xx10^(-6)`
B. `6.25xx10^(-4)`
C. `1.25xx10^(-4)`
D. `1.25xx10^(-5)`

Answer

D

Explanation

Degree of dissociation :
`alpha=(Lamda)/(Lamda_(oo))`
`=(8.00)/(400)=2xx10^(-2)`
`K_(a)=(Calpha^(2))/((1-alpha))~~Calpha^(2)`
`=(1)/(32)xx(2xx10^(-2))^(2)`
`=1.25xx10^(-5)`.