The dissociation constant of a monobasic acid which is `3.5%` dissociated in `(N)/(20)` solution at `20^(@)C` is

Options

A. `3.5xx10^(-2)`
B. `5xx10^(-3)`
C. `6.34xx10^(-5)`
D. `6.75xx10^(-2)`

Answer

C

Explanation

Concentration of acid `=(N)/(20)=0.05N`
Out of `100` molecules, `3.5` molecules has been dissociated
`:.` Out of `1` molecule the no. of disscoiated molecules
`=(35)/(100)=0.035=alpha`
`K_(a)=(c alpha^(2))/((1-alpha))=(0.05xx(0.035)^(2))/(1-0.035)`
`=6.34xx10^(-5)`