A monoprotic acid in a 0.1 M solution ionizes to `0.001%`. Its ionisation constant is

Options

A. `1.0 xx 10^(-3)`
B. `1.0 xx 10^(-6)`
C. `1.0 xx 10^(-8)`mol `L^(-1)`
D. `1.0 xx 10^(-11)`

Answer

D

Explanation

Ionisation constant , `K_(a)=(a^(2)xx c)/(1-a)`

`alpha=` degree of dissociation `=0.001%`
`=(0.001)/(100)=1 xx 10^(-5)`
C `= ` concentration `=0.1 M`
When `alpha` is very small
`K=alpha^(2)xx C`
`=(1 xx 10^(-5))^(2) xx 0.1 =1 xx 10^(-11)`