pH of a saturated solution of `Ba(OH)_2` is 12. The value of solubility product (`K_(sp)`) of `Ba(OH)_2` is-

Options

A. `4.0xx10^(-6)`
B. `5.0xx10^(-6)`
C. `3.3xx10^(-7)`
D. `5.0xx10^(-7)`

Explanation

`pOH=14-pH=14-12=2`
`:.[OH^-]=10^(-2)(M)`
`Ba(OH)_2(s) `:.[Ba^(2+)]=1/2[OH^-]=1/2xx10^(-2)=5xx10^(-2)=5xx10^(-3)(M)`
`K_(sp)=[Ba^(2+)][OH^-]^2=5xx10^(-3)xx(10^(-2))^2`
`=5xx10^(-7)`