`pH` of 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)`

Answer

D

Explanation

`Ba(OH)_(2)`
pH=12
`therefore pOH=2`
`because [OH]^(-)=2S=10^(2)`
`therefore S=(1)/(2)xx10^(-2)=5xx10^(-3)`
`K_(SP)=4S^(3)=4(5xx10^(-3))^(3)`
`=4xx125xx10^(-9)`
`=500xx10^(-9)`
`=5xx10^(-7)`