`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

`pH=12 :. pOH=2`
`:. [OH^(-)]=10^(-2)`
Now `Ba(OH)_(2)=Ba^(2+)+2OH^(-)`
`:. K_(sp)=[Ba^(2+)][OH^(-)]^(2)`
`=[10^(-2)/(2)][10^(-2)]^(2)`
`:.[(Ba^(2+))=(1)/(2)xx(OH^(-))]=5xx10^(-7)`