`pH` of saturated solution of `Ba(OH)_(2)` is `12`. The value of solubility product `(K_(sp))` of `Ba(OH)_(2)` is

Options

A. `10^(-6) M^(3)`
B. `4xx10^(-6) M^(3)`
C. `0.5xx10^(-7) M^(3)`
D. `5xx10^(-7) M^(3)`

Answer

D

Explanation

Since `pH=12 :. pOH=14-12=2`
`:. [OH^(-)]=10^(-2) M`
We know `Ba(OH)_(2)hArrBa^(++)+2OH^(-)`
`:.[Ba^(++)]=M`
`:. K_(sp)=[Ba^(++)][OH^(-)]^(2)=((10^(-2))/(2))xx(10^(-2))^(2)`