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

Options

A. `5.0 xx 10^(-7) m^(3)`
B. `0.6 xx 10^(-12) M^(3)`
C. `4.0 xx 10^(-8)M^(3)`
D. `5.0 xx 10^(-9) M^(3)`

Answer

A

Explanation

Solubility expressed in terms of molarity
`[OH^(-)] = 10^(-2)`
`[Ba^(+2)] = 0.5 xx 10^(-2)`
`K_(SP) = (0.5 xx 10^(-2)) (10^(-2))^(2)`
`K_(SP) = 5 xx 10^(-7)`