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

Options

A. `3.3 xx 10^(-7)`
B. `5.0 xx 10^(-7)`
C. `4.0 xx 10^(-6)`
D. `5.0 xx 10^(-6)`

Answer

B

Explanation

`Ba(OH)_(2)hArr underset (s)(Ba^(2+)) +underset (2s)(2OH^(-))`
`[OH^(-)] = 10^(-2)`
`2s= 10^(-2) " " rArr s = (10^(-2))/2 `
`Ksp = 4s^(3) = 4 xx ((10^(-2))/2)^(3) = 5 xx 10^(-7)`