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

Options

A. `4.00xx10^(-6)m^(3)`
B. `4.00xx10^(-7)M^(3)`
C. `5.00xx10^(-6)M^(3)`
D. `5.00xx10^(-7)M^(3)`

Answer

D

Explanation

At 25 C pH+pOH=14
`therefore pOH=2`
`therefore[OH]=10^(-2)M`
Now Let solulity of `Ba(OH)_(2)` be S
`{:(Ba(OH)_(2)to, Ba^(+2),+2OH^(-2)),(S,S,2S):}`
`[OH^(-)]=2S=10^(-2)`
[solubility of `Ba(OH)_(2)]S=(10^(-2))/(2)=5xx10^(-3)mol//L`
Now Ksp for `Ba(OH)_(2)=4s^(3)`
`=4xx5xx10^(-3))^(-3)=5xx10^(-7)M^(-3)`