`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^(-7)M^(3)`
D. `5.00xx10^(-6)M^(3)`

Answer

D

Explanation

Given, pH of `Ba(OH)_(2)=12`
So, pOH = 2
`therefore [H^(+)]=[1xx10^(-12)]`
`K_(w)=(H^(+))(OH^(-))`
`K_(w)=1xx10^(-14)`
`OH^(-)=(K_(omega))/(H^(+))`
and `[OH^(-)]=(1xx10^(-14))/(1xx10^(-12))[because[H^(+)][OH^(-)]=1xx10^(-14)]`
`=1xx10^(-2)mol//L`
`Ba(OH)_(2)rarr underset(S)(Ba^(2+)+underset(2S)(2OH^(-))`
`K_(sp)=[Ba^(2+)][OH^(-)]^(2)=[S][2S]^(2)`
`=[(1xx10^(-2))/(2)](1xx10^(-2))^(2)`
`=0.5xx10^(-6)=5.0xx10^(-6)M^(3)`