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

Options

A. `3.3xx10^(-7)`
B. `5.0xx10^(-7)`
C. `4.0xx10^(-6)`
D. `5.0xx10^(-6)`

Answer

B

Explanation

Given, pH of `Ba(OH)_(2)=12`
`therefore pOH=14-pH`
= 14 - 12 = 2
We know that,
`pOH=-log[OH^(-)]`
`2=-log[OH^(-)]`
`[OH^(-)]` = antilog (-2)
`[OH^(-)]=1xx10^(-2)`
`Ba(OH)_(2)` dissolves in water as
`underset("S mol L"^(-1))(Ba(OH)_(2))(S)hArr underset(S)(Ba^(2+))+underset(2S)(2OH^(-))`
`therefore [OH^(-)]=2S=1xx10^(-2)`
`S=([OH^(-)])/(2) " " [Ba^(2+)=S]`
`[Ba^(2+)]=([OH^(-)])/(2)=(1xx10^(-2))/(2)`
`K_(sp)=[Ba^(2+)][OH^(-)]^(2)`
`=((1xx10^(-2))/(2))(1xx10^(-2))^(2)`
`=0.5xx10^(-6)=5xx10^(-7)`