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

Options

A. `5.06 xx 10^(-7) M^(3)`
B. `4.00xx10^(-6)M^(3)`
C. `4.00xx 10^(-7) M^(3)`
D. `5.00 xx 10^(-6) M^(3)`

Answer

B

Explanation

`pH =12` means `|H^(+)|=10^(-12)M`
`:. [OH^(-1)]=10^(-2)`
`Ba(OH)_(2) hArr Ba^(2+)+2OH^(-) `
`pH =10^(-2) M :. Ba^(2+)=(10^(-2))/(2)M`
`=5 xx 10^(-3)M`
`K_(sp)=|Ba^(2+)|" "|OH^(+)|^(2)=(5 xx 10^(-3))(10^(-2))^(2)`
`=5 xx 10^(-7) M^(3)`