If pH of a saturated solution of `Ba(OH)_(2)` is 12, the value of its `K_((SP))` is

Options

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

Answer

B

Explanation

Given, pH of `Ba(OH)_(2)=12`
`because pH= pOH=14`
`therefore pOH= 14-pH= 14-12=2`
Now, `pOH= -log [OH^(-)]`
`rArr 2= - log [OH^(-)]`
`[OH^(-)]=10^(-2)`
`underset(S mol^(-1))(Ba(OH)_(2) (s)) hArr underset(S)(Ba^(2+)) + underset(2S)(2OH^(-))`
`[OH^(-)] =2S =10^(-2)`
`therefore S=10^(-1)`
`K_(sp)= [Ba^(2+)] [OH^(-)]^(2)`
`rArr [10^(-1)] [10^(-2)]^(2)=0.5 xx 10^(-6) =5 xx 10^(-7)M`