pH of a saturated solution of `Ca(OH)_(2)` is 9. the solubility product `(K_(sp))` of `Ca(OH)_(2)` is

Options

A. `0.5xx10^(-10)`
B. `0.5xx10^(-15)`
C. `0.25xx10^(-10)`
D. `0.125xx10^(-15)`

Answer

B

Explanation

pH of saturated solution of `Ca(OH)_(2)=9`
`pOH` of saturated of `Ca(OH)_(2)=5`
`OH^(-)=10^(-5)`
`Ca(OH)_(2)hArrunderset(1//2)xx10^(-5)underset(S)(Ca^(2+))+underset(10^(-5))underset(2S)(2OH^(-))`
solubility `(s)=1//2xx2" "10^(-5)`
`K_(ap)=[Ba2+][OH^(-)]^(2)`
`[1//2xx10^(-8)][10^(-5)]^(2)=0.5xx10^(-15)`