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

Answer

A

Explanation

`Ca(OH)_2 harr Ca^(2+) + 2OH^(-)`
pH + pOH = 14
`pH=9`, thus pOH = 14-9=5
`[OH^(-)]=10^(-5)` M
Thus `[Ca^(2+)]=(10^(-5))/(2)`
So, `K_(sp)=[Ca^(2+)][OH^(-)]^2`
=`(10^(-5))/(2)(10^(-5))^2`
`=0.5xx10^(-15)`