pH of saturated solution of `Ca(OH)_2` is 9. The solubility product `(K_(sp))` of `Ca(OH)_2`

Options

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

Answer

A

Explanation

`0.5xx10^(-15)`
`Ca(OH)_(2)hArr Ca^(2+)+2OH^(-)`
Given that `pH=9`
`pOH=14-9=5`
`[pOH=-log_(10)[OH^(-)]]`
`:.[OH^(-)]=10^(-pOH)`
`[OH^(-)]=10^(-5)M`
`K_(sp)=[Ca^(2+)][OH^(-)]^(2)`
`=(10^(-5))/(2)xx(10^(-5))^(2)=0.5xx10^(-15)`