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

Options

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

Answer

B

Explanation

pH of saturated solution of `Ca(OH)_(2) = 9`
pOH of saturated solution of `Ca(OH)_(2) = 5`
`OH^(-) = 10^(-5)`
`{:(Ca(OH)_(2),hArr,Ca^(2+),+,2OH^(-)),(,,S,,2S),(,,1//2xx10^(-5),,10^(-5)):}`
Solubility (s) = `1//2 xx 2 xx 10^(-5)`
`K_(sp) = [Ba^(2+)][OH^(-)]^(2)`
`[1//2 xx 10^(-5)] [10^(-5)]^(2) = 0.5 xx 10^(-15)`