The `K_(SP)of Ca(OH)_(2)is 4.42xx10^(-5)at 25^(@)C`. A 500 mL of saturated solution of `Ca(OH)_(2)` is mixed with equal volume of `0.4M NaOH`. How much `Ca(OH)_(2)` in mg is preciptated ?

Explanation

500 mL of `0.4M NaOH` are mixed with 500 mL of `Ca(OH)_(2)`, a saturated solution having `Ca(OH)_(2)` solubility as S M.
For `Ca(OH)_(2)hArr Ca^(2+)+2OH^(-)`
`K_(SP)=Sxx(2S)^(2)=4S^(3)`
Then, `4S^(3)=4.42xx10^(-5)`
`:. S= 3sqrt(((4.42xx10^(-5))/(4)))=0.0223M`
Now `Ca(OH)_(2)+NaOH` are mixed.
`:.` Solution has `Ca^(2+)` and `OH^(-)`, out of which some `Ca^(2+)` are precipitated.
On mixing, `[Ca^(2+)]=(0.0223xx500)/(1000)`
`=0.01115=111.5xx10^(-4)M`
`(0.0223xx2xx500)/(underset(["from" Ca(OH)_(2)])(1000))+(500xx0.4)/(underset(["from" Ca(OH)_(2)])(1000))=0.2223 M`
`:. [Ca^(2+)][OH^(-)]^(2)=K_(SP)`
`[Ca^(2+)]_("left")[0.2223]^(2)=4.42xx10^(-5)`
`[Ca^(2+)]_("left")=(4.42xx10^(-5))/([0.2223]^(2))=8.94xx10^(-4)"mol litre"^(-1)`
`:' "Mole of" Ca(OH)_(2)` precipitated = Mole of `[Ca^(2+)]` precipitated
`=111.5xx10^(-4)-8.94xx10^(-4)=102.46xx10^(-4)`
`:.` Wt. of `Ca(OH)_(2)` precipitated form `Ca(OH)_(2)` sollution
`=102.46xx10^(-4)xx74`
`=7582.04xx10^(-4)g= 758.2mg`