The `K_(sp)` of `Ca(OH)_(2)` is `4.42 xx 10^(-5)` at `25^(@)C`. A 500 ml of saturated solution of `Ca(OH)_(2)` is mixed with an equal volume of `0.4M NaOH`. How much `Ca(OH)_(2)` in mg is precipitated ?

Explanation

Suppose the solubility of `Ca(OH)_(2)` in saturated solution is S mol `L^(-1)`.
`{:(Ca(OH)_(2),hArr,Ca^(2+),+,2OH^(-),),(,,S,,2S,):}`
`K_(sp) ` of `Ca(OH)_(2) hArr [Ca^(2+) ] [ OH^(-)]^(2)`
`4.42xx10^(-5) = S xx (2S)^(2) = 4S^(3)`
or`S=((4.42xx10^(-5))/(4))^(1//3) = 0.223 ` mol `L^(-1)`
After mixing the two solutions, total volume becomes = 500 + 500 = 100 mL
Now, concentration will be
`[Ca^(2+) ] = (0.0223)/(1000) xx 500 = 0.01115 ` mol `L^(-1)`
`{:([OH^(-)]=(0.0223xx2xx500)/(1000)+(0.4xx500)/(1000)=0.2223 " mol "L^(-1)),(" "["From"Ca(OH)_(2)]"" ["From" NaOH]):}`
On adding NaOH solution to saturated solution of `Ca(OH)_(2)` some `Ca(OH)_(2)` will be precipitated (due to common ion effect).
Now, as `Ca^(2+)` ion left are still present in the left out saturated solution, `[ Ca^(2+) ] _("left") [OH^(-)]^(2) = K_(sp)`
`:. [Ca^(2+) ]_("left") = (K_(sp))/([OH^(-)]^(2))=(4.42xx10^(-5))/((0.2223)^(2))=8.94xx10^(-4)` mol `L^(-1)`
Moles of `Ca(OH)_(2)` precipitated = Moles of `Ca^(2+)` ions precipitated `= [Ca^(2+) ] _("initial")-[Ca^(2+)]_("left")`
`=0.01115-8.94xx10^(-4) = 111.5xx10^(-4) - 8.94xx10^(-4) = 102.56xx10^(-4)` moles
`=102.56xx10^(-4) xx 74 g = 7589.44 xx 10^(-4) g = 758.944 ` mg
Initial `[Ca(OH)_(2)] = 0.01115 ` mol `L^(-1) = 111.5xx10^(-4) ` mol `L^(-1)`
`Ca(OH)_(2)` precipitated `= 102.56xx10^(-4) ` mol `L^(-1)`
`:. ` % `Ca(OH)_(2)` precipitated `= (Ca(OH)_(2)"precipitated")/("Initial conc. ")xx100=(102.56xx10^(-4))/(111.5xx10^(-4))xx100=91.98% = 92 %`
Note. The above example illustrates the method of calculation of amount precipitated and amount left after precipitation when to a saturated solution of a sparingly soluble salt present in excess, some solution containing a common ion is added .