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

First find the concentration of `Ca^(2+)` ions at saturation using `K_(sp)` of `Ca(OH)_(2)`.
`Ca(OH)_(2) hArr Ca^(2+) + 2overset(Θ)OH`
`rArr K_(sp) = [Ca^(2+)] [overset(Θ)OH]^(2)`
Let `[Ca^(2+)] = x mol L^(-1) rArr [overset(Θ)OH]^(2) = 2x`
`rArr K_(sp) = 4x^(3) or x = 3sqrt((K_(sp))/(4)) = 0.01M`
`rArr [Ca^(2+)] = 0.01M` and `[overset(Θ)OH] = 0.02M`
As equal volumes of saturated solution and `.4M NaOH` are mixed:
`[Ca^(2+)] = (0.01)/(2) = 5 xx10^(-3)M`
and `[overset(Θ)OH]_("Total") = (0.02)/(2) + (0.4)/(2) = 0.21M`
[`:'[overset(Θ)OH]_("Total") = [overset(Θ)OH]_(NaOH) +[overset(Θ)OH]_(From) Ca(OH)_(2)]`
Now, calculate ionic product of `[Ca(OH)_(2)] = [Ca^(2+)] [overset(Θ)OH]^(2)`
`= (5 xx 10^(-3)) (0.21)^(2) = 2.2 xx 10^(-4) gt K_(sp)`
Since concentration of `overset(Θ)OH` is quite high, `Ca^(2+)` will be precipitated till a new saturation state is reached.
Let at new saturated state, `[overset(Θ)OH] = 0.21M`
(Assuming no change in `[overset(Θ)OH]`
`[Ca^(2+)]_("left") = (K_(sp))/([overset(Θ)OH]^(2)) = ((4xx10^(-6)))/((0.21)^(2))= 9.07 xx 10^(-5)M`
`rArr [Ca^(2+)]` precipitated `= 5 xx 10^(-3) - 9.07 xx 10^(-5)M`
`= 4.91 xx 10^(-3)M`
`rArr` Amount of `Cr^(2+) = 4.91 xx 10^(-3) xx 74 xx 10^(3) mg L^(-1)`
`= 363.3 mg L^(-1)`
Check the approximation: `[overset(Θ)OH]_("left") ~~ 0.21 M`. How ?
Find out the `[overset(Θ)OH]_(used) = 2[Ca^(2+)]_("used")`
`rArr [overset(Θ)OH]_("left") = 0.21 - 9.8 xx 10^(-3) M = 0.2M ~~0.21M`
Alternative method:

Now, `IP = K_(sp)` at equilibrium
`(5 xx 10^(-3)-x) (0.21-2x)^(2) = 4 xx 10^(6) rArr` This equaiton in `x` si clearly a cubic.
To solve the above equaiton approximately, assume `0.21 - 2 x ~~0.21`, which is what has been done in the previous approach