The solubility product `K_(sp)` of Ca `(OH)_2 " at " 25^(@) C " is " 4.42 xx 10 ^(-5)` A 500 mL of saturated solution of `Ca(OH)_2 ` is mixed with equal volume of 0.4 M NaOH . How much `Ca(OH)_2` in milligrams is precipitated?

Explanation

`K_(sp) = 4s^(3) = 4.42 xx 10^(-5)`
s = 0.022 M
mmol of `Ca(OH)_(2)` in 500 mL saturated solution = 11
mmol of NaOH in 500 mL 0.40 M solution = 200
Total mmol of `OH^(-)` = 200 + 2 xx 11 = 222
`[OH^(-)] = 0.222 M`
Solubility in presence of NaOH = `K_(sp)/[OH]^(-2)`
mmol of `Ca^(2+)` remaining in solution = 0.9
mmol of `Ca(OH)_(2)` precipitated = 10.1
mg of `Ca(OH)_(2)` precipitated = `10.1 xx 7.4 = 747.4 mg`