Question

The solubility product $ ( K_{ sp})$ of $ Ca(OH)_2$ at $ 25^{\circ } C, \, 4.42 \times 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?

Answer 1

$ K_{ sp} = 4S^3 = 4.42 \times 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 \times 11 = 222 $
$ [OH^- ]= 0.222 \,M$
Solubility in presence of $NaOH = \frac{ K_{ sp}}{ [ OH^- ]^2 } $
$ = \frac{ 4.42 \times 10^{ - 5 }}{ ( 0.222)^2 } = 9 \times 10^{ - 4 } \,M$
mmol of $Ca^{2+}$ remaining in solution $= 0.9$
mmol of $Ca(OH)_2 $ precipitated $=10.1$
mg of $Ca(OH)_2$ precipitated $= 10.1 \times 7.4 = 747.4\, mg$