Question

The solubility product $(K_{sp})$ of $Ca(OH{)}_2$ at $25^{\circ }C,4.42\times 10^{-5}.$ A $500mL$ of saturated solution of $Ca(OH{)}_2$ is mixed with equal volume of $0.4MNaOH$. How much $Ca(OH{)}_2$ in milligrams is precipitated?

Answer 1

$K_{sp}=4S^3=4.42\times 10^{-5}$
$S=0.022M$
mmol of $Ca(OH{)}_2$ in $500mL$ saturated solution $=11$
mmol of $NaOH$ in $500mL0.40M$ solution $=200$
Total mmol of $O{H}^{-}=200+2\times 11=222$
$[O{H}^{-}]=0.222M$
Solubility in presence of $NaOH=\frac{K_{sp}}{[O{H}^{-}{]}^2}$
$=\frac{4.42\times 10^{-5}}{(0.222{)}^2}=9\times 10^{-4}M$
mmol of $C{a}^{2+}$ remaining in solution $=0.9$
mmol of $Ca(OH{)}_2$ precipitated $=10.1$
mg of $Ca(OH{)}_2$ precipitated $=10.1\times 7.4=747.4mg$