Chemical and Ionic Equilibrium 2 Question 66
66. The solubility product $\left(K_{\mathrm{sp}}\right)$ of $\mathrm{Ca}(\mathrm{OH}){2}$ at $25^{\circ} \mathrm{C}$ is $4.42 \times 10^{-5}$. A $500 \mathrm{~mL}$ of saturated solution of $\mathrm{Ca}(\mathrm{OH}){2}$ is mixed with equal volume of $0.4 \mathrm{M} \mathrm{NaOH}$. How much $\mathrm{Ca}(\mathrm{OH})_{2}$ in milligrams is precipitated?
$(1992,4 \mathrm{M})$
Show Answer
Solution:
- $K_{\text {sp }}=4 S^{3}=4.42 \times 10^{-5}$
$S=0.022 \mathrm{M}$
$\mathrm{mmol}$ of $\mathrm{Ca}(\mathrm{OH})_{2}$ in $500 \mathrm{~mL}$ saturated solution $=11$ $\mathrm{mmol}$ of $\mathrm{NaOH}$ in $500 \mathrm{~mL} 0.40 \mathrm{M}$ solution $=200$
Total $\mathrm{mmol}$ of $\mathrm{OH}^{-}=200+2 \times 11=222$
$$ \left[\mathrm{OH}^{-}\right]=0.222 \mathrm{M} $$
Solubility in presence of $\mathrm{NaOH}=\frac{K_{\mathrm{sp}}}{\left[\mathrm{OH}^{-}\right]^{2}}$
$$ =\frac{4.42 \times 10^{-5}}{(0.222)^{2}}=9 \times 10^{-4} \mathrm{M} $$
mmol of $\mathrm{Ca}^{2+}$ remaining in solution $=0.9$ mmol of $\mathrm{Ca}(\mathrm{OH})_{2}$ precipitated $=10.1$
$\mathrm{mg}$ of $\mathrm{Ca}(\mathrm{OH})_{2}$ precipitated $=10.1 \times 7.4=747.4 \mathrm{mg}$
