Chemical and Ionic Equilibrium - Result Question 122
####66. The solubility product $\left(K _{sp}\right)$ of $Ca(OH) _2$ at $25^{\circ} C$ is $4.42 \times 10^{-5}$. 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?
$(1992,4 M)$
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Solution:
- $K _{\text {sp }}=4 S^{3}=4.42 \times 10^{-5}$
$S=0.022 M$
$mmol$ of $Ca(OH) _2$ in $500 mL$ saturated solution $=11$ $mmol$ of $NaOH$ in $500 mL 0.40 M$ solution $=200$
Total $mmol$ of $OH^{-}=200+2 \times 11=222$
$$ \left[OH^{-}\right]=0.222 M $$
Solubility in presence of $NaOH=\frac{K _{sp}}{\left[OH^{-}\right]^{2}}$
$$ =\frac{4.42 \times 10^{-5}}{(0.222)^{2}}=9 \times 10^{-4} M $$
mmol of $Ca^{2+}$ remaining in solution $=0.9$ mmol of $Ca(OH) _2$ precipitated $=10.1$
$mg$ of $Ca(OH) _2$ precipitated $=10.1 \times 7.4=747.4 mg$
