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)
Show Answer
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 $NaOH=\frac{K _{sp}}{\left[OH^{-}\right]^{3}}$
$=\frac{4.42 \times 10^{-5}}{(0.222)^{2}}=9.0 \times 10^{-4} M$
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 $ $mg$
BETA
