Chemical and Ionic Equilibrium - Result Question 122

Solution:

66. $K _{\text {sp }}=4 S^{3}=4.42 \times 10^{-5}$

$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 $ $mg$