SATHEE: Chemical and Ionic Equilibrium 2 Question 66

Chemical and Ionic Equilibrium 2 Question 66

66. The solubility product $(K_{sp})$ of $\mathrm{Ca}(\mathrm{OH}){2} $a t$ 25^{\circ} \mathrm{C} $i s$ 4.42 \times 10^{-5} $. A$ 500 \mathrm{~mL} $o f s a t u r a t e d s o l u t i o n o f$ \mathrm{Ca}(\mathrm{OH}){2} $i s m i x e d w i t h e q u a l v o l u m e o f$ 0.4 \mathrm{M} \mathrm{NaOH} $. H o w m u c h$ \mathrm{Ca}(\mathrm{OH})_{2}$ in milligrams is precipitated?

$(1992, 4 M)$

Show Answer

Solution:

$K_{sp} = 4 S^{3} = 4.42 \times 10^{- 5}$

$S = 0.022 M$

$mmol$ of $Ca (OH)_{2}$ in $500 mL$ saturated solution $= 11$ $mmol$ of $NaOH$ in $500 mL 0.40 M$ solution $= 200$

Total $mmol$ of ${OH}^{-} = 200 + 2 \times 11 = 222$

$[{OH}^{-}] = 0.222 M$

Solubility in presence of $NaOH = \frac{K_{sp}}{{[{OH}^{-}]}^{2}}$

$= \frac{4.42 \times 10^{- 5}}{(0.222)^{2}} = 9 \times 10^{- 4} M$

mmol of ${Ca}^{2 +}$ remaining in solution $= 0.9$ mmol of $Ca (OH)_{2}$ precipitated $= 10.1$

$mg$ of $Ca (OH)_{2}$ precipitated $= 10.1 \times 7.4 = 747.4 mg$

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