SATHEE: Chemical and Ionic Equilibrium

Chemical and Ionic Equilibrium - Result Question 122

####66. The solubility product $(K_{s p})$ of $C a (O H)_{2}$ at $25^{\circ} C$ is $4.42 \times 10^{- 5}$ . A $500 m L$ of saturated solution of $C a (O H)_{2}$ is mixed with equal volume of $0.4 M N a O H$ . How much $C a (O H)_{2}$ in milligrams is precipitated?

$(1992, 4 M)$

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Solution:

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

$S = 0.022 M$

$m m o l$ of $C a (O H)_{2}$ in $500 m L$ saturated solution $= 11$ $m m o l$ of $N a O H$ in $500 m L 0.40 M$ solution $= 200$

Total $m m o l$ of $O H^{-} = 200 + 2 \times 11 = 222$

$[O H^{-}] = 0.222 M$

Solubility in presence of $N a O H = \frac{K_{s p}}{{[O H^{-}]}^{2}}$

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

mmol of $C a^{2 +}$ remaining in solution $= 0.9$ mmol of $C a (O H)_{2}$ precipitated $= 10.1$

$m g$ of $C a (O H)_{2}$ precipitated $= 10.1 \times 7.4 = 747.4 m g$

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Chemical and Ionic Equilibrium - Result Question 122

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