500 mL of saturated solution of $C a {(O H)}_{2}$ is mixed with equal volume of aqueous $N a O H$ solution such that resultant solution has a concentration of $O H^{-}$ ion is 0.04 M. Calculate the no. of mill moles of $C a {(O H)}_{2}$ is precipitated $(g i v e n K s p o f C a {(O H)}_{2} = 4 \times 10^{- 6})$ $(p k_{w = 14})$

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answer is 2.5.

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Detailed Solution

$\begin{array}{l} K_{s p} = (C a {(O H)}_{2}) = 4 s^{3} = 4 \times 10^{- 6} \\ s = 0.01 M \end{array}$
Mole of $C a^{2 +} = \frac{0.01 \times 500}{1000} = 0.005$
After mixing $[C a^{2 +}] = \frac{0.01 \times 500}{1000} = 0.005$
$\begin{array}{l} K_{s p} = {[C a^{2 +}]}_{l e f t} {[O H^{-}]}^{2} \\ 4 \times 10^{- 6} = {[C a^{2 +}]}_{l e f t} {(0.04)}^{2} \\ {[C a^{2 +}]}_{l e f t} = 2.5 \times 10^{- 3} M \end{array}$
Moles of $C a {(O H)}_{2}$ precipitated
$\begin{array}{l} = {[C a^{2 +}]}_{I n i t i a l} - {[C a^{2 +}]}_{l e f t} \\ = 0.005 - 2.5 \times 10^{- 3} = 2.5 \times 10^{- 3} \end{array}$