Question

The $K_{sp}$ of $Ca{\left(OH\right)}_2$ is $4.42\times {10}^{-5}$ at ${25}^{o}C$. A $500mL$ of saturated solution of $Ca{\left(OH\right)}_2$ is mixed with equal volume of $0.4M$ $NaOH$. How much $Ca{\left(OH\right)}_2$ in $mg$ is precipitated?

• A. $659.2mg$
• B. $759.2mg$
• C. $578.2mg$
• D. $785.2mg$

Answer 1

The correct option is B $759.2mg$

For, $Ca{\left(OH\right)}_2$:

$Ca{\left(OH\right)}_2\rightleftharpoons {Ca}^{2+}+2{OH}^{-}$

$\therefore$ $K_{sp}=\left[{Ca}^{2+}\right]{\left[{OH}^{-}\right]}^2=4S^3$

where, $S$ is solubility of $Ca{\left(OH\right)}_2$ in $mol$ ${litre}^{-1}$

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

$\therefore$ $S=0.0223M$

In presence of $NaOH$, $\left[{OH}^{-}\right]$ increases and thus, some of the ${Ca}^{2+}$ ions are settled down as $Ca{\left(OH\right)}_2$ to have $K_{sp}$ constant.

On mixing
$\left[{OH}^{-}\right]=\frac{500\times 0.4}{1000}+\frac{0.0223\times 2\times 500}{1000}=0.223M$

From $NaOH$ From $Ca{\left(OH\right)}_2$

$\left[{Ca}^{2+}\right]=\frac{0.0223\times 500}{1000}=0.01115M=111.5\times {10}^{-4}M$

Also ${\left[{Ca}^{2+}\right]}_{left}{\left[{OH}^{-}\right]}^2=K_{sp}$

$\therefore {\left[{Ca}^{2+}\right]}_{left}{\left[0.2223\right]}^2=4.42\times {10}^{-5}$

$\therefore {\left[{Ca}^{2+}\right]}_{left}=8.94\times {10}^{-4}$ $mol$ ${litre}^{-1}$

$\because$ Mole of $Ca{\left(OH\right)}_2$ precipitated$=$ Mole of ${Ca}^{2+}$ precipitated

$=111.5\times {10}^{-4}-8.9\times {10}^{-4}$

$=102.6\times {10}^{-4}$

$\therefore$ Mass of $Ca{\left(OH\right)}_2$ precipitated $=102\times {10}^{-4}\times 74g$

$=7592.4\times {10}^{-4}g=759.2mg$