Question

The $Ca{\left(OH\right)}_2$ is $4.42\times {10}^{-5}$ at ${25}^{o}C$ with $500mL$ 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?

Answer 1

Let $S$ be the solubility of $Ca{\left(OH\right)}_2$ in saturated solution
$Ca{\left(OH\right)}_2\rightleftharpoons {Ca}^{2+}+2{OH}^{-}$

$2S$ $S$

$K_{sp}Ca{\left(OH\right)}_2=2{\left[{OH}^{-}\right]}^2\left[{Ca}^{2+}\right]$
$4.42\times {10}^{-5}=S\times 4S^2=4S^3$
$S=0.0223mol{L}^{-1}$
After mixing the two solutions the total volume become $1$ litre
$\left[{Ca}^{2+}\right]=\frac{0.0223}{1000}\times 500=0.01115mol{L}^{-1}$
$\left[{OH}^{-}\right]=\frac{0.0223\times 2\times 500}{1000}+\frac{0.4\times 500}{1000}=0.2223mol{L}^{-1}$
Under the high concentration of ${OH}^{-1}$ ions, some $Ca{\left(OH\right)}_2$ will be precipitated.
${\left[{Ca}^{2+}\right]}_{left}{\left[{OH}^{-}\right]}^2=K_{sp}$
${\left[{Ca}^{2+}\right]}_{left}=8.94\times {10}^{-4}mol{L}^{-1}$
Moles of $Ca{\left(OH\right)}_2$ precipitated$=$ Moles of ${Ca}^{2+}$ precipitated
$={\left[{Ca}^{2+}\right]}_{initial}-{\left[{Ca}^{2+}\right]}_{left}=0.01115-8.94\times {10}^{-4}=758.944mg$