Question

The solubility product $\left(k_{sp}\right)$ of $Ca(OH{)}_2$ at ${25}^{0}C$ is $4.42\times {10}^{-5}$. A 500 mL of saturated solution of $Ca{\left(OH\right)}_2$ is mixed with equal volume of 0.4 M NaOH. How much $Ca{\left(OH\right)}_2$ in milligrams is precipitated ?

Answer 1

Step I : Calculation of amount of $Ca(OH{)}_2$ in $500mL$ od saturated solution. Suppose solubility of $Ca(OH{)}_2=xmol{L}^{-1}$
$Ca(OH{)}_2\to C{a}^{2+}+2O{H}^{-}$
$\therefore$ $K_{sp}=\left[C{a}^{2+}\right][O{H}^{-}{]}^2$
$=x\times (2x{)}^2=4x^3$
$\therefore$ $4x^3=4.42\times {10}^{-5}$
or $x^3=1.105\times {10}^{-5}$
$3logx=log(1.105\times {10}^{-5})$
$=0.0434-5=-4.9566$
$logx=-1.6522=\overline{2}.3478$
or $x=2.227\times {10}^{-2}mol{L}^{-1}$
$\therefore$ Amount of $Ca(OH{)}_2$ present in $500mL$
$=\frac{2.227\times {10}^{-2}}{2}\times 74g$
$82.39\times {10}^{-2}=823.9mg$
Step II : Calculation of the amount of $Ca(OH{)}_2$ in solution after mixing. As equal volumes of $Ca(OH{)}_2$ solution and $0.4MNaOH$ solution have been mixed
Conc. of $NaOH$ in the mixture $=\frac{0.4}{2}=0.2M$
$\left[O{H}^{-}\right]=\left[NaOH\right]=0.2M$
$K_{sp}$ for $Ca(OH{)}_2=[C{a}^{2+}\left]\right[O{H}^{-}{]}^2$
$\therefore$ $\left[C{a}^{2+}\right]=\frac{4.42\times {10}^{-5}}{(0.2{)}^2}$
$=1.105\times {10}^{-3}mol{L}^{-1}$
Total volume of the solution after mixing
$=500+500=1000mL$
$\therefore$ Amount of $Ca(OH{)}_2$ in the mixture solution.
$=1.105\times {10}^{-3}\times 74g=0.818g=81.8mg$
$\therefore$ Amount of $Ca(OH{)}_2$ precipitated
$=823.9-81.8=742.1mg$