Question

The solubility product of $Ca(OH{)}_{2}at{25}^{\circ }Cis4.42\times {10}^{-5}$. A 500 ml of saturated solution of $Ca(OH{)}_2$ is mixed with equal volume of 0.4M NaOH solution. How much $Ca(OH{)}_2$ in milligrams is precipitated

• A. 74 mg
• B. 73 mg
• C. 47 mg
• D. 77 mg

Answer 1

The correct option is A 74 mg

Let the solubility of $Ca(OH{)}_2$ in pure water
= S moles/litre
$Ca(OH{)}_2\rightleftharpoons C{a}^{2+}+2O{H}^{-}$
S 2S
$Then{K}_{sp}=\left[C{a}^{2+}\right]\left[O{H}^{-}\right]$
$4.42\times {10}^{-5}=S\times (2S{)}^2$
$4.42\times {10}^{-5}=4S^3$
$S=2.224\times {10}^{-2}=0.0223moleslitr{e}^{-1}$
$\therefore$ No. of moles of $C{a}^{2+}$ ions in 500 ml of solutions = 0.011. Now when 500 ml of saturated solution is mixed with 500 ml of 0.4 M NaOH, the resultant volume is 1000 ml. The molarity of $O{H}^{-}$ ions on the resultant solution would therefore be 0.2 M
Maximum $\left[C{a}^{2+}\right]$ that can be tolerated $=\frac{ksp}{{\left[O{H}^{-}\right]}^2}$
$=\frac{4.42\times {10}^{-5}}{(0.2{)}^2}=0.0011M$
Thus number of moles of $C{a}^{2+}$ or $Ca(OH{)}_2$ precipitated $=0.011-0.001=0.010$
Mass of $Ca(OH{)}_2$ precipitated $=0.010\times 74=0.74g=74mg$