Question

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

• A. 760gm
• B. 743 mg
• C. 711 gm
• D. 710 mg

Answer 1

The correct option is B 743 mg

Let S M be the solubility of $Ca(OH{)}_2$ in pure water.

$\underset{S}{Ca(OH{)}_2}\rightleftharpoons \underset{S}{C{a}^{2+}}+\underset{2S}{O{H}^{-}}$

$K_{sp}=\left[C{a}^{2+}\right][O{H}^{-}{]}^2$

Substitute values in the above expression.

$4.42\times {10}^{-5}=S(2S{)}^2$ or $S=0.0223M$

500 ml of solution contains 0.01115 moles of calcium ions.

The molarity of $NaOH$ solution is 0.4 M. When it is mixed with equal volume of calcium hydroxide solution, its molarity is reduced to half, i.e., 0.2M.

$\left[O{H}^{-}\right]=0.2M$

$\left[C{a}^{2+}\right]=\frac{K_{sp}}{[O{H}^{-}{]}^2}=\frac{4.42\times {10}^{-5}}{(0.2{)}^2}=0.001105M$.

The number of moles of calcium ion precipitated $=0.01115-0.001105=0.010045$.

The number of milimoles of calcium hydroxide precipitated $=0.010045\times 74\times 1000=743mg$.