Question

What is the solubility of $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$ water? (Given the solubility product of $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$ water$=5.5\times {10}^{-6}$)

• A. $1.11\times {10}^{-6}$
• B. $1.77\times {10}^{-6}$
• C. $1.77\times {10}^{-2}$
• D. $1.11\times {10}^{-2}$

Answer 1

The correct option is D

$1.11\times {10}^{-2}$

Explanation for the correct options

(D) $1.11\times {10}^{-2}$

Step 1:

• $$\begin{gathered} \mathrm{Ca}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Ca}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ \left(\mathrm{s}\right)\left(2\mathrm{s}\right) \end{gathered}$$

Step 2:

• Expression used for the calculation of the solubility product:
• $\begin{array}{rcl}{\mathrm{K}}_{\mathrm{sp}}& =& \left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{OH}}^{-}\right]\\ & =& \left(\mathrm{s}\right){\left(2\mathrm{s}\right)}^2=4{\mathrm{s}}^3\\ 5.5\times {10}^{-6}& =& 4{\mathrm{s}}^3\\ s& =& {\left(\frac{5.5}{4}\right)}^{\frac{1}{3}}\times {10}^{-2}\\ & =& 1.11\times {10}^{-2}\end{array}$

Final Answer: Hence option (D) is correct.