To calculate the molar solubility of calcium hydroxide [tex]\(\text{Ca(OH)}_2\)[/tex] in water, we will use the solubility product constant ([tex]\(K_{\text{sp}}\)[/tex]).
Given:
[tex]\[ K_{\text{sp}} \text{ for } \text{Ca(OH)}_2 = 5.02 \][/tex]
The dissociation of calcium hydroxide in water is represented by the equation:
[tex]\[ \text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq) \][/tex]
Let [tex]\( s \)[/tex] be the molar solubility of [tex]\(\text{Ca(OH)}_2\)[/tex]. At equilibrium, the concentrations of the ions in solution will be:
[tex]\[ [\text{Ca}^{2+}] = s \][/tex]
[tex]\[ [\text{OH}^-] = 2s \][/tex]
Given the expression for the solubility product constant:
[tex]\[ K_{\text{sp}} = [\text{Ca}^{2+}][\text{OH}^-]^2 \][/tex]
Substitute the expressions for the ion concentrations:
[tex]\[ K_{\text{sp}} = s \cdot (2s)^2 \][/tex]
[tex]\[ K_{\text{sp}} = s \cdot 4s^2 \][/tex]
[tex]\[ K_{\text{sp}} = 4s^3 \][/tex]
Now, solve for [tex]\( s \)[/tex]:
[tex]\[ 4s^3 = K_{\text{sp}} \][/tex]
[tex]\[ s^3 = \frac{K_{\text{sp}}}{4} \][/tex]
[tex]\[ s = \left( \frac{5.02}{4} \right)^{1/3} \][/tex]
To find the numeric value of molar solubility [tex]\( s \)[/tex]:
[tex]\[ s \approx 1.0786517240005968 \, \text{M} \][/tex]
Therefore, the molar solubility of [tex]\(\text{Ca(OH)}_2\)[/tex] in water is closest to:
[tex]\[ 1.08 \times 10^{-2} \, \text{M} \][/tex]
Thus, the correct answer is:
[tex]\[ 1.08 \times 10^{-2} \, \text{M} \][/tex]
