Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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]
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]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.