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The solubility product constant, denoted as \(K_{\text{sp}}\), is a measure of the solubility of a compound. Specifically, it is an equilibrium constant for the dissociation of a solid substance into its aqueous ions. For a general salt \( \text{AB}_2 \), which dissociates as \( \text{AB}_2(s) \rightarrow \text{A}^{2+}(aq) + 2\text{B}^-(aq) \), the \(K_{\text{sp}}\) expression would be:
\[ K_{\text{sp}} = [\text{A}^{2+}][\text{B}^-]^2 \]
The value of \( K_{\text{sp}} \) indicates how much of the solid will dissolve in water. A higher \( K_{\text{sp}} \) means the compound is more soluble. In the given exercise, the \( K_{\text{sp}} \) values of \[ K_{\text{sp}}(\text{Mg(OH)}_2) = 1.8 \times 10^{-11} \text{ and } K_{\text{sp}}(\text{Ca(OH)}_2) = 5.5 \times 10^{-6} \] are used to determine the solubility and potential for precipitation.

The ion product, \( Q \), is similar to the solubility product constant but applies to any given set of initial concentrations, not just at equilibrium.
For a compound like \( \text{Mg(OH)}_2 \), which dissociates as \( \text{Mg(OH)}_2(s) \rightarrow \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \), the ion product is calculated as:
\[ Q = [\text{Mg}^{2+}][\text{OH}^{-}]^2 \]
By comparing \( Q \) to \( K_{\text{sp}} \), one can predict whether precipitation will occur:

• If \( Q < K_{\text{sp}} \): the solution is unsaturated, and no precipitation occurs.
• If \( Q = K_{\text{sp}} \): the solution is saturated, and at equilibrium.
• If \( Q > K_{\text{sp}} \): the solution is supersaturated, leading to precipitation.

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of the reactants and products remain constant over time. This concept is crucial in understanding both the solubility product constant (\( K_{\text{sp}} \)) and the ion product (\( Q \)).
In a saturated solution of \( \text{Mg(OH)}_2 \), the following equilibrium exists:
\[ \text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \]
At equilibrium, \[ K_{\text{sp}} = [\text{Mg}^{2+}][\text{OH}^{-}]^2 \]
This concept helps explain the balance between dissolved ions and undissolved solid, which is effectively used to solve problems of precipitation.

Molar concentration, also known as molarity (M), is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution.
In the context of precipitation reactions, molar concentration helps in calculating the ion product \( Q \) and comparing it to the solubility product constant \( K_{\text{sp}} \). For instance, the initial molar concentration of \( \text{Mg}^{2+} \) in seawater given in the problem is 0.052 M.
The concentration of hydroxide ions, \( \text{OH}^- \), from saturated \( \text{Ca(OH)}_2 \) can be found using its solubility:
\[ [\text{OH}^-] = 2 \times (\text{solubility of } \text{Ca(OH)}_2) \ \]
Thus, the molarity of various ions is crucial in calculating the likelihood of precipitation.

Precipitation reactions occur when ions in a solution combine to form an insoluble compound or precipitate. This happens when the ion product, \( Q \), of the insoluble compound exceeds its solubility product constant, \( K_{\text{sp}} \).
For example, in the given exercise, \( \text{Mg(OH)}_2 \) precipitates from seawater when hydroxide ions are introduced from \( \text{Ca(OH)}_2 \). The steps are:

• Calculate the hydroxide ion concentration from \( \text{Ca(OH)}_2 \)
• Calculate the ion product \( Q \) for \( \text{Mg(OH)}_2 \)
• Compare \( Q \) to \( K_{\text{sp}} \) of \( \text{Mg(OH)}_2 \)
• If \( Q > K_{\text{sp}} \), \( \text{Mg(OH)}_2 \) will precipitate.

Understanding precipitation reactions is vital in predicting the formation of precipitates in chemical solutions.