For which of the following reactions is the equilibrium constant called a solubility product? $$ \begin{array}{l} \text { (a) } \mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q) \rightleftharpoons \\ \text { (b) } 3 \mathrm{Ca}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons \mathrm{Ca}_{3}(\mathrm{OH})_{4}^{2-}(a q) \end{array} $$ (c) \(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightleftharpoons\) $$ \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) $$ (d) \(\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)\)

Understanding the equilibrium constant is essential for interpreting the behavior of substances in reversible chemical reactions. The equilibrium constant (\( K \text{eq} \) or simply, K) is used to describe the ratio of concentrations of products to reactants, raised to their respective stoichiometric coefficients, at equilibrium.

In the context of solubility, we refer to a special type of equilibrium constant known as the solubility product constant (\( K_{sp} \)). This value is exclusive to the dissolution of sparingly soluble ionic compounds. When substances such as lead iodide (\( PbI_2 \) in reaction (d)) dissolve, they do so by dissociating into their constituent ions. The solubility product is then defined by the concentrations of these ions at equilibrium, each raised to the power of their coefficient in the balanced dissolution equation.

For example, for lead iodide dissolving in water: \( PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq) \) the solubility product (\( K_{sp} \) would be given by \( K_{sp} = [Pb^{2+}][I^-]^2 \). It’s important for students to note that solids and pure liquids do not appear in the expression for the equilibrium constant.

Dissolution is the process through which substances, typically solids, disintegrate to form a solution. With ionic compounds, dissolution involves the compound breaking down into its constituent ions.

When an ionic compound like lead iodide (\( PbI_2 \)) is introduced to a solvent like water, it dissociates into lead ions (\( Pb^{2+} \) and iodide ions (\( I^- \) (as shown in reaction (d)). The solubility of the substance determines the extent to which it will dissolve, with sparingly soluble compounds establishing a dynamic equilibrium between the solid and its dissolved ions.

This dissolution process can be represented by a reversible chemical equation with the undissolved solid on one side and the dissolved ions on the other. In an educational setting, categorizing reactions incorrectly – like mistakenly identifying reaction (a) as a dissolution process – can lead to misunderstanding the nature of solubility equilibria.

Solubility in chemistry refers to the ability of a substance, known as the solute, to dissolve in a solvent to form a homogeneous mixture known as a solution. The degree of solubility varies greatly depending on both the solute and the solvent, as well as temperature and pressure.

For example, the dissolution of calcium carbonate (\( CaCO_3 \) in reaction (c)) in a solution containing hydrogen ions (\( H^+ \) is complex as it results in the formation of water (\( H_2O \) and carbon dioxide gas (\( CO_2(g) \)), rather than simply dissolving into its constituent ions. Hence, it does not qualify as a simple dissolution process as required for calculating a solubility product.

Soluble compounds readily dissolve and exist predominantly as ions in solution, whereas insoluble compounds do not dissolve appreciably and contribute very little to the ionic makeup of the solution. Understanding solubility helps predict the outcome of reactions and the solubility product provides a quantitative measure for the solubility of sparingly soluble compounds.