Determine \(K_{\mathrm{sp}}\) for each of the following sparingly soluble compounds, given their molar solubilities: (a) \(\mathrm{AgI}\), \(9.1 \times 10^{-9} \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (b) \(\mathrm{Ca}(\mathrm{OH})_{2}, 0.011 \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (c) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\), \(2.7 \times 10^{-6} \mathrm{~mol} \cdot \mathrm{L}^{-1}\); (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, 5.2 \times 10^{-7} \mathrm{~mol} \cdot \mathrm{L}^{-1}\). For the purpose of this calculation, ignore any reaction of the anion with water and the autoprotolysis of water.

Understanding the solubility product constant, often abbreviated as Ksp, requires a grasp of the concept of chemical equilibrium. Chemical equilibrium occurs when a reversible chemical reaction reaches a state where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time.

In the context of sparingly soluble substances, equilibrium exists between the solid compound and its ions in solution. The Ksp is the product of the concentrations of the ions at equilibrium. Each ion concentration is raised to the power of its coefficient in the balanced dissociation equation. This constant is of immense importance as it provides insight into how much of a compound will dissolve in a solution before reaching saturation.

The calculation of Ksp from molar solubility, as illustrated in the given exercise, is a crucial skill in chemistry, specifically for predicting the solubility of ionic compounds and for determining the extent of precipitation reactions. Remember that when calculating the concentrations of ions in solution, stoichiometry plays a key role as it dictates the number of ions produced from the dissociation of one unit of the compound.

Molar solubility refers to the number of moles of a solute that can be dissolved per liter of solution before the solution becomes saturated. It is a measurement of a substance's solubility and is usually expressed in mol/L. Molar solubility is particularly relevant for substances that do not fully dissolve in a solvent, characterized as sparingly soluble or slightly soluble compounds.

To calculate the molar solubility, it's important to carefully examine the dissociation of the compound in water, keeping in mind the stoichiometry of the reaction, as the ratio of ions produced per formula unit of the compound can affect the final calculation. For instance, for a compound that dissociates to produce more than one ion of a particular type, the concentration of that ion will be a multiple of the compound's molar solubility. In the textbook exercises provided, the molar solubility played a critical role in determining the Ksp of each compound.

Dissociation equations are vital to understanding how ionic compounds interact with solvents, such as water. These equations express the chemical reaction that occurs when an ionic compound separates into its constituent ions upon dissolving. The equations are balanced according to the principle of conservation of mass and charge.

For example, in the case of AgI dissolving in water, the dissociation equation is written as \( \text{AgI(s)} \rightleftharpoons \text{Ag}^+(aq) + \text{I}^−(aq) \). This demonstrates that each formula unit of solid AgI yields one silver ion and one iodide ion when it dissolves. Understanding these dissociation reactions is essential for representing the process by which salts and other ionic compounds dissolve and for calculating the corresponding solubility product constant (Ksp). When writing or interpreting dissociation equations, be sure to consider the charges and the physical states of all reactants and products for a correct portrayal of the solubility phenomenon.