Write the ionic equation for the dissolution and the \(K_{\mathrm{sp}}\) expression for each of the following slightly soluble ionic compounds: (a) \(\mathrm{LaF}_{3}\) (b) \(\mathrm{CaCO}_{3}\) (c) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{Pb}(\mathrm{OH})_{2}\)

In chemistry, the ionic equation is a method used to represent the dissociation of electrolytes—substances that dissolve in water to produce ions. This type of equation shows the individual ions that are released when a compound goes into solution.

For instance, the dissolution of \textbf{Lanthanum fluoride (\textbf{LaF₃})} in water is represented by the equation: $$\text{LaF}_{3(s)} \rightarrow \text{La}^{3+}_{(aq)} + 3\text{F}^{-}_{(aq)}$$
Here, \textbf{(s)} signifies a solid and \textbf{(aq)} denotes that the ion is in aqueous or dissolved form. The equation tells us that each unit of solid LaF₃ yields one lanthanum ion and three fluoride ions in the solution. Similarly, writing the ionic equations for other compounds like calcium carbonate (CaCO₃), silver sulfate (Ag₂SO₄), and lead(II) hydroxide (Pb(OH)₂) provides a clear picture of the ions produced during their dissolution.

A dissolution reaction refers to the process by which a solid dissolves in a solvent, such as water, to form a homogeneous mixture or solution. When an ionic compound dissolves, it separates into its constituent ions.

To improve comprehension, let's visualize the dissolution of \textbf{calcium carbonate (CaCO₃)}: $$\text{CaCO}_{3(s)} \rightarrow \text{Ca}^{2+}_{(aq)} + \text{CO}_{3}^{2-}_{(aq)}$$
This equation shows that solid calcium carbonate separates into calcium ions (Ca²⁺) and carbonate ions (CO₃²⁻) when it dissolves in water. The dissolution is essential because it sets the stage for the concept of solubility equilibrium, determining how much of the compound can dissolve at a given temperature.

Solubility equilibrium is a state where a chemical compound in the solid state is in a dynamic balance with a solution of that compound. At this point, the rate at which the compound dissolves in the solvent equals the rate at which it precipitates out of the solution.

The solubility product constant (\textbf{Kₛₚ}) is a value that helps us understand this equilibrium. It is calculated as the product of the molar concentrations of the ions involved in the dissolution, each raised to the power of their coefficient in the balanced equation. For example, for lead(II) hydroxide, the \textbf{Kₛₚ} expression is given by:$$K_{\text{sp}} = [\text{Pb}^{2+}] [\text{OH}^{-}]^{2}$$
This signifies that the product of the concentration of lead ions squared and the concentration of hydroxide ions squared equals a constant value at equilibrium. The \textbf{Kₛₚ} value varies depending on the substance and the conditions of the experiment, like temperature, therefore, offering insight into the solubility of the compound.