Question

Write an expression for the solubility-product constant of \(A_{2} B(s)\). Assume that an ion of \(B\) has a charge of \(2-\) (that is, \(B^{2-}\) ).

Short answer

\(K_{sp} = [A^{+}]^{2}[B^{2-}]\)

Step-by-step solution

Write the Dissolution Reaction

First, write the chemical dissolution reaction of the solid compound in water. For the compound \(A_{2}B(s)\), it dissociates into its respective ions: \[A_{2}B(s) \longrightarrow 2A^{+} + B^{2-}\]

Write the Expression for the Solubility Product Constant

Next, write the expression for the solubility-product constant (\(K_{sp}\)) based on the dissociation reaction. The solubility-product constant is the product of the concentrations of the ions, each raised to the power of its coefficient in the balanced equation: \[K_{sp} = [A^{+}]^{2}[B^{2-}]\] Here, the concentration of \(A^{+}\) is squared because there are two \(A^{+}\) ions produced for every formula unit of \(A_{2}B\) that dissolves.

Key concepts

Dissolution Reaction

The process where a solid, liquid or gas forms a solution in a solvent is known as a dissolution reaction. In the context of ionic compounds, this specifically refers to the process by which a solid ionic compound disperses into its individual ions when dissolved in water.

For our compound, \(A_{2}B(s)\) , the dissolution reaction can be represented as follows:
\[A_{2}B(s) \rightarrow 2A^{+} + B^{2-}\] During dissolution, the solid (solute) is breaking apart into its ions. These ions spread out within the solvent, usually water, creating a homogeneous mixture known as a solution. It's essential for students to understand that the coefficients in the balanced chemical equation represent not only the molar ratio in which substances react or form but also the stoichiometry of the ions in the solution. Thus, when \(A_{2}B\) dissolves, for every one unit that breaks down, two \(A^{+}\) ions and one \(B^{2-}\) ion are released.

Chemical Equilibrium

When a dissolution reaction occurs, it may reach a point where the rate of dissolution equals the rate of precipitation, a state called chemical equilibrium. At this stage, the solution becomes saturated, meaning no more solute can dissolve at the given temperature and pressure conditions. The concentrations of each species remain constant over time.

The solubility-product constant, \(K_{sp}\), comes into play here as a particular type of equilibrium constant. It only applies to the dissolving of sparingly soluble salts. We express \(K_{sp}\) using the equilibrium concentrations of the ions involved in the dissolution reaction. For the reaction \[A_{2}B(s) \rightarrow 2A^{+} + B^{2-}\], the equilibrium expression for the solubility-product constant is \[K_{sp} = [A^{+}]^{2}[B^{2-}]\].

Understanding chemical equilibrium is crucial for predicting whether a precipitate will form when solutions of different ions are mixed and for calculating the concentrations of ions in a solution at equilibrium.

Ionic Compounds

An ionic compound is made up of positively charged ions called cations and negatively charged ions called anions. These ions are held together by strong electrostatic forces of attraction known as ionic bonds. Examples of such compounds include salts, such as sodium chloride (NaCl).

The formula unit of an ionic compound provides the simplest ratio of the ions that are present in the structure. In the case of our example, \(A_{2}B\), the formula unit suggests that for every two cations (\(A^{+}\)), there is one anion (\(B^{2-}\)).

It is also important to note that ionic compounds have distinct properties: they tend to have high melting and boiling points, are typically solid at room temperature, and are good conductors of electricity when dissolved in water. This last property is due to the mobility of the ions in the solution, which is essential for the dissolution reaction and the establishment of equilibrium in a saturated solution.