Problem 51 - Chemistry The Central Science In Si Units

Question

Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{BaCrO}_{4}, \mathrm{CuS}, \mathrm{PbCl}_{2}\) and \(\mathrm{LaF}_{3}\).

Step-by-Step Solution

Answer

For \( \mathrm{BaCrO}_{4} \), \( K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{CrO}_{4}^{2-}] \). For \( \mathrm{CuS} \), \( K_{sp} = [\mathrm{Cu}^{2+}][\mathrm{S}^{2-}] \). For \( \mathrm{PbCl}_{2} \), \( K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^{2} \). For \( \mathrm{LaF}_{3} \), \( K_{sp} = [\mathrm{La}^{3+}][\mathrm{F}^{-}]^{3} \).

Step 1: Understanding the Solubility Product Constant

The solubility product constant, denoted as \( K_{sp} \), is a constant for a given substance at a specific temperature. It represents the level at which a solute dissolves in solution. The expression for \( K_{sp} \) is derived from the equilibrium concentrations of the ions produced when the compound dissolves in water.

Step 2: Write the Dissolution Equation for BaCrO₄

The dissolution of \( \mathrm{BaCrO}_{4} \) in water can be represented as:\[ \mathrm{BaCrO}_{4} (s) \rightleftharpoons \mathrm{Ba}^{2+} (aq) + \mathrm{CrO}_{4}^{2-} (aq) \]The solubility-product expression for this is:\[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{CrO}_{4}^{2-}] \]

Step 3: Write the Dissolution Equation for CuS

The dissolution of \( \mathrm{CuS} \) in water is given by:\[ \mathrm{CuS} (s) \rightleftharpoons \mathrm{Cu}^{2+} (aq) + \mathrm{S}^{2-} (aq) \]Its solubility-product expression is:\[ K_{sp} = [\mathrm{Cu}^{2+}][\mathrm{S}^{2-}] \]

Step 4: Write the Dissolution Equation for PbCl₂

The dissolution of \( \mathrm{PbCl}_{2} \) can be represented as:\[ \mathrm{PbCl}_{2} (s) \rightleftharpoons \mathrm{Pb}^{2+} (aq) + 2\mathrm{Cl}^{-} (aq) \]The solubility-product expression for this equilibrium is:\[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^{2} \]

Step 5: Write the Dissolution Equation for LaF₃

The dissolution of \( \mathrm{LaF}_{3} \) is described by:\[ \mathrm{LaF}_{3} (s) \rightleftharpoons \mathrm{La}^{3+} (aq) + 3\mathrm{F}^{-} (aq) \]Therefore, the solubility-product expression is:\[ K_{sp} = [\mathrm{La}^{3+}][\mathrm{F}^{-}]^{3} \]

Key Concepts

Ionic Compounds

Ionic compounds are substances formed from the combination of positively and negatively charged ions. These ions are held together by strong electrostatic forces known as ionic bonds. This formation typically occurs between metals and non-metals. For instance, in the compound \( \text{BaCrO}_4 \), barium \( (\text{Ba}^{2+}) \) is a metal, and chromate \( (\text{CrO}_4^{2-}) \) is a polyatomic ion consisting of non-metals.
These compounds dissolve in water to release their constituent ions. In solution, ionic compounds dissociate completely or partially into their ions, depending on their solubility. The behavior and solubility of these compounds in water are crucial for determining the concentration of the ions, which directly influences the solubility product constant \( (K_{sp}) \). Understanding this concept is important when considering reactions in aqueous solutions in various chemical processes.

Dissolution Equation

A dissolution equation represents the process where an ionic compound dissolves in water to form its constituent ions. This equation is essential in visualizing the equilibrium that exists between the undissolved solid and the ions in solution.
For instance, for the compound \( \text{BaCrO}_4 \), the dissolution equation is: \[ \text{BaCrO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \]
This equation signifies that the solid barium chromate \( \text{(BaCrO}_4 \text{)} \) dissolves to form barium ions \( \text{(Ba}^{2+} \text{)} \) and chromate ions \( \text{(CrO}_4^{2-} \text{)} \). Each dissolution equation varies depending on the compound in question, showcasing the unique stoichiometry and resulting ions.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of ions in a solution when a dissolution reaction reaches equilibrium. At equilibrium, the rate at which the solid dissolves to form ions equals the rate at which ions recombine to form the solid.
Let's take \( \text{PbCl}_2 \) as an example: \[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^{-} (aq) \]
Here, the equilibrium concentrations of the ions \([\text{Pb}^{2+}]\) and \([\text{Cl}^{-}]\) are used to write the solubility product constant expression, which helps to determine the extent of dissolution. Specifically, equilibrium concentrations are crucial in calculating the \( K_{sp} \), the solubility product constant, which quantifies this balance.

Chemical Equilibrium

Chemical equilibrium is the state in a chemical reaction where the rates of the forward and reverse reactions are equal. For dissolution reactions like those of ionic compounds, equilibrium signifies that the process of dissolving and precipitating ions occurs at the same rate. This balance is key to understanding reactions in closed systems.
Take the dissolution of \( \text{LaF}_3 \) for instance: \[ \text{LaF}_3 (s) \rightleftharpoons \text{La}^{3+} (aq) + 3\text{F}^{-} (aq) \]
Once this reaction reaches chemical equilibrium, no net change in concentration occurs for \( \text{La}^{3+} \) and \( \text{F}^{-} \) ions. They continue to form and dissolve at equal rates, maintaining their concentrations constant over time. Understanding equilibrium is crucial for predicting reaction behavior and calculating the solubility product constant \( K_{sp} \). This concept applies broadly across numerous chemical reactions and solutions.