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 \(\mathrm{LaF}_{3} .\)

Short answer

The solubility-product constant expressions for the given ionic compounds are: - BaCrO\(_4\): \(K_\mathrm{sp} = [\mathrm{Ba^{2+}}][\mathrm{CrO_4^{2-}}]\) - CuS: \(K_\mathrm{sp} = [\mathrm{Cu^{2+}}][\mathrm{S^{2-}}]\) - PbCl\(_2\): \(K_\mathrm{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2\) - LaF\(_3\): \(K_\mathrm{sp} = [\mathrm{La^{3+}}][\mathrm{F^-}]^3\)

Step-by-step solution

1. Understanding solubility-product constant (K\(_{sp}\))

The solubility-product constant, K\(_{sp}\), is an equilibrium constant that measures the solubility of a sparingly soluble ionic compound in a solution. It is represented by the product of the equilibrium concentrations of the constituent ions of the ionic compound, each raised to the power of its stoichiometric coefficient.

2. Write the dissociation equation for BaCrO\(_4\)

First, let's write the dissociation equation for the ionic compound in water: \[ \mathrm{BaCrO_4}(s) \rightleftharpoons \mathrm{Ba^{2+}}(aq) + \mathrm{CrO_4^{2-}}(aq) \]

3. Write the K\(_{sp}\) expression for BaCrO\(_4\)

Now, we can write the solubility-product constant using the concentrations of the ions and their stoichiometric coefficients according to the balanced equation: \[ K_\mathrm{sp} = [\mathrm{Ba^{2+}}][\mathrm{CrO_4^{2-}}] \]

4. Write the dissociation equation for CuS

Let's write the dissociation equation for CuS: \[ \mathrm{CuS}(s) \rightleftharpoons \mathrm{Cu^{2+}}(aq) + \mathrm{S^{2-}}(aq) \]

5. Write the K\(_{sp}\) expression for CuS

The solubility-product constant for CuS can be written as: \[ K_\mathrm{sp} = [\mathrm{Cu^{2+}}][\mathrm{S^{2-}}] \]

6. Write the dissociation equation for PbCl\(_2\)

Now, let's write the dissociation equation for PbCl\(_2\): \[ \mathrm{PbCl_2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2\mathrm{Cl^{-}}(aq) \]

7. Write the K\(_{sp}\) expression for PbCl\(_2\)

The solubility-product constant for PbCl\(_2\) can be written as: \[ K_\mathrm{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2 \]

8. Write the dissociation equation for LaF\(_3\)

Finally, let's write the dissociation equation for LaF\(_3\): \[ \mathrm{LaF_3}(s) \rightleftharpoons \mathrm{La^{3+}}(aq) + 3\mathrm{F^{-}}(aq) \]

9. Write the K\(_{sp}\) expression for LaF\(_3\)

The solubility-product constant for LaF\(_3\) can be written as: \[ K_\mathrm{sp} = [\mathrm{La^{3+}}][\mathrm{F^-}]^3 \]

Summary

We've written the solubility-product constant expressions for the given ionic compounds as follows: - BaCrO\(_4\): \(K_\mathrm{sp} = [\mathrm{Ba^{2+}}][\mathrm{CrO_4^{2-}}]\) - CuS: \(K_\mathrm{sp} = [\mathrm{Cu^{2+}}][\mathrm{S^{2-}}]\) - PbCl\(_2\): \(K_\mathrm{sp} = [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2\) - LaF\(_3\): \(K_\mathrm{sp} = [\mathrm{La^{3+}}][\mathrm{F^-}]^3\)

Key concepts

Ionic Equilibrium

In chemistry, ionic equilibrium refers to the state when a chemical reaction involving ions does not proceed any further because both the forward and reverse reactions occur at the same rate. This balance happens in a saturated solution of an ionic compound. At this point, the solution holds the maximum concentration of ions that can exist in equilibrium with the solid form of the compound.

• Saturated Solution: A solution in which no more solute can dissolve at a given temperature.
• Sparingly Soluble Compounds: Some compounds do not dissolve well in water, thus achieving equilibrium quickly with their solid forms.

Ionic equilibrium is significant in calculating the solubility of a compound using the solubility product constant, often denoted as \(K_{sp}\). This value helps understand how much of the ion can exist in solution, explaining the upper limit to the ion concentration in aqueous solutions.

Learning to express equations with \(K_{sp}\) is essential to predict the solubility product and understand how other solutions or stressors may impact the equilibrium.

Chemical Equilibrium

Chemical equilibrium is a fundamental concept that describes the state in which both the reactants and products of a reversible chemical reaction are present in concentrations that have no further tendency to change with time. When a system reaches chemical equilibrium, the rates of the forward reaction (reactants to products) and the reverse reaction (products to reactants) are equal, thus the concentrations remain constant.

• Dynamic Equilibrium: Although reactions continue to occur, there is no net change in concentration.
• Reversible Reactions: Reactions that can proceed in both forward and backward directions.

The expression for chemical equilibrium is known as the equilibrium constant \(K_{eq}\). For solubility equilibria, this is referred to as the \(K_{sp}\) which specifically applies to the dissolution of sparingly soluble salts.

Understanding this concept helps in predicting how changes in conditions such as concentration, pressure, and temperature affect the system’s equilibrium, according to Le Chatelier's Principle. This situation can help chemists manipulate conditions to favor the formation of a desired product.

Dissociation Equations

Dissociation equations are very helpful tools in chemistry that show how compounds break down into ions in a solution. When a solid ionic compound dissolves in water, it separates into its cations and anions. These equations represent the process and help to understand what's happening at the molecular level.

Writing dissociation equations correctly is the first step in determining how much of each ion is present in the solution, vital for composing the solubility product expression.

• Ions: Charged particles that result from an atom losing or gaining electrons.
• Stoichiometry: This concept ensures that chemical equations are balanced with respect to both mass and charge, showing the correct number of ions.

Each dissociation equation is typically unique to a specific compound. For example:- For \(\mathrm{BaCrO_4}\), the dissociation is \(\mathrm{BaCrO_4\:(s) \rightleftharpoons Ba^{2+}\:(aq) + CrO_4^{2-}\:(aq)}\)- For \(\mathrm{CuS}\), it is \(\mathrm{CuS\:(s) \rightleftharpoons Cu^{2+}\:(aq) + S^{2-}\:(aq)}\)
These equations lay the groundwork for forming expressions of \(K_{sp}\), allowing chemists to quantify the solubility and stability of ionic substances in solutions.