Question

The solubility of \(\mathrm{A}_{2} \mathrm{X}_{3}\) is \(\mathrm{y} \mathrm{mol} \mathrm{dm}^{-3} .\) Its solubility product is: (a) \(6 \mathrm{y}^{4}\) (b) \(64 \mathrm{y}^{4}\) (c) \(36 \mathrm{y}^{5}\) (d) \(108 \mathrm{y}^{5}\)

Short answer

The solubility product is (d) \(108y^5\).

Step-by-step solution

Write the Dissolution Equation

The compound \( \text{A}_2\text{X}_3 \) dissolves in water to form its ions as follows: \[ \text{A}_2\text{X}_3 (s) \rightleftharpoons 2\text{A}^{3+} (aq) + 3\text{X}^{2-} (aq) \]

Define the Solubility

Let \( y \) be the solubility of \( \text{A}_2\text{X}_3 \) in \( \text{mol dm}^{-3} \). This means that in solution, the concentration of \( \text{A}^{3+} \) ions is \( 2y \) and the concentration of \( \text{X}^{2-} \) ions is \( 3y \).

Write the Solubility Product Expression

The solubility product expression \( K_{sp} \) for \( \text{A}_2\text{X}_3 \) is defined as: \[ K_{sp} = [\text{A}^{3+}]^2 [\text{X}^{2-}]^3 \]

Substitute the Ion Concentrations

Substitute the concentrations from Step 2 into the solubility product expression: \[ K_{sp} = (2y)^2 (3y)^3 \]

Calculate the Solubility Product

Expand the expression: \[ K_{sp} = 4y^2 \times 27y^3 = 108y^5 \] This results in \( K_{sp} = 108y^5 \).

Match the Solubility Product to the Options

Compare the calculated \( K_{sp} = 108y^5 \) with the given options. The correct match is (d) \( 108y^5 \).

Key concepts

Dissolution Equation

The concept of dissolution is pivotal when studying reactions in chemistry that involve solubility. Specifically, the dissolution equation represents how a solid compound breaks down into its ions in a solution. For the compound \( \text{A}_2\text{X}_3 \), the process can be shown through the equation: \[ \text{A}_2\text{X}_3 (s) \rightleftharpoons 2\text{A}^{3+} (aq) + 3\text{X}^{2-} (aq) \] Here, the solid \( \text{A}_2\text{X}_3 \) dissociates into two cations \( \text{A}^{3+} \) and three anions \( \text{X}^{2-} \). The equation illustrates the dynamic equilibrium between the dissolved ions and the undissolved solid particles in a saturated solution. Recognizing and writing the dissolution equation correctly is crucial for further calculations involving solubility, as it sets the foundation for ion concentration and solubility product expressions.

Solubility Expression

To express solubility correctly, we start by understanding what it represents. Solubility is the maximum amount of a substance that can dissolve in a given volume of solvent at a specific temperature to form a saturated solution. For \( \text{A}_2\text{X}_3 \), if \( y \) denotes its solubility in \( \text{mol dm}^{-3} \), the dissolved ions are present in specific ratios according to the dissolution equation:

• Each molecule of \( \text{A}_2\text{X}_3 \) gives 2 ions of \( \text{A}^{3+} \), hence \([\text{A}^{3+}] = 2y \).
• Similarly, it gives 3 ions of \( \text{X}^{2-} \), thus \([\text{X}^{2-}] = 3y \).

This step is essential for determining the concentrations needed for further calculations, especially when calculating the solubility product.

Ion Concentration Calculation

Calculating the concentration of ions in solution is a straightforward but essential process in understanding solubility relationships. Given the solubility \( y \) of \( \text{A}_2\text{X}_3 \), we must find the concentrations of the resulting ions:

• The concentration of \( \text{A}^{3+} \) ions is determined as \( 2y \), as each unit of \( \text{A}_2\text{X}_3 \) results in two \( \text{A}^{3+} \) ions.
• Similarly, the concentration of \( \text{X}^{2-} \) ions becomes \( 3y \), due to the three \( \text{X}^{2-} \) ions yielded per formula unit of the dissolving compound.

These concentrations are not just critical for understanding the chemistry in the solution, they are also directly applied in calculating the solubility product. Making sure these values are correct helps in ensuring accurate calculations and understanding the potential of the dissolution.

Solubility Product Expression

The solubility product constant, \( K_{sp} \), is a key expression in predicting the solubility of ionic compounds in solution. For \( \text{A}_2\text{X}_3 \), its \( K_{sp} \) is determined by the equilibrium concentrations of its ions: \[ K_{sp} = [\text{A}^{3+}]^2 [\text{X}^{2-}]^3 \] Upon substituting the ion concentrations from previous calculations, \( 2y \) and \( 3y \), we find: \[ K_{sp} = (2y)^2 (3y)^3 \] This computation expands to \[ K_{sp} = 4y^2 \times 27y^3 = 108y^5 \]. This expression provides the quantitative measure of the product of the concentrations of the ions raised to their stoichiometric coefficients, confirming the answer as \( 108y^5 \). Understanding this expression is crucial in solving solubility problems as it links the concept of dissolution to observable chemical quantities.