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 \(108y^5\), option (d).

Step-by-step solution

Write the Dissociation Equation

The compound \( \mathrm{A}_{2} \mathrm{X}_{3} \) dissociates in an aqueous solution as \( \mathrm{A}_{2} \mathrm{X}_{3} \rightarrow 2\mathrm{A}^{3+} + 3\mathrm{X}^{2-} \).

Express Concentrations in Terms of Solubility y

If the solubility of \( \mathrm{A}_{2} \mathrm{X}_{3} \) is \( y \), then the concentration of \( \mathrm{A}^{3+} \) ions is \( 2y \) and the concentration of \( \mathrm{X}^{2-} \) ions is \( 3y \).

Write the Solubility Product Expression

The solubility product, \( K_{sp} \), for the dissociation is expressed as: \( K_{sp} = [\mathrm{A}^{3+}]^2[\mathrm{X}^{2-}]^3 \).

Substitute Concentrations in Ksp Expression

Substitute the ionic concentrations to get: \( K_{sp} = (2y)^2 (3y)^3 \).

Calculate the Solubility Product

Calculate \( K_{sp} \) as: \( K_{sp} = 4y^2 \times 27y^3 = 108y^5 \).

Review Options and Choose the Correct Answer

Based on the calculations, the correct option is (d) \( 108\mathrm{y}^5 \).

Key concepts

Ionic Equilibrium

Ionic equilibrium in chemistry refers to the state of balance between the ions and the unionized molecules in a solution where a reversible chemical reaction occurs. This equilibrium is crucial because, during the dissolution process, ionic compounds dissociate into their constituent ions in a solvent, like water.

For example, consider the dissociation of a generic salt, \[ \text{Salt} \rightleftharpoons \text{Cat}^{+} + \text{An}^{-}. \]This process involves establishing an equilibrium between the remaining undissolved salt and the ions in the solution.
The concentration of ions will remain constant at equilibrium as long as the solution conditions remain unchanged.

Understanding ionic equilibrium is important for calculating the solubility product, which shows how much of a compound will dissolve under equilibrium conditions.

Dissociation Equation

The dissociation equation describes how a particular ionic compound breaks down into its ions when it dissolves. This is crucial for calculating the solubility product of a compound. In the example, \[ \text{A}_2\text{X}_3 \rightarrow 2\text{A}^{3+} + 3\text{X}^{2-}, \]the solid \(\text{A}_2\text{X}_3\) dissociates into its ion components.
By understanding the stoichiometry of the dissociation reaction, we can determine the concentration of each ion in terms of the solubility \(y\).

Each mole of \(\text{A}_2\text{X}_3\) produces two moles of \(\text{A}^{3+}\) and three moles of \(\text{X}^{2-}\).
These ratios are fundamental when deriving expressions and performing calculations concerning chemical equilibrium.

Concentration Terms

Concentration terms help us express how much solute is present in a solution. It's crucial to understand these terms to perform chemical calculations effectively.
For the compound \(\text{A}_2\text{X}_3\) dissociating in the solution, if the solubility is given as \(y\), we can calculate the ion concentrations:

• Concentration of \(\text{A}^{3+}\) ions is \(2y\).
• Concentration of \(\text{X}^{2-}\) ions is \(3y\).

Using these expressions is vital for deriving the solubility product, as they represent the amounts of ions available for participating in equilibrium. Precise measurement and understanding of such concentrations form the crux of determining the behavior of ions in a chemical system.

Chemical Calculations

Chemical calculations often involve using established equations to determine unknown quantities based on known data. In this case, calculating the solubility product involves substituting the concentrations of ions.
For \(\text{A}_2\text{X}_3\), the solubility product expression is:\[ K_{sp} = [\text{A}^{3+}]^2[\text{X}^{2-}]^3. \]Substituting the known concentrations, this becomes:\[ K_{sp} = (2y)^2 (3y)^3 = 4y^2 \times 27y^3 = 108y^5. \]Through these calculations, we find that the solubility product, which indicates the ease of dissolution, is \(108y^5\).
This step-by-step calculation is essential for chemistry problems involving equilibrium as it links theoretical understanding with practical data.