Problem 63 - A Complete Resource Book in Chemistry for JEE Main

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) \(108 \mathrm{y}^{5}\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

The exercise asks us to determine the solubility product (K_{ ext{sp}}) of a compound A_{2}X_{3} given its solubility is y mol dm^{-3}. We must express K_{ ext{sp}} in terms of y.

Step 2: Write the Dissociation Equation

Write the dissociation equation for A_{2}X_{3}: A_{2}X_{3 (s)} ightleftharpoons 2A^{3+}_{(aq)} + 3X^{2-}_{(aq)}.

Step 3: Express the Ion Concentrations in Terms of Solubility

If the solubility of A_{2}X_{3} is y mol dm^{-3}, then at equilibrium, the concentration of A^{3+} will be 2y mol dm^{-3} and the concentration of X^{2-} will be 3y mol dm^{-3}.

Step 4: Write the Expression for the Solubility Product

The solubility product K_{ ext{sp}} is given by the expression K_{ ext{sp}} = [A^{3+}]^2 imes [X^{2-}]^3. Substituting the ion concentrations gives K_{ ext{sp}} = (2y)^2 imes (3y)^3.

Step 5: Simplify the Expression

Simplify the expression from Step 4: \( K_{\text{sp}} = 4y^2 \times 27y^3 \). Therefore, \( K_{\text{sp}} = 108y^5 \).

Step 6: Choose the Correct Answer

The simplified expression is K_{ ext{sp}} = 108y^5, which matches option (d).

Key Concepts

Solubility

Solubility refers to the maximum amount of a solute that can dissolve in a solvent at a given temperature, resulting in a saturated solution. In simple terms, it's the point at which a substance stops dissolving. For example, if you're making lemonade, the point when no more sugar can dissolve in water is the solubility limit.

In this exercise, the solubility of the compound \( \mathrm{A}_2 \mathrm{X}_3 \) is given as \( y \ mol \ dm^{-3} \). This means at equilibrium, the solution contains \( y \) moles of the solute per liter of solution. Understanding solubility helps us determine the concentration of ions in a solution when a slightly soluble salt is dissolved.

Dissociation Equation

A dissociation equation describes how a compound separates into its ions in solution. This process is crucial in understanding how substances interact in a solvent.

For the compound \( \mathrm{A}_2 \mathrm{X}_3 \), the dissociation equation is given by:

• \( \mathrm{A}_2 \mathrm{X}_3(s) \rightleftharpoons 2\mathrm{A}^{3+}_{(aq)} + 3\mathrm{X}^{2-}_{(aq)} \)

This equation tells us that one formula unit of \( \mathrm{A}_2 \mathrm{X}_3 \) dissociates into two \( \mathrm{A}^{3+} \) ions and three \( \mathrm{X}^{2-} \) ions in the solvent. The stoichiometry of the dissociation is important for calculating the ion concentrations later in the problem.

Ion Concentration

Ion concentration refers to the amount of a specific ion present in a solution, generally expressed in moles per liter, denoted as \( mol \ dm^{-3} \). After dissociation, the concentration of ions depends on the stoichiometry of the equation.

Given that the solubility of \( \mathrm{A}_2 \mathrm{X}_3 \) is \( y \ mol \ dm^{-3} \), this means:

• The concentration of \( \mathrm{A}^{3+} \) ions in the solution is \( 2y \ mol \ dm^{-3} \) since two \( \mathrm{A}^{3+} \) ions are produced per unit of solute.
• The concentration of \( \mathrm{X}^{2-} \) ions is \( 3y \ mol \ dm^{-3} \), as three \( \mathrm{X}^{2-} \) ions are formed from each unit.

Understanding these concentrations is vital to calculating the solubility product.

Equilibrium Chemistry

Equilibrium chemistry involves the study of reactions that reach a state where the forward and backward reactions occur at the same rate. This state is known as chemical equilibrium, often seen in the context of solubility and dissociation.

For the equilibrium of \( \mathrm{A}_2 \mathrm{X}_3 \) in solution, the solubility product \( K_{\text{sp}} \) is an important concept:

• \( K_{\text{sp}} \) is calculated from the equilibrium concentrations of the ions: \( [\mathrm{A}^{3+}]^2 \times [\mathrm{X}^{2-}]^3 \).
• Substituting the ion concentrations \((2y)^2 (3y)^3\) into this expression gives \( K_{\text{sp}} = 4y^2 \times 27y^3 = 108y^5 \).

This result explains how the concentrations of ions contribute to the overall equilibrium, leading to the solubility product that characterizes the compound's solubility in water.