Problem 100 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility product of \(\mathrm{A}_{2} \mathrm{X}_{3}\) is \(1.08 \times 10^{-23}\). Its solubility will be (a) \(1.0 \times 10^{-3} \mathrm{M}\) (b) \(1.0 \times 10^{-4} \mathrm{M}\) (c) \(1.0 \times 10^{-5} \mathrm{M}\)

Step-by-Step Solution

Answer

The solubility of \( \mathrm{A}_{2} \mathrm{X}_{3} \) is \( 1.0 \times 10^{-5} \mathrm{M} \).

Step 1: Write the Dissociation Equation

Write the dissociation equation for the salt \( \mathrm{A}_{2} \mathrm{X}_{3} \). It dissociates into 2 moles of \( \mathrm{A}^{3+} \) and 3 moles of \( \mathrm{X}^{2-} \) ions:\[ \mathrm{A}_{2} \mathrm{X}_{3} \rightleftharpoons 2\mathrm{A}^{3+} + 3\mathrm{X}^{2-} \]

Step 2: Express Solubility in Terms of s

Let \( s \) be the solubility of \( \mathrm{A}_{2} \mathrm{X}_{3} \) in mol/L. Therefore, the concentration of \( \mathrm{A}^{3+} \) will be \( 2s \), and the concentration of \( \mathrm{X}^{2-} \) will be \( 3s \).

Step 3: Write the Expression for Ksp

The solubility product (\(K_{sp}\)) is given by\[ K_{sp} = [\mathrm{A}^{3+}]^2 [\mathrm{X}^{2-}]^3 \]Substitute the concentrations in terms of \( s \):\[ K_{sp} = (2s)^2 (3s)^3 \]

Step 4: Simplify the Expression

Simplify the expression for \( K_{sp} \):\[ K_{sp} = 4s^2 \times 27s^3 = 108s^5 \]

Step 5: Calculate s Using the Ksp Value

Given \( K_{sp} = 1.08 \times 10^{-23} \), solve for \( s \):\[ 108s^5 = 1.08 \times 10^{-23} \]\[ s^5 = \frac{1.08 \times 10^{-23}}{108} \]Calculate \( s \):\[ s = \left( \frac{1.08 \times 10^{-23}}{108} \right)^{1/5} \]

Step 6: Approximate Calculation Result

Perform the calculation to approximate \( s \):\[ s = \left( 10^{-25} \right)^{1/5} = 10^{-5} \]Thus, the solubility of \( \mathrm{A}_{2} \mathrm{X}_{3} \) is \( 1.0 \times 10^{-5} \mathrm{M} \).

Key Concepts

Dissociation Equation

A dissociation equation shows how a compound breaks apart into its ions when it dissolves in water. In this example, we look at the compound \( \mathrm{A}_{2} \mathrm{X}_{3} \), which dissociates into ions. Each formula unit of \( \mathrm{A}_{2} \mathrm{X}_{3} \) separates into two \( \mathrm{A}^{3+} \) ions and three \( \mathrm{X}^{2-} \) ions.The equation is written as:

• \[ \mathrm{A}_{2} \mathrm{X}_{3} \rightleftharpoons 2\mathrm{A}^{3+} + 3\mathrm{X}^{2-} \]

This shows the equilibrium between the solid \( \mathrm{A}_{2} \mathrm{X}_{3} \) and its ions in solution. It's essential to represent the stoichiometry correctly, as it tells us the ratio of ions released per formula unit of compound.Understanding how compounds dissociate is crucial for solving problems related to solubility and chemical equilibrium. It forms the foundation for further calculations and gives insight into how concentrations of ions affect solubility.

Solubility Calculation

Calculating solubility involves expressing the amounts of ions derived from a compound as it dissolves. Let's define solubility as \( s \), representing the maximum concentration of the solute that dissolves in a solution at equilibrium.For the compound \( \mathrm{A}_{2} \mathrm{X}_{3} \), its dissociation provides:

• Concentration of \( \mathrm{A}^{3+} \) as \( 2s \)
• Concentration of \( \mathrm{X}^{2-} \) as \( 3s \)

Once these concentrations are established, the solubility product constant, \( K_{sp} \), is used:

• \[ K_{sp} = [\mathrm{A}^{3+}]^2 [\mathrm{X}^{2-}]^3 \]

Substitute \( 2s \) for \( \mathrm{A}^{3+} \) and \( 3s \) for \( \mathrm{X}^{2-} \), the equation becomes:

• \[ K_{sp} = (2s)^2 (3s)^3 = 108s^5 \]

This equation is pivotal. Solving it for \( s \) gives the solubility of the compound. Using the provided \( K_{sp} \), you can compute the value of \( s \), indicating how much substance dissolves to reach saturation.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction (dissolution) equals the rate of the reverse reaction (precipitation). For salts like \( \mathrm{A}_{2} \mathrm{X}_{3} \), equilibrium between the dissolved ions and solid form defines its solubility.At equilibrium, the concentration of ions remains stable, and this dynamic balance is described by the solubility product constant, \( K_{sp} \). For this compound, equilibrium is expressed through:

• \[ \mathrm{A}_{2} \mathrm{X}_{3} \rightleftharpoons 2\mathrm{A}^{3+} + 3\mathrm{X}^{2-} \]
• \[ K_{sp} = [\mathrm{A}^{3+}]^2 [\mathrm{X}^{2-}]^3 \]

The constant \( K_{sp} \) quantifies the degree to which the compound dissolves before reaching saturation.Understanding chemical equilibrium helps predict the saturation point of a solution and how changes in conditions (like temperature or pressure) could affect solubility. Properly grasping equilibrium allows students to apply principles of chemistry to real-world scenarios, like understanding mineral deposits or the behavior of medicines in the body.