Problem 263 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility in water of a sparingly soluble salt \(\mathrm{AB}_{2}\) is \(1.0 \times 10^{-5} \mathrm{molL}^{-1}\). Its solubility product number will be \(\quad[\mathbf{2 0 0 3}]\) (a) \(4 \times 10^{-15}\) (b) \(4 \times 10^{-10}\) (c) \(1 \times 10^{-15}\) (d) \(1 \times 10^{-10}\)

Step-by-Step Solution

Answer

The solubility product is \(4 \times 10^{-15}\), choice (a).

Step 1: Understand the Solubility Concept

The solubility of a salt defines how much of that salt can be dissolved in a solution to reach equilibrium. Here, we have \(\mathrm{AB}_2\), which means one cation A and two anions B are released per molecule.

Step 2: Write the Dissociation Equation

For the salt \(\mathrm{AB}_{2}\), the dissociation in water can be given by the equation: \[ \mathrm{AB}_{2}(s) \rightleftharpoons \mathrm{A}^{2+} (aq) + 2\mathrm{B}^{-} (aq) \]

Step 3: Express Concentrations at Equilibrium

Let \(s\) be the solubility of \(\mathrm{AB}_2\), which is \(1.0 \times 10^{-5} \ \mathrm{molL}^{-1}\). At equilibrium, the concentrations are: \( \text{[A}^{2+}] = s = 1.0 \times 10^{-5} \) and \( \text{[B}^-] = 2s = 2.0 \times 10^{-5} \).

Step 4: Calculate the Solubility Product \(K_{sp}\)

Use the formula for the solubility product, \( K_{sp} = [\mathrm{A}^{2+}][\mathrm{B}^{-}]^2 \). Substituting the equilibrium concentrations: \[ K_{sp} = (1.0 \times 10^{-5})(2.0 \times 10^{-5})^2 = (1.0 \times 10^{-5})(4.0 \times 10^{-10}) = 4.0 \times 10^{-15} \]

Step 5: Determine the Correct Choice

Compare the calculated \(K_{sp}\) value \(4.0 \times 10^{-15}\) with the options provided in the exercise. The correct answer is (a) \(4 \times 10^{-15}\).

Key Concepts

Sparingly Soluble Salts

Sparingly soluble salts are compounds that only dissolve in water to a small extent. Unlike highly soluble salts which dissolve completely, sparingly soluble salts reach an equilibrium between dissolved ions and the undissolved solid. When these salts are added to water, only a tiny amount dissolves to achieve saturation.

• What makes them sparingly soluble? The attraction between the ions in the solid is stronger than their attraction to water molecules.
• Examples: Calcium sulfate and silver chloride.

Understanding sparingly soluble salts is crucial because it helps us determine how much of a compound will be present in its ionic form in a solution. Knowing its solubility is essential for predicting reactions and product formation in various chemistry scenarios.
Solubility data can be crucial for industries like pharmaceuticals and mining.

Dissociation Equation

To understand what happens when a sparingly soluble salt dissolves, we use a dissociation equation. This chemical equation represents how the solid ionic compound breaks apart into its individual ions in water. For our particular salt, For the salt \(\mathrm{AB}_{2}\), the dissociation in water can be given by:\[ \mathrm{AB}_{2}(s) \rightleftharpoons \mathrm{A}^{2+} (aq) + 2\mathrm{B}^{-} (aq) \] A dissociation equation is helpful because:

• It shows the stoichiometry, which means the proportion of ions produced.
• The convertible equilibrium sign \(\rightleftharpoons\) indicates that the reaction can proceed forwards and backwards, meaning the reactants and products are continually interconverted.

Understanding the dissociation process allows us to predict how changes in concentration, pressure, or temperature might affect the solution.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of ions present in a solution when a reaction, such as the dissolution of a salt, reaches equilibrium. At this point, the rate of dissolution of the salt equals the rate of precipitation.

• Initial solubility: As given, the solubility of our salt \(\mathrm{AB}_{2}\) is \(1.0 \times 10^{-5} \ \mathrm{molL}^{-1}\).
• Determining concentrations: The solution equilibrates with concentrations of \([\mathrm{A}^{2+}] = s\) and \([\mathrm{B}^{-}] = 2s\).

These concentrations are then used to calculate the solubility product \(K_{sp}\). This is calculated as \[K_{sp} = [\mathrm{A}^{2+}][\mathrm{B}^{-}]^2\]By substituting the known concentrations, the \(K_{sp}\) value gives us the product of the concentrations of the ions at equilibrium, raised to the power of their coefficients in the dissociation equation. Understanding these equilibrium concentrations is invaluable to predict whether a precipitate will form under given conditions.