Problem 207 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility product of \(\mathrm{AgCl}\) is \(1.8 \times 10^{-10}\) at \(298 \mathrm{~K}\). The solubility of \(\mathrm{AgCl}\) in \(0.01 \mathrm{M} \mathrm{HCl}\) solution is (a) \(1.8 \times 10^{-8} \mathrm{M}\) (b) \(1.4 \times 10^{-6} \mathrm{M}\) (c) \(1.8 \times 10^{-6} \mathrm{M}\) (d) \(1.4 \times 10^{-4} \mathrm{M}\)

Step-by-Step Solution

Answer

The solubility of AgCl in 0.01 M HCl is \(1.8 \times 10^{-8}\) M, so option (a) is correct.

Step 1: Write the Dissociation Equation

The dissociation of AgCl in water is represented by the equation:\[ \text{AgCl} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

Step 2: Introduce the Solubility Expression

The solubility product constant (\(K_{sp}\)) for AgCl is given by:\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]Given \(K_{sp}\) is \(1.8 \times 10^{-10}\).

Step 3: Consider Effect of Common Ion (Cl-) from HCl

In a 0.01 M HCl solution, the chloride ion concentration \([\text{Cl}^-]\) is approximately 0.01 M because HCl fully dissociates in water. This affects the solubility of AgCl.

Step 4: Set Up Expression for [Ag⁺]

Let \(s\) be the solubility of AgCl in M. In the presence of 0.01 M Cl⁻, the concentration of Cl⁻ is primarily from HCl. Thus:\[ K_{sp} = [\text{Ag}^+] (0.01) \]\[ 1.8 \times 10^{-10} = s \times 0.01 \]

Step 5: Solve for Solubility (s)

Rearrange the equation to solve for \(s\):\[ s = \frac{1.8 \times 10^{-10}}{0.01} = 1.8 \times 10^{-8} \text{ M} \]

Step 6: Compare and Choose the Correct Option

The solubility of AgCl in 0.01 M HCl is \(1.8 \times 10^{-8}\) M. Therefore, option (a) is the correct answer.

Key Concepts

Common Ion Effect

The Common Ion Effect refers to the phenomenon where the solubility of a salt is reduced because of the presence of a common ion in the solution. In this context, the common ion is usually provided by another dissolved substance, affecting the equilibrium of the dissociation reaction.
This is observed, for instance, when you add a salt like sodium chloride (NaCl) to a solution already containing chloride ions (Cl⁻). The additional chloride ions shift the equilibrium of a sparingly soluble salt containing chloride ions, like silver chloride (AgCl), backwards, thus reducing its solubility.

• This shift occurs because the salt already contains the chloride ion, and its common presence in solution decreases the extent to which the salt can dissociate.
• In practice, this means we can predict how much less a salt will dissolve in a solution with a common ion compared to pure water.

In the exercise provided, AgCl was less soluble in the presence of HCl due to the chloride ions introduced by HCl, illustrating this effect.

Dissociation Equation

A Dissociation Equation helps us understand how a compound breaks into its ions in a solution. This is critical for calculating the solubility and the solubility product (Ksp).
For example, sodium chloride (NaCl) dissociates in water into Na⁺ and Cl⁻ ions. Similarly, the dissociation of AgCl in water can be represented by the following equation:\[ \text{AgCl} (s) \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

• When writing these equations, ensure the conservation of mass and charge.
• The double arrow indicates a reversible reaction, establishing the concept of dynamic equilibrium.

This equation shows that solid AgCl can dissolve to form Ag⁺ and Cl⁻ ions, and the equation is foundational for calculating the solubility product.

Solubility in Solutions

Solubility in Solutions determines how much of a solute can be dissolved in a solvent at a given temperature. It's important to understand that solubility is not just a property of the solute or solvent but also often affected by other ions, as seen with the common ion effect.
In solutions containing additional ions, like HCl, solubility can be altered significantly. Let's break it down:

• In pure water, AgCl dissociates to a certain extent until it reaches its Ksp.
• In 0.01 M HCl, due to additional common chloride ions, the solubility of AgCl decreases.

The knowledge of how different ions and concentrations influence solubility helps us predict outcomes when mixing various substances.

Ksp Calculations

Ksp Calculations are essential for determining the solubility of sparingly soluble salts in a solution. The solubility product constant (Ksp) quantifies the extent to which a compound can dissolve, and it is specific to each compound at a given temperature.
For AgCl, with a known Ksp of 1.8 \times 10^{-10}, we use the following expression:\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]

• This equation is derived from the balanced dissociation equation and represents the concentrations of the ions in solution.
• To solve for the solubility in the presence of additional ions (like in 0.01 M HCl), the concentration of the common ion (Cl^-) is included to find [Ag^+].

Mathematically solving for the solubility (s) uses this principle: \[ s = \frac{K_{sp}}{\text{Common Ion Concentration}} = \frac{1.8 \times 10^{-10}}{0.01} = 1.8 \times 10^{-8} \text{ M} \]These calculations provide a quantitative measure of solubility under specific conditions, allowing for practical applications in labs and industry.