Problem 89 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility of \(\mathrm{AgCl}\) in \(0.2 \mathrm{M} \mathrm{NaCl}\) is \([\mathrm{Ksp} \mathrm{AgCl}=\) \(\left.1.8 \times 10^{-10}\right]\) (a) \(1.8 \times 10^{-11} \mathrm{M}\) (b) \(9 \times 10^{-10} \mathrm{M}\) (c) \(6.5 \times 10^{-12} \mathrm{M}\) (d) \(5.6 \times 10^{-11} \mathrm{M}\)

Step-by-Step Solution

Answer

The solubility of \( \mathrm{AgCl} \) in 0.2 M \( \mathrm{NaCl} \) is \( 9 \times 10^{-10} \mathrm{M} \). Option (b).

Step 1: Understand the Concept

The solubility of a compound in a solution already containing one of its ions is affected by the common ion effect. Here, the common ion is chloride source from NaCl, which already contributes 0.2 M to the chloride concentration.

Step 2: Set Up the Equilibrium Expression

The dissolution of AgCl can be described by the equation: \[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]The solubility product, \( K_{sp} \), is given by \[ K_{sp} = [\text{Ag}^+] \times [\text{Cl}^-] \]\[ K_{sp} = 1.8 \times 10^{-10} \]

Step 3: Apply the Common Ion Effect

Since the solution already contains 0.2 M Cl⁻ from NaCl, the equilibrium expression becomes \[ K_{sp} = s \times (0.2 + s) \]where \( s \) is the solubility of AgCl in this solution.

Step 4: Simplify the Equation

Because \( s \) is very small compared to 0.2 M, we can approximate the equation as:\[ K_{sp} = s \times 0.2 \]\( s \) is the solubility of AgCl in the solution.

Step 5: Solve for s

Rearrange the equation to find \( s \):\[ s = \frac{K_{sp}}{0.2} \]\[ s = \frac{1.8 \times 10^{-10}}{0.2} = 9 \times 10^{-10} \text{ M} \]

Step 6: Select the Correct Answer

Compare the calculated solubility with the given options: (a) \( 1.8 \times 10^{-11} \mathrm{M} \), (b) \( 9 \times 10^{-10} \mathrm{M} \), (c) \( 6.5 \times 10^{-12} \mathrm{M} \), (d) \( 5.6 \times 10^{-11} \mathrm{M} \).The correct option is (b) \( 9 \times 10^{-10} \mathrm{M} \).

Key Concepts

Understanding the Common Ion Effect

The common ion effect plays a vital role when considering the solubility of ionic compounds. It occurs when a solution already contains an ion that is part of the dissociation of another compound being added.
In our given example, we have a solution of AgCl, which dissociates into Ag⁺ and Cl⁻ ions.
However, the solution originally contains NaCl, providing an additional source of Cl⁻ ions. Thus, the concentration of Cl⁻ already in the solution affects the equilibrium by reducing the solubility of AgCl. - This is because the system reaches equilibrium at lower concentrations of dissolved Ag⁺ and Cl⁻ than it would in pure water. - As a result, the presence of a common ion, Cl⁻ from NaCl, decreases the solubility of AgCl. Ultimately, the common ion effect is why AgCl is less soluble in NaCl than in a pure solvent.

Exploring Ksp (Solubility Product Constant)

The solubility product constant, Ksp, is essential for determining how much of a compound can dissolve in a solution before it reaches equilibrium.
Ksp is the product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced equation.For AgCl, which dissociates as Ag⁺ and Cl⁻, Ksp is calculated using the equation:\[ K_{sp} = [\text{Ag}^+] \times [\text{Cl}^-] \]- The Ksp value indicates the degree of solubility for a sparingly soluble compound.
- Lower Ksp values suggest the compound does not dissolve well, as is the case with AgCl.In our problem, the Ksp of AgCl is given as \( 1.8 \times 10^{-10} \), meaning under normal conditions, very little AgCl dissolves into the solution. By knowing the Ksp, we assess how much AgCl dissolves in the presence of the common ion and ultimately solve for its solubility.

The Role of Dissolution Equilibrium

Dissolution equilibrium is a state where a solid phase is in balance with its dissolved ions in a solution. It involves constant dissolution and precipitation of the substance.
When a compound like AgCl is added to a solvent, it begins to dissolve.
However, it also reaches a point where the rate of dissolution equals the rate of precipitation.For AgCl, the equilibrium can be illustrated by:\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]At this point, we apply the solubility product constant, Ksp, to maintain the balance of ions in the solution.- The presence of additional ions from NaCl alters this balance because the solution now has more Cl⁻ ions than it would otherwise.The system responds by dissolving less AgCl to maintain equilibrium, thus exemplifying dissolution equilibrium. Understanding this concept helps predict the behavior of ionic compounds in different environments and solve solubility problems by considering both the compound's nature and the influence of common ions.