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 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 is \(1.8 \times 10^{-8} \mathrm{M}\). Option (a).

Step 1: Understand the Concept of Solubility Product

The solubility product, denoted as \( K_{sp} \), is a constant for a given compound at a specific temperature. For \( \mathrm{AgCl} \), the equation at equilibrium is \( \mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^+(aq) + \mathrm{Cl}^-(aq) \). The solubility product is expressed as \( K_{sp} = [\mathrm{Ag}^+][\mathrm{Cl}^-] \). This implies that the product of the concentrations of the dissolved ions equals the \( K_{sp} \) value.

Step 2: Set Up the Ionization Equilibrium in the Presence of Common Ion

In a \(0.01 \mathrm{M} \mathrm{HCl}\) solution, \( \mathrm{HCl} \) fully dissociates to give \( \mathrm{Cl}^- \) ions, affecting the solubility of \( \mathrm{AgCl} \). Let \( s \) be the solubility of \( \mathrm{AgCl} \) in \(0.01 \mathrm{M} \mathrm{HCl}\), thus the concentration of \( \mathrm{Ag}^+ \) is \( s \) and \( \mathrm{Cl}^- \) concentration becomes \( s + 0.01 \).

Step 3: Use the Solubility Product Expression

Since \( K_{sp} = [\mathrm{Ag}^+][\mathrm{Cl}^-] \), we substitute the equilibrium concentrations into the expression: \[K_{sp} = s(s + 0.01)\] Given that \( K_{sp} = 1.8 \times 10^{-10} \).

Step 4: Assume Simplification for Small Solubility

Assume that \( s \), the solubility of \( \mathrm{AgCl} \), is much less than \( 0.01 \), so \( s + 0.01 \approx 0.01 \). Substitute to get: \[1.8 \times 10^{-10} = s \times 0.01\] which gives \[s = \frac{1.8 \times 10^{-10}}{0.01}\]

Step 5: Calculate the Solubility

Solving the equation gives: \[s = 1.8 \times 10^{-8} \]Thus, the solubility of \( \mathrm{AgCl} \) in \(0.01 \mathrm{M} \mathrm{HCl}\) is \( 1.8 \times 10^{-8} \mathrm{M}\).

Key Concepts

Understanding AgCl

Silver chloride, denoted as \( \mathrm{AgCl} \), is a white crystalline compound well-known for its low solubility in water. The process of dissolving \( \mathrm{AgCl} \) in water is represented by the dissociation into its ions: \( \mathrm{AgCl(s)} \rightleftharpoons \mathrm{Ag}^+(aq) \) and \( \mathrm{Cl}^-(aq) \). This equilibrium is extremely crucial, as it defines the solubility of \( \mathrm{AgCl} \), meaning how much \( \mathrm{AgCl} \) can dissolve into the solution before it reaches a point where no more can be dissolved.

In terms of solubility product, or \( K_{sp} \), this is expressed as the product of the concentrations of silver, \( \mathrm{Ag}^+ \), and chloride ions, \( \mathrm{Cl}^- \). For \( \mathrm{AgCl} \), the \( K_{sp} \) is a constant value \( 1.8 \times 10^{-10} \) at 298 K, which helps determine how much \( \mathrm{AgCl} \) dissolves under certain conditions.

Role of HCl Solution

Hydrochloric acid, noted as \( \mathrm{HCl} \), plays a unique role in solubility reactions due to its dissociation into hydrogen ions \( \mathrm{H}^+(aq) \) and chloride ions \( \mathrm{Cl}^-(aq) \) when in solution. It is highly soluble and provides a common source of chloride ions in the solution.

When \( \mathrm{AgCl} \) is placed in an \( 0.01 \mathrm{M} \mathrm{HCl} \) solution, the chloride ions from \( \mathrm{HCl} \) affects the equilibrium of \( \mathrm{AgCl} \). This increase in chloride ions in the solution shifts the equilibrium, reducing the solubility of \( \mathrm{AgCl} \). This happens because additional chloride ions from the \( \mathrm{HCl} \) already satisfy part of the equilibrium requirement, thus less \( \mathrm{AgCl} \) needs to dissolve to reach the \( K_{sp} \).

Influence of Common Ion Effect

The common ion effect is a fundamental concept in understanding solubility equilibria. It describes the phenomenon where the solubility of a salt is reduced in a solution that already contains one of its ions.

In the case of \( \mathrm{AgCl} \), the "common ion" is \( \mathrm{Cl}^- \), introduced into the solution through the dissociation of \( \mathrm{HCl} \). This addition of chloride ions decreases the solubility of \( \mathrm{AgCl} \) because the additional \( \mathrm{Cl}^- \) ions shift the equilibrium towards the undissolved \( \mathrm{AgCl} \), according to Le Chatelier's principle. Consequently, the solubility product balance is maintained, and less silver chloride dissolves.

Equilibrium Concentrations

Equilibrium concentration refers to the concentrations of ions in a solution when the dissolution or precipitation of a compound has reached equilibrium. For \( \mathrm{AgCl} \) in a solution containing \( 0.01 \mathrm{M} \mathrm{HCl} \), understanding these concentrations is key to determining solubility in the presence of a common ion.

Given \( K_{sp} = [\mathrm{Ag}^+][\mathrm{Cl}^-] \), with \( [\mathrm{Ag}^+] = s \) (the solubility of \( \mathrm{AgCl} \)) and \( [\mathrm{Cl}^-] = s + 0.01 \approx 0.01 \), the equation simplifies under the assumption that \( s \) is very small relative to 0.01. The calculated \( s \) results directly from solving the simplified equilibrium concentration equation: \[1.8 \times 10^{-10} = s \times 0.01\]. This results in a solubility \( s \) of \( 1.8 \times 10^{-8} \mathrm{M} \), indicating the minimal amount of \( \mathrm{AgCl} \) that will dissolve under these conditions.