Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

The solubility product of \(\mathrm{AgI}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-16}\) \(\mathrm{mol}^{2} \mathrm{~L}^{-2}\). The solubility of \(\mathrm{AgI}\) in \(10^{-4} \mathrm{~N}\) solution of \(\mathrm{KI}\) at \(25^{\circ} \mathrm{C}\) is \(\left(\right.\) in \(\left.\mathrm{mol} \mathrm{L}^{-1}\right)\) (a) \(1.0 \times 10^{-10}\) (b) \(1.0 \times 10^{-8}\) (c) \(1.0 \times 10^{-16}\) (d) \(1.0 \times 10^{-12}\)

Step-by-Step Solution

Answer

The solubility of \( \mathrm{AgI} \) is \( 1.0 \times 10^{-12} \mathrm{~mol~L}^{-1} \), option (d).

Step 1: Understanding the problem

The problem is asking for the solubility of \( \mathrm{AgI} \) in a \( 10^{-4} \mathrm{~N} \) solution of \( \mathrm{KI} \) at \( 25^{\circ} \mathrm{C} \). The presence of \( \mathrm{KI} \) affects the solubility due to the common ion effect of \( \mathrm{I^-} \). Given the solubility product \( K_{sp} = 1.0 \times 10^{-16} \).

Step 2: Setting up the expression for solubility product

For \( \mathrm{AgI} \), the dissolution can be represented by the equation:\[ \mathrm{AgI (s) \rightleftharpoons Ag^+ (aq) + I^- (aq)} \]The solubility product expression is:\[ K_{sp} = [\mathrm{Ag^+}][\mathrm{I^-}] \]

Step 3: Determining ion concentrations due to common ion effect

In a \( 10^{-4} \mathrm{~N} \) solution of \( \mathrm{KI} \), the initial concentration of \( \mathrm{I^-} \) is \( 10^{-4} \mathrm{~mol~L}^{-1} \). This is the effect of the common ion present in the solution.

Step 4: Solving for Ag+ concentration

Assume that the solubility of \( \mathrm{AgI} \) is \( s \) mol/L. Then, at equilibrium, \([\mathrm{Ag^+}] = s\) and \([\mathrm{I^-}] = 10^{-4} + s\). For dilute solutions, \( s \ll 10^{-4} \), thus \([\mathrm{I^-}] \approx 10^{-4} \).

Step 5: Substituting into Ksp expression

Substitute into the solubility product expression:\[ 1.0 \times 10^{-16} = s \times 10^{-4} \]Solve for \( s \):\[ s = \frac{1.0 \times 10^{-16}}{10^{-4}} = 1.0 \times 10^{-12} \]

Step 6: Selecting the correct option

The calculated solubility \( s = 1.0 \times 10^{-12} \mathrm{~mol~L}^{-1} \) corresponds to option (d) \( 1.0 \times 10^{-12} \).

Key Concepts

Common Ion Effect

When a salt like silver iodide (AgI) is dissolved in a solution that already contains one of its ions, such as iodide (I^-), its solubility decreases. This phenomenon is known as the **common ion effect**. It occurs because adding an ion common to the dissolving compound shifts the equilibrium position of the dissolution reaction. This shift results in a reduced concentration of the other ion in the solution, in this case, Ag^+.

The common ion effect is particularly significant in solutions where the solute has a very low solubility in pure water. When more of the iodide ion is present from a compound like potassium iodide (KI), the excess I^- makes it difficult for additional AgI to dissolve since the product of the concentrations of Ag^+ and I^- ions has to equal the solubility product constant (Ksp) for equilibrium.

Ionic Equilibrium

Ionic equilibrium involves a balance between the ions in a dissolved solution, which establishes a state of equilibrium between the dissolved phase and the undissolved phase. In the case of silver iodide solubility, the equilibrium can be represented by:

• The dissolution reaction of AgI:
  1. \[\mathrm{AgI (s) \rightleftharpoons Ag^+ (aq) + I^- (aq)}\]

This equation implies that at equilibrium the rate of dissolution (\rightarrow) equals the rate of precipitation (\leftarrow), leading to a stable concentration of ions in the solution.

To maintain this equilibrium, the concentrations of the ions must satisfy the relationship given by the solubility product (Ksp). If the ionic product exceeds Ksp, precipitation occurs, thereby decreasing the concentration of the ions in the solution back to equilibrium values.

Silver Iodide Solubility

The solubility of silver iodide is affected by the presence of other ions in the solution, as demonstrated in a practical experiment with KI.

In a solution containing KI, the initial concentration of I^- is already established at 10^{-4} mol/L due to the complete dissociation of KI. This existing high concentration of the iodide ion reduces the solubility of AgI, which is calculated using the formula:

• Solving for AgI solubility when [I^-] = 10^{-4} mol/L:
  1. Given \[ K_{sp} = [\mathrm{Ag^+}][\mathrm{I^-}] = 1.0 \times 10^{-16} \]
  2. Assume solubility of AgI is s. Hence, \[ s \times 10^{-4} = 1.0 \times 10^{-16} \]
  3. Solve: \[ s = \frac{1.0 \times 10^{-16}}{10^{-4}} = 1.0 \times 10^{-12} \] mol/L.

Thus, the solubility of AgI in the presence of a common ion is much lower than it would be in pure water.

Chemistry Problem-Solving

Chemistry problem-solving often requires translating word problems into mathematical equations and applying known principles, like the common ion effect and equilibrium, to find solutions.

When solving these types of problems:

• **Identify the given data:** Understand the relevance of each piece of information—like the Ksp value and ion concentrations.
• **Set up the equation:** Write the expression for the solubility product and relate it to the given concentrations.
• **Assumptions for calculations:** Assume negligible changes in ion concentration when reasonable (e.g., s is negligible compared to initial I- concentration).
• **Solve and verify:** Calculate the unknowns, then verify by checking if your solution respects all chemical principles involved.

Breaking down the problem into these manageable steps leads to accurate and confident solutions in chemistry.