Problem 192 - A Complete Resource Book in Chemistry for JEE Main

Question

When equal volumes of the following solutions are mixed, the precipitation of \(\mathrm{AgCl}\left(\mathrm{K}_{s p}=1.8 \times 10^{-10}\right)\) will occur with (a) \(10^{-5} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-3} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (b) \(10^{-4} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-4} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (c) \(10^{-5} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-1} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (d) \(10^{-5} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-5} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\)

Step-by-Step Solution

Answer

Precipitation occurs in scenarios (a), (b), and (c).

Step 1: Identify Scenario for Q Calculation

When mixing equal volumes of solutions, the concentration of each ion will be halved. For each pair, calculate the final concentrations of \( \text{Ag}^+ \) and \( \text{Cl}^- \) ions once mixed.

Step 2: Calculate Ion Product Q

Calculate the ion product \( Q \) for each pair using \( Q = [\text{Ag}^+][\text{Cl}^-] \). Be sure to use the halved concentrations from Step 1.

Step 3: Scenario (a) Analysis

For (a), solutions mixed will have \( [\text{Ag}^+] = 5 \times 10^{-6} \) M and \( [\text{Cl}^-] = 5 \times 10^{-4} \) M. Calculate \( Q: Q = (5 \times 10^{-6})(5 \times 10^{-4}) = 2.5 \times 10^{-9} \).

Step 4: Scenario (b) Analysis

For (b), solutions mixed will have \( [\text{Ag}^+] = 5 \times 10^{-5} \) M and \( [\text{Cl}^-] = 5 \times 10^{-5} \) M. Calculate \( Q: Q = (5 \times 10^{-5})(5 \times 10^{-5}) = 2.5 \times 10^{-9} \).

Step 5: Scenario (c) Analysis

For (c), solutions mixed will have \( [\text{Ag}^+] = 5 \times 10^{-6} \) M and \( [\text{Cl}^-] = 5 \times 10^{-2} \) M. Calculate \( Q: Q = (5 \times 10^{-6})(5 \times 10^{-2}) = 2.5 \times 10^{-7} \).

Step 6: Scenario (d) Analysis

For (d), solutions mixed will have \( [\text{Ag}^+] = 5 \times 10^{-6} \) M and \( [\text{Cl}^-] = 5 \times 10^{-6} \) M. Calculate \( Q: Q = (5 \times 10^{-6})(5 \times 10^{-6}) = 2.5 \times 10^{-11} \).

Step 3: Compare Q with Ksp

Compare the calculated \( Q \) to the \( K_{sp} \) for \( \text{AgCl} \), which is \( 1.8 \times 10^{-10} \). Precipitation occurs if \( Q > K_{sp} \).

Step 8: Precipitation Analysis

From previous steps: \( Q = 2.5 \times 10^{-9} \) for (a) and (b), \( 2.5 \times 10^{-7} \) for (c), and \( 2.5 \times 10^{-11} \) for (d). In (a), (b), and (c), \( Q > K_{sp} \). Thus, all these scenarios will result in precipitation.

Key Concepts

Precipitation Reaction

When certain ions in a solution come together to form an insoluble compound, it results in what is known as a precipitation reaction. It often happens when two aqueous solutions containing soluble salts are mixed, and their ions interact to form a solid precipitate.
In the case of Silver Chloride (AgCl), a classic example of precipitation occurs. When solutions containing silver ions (Ag^+) and chloride ions (Cl^-) are mixed, they may form a white, insoluble product: AgCl.
Understanding precipitation reactions is crucial in predicting whether a particular mixture will create a precipitate based on the solubility rules and product thresholds.

Ion Product Q

The ion product (Q) is a helpful concept used to predict whether a precipitate will form when two ionic solutions are mixed. Q is calculated as:

• Q = [Ag^+] BCl^-]

When mixing solutions, the concentration of each ion can change, which influences Q. In essence, Q compares to the solubility product constant (K_{sp}) of a substance to decide on precipitation.

• If Q > K_{sp}, a precipitate forms.
• If Q = K_{sp}, the solution is at equilibrium (saturated).
• If Q < K_{sp}, no precipitate forms as the solution is unsaturated.

This simple comparison helps determine the likelihood of a precipitation reaction.

Chemical Equilibrium

In chemistry, equilibrium refers to a state where the concentrations of all reactants and products remain constant over time. For precipitation reactions, when a solution reaches its solubility limit, an equilibrium is established between the dissolved ions and the precipitate.
This concept of equilibrium is reflected in the solubility product constant (K_{sp}), a specific type of equilibrium constant applied to sparingly soluble compounds. K_{sp} helps in predicting whether a salt will precipitate out from the solution, especially when mixed with other ionic solutions.
It is important to remember that being at chemical equilibrium does not necessarily mean that the concentrations are equal, but rather that the process of forward (dissolution) and reverse (precipitation) reactions occur at the same rate.

Silver Chloride (AgCl)

Silver Chloride, with its chemical formula AgCl, is quite famous for its low solubility in water, making it a common subject in studies of solubility and precipitation. AgCl forms a white precipitate when its ions are mixed in solution, provided that the Q value exceeds its K_{sp} value of 0^{-10}.
Silver Chloride precipitates are also involved in a significant number of laboratory experiments. Understanding its precipitation reactions can lend insight into broader analytical processes such as qualitative analysis of mixtures.
Moreover, the behavior of AgCl can be explored by experimenting with factors like ionic strength, temperature, and the presence of complexing agents, which can all alter its properties and solubility.