Problem 184 - A Complete Resource Book in Chemistry for JEE Main

Question

When equal volume of the following solutions are mixed, which of the following gives maximum precipitate? \(\left(\mathrm{K}_{\text {sp }}\right.\) of \(\mathrm{AgCl}=10^{-12}\) ) (a) \(10^{-4} \mathrm{M} \mathrm{Ag}^{+}\)and \(10^{-4} \mathrm{M} \mathrm{Cl}^{-}\) (b) \(10^{-3} \mathrm{M} \mathrm{Ag}^{+}\)and \(10^{-3} \mathrm{M} \mathrm{Cl}^{-}\) (c) \(10^{-5} \mathrm{M} \mathrm{Ag}^{+}\)and \(10^{-5} \mathrm{M} \mathrm{Cl}^{-}\) (d) \(10^{-6} \mathrm{M} \mathrm{Ag}^{+}\)and \(10^{-6} \mathrm{M} \mathrm{Cl}^{-}\)

Step-by-Step Solution

Answer

Mixture (b) gives maximum precipitate.

Step 1: Write the Precipitation Reaction

The reaction between silver ions \( \mathrm{Ag}^+ \) and chloride ions \( \mathrm{Cl}^- \) to form the precipitate silver chloride \( \mathrm{AgCl} \) is given by the following equation: \[ \mathrm{Ag}^+ (aq) + \mathrm{Cl}^- (aq) \rightarrow \mathrm{AgCl} (s) \] This precipitation occurs when the product of the concentrations of the ions exceeds the solubility product \( \mathrm{K}_{\text{sp}} \).

Step 2: Calculate the Ion Product for Precipitation

For a given solution, the ion product \( Q \) is calculated as: \[ Q = [\mathrm{Ag}^+][\mathrm{Cl}^-] \] We need to calculate \( Q \) for each combination to determine if it exceeds \( \mathrm{K}_{\text{sp}} \) and by how much.

Step 3: Apply Changes in Concentrations

When equal volumes of two solutions are mixed, the concentration of each ion in the resulting mixture is halved. Therefore, the concentrations become: - For solution (a): \( [\mathrm{Ag}^+] = 5 \times 10^{-5} \ M \) and \( [\mathrm{Cl}^-] = 5 \times 10^{-5} \ M \) - For solution (b): \( [\mathrm{Ag}^+] = 5 \times 10^{-4} \ M \) and \( [\mathrm{Cl}^-] = 5 \times 10^{-4} \ M \) - For solution (c): \( [\mathrm{Ag}^+] = 5 \times 10^{-6} \ M \) and \( [\mathrm{Cl}^-] = 5 \times 10^{-6} \ M \) - For solution (d): \( [\mathrm{Ag}^+] = 5 \times 10^{-7} \ M \) and \( [\mathrm{Cl}^-] = 5 \times 10^{-7} \ M \)

Step 4: Calculate Q for Each Mixture

Calculate \( Q \) for each mixture using the new concentrations: - For (a): \( Q = (5 \times 10^{-5})(5 \times 10^{-5}) = 2.5 \times 10^{-9} \) - For (b): \( Q = (5 \times 10^{-4})(5 \times 10^{-4}) = 2.5 \times 10^{-7} \) - For (c): \( Q = (5 \times 10^{-6})(5 \times 10^{-6}) = 2.5 \times 10^{-11} \) - For (d): \( Q = (5 \times 10^{-7})(5 \times 10^{-7}) = 2.5 \times 10^{-13} \)

Step 5: Compare Q with Ksp

\( \mathrm{K}_{\text{sp}} \) of \( \mathrm{AgCl} \) is \( 1 \times 10^{-12} \). Compare \( Q \) from each mixture:- For (a), \( Q = 2.5 \times 10^{-9} > 10^{-12} \)- For (b), \( Q = 2.5 \times 10^{-7} > 10^{-12} \)- For (c), \( Q = 2.5 \times 10^{-11} > 10^{-12} \)- For (d), \( Q = 2.5 \times 10^{-13} < 10^{-12} \)

Step 6: Determine Maximum Precipitation

Since precipitate forms when \( Q > \mathrm{K}_{\text{sp}} \), mixtures (a), (b), and (c) all form precipitate. However, (b) has the highest \( Q \) value (largest exceedance over \( \mathrm{K}_{\text{sp}} \)), indicating that it will produce the maximum precipitate.

Key Concepts

Solubility Product (Ksp)

The solubility product, often denoted as \( K_{sp} \), is a constant that represents the maximum amount of a compound that can dissolve in a solution before it starts precipitating. It is unique to each compound at a specific temperature. For the formation of silver chloride (AgCl), its \( K_{sp} \) value is \( 10^{-12} \).

The \( K_{sp} \) value helps us predict whether a precipitate will form when we mix two solutions. If the ion product \( Q \), calculated as \([\mathrm{Ag}^+][\mathrm{Cl}^-]\), exceeds \( K_{sp} \), the solution is supersaturated, resulting in the formation of a precipitate. Understanding the \( K_{sp} \) allows us to determine the conditions under which silver chloride starts separating out as a solid from its ions in solution.

Ion Concentration

Ion concentration is crucial when predicting the behavior of ions in a solution. Concentration determines how much of a substance is present in a given volume. This is typically measured in moles per liter (M).To find out if a precipitation reaction occurs, we need to look at the concentrations of the ions involved.

In a solution where a precipitation can happen, knowing the concentration of the ions gives insight into calculating \( Q \), the ion product. For example, if equal volumes of \( 10^{-3} \, M \, \mathrm{Ag}^+ \) and \( 10^{-3} \, M \, \mathrm{Cl}^- \) are mixed, the concentrations become \( 5 \times 10^{-4} \, M \) for each ion.

These adjusted concentrations are then used to find \( Q \), which reveals if the precipitate will form by comparing it to the \( K_{sp} \). Thus, calculating and understanding ion concentration allows us to anticipate the changes that occur during chemical reactions.

Silver Chloride (AgCl)

Silver chloride \( (\mathrm{AgCl}) \) is an example of a compound that forms through a precipitation reaction. When ions of silver \( (\mathrm{Ag}^+) \) and chloride \( (\mathrm{Cl}^-) \) come together in a solution, they can create solid silver chloride if their concentrations exceed the \( K_{sp} \) value.

This compound is known for its low solubility in water, which means even small amounts of \( \mathrm{Ag}^+ \) and \( \mathrm{Cl}^- \) ions can lead to precipitation. The balanced reaction for forming \( \mathrm{AgCl} \) is typically written as: \[ \mathrm{Ag}^+ (aq) + \mathrm{Cl}^- (aq) \rightarrow \mathrm{AgCl} (s) \]

Practically, this means that in an experiment, if a solution contains enough of these ions to surpass the solubility product \( (10^{-12}) \), silver chloride will begin to precipitate out of the solution as a solid. Understanding \( \mathrm{AgCl} \) formation helps in applications such as photography and analytical chemistry, where precise manipulation or observation of reactions is critical. Thus, \( \mathrm{AgCl} \) serves as an important illustration of how ionic interactions lead to observable chemical changes.