Problem 30 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

When equal volumes of the following solutions are mixed, precipitation of \(\mathrm{AgCl}\left(\mathrm{K}_{\mathrm{sp}}=1.8 \times 10^{-10}\right)\) will occur only with (a) \(10^{-4} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-4} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (b) \(10^{-5} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-5} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (c) \(10^{-6} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-6} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (d) \(10^{-10} \mathrm{M}\left(\mathrm{Ag}^{+}\right)\)and \(10^{-10} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\)

Step-by-Step Solution

Answer

Precipitation occurs only with option (a) 10⁻⁴ M [Ag⁺] and 10⁻⁴ M [Cl⁻].

Step 1: Understand the Concept of Precipitation

Precipitation occurs when the product of the concentrations of ions in a solution exceeds the solubility product constant (\( K_{sp} \)). For AgCl, the ionic concentrations must satisfy the condition \[ [Ag^+][Cl^-] > K_{sp} \] for precipitation to occur.

Step 2: Calculate Ion Concentration after Mixing

When equal volumes of two solutions are mixed, their concentrations are halved. Therefore, after mixing solutions with initial concentrations \( x \), new concentrations will be \( x/2 \). Hence, the mixed concentrations will be:(a) \( [Ag^+] = 5 \times 10^{-5} \, M \) and \( [Cl^-] = 5 \times 10^{-5} \, M \) (b) \( [Ag^+] = 5 \times 10^{-6} \, M \) and \( [Cl^-] = 5 \times 10^{-6} \, M \) (c) \( [Ag^+] = 5 \times 10^{-7} \, M \) and \( [Cl^-] = 5 \times 10^{-7} \, M \) (d) \( [Ag^+] = 5 \times 10^{-11} \, M \) and \( [Cl^-] = 5 \times 10^{-11} \, M \).

Step 3: Compute Ion Product for Mixed Solutions

Calculate the ion product for each pair to determine if they exceed the \( K_{sp} \).(a) Ion Product = \( (5 \times 10^{-5}) \times (5 \times 10^{-5}) = 2.5 \times 10^{-9} \) (b) Ion Product = \( (5 \times 10^{-6}) \times (5 \times 10^{-6}) = 2.5 \times 10^{-11} \) (c) Ion Product = \( (5 \times 10^{-7}) \times (5 \times 10^{-7}) = 2.5 \times 10^{-13} \) (d) Ion Product = \( (5 \times 10^{-11}) \times (5 \times 10^{-11}) = 2.5 \times 10^{-21} \).

Step 4: Compare Ion Products with Ksp

Compare each ion product with the given \( K_{sp} \) of \( 1.8 \times 10^{-10} \):(a) Ion product \(2.5 \times 10^{-9} > K_{sp} = 1.8 \times 10^{-10} \) (Precipitation occurs) (b) Ion product \(2.5 \times 10^{-11} < K_{sp} \) (c) Ion product \(2.5 \times 10^{-13} < K_{sp} \) (d) Ion product \(2.5 \times 10^{-21} < K_{sp} \).Therefore, precipitation occurs only in option (a).

Key Concepts

Solubility Product Constant

The solubility product constant, often symbolized as \( K_{sp} \), is a crucial concept in precipitation chemistry. It helps in predicting whether a salt will precipitate from a solution under certain conditions. The \( K_{sp} \) is specific for each ionic compound and represents the maximum product of ion concentrations that can exist together in a solution without forming a precipitate.
For example, for silver chloride (AgCl), the \( K_{sp} \) is given as \( 1.8 \times 10^{-10} \). This value indicates how saturated the solution can get before the salt begins to form a solid.

• A higher \( K_{sp} \) means the compound is more soluble in water.
• A lower \( K_{sp} \) indicates that it is less soluble and more likely to precipitate out of the solution.

Understanding and using \( K_{sp} \) allows chemists to control and predict precipitation reactions, which is particularly useful in fields like analytical chemistry and environmental science.

Ionic Concentration

Ionic concentration refers to the abundance of individual ions in a solution. When two solutions of ionic compounds are mixed, like solutions of \( \text{Ag}^+ \) and \( \text{Cl}^- \) ions, their concentrations change based on their initial amounts and how they are combined.
In the given exercise, the initial concentrations are halved after mixing equal volumes. This is because the number of ions available is now distributed over a greater volume.

• Initial concentration of \( \text{Ag}^+ \): \( 10^{-4} \text{ M} \)
• Initial concentration of \( \text{Cl}^- \): \( 10^{-4} \text{ M} \)
• After mixing these become \( 5 \times 10^{-5} \text{ M} \) each

Knowing the concentration of each ion post-mixing lets us calculate the ion product, which is a decisive factor in predicting precipitation.

Ion Product Calculation

Ion product calculation is a straightforward method used to predict whether a precipitation will occur when two ionic solutions are mixed. It involves multiplying the concentrations of the ions present in the solution.
For instance, if equal volumes of two solutions with ions \( [Ag^+] \) and \( [Cl^-] \) are mixed:

• Post-mixing concentrations are \( 5 \times 10^{-5} \text{ M} \) each.
• The ion product is \((5 \times 10^{-5}) \times (5 \times 10^{-5}) = 2.5 \times 10^{-9}\)

The ion product is then compared to the solubility product constant, \( K_{sp} \).
A higher ion product than \( K_{sp} \) indicates that the solution is supersaturated, and precipitation will occur. On the other hand, if the ion product is less than \( K_{sp} \), the solution remains undersaturated, and no precipitation happens. This calculation method is essential in predicting and controlling reactions leading to precipitation in various chemical processes.