Problem 115 - A Complete Resource Book in Chemistry for JEE Main

Question

Equal volumes of the following \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\)solutions are mixed. In which of the solutions will precipitation occurs? \(\left[\mathrm{Ksp}\right.\) of \(\left.\mathrm{CaF}_{2}=1.7 \times 10^{-10}\right]\) 1\. \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-5} \mathrm{M} \mathrm{F}^{-}\) 2\. \(10^{-3} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{M} \mathrm{F}^{-}\) 3\. \(10^{-4} \mathrm{M} \mathrm{Ca}^{2+}+10^{-2} \mathrm{M} \mathrm{F}^{-}\) 4\. \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{M} \mathrm{F}^{-}\) Select the correct answer using the codes given below: (a) in 4 only (b) in 1 and 2 (c) in 3 and 4 (d) in 2,3 and 4

Step-by-Step Solution

Answer

Precipitation does not occur in any of the solutions.

Step 1: Understand the Problem

We need to determine in which mixtures of calcium ions (\(\text{Ca}^{2+}\)) and fluoride ions (\(\text{F}^-\)) precipitation of calcium fluoride (\(\text{CaF}_2\)) occurs. The solubility product \(\left(K_{sp}\right)\) for \(\text{CaF}_2\)) is given as \(1.7 \times 10^{-10}\).

Step 2: Recognize the Precipitation Condition

Precipitation occurs if the ionic product \(Q\) of \(\text{CaF}_2\) is greater than the solubility product \(K_{sp}\). The equation for the ionic product of \(\text{CaF}_2\) is: \( Q = [\text{Ca}^{2+}][\text{F}^-]^2 \).

Step 3: Calculate Ionic Product for Each Mixture

For a 1:1 mixture, the concentrations of \(\text{Ca}^{2+}\) and \(\text{F}^-\) become half of their initial values because they mix in equal volumes.1. \([\text{Ca}^{2+}] = 5 \times 10^{-3} \text{ M}\), \([\text{F}^-] = 5 \times 10^{-6} \text{ M}\)\(Q = 5 \times 10^{-3} \times (5 \times 10^{-6})^2 = 1.25 \times 10^{-13}\)2. \([\text{Ca}^{2+}] = 5 \times 10^{-4} \text{ M}\), \([\text{F}^-] = 5 \times 10^{-4} \text{ M}\)\(Q = 5 \times 10^{-4} \times (5 \times 10^{-4})^2 = 1.25 \times 10^{-10}\)3. \([\text{Ca}^{2+}] = 5 \times 10^{-5} \text{ M}\), \([\text{F}^-] = 5 \times 10^{-3} \text{ M}\)\(Q = 5 \times 10^{-5} \times (5 \times 10^{-3})^2 = 1.25 \times 10^{-10}\)4. \([\text{Ca}^{2+}] = 5 \times 10^{-3} \text{ M}\), \([\text{F}^-] = 5 \times 10^{-4} \text{ M}\)\(Q = 5 \times 10^{-3} \times (5 \times 10^{-4})^2 = 1.25 \times 10^{-10}\)

Step 4: Compare Ionic Products with Ksp

Compare each calculated ionic product \(Q\) with the given \(K_{sp} = 1.7 \times 10^{-10}\):1. \(Q = 1.25 \times 10^{-13} < 1.7 \times 10^{-10}\) (No precipitation)2. \(Q = 1.25 \times 10^{-10} < 1.7 \times 10^{-10}\) (No precipitation)3. \(Q = 1.25 \times 10^{-10} < 1.7 \times 10^{-10}\) (No precipitation)4. \(Q = 1.25 \times 10^{-10} < 1.7 \times 10^{-10}\) (No precipitation)

Step 5: Synthesize Results

Check each scenario, noting that in none of the solutions the ionic product \(Q\) exceeds \(K_{sp}\). Therefore, precipitation does not occur in any of the given solutions.

Key Concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, commonly referred to as \( K_{sp} \), is a fundamental concept in understanding chemical equilibrium, particularly in the realm of slightly soluble salts. \( K_{sp} \) is defined as the product of the concentrations of the ions of a particular salt in a saturated solution, each raised to the power of their stoichiometric coefficients. For instance, for calcium fluoride, CaF\(_2\), which dissociates into Ca\(^{2+}\) and two F\(^{-}\) ions, the solubility product expression is given by:

• \( K_{sp} = [Ca^{2+}][F^-]^2 \)

This value is constant at a given temperature and is unique for each ionic compound.
Understanding \( K_{sp} \) allows us to predict whether a precipitate will form under certain conditions. If the concentration of ions in a solution exceeds the \( K_{sp} \), the solution becomes supersaturated, and the excess ions will typically precipitate out as a solid, restoring equilibrium.
Conversely, if the concentration of ions is less than \( K_{sp} \), no precipitation will occur, and the solution remains unsaturated.

Precipitation Reactions

Precipitation reactions are crucial for understanding how solubility and chemical equilibrium work together. These reactions occur when two soluble salts in aqueous solutions are mixed, resulting in the formation of an insoluble compound known as a precipitate. This happens when the product of the ionic concentrations in solution exceeds the compound’s \( K_{sp} \). This critical point marks the transition from dissolved ions to solid precipitate.
Checking whether a precipitation reaction will occur involves calculating the ionic product \( Q \) when the solutions are mixed. If \( Q \) is greater than the \( K_{sp} \) of the compound in question, a precipitate will form as the solution seeks to reduce the excess ions.
Precipitation is an important process in many natural and industrial applications, including water purification, mineral extraction, and even in biological systems where certain minerals need to be kept in solution or precipitated out as needed.

Ionic Product

The concept of ionic product \( Q \) is pivotal in determining the course of precipitation reactions in chemistry. It serves as a snapshot of the current state of a solution, reflecting the product of the molar concentrations of the ions at any point in time, each raised to their respective powers as given by the reaction equation.
The ionic product \( Q \) is calculated using the initial concentrations of the ions when two solutions are mixed, specifically in equal volumes without assuming any precipitation has occurred:

• \( Q = [Ca^{2+}][F^-]^2 \)

By comparing \( Q \) with \( K_{sp} \), we can foresee if the conditions favor the precipitation of the solute. If \( Q \) surpasses \( K_{sp} \), the system is beyond its saturation point, leading to precipitation as the solution tries to return to its stable state.
However, if \( Q \) is less than or equal to \( K_{sp} \), the solution can accommodate more ions without forming a precipitate, remaining either saturated or unsaturated. This comparison is vital for predicting and controlling reactions in laboratory and real-world settings, ensuring precise manipulation of concentrations in solutions.