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

Option (d) in 2, 3, and 4 is correct; however, with deeper analysis, no precipitation occurs.

Step 1: Initial Concentrations After Mixing

Since we are mixing equal volumes of two solutions, each concentration is halved. After mixing, the concentration of \( \mathrm{Ca}^{2+} \) ions will become half of its original concentration as well as \( \mathrm{F}^{-} \) ions. For example, for the solution 1, \([\mathrm{Ca}^{2+}] = \frac{10^{-2}}{2}= 5 \times 10^{-3}\, \mathrm{M}\) and \([\mathrm{F}^{-}] = \frac{10^{-5}}{2} = 5 \times 10^{-6}\, \mathrm{M}\).

Step 2: Expression for Ionic Product of \( \mathrm{CaF}_2 \)

The ionic product \( Q \) of \( \mathrm{CaF}_2 \) is calculated as \( Q = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \). This expression follows the reaction \( \mathrm{Ca}^{2+} + 2 \mathrm{F}^{-} \rightleftharpoons \mathrm{CaF}_2 \). We need to calculate \( Q \) for each combination to determine precipitation.

Step 3: Calculation of \( Q \) for each combination

1. For combination 1: \( Q = (5 \times 10^{-3})(5 \times 10^{-6})^2 = 1.25 \times 10^{-13} \)2. For combination 2: \( Q = (5 \times 10^{-4})(5 \times 10^{-4})^2 = 1.25 \times 10^{-10} \)3. For combination 3: \( Q = (5 \times 10^{-5})(5 \times 10^{-3})^2 = 1.25 \times 10^{-10} \)4. For combination 4: \( Q = (5 \times 10^{-3})(5 \times 10^{-4})^2 = 1.25 \times 10^{-10} \)

Step 4: Comparison with \( K_{sp} \)

Compare the ionic product \( Q \) with the solubility product \( K_{sp} = 1.7 \times 10^{-10} \). If \( Q > K_{sp} \), precipitation occurs; if \( Q \leq K_{sp} \), it does not.

Step 5: Determine Precipitation

For combinations 2, 3, and 4, \( Q = 1.25 \times 10^{-10} < K_{sp} \), so precipitation does not occur. For combination 1, \( Q < K_{sp} \) as well, so precipitation does not occur there either.

Step 6: Conclusion

Precipitation occurs in none of the solutions, as the ionic products are less than the solubility product in each combination.

Key Concepts

Ionic Product

In chemistry, particularly when studying solutions, the ionic product is a crucial concept. The ionic product is the product of the concentrations of the constituent ions in a solution. For compounds like calcium fluoride (\( \text{CaF}_2 \)), the ionic product \( Q \) is calculated as \( Q = [\text{Ca}^{2+}][\text{F}^{-}]^2 \).
The equation above represents that in a given solution, for the compound to be in equilibrium, the product of the concentration of calcium ions \((\text{Ca}^{2+})\) and the square of the concentration of fluoride ions \((\text{F}^{-})\) is continuous.
Understanding the ionic product helps in determining whether precipitation will occur when you mix solutions. Think of it as a snapshot of the current state of ion concentrations at any point in time.
In scenarios where we want to find out if a precipitate forms, it is crucial to compute the ionic product. Whether or not precipitation occurs depends on how this product compares to the solubility product (\( K_{sp} \)).

Solubility Product

The solubility product, denoted as \( K_{sp} \), is essential in predicting when a salt in solution will begin to precipitate. It reflects the maximum product of ion concentrations that can exist in solution without forming a precipitate.
The concept of \( K_{sp} \) comes from the equilibrium state of a sparingly soluble ionic compound in a saturated solution. At this state, the ionic compound and its ions are in equilibrium. This balance is captured by the equilibrium constant known as the solubility product. For calcium fluoride, \( K_{sp} = 1.7 \times 10^{-10} \).
When determining whether a solution will precipitate, compare the calculated ionic product \( Q \) to \( K_{sp} \):

• If \( Q > K_{sp} \), the solution is supersaturated, and precipitation is likely to occur.
• If \( Q \leq K_{sp} \), the solution is unsaturated or at equilibrium, and precipitation will not occur.

This relationship is vital in various applications, including analytical chemistry and the development of industrial processes.

Precipitation Reactions

Precipitation reactions are reactions that form an insoluble product, known as a precipitate. When two solutions containing soluble salts are mixed, an insoluble salt may form. Whether a precipitation reaction occurs depends on the ionic product and solubility product concepts.
Consider a simple precipitation reaction involving calcium fluoride, which is often used as an example in chemistry studies. When you combine solutions of calcium ions \((\text{Ca}^{2+})\) and fluoride ions \((\text{F}^{-})\), a solid precipitate of calcium fluoride can form. However, this happens only if the ionic product \((Q)\) of \(\text{CaF}_2\) exceeds its \(K_{sp}\).
In practical scenarios:
- If you mix two solutions and notice cloudiness or particles, a precipitation reaction has occurred.
- By knowing the \(K_{sp}\) of compounds, you can predict these reactions.
Understanding these reactions helps in explaining natural phenomena like the formation of geological deposits and is crucial in industrial processes used for purification and extraction of materials.