Problem 59 - Chemistry The Central Science In Si Units

Question

Consider a beaker containing a saturated solution of CaF \(_{2}\) in equilibrium with undissolved \(\mathrm{CaF}_{2}(s)\). Solid \(\mathrm{CaCl}_{2}\) is then added to the solution. (a) Will the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Ca}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of F ions in solution increase or decrease?

Step-by-Step Solution

Answer

(a) Increase; (b) Increase initially, then stabilizes; (c) Decrease.

Step 1: Understanding the Initial Setup

Initially, we have a saturated solution of \( \text{CaF}_2 \) with some solid \( \text{CaF}_2 \) at the bottom of the beaker, indicating an equilibrium between the dissolved ions and the solid. The solubility product expression for this system is given by \( K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \).

Step 2: Analyzing the Addition of CaCl2

When \( \text{CaCl}_2 \) is added to the beaker, it dissociates into \( \text{Ca}^{2+} \) and \( \text{Cl}^- \) ions. The concentration of \( \text{Ca}^{2+} \) increases due to this addition, affecting the equilibrium of the saturated \( \text{CaF}_2 \) solution.

Step 3: Applying Le Chatelier's Principle

According to Le Chatelier's Principle, an increase in the concentration of \( \text{Ca}^{2+} \) will shift the equilibrium to the left to counteract this change. This means more \( \text{CaF}_2 \) will precipitate out of the solution to decrease the \( \text{Ca}^{2+} \) concentration.

Step 4: Effect on Solid CaF2

As the equilibrium shifts left, more solid \( \text{CaF}_2 \) is formed, increasing the amount of undissolved \( \text{CaF}_2 \) at the bottom of the beaker.

Step 5: Change in Concentration of Ca2+ Ions

Although initially the concentration of \( \text{Ca}^{2+} \) increases due to \( \text{CaCl}_2 \) addition, the shift in equilibrium and the precipitation of \( \text{CaF}_2 \) helps restore the \( \text{Ca}^{2+} \) concentration towards its original equilibrium level, though it may still be slightly elevated compared to the initial state without \( \text{CaCl}_2 \).

Step 6: Change in Concentration of F- Ions

The shift in equilibrium towards the formation of solid \( \text{CaF}_2 \) reduces the concentration of \( \text{F}^- \) ions in the solution since fluoride ions are being incorporated into the precipitated solid.

Key Concepts

Solubility Product

The solubility product, often denoted as \( K_{sp} \), is a constant at a given temperature that quantifies the saturated solution equilibrium of a sparingly soluble salt. In our scenario, it applies to calcium fluoride, \( \text{CaF}_2 \). When \( \text{CaF}_2 \) is in a saturated solution, it stands in balance with its ions in the water, specifically calcium ions \([\text{Ca}^{2+}]\) and fluoride ions \([\text{F}^-]^2\). The solubility product expression for \( \text{CaF}_2 \) is thus formulated as:

• \( K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \)

This equilibrium expression indicates that the product of the concentration of the ions raised to the power of their coefficients in the balanced equation remains constant. When this equilibrium is disturbed, for example, by adding other salts that affect ion concentrations, the system will react to establish equilibrium again. Thus, understanding \( K_{sp} \) is essential in predicting the behavior of solutions when dynamic changes occur.

Equilibrium Shift

An equilibrium shift, as described by Le Chatelier's Principle, occurs when a system at equilibrium is disturbed. In our example, adding \( \text{CaCl}_2 \) to a saturated \( \text{CaF}_2 \) solution increases the \([\text{Ca}^{2+}]\) in the solution. According to Le Chatelier's Principle:

• If a change is applied to a system at equilibrium, the system will shift to counteract that change.

When the concentration of \( \text{Ca}^{2+} \) increases, the equilibrium of the saturated \( \text{CaF}_2 \) solution will shift left, favoring the formation of more solid \( \text{CaF}_2 \) to reduce the excess ions in the solution. This shift re-establishes the balance, causing more precipitate to form at the bottom of the beaker, showing a tangible model of how equilibrium responds to external stresses.

Precipitation Reaction

A precipitation reaction occurs when two solutions react to form an insoluble solid, known as a precipitate. In this exercise, when extra \( \text{Ca}^{2+} \) is introduced from \( \text{CaCl}_2 \), a precipitation reaction is prompted to restore equilibrium by forming additional \( \text{CaF}_2 \) solids. Key takeaways for precipitation reactions:

• The system tries to minimize disturbances by forming solids out of the ions in solution.
• Precipitation reactions serve as visual evidence of equilibrium shifts.

In this case, the formation of a greater amount of solid \( \text{CaF}_2 \) correctly illustrates the concept of precipitation to balance the concentration of ions. Consequently, the formation of a solid precipitate emphasizes the direct application of Le Chatelier's Principle and the solubility product in real-world chemical balances.