Problem 104 - Chemistry: An Atoms-Focused Approach

Question

In which of the following solutions will \(\mathrm{CaF}_{2}\) be most soluble? (a) \(0.010 M \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ;\) (b) \(0.01 M \mathrm{NaF}\) (c) \(0.001 M\) NaF; (d) \(0.10 M \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)

Step-by-Step Solution

Answer

Answer: CaF2 will be most soluble in solution (c) 0.001 M NaF.

Step 1: Determine the solubility product constant (Ksp) of CaF2

The general dissolution reaction for CaF2 is as follows: CaF2 (s) ⇌ Ca²⁺ (aq) + 2F⁻ (aq) The Ksp expression for CaF2 can be written as: Ksp = [Ca²⁺][F⁻]² We don't have the Ksp value provided in the problem, so let's assume it as a constant K.

Step 2: Analyze each solution and determine the effect on CaF2 solubility

(a) 0.010 M Ca(NO3)2: Because this solution contains Ca²⁺, the common ion effect will cause the solubility of CaF2 to decrease. Assume the solubility at equilibrium is "x" K = (0.010 + x)(2x)² (b) 0.01 M NaF: This solution contains F⁻, so the common ion effect will also cause a decrease in the solubility of CaF2. Assume the solubility at equilibrium is "y." K = (y)(0.01 + 2y)² (c) 0.001 M NaF: Similar to (b), this solution contains F⁻. Assume the solubility at equilibrium is "z." K = (z)(0.001 + 2z)² (d) 0.10 M Ca(NO₃)₂: This solution contains Ca²⁺, so the common ion effect will cause a decrease in the solubility of CaF₂. Assume the solubility at equilibrium is "w." K = (0.10 + w)(2w)²

Step 3: Compare the solubility of CaF2 in each solution

We can now compare the solubility of CaF2 in each solution. Solubility will be higher in solutions with a larger K value. K = (0.010 + x)(2x)² for solution (a) K = (y)(0.01 + 2y)² for solution (b) K = (z)(0.001 + 2z)² for solution (c) K = (0.10 + w)(2w)² for solution (d) Since K is a constant, a higher concentration of Ca²⁺ implies lower F⁻ concentration, and vice versa. Therefore, we can compare the solutions based on the concentration of the common ion present. Solution (c) has the lowest concentration of the common ion F⁻, so it will have the least impact on the solubility of CaF2.

Step 4: Conclude the solution

Based on our analysis of the four given solutions, CaF2 will be most soluble in solution (c) 0.001 M NaF. This is because it has the lowest concentration of the common ion F⁻, leading to less of an impact on CaF2 solubility due to the common ion effect.

Key Concepts

Common Ion Effect

The common ion effect is an important concept in chemistry that affects the solubility of ionic compounds. It occurs when an ion present in a solution is the same as an ion in the dissolved compound. This overlap causes the dissolution equilibrium to shift according to Le Chatelier's principle, leading to a decrease in the solubility of the compound.
The common ion effect can be seen in the context of the exercise with calcium fluoride \( \text{CaF}_2 \). When \( \text{CaF}_2 \) is added to a solution already containing \( \text{Ca}^{2+} \) or \( \text{F}^- \) ions, its solubility decreases. This is because additional common ions shift the equilibrium of the dissolution reaction \( \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \) towards the left, indicating the formation of more solid calcium fluoride.
In the solutions provided in the exercise, solutions (a) and (d) contain \( \text{Ca}^{2+} \) ions, and solutions (b) and (c) contain \( \text{F}^- \) ions. The presence of these ions imposes the common ion effect, reducing the solubility of \( \text{CaF}_2 \) more than in solutions lacking these ions.

Solubility Product Constant (Ksp)

The solubility product constant, represented as \( K_{sp} \), is a measure of the solubility of an ionic compound under a given set of conditions. It quantifies the extent to which a compound can dissolve in water, forming ions in the solution. For any general dissolution reaction like \( \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \), the expression for \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
This expression indicates that \( K_{sp} \) is the product of the molar concentrations of the dissolved ions, each raised to the power of their coefficients in the balanced chemical equation.
In exercises like the one provided, you assumed a solubility value ("like x, y, z, w") for each solution and set these up in the \( K_{sp} \) expression. By comparing these expressions, you can determine which solution allows for more solubility. A solution with lower amounts of common ions, as calculated in solution (c), implies greater solibility of \( \text{CaF}_2 \) with a constant \( K_{sp} \) value.

Calcium Fluoride (CaF2)

Calcium fluoride \( \text{CaF}_2 \) is an ionic compound composed of calcium and fluoride ions. It often appears in the form of a crystalline solid naturally occurring as the mineral fluorite. Being sparingly soluble in water, it creates equilibrium between its solid and dissolved ions when placed in an aqueous solution.
The solubility of \( \text{CaF}_2 \) is influenced by several factors, including temperature and the presence of common ions in the solution. In academic exercises, as presented, we focus mainly on the effect of common ions.
When calculating the dissolution of \( \text{CaF}_2 \) in various solutions, it is vital to understand how adding ions like \( \text{Ca}^{2+} \) or \( \text{F}^- \) externally affects its solubility. The mathematical representation involving \( K_{sp} \) helps illustrate these changes and allows you to ascertain which conditions promote higher solubility, such as the minimal presence of common ions as seen with solution (c) in the exercise.