Problem 107 - Chemistry The Science in Context

Question

Will calcium fluoride precipitate when \(125 \mathrm{mL}\) of \(0.375 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(245 \mathrm{mL}\) of \(0.255 M \mathrm{NaF}\) at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Answer: __________

Step 1: Calculate the moles of Ca²⁺ and F⁻ ions

First, we need to determine the moles of Ca²⁺ ions from the \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) solution and F⁻ ions from the \(\mathrm{NaF}\) solution. Moles of Ca²⁺ = Volume of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) solution × Concentration of Ca²⁺ = \(125\;\mathrm{mL} \times 0.375\;\mathrm{M}\) Moles of F⁻ = Volume of \(\mathrm{NaF}\) solution × Concentration of F⁻ = \(245\;\mathrm{mL} \times 0.255\;\mathrm{M}\)

Step 2: Calculate the final volume of the mixture

We need to find the final volume of the mixture to calculate the concentrations of Ca²⁺ and F⁻ ions in the mixture. Final volume of mixture = Volume of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) solution + Volume of \(\mathrm{NaF}\) solution = \(125\;\mathrm{mL} + 245\;\mathrm{mL}\)

Step 3: Calculate the concentrations of Ca²⁺ and F⁻ ions in the final mixture

Now, we will calculate the new concentrations of the ions in the mixture. Concentration of Ca²⁺ = Moles of Ca²⁺ / Final volume of mixture Concentration of F⁻ = Moles of F⁻ / Final volume of mixture

Step 4: Calculate the reaction quotient (Q) for the mixture

The reaction quotient (Q) for the formation of calcium fluoride (CaF₂) can be calculated using the formula: Q = [Ca²⁺] × [F⁻]²

Step 5: Compare Q with the Ksp of CaF₂

Now, we will compare the calculated Q value with the known Ksp value for calcium fluoride (CaF₂) at \(25^{\circ} \mathrm{C}\) (Ksp = \(3.9 \times 10^{-11}\)). If Q is greater than the Ksp, precipitation will occur; otherwise, it will not.

Key Concepts

Calcium Fluoride and Its Role in Precipitation Reactions

Calcium fluoride, known chemically as \( \text{CaF}_2 \), is a compound that is often studied in chemistry for its behavior in solutions. It consists of calcium ions \( (\text{Ca}^{2+}) \) and fluoride ions \( (\text{F}^-) \). Calcium fluoride is an example of an ionic compound that can undergo a process called precipitation when it is in solution. Precipitation occurs when the concentration of the ions exceeds the limits of their solubility, forming a solid, also known as a precipitate. This formation can be predicted using the solubility product constant (Ksp), which reflects the degree to which the compound can dissolve in water.
When solutions containing calcium and fluoride ions are mixed, the possibility of calcium fluoride precipitating depends on the concentration of these ions. For instance, when \( \text{Ca(NO}_3\text{)}_2 \) is combined with \( \text{NaF} \), the calcium ions from the former and the fluoride ions from the latter can react to potentially form calcium fluoride. The precipitation of calcium fluoride can remove the ions from the solution because they form a solid that is not soluble under the given conditions.

Understanding the Reaction Quotient: Q

The reaction quotient \( Q \) is a valuable concept that helps chemists understand the direction in which a chemical reaction will proceed. It is computed similarly to the equilibrium constant \( K \), but unlike \( K \), it applies to the system at any moment, not just at equilibrium. For the precipitation of calcium fluoride \( \text{CaF}_2 \), \( Q \) is calculated by multiplying the concentration of the calcium ions \( [\text{Ca}^{2+}] \) by the square of the concentration of the fluoride ions \( [\text{F}^-]^2 \).
The formula for \( Q \) in this reaction is:

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

The values used in calculating \( Q \) are the concentrations of ions in the mixture at any given time. By comparing \( Q \) with the solubility product constant \( \text{Ksp} \) of calcium fluoride, we can determine whether the solution is saturated, unsaturated, or supersaturated. If the value of \( Q \) is greater than \( \text{Ksp} \), the solution is supersaturated, leading to the formation of a precipitate.

Decoding the Solubility Product Constant (Ksp)

The solubility product constant, denoted as \( \text{Ksp} \), is an important concept that characterizes the solubility of a slightly soluble compound in a solvent, typically water. It provides a quantitative measure of the compound's propensity to dissolve or remain solid by establishing an equilibrium between dissolved ions and the solid precipitate.
For calcium fluoride \( \text{CaF}_2 \), the \( \text{Ksp} \) value is crucial. It is defined by the equilibrium expression:

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

The \( \text{Ksp} \) value is specific to the temperature at which it is measured. For example, at \( 25^{\circ} \text{C} \), the \( \text{Ksp} \) of calcium fluoride is typically \( 3.9 \times 10^{-11} \).
This value serves as a benchmark to predict whether a precipitate will form. If the reaction quotient \( Q \) of a solution exceeds \( \text{Ksp} \), the ions in the solution exceed their soluble limit and begin to crystallize into a solid. Conversely, if \( Q \) is less than \( \text{Ksp} \), no precipitation occurs since the solution can still accommodate more of the ionic species in its dissolved form.