Problem 104 - Chemistry The Molecular Science

Question

Calcium fluoride, \(\mathrm{CaF}_{2}\), is used to fluoridate a municipal water supply. The water is extremely hard with a \(\mathrm{Ca}^{2+}\) concentration of \(0.070 \mathrm{M}\). Calculate the fluoride con- centration in this solution. Calcium fluoride has \(K_{\mathrm{sp}}=5.3 \times 10^{-9}\)

Step-by-Step Solution

Answer

The fluoride concentration is \(8.7 \times 10^{-4} \text{ M}\.\)

Step 1: Understanding the Task

The task is to find the concentration of fluoride ions \( \text{[F}^-\text{]} \) provided the concentration of calcium ions \( \text{[Ca}^{2+}\text{]} = 0.070 \, M \) and the solubility product \( K_{\text{sp}} \) for calcium fluoride \( \text{CaF}_2 \).

Step 2: Write the Dissolution Equation and Expression for Ksp

The dissolution of calcium fluoride \( \text{CaF}_2 \) can be expressed as:\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) \]The solubility product constant \( K_{\text{sp}} \) for this dissolution is given by:\[ K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^-]^2 \]

Step 3: Substitute Known Values

Substitute \( \text{[Ca}^{2+}\text{]} \) into the \( K_{\text{sp}} \) expression:\[ K_{\text{sp}} = (0.070)(x)^2 \] where \( x \) represents the fluoride concentration.

Step 4: Solve for Fluoride Concentration

Rearrange the equation to solve for \( x \):\[ 5.3 \times 10^{-9} = (0.070)(x)^2 \]\[ x^2 = \frac{5.3 \times 10^{-9}}{0.070} \]\[ x^2 = 7.57 \times 10^{-8} \]Take the square root to find \( x \):\[ x = \sqrt{7.57 \times 10^{-8}} \approx 8.7 \times 10^{-4} \text{ M} \]

Step 5: Conclusion

The concentration of fluoride ions \( \text{[F}^-\text{]} \) in the solution is approximately \( 8.7 \times 10^{-4} \text{ M} \). This accounts for the equilibrium established in the presence of the given calcium ion concentration.

Key Concepts

Calcium Fluoride Dissolution

When calcium fluoride ( CaF_2 ) is added to water, it undergoes a process called dissolution. This means it separates into ions and distributes them into the solution. For calcium fluoride, the dissolution can be represented by the following equation:
\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) \]This equation shows that one unit of calcium fluoride dissolves to produce one calcium ion (\(\text{Ca}^{2+}\)) and two fluoride ions (\(\text{F}^-\)) in the solution.

• Calcium ions result from the cation part of the compound.
• Fluoride ions come from the anionic part.

The dissolution is represented by the two-sided arrow, indicating an equilibrium state where the rates of dissolution and crystallization are equal. This is important for understanding how much of each ion is present in the solution.

Fluoride Ion Concentration

In any solution containing dissolved CaF_2, the concentration of fluoride ions, represented as \(\text{[F}^-\text{]}\), can be determined using the solubility product constant, \(K_{\text{sp}}\). The \(K_{\text{sp}}\) value for calcium fluoride is given as \(5.3 \times 10^{-9}\).
This constant helps us understand how the ions are concentrated when the solution is at equilibrium.

• Given [\(\text{Ca}^{2+}\)] = 0.070 M, we substitute this into the \(K_{\text{sp}}\) expression:
• \(K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^-]^2\)
• Solving \(5.3 \times 10^{-9} = (0.070) \times (x)^2\), where \(x\) is the fluoride concentration.

Through this calculation, the fluoride ion concentration is determined to be approximately \(8.7 \times 10^{-4}\) M. This reflects the equilibrium state where the dissolved ions are constantly interchanging with the solid form of the compound.

Equilibrium Chemistry

Equilibrium chemistry examines reactions that occur both forwards and backwards, reaching a balance point. This is critical when understanding the solubility in solutions like CaF_2, as it explains the dynamic between dissolving into ions and reforming the solid compound.
The concept of equilibrium is depicted by the equation:

• \( K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^-]^2 \)
• This helps determine the concentrations of ions in solution.

At equilibrium, the concentration of the dissolved species does not change, accounting for the constant back and forth shift.
In this scenario, the concentrations of ions in the water achieve a steady state, indicating the solubility limit of CaF_2 in the given conditions.