Problem 118 - Chemistry: The Central Science in SI Units, Global Edition

Question

Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically. the \(\mathrm{F}^{-}\) ion concentration is adjusted to about \(1 \mathrm{ppm}\). Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is \(8 \mathrm{ppm}\). Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

Step-by-Step Solution

Answer

In the given solution with F- ion concentration of \(1 \, ppm\) and Ca2+ ion concentration of \(8 \, ppm\), the ion product (Q) is found to be \(5.503 \times 10^{-13}\), which is less than the solubility product constant (Ksp) of CaF2, \(3.9 \times 10^{-11}\). As \(Q < Ksp\), the solution is not saturated, and a precipitate of CaF2 will not form under these conditions.

Step 1: Convert ppm to molar concentration

Ppm (parts per million) is a unit of concentration that expresses the ratio of the solute to the solvent. To convert ppm to molarity, we need to know the molar mass of the ion and the density of water. The molar mass of F- ion: \(m_{F^-} = 19.00 \, g/mol\) The molar mass of Ca2+ ion: \(m_{Ca^{2+}} = 40.08 \, g/mol\) Density of water: \(\rho_{water} = 1 \, g/cm^3\) For F- ion: \[C_{F^-} = \dfrac{1 \, ppm \cdot \rho_{water}}{m_{F^-}}\] For Ca2+ ion: \[C_{Ca^{2+}} = \dfrac{8 \, ppm \cdot \rho_{water}}{m_{Ca^{2+}}}\] Keep in mind that \(1 \, ppm = 1 \frac{mg}{L}\), so we have:

Step 2: Calculate the molar concentration of each ion

For F- ion: \[C_{F^-} = \dfrac{1 \frac{mg}{L} \cdot 1 \frac{g}{1000 \, mg}}{19.00 \frac{g}{mol}} = 5.26 \times 10^{-5} \, M\] For Ca2+ ion: \[C_{Ca^{2+}} = \dfrac{8 \frac{mg}{L} \cdot 1 \frac{g}{1000 \, mg}}{40.08 \frac{g}{mol}} = 1.997 \times 10^{-4} \, M\]

Step 3: Calculate the ion product (Q)

Next, we need to calculate the ion product (Q) by multiplying the molar concentrations of the ions: \[Q = [Ca^{2+}][F^-]^2]\] Plugging in the values: \[Q = (1.997 \times 10^{-4} M)(5.26 \times 10^{-5} M)^2 = 5.503 \times 10^{-13}\]

Step 4: Compare Q with Ksp

The solubility product constant (Ksp) of CaF2 is \(3.9 \times 10^{-11}\). Now we compare the ion product (Q) with the Ksp: \(Q = 5.503 \times 10^{-13}\) \(Ksp = 3.9 \times 10^{-11}\) Since \(Q < Ksp\), the solution is not saturated, and a precipitate of CaF2 will not form under these conditions.

Key Concepts

Calcium Hardness

Calcium hardness in water refers to the concentration of calcium ions (\( \text{Ca}^{2+} \)). These ions come from dissolved calcium salts. Natural sources like limestone often contribute to calcium hardness.
Calcium ions have properties that make water "hard," which means it can interfere with soap's ability to lather. Additionally, hard water can cause scale build-up in pipes and appliances. Understanding water hardness helps in managing its effects in daily use.
In the context of water fluoridation, calcium hardness could affect the formation of certain precipitates. In situations where fluoride ions (\( \text{F}^- \)) are added, calcium can combine with fluoride to potentially form calcium fluoride (\( \text{CaF}_2 \)), which might lead to precipitation if certain conditions are met.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions mix to form an insoluble solid, called the precipitate. This results from the combination of ions that form a compound with low solubility in water.
For example, in our case, when calcium ions (\( \text{Ca}^{2+} \)) and fluoride ions (\( \text{F}^- \)) come together, they can form calcium fluoride (\( \text{CaF}_2 \)). Whether this compound precipitates depends on the concentration of the ions and the solubility rules.
Identifying and predicting precipitation reactions are crucial in various fields such as chemistry and environmental science, especially when managing water quality.

Solubility Product Constant

The solubility product constant (\( K_{sp} \)) is a value that indicates the solubility of a compound. It helps predict whether a precipitate will form in a solution.
The \( K_{sp} \) value for a substance like calcium fluoride (\( \text{CaF}_2 \)) represents the product of the concentrations of its ions, each raised to the power of their stoichiometric coefficients.
The formula used is:

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

By comparing \( Q \) (the ion product) to the \( K_{sp} \), we determine if the solution is saturated:

• If \( Q < K_{sp} \), the solution is unsaturated, and no precipitate forms.
• If \( Q > K_{sp} \), the solution is supersaturated, and precipitation occurs.

Understanding this concept helps in predicting and controlling reactions in water treatment processes.

Ion Concentration Conversion

Converting ion concentration from parts per million (ppm) to molarity is essential for many calculations in chemistry.
Ppm is a mass-based unit often used to express tiny concentrations, whereas molarity (\( M \)) is the number of moles of solute per liter of solution.
To convert ppm to molarity for a particular ion, you need:

• The molar mass of the ion
• The density of water (assuming it is similar to 1 g/mL)

For example, converting \( 8 \text{ ppm} \) of \( \text{Ca}^{2+} \):

• \( C_{Ca^{2+}} = \frac{8 \frac{mg}{L} \times 1 \frac{g}{1000 \, mg}}{40.08 \frac{g}{mol}} = 1.997 \times 10^{-4} \, M \)

Mastering this conversion technique is crucial for solving chemical equilibrium problems, especially in aqueous solutions.