Question

Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically. the F- ion concentration is adjusted to about 1 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 ppm. Could a precipitate of ${\mathrm{CaF}}_2$ form under these conditions? (Make any necessary approximations.)

Short answer

A precipitate of ${\mathrm{CaF}}_2$ will not form under the given conditions, as the ion product (Q) is less than the solubility product constant (Ksp).

Step-by-step solution

Convert concentration from ppm to mol/L

First, we need to convert the given concentrations from ppm (parts per million) to mol/L (moles per liter). To do this conversion, we can use the following relation for each ion: given concentration (ppm) × (1 mol / atomic or molecular weight (g)) × (1 g / 1,000,000 mL) × (1,000 L / 1,000,000 mL) For F⁻ ion concentration (considering atomic weight of F = 19 g/mol): 1 ppm × (1 mol/ 19 g) × (1 g / 1,000,000 mL) × (1,000 L / 1,000,000 mL)= $\frac{1}{19,000}$ mol/L For Ca²⁺ ion concentration (considering atomic weight of Ca = 40 g/mol): 8 ppm × (1 mol / 40 g) × (1 g / 1,000,000 mL) × (1,000 L / 1,000,000 mL) = $\frac{1}{5,000}$ mol/L

Calculate ion product (Q) under given conditions

The ion product (Q) for ${\mathrm{CaF}}_2$ can be calculated by multiplying the concentrations of the ions in the solution: Q = [Ca²⁺] × [F⁻]² = ($\frac{1}{5,000}$) × ($\frac{1}{19,000}$)²

Compare Q with Ksp

Now, we'll compare the ion product (Q) with the solubility product constant (Ksp) for ${\mathrm{CaF}}_2$, which is $3.9\times {10}^{-11}$. If Q > Ksp, a precipitate will form, otherwise, it won't. $Q=\frac{1}{5,000}\times {\left(\frac{1}{19,000}\right)}^2=2.77\times {10}^{-12}$ As $2.77\times {10}^{-12}<3.9\times {10}^{-11}$, which means Q < Ksp.

Conclusion

A precipitate of ${\mathrm{CaF}}_2$ will not form under the given conditions, as the ion product (Q) is less than the solubility product constant (Ksp).

Key concepts

Chemistry of Solutions

Solutions are homogeneous mixtures composed of two or more substances. The substance present in the largest amount is usually the solvent, and the substances present in smaller amounts are solutes. Understanding the chemistry of solutions is essential, as it explains how substances interact and dissolve in liquids.

When talking about the chemistry of solutions, concentrations are crucial. Concentration refers to the amount of solute in a given volume of solution. It's commonly expressed in molarity (mol/L), but it can also be expressed in parts per million (ppm), which is often used for very dilute solutions like those found in water treatment.

In this context, fluoridation refers to adding small amounts of fluoride ions to water. Fluoride adjusts these solutions' properties to help with health benefits, such as reducing tooth decay. It is essential to carefully control concentrations in these solutions to avoid unwanted reactions or precipitation.

Precipitation Reactions

Precipitation reactions occur when two soluble substances in a solution react to form an insoluble solid, known as a precipitate. This process is a type of double displacement reaction. It happens when the product of the reaction has low solubility in the solvent.

Let's think about how precipitation works. For example, when calcium ions (${\text{Ca}}^{2+}$) and fluoride ions (${\text{F}}^{-}$) are present in water at sufficient concentrations, they can form calcium fluoride (${\text{CaF}}_2$) as a precipitate.

Precipitation can be visualized as the opposite of dissolving. Instead of solute particles dispersing, they come together to form a solid. It is important in various fields such as water treatment, mineral processing, and even in crafting certain types of medications.

Solubility Product Constant

The solubility product constant, often denoted as (K_{sp}), is a crucial value in chemistry that determines the solubility of a compound. It is the product of the concentrations of the ions in a saturated solution raised to the power of their stoichiometric coefficients.

For example, in the case of calcium fluoride (${\text{CaF}}_2$), which dissociates into calcium ions and fluoride ions, the solubility product is expressed as:$$K_{sp}=[{\text{Ca}}^{2+}][{\text{F}}^{-}{]}^2$$

Knowing the (K_{sp}) allows us to predict whether a precipitate will form under certain conditions. By calculating the ion product (Q) of a solution and comparing it to the (K_{sp}), we can determine if the solution is undersaturated, saturated, or supersaturated. If (Q < K_{sp}), no precipitate forms. If (Q > K_{sp}), precipitation occurs. This concept is vital in understanding and managing chemical processes in solutions.