Problem 84 - Chemistry

Question

The U.S. Public Health Service recommends the fluoridation of water as a means for preventing tooth decay. The recommended concentration is 1 \(\mathrm{mg} \mathrm{F}^{-}\) per liter. The presence of calcium ions in hard water can precipitate the added fluoride. What is the maximum molarity of calcium ions in hard water if the fluoride concentration is at the USPHS recommended level? \(\left(K_{\mathrm{sp}} \text { for } \mathrm{CaF}_{2}=4.0 \times 10^{-11}\right)\)

Step-by-Step Solution

Answer

The maximum molarity of calcium ions in hard water, with the fluoride concentration at the USPHS recommended level, is \(1.45 \times 10^{-2}\) M.

Step 1: Write the balanced chemical equation for the dissolution of CaF₂

First, we need to write the balanced chemical equation for the dissolution of calcium fluoride (CaF₂) in water: \[ \mathrm{CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2 F^{-} (aq)}\]

Step 2: Write the expression for Ksp

Next, we need to write the expression for the solubility product constant (Ksp) of CaF₂. From the balanced chemical equation, we can write the Ksp expression as: \[ K_\mathrm{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^{-}}]^2\] We are given the value of Ksp for CaF₂: \[ K_\mathrm{sp} = 4.0\times10^{-11}\]

Step 3: Use the recommended fluoride concentration

We know that the recommended concentration of fluoride ions is 1 mg F⁻ per liter. First, we need to convert this concentration into moles per liter (molarity). The molar mass of F⁻ is 19 g/mol. Since 1 mg = 0.001 g, we can write the molarity of fluoride ions as: \[ [\mathrm{F^{-}}] = \frac{0.001\ \text{g}}{19\ \text{g/mol}} = 5.26\times10^{-5}\ \text{M}\]

Step 4: Find the maximum molarity of calcium ions

Now, substitute the given Ksp and the molarity of fluoride ions into the Ksp expression to find the maximum molarity of calcium ions: \[ K_\mathrm{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^{-}}]^2\] \[ 4.0\times10^{-11} = [\mathrm{Ca^{2+}}](5.26\times10^{-5})^2\] Now, solve for the molarity of calcium ions: \[ [\mathrm{Ca^{2+}}] = \frac{4.0\times10^{-11}}{(5.26\times10^{-5})^2} = 1.45\times10^{-2}\ \text{M}\] So, the maximum molarity of calcium ions in hard water, with the fluoride concentration at the USPHS recommended level, is 1.45 × 10⁻² M.

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. This leads to the concentrations of reactants and products remaining constant over time. In the context of solubility and equilibrium, the dissolution of sparingly soluble salts, like calcium fluoride (\(\text{CaF}_2\)), can be described by an equilibrium state:

• At equilibrium, the ions dissolve into the solution until the maximum concentration (saturation) is reached.
• The solution then maintains a dynamic balance, continually dissolving and precipitating ions at equilibrium rates.

The effectiveness of a reaction reaching equilibrium depends on temperatures, pressure, and the nature of the substances involved. A saturated solution at equilibrium contains undissolved calcium fluoride at the bottom because the solubility point is reached.

Calcium Fluoride

Calcium fluoride is an inorganic compound with the formula \(\text{CaF}_2\). It's a white, insoluble solid that occurs naturally as fluorite. Calcium fluoride is used in several industries:

• In the production of hydrogen fluoride (HF)
• As a component in the manufacture of aluminum fluoride

When calcium fluoride dissolves in water, it dissociates into calcium ions (\(\text{Ca}^{2+}\)) and fluoride ions (\(\text{F}^{-}\)). Although it has low solubility, this dissolution process is crucial for understanding equilibrium in saturated solutions. It signifies how much fluoride and calcium ions are available in the solution. The solubility product constant, \(K_{sp}\), helps to quantify this equilibrium, indicating at what point the reaction reaches saturation.

Fluoride Concentration

Fluoride concentration in water is critical, especially in dental health. The optimal concentration can prevent tooth decay by enhancing tooth enamel's resilience to acids produced by bacteria.
The recommended fluoride concentration in drinking water by health bodies is about 1 mg/L, which is equivalent to \(5.26 \times 10^{-5}\) Molar.

• Ensuring this concentration is met is essential for effective water fluoridation and public health.
• The right fluoride concentration aids in remineralizing tooth surfaces and demineralizing bacterial plaque.

To maintain this balance, calculations like the solubility of calcium fluoride in hard water must be done accurately to avoid excess calcium, which can precipitate fluoride as insoluble calcium fluoride.

Water Fluoridation

Water fluoridation is the process of adjusting fluoride levels in drinking water to help reduce tooth decay. The effort is particularly significant in communities with naturally low fluoride levels. Here are some key points:

• Fluoridation has been shown to reduce tooth decay by 25% in children and adults.
• Maintaining fluoride concentration at 1 mg/L prevents decay without causing fluorosis (a condition caused by excess fluoride).

Communities must ensure proper water hardness to not only provide the needed fluoride levels but to also avert unwanted chemical reactions. When calcium ion concentration increases, it reacts with fluoride ions, reducing availability in water due to precipitation. Calculations involving \(K_{sp}\) are therefore essential to predict the compatibility of fluoride with available calcium, ensuring optimal public health benefits from fluoridation.