Question

Fluoridated drinking water contains about 1 part per million (ppm) of \(\mathrm{F}^{-}\). Is \(\mathrm{CaF}_{2}\) sufficiently soluble in water to be used as the source of fluoride ion for the fluoridation of drinking water? Explain. [Hint: Think of 1 ppm as signifying \(1 \mathrm{g} \mathrm{F}^{-}\) per \(10^{6} \mathrm{g}\) solution.

Short answer

No, \( \mathrm{CaF}_{2} \) is not sufficiently soluble to provide the required fluoride ion concentration for fluoridation of drinking water.

Step-by-step solution

Calculate Molar Mass of \( \mathrm{CaF}_{2} \) and \( \mathrm{F}^{-} \)

First, calculate the molar mass of \( \mathrm{CaF}_{2} \) which is \(40.08 g/mol + 2*18.9984 g/mol = 78.0768 g/mol\). Molar mass of fluoride ion \( \mathrm{F}^{-} \) is \(18.9984 g/mol\).

Understanding \( \mathrm{CaF}_{2} \) Dissociation

When 1 mole of \( \mathrm{CaF}_{2} \) dissolves in water, it dissociates into 1 mole of \( \mathrm{Ca}^{2+} \) and 2 moles of \( \mathrm{F}^{-} \). Therefore, for 1 mole of \( \mathrm{CaF}_{2} \), we get 2 moles of \( \mathrm{F}^{-} \).

Calculate \( \mathrm{F}^{-} \) Concentration

So if we dissolve 1 ppm (1g) of \( \mathrm{CaF}_{2} \) in \(10^{6} \mathrm{g}\) of water, it would provide \( \frac{1}{78.0768} = 0.0128 mol \) of \( \mathrm{CaF}_{2} \), and thus \( 2*0.0128 = 0.0256 mol \) of \( \mathrm{F}^{-} \) or 25.6 millimoles of \( \mathrm{F}^{-} \).

Compare with Required Concentration

The required concentration for fluoridation of water is 1 ppm, which signifies \( \frac{1 g}{18.9984 g/mol} = 0.0526 mol \) or 52.6 millimoles of \( \mathrm{F}^{-} \) per million grams of water.

Concluding Remarks

Comparing the obtained and required concentrations, \( \mathrm{CaF}_{2} \) is not sufficiently soluble in water to be used as the source of fluoride ion. As the obtained concentration is less than half of the required.

Key concepts

Fluoridation of Drinking Water

Fluoridation of drinking water is a widespread practice aimed at improving public dental health. Adding small amounts of fluoride ions to drinking water helps to reduce tooth decay. Typically, water is fluoridated to a concentration of about 1 part per million (ppm), which is equivalent to 1 gram of fluoride ions per million grams of water.

This low concentration is considered safe and effective for strengthening tooth enamel and reducing dental cavities. Fluoridation is managed carefully to maintain the balance between beneficial and excessive fluoride levels. In many regions, regulations ensure that fluoride levels in water remain within the recommended range to prevent potential adverse effects from overexposure.

Individuals often receive additional fluoride from other sources such as toothpaste or dietary supplements. Thus, it is essential to evaluate the total fluoride exposure to ensure dental health benefits without negative consequences.

Fluoride Ion Concentration

Fluoride ion concentration is crucial when it comes to water fluoridation. The concentration level of 1 ppm in drinking water is carefully selected for its effectiveness in dental health. However, to achieve this concentration, the solubility of the source compound must be adequate.

In our exercise, Calcium Fluoride ( CaF_2 ) serves as a potential source of fluoride ions. This compound releases fluoride ions when dissolved in water, but its solubility directly impacts its effectiveness in water fluoridation.

The concentration of fluoride ions achievable from dissolving CaF_2 was calculated at approximately 25.6 millimoles per million grams of water. This is less than half of the required 52.6 millimoles needed for 1 ppm. This highlights the importance of selecting a highly soluble fluoride compound for water fluoridation to meet the required ion concentration effectively.

Chemical Dissolution Process

Understanding the chemical dissolution process is key to using compounds like CaF_2 for water fluoridation. Dissolution involves the breaking down of solid ionic compounds into ions in a solution. For CaF_2 , this means splitting one mole of CaF_2 into one mole of Ca^{2+} ions and two moles of F^{-} ions when in water.

This dissociation process is crucial because the amount of fluoride ions depends on how much CaF_2 can dissolve. However, CaF_2 is not very soluble, which limits the fluoride ions available in solution. The low solubility results in a concentration that fails to reach the desired level for effective water fluoridation.

The choice of compound dramatically affects the solubility and, subsequently, the effectiveness of the water fluoridation process. That's why more soluble compounds, such as sodium fluoride, are often preferred in practice, ensuring that sufficient fluoride ions are present to aid in dental health.