Question

Fluoridation of drinking water is employed in many places to aid in the prevention of dental caries. Typically the \(\mathrm{F}^{-}\) ion concentration is adjusted to about \(1 \mathrm{ppb}\). 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 ppb. 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 these conditions, as the ion product (\(2.77 \times 10^{-22}\)) is much smaller than the \(K_\mathrm{sp}\) value (\(3.9 \times 10^{-11}\)).

Step-by-step solution

Write the balanced dissolution equation and the equilibrium expression for the solubility product constant

The dissolution equation for calcium fluoride (\(\mathrm{CaF}_{2}\)) is: \[ \mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^{-}(aq) \] The solubility product constant (\(K_\mathrm{sp}\)) expression is: \[ K_\mathrm{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \]

Find the \(K_\mathrm{sp}\) value of calcium fluoride

From reference tables or textbooks, the \(K_\mathrm{sp}\) value for \(\mathrm{CaF}_{2}\) is found to be approximately \(3.9 \times 10^{-11}\).

Convert ppb (parts per billion) concentrations to molar concentrations

1 ppb is equal to 1 microgram per liter (\(1 \mu g/L\)). We will convert the 1 ppb \(\mathrm{F}^-\) and 8 ppb \( \mathrm{Ca}^{2+}\) concentrations to molar concentrations using their respective molar masses. For \(\mathrm{F}^-\): Molar mass of Fluorine is approximately 19.00 g/mol \(1 \mu g/L = 1\times10^{-6} g/L \) \[ \text{Molar concentration of F}^- = \frac{1\times10^{-6}\,g/L}{19.00\,g/mol} \approx 5.26\times10^{-8}\,M \] For \(\mathrm{Ca}^{2+}\): Molar mass of Calcium is approximately 40.08 g/mol \(8 \mu g/L = 8\times10^{-6} g/L \) \[ \text{Molar concentration of Ca}^{2+} = \frac{8\times10^{-6}\,g/L}{40.08\,g/mol} \approx 2.00\times10^{-7}\,M \]

Calculate the ion product from the given ion concentrations

The ion product is calculated by plugging the molar concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^-\) into the equilibrium expression: Ion product = \([\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2\) \[ \text{Ion product} = (2.00\times10^{-7})(5.26\times10^{-8})^2 \approx 2.77\times10^{-22} \]

Compare the ion product to the solubility product constant

Now we compare the ion product to \(K_\mathrm{sp}\) to determine if the precipitation of \(\mathrm{CaF}_{2}\) will occur. If the ion product > \(K_\mathrm{sp}\), precipitation occurs; otherwise no precipitation. In this case, the ion product (\(2.77 \times 10^{-22}\)) is much smaller than the \(K_\mathrm{sp}\) value (\(3.9 \times 10^{-11}\)). Therefore, a precipitate of \(\mathrm{CaF}_{2}\) will not form under these conditions.

Key concepts

Fluoridation of Drinking Water

Fluoridation is a controlled addition of fluoride to a public water supply to reduce tooth decay. Municipal water fluoridation mimics the natural occurrence of fluoride in water, adjusting the concentration to a level that is effective for reducing cavities. Generally, this level is maintained around 1 part per billion (ppb), which is equivalent to 1 microgram of fluoride per liter of water. The process is supported by numerous health organizations and is considered one of the major public health achievements of the 20th century.

Despite its benefits for dental health, the process must be carefully managed to avoid the excessive presence of fluoride, which could potentially lead to dental fluorosis - a condition that can cause discoloration or pitting of the teeth. To strike the right balance, water authorities rely on precise equilibrium calculations to ensure that fluoride levels are effective without causing harm. In contexts where water is 'hard', or contains substantial amounts of minerals like calcium, special attention must be given, as fluoride can interact with these minerals and affect both its efficacy and solubility.

Hard Water

Hard water is characterized by its high mineral content, primarily calcium and magnesium ions. These minerals are naturally dissolved from rock and soil and end up in water supplies. Although hard water poses no health risk, it can cause practical issues. For instance, it interferes with the action of soap, leading to less lather and the formation of soap scum. Moreover, over time, these minerals can deposit and form scale within pipes, water heaters, and kettles, which can hinder the efficiency of household and industrial appliances.

The hardness of water is typically reported in parts per million (ppm) or parts per billion (ppb) of calcium carbonate equivalent. When dealing with problems like fluoridation in hard water, it is important to understand how calcium ions present in the water can affect the solubility and effectiveness of fluoride, as in the case of potential calcium fluoride (CaF2) precipitation. Here, knowledge of solubility equilibria and the solubility product constant (Ksp) is crucial.

Equilibrium Calculation

Equilibrium calculations are at the core of understanding various chemical processes, including those that involve the solubility of compounds in solution. When a salt like calcium fluoride dissolves in water, it reaches a point of dynamic equilibrium where the rate of dissolving equals the rate of precipitation. The concentration of ions at this point is described by the solubility product constant (Ksp), which is unique for each compound under a given set of conditions, such as temperature and pressure.

In the case of fluoridation in hard water, an equilibrium calculation can help predict whether or not a precipitate will form. By comparing the ion product (IP), which is the product of the molar concentrations of the dissolved ions, to the Ksp value, one can determine the likelihood of precipitation. If IP exceeds Ksp, the solution is supersaturated, and precipitation is expected. However, if IP is less than Ksp, the solution is unsaturated, indicating that no precipitate will form. This type of calculation is vital in environmental chemistry, water treatment, and many industrial processes.