Question

A town adds 2.0 ppm of \(\mathrm{F}^{-}\) ion to fluoridate its water supply. (Fluoridation of water reduces the incidence of dental caries). If the concentration of \(\mathrm{Ca}^{2+}\) in the water is \(3.5 \times 10^{-4} \mathrm{M}\), will a precipitate of \(\mathrm{CaF}_{2}\) form when the water is fluoridated?

Short answer

Short Answer: No, the addition of fluoride ions to the water supply will not cause the precipitation of calcium fluoride, as the ion product (Qsp) is less than the solubility product constant (Ksp).

Step-by-step solution

Finding the Ksp of CaF2

To solve this problem, first, we need to look up the Ksp value of calcium fluoride (CaF2). The Ksp of CaF2 is 3.9 x 10^-11.

Write the balanced equation for the dissolution of CaF2

The balanced equation for the dissolution of CaF2 is as follows: CaF2 (s) <=> Ca2+ (aq) + 2F- (aq)

Calculate the concentration of F- after fluoride addition

The problem states that 2.0 ppm of F- ions are added to the water supply. We need to convert this value to molarity to compare it with the Ca2+ concentration. 1 ppm is equal to 1mg/L. So, the concentration of F- after the fluoridation can be calculated as follows: 2.0 mg/L * (1 mol/18.998 g) * (1/1000 L) = 1.05 x 10^-4 M

Calculate the ion product (Qsp)

Now, we need to calculate the ion product (Qsp) using the given concentrations of Ca2+ and F-. According to the balanced equation, the stoichiometry is 1 mole of Ca2+ for every 2 moles of F-. The Qsp equation is: Qsp = [Ca2+] * [F-]^2 Using the given concentrations: Qsp = (3.5 x 10^-4 M) * (1.05 x 10^-4 M)^2 = 3.87 x 10^-12

Compare Qsp to Ksp

Finally, we need to compare the ion product (Qsp) to the solubility product constant (Ksp) to determine if a precipitate will form. The Ksp of CaF2 is 3.9 x 10^-11 and our calculated Qsp is 3.87 x 10^-12. Since Qsp < Ksp, the solution stays unsaturated, and no precipitate of CaF2 will form when the water is fluoridated.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, is a special type of equilibrium constant that measures the extent to which a compound can dissolve in water. Specifically, it tells us the maximum concentration of ions that can be present in a solution before the compound starts to precipitate. The Ksp value is specific to each solid substance and is temperature-dependent.

In mathematical terms, for a compound like calcium fluoride (\text{CaF}_2), which dissociates into \text{Ca}^{2+} and \text{F}^{-} ions in water, the Ksp expression would be written as follows:\[ Ksp = [Ca^{2+}][F^{-}]^2 \]
Here, the square brackets indicate the molar concentration of the ions at equilibrium. The superscript ‘2’ for the fluoride ion accounts for its stoichiometry in the dissolution reaction, as there are two fluoride ions for every calcium ion. Knowing the Ksp value allows us to predict whether a precipitate will form when we mix two solutions that contain these ions.

Ion Product (Qsp)

The ion product (Qsp) is similar to the Ksp but it is not an equilibrium constant. Instead, it represents the product of the concentrations of the ions at any moment in time, which might not be at equilibrium. The Qsp is particularly useful for predicting whether a precipitate will form under given conditions. The formula for the ion product is the same as the Ksp expression, but with the actual, non-equilibrium concentrations of the ions.

When comparing Qsp and Ksp:

• If Qsp < Ksp, the solution is unsaturated, and no precipitate forms.
• If Qsp = Ksp, the solution is at equilibrium, and the system is saturated without any excess precipitate.
• If Qsp > Ksp, the solution is supersaturated, and a precipitate will form until equilibrium is reached.

The concept of ion product gives us a powerful tool to prevent or encourage the formation of precipitates in industrial processes, water treatment, and other applications.

Precipitation Reactions

Precipitation reactions occur when ions in solution combine to form an insoluble compound or precipitate. These reactions are governed by the solubility rules of the ions involved. The solubility rules help predict which combinations of ions will produce a precipitate and which will remain in solution.

Generally, a precipitation reaction can be represented by a chemical equation.For instance, the precipitation reaction for calcium fluoride is:\[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^{-} (aq) \]
It is important to note that not all reactions where ions come together will result in a precipitate. We must always compare the ion product of the reactants' concentrations with the solubility product constant to determine whether the reaction leads to precipitation. Understanding these reactions is crucial for applications ranging from analytical chemistry to environmental science.

Molar Concentration

Molar concentration, also known as molarity, is expressed as the number of moles of a substance per liter of solution. This unit of concentration is widely used in chemistry because it directly relates the volume of the solution to the amount of dissolved substance. The formula to calculate molarity (M) is:\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
In the context of water fluoridation, converting parts per million (ppm) to molarity is important for determining the concentration of fluoride ions added to the water supply. By knowing the molar concentration, we can then use stoichiometry to relate the reactants and predict possible changes in the solution, such as the potential formation of precipitates.