Problem 76 - Chemistry and Chemical Reactivity

Question

Calcium fluoride can be used in the fluoridation of municipal water supplies. If you want to achieve a fluoride ion concentration of \(2.0 \times 10^{-5} \mathrm{M},\) what mass of \(\mathrm{CaF}_{2}\) must you use for \(1.0 \times 10^{6} \mathrm{L}\) of water? \(\left(K_{\mathrm{sp}}\) for \right. \(\left.\mathrm{CaF}_{2} \text { is } 5.3 \times 10^{-11} .\right)\)

Step-by-Step Solution

Answer

780.8 grams of \(\mathrm{CaF}_{2}\) is needed.

Step 1: Understand the Dissociation of Calcium Fluoride

Calcium fluoride ( \(\mathrm{CaF}_{2} \)) dissociates in water according to the equation: \(\mathrm{CaF}_{2} \rightleftharpoons \mathrm{Ca}^{2+} + 2\mathrm{F}^{-} \). This tells us that each mole of \(\mathrm{CaF}_{2}\) produces one mole of calcium ions and two moles of fluoride ions.

Step 2: Relate Moles of Fluoride Ions to Moles of Calcium Fluoride

To find the number of moles of \(\mathrm{CaF}_{2}\) correspond with a given fluoride ion concentration, use the stoichiometry:\[ [\mathrm{F}^{-}] = 2x \],where \( x \) is the molarity of \(\mathrm{CaF}_{2}\). Therefore, the concentration of \( \mathrm{CaF}_{2} \) needed is half that of the fluoride ion concentration:\[ x = \frac{[\mathrm{F}^{-}]}{2} = \frac{2.0 \times 10^{-5}}{2} = 1.0 \times 10^{-5} \].

Step 3: Calculate the Volume's Moles of Calcium Fluoride

Use the concentration of \(\mathrm{CaF}_{2}\) in moles per liter (molarity) to find the moles in the given volume:\[ \text{Moles of } \mathrm{CaF}_{2} = [\mathrm{CaF}_{2}] \times \text{Volume} = 1.0 \times 10^{-5} \times 1.0 \times 10^6 = 10 \mbox{ moles} \].

Step 4: Convert Moles to Mass of Calcium Fluoride

First, find the molar mass of \(\mathrm{CaF}_{2}\): Calcium (Ca) is about 40.08 g/mol and each Fluorine (F) is about 19.00 g/mol. Therefore:\[ \text{Molar mass of } \mathrm{CaF}_{2} = 40.08 + 2(19.00) = 78.08 \text{ g/mol} \].Now, convert moles to mass:\[ \text{Mass} = \text{Moles} \times \text{Molar mass} = 10 \times 78.08 = 780.8 \text{ grams} \].

Step 5: Verify the Solubility Product Is Not Exceeded

The \(K_{sp}\) for \(\mathrm{CaF}_{2}\) is given as \(5.3 \times 10^{-11}\).Ensure the fluoride concentration used doesn't exceed the solubility product:\[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 = x(2x)^2 = 4x^3.\]With \(x = 1.0 \times 10^{-5} \):\[ 4(1.0 \times 10^{-5})^3 = 4 \times 10^{-15} \leq 5.3 \times 10^{-11} \].So, the calculated fluoride level is below the \(K_{sp}\) limit.

Key Concepts

Stoichiometry

Stoichiometry plays a crucial role when it comes to calculating the amount of substances involved in chemical reactions. In the case of calcium fluoride (\(\mathrm{CaF}_2\)), when it dissolves in water, it breaks down into calcium ions (\(\mathrm{Ca}^{2+}\)) and fluoride ions (\(\mathrm{F}^{-}\)). This is depicted by the equation: \[\mathrm{CaF}_2 \rightleftharpoons \mathrm{Ca}^{2+} + 2\mathrm{F}^{-}\]. Each mole of \(\mathrm{CaF}_2\) yields one mole of \(\mathrm{Ca}^{2+}\) and two moles of \(\mathrm{F}^{-}\). From the stoichiometry of this reaction, we can see the relationship between \(\mathrm{CaF}_2\) and \(\mathrm{F}^{-}\) ions.

• 1 mole of \(\mathrm{CaF}_2\) produces 1 mole of \(\mathrm{Ca}^{2+}\).
• 1 mole of \(\mathrm{CaF}_2\) yields 2 moles of \(\mathrm{F}^{-}\).

This means the concentration of \(\mathrm{CaF}_2\) in solution is half that of \(\mathrm{F}^{-}\) ions. To get the required number of moles for a given \(\mathrm{F}^{-}\) concentration, we divide the fluoride concentration by two. This understanding ensures accurate calculations of the chemicals needed in the solution.

Solubility Product (Ksp)

The solubility product constant, \(K_{sp}\), helps us understand the extent to which calcium fluoride will dissolve in water. It is an equilibrium constant specifically for ionic compounds that are sparingly soluble. For \(\mathrm{CaF}_2\), the \(K_{sp}\) is given as \(5.3 \times 10^{-11}\).The expression for the solubility product of \(\mathrm{CaF}_2\) is:\[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^2 \].Since each mole of \(\mathrm{CaF}_2\) produces 1 mole of \(\mathrm{Ca}^{2+}\) and 2 moles of \(\mathrm{F}^{-}\):\[ [\mathrm{Ca}^{2+}] = x \quad \text{and} \quad [\mathrm{F}^{-}] = 2x \].Substituting these into the \(K_{sp}\) equation gives us:\[ K_{sp} = x(2x)^2 = 4x^3 \].This equation tells us how the concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^{-}\) ions relate at equilibrium when \(\mathrm{CaF}_2\) dissolves. For the fluoride concentration to not exceed solubility limits, the calculated value must be less than or equal to the \(K_{sp}\). This ensures that no more \(\mathrm{CaF}_2\) will dissolve once we reach the equilibrium point.

Molar Concentration

Molar concentration, also known as molarity, refers to the number of moles of a solute present in one liter of solution. It's a key concept when determining how much calcium fluoride is needed to achieve a specific concentration of fluoride ions in water.For a desired fluoride ion concentration of \(2.0 \times 10^{-5} \mathrm{M}\), we start by establishing the molarity of \(\mathrm{CaF}_2\), which is half the fluoride concentration due to the stoichiometry:\[ x = \frac{2.0 \times 10^{-5}}{2} = 1.0 \times 10^{-5} \mathrm{M} \].If the volume of water is \(1.0 \times 10^6 \text{ L}\), the moles of \(\mathrm{CaF}_2\) needed are calculated using: \[ \text{Moles of } \mathrm{CaF}_2 = [\mathrm{CaF}_2] \times \text{Volume} = 10 \text{ moles} \].This highlights that molarity is foundational in determining how much of a solute is required. Understanding molar concentration ensures you mix accurate solutions for practical applications, like water fluoridation.