Problem 102 - 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 \(1.0 \times 10^{6}\) L of water? \(\left(K_{s p} \text { for } \operatorname{CaF}_{2}\) is \right. \(\left.5.3 \times 10^{-11} .\right)\)

Step-by-Step Solution

Answer

780.76 g of \(\text{CaF}_2\) is required.

Step 1: Understanding the Dissolution of Calcium Fluoride

Calcium fluoride, represented by the chemical formula \( \text{CaF}_2 \), dissociates in water to release calcium and fluoride ions. The balanced dissociation equation is: \[ \text{CaF}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2\text{F}^- (aq) \] This means one mole of \( \text{CaF}_2 \) produces two moles of fluoride ions.

Step 2: Using the Solubility Product Constant

The solubility product constant (\( K_{sp} \)) for \( \text{CaF}_2 \) is given as \( 5.3 \times 10^{-11} \). This can be expressed in terms of the concentrations of the dissociated ions: \[ K_{sp} = [\text{Ca}^{2+}] [\text{F}^-]^2 \]We will set \( [\text{Ca}^{2+}] = x \) and \( [\text{F}^-] = 2x \) based on the stoichiometry of the dissolution.

Step 3: Solve for Fluoride Ion Concentration

The desired fluoride ion concentration is \( 2.0 \times 10^{-5} \text{ M} \). Using \( [\text{F}^-] = 2x \), we find that \[ 2x = 2.0 \times 10^{-5} \] Thus, \[ x = 1.0 \times 10^{-5} \text{ M} \], which is the concentration of calcium ions.

Step 4: Calculate Moles of Calcium Fluoride Required

Since the ratio between \( \text{CaF}_2 \) and \( \text{Ca}^{2+} \) ions is 1:1 from the dissolution equation, the moles of \( \text{CaF}_2 \) needed are equal to the moles of calcium ions in the water. Using the solution volume of \( 1.0 \times 10^6 \) L: \[ \text{Moles of } \text{Ca}^{2+} = 1.0 \times 10^{-5} \text{ mol/L} \times 1.0 \times 10^6 \text{ L} = 10 \text{ mol} \]

Step 5: Calculate Mass of Calcium Fluoride

The molar mass of \( \text{CaF}_2 \) is calculated by adding the atomic masses of one calcium (40.08 g/mol) and two fluorines (2x18.998 g/mol): \[ \text{Molar Mass of CaF}_2 = 40.08 + 2 imes 18.998 = 78.076 \text{ g/mol} \] The mass of \( \text{CaF}_2 \) needed is then: \[ \text{Mass} = \text{moles} \times \text{molar mass} = 10 \text{ mol} \times 78.076 \text{ g/mol} = 780.76 \text{ g} \]

Key Concepts

Fluoride Ion Concentration

When calcium fluoride (CaF_2) dissolves in water, it dissociates into calcium ions (Ca^{2+}) and fluoride ions (F^-). Understanding this behavior is crucial for determining the concentration of fluoride ions in a solution.
The dissolution equation is given as:\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) \]This means for every mole of CaF_2 dissolved, two moles of fluoride ions are produced. The fluoride ion concentration is then twice the number of moles of CaF_2 ions entering the solution.
If you need to maintain a specific fluoride ion concentration, such as 2.0 \times 10^{-5} M, you have to ensure that the dissolution results in the desired amount of fluoride ions. This concentration is achieved by manipulating the amount of calcium fluoride added to the water, acknowledging that the fluoride ions directly affect the fluoride ion concentration in the water.

Solubility Product Constant

The solubility product constant, abbreviated K_{sp}, refers to the equilibrium constant for a solid substance dissolving in an aqueous solution. This value indicates how much of the ionic compound can dissolve to yield ions in water.
For CaF_2, the solubility product constant (K_{sp}) is given as 5.3 \times 10^{-11}. This shows that CaF_2 is not very soluble in water, meaning only a small amount can dissolve.
Mathematically, this expression is shown as:\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]Where:

• [\text{Ca}^{2+}] is the concentration of calcium ions,
• [\text{F}^-] is the concentration of fluoride ions,

Given the stoichiometry of the dissolution, the Ca^{2+} concentration is x, and F^- concentration is 2x. Applying these values to the K_{sp} equation helps calculate the exact concentrations of ions that will reach equilibrium in the solution.

Molar Mass Calculation

The molar mass of a compound is the mass of one mole of that substance. It is calculated by summing the atomic masses of all the atoms in a molecule of the compound. Applying this to calcium fluoride:

• The atomic mass of calcium (Ca) is approximately 40.08 g/mol.
• Each fluorine (F) atom has an atomic mass of about 18.998 g/mol.

Since CaF_2 contains one calcium atom and two fluorine atoms, its molar mass is:\[ \text{Molar Mass of CaF}_2 = 40.08 + 2 \times 18.998 = 78.076 \text{ g/mol} \]Understanding the molar mass is essential when calculating the mass of CaF_2 needed to achieve a desired fluoride ion concentration, as it allows you to convert between moles of the compound and grams, which is a practical unit for measuring substances for chemical reactions.