Problem 47 - Chemistry: Principles and Reactions

Question

What is the concentration of fluoride ion in a water solution saturated with \(\mathrm{BaF}_{2}, K_{s \mathrm{p}}=1.8 \times 10^{-7}\) ?

Step-by-Step Solution

Answer

Answer: The concentration of fluoride ions in the saturated water solution is approximately \(5.88 \times 10^{-3}\,\text{M}\).

Step 1: Write the dissolution equilibrium equation

Write the balanced chemical equation for the dissolution of BaF2 in water: \[\text{BaF}_2(s) \rightleftharpoons \text{Ba}^{2+}(aq) + 2\text{F}^-(aq)\]

Step 2: Write the solubility product expression

Write the expression for Ksp for BaF2: \[K_\text{sp} = [\text{Ba}^{2+}][\text{F}^-]^2\]

Step 3: Define variables for ion concentrations

Let x be the concentration of Ba2+ ions at equilibrium. Then, the concentration of F- ions at equilibrium will be 2x because there are two F- ions for each Ba ion: \[[\text{Ba}^{2+}] = x\] \[[\text{F}^-] = 2x\]

Step 4: Substitute concentrations into the Ksp expression and solve for x

Insert the expressions for the ion concentrations into the Ksp expression, and then solve for x: \[K_\text{sp} = [\text{Ba}^{2+}][\text{F}^-]^2 = (x)(2x)^2\] \[K_\text{sp} = 4x^3\] \[x = \sqrt[3]{\frac{K_\text{sp}}{4}}\]

Step 5: Calculate the value of x

Substitute the given value for Ksp into the equation and solve for x: \[x = \sqrt[3]{\frac{1.8 \times 10^{-7}}{4}}\] \[x ≈ 2.94 \times 10^{-3}\,\text{M}\]

Step 6: Calculate the concentration of fluoride ions

Now that we have the concentration of Ba2+ ions at equilibrium (x), we can find the concentration of F- ions by multiplying x by 2: \[[\text{F}^-] = 2x = 2 \times 2.94 \times 10^{-3}\,\text{M} \approx 5.88 \times 10^{-3}\,\text{M}\] In conclusion, the concentration of fluoride ions in the saturated water solution is approximately \(5.88 \times 10^{-3}\,\text{M}\).

Key Concepts

Dissolution Equilibrium

The concept of dissolution equilibrium is fundamental in understanding how compounds dissolve in solutions. Dissolution involves the dissolving of a solute in a solvent to form a solution. When a solid like barium fluoride (\( \text{BaF}_2 \)) is added to water, it reaches a point where it cannot dissolve any further, and a saturated solution is formed. At this point, the rate at which \( \text{BaF}_2 \) dissolves into its ions equals the rate at which the ions come together to form \( \text{BaF}_2 \) again.

In a chemical equation, dissolution equilibrium can be represented as:\[ \text{BaF}_2(s) \rightleftharpoons \text{Ba}^{2+}(aq) + 2\text{F}^-(aq) \]

• The double arrow (↔) signifies that the reaction can go both ways, emphasizing equilibrium.
• Each side of the equation has equal rates of reaction, implying a balance in concentrations over time.

This balance allows for the calculation of different ion concentrations in a solution and is fundamental to predicting how much solute can be dissolved under certain conditions.

Ion Concentration

Ion concentration is a measure of how much of a particular ion is present in a solution. For compounds like \( \text{BaF}_2 \), dissolution leads to the formation of ions in specific ratios. When \( \text{BaF}_2 \) dissolves, it splits into one barium ion (\( \text{Ba}^{2+} \)) and two fluoride ions (\( \text{F}^- \)).

The goal is often to find these concentrations at equilibrium. In this scenario, if \( x \) is the concentration of \( \text{Ba}^{2+} \), then \( 2x \) would be the concentration of \( \text{F}^- \) due to its stoichiometry:

• The stoichiometric coefficients (from the balanced equation) guide us in determining the relational expression for ion concentrations.
• The solubility product equation helps relate the concentrations: \( K_\text{sp} = [\text{Ba}^{2+}][\text{F}^-]^2 \)

This relationship is crucial in determining how much of a solute can form ions before they precipitate out of the solution.

Fluoride Ion Concentration

Fluoride ion concentration specifically refers to the amount of \( \text{F}^- \) present in a solution at equilibrium. When calculating this, it helps in predicting the solubility and the potential effects of fluoride in a solution.

For \( \text{BaF}_2 \), the balanced dissolution equation indicates that there are two fluoride ions for every barium ion released:

• The concentration of \( \text{F}^- \) is \( 2x \), where \( x \) is the concentration of barium ions at equilibrium.
• By solving the expression \( 4x^3 = K_\text{sp} \), we can find \( x \) and thereafter \( 2x \), which gives us fluoride concentration.

In the exercise, substituting the given \( K_\text{sp} \) value, we arrive at a fluoride concentration of approximately \( 5.88 \times 10^{-3} \,\text{M} \).This calculation is important in fields like chemistry and environmental science, where it's necessary to understand solubility dynamics and the safety or toxicity of fluoride levels.