Problem 78 - Chemistry and Chemical Reactivity

Question

A sample of hard water contains about \(2.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ca}^{2+} .\) A soluble fluoride-containing salt such as NaF is added to "fluoridate" the water (to aid in the prevention of dental cavities). What is the maximum concentration of \(\mathrm{F}^{-}\) that can be present without precipitating \(\mathrm{CaF}_{2} ?\) (IMAGE CAN'T COPY)

Step-by-Step Solution

Answer

The maximum concentration of \( \mathrm{F}^{-} \) without precipitating \( \mathrm{CaF}_{2} \) is \(1.4 \times 10^{-4}\, \mathrm{M}.\)

Step 1: Understand the Problem

We are tasked with determining the maximum concentration of fluoride ions (\( \mathrm{F}^{-} \)) in a water sample that already contains calcium ions (\(\mathrm{Ca}^{2+}\)) without causing the calcium fluoride (\(\mathrm{CaF}_{2}\)) to precipitate. This involves using the solubility product constant \(K_{sp}\) for \(\mathrm{CaF}_{2}\).

Step 2: Write the Equilibrium Expression

The solubility of calcium fluoride in water is described by the equilibrium \[\mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\, \mathrm{F}^{-}(aq)\] where the solubility product \(K_{sp}\) is given by \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2}\).

Step 3: Look Up or Determine \(K_{sp}\) for \(\mathrm{CaF}_{2}\)

From the table of solubility product constants, we find that the \(K_{sp}\) for \(\mathrm{CaF}_{2}\) is \(3.9 \times 10^{-11}\).

Step 4: Calculate Maximum \([\mathrm{F}^{-}]\) Concentration

Using the equation from Step 2: \[\begin{align*}K_{sp} &= [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2} \3.9 \times 10^{-11} &= (2.0 \times 10^{-3})[\mathrm{F}^{-}]^{2}\end{align*}\]Solve for \([\mathrm{F}^{-}]\):\[[\mathrm{F}^{-}] = \sqrt{\frac{3.9 \times 10^{-11}}{2.0 \times 10^{-3}}}\]

Step 5: Compute the Result

Calculate the fluoride ion concentration:\[[\mathrm{F}^{-}] = \sqrt{\frac{3.9 \times 10^{-11}}{2.0 \times 10^{-3}}} = \sqrt{1.95 \times 10^{-8}} \approx 1.4 \times 10^{-4}\] Therefore, the maximum concentration of \([\mathrm{F}^{-}]\) is approximately \(1.4 \times 10^{-4}\).

Step 6: Conclusion: Interpret the Result

The calculation shows that the maximum allowable concentration of fluoride ions before precipitation of \(\mathrm{CaF}_{2}\) occurs is \(1.4 \times 10^{-4}\) mol/L.

Key Concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, commonly known as Ksp, is a special equilibrium constant used in the context of sparingly soluble salts. Unlike regular equilibrium constants used for reactions in solutions, the Ksp represents the point at which a salt dissolves in water until it reaches a state of dynamic equilibrium. When an ionic compound dissolves in water, it splits into its respective ions. Let's take calcium fluoride (CaF₂) for example. When it dissolves, it separates into calcium ions (Ca²⁺) and fluoride ions (F⁻). This dissociation is represented by the equation:

• CaF₂(s) ⇌ Ca²⁺(aq) + 2 F⁻(aq)

The expression for the solubility product constant is derived from this equation. It's calculated as:

• Ksp = [Ca²⁺][F⁻]²

Given the specific Ksp value for CaF₂, which is 3.9 x 10⁻¹¹, we use this in calculations to determine concentrations of ions in solution at equilibrium. Understanding Ksp helps predict whether a compound will precipitate from a solution or remain dissolved.

Precipitation Reactions

Precipitation reactions occur when solutions mix, and an insoluble substance forms as a result of the reaction. This solid, known as a precipitate, differs from both the reactants in the solution. For instance, in our scenario where we want to avoid calcium fluoride (CaF₂) precipitation, we must understand the conditions under which CaF₂ becomes insoluble in water. Precipitation occurs when the product of the ion concentrations exceeds their Ksp.
In the context of CaF₂, precipitation will occur if the product of the concentration of Ca²⁺ ions and the square of the concentration of F⁻ ions surpasses 3.9 x 10⁻¹¹. Think of precipitation like a tipping point: if too many dissolved ions are present, they "tip" out of solution, forming a solid. We use the Ksp value as a guide to ensure that the ion concentrations remain below this tipping point, maintaining a clear, homogenous solution.

Ion Concentration Calculations

Calculating ion concentrations is essential to manage the delicate balance between a clear solution and precipitation. In our scenario, the objective is to find the maximum concentration of fluoride ions (F⁻) in a solution that contains a known concentration of calcium ions (Ca²⁺) without forming a precipitate. Given that the concentration of Ca²⁺ is 2.0 x 10⁻³ M, and using the equation for Ksp of CaF₂:

• Ksp = [Ca²⁺][F⁻]² = 3.9 x 10⁻¹¹

We rearrange this equation to solve for the concentration of F⁻:

• [F⁻]² = Ksp / [Ca²⁺] = 3.9 x 10⁻¹¹ / 2.0 x 10⁻³
• [F⁻] = √(1.95 x 10⁻⁸)
• [F⁻] ≈ 1.4 x 10⁻⁴ M

This calculation outlines the threshold for fluoride ions before the formation of a calcium fluoride precipitate, ensuring that your water remains free of unwanted solids.