Question

Calculate the \(\mathrm{pH}\) of a \(0.082 \mathrm{M} \mathrm{NaF}\) solution. \(\left(K_{a}\right.\) for \(\left.\mathrm{HF}=7.1 \times 10^{-4} .\right)\)

Short answer

The pH of the 0.082 M NaF solution is 8.03.

Step-by-step solution

Understand the problem

We are asked to find the pH of a 0.082 M NaF solution. Sodium fluoride (NaF) dissociates completely in water into sodium ions (Na⁺) and fluoride ions (F⁻). The fluoride ion is a conjugate base of hydrofluoric acid (HF) and can hydrolyze in water to produce hydroxide ions, affecting the pH.

Write the hydrolysis reaction

Write the chemical equation for the hydrolysis of fluoride ions in water: \[ \mathrm{F}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HF} + \mathrm{OH}^- \] This reaction generates hydroxide ions, which will affect the pH of the solution.

Calculate the \\(K_b\\) for F⁻

Use the relation between the ionization constant of water \( K_w \) and the acid dissociation constant \( K_a \) of HF to calculate the base dissociation constant \( K_b \) for F⁻: \[ K_w = 1.0 \times 10^{-14}, \quad K_a = 7.1 \times 10^{-4} \]\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{7.1 \times 10^{-4}} = 1.41 \times 10^{-11} \]

Set up the equilibrium expression for hydrolysis

For the hydrolysis reaction of fluoride, set up the expression for \( K_b \): \[ K_b = \frac{[\mathrm{HF}][\mathrm{OH}^-]}{[\mathrm{F}^-]} \] Assuming initial concentration of F⁻ is 0.082 M, at equilibrium let the concentration of OH⁻ be \( x \), then: \[ K_b = \frac{x^2}{0.082 - x} \approx \frac{x^2}{0.082} \]

Solve for hydroxide ion concentration

Assume \( x \ll 0.082 \) and solve for \( x \): \[ 1.41 \times 10^{-11} = \frac{x^2}{0.082} \]\[ x^2 = 1.41 \times 10^{-11} \times 0.082 \]\[ x = \sqrt{1.16 \times 10^{-12}} \approx 1.08 \times 10^{-6} \] Thus, the concentration of hydroxide ions \([\mathrm{OH}^-] = 1.08 \times 10^{-6} \, M\).

Calculate pOH

Calculate the pOH of the solution: \[ \mathrm{pOH} = -\log([\mathrm{OH}^-]) = -\log(1.08 \times 10^{-6}) \approx 5.97 \]

Calculate pH

Use the relation between pH and pOH to find the pH: \[ \mathrm{pH} + \mathrm{pOH} = 14 \]\[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 5.97 = 8.03 \]

Key concepts

Hydrolysis Reaction

Hydrolysis reactions are interactions between water and another chemical that result in the decomposition of that chemical. In our example, the fluoride ion (\(\text{F}^-\)) undergoes a hydrolysis reaction when it reacts with water.
The balanced equation for this reaction is:\[\text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^- \]This reaction is significant because it produces hydroxide ions (\(\text{OH}^-\)), which increase the solution's pH. The production of hydroxide ions is key to understanding how a seemingly neutral salt like sodium fluoride can create a basic solution.
This is a clear example of how salts can participate in hydrolysis reactions to affect the pH of a solution.

Acid-Base Equilibrium

Acid-base equilibrium involves the balance between hydrogen ions (\(\text{H}^+\)) and hydroxide ions (\(\text{OH}^-\)) in a solution.
The equilibrium portrays how acid and base dissociation or association affects the overall hydrogen ion concentration.

• In our scenario, the equilibrium is described by the reaction:\[\text{HF} \rightleftharpoons \text{H}^+ + \text{F}^-\]while also considering its reverse via hydrolysis.

For fluoride, since it is a conjugate base of hydrofluoric acid, it tends to interact with water, pushing the equilibrium to generate more \(\text{OH}^-\), resulting in a base-dominated environment.

This balance directs the equilibrium calculation and allows us to understand why and how pH adjustments occur in solutions with weak acids and bases.

Sodium Fluoride

Sodium fluoride (\(\mathrm{NaF}\)) is an ionic compound that readily dissociates in water. Upon dissolution, it separates into sodium ions (\(\text{Na}^+\)) and fluoride ions (\(\text{F}^-\)). Being an ionic solid, it has a strong crystal lattice that releases these ions easily.

• Sodium (\(\text{Na}^+\)) itself is a spectator ion. It doesn't participate in pH-modifying reactions in this context.
• However, the fluoride ion can affect solution pH through hydrolysis.

Understanding sodium fluoride's behavior in water is crucial as it demonstrates how ionic compounds can have unexpected effects on the pH, despite initially appearing neutral.

Fluoride Ion

The fluoride ion (\(\text{F}^-\)) is the conjugate base of hydrofluoric acid (\(\text{HF}\)). It is a key player in this pH calculation.

When it dissolves in water, as highlighted in the hydrolysis reaction, it reacts to form hydrogen fluoride and hydroxide ions:\[\text{F}^- + \text{H}_2\text{O} \rightleftharpoons \text{HF} + \text{OH}^- \]

Fluoride ions are interesting because they can make a solution basic by increasing the \(\text{OH}^-\) concentration.

Adding the fluoride ion to water enhances its basicity, which affects the pH balance, making the solution less acidic and more basic. This underscores the concept of conjugate bases and their influence on the pH of solutions.