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Calculate Oh and Ph of 0.105 M Naf

Reviewed by Calculator Editorial Team

This guide explains how to calculate the hydroxide concentration ([OH⁻]) and pH of a 0.105 M sodium fluoride (NaF) solution. We'll cover the chemical equilibrium, calculation method, and practical applications of this calculation.

Introduction

When sodium fluoride (NaF) dissolves in water, it dissociates completely into sodium ions (Na⁺) and fluoride ions (F⁻). The fluoride ions then react with water to form hydrofluoric acid (HF) and hydroxide ions (OH⁻):

F⁻ + H₂O ⇌ HF + OH⁻

This reaction is weak, so the equilibrium constant (Kb) is small. The equilibrium expression for this reaction is:

Kb = [HF][OH⁻] / [F⁻]

For a 0.105 M NaF solution, we can calculate the [OH⁻] concentration and then determine the pH using the relationship between [OH⁻] and pH:

pH = 14 - pOH

pOH = -log[OH⁻]

Calculation Method

To calculate the [OH⁻] concentration and pH of a 0.105 M NaF solution:

  1. Determine the initial concentration of F⁻ ions (equal to the NaF concentration since NaF is a strong electrolyte).
  2. Use the equilibrium constant (Kb) for the reaction between F⁻ and H₂O to form HF and OH⁻.
  3. Set up the equilibrium expression and solve for [OH⁻].
  4. Calculate pOH from [OH⁻] and then determine pH.

The equilibrium constant Kb for the reaction between F⁻ and H₂O is approximately 1.6 × 10⁻³ at 25°C.

Example Calculation

Let's calculate the [OH⁻] concentration and pH for a 0.105 M NaF solution:

  1. Initial [F⁻] = 0.105 M
  2. Let x = [OH⁻] = [HF] (since the reaction goes to completion)
  3. At equilibrium, [F⁻] = 0.105 M - x
  4. Set up the equilibrium expression:

Kb = (x)(x) / (0.105 - x)

1.6 × 10⁻³ = x² / (0.105 - x)

Solving this quadratic equation gives x ≈ 0.0127 M.

Therefore, [OH⁻] ≈ 0.0127 M.

pOH = -log[OH⁻] ≈ -log(0.0127) ≈ 1.90

pH = 14 - pOH ≈ 14 - 1.90 ≈ 12.10

Results for 0.105 M NaF

[OH⁻] ≈ 0.0127 M
pH ≈ 12.10

This solution is basic due to the formation of hydroxide ions from the reaction between fluoride ions and water.

Frequently Asked Questions

Why is the pH of a 0.105 M NaF solution higher than 7?

The pH is higher than 7 because fluoride ions react with water to form hydroxide ions, making the solution basic.

What is the equilibrium constant Kb for this reaction?

The equilibrium constant Kb is approximately 1.6 × 10⁻³ at 25°C.

How does the concentration of NaF affect the pH?

Higher concentrations of NaF will result in higher [OH⁻] concentrations and lower pH values (more basic).