Calculate H3o of A 0.150 M Acetic Acid Solution

Introduction

When acetic acid (CH₃COOH) dissolves in water, it partially dissociates into acetate ions (CH₃COO^-) and hydronium ions (H₃O⁺). The concentration of H₃O⁺ ions is crucial for understanding the acidity of the solution.

The dissociation constant (K_a) for acetic acid is 1.8 × 10^-5 at 25°C. This value helps calculate the equilibrium concentration of H₃O⁺ in a given solution.

Formula

The concentration of H₃O⁺ in a weak acid solution can be calculated using the following steps:

1. Assume the initial concentration of H₃O⁺ is equal to the initial concentration of acetic acid (x = [CH₃COOH]).
2. Set up the equilibrium expression for the dissociation of acetic acid:
   Kₐ = [H₃O⁺][CH₃COO⁻] / [CH₃COOH]
3. Since the initial concentration of CH₃COO^- is zero, the equilibrium concentration of H₃O⁺ is equal to the equilibrium concentration of CH₃COO^-.
4. Solve the quadratic equation to find the equilibrium concentration of H₃O⁺:
   x² + Kₐx - Kₐ[CH₃COOH] = 0

Example Calculation

Let's calculate the concentration of H₃O⁺ in a 0.150 M acetic acid solution:

1. Given: [CH₃COOH] = 0.150 M, K_a = 1.8 × 10^-5.
2. Assume x = [H₃O⁺] = [CH₃COO^-].
3. Set up the quadratic equation:
   x² + (1.8 × 10⁻⁵)x - (1.8 × 10⁻⁵)(0.150) = 0
4. Solve for x using the quadratic formula:
   x = [-b ± √(b² - 4ac)] / 2a
   Where a = 1, b = 1.8 × 10^-5, and c = -2.7 × 10^-6.
5. Calculate the discriminant:
   √[(1.8 × 10⁻⁵)² - 4(1)(-2.7 × 10⁻⁶)] = √(3.24 × 10⁻¹⁰ + 1.08 × 10⁻⁵) ≈ √(1.08 × 10⁻⁵) ≈ 3.29 × 10⁻³
6. Find the positive root:
   x = [-1.8 × 10⁻⁵ + 3.29 × 10⁻³] / 2 ≈ 1.645 × 10⁻³ M

The concentration of H₃O⁺ in this solution is approximately 1.65 × 10^-3 M.

Interpreting Results

The calculated concentration of H₃O⁺ provides insight into the acidity of the solution. A higher concentration indicates a more acidic solution. The pH of the solution can be calculated using the formula:

For a 0.150 M acetic acid solution, the pH is approximately 2.78, indicating a weak acid solution.