Understanding the Henderson-Hasselbalch Equation: Buffer pH Made Easy

Buffers are solutions that maintain pH stability even when small amounts of acid or base are added. The Henderson-Hasselbalch equation is a powerful tool that allows us to quickly calculate the pH of these buffer solutions. In this guide, we break down the Henderson-Hasselbalch equation, explore its components, and demonstrate its practical application with examples.

What Is the Henderson-Hasselbalch Equation?

The Henderson-Hasselbalch equation is used to estimate the pH of a buffer solution:

$$\text{pH} = \text{p}K_a + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$

Key Variables:

• pH: The measure of acidity in the solution.
• pKa: The negative logarithm of the acid dissociation constant (Ka), representing the strength of the weak acid.
• [A-]: The concentration of the conjugate base.
• [HA]: The concentration of the weak acid.

This equation is particularly useful for buffer solutions because it relates the pH of a solution to the ratio of the concentrations of the conjugate base ([A-]) and the weak acid ([HA]).

Why Buffers Matter

Buffers are composed of a weak acid and its conjugate base (or a weak base and its conjugate acid). These components allow the solution to resist changes in pH when small amounts of acid or base are added. The Henderson-Hasselbalch equation explains why buffers are most effective when the concentrations of the acid and its conjugate base are equal. In this case, the logarithmic term becomes zero, and the pH is equal to the pKa.

Breaking Down the Equation

1. pH: Represents how acidic or basic a solution is.
2. pKa: Measures the strength of the weak acid; a lower pKa value means a stronger acid.
3. $\log(\frac{[A^-]}{[HA]})$: Describes the ratio of the conjugate base to the acid. When [A-] = [HA], the logarithmic term becomes zero, and pH = pKa.

Example Problems Using the Henderson-Hasselbalch Equation

Example 1: Directly Stated Buffer

Problem:
Calculate the pH of a buffer containing 0.5M CH₃COOH (acetic acid) mixed with 0.25M CH₃COONa (sodium acetate), given Ka = 1.8 × 10⁻⁵.

Solution:

1. Calculate the pKa of CH₃COOH: $$\text{pKa} = -\log(1.8 \times 10^{-5}) \approx 4.74$$
   
2. Apply the Henderson-Hasselbalch equation: $$\text{pH} = 4.74 + \log\left(\frac{0.25}{0.5}\right) = 4.74 + \log(0.5) = 4.74 – 0.30 \approx 4.44$$
   

Example 2: Using the Henderson-Hasselbalch Equation During a Titration

Problem:
Calculate the pH during the titration of 25.0 mL of 0.100M acetic acid with 15.0 mL of 0.100M NaOH.

Solution:

1. Write the net ionic equation for the reaction:

   $$\text{CH₃COOH} + \text{OH}^- \rightarrow \text{CH₃COO}^- + \text{H₂O}$$
   

2. Calculate mmol of reactants:

   • CH₃COOH: $25 \times 0.100 = 2.5$ mmol
   • OH⁻: $15 \times 0.100 = 1.5$ mmol

3. After reaction:

   • Remaining CH₃COOH: $2.5 – 1.5 = 1.0$ mmol
   • Produced CH₃COO⁻: $1.5$ mmol

4. Use the Henderson-Hasselbalch equation:

   • Convert mmol to molarity (since volumes cancel out, use the ratio): $$\text{pH} = \text{pKa} + \log\left(\frac{1.5}{1.0}\right) = 4.74 + \log(1.5) \approx 4.74 + 0.18 = 4.92$$
     

Practical Applications of the Henderson-Hasselbalch Equation

• Buffers in Biology: Maintaining stable pH in blood and cellular fluids.
• Industrial Processes: Ensuring consistent pH levels in chemical reactions.
• Titrations: Determining pH changes and buffer capacities during acid-base titrations.

Quick Recap

• The Henderson-Hasselbalch equation relates pH, pKa, and the ratio of the concentrations of a conjugate base and weak acid.
• Buffers work best when [A-] = [HA], making pH = pKa.