The Henderson–Hasselbalch Equation
Simple Explanation
The Henderson–Hasselbalch equation calculates a buffer's pH directly from the ratio of its conjugate base to weak acid concentrations, without needing to solve a full equilibrium problem from scratch.
Why Do We Need It?
This equation is the practical, everyday tool for designing a buffer with a specific target pH — by choosing the right weak acid (based on its pKa) and adjusting the ratio of its two forms, chemists and biologists can create a buffer for almost any desired pH.
Formula
The Henderson–Hasselbalch Equation
pH = pKa + log₁₀([A⁻] / [HA])
Calculates the pH of a buffer solution directly from the ratio of conjugate base to weak acid, without needing a full equilibrium calculation.
- pH
- — the pH of the buffer solution
- pKa
- — −log₁₀(Ka) of the weak acid used in the buffer
- [A⁻]
- — concentration of the conjugate base (e.g. from a salt) in the buffer, in mol/L
- [HA]
- — concentration of the weak acid in the buffer, in mol/L
When to use it: Whenever you need the pH of a buffer solution made from a weak acid and its conjugate base (or a weak base and its conjugate acid).
Worked Example
Calculate a buffer's pH
A buffer is made from acetic acid (Ka = 1.8 × 10⁻⁵, so pKa ≈ 4.74) with [CH₃COOH] = 0.20 mol/L and [CH₃COO⁻] = 0.20 mol/L. Find the pH.
Why Does This Work?
This equation is just a rearranged form of the Ka expression — taking the negative log of both sides of Ka = [H⁺][A⁻]/[HA] and rearranging gives exactly this relationship. It also reveals a useful shortcut: whenever [A⁻] = [HA] (a 1:1 ratio), the log term is zero, so the buffer's pH exactly equals the weak acid's pKa.
Real-Life Example
Designing a buffer for a biology experiment
A biologist needs a buffer that holds steady at pH 7.4 to keep enzymes stable during an experiment.
Using the Henderson–Hasselbalch equation, the biologist picks a weak acid with a pKa close to 7.4, then calculates the exact ratio of acid to conjugate base needed to hit that target pH precisely.
Practice
A buffer has pKa = 4.74, with [A⁻] = 0.40 mol/L and [HA] = 0.20 mol/L. Find the pH (log₁₀2 ≈ 0.30).
HardCommon mistake
Forgetting that the Henderson–Hasselbalch equation only works well for buffers where significant amounts of BOTH the weak acid and conjugate base are present — it breaks down as an approximation if either component is nearly used up.
Quick Review
- pH = pKa + log₁₀([A⁻]/[HA]).
- When [A⁻] = [HA], pH = pKa exactly (log of 1 is 0).
- Lets chemists design a buffer for a specific target pH.