The Henderson–Hasselbalch equation
Master the Henderson–Hasselbalch equation pH = pKa + log([A⁻]/[HA]): its derivation, how to use it, and its limits.
Quick answer
The Henderson–Hasselbalch equation gives the pH of a conjugate-pair solution: pH = pKa + log([base]/[acid]), where [base] is the conjugate base concentration and [acid] the weak acid concentration.
Example: An acetate buffer with [CH₃COO⁻] = 0.10 M and [CH₃COOH] = 0.10 M (pKa = 4.74): pH = 4.74 + log(0.10/0.10) = 4.74 + 0 = 4.74. With [CH₃COO⁻] = 0.25 M and [CH₃COOH] = 0.10 M: pH = 4.74 + log(2.5) = 4.74 + 0.40 = 5.14.
What you will learn
- Write the Henderson–Hasselbalch equation and identify each term.
- Derive it from the Ka expression.
- Calculate buffer pH from a concentration ratio, and the ratio from a target pH.
- State the approximations behind the equation and when they fail.
Derivation from Ka
Start from the Ka expression for HA ⇌ H⁺ + A⁻:
Solve for [H⁺], take the negative logarithm of both sides, and rearrange:
That is the entire derivation — the Henderson–Hasselbalch equation is the Ka expression wearing logarithmic clothes. [A⁻] is the conjugate base, [HA] the weak acid. No new chemistry was introduced; the power is in the convenient form.
Using the equation: two worked examples
Example 1 — equal partners. An acetate buffer with [CH₃COO⁻] = 0.10 M and [CH₃COOH] = 0.10 M (pKa = 4.74):
Example 2 — unequal partners. [CH₃COO⁻] = 0.25 M, [CH₃COOH] = 0.10 M:
More base than acid pushes pH above pKa; more acid than base pulls it below. Each tenfold change in the ratio moves pH by exactly one unit.
Designing a buffer: solving for the ratio
The equation also runs backward — designing buffers. Want pH 5.00 with an acetate buffer (pKa 4.74)?
Mix acetate and acetic acid in a 1.8:1 ratio. This is the standard buffer-preparation workflow: pick an acid with pKa near the target pH, then dial the ratio.
Assumptions and limits
The equation assumes equilibrium concentrations equal the amounts you mixed — valid when both partners are present in moderate concentration and the ratio is not extreme (roughly 0.1 to 10). It breaks down for very dilute buffers, where dissociation shifts the numbers significantly, and it cannot handle strong acids or bases (no equilibrium to describe).
Also remember: concentrations should technically be activities, and volumes must match (using moles works only if both are in the same solution volume — though the volume cancels in the ratio, so moles are fine).
Variables and units
| Symbol | Name | Unit | Meaning |
|---|---|---|---|
| pKa | unitless | Negative log of the acid dissociation constant for the pair's acid | |
| Base concentration | M | Equilibrium concentration of the conjugate base | |
| Acid concentration | M | Equilibrium concentration of the weak acid |
Common mistakes
It works for any solution containing a conjugate pair in known amounts — buffer or not. It is just the Ka expression rewritten. The 'buffer' association exists because that is where it is most useful.
The equation uses equilibrium concentrations. Equating them to initial amounts is an approximation that fails for very dilute solutions or extreme ratios. It is excellent in the normal buffer range and shaky outside it.
No — inverting the ratio flips the sign: pH = pKa − log([acid]/[base]). The base always goes on top in the standard form. Writing the ratio upside-down is the most common Henderson–Hasselbalch error.
Real-world applications
Buffer preparation: computing the exact base/acid ratio for a target pH.
Biochemistry: predicting the charge state of amino acids at a given pH.
Pharmacology: estimating how much of a drug is ionized (absorbable) at blood pH.
Analytical chemistry: quick pH estimates for any conjugate-pair mixture.
Frequently asked questions
Where does the equation come from?
Take −log of both sides of Ka = [H⁺][A⁻]/[HA] and rearrange: −log[H⁺] = −log Ka + log([A⁻]/[HA]), which is pH = pKa + log([base]/[acid]). It is the Ka expression in logarithmic disguise — no new chemistry, just algebra.
Is it exact or an approximation?
No — it assumes the equilibrium concentrations equal the prepared (initial) concentrations, i.e., that dissociation shifts them negligibly. That holds when both partners are present in reasonable amounts. For very dilute buffers or extreme ratios, solve the equilibrium properly.
Which pKa do I use for a base-centered buffer like NH₄⁺/NH₃?
Use pKa of the acid form in your pair. For the ammonium/ammonia pair (NH₄⁺/NH₃), use pKa(NH₄⁺) = 9.26 — not 'pKb of ammonia' directly (though pKa + pKb = 14 converts between them).
What happens when [base] = [acid]?
When the ratio is 1, log(1) = 0 and pH = pKa exactly. This is why buffers are made with pKa near the target pH — and why pKa can be read off a titration curve at the half-equivalence point.