How to Use the Henderson-Hasselbalch Equation

Buffers run on a ratio, not absolute concentrations

The Henderson-Hasselbalch equation tells you something powerful: a buffer’s pH depends on the ratio of conjugate base to acid, not on either concentration alone. Double both, halve both, dilute the whole buffer 10x — the pH stays put. What changes is buffer capacity (how much added acid or base it can absorb), not the pH itself.

pH = pKa + log₁₀([A⁻] / [HA])

The equation falls straight out of the Ka expression after you take the negative log of both sides. The shortcut is so useful that buffer prep, drug formulation, and almost every biochemistry experiment lean on it.

How to use it

Step 1. Identify HA (the weak acid form) and A⁻ (the conjugate base form). Look up pKa. For acetate buffer it’s the pKa of acetic acid (4.74); for phosphate at physiological pH it’s the pKa₂ of H₂PO₄⁻/HPO₄²⁻ (7.20).

Step 2. Take the ratio [A⁻]/[HA]. Note: A⁻ is on top. Inverting the ratio is the single most common mistake.

Step 3. Plug in: pH = pKa + log(ratio).

Step 4. Sanity check the buffer’s working range. A buffer is effective from roughly pKa − 1 to pKa + 1. Outside that window you’re working at a 10:1 ratio or worse, and the buffer has almost no capacity left in one direction.

Worked examples

1. Acetate buffer pH. 0.30 M acetic acid + 0.50 M sodium acetate, pKa = 4.74.

pH = 4.74 + log(0.50/0.30) = 4.74 + 0.222 = 4.96. Inside [3.74, 5.74], so the buffer works.

2. Designing a buffer at a target pH. Phosphate buffer at pH 7.00 with pKa = 7.20.

log([A⁻]/[HA]) = 7.00 − 7.20 = −0.20 → ratio = 10⁻⁰·²⁰ = 0.631. So 0.63 mol HPO₄²⁻ for every 1.00 mol H₂PO₄⁻. The buffer is mildly acid-heavy because the target pH sits below the pKa.

3. Buffer prep recipe. 1.00 L of pH 5.00 acetate buffer, total acetate = 0.200 M, pKa = 4.74.

Ratio: 10^(5.00 − 4.74) = 10^0.26 = 1.820. Let [HA] = x, [A⁻] = 1.820x. Then x + 1.820x = 0.200 → x = 0.0709. So [HA] = 0.0709 M, [A⁻] = 0.1291 M. For 1 L: 0.0709 × 60.05 = 4.26 g acetic acid and 0.1291 × 82.03 = 10.59 g sodium acetate. Dissolve, top to volume, verify pH with a meter (your calculation gets you within 0.05 pH units; the meter handles the rest).

4. Strong acid challenge. 0.010 mol HCl is added to 1.00 L of buffer with [HOAc] = [OAc⁻] = 0.100 M.

The strong acid reacts stoichiometrically: H⁺ + OAc⁻ → HOAc. New concentrations: [OAc⁻] = 0.100 − 0.010 = 0.090, [HOAc] = 0.100 + 0.010 = 0.110. New pH = 4.74 + log(0.090/0.110) = 4.74 − 0.087 = 4.65.

The pH dropped only 0.09 units. Same amount of HCl in a liter of pure water would have crashed the pH from 7 to 2 — that’s the buffer doing its job. Always do the stoichiometric neutralization first, then plug the post-neutralization concentrations into Henderson-Hasselbalch.

5. Basic buffer. Ammonia buffer with 0.20 M NH₃ + 0.30 M NH₄Cl. Use the pKa of NH₄⁺ (9.26). Here NH₃ is the conjugate base, NH₄⁺ is the acid.

pH = 9.26 + log(0.20/0.30) = 9.26 − 0.176 = 9.08.

Same answer if you go through pKb and pOH, but using pKa of the conjugate acid keeps you in pH-space and avoids one round of arithmetic.

Traps to watch for

Inverted ratio. [A⁻] (conjugate base) on top. If you accidentally put HA on top, you’ll get a pH on the wrong side of the pKa — a useful sanity check is to ask: “is my buffer mostly acid (pH < pKa) or mostly base (pH > pKa)?” and confirm the result matches.

Skipping the stoichiometry step when acid or base is added. Henderson-Hasselbalch describes the equilibrium after any added strong acid/base has been neutralized. You can’t just plug the original concentrations and the moles of HCl into the equation — do the subtraction/addition first.

Using it outside its valid range. The derivation assumes [HA] and [A⁻] from the salts dominate over self-dissociation. That breaks when (a) the ratio is more extreme than ~10:1, (b) total concentration drops below ~0.01 M, or (c) you’re trying to buffer near pH 0–2 or 12–14 with a weak acid. At those extremes you need to do the full ICE-table treatment.

Ignoring temperature and ionic strength. pKa values shift with both. Tris is famous for this — Tris pKa drops by ~0.03 per °C, so a buffer prepared to pH 8.0 at 25 °C is actually pH ~7.6 at 37 °C. For protein work, prep your buffer at the temperature you’ll use it.

Practice problems (verify with the Buffer pH Calculator):

1. pH of 0.15 M HF + 0.25 M NaF (Ka = 6.8 × 10⁻⁴)?
2. NH₃/NH₄Cl ratio for pH 10.00 (Kb = 1.8 × 10⁻⁵)?
3. pH after adding 0.020 mol NaOH to 500 mL of 0.200 M HOAc / 0.200 M NaOAc?
4. Tris buffer at pH 7.50, total 0.100 M, 500 mL volume — how much Tris and Tris-HCl? (pKa = 8.07)
5. What’s the pH of an equimolar buffer in terms of pKa?