Henderson–Hasselbalch Equation: Buffer pH in Three Steps
A buffer resists change in pH when small amounts of acid or base are added. Your blood is a buffer; so is the solution in almost every biochemistry experiment. The Henderson–Hasselbalch equation is the shortcut that turns a buffer question from a quadratic equilibrium calculation into one line of arithmetic — provided you know when you are allowed to use it.
The equation
pH = pKa + log₁₀ ( [conjugate base] / [weak acid] ) = pKa + log ([A⁻]/[HA])
Basic buffer (weak base + its salt):
pOH = pKb + log₁₀ ( [conjugate acid] / [weak base] ) = pKb + log ([BH⁺]/[B])
then pH = 14 − pOH (at 25 °C)
What each term means
| Term | Meaning | Unit |
|---|---|---|
| pKa | −log₁₀ Ka, a fixed property of the weak acid at that temperature | no unit |
| [HA] | Concentration (or moles) of the weak acid — the "salt-donor" | mol L⁻¹ or mol |
| [A⁻] | Concentration (or moles) of its conjugate base, supplied by the salt | mol L⁻¹ or mol |
| pKb | −log₁₀ Kb, for the basic-buffer form | no unit |
A useful consequence: because the two concentrations sit in a ratio, the volume of the solution cancels. You may put moles straight into the log — no need to convert to molarity at all, as long as both species are in the same flask.
The three steps
- Identify the pair — which species is the weak acid and which is its conjugate base — and convert Ka to pKa.
- Work out the ratio [A⁻]/[HA], using moles if the question gives volumes.
- Add the log to pKa. That is the pH.
Worked example 1 — The equal-concentration buffer
A buffer contains 0.100 M CH₃COOH and 0.100 M CH₃COONa. Ka = 1.8 × 10⁻⁵. Find the pH.
pKa = −log(1.8 × 10⁻⁵) = 5 − log 1.8 = 5 − 0.255 = 4.74
Ratio [A⁻]/[HA] = 0.100 / 0.100 = 1, and log 1 = 0
pH = 4.74 + 0 = 4.74
Key result to remember: when acid and salt are equal, pH = pKa. That is also the half-equivalence point of a weak-acid titration, and it is where the buffer works best.
Worked example 2 — Unequal concentrations
0.200 M sodium acetate with 0.100 M acetic acid. Find the pH.
Ratio = 0.200 / 0.100 = 2.00, and log 2.00 = 0.301
pH = 4.74 + 0.301 = 5.04
More conjugate base means a higher pH — the log term is positive. If the acid had been in excess instead, the log would be negative and the pH would fall below pKa.
Worked example 3 — Given volumes, use moles
50.0 mL of 0.200 M CH₃COOH is mixed with 30.0 mL of 0.200 M CH₃COONa. Find the pH.
Moles of acid = 0.200 × 0.0500 = 0.0100 mol
Moles of salt = 0.200 × 0.0300 = 0.00600 mol
Ratio = 0.00600 / 0.0100 = 0.600, and log 0.600 = −0.222
pH = 4.74 − 0.222 = 4.52
Notice we never needed the total volume of 80.0 mL. Both species share it, so it cancels.
Worked example 4 — A basic buffer
0.100 M NH₃ with 0.0500 M NH₄Cl. Kb(NH₃) = 1.8 × 10⁻⁵. Find the pH.
Method A (pKb form):
pKb = 5 − 0.255 = 4.74
pOH = 4.74 + log(0.0500 / 0.100) = 4.74 + log 0.500 = 4.74 − 0.301 = 4.44
pH = 14 − 4.44 = 9.56
Method B (pKa of the conjugate acid):
pKa(NH₄⁺) = 14 − 4.74 = 9.26
pH = 9.26 + log(0.100 / 0.0500) = 9.26 + 0.301 = 9.56 ✔ same answer.
Worked example 5 — Buffer action, proved with numbers
This is the calculation that shows why buffers matter.
1.00 L of buffer contains 0.100 mol CH₃COOH and 0.100 mol CH₃COONa (pH 4.74). Add 0.0100 mol of HCl. What is the new pH?
The added H⁺ converts acetate into acetic acid:
Acid: 0.100 + 0.0100 = 0.110 mol
Base: 0.100 − 0.0100 = 0.0900 mol
Ratio = 0.0900 / 0.110 = 0.818, and log 0.818 = −0.087
pH = 4.74 − 0.087 = 4.65
The pH fell by only 0.09 units. Add the same 0.0100 mol of HCl to 1.00 L of pure water and [H⁺] becomes 0.0100 M, so the pH crashes from 7.00 to 2.00 — a change of five units. That contrast is the whole point of a buffer.
When the equation is NOT valid
- Strong acids and strong bases have no buffer pair. There is no Ka to speak of; use pH = −log[H⁺] directly.
- Ratios far from 1. The useful range is roughly pKa ± 1, i.e. a ratio between 0.1 and 10. Outside that, the buffer has almost no capacity left and the approximation degrades.
- Very dilute buffers. The derivation assumes the acid and salt concentrations are much larger than the amount of ionisation and than 10⁻⁷ M. Below about 10⁻³ M, water's own ions matter and you must solve the full equilibrium.
- After the equivalence point. If added strong acid or base consumes one component entirely, no buffer pair remains and the equation no longer applies.
Common mistakes that cost marks
- Inverting the ratio. Base on top, acid on the bottom. Getting it upside down flips the sign of the log and moves the pH the wrong way by twice the log value.
- Using Ka instead of pKa. Take the negative log first.
- Mixing pKb into the pH form. pKb belongs with pOH. If you want to work in pH for a basic buffer, convert to pKa = 14 − pKb.
- Forgetting to update both species after adding acid or base. One goes up by exactly the amount the other goes down.
- Converting to molarity unnecessarily. Moles are fine — but only because both species share the same volume. Never mix a mole value with a molarity value in the same ratio.
- Ignoring the ± 1 range. A "buffer" of pKa 4.74 asked to hold pH 8 is not a buffer at all.
Where buffers appear in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11–12 | Ionic equilibrium: buffer pH, buffer action, blood and biological buffers |
| JEE / NEET | pH after adding strong acid/base to a buffer, half-equivalence point |
| IIT-JAM / CUET-PG | Buffer capacity, choosing a buffer for a given pH, titration curves |
| GATE / CSIR-NET | Biochemistry buffers, amino-acid isoelectric points, analytical method design |
Check any buffer in one line. The Buffer (Henderson–Hasselbalch) calculator takes pKa or Ka with the acid and salt amounts and returns the pH, so you can confirm every step of a long ionic-equilibrium question.
Open the Buffer / Henderson–Hasselbalch Calculator →Buffers are a reliable source of marks once the logic clicks. ABC Chemistry teaches Class 11–12 chemistry at the Gurugram coaching centre and through online classes across India, with home tuition available in Delhi-NCR — see abcchemistry.in.