The pH of a buffer — flat and stubborn near the pKa. For a weak acid HA and its conjugate base A−, the pH is fixed by a single ratio: pH = pKa + log₁₀([A−]/[HA]). Pour in acid or base and the plateau barely moves — the buffer resists.
source Henderson, L.J. (1908), The Theory of Neutrality Regulation in the Animal Organism, Am. J. Physiol. 21(4):427 — doi:10.1152/ajplegacy.1908.21.4.427; logarithmic form: Hasselbalch, K.A. (1917), Biochem. Z. 78:112 amber (no stable DOI).
Rendered, not quoted — the engine below computes pH from the equilibrium law live; no numbers are transcribed.
Dissolve a weak acid HA; it partly dissociates: HA ⇌ H⁺ + A−. At equilibrium the acid constant holds:
Take −log₁₀ of both sides. With pH = −log[H⁺] and pKa = −log Ka, the [H⁺] term becomes pH, the constant becomes pKa, and the ratio survives:
Only the ratio of base to acid matters — not the absolute amounts, and not dilution (to first order).
This is the-equilibrium-constant of a weak-acid solution, solved for pH. Guldberg & Waage give Ka = [H⁺][A−]/[HA]; Henderson-Hasselbalch simply takes its logarithm.
Neighbour sphere: THE MASS-ACTION LAW hands over Ka; this sphere reads pH straight off it. Dead flat at the pKa, where base and acid are equal.
Live re-check of the four laws below. Confirms green; flips red the instant the RED TEAM tamper (window 6) inverts the ratio.
witness idle
A weak acid with pKa, and a base:acid ratio set by titration. Drag to walk the curve.
pH computed live from pKa + log₁₀([A−]/[HA]). Titration curve: pH vs fraction titrated. Note the flat buffer plateau straddling the pKa (marked).
At half-equivalence ([A−]=[HA]) the log is exactly 0, so pH = pKa. Each 10× in the ratio moves pH by exactly +1. Proven at boot:
unverified
"The equation is exact." It is not. Henderson-Hasselbalch assumes activities = concentrations (ignores ionic strength), assumes [A−] and [HA] equal their formal amounts (ignores the acid's own dissociation and water's autoionization), and breaks when the buffer is very dilute or the ratio is extreme. It is an approximation that is excellent within roughly one pH unit of the pKa.
"More total buffer raises the pH."
→ No. pH depends only on the ratio. Doubling both HA and A− leaves pH unchanged — more buffer raises capacity, not pH.
"Buffering is strongest far from the pKa."
→ Reversed. Capacity is maximal at the pKa and collapses past ±1 unit, where the curve steepens.
"pH = pKa + log([HA]/[A−])."
→ Ratio inverted — base over acid, never acid over base. This is exactly the tamper below.
Invert the ratio — write log([HA]/[A−]). pH now moves the wrong way as base is added: pKa−1 at a 10:1 base:acid ratio. The WITNESS (7) catches it instantly.