The Henderson–Hasselbalch equation gives the pH of a buffer — a mixture of a weak acid and its conjugate base — from one logarithm: pH = pKa + log([A⁻]/[HA]). The pKa fixes the buffer's centre, and the base-to-acid ratio nudges the pH up or down from there. Because only the ratio appears, the pH is set by proportions, not absolute amounts, which is why a buffer holds its pH when diluted and why buffers are chosen with a pKa close to the pH they need to maintain.
Reviewed: June 19, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the Henderson–Hasselbalch form of the weak-acid equilibrium.
The buffer equations
[A⁻] is the conjugate base and [HA] the weak acid. When they are equal, log(1) = 0 and pH = pKa. Each tenfold change in the ratio moves the pH by exactly one unit, because log(10) = 1. To prepare a buffer at a chosen pH, rearrange to find the ratio: [A⁻]/[HA] = 10^(pH − pKa), then mix the two components in that proportion.
Worked example — an acetate buffer
Scenario: You mix 0.1 M sodium acetate (the conjugate base) with 0.1 M acetic acid (pKa 4.76). What is the buffer pH, and what happens if you double the acetate?
At equal concentrations the buffer sits right at its pKa of 4.76 — its strongest point. Doubling the conjugate base raises the pH by log(2) ≈ 0.30 to 5.06. Push the ratio to 10:1 and you reach 5.76, the practical upper edge of this buffer; beyond that the acetic acid is too depleted to neutralise added base, and the pH would start to climb sharply.
Frequently Asked Questions
pH = pKa + log([A⁻]/[HA]). [A⁻] is conjugate base, [HA] the weak acid. Equal amounts → pH = pKa.
Add log of the base/acid ratio to pKa. Acetate pKa 4.76 at 0.2/0.1 → 4.76 + log2 = 5.06.
log(1) = 0, so pH = pKa exactly — the point of maximum buffer capacity.
About pKa ± 1. At 10:1 the pH is pKa + 1; at 1:10 it's pKa − 1. Outside that it buffers poorly.
Barely — only the ratio matters, and dilution scales both equally. pH stays essentially constant.