pH = pKa + log₁₀([A⁻] ÷ [HA]) — fill in every field except the one you want; the empty field is solved for.
Buffer system (pKa at 25 °C):
pH (pH) = 4.9361
- Rearrange:
pH = pKa + log₁₀([A⁻] ÷ [HA]) - Substitute:
(4.76) + log₁₀((0.15 M) ÷ (0.1 M))=4.9361
How it works
The Henderson–Hasselbalch equation, pH = pKa + log₁₀([A⁻] ÷ [HA]), follows from the acid dissociation constant. Only the ratio of the two concentrations matters, so you can enter them in any concentration unit — or even as moles, if both are in the same final volume.
The equation is most accurate when both concentrations are much larger than [H⁺] and [OH⁻] and the ratio is between about 0.1 and 10, which is also the useful buffering range (pH = pKa ± 1). The presets give pKa values at 25 °C; they shift with temperature and ionic strength (Tris especially).
Frequently asked questions
- How do I make a buffer of a given pH?
- Pick a weak acid with a pKa within 1 unit of the target pH, then set the base-to-acid ratio to 10^(pH − pKa). For pH 5.0 with acetic acid (pKa 4.76), the ratio is 10^0.24 ≈ 1.74 : 1 acetate to acetic acid.
- What happens when [A⁻] = [HA]?
- log(1) = 0, so pH = pKa. This is the point of maximum buffer capacity.
- Why does a buffer resist pH change?
- Added acid is mopped up by the conjugate base (A⁻ + H⁺ → HA) and added base by the weak acid (HA + OH⁻ → A⁻ + H₂O), so the ratio — and the pH — changes only a little.
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