Henderson-Hasselbalch Calculator
pH = pKa + log10([A-]/[HA]). Enter the pKa and the two concentrations; the Henderson-Hasselbalch calculator also solves the full equilibrium and flags where the equation stops being trustworthy.
pH
- pH (Henderson-Hasselbalch) pH = pKa + log10([A-]/[HA]) = 4.76 + log10(1.50 x 10^-1 / 1.00 x 10^-1) = 4.76 + 0.18.
- 4.94
- pH (full equilibrium) Agrees with Henderson-Hasselbalch to 0.00. The equation is reliable here because both concentrations are large next to [H+] and [OH-].
- 4.94
- Ratio [A-]/[HA] Within 0.1 to 10, so the pH sits within one unit of the pKa, the range where a buffer has both partners in useful amounts.
- 1.5
- pOH 14.00 - pH, from the Henderson-Hasselbalch figure.
- 9.06
- [H+] 10^-pH.
- 1.16 x 10^-5 mol/L
- Acid [A-] 0.15 mol/L, [HA] 0.1 mol/L. By Henderson-Hasselbalch, equal amounts give pH = pKa, which is why pKa is the center of the buffer's range.
- Ka 1.74 x 10^-5 (pKa 4.76)
- Acidic, neutral or basic Relative to pH 7.00, which is neutral at 25 C.
- acidic
| [A-]/[HA] | log10 ratio | pH | Within pKa +/- 1 |
|---|---|---|---|
| 0.1 | -1.00 | 3.76 | yes |
| 0.25 | -0.60 | 4.16 | yes |
| 0.5 | -0.30 | 4.46 | yes |
| 1 | 0.00 | 4.76 | yes |
| 2 | 0.30 | 5.06 | yes |
| 4 | 0.60 | 5.36 | yes |
| 10 | 1.00 | 5.76 | yes |
- Henderson-Hasselbalch uses the amounts as mixed, assuming neither partner reacts with water to any meaningful degree. The full-equilibrium figure drops that assumption, and where the two disagree the full equilibrium is the one to use.
- All figures assume 25 C, where Kw = 1.0 x 10^-14 and pH + pOH = 14. Kw depends on temperature, so at any other temperature both the neutral point and the pH + pOH sum move.
- pH is a logarithm, so its significant figures live in the decimal places: a concentration known to two significant figures gives a pH to two decimal places. The digits before the point only locate the power of ten.
How the pH Calculator works
The Henderson-Hasselbalch equation is pH = pKa + log10([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] the weak acid. For a buffer of 0.15 mol/L acetate and 0.10 mol/L acetic acid (pKa 4.76), the ratio is 1.5, log10(1.5) = 0.18, and the pH is 4.76 + 0.18 = 4.94.
It comes straight from the acid's equilibrium constant. Ka = [H+][A-]/[HA]; take -log10 of both sides and rearrange, and the equation falls out. Nothing is lost in that step. The approximation comes afterward, when the concentrations you mixed are put in for the concentrations at equilibrium. That is close whenever only a small fraction of either partner reacts with water, which is true when both are much larger than [H+] and [OH-].
Three things follow that are worth knowing without a calculator. Equal amounts of acid and base give pH = pKa exactly, because log10(1) = 0. That is how pKa is often measured, at the half-equivalence point of a titration. Each tenfold change in the ratio moves the pH by one unit. And only the ratio matters, so diluting a buffer leaves the Henderson-Hasselbalch pH unchanged; what dilution does reduce is the buffer's capacity to absorb added acid or base.
The equation's limits are where it gets misused. A buffer is effective within about one pH unit of its pKa, a ratio between 0.1 and 10; beyond that one partner is nearly used up and the mixture resists change badly in one direction. So choosing a buffer starts with choosing an acid whose pKa is near the pH you want, then setting the ratio. The second limit is dilution and extreme pKa. For a buffer of 1 mM acid and 1 mM conjugate base with pKa 2.00, the acid ionizes heavily on its own and the mixed amounts are nothing like the equilibrium amounts: Henderson-Hasselbalch says pH 2.00, and the full equilibrium gives 3.07. That is why this calculator solves the full equilibrium (the acid's Ka, water's Kw and the charge balance together) next to the equation, and says when they part.
Enter concentrations in any one unit; since only the ratio enters the equation, the unit cancels from it, though the full-equilibrium check needs the real values. The figures assume 25 C.
pH Calculator FAQ
What is the Henderson-Hasselbalch equation?
pH = pKa + log10([A-]/[HA]), relating a buffer's pH to the acid's pKa and the ratio of conjugate base to acid. It is the acid's Ka expression in logarithmic form, with the mixed concentrations used in place of the equilibrium ones.
What happens when [A-] equals [HA]?
The ratio is 1, its logarithm is 0, and the pH equals the pKa. That is the half-equivalence point of a weak-acid titration and the center of the buffer's useful range.
Does diluting a buffer change its pH?
Not by Henderson-Hasselbalch, because only the ratio enters it and dilution leaves the ratio unchanged. In practice a strongly diluted buffer does drift, because the equation's assumption weakens, and it has less capacity to absorb added acid or base.
How do I choose a buffer for a target pH?
Pick a weak acid whose pKa is within one unit of the target, ideally close to it, then set [A-]/[HA] = 10^(pH - pKa). For pH 5.00 with pKa 4.76 that is a ratio of 1.74.