Henderson-Hasselbalch Calculator
Find the pH of a buffer from its pKa and the conjugate-base-to-acid ratio.
[A⁻] ÷ [HA] = 1 — equal concentrations, so pH = pKa
The Henderson-Hasselbalch equation gives a buffer’s pH as pH = pKa + log₁₀([A⁻] ÷ [HA]), where [A⁻] is the conjugate base and [HA] the weak acid. For acetic acid (pKa 4.76) with equal concentrations, the ratio is 1, log₁₀1 = 0, so pH = pKa = 4.76.
What the equation does
The Henderson-Hasselbalch equation predicts the pH of a buffer — a mixture of a weak acid (HA) and its conjugate base (A⁻). Instead of solving the full equilibrium, it links pH directly to the acid’s pKa and the ratio of the two species. Because the relationship is logarithmic, only the ratio of base to acid matters, not the absolute amounts.
[A⁻] = conjugate base concentration, [HA] = weak acid concentration (same units)
Worked example
Find the pH of an acetate buffer (pKa 4.76) with [A⁻] = 1.0 mol/L and [HA] = 0.1 mol/L.
- 1 Identify pKa and the two concentrations. pKa = 4.76, [A⁻] = 1.0 mol/L, [HA] = 0.1 mol/L.
- 2 Form the base-to-acid ratio. [A⁻] ÷ [HA] = 1.0 ÷ 0.1 = 10.
- 3 Take the base-10 logarithm. log₁₀(10) = 1.
- 4 Add it to the pKa. pH = 4.76 + 1 = 5.76. When the ratio is instead 1, log₁₀1 = 0 and pH = pKa exactly.
How the ratio shifts pH from pKa
The log term moves pH one unit per tenfold change in the [A⁻]/[HA] ratio.
| [A⁻] : [HA] ratio | log₁₀(ratio) | pH relative to pKa |
|---|---|---|
| 1 : 10 | −1 | pKa − 1 |
| 1 : 1 | 0 | pKa (equal concentrations) |
| 10 : 1 | +1 | pKa + 1 |
Reading the result
Buffers resist pH change. Adding a little acid or base shifts the [A⁻]/[HA] ratio only slightly, so the log term — and the pH — barely moves. That is what makes a buffer useful.
It works best near the pKa. The equation is most reliable when the ratio stays between about 1:10 and 10:1, i.e. within one pH unit of the pKa. Outside that window the buffer has little capacity and the approximation weakens.
Assumptions. The formula treats the equilibrium concentrations as equal to the amounts you mixed, which holds when both are much larger than the H⁺ produced. It also assumes dilute, ideal behavior — it does not correct for activity coefficients or ionic strength.