Buffer pH Calculator
Henderson–Hasselbalch: pH from pKa and the base-to-acid ratio.
pKa = 4.74 · [A⁻]/[HA] = 1
A buffer’s pH follows the Henderson–Hasselbalch equation, pH = pKa + log₁₀([A⁻] ÷ [HA]). An acetate buffer (pKa 4.76) with 0.20 M conjugate base and 0.10 M acid has pH = 4.76 + log₁₀(2.0) = 4.76 + 0.30 = 5.06.
The Henderson–Hasselbalch equation
A buffer resists pH change because it contains both a weak acid and its conjugate base. Its pH is set by the acid’s pKa and the ratio of the two species. Adding a little acid or base shifts that ratio only slightly, so the pH barely moves.
[A⁻] is the conjugate base, [HA] is the weak acid; pKa = −log₁₀(Ka)
Worked example
Find the pH of an acetate buffer (pKa = 4.76) with 0.20 M conjugate base and 0.10 M weak acid.
- 1 Form the base-to-acid ratio. [A⁻] / [HA] = 0.20 ÷ 0.10 = 2.0.
- 2 Take the logarithm. log₁₀(2.0) = 0.30, so the buffer sits 0.30 units above its pKa.
- 3 Add to pKa. pH = 4.76 + 0.30 = 5.06 — slightly basic of the pKa because base outweighs acid.
pKa of common buffer acids
At 25 °C; a buffer works best within about ±1 pH unit of its pKa.
| Buffer system | Weak acid | pKa |
|---|---|---|
| Acetate | Acetic acid (CH₃COOH) | 4.76 |
| Carbonate | Carbonic acid (H₂CO₃) | 6.35 |
| Phosphate | H₂PO₄⁻ | 7.21 |
| TRIS | TRIS-H⁺ | 8.07 |
| Ammonium | NH₄⁺ | 9.25 |
| Bicarbonate | HCO₃⁻ | 10.33 |
Choosing and reading a buffer
When base equals acid. If [A⁻] = [HA], the log term is zero and pH equals pKa exactly — the point of maximum buffering capacity.
Pick a pKa near your target pH. A buffer is effective within roughly ±1 unit of its pKa, where the ratio stays between about 1:10 and 10:1. To buffer at pH 7.2, the phosphate system is ideal.
Limits and mistakes. The equation assumes a weak acid and concentrations that aren’t extremely dilute, so the equilibrium ratio stays close to the initial ratio. Flipping [A⁻] and [HA] is a frequent error — it changes the sign of the log term and moves the pH the wrong way.