Buffer Solution pH Calculator (Henderson-Hasselbalch)
Calculate the pH of a buffer solution using the Henderson-Hasselbalch equation.
Enter the acid pKa and concentrations of the conjugate base and weak acid.
The Henderson-Hasselbalch Equation
A buffer solution resists changes in pH when small amounts of acid or base are added. Buffers are essential in chemistry, biochemistry, medicine, and industrial processes. They work by containing a weak acid and its conjugate base (or a weak base and its conjugate acid) in equilibrium.
The Henderson-Hasselbalch equation:
pH = pKa + log₁₀([A⁻] / [HA])
Where:
- pH = the acidity/alkalinity of the solution (scale 0–14)
- pKa = the acid dissociation constant of the weak acid (lower pKa = stronger acid)
- [A⁻] = molar concentration of the conjugate base (the deprotonated form)
- [HA] = molar concentration of the weak acid (the protonated form)
What pKa means: pKa = -log₁₀(Ka). When [A⁻] = [HA] (equal concentrations), log(1) = 0, so pH = pKa exactly. This is the half-equivalence point — and it’s the center of the buffer’s effective range.
Effective buffer range: A buffer works effectively within pKa ± 1 pH unit. Outside this range, one component dominates and the buffer loses its capacity to resist pH changes.
Common buffers and their pKa values:
| Buffer System | pKa | Useful pH Range | Application |
|---|---|---|---|
| Acetic acid / Acetate | 4.76 | 3.8 – 5.8 | Food chemistry, lab buffers |
| Citric acid / Citrate | 3.13, 4.76, 6.40 | 3.0 – 6.2 | Food, pharmaceuticals |
| Phosphate (H₂PO₄⁻/HPO₄²⁻) | 7.20 | 6.2 – 8.2 | Biology, blood, lab |
| Bicarbonate (H₂CO₃/HCO₃⁻) | 6.35 | 5.4 – 7.4 | Blood pH regulation |
| Tris | 8.06 | 7.0 – 9.0 | Molecular biology |
| Boric acid / Borate | 9.24 | 8.2 – 10.2 | Electrophoresis |
Blood pH regulation: Human blood is maintained at pH 7.35–7.45 primarily through the bicarbonate buffer system (pKa 6.35), combined with respiratory control of CO₂ and kidney regulation of bicarbonate reabsorption.
Worked example: Acetate buffer — pKa = 4.76, [A⁻] = 0.2 mol/L, [HA] = 0.1 mol/L: pH = 4.76 + log(0.2/0.1) = 4.76 + log(2) = 4.76 + 0.301 = 5.06