The Henderson–Hasselbalch equation links the pH of a solution to the pKa of a weak acid and the ratio of its two forms: the protonated acid HA and the deprotonated conjugate base A⁻. It comes straight from the expression for the acid dissociation constant, Ka = [H⁺][A⁻]/[HA], by taking the negative logarithm of both sides. Lawrence Henderson wrote the relation for blood buffering in 1908 and Karl Hasselbalch put it into logarithmic form in 1917.
The equation can be read in several directions. Knowing the composition, it gives the pH of a buffer. Knowing the pH you want, it gives the base-to-acid ratio you need to mix. Knowing the pH and the ratio, it gives the pKa – which is how pKa values are measured from titration curves, since at the half-equivalence point pH = pKa. It also tells you what fraction of a drug, amino acid side chain or indicator is ionised at a given pH, which controls solubility, membrane crossing and colour.
It is an approximation: it treats concentrations as activities and assumes neither form is consumed significantly by water dissociation, so it works best for ratios between 0.1 and 10 and concentrations above about 1 mM.
How to use the Henderson–Hasselbalch Calculator
- Solve for
- Acid strength given as
- pKa
- Ka
- pH
- Conjugate base concentration [A⁻]
- Weak acid concentration [HA]
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Choose the unknown. The calculator converts Ka to pKa when needed, then rearranges pH = pKa + log₁₀([A⁻]/[HA]) for that unknown and reports the ratio and the percentage of the acid present as A⁻.
Worked example
A buffer contains 0.7 M conjugate base and 0.5 M weak acid with Ka = 1.4 × 10⁻⁵. pKa = −log₁₀(1.4 × 10⁻⁵) = 4.854. pH = 4.854 + log₁₀(0.7/0.5) = 4.854 + 0.146 = 5.000. At this pH, 0.7 / 1.2 = 58.3% of the acid is present as A⁻.
Rearranged: to make a pH 5.30 buffer from the same acid with 0.5 M HA, you need [A⁻] = 0.5 × 10^(5.30 − 4.854) = 0.5 × 2.79 = 1.40 M.
How the ratio changes with pH
| pH − pKa | [A⁻]/[HA] | % as A⁻ |
|---|---|---|
| −2 | 0.01 | 0.99% |
| −1 | 0.1 | 9.1% |
| 0 | 1 | 50% |
| +1 | 10 | 90.9% |
| +2 | 100 | 99.0% |
Tips
- Use the final volume of the solution, not the volume of solvent added, for molarity.
- For dilutions, C₁V₁ = C₂V₂ only holds when both concentrations use the same unit.
- Add acid to water, never water to acid.
- Everything is calculated in your browser.
Common mistakes to avoid
- Mixing w/w, w/v and v/v percentages.
- Forgetting that pH is logarithmic – one unit is a tenfold change in [H⁺].
- Using pKa where Ka is asked for (or the other way round).
Inputs explained
| Field | What to enter |
|---|---|
| Solve for | Choose one: pH, pKa, Conjugate base concentration [A⁻], Weak acid concentration [HA] … |
| Acid strength given as | Choose one: pKa, Ka |
| pKa | A number |
| Ka | A number |
| pH | A number |
| Conjugate base concentration [A⁻] | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
| Weak acid concentration [HA] | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
What each result means
- Ka = 10^(-[pKa])
- Conjugate base [A⁻] = [Weak acid concentration [HA]]×10^([pH] – [pKa])
- Weak acid [HA] = [Conjugate base concentration [A⁻]]×10^([pKa] – [pH])
- Ratio [A⁻]/[HA] = 10^([pH] – [pKa])
- Ratio [A⁻]/[HA] = 10^([pH] – [pKa])
- Fraction deprotonated (as A⁻) = (1÷(1 + 10^([pKa] – [pH])))×100
- Fraction protonated (as HA) = (1÷(1 + 10^([pH] – [pKa])))×100
Formula
pH = pKa + log₁₀([A⁻] ÷ [HA]) · pKa = −log₁₀ Ka · [A⁻] = [HA]·10^(pH − pKa) · [HA] = [A⁻]·10^(pKa − pH) · fraction deprotonated = 1 ÷ (1 + 10^(pKa − pH))Worked example
Inputs — Solve for: pH; Acid strength given as: Ka; Ka: 1.4e-05; Conjugate base concentration [A⁻]: 0.7 mol/L (M); Weak acid concentration [HA]: 0.5 mol/L (M).
Results — pH = 5; Ka = 0.000014; Ratio [A⁻]/[HA] = 1.4; Fraction deprotonated (as A⁻) = 58.33 %; Fraction protonated (as HA) = 41.67 %.
Key terms
- Weak acid (HA)
- An acid that only partly dissociates in water.
- Conjugate base (A⁻)
- What remains after the weak acid loses a proton.
- Ka
- Acid dissociation constant, [H⁺][A⁻]/[HA].
- Half-equivalence point
- The point in a titration where half the acid is neutralised and pH = pKa.
Frequently asked questions
What is the Henderson–Hasselbalch equation?
pH = pKa + log₁₀([A⁻]/[HA]). It gives the pH of a weak acid / conjugate base mixture from the pKa and the ratio of the two forms.
How do I calculate pH with Ka instead of pKa?
Convert first: pKa = −log₁₀ Ka. For Ka = 1.4 × 10⁻⁵, pKa = 4.854; with [A⁻] = 0.7 M and [HA] = 0.5 M, pH = 4.854 + log₁₀ 1.4 = 5.00.
What ratio do I need for a buffer at pH 5.0 with acetic acid?
[A⁻]/[HA] = 10^(5.0 − 4.76) = 1.74, so 1.74 mol acetate per mol acetic acid.
Why is pH equal to pKa at the half-equivalence point?
Half the acid has been converted to its conjugate base, so [A⁻] = [HA], the log term is zero and pH = pKa.
How do I find the ionised fraction of a drug?
For an acid the ionised fraction is 1 / (1 + 10^(pKa − pH)). Aspirin (pKa about 3.5) is about 99.9% ionised at blood pH 7.4.
Can I use it for bases?
Yes – use the pKa of the conjugate acid (e.g. 9.25 for NH₄⁺) with [A⁻] = [base] and [HA] = [conjugate acid].
When does the equation fail?
For very dilute buffers, for strong acids, and when the ratio is more extreme than about 1:10 or 10:1, because the dissociation of water and of the acid itself is no longer negligible.
What formula does the Henderson–Hasselbalch Calculator use?
pH = pKa + log₁₀([A⁻] ÷ [HA]) · pKa = −log₁₀ Ka · [A⁻] = [HA]·10^(pH − pKa) · [HA] = [A⁻]·10^(pKa − pH) · fraction deprotonated = 1 ÷ (1 + 10^(pKa − pH)).