Henderson–Hasselbalch Buffer Calculator pH = pKa + log10([A⁻]/[HA])
Solve buffer pH, conjugate base-to-acid ratio, or pKa — and prepare laboratory buffers with exact gram recipes. Includes a 34-buffer pKa table standardised to 25 °C, a buffer-capacity meter, and fully worked examples.
Buffer preparation mode: enter a target pH and total concentration to get the exact base/acid ratio plus a weigh-out recipe in grams.
Buffer pKa Table (25 °C)
34 common laboratory buffers, standardised to 25 °C — use it as a buffer pKa chart for quick reference. The classic Good et al. (1966) values were tabulated at 20 °C — for example HEPES 7.55 at 20 °C becomes 7.48 at 25 °C. The ΔpKa/°C column lets you correct for your working temperature: pKa(T) ≈ pKa(25 °C) + ΔpKa/°C × (T − 25).
| Buffer | pKa (25 °C) | Useful range | ΔpKa/°C |
|---|
Sources: Good et al., Biochemistry 5, 467–477 (1966) via Kyoto Univ. buffer reference (20 °C values with temperature coefficients); Sigma-Aldrich biological buffers table (25 °C column); standard inorganic pKa values (CRC Handbook). Polyprotic acids list each dissociation step.
Worked Examples
Example 1 — pH of an acetate buffer
An acetate buffer has pKa = 4.76 and a base-to-acid ratio [A⁻]/[HA] = 2.
Try it: open the Solve pH tab — it loads with exactly these values.
Example 2 — preparing 1 L of 0.1 M phosphate buffer, pH 7.4
Use the phosphate pKa of 7.21 (the H2PO4⁻/HPO4²⁻ step).
Dissolve both salts in about 800 mL of distilled water, check the pH, adjust with small additions of NaOH or HCl if needed, then make up to exactly 1 L. Try it: the Buffer prep tab loads with these values.
Derivation of the Henderson–Hasselbalch Equation
Start from the acid dissociation equilibrium of a weak acid HA:
Take the base-10 logarithm of both sides:
Multiply by −1 and use the definitions pH = −log₁₀[H⁺] and pKa = −log₁₀(Ka):
Two rearrangements cover the other calculator modes:
Assumptions behind the formula
The derivation assumes: (1) activities can be replaced by concentrations (valid in dilute solution), (2) the temperature is constant (pKa is temperature-dependent), (3) only the stated conjugate pair contributes significantly to the proton balance, and (4) the acid dissociation is the dominant equilibrium. Violate any of these and the equation fails — see the limitations section below.
Why buffers resist pH change
When pH = pKa, the ratio is 1 and the solution holds equal amounts of acid and base — added H⁺ is consumed by A⁻ and added OH⁻ by HA. Buffer capacity β = 2.303 × C × r/(1+r)² (C = total concentration, r = [A⁻]/[HA]) is maximal here: a 0.1 M buffer at pH = pKa has β = 57.6 mM per pH unit. Move one pH unit away and capacity drops to 33% of maximum — the origin of the pKa ± 1 effective-range rule.
When the Equation Breaks Down
The Henderson–Hasselbalch equation is an approximation. It assumes activities equal concentrations and that the buffer pair dominates the proton balance. In short, this is when the Henderson–Hasselbalch equation fails — treat results with caution when:
- Very dilute buffers (below ~1 mM — extreme dilution): water autoionisation contributes a comparable number of protons; the simple ratio no longer controls pH.
- Very concentrated solutions (above ~0.5–1 M): activity coefficients deviate significantly from 1, especially for multiply-charged ions like HPO4²⁻.
- Extreme pH (near 0 or 14): [H⁺] or [OH⁻] from water itself becomes significant.
- Overlapping pKa values: for polyprotic acids with steps closer than ~2 pH units, the neighbouring equilibrium interferes — solve the full equilibrium instead.
- Wrong temperature: pKa shifts with temperature (see the ΔpKa/°C column). A Tris buffer set at room temperature can be off by more than half a pH unit in the cold room.
Frequently Asked Questions
What is the Henderson–Hasselbalch equation?
The Henderson–Hasselbalch equation relates the pH of a buffer solution to the pKa of its weak acid and the ratio of conjugate base to acid: pH = pKa + log10([A−]/[HA]), where [A−] is the conjugate base concentration and [HA] is the weak acid concentration. It is derived from the acid dissociation equilibrium Ka = [H+][A−]/[HA] by taking the negative logarithm of both sides.
Worked example: what is the pH of an acetate buffer with pKa 4.76 and [A−]/[HA] = 2?
pH = 4.76 + log10(2) = 4.76 + 0.301 = 5.06. Doubling the base-to-acid ratio always raises the pH by log10(2), about 0.30 pH units, regardless of which buffer you use.
How do I prepare 1 litre of 0.1 M phosphate buffer at pH 7.4?
Use the phosphate pKa of 7.21. The required ratio is 10^(7.4 − 7.21) = 1.55, so [HPO42−] = 0.0608 M and [H2PO4−] = 0.0392 M. For 1 L that is 60.77 mmol of disodium hydrogen phosphate (8.63 g of Na2HPO4) plus 39.23 mmol of sodium dihydrogen phosphate monohydrate (5.41 g of NaH2PO4·H2O). Dissolve in about 800 mL of water, then make up to 1 L.
What is buffer capacity and how is it calculated?
Buffer capacity (β) measures how much strong acid or base a buffer absorbs per unit pH change: β = 2.303 × C × r/(1+r)², where C is the total buffer concentration and r = [A−]/[HA]. Capacity is maximal when pH = pKa (r = 1): a 0.1 M buffer there has β = 57.6 mM per pH unit.
Over what pH range is a buffer effective?
The practical rule is pKa ± 1 pH unit, corresponding to base-to-acid ratios from 1:10 to 10:1. At the edge of this range the buffer capacity has already fallen to 33% of its maximum, so for critical work stay within pKa ± 0.5 when possible.
Why does this calculator list HEPES pKa as 7.48 instead of 7.55?
Both numbers are correct — at different temperatures. Good et al. (1966) tabulated HEPES pKa as 7.55 at 20 °C; at 25 °C it is 7.48. This calculator standardises every value to 25 °C and shows each buffer's temperature coefficient so you can correct for your working temperature.
When does the Henderson–Hasselbalch equation break down?
The equation assumes activities equal concentrations and that the buffer dominates the proton balance. It becomes unreliable for very dilute buffers (below about 1 mM, where water autoionisation matters), very concentrated solutions (activity coefficients deviate from 1), extreme pH values, and polyprotic acids whose pKa values are closer than about 2 units apart.
How do I calculate the [A−]/[HA] ratio from pH and pKa?
Rearrange the equation: ratio = 10^(pH − pKa). For a phosphate buffer at pH 7.4 with pKa 7.21, the ratio is 10^0.19 = 1.55, meaning about 61% of the phosphate is in the base form (HPO42−) and 39% in the acid form (H2PO4−).
If I double the base-to-acid ratio, how much does the pH change?
By exactly log10(2) = 0.30 pH units. Because the equation is logarithmic, multiplying the ratio by any factor n shifts the pH by log10(n) — a tenfold change in ratio moves the pH by exactly 1 unit.
Can I use the Henderson–Hasselbalch equation for polyprotic acids?
Yes — apply it to the dissociation step whose pKa is closest to your target pH. Phosphate has pKa values 2.15, 7.21 and 12.32; a pH 7.4 buffer uses the 7.21 step (H2PO4−/HPO42−). Citric acid (3.13, 4.76, 6.40) and carbonate (6.37, 10.33) work the same way.
What is the difference between pKa and pH?
pKa is a property of the acid — the pH at which it is exactly half dissociated ([A−] = [HA]). pH is a property of the solution — the actual acidity right now. The Henderson–Hasselbalch equation connects them: when the solution pH equals the acid's pKa, base and acid are present in equal amounts.
Does temperature affect the Henderson–Hasselbalch calculation?
Yes. Every pKa shifts with temperature; Good's buffers change by about 0.01–0.02 pH units per °C (for example HEPES −0.015/°C, MOPS −0.011/°C). Tris is strongly temperature-dependent at roughly −0.03/°C, so a Tris buffer adjusted to pH 8.0 at 25 °C will read about 8.6 at 4 °C.
How is this different from a titration calculation?
The Henderson–Hasselbalch equation gives the equilibrium pH of a buffer already containing both conjugate partners. A titration calculation tracks pH continuously as strong acid or base is added, including regions far from any pKa where this equation does not apply.
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