Henderson–Hasselbalch Calculator (Acid & Base Buffers)
Calculate buffer pH, the base-to-acid ratio, or pKa/pKb using the Henderson–Hasselbalch equation — for acid buffers or base buffers. See exactly where your buffer sits on the pH scale, whether it's within its optimal buffering range, and what percentage of each species is present, with a built-in reference table of common lab and physiological buffer systems.
Background
The Henderson–Hasselbalch equation estimates buffer pH from a weak acid or base's dissociation constant and the ratio of conjugate base to acid in solution. It's most accurate when that ratio is between 0.1 and 10 (within one pH unit of the pKa) — outside that range, a buffer still works, but its capacity to resist pH change weakens, or in some physiological systems, is continuously reinforced by other regulation (more on that in the examples below).
How to use this calculator
- Choose Acid buffer (weak acid + conjugate base) or Base buffer (weak base + conjugate acid).
- Choose what you're solving for: pH, the required ratio, or the pKa/pKb.
- Enter the two known values, or click a row in the buffer reference table to load a real system's pKa.
- Click Calculate to see the result, a pH-scale visual showing your buffer's effective range, a composition chart, and full step-by-step math.
How Henderson–Hasselbalch works
Acid buffers: pH = pKₐ + log₁₀([A⁻]/[HA]) — when the conjugate base and weak acid are present in equal amounts, pH equals pKa exactly. More base than acid pushes pH up; more acid than base pulls it down.
Base buffers: the same logic applies in pOH terms: pOH = pKᵦ + log₁₀([BH⁺]/[B]), then pH = 14 − pOH at 25 °C.
Effective buffering range: a buffer resists pH change best when the ratio is between 0.1 and 10 — i.e., within one pH unit of its pKa. Outside that window, one component is nearly used up, and the buffer has much less capacity left in that direction.
Formula & Equations Used
Acid buffer pH: pH = pKₐ + log₁₀([A⁻]/[HA])
Base buffer pH: pOH = pKᵦ + log₁₀([BH⁺]/[B]), then pH = 14 − pOH
Solve for ratio: Acid: ratio = 10^(pH − pKₐ). Base: ratio = 10^((14 − pH) − pKᵦ)
Solve for pK: Acid: pKₐ = pH − log₁₀(ratio). Base: pKᵦ = (14 − pH) − log₁₀(ratio)
Percent composition: % base form = ratio ÷ (1 + ratio) × 100; the remainder is the acid form.
Common Buffer Systems (Quick Reference)
| Buffer system | Relevant pKa (25 °C) | Typical use |
|---|---|---|
| Acetate (acetic acid/acetate) | 4.76 | General lab buffer, pH 3.7–5.7 |
| Citrate (2nd pKa) | 4.76 | Food science, pharma formulations |
| MES | 6.15 | Cell culture, plant biology |
| Bicarbonate (physiological, apparent) | 6.10 | Blood plasma pH regulation |
| PIPES | 6.76 | Cell/tissue culture |
| Phosphate (2nd pKa) | 7.21 | Physiological/IV buffers, biochemistry |
| HEPES | 7.48 | Cell culture (low metal/enzyme interference) |
| Tris | 8.06 | Molecular biology (DNA/protein buffers) |
| Borate | 9.24 | Electrophoresis, cosmetics |
| Ammonia / Ammonium (pKb shown) | 4.75 (pKᵦ) | Classic base-buffer teaching example |
| Glycine (amino group) | 9.60 | Electrophoresis, protein buffers |
Values are commonly cited approximations for teaching and lab planning — always confirm against a current reference (e.g., USP or a supplier's certificate of analysis) before formulating anything intended for clinical or regulated use.
Example Problems & Step-by-Step Solutions
Example 1 — Acid buffer, solve pH
An acetate buffer has pKₐ = 4.76 with equal concentrations of acetate and acetic acid (ratio = 1.0).
Step: pH = 4.76 + log₁₀(1.0) = 4.76 + 0 = 4.76.
Why it matters: whenever the ratio is exactly 1, pH always equals pKa — the buffer's "center point."
Example 2 — Acid buffer, solve ratio
You need a phosphate buffer (pKₐ = 7.21) at pH 7.40 for a cell culture experiment.
Step: ratio = 10^(7.40 − 7.21) = 10^0.19 ≈ 1.55.
Result: you need about 1.55 parts HPO₄²⁻ for every 1 part H₂PO₄⁻ — roughly 61% base form, 39% acid form.
Example 3 — Base buffer, solve pH
An ammonia buffer has pKᵦ = 4.75 with equal concentrations of NH₃ and NH₄⁺.
Step: pOH = 4.75 + log₁₀(1.0) = 4.75, so pH = 14.00 − 4.75 = 9.25.
Note: this matches the known pKa of NH₄⁺ (≈9.25) — pKa + pKb = 14 for a conjugate acid/base pair at 25 °C.
Example 4 — Physiological blood buffer
Blood plasma holds pH ≈ 7.40 using the bicarbonate system, with an apparent pKa' ≈ 6.10 and a bicarbonate-to-carbonic-acid ratio of about 20:1.
Step: pH = 6.10 + log₁₀(20) = 6.10 + 1.30 = 7.40.
Why this ratio "shouldn't" work: 20:1 is far outside the usual 0.1–10 optimal range — yet the body maintains it perfectly because breathing (removing CO₂) and the kidneys (adjusting HCO₃⁻) continuously replenish both sides. It's an open, regulated system, not a fixed lab buffer.
Frequently Asked Questions
What's the difference between acid and base buffers here?
Acid buffers use pH = pKₐ + log₁₀([A⁻]/[HA]) directly. Base buffers work in pOH first — pOH = pKᵦ + log₁₀([BH⁺]/[B]) — then convert with pH = 14 − pOH at 25 °C.
When is Henderson–Hasselbalch accurate?
Best when the ratio is between 0.1 and 10 and both species are present at reasonable concentrations. At high ionic strength or very dilute solutions, activity effects introduce error the simple equation doesn't capture.
Can I enter concentrations instead of a ratio?
Yes — just divide the base-form concentration by the acid-form concentration first: [A⁻]/[HA] for acid buffers, [BH⁺]/[B] for base buffers, then enter that number as the ratio.
Does temperature affect the result?
Yes. pKa and pKb are temperature-dependent, and pH + pOH = 14.00 is only exact at 25 °C. For work at body temperature (37 °C) or other conditions, use pK values measured at that temperature.
Why does the reference table matter for pharma and physiology students?
Real formulation and physiology questions rarely hand you a pKa — you're expected to know (or look up) that phosphate buffers work near pH 7.2, or that blood uses bicarbonate near pH 7.4. The reference table exists so you can go straight from "which buffer system" to a calculated answer without a separate lookup.
Why is the bicarbonate example ratio (20:1) outside the "optimal" range?
Because the body isn't a closed, fixed-composition buffer like a beaker in lab — it's continuously regulated. The lungs and kidneys keep replenishing both sides of the reaction, so the buffer never actually runs low the way a sealed lab buffer would. That's a meaningfully different system from the ones the 0.1–10 rule of thumb was built around.
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