Henderson-Hasselbalch Calculator
Solve buffer pH, pKa, conjugate base, or weak acid amount with acid/base ratio insight and step-by-step Henderson-Hasselbalch work.
What can you solve?
The Henderson-Hasselbalch equation connects buffer pH to the acid dissociation constant and the ratio of conjugate base to weak acid. Fill any three values and leave one blank.
| Input | Meaning | Use it for |
|---|---|---|
| pH | Target or measured buffer pH | Leave blank to calculate final buffer pH |
| pKa | Acid strength of the HA/A- pair | Choose the value for your buffer system and temperature |
| [A-] | Conjugate base amount | Can be molarity, moles, or mmol if [HA] uses the same unit |
| [HA] | Weak acid amount | Can be molarity, moles, or mmol if [A-] uses the same unit |
Formula used
[A-] / [HA] = 10^(pH - pKa)
When [A-] equals [HA], the ratio is 1 and log10(1) is 0, so pH equals pKa. If the conjugate base is larger than the weak acid, pH rises above pKa. If the weak acid is larger, pH falls below pKa.
Worked examples
Equal acetate buffer: pKa = 4.76, [A-] = 0.10 M, [HA] = 0.10 M. Ratio = 1, so pH = 4.76 + log10(1) = 4.76.
Base-rich acetate buffer: pKa = 4.76, [A-] = 0.30 M, [HA] = 0.10 M. Ratio = 3, so pH = 4.76 + log10(3) = 5.237.
Target pH planning: For pH 5.20 with pKa 4.76, [A-]/[HA] = 10^(5.20 - 4.76) = 2.754. You need about 2.75 times as much conjugate base as weak acid.
Built-in examples
The quick example dropdown includes common classroom setups for acetate and phosphate buffers plus rearrangement practice for pH, pKa, [A-], and [HA].
| Example | Known values | Blank value | Lesson |
|---|---|---|---|
| Acetate equal acid/base | pKa 4.76, [A-] 0.10, [HA] 0.10 | pH | Shows why pH = pKa |
| Acetate base-rich | pKa 4.76, [A-] 0.30, [HA] 0.10 | pH | Shows pH above pKa |
| Phosphate near pH 7.40 | pH 7.40, pKa 7.21, [HA] 0.050 | [A-] | Finds base amount for a target pH |
| Find pKa | pH 6.10, [A-] 0.020, [HA] 0.050 | pKa | Works backward from measured buffer data |
| Find acid amount | pH 4.50, pKa 4.76, [A-] 0.20 | [HA] | Calculates acid needed when base is fixed |
How to interpret the ratio
| [A-]/[HA] | pH relationship | Meaning |
|---|---|---|
| 0.1 | pH = pKa - 1 | More weak acid than conjugate base; lower end of common buffer range |
| 1 | pH = pKa | Balanced acid/base mixture; strong buffer balance |
| 10 | pH = pKa + 1 | More conjugate base than weak acid; upper end of common buffer range |
| Below 0.1 or above 10 | More than 1 pH unit from pKa | Buffer still may exist, but capacity is usually much less balanced |
When the equation works best
The Henderson-Hasselbalch equation works best when the solution contains appreciable amounts of both a weak acid HA and its conjugate base A-. It is widely used for buffer problems, titration-region estimates, biochemical pH checks, and lab buffer preparation when activities are approximated by concentrations.
It becomes less reliable for very dilute solutions, very concentrated solutions, systems with strong acid or strong base left over, mixtures far from the pKa region, or cases where activity coefficients matter. For most school and introductory lab problems, the concentration-ratio form is the expected model.
Buffer planning tips
- Choose a weak acid with pKa close to the target pH, ideally within about 1 pH unit.
- Use the same unit basis for [A-] and [HA]. Molarity with molarity is fine; mmol with mmol is also fine.
- After calculating the ratio, decide the total buffer concentration separately if your lab procedure requires a specific capacity.
- Temperature can shift pKa, so use the pKa table value that matches your experiment when precision matters.
- For buffers made by partial neutralization, first do the stoichiometry, then use Henderson-Hasselbalch with the remaining HA and produced A-.
Common mistakes
- Reversing the ratio as [HA]/[A-] instead of [A-]/[HA].
- Using Ka directly in the pH equation instead of pKa.
- Mixing units for acid and base amounts.
- Using Henderson-Hasselbalch before neutralization stoichiometry is complete.
- Assuming the buffer is strongest when the ratio is extremely high or extremely low.
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Henderson-Hasselbalch Calculator FAQs
What is the Henderson–Hasselbalch equation?
The Henderson–Hasselbalch equation is a powerful relationship that gives the pH of a buffer solution from the pKa of the weak acid and the ratio of conjugate base to acid concentrations. It is derived from the Ka expression by taking the negative logarithm. The equation is widely used by chemists, biochemists and pharmacists to design and analyse buffers — and it is the conceptual basis for physiological systems like the bicarbonate buffer in blood. pH = pKa + log10 ([A-] / [HA])
When to use Henderson–Hasselbalch equation?
Use it whenever you have a buffer solution — that is, a mixture of a weak acid and its conjugate base (or weak base and its conjugate acid) in significant amounts. Conditions: (i) both acid and conjugate base are present; (ii) neither is so dilute that water autoionisation matters; (iii) we are not dealing with very strong acids or bases. Typical applications: calculating the pH of a buffer, choosing a weak acid for a target pH, or interpreting biological buffers in living systems.
Does Henderson–Hasselbalch equation work for bases?
Yes — it can be adapted for buffers based on weak bases. The analogous form is pOH = pKb + log([BH+]/[B]), where B is the weak base and BH+ is its conjugate acid. Then convert pOH to pH using pH + pOH = 14 (at 25 °C). Alternatively, convert Kb to Ka of the conjugate acid (KaKb = Kw) and use the original Henderson–Hasselbalch equation directly. Both approaches give the same answer. pOH = pKb + log10 ([BH+] / [B])
Is Henderson–Hasselbalch equation only for buffers?
It is most accurate and most useful for buffers, but the formula itself is just a rearrangement of the Ka expression and applies whenever both HA and A- are present in the solution at significant concentrations. So it works during a weak-acid–strong-base titration up until the equivalence point — exactly because the solution becomes a buffer along the way. It does not work for pure weak acid or pure weak base solutions, where you must solve the full Ka or Kb expression with the ICE table.
Can you use moles in Henderson–Hasselbalch equation?
Absolutely yes! Because the volume is the same for both HA and A- in a buffer, the ratio of concentrations equals the ratio of moles: [A-]/[HA] = nA-/nHA. So you can plug in moles directly in the log term without converting to molarity. This is especially handy when adding strong acid or base to a buffer — calculate the new moles after the neutralisation reaction, then use the equation to find the new pH.
Can you use Henderson–Hasselbalch for titrations?
Yes — the equation is the easiest way to compute pH at any point along a weak-acid vs. strong-base titration before the equivalence point. After each addition of strong base, work out moles of HA remaining and A- formed, then pH = pKa + log(A-/HA). At the half-equivalence point, [A-] = [HA], so log term = 0 and pH = pKa — a useful way to determine pKa experimentally. At/after the equivalence point, the equation no longer applies.