Henderson Hasselbalch Calculator
You have the pair; the page reads its pH: pKa plus the logarithm of the mole ratio, with the ionized fraction alongside.
Henderson Hasselbalch Calculator
Results recalculate instantly on every keystroke. Nothing you type is transmitted.
What this result does not account for
- One weak-acid pair; no polyprotic sums
- Activity ignored — concentration stands in
In short: 0.050000 mol of acetic acid with 0.050000 mol of acetate (pKa 4.756): pH = 4.756 + log10(1.000000) = 4.756000 — the pair sits exactly on its pKa, 50% ionized. Lean the ratio to ten — 0.050000 against 0.500000 — and the pH climbs exactly one unit to 5.756000 with 90.909091% of the pair in base form: each tenfold of ratio is one pH unit, that is the whole ladder.
Formula
pH = pKa + log10([A⁻]/[HA]) · ratio = n(A⁻)/n(HA) · ionized % = ratio/(1+ratio) × 100
The Henderson–Hasselbalch equation is the acid dissociation equilibrium wearing a log ruler: pH sits at the pKa when the pair is even, and every tenfold excess of base form lifts it exactly one unit. Because the equation reads a RATIO, diluting the whole buffer leaves the pH where it was — what thins is the capacity, not the reading.
Worked Example
- Enter the moles of the acid form.
- Enter the moles of the conjugate base.
- Enter the pair’s pKa.
- Read the pH, the ratio, and the ionized fraction.
Defaults: 0.050000/0.050000, pKa 4.756 → pH 4.756000, 50% ionized. Tenfold base: pH 5.756000, 90.909091%. TRIS at pH 7.5 wants ratio 0.275423.
Strengths & Limits Of This Model
Where this engine is strong
- Ratio and ionized fraction together
- The dilution doctrine printed, not implied
Where it stops
- No temperature correction of pKa
- No ionic-strength terms
Practical Use Cases
Checking a made-up bath
moles in, pH verdict out
Designing the split
how far off pKa a target sits
Teaching
the log ruler, one ratio at a time
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Henderson Hasselbalch Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
Why does an equal pair sit exactly on the pKa?
The log of 1 is 0, so an even pair donates nothing to the equation and pH lands on pKa. The pKa IS the even-pair pH — that is the cleanest way to remember what the table value means, and the buffer-preparation page builds its whole 50/50 recipe on it.
Why does ten times the base lift the pH exactly one unit?
Because the ruler is logarithmic: log10(10) = 1. A hundredfold lifts two units, a tenthfold drops one. The same rung structure as the decibel and the pH scale — chemistry counts in powers of ten wherever ratios span orders of magnitude.
Does diluting the buffer change its pH?
The equation says no — both numerator and denominator shrink together and the ratio survives. What dies is the CAPACITY: a tenth-strength pair absorbs a tenth of the acid before it drifts. The card prints both truths because the bench keeps confusing them.
Why must both forms be present?
The logarithm divides by the acid form and is cancelled by the base form — either one at zero sends the pH to an infinite ledge that no real solution sits on. A bottle of weak acid is not a buffer; the pair is the buffer, and the page refuses to price half a pair.
What does the ionized fraction mean here?
The share of the pair currently wearing the base form: ratio/(1+ratio). Even pair → 50%. Tenfold base → 90.909091%. It is the same arithmetic the drug-absorption literature runs for ionized versus membrane-crossing fractions, one page earlier in the chain.
Can I use this for a weak base pair like ammonia?
Yes — run the acid form of the pair (ammonium, pKa 9.25) and its conjugate base (ammonia). Every weak base is the conjugate base of some weak acid, and the equation only ever sees the pair. The pKa table is the wardrobe.
How accurate is a pH computed from weighed moles?
Good to a few hundredths for dilute work — then activity coefficients, temperature and the acid’s own purity take over. The equation prices the ideal; the meter prices the bath. Adjust to the electrode and let the arithmetic explain the drift.
How does this feed the buffer-preparation page?
It is the same logarithm walked the other way: there you name the pH and receive the split; here you weigh the split and receive the pH. The titration page joins them — at half-neutralization the flask IS an equal pair and the pH reads the pKa straight off the burette.