Acid-Base Titration

Trace the pH curve as base is added. Compare the sharp jump of a strong acid with the buffer shoulder of a weak acid.

Worked example — Acid-Base Titration

pH	2.88
[H⁺]	1.32 × 10⁻³ mol/L
[OH⁻]	7.59 × 10⁻¹² mol/L
Equiv. volume V_eq	25.00 mL
Region	Before equivalence

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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The buffer equation only works in the buffer region 🖖

The Henderson-Hasselbalch equation pH = pKa + log([A⁻]/[HA]) is one of the most used equations in biochemistry. Blood plasma is buffered at pH 7.4 by the carbonic acid/bicarbonate system (pKa 6.1): the 20:1 ratio of bicarbonate (24 mmol/L) to dissolved CO₂ (1.2 mmol/L) is what puts pH at 6.1 + log 20 = 7.40, and it is steeply leveraged — add 1 mmol/L of dissolved CO₂ and the ratio falls to 24/2.2, dropping pH to 7.14, well into dangerous acidosis. Pharmaceutical formulations, enzyme assays, and electrophoresis all depend on buffers tuned to the right pKa. The equation is only valid in the buffer region (roughly pKa ± 1.5 pH units) — outside that range the approximation breaks down and exact equilibrium expressions must be used.

Finding a hidden concentration 🖖

A titration answers a simple question: how much acid is in this solution? You add base drop by drop until it exactly neutralises the acid — the equivalence point — and because you know the base's concentration and volume, you can work backwards to the acid's. The trick is the pH: right around that point one extra drop swings the pH by several units, so an indicator's colour change pinpoints the endpoint. Drag the V_b slider to watch the swing.

Some acids are too weak to titrate 🖖

Raise the pKa slider and watch the jump at the equivalence point flatten out. As a rule, an acid gives a sharp, usable endpoint in water only if its Ka is above about 10⁻⁸ (pKa below ~8); weaker than that, the conjugate base grabs so many protons back from water that the curve has no clear step for any indicator to catch. Chemists get around this by titrating such acids in non-aqueous solvents that don't compete with the reaction.

Problem solved in full

A 0.1 M solution of a weak acid with pK a 4.76 5 steps
A 0.1 M solution of a weak acid with pK_a 4.76 has a pH of 2.88. A strong acid at the same concentration would be pH 1.00. Work out where the difference goes.
1. A weak acid does not fully dissociate. The equilibrium constant fixes how far it gets, and 10^−4.76 is a small number, so it does not get far.
  
  $K_{a} = 1 0^{- 4.76} = 1.738 \times 1 0^{- 5}, C_{a} = 0.1 M$
2. Writing the equilibrium with x for the hydrogen ion concentration gives a quadratic, and the standard approximation is to neglect x against the initial concentration. Step 5 checks that this was allowed.
  
  $K_{a} = \frac{x ^{2}}{C _{a} - x} \approx \frac{x ^{2}}{C _{a}} (x ≪ C_{a})$
3. So the hydrogen ion concentration is the geometric mean of K_a and the concentration — the square root is why pH lands halfway between the two on a log scale.
  
  $x = [H^{+}] = K_{a} C_{a} = 1.738 \times 1 0^{- 6} = 1.32 \times 1 0^{- 3}$
4. Taking the negative logarithm gives 2.88.
  
  $pH = - lo g_{10} (1.32 \times 1 0^{- 3}) = 2.88$
5. Only 1.32% of the acid has ionised, which justifies the approximation and answers the question: a strong acid at 0.1 M gives 0.1 M of hydrogen ions and pH 1.00, so the weak one supplies 76 times fewer.
  
  $\frac{1.32 \times 1 0 ^{- 3}}{0.1} = 1.32% ionised; a strong 0.1 M acid gives pH 1.00$
Answer

The tool prints pH 2.88, [H⁺] = 1.32 × 10⁻³ and an equivalence volume of 25.00 mL. Two consequences follow. The pH sits between the strong-acid value and neutral because the square root halves the exponent — which is what the formula pH = ½(pK_a − log C) is saying. And the equivalence volume does not care about strength at all: it is set by moles alone, so a weak acid and a strong one at the same concentration need exactly the same 25 mL of base. Strength changes the shape of the curve and where the jump sits, never where it ends. Set the volume of base to half the equivalence volume and the pH reads 4.76 — the pK_a exactly.

$pH = \frac{1}{2} (p K_{a} - lo g_{10} C_{a}) = 2.88; at half-equivalence pH = p K_{a}$

References (3)

The equation named in the first block: L. J. Henderson, "Concerning the relationship between the strength of acids and their capacity to preserve neutrality." American Journal of Physiology 21(2), 173–179, 1908.
The 20:1 bicarbonate-to-CO₂ ratio and the pH it fixes: J. Feher, "Acid–Base Physiology I: The Bicarbonate Buffer System and Respiratory Compensation", in Quantitative Human Physiology, 929–941. Elsevier.
Buffer range, titration curves and when the approximation fails: D. C. Harris, Quantitative Chemical Analysis, 9th ed., ch. 8–11. W. H. Freeman, 2015. ISBN 978-1-4641-3538-5.

By the 42Math team · Updated 2026-07-11

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

HCl + NaOH - Strong acid + strong base: sharp equivalence at pH 7
Acetic acid - Same 0.1 mol/L and the same 25 mL as the HCl preset, but the beaker starts at pH 2.88. Acetic acid has released 1.32 × 10⁻³ mol/L of hydrogen ion out of a possible 0.1 — about one molecule in seventy-five. The equivalence volume is unchanged at 25.00 mL, because you titrate all the acid, dissociated or not.
Ammonium - Ammonium is the weakest acid on this row, pKa 9.25, and 0.1 mol/L of it is still acidic: pH 5.13, with [H⁺] at 7.50 × 10⁻⁶ mol/L. Look at the formula the panel is using, pH = ½(pKa − log Ca). It is an average, so pKa moves the starting pH at half rate: raise 9.25 by two and the pH climbs by one.
Phosphoric (1st) - Phosphoric acid’s first proton is barely weak at all: at 0.1 mol/L the panel opens at pH 1.57 with [H⁺] of 2.66 × 10⁻² mol/L, so better than a quarter of it has let go before a single drop of base. The label says first because there are three, each needing its own 25.00 mL, and this curve draws only the first.