IB Chemistry SL Topic 6 — Proton Transfer Paper 1 & 2 Practical skill ~14 min read

pH Titration Curves

Add alkali to acid a drop at a time and plot the pH. Almost nothing happens, almost nothing happens, and then the pH leaps by seven units in the space of a single drop. That leap is the whole point of the graph.

📚 What you need to know

How the curve is produced

Put a known volume of acid in a flask with a pH probe in it, add alkali from a burette in small portions, and record the pH after each addition. Near the equivalence point you switch to adding a drop at a time, because that is where everything happens.

A STRONG ACID TITRATED WITH A STRONG BASE25.0 cm³ of 0.100 mol dm⁻³ HCl with 0.100 mol dm⁻³ NaOH0102030405002468101214equivalence point at 25.0 cm³pH 7.00 for strong acid + strong basea jump of about 7 pH unitsat the start there is only acidso pH = −log(0.100) = 1.00excess NaOH from here onvolume of NaOH added / cm³pH
Every point on this curve was calculated from the amounts of acid and base present. The near-vertical section spans roughly one drop of solution.

Reading the four regions

The steep part is not magic, it is arithmetic. Just before equivalence there might be 0.0001 mol of acid left in the flask; one more drop of alkali removes it entirely and leaves 0.0001 mol of alkali instead. Because pH is logarithmic, going from a tiny excess of one to a tiny excess of the other swings the reading enormously.

🧩 Calculating the pH at any point

  1. Find moles of acid and moles of base added so far.
  2. Subtract to find which is in excess, and by how much.
  3. Divide by the total volume — both solutions, added together.
  4. If acid is in excess, take –log[H+]. If base is in excess, find pOH first and subtract from 14.00.
  5. At exact equivalence with a strong acid and strong base, the answer is simply 7.00.
WORKED EXAMPLE

25.0 cm3 of 0.100 mol dm–3 HCl is titrated with 0.100 mol dm–3 NaOH. Calculate the pH after adding (a) 10.0 cm3 (b) 24.0 cm3 (c) 26.0 cm3 of NaOH.

Moles of acid at the start n(HCl) = 0.0250 × 0.100 = 2.50 × 10⁻³ mol (a) after 10.0 cm³ n(NaOH) = 0.0100 × 0.100 = 1.00 × 10⁻³ excess acid = 1.50 × 10⁻³ mol in 0.0350 dm³ [H⁺] = 0.04286 → pH = 1.37 pH = 1.37 (b) after 24.0 cm³ excess acid = 2.50 × 10⁻³ − 2.40 × 10⁻³ = 1.00 × 10⁻⁴ mol [H⁺] = 1.00 × 10⁻⁴ ÷ 0.0490 = 2.04 × 10⁻³ pH = 2.69 (c) after 26.0 cm³ excess base = 2.60 × 10⁻³ − 2.50 × 10⁻³ = 1.00 × 10⁻⁴ mol [OH⁻] = 1.00 × 10⁻⁴ ÷ 0.0510 = 1.96 × 10⁻³ pOH = 2.71, so pH = 14.00 − 2.71 pH = 11.29 Look at (b) and (c). Two cubic centimetres of alkali — roughly forty drops — move the pH by more than eight units. That is the vertical section.

The other three combinations

The shape survives, but the equivalence point moves. The reason is the salt left in the flask: it is not always neutral.

THE FOUR COMBINATIONSeach one is 25.0 cm³ of 0.100 mol dm⁻³ acid, titrated with 0.100 mol dm⁻³ baseSTRONG ACID + STRONG BASE7STRONG ACID + WEAK BASE7equivalence at pH 7 — neutral saltequivalence below 7 — acidic saltWEAK ACID + STRONG BASE7WEAK ACID + WEAK BASE7equivalence above 7 — alkaline saltbarely any vertical section at allthe vertical section is what makes a titration usable — the last combination has almost none
All four cross the same equivalence volume of 25.0 cm3, because the stoichiometry is identical. What differs is the pH there, and how sharply the curve rises through it.
CombinationpH at equivalenceWhyVertical section
strong acid + strong base7the salt is neutral; neither ion reacts with watervery long, about pH 3 to 11
strong acid + weak basebelow 7the salt contains the conjugate acid of a weak base, e.g. NH4+shorter, at the acidic end
weak acid + strong baseabove 7the salt contains the conjugate base of a weak acid, e.g. CH3COOshorter, at the alkaline end
weak acid + weak basearound 7the two effects roughly cancelalmost none — avoid this titration
Notice that a weak acid curve does not start at pH 1. Ethanoic acid at 0.100 mol dm–3 begins near pH 2.9, because only a fraction of it has dissociated. If a question shows you a curve starting around 3, the acid is weak before you read anything else.

Choosing an indicator

An indicator is a weak acid whose two forms have different colours, and it changes over a range of about two pH units. It is only useful if that range lies inside the vertical section, because that is the only way one drop can flip the colour.

IndicatorpH range of colour changeSuitable for
methyl orange3.1 – 4.4strong acid + strong base; strong acid + weak base
phenolphthalein8.3 – 10.0strong acid + strong base; weak acid + strong base
Keep two terms apart. The equivalence point is where the amounts match the equation — a fact about the chemistry. The end point is where the indicator changes colour — a fact about what you can see. A well-chosen indicator makes them close enough to be treated as the same.
WORKED EXAMPLE

A pH curve for 25.0 cm3 of sodium hydroxide titrated with 0.100 mol dm–3 hydrochloric acid has its equivalence point at 22.5 cm3. Calculate the concentration of the sodium hydroxide.

Step 1 — moles of acid at equivalence n(HCl) = 0.0225 × 0.100 = 2.25 × 10⁻³ mol Step 2 — the ratio is 1 : 1 n(NaOH) = 2.25 × 10⁻³ mol Step 3 — divide by the volume in the flask c = 2.25 × 10⁻³ ÷ 0.0250 0.0900 mol dm⁻³ The equivalence volume is read from the middle of the vertical section, not from where it begins or ends.
WORKED EXAMPLE

A titration curve starts at pH 2.9, rises gradually, and has a vertical section running from about pH 7 to pH 11. Identify the acid and base as strong or weak, and choose a suitable indicator.

The starting pH A 0.1 mol dm⁻³ strong acid would start at pH 1. Starting at 2.9 means only a fraction has dissociated. the acid is weak The equivalence pH The midpoint of the vertical section is around pH 9, which is alkaline. the base is strong The indicator phenolphthalein, range 8.3 – 10.0 Methyl orange would change colour at pH 3–4, long before the equivalence point, and would give a badly wrong titre.

💡 Exam tip

⚠️ Common mix-up

Up next: Electron Transfer Reactions — that completes proton transfer. The other great family of reactions moves electrons instead of protons, and almost everything you have just learned has a mirror image there.

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