IB Chemistry HL Topic 6 — Proton Transfer Paper 1 & 2 HL only ~12 min read

Interpreting pH Curves (HL)

Hand an examiner four unlabelled titration curves and they can tell you instantly which acid met which base. The clues are all in the shape: where the curve starts, whether it has a flat shoulder near the bottom, how tall the cliff is, and where the middle of that cliff sits. Learn to read those four things and this becomes one of the easiest HL topics.

📘 What you need to know

All four curves side by side

Each of these is 25.0 cm3 of a 0.100 mol dm−3 solution in the flask, with 0.100 mol dm−3 of the other reagent added from the burette. The equivalence volume is 25.0 cm3 every single time, because the moles always match at the same point. Only the pH at that point moves.

The four titration curves you must recognise Same volumes, same concentrations. Only the strengths change. STRONG ACID + STRONG BASE WEAK ACID + STRONG BASE WEAK BASE + STRONG ACID WEAK ACID + WEAK BASE 14 7 0 14 7 0 14 7 0 14 7 0 buffer region starts alkaline no steep bit equivalence at pH 7.00 equivalence at pH 8.73 equivalence at pH 5.28 no clear equivalence point The dashed line across each box is pH 7, for comparison. All four go from 0 to 50 cm³ added, with equivalence at 25.0 cm³.
These are solved from the charge-balance equations rather than sketched, so the heights and the positions of the cliffs are the real ones you would measure in the lab.

Reading the four clues

1. Where does it start?

The very first point, before you add anything, tells you the strength of what is in the flask. A 0.100 mol dm−3 strong acid starts at pH 1.00, because it is fully split. The same concentration of weak acid starts near pH 2.9, because only about 1% of it has split. Weak bases start near pH 11 rather than the 13 of a strong base.

2. Is there a shoulder?

Look just after the start. On the two weak curves the line rises quickly for a moment and then flattens into a shoulder that lasts for a long stretch. That shoulder is a buffer region: the flask now contains both the weak acid and its conjugate base, which soak up whatever you add.

Strong-strong curves have no shoulder, because there is no conjugate pair in the flask to do the soaking.

3. How tall is the cliff?

Strong with strong gives the tallest vertical section, roughly pH 4 to 10. Each weak partner you introduce chops a piece off one end. Two weak partners and there is barely a cliff left at all, which is why weak-with-weak titrations are almost never done.

4. Where is the middle of the cliff?

That is the equivalence point, and its pH depends entirely on what salt is now sitting in the flask.

The salt decides the pH at equivalence At equivalence the acid is completely used up. Only the salt is left. STRONG + STRONG WEAK ACID + STRONG BASE STRONG ACID + WEAK BASE NaCl CH₃COONa NH₄Cl Neither Na⁺ nor Cl⁻ reacts with water. CH₃COO⁻ grabs H⁺ from water, leaving OH⁻. NH₄⁺ gives H⁺ to water, making H₃O⁺. pH = 7 pH above 7 pH below 7 A weak partner always leaves behind an ion that reacts with water. That is salt hydrolysis, and it is coming up in a few pages.
This is why the equivalence point is not at pH 7 for the weak cases. The neutralisation itself is complete — it is the salt that then shifts the pH.
Students often assume the equivalence point of a weak acid is above 7 “because the acid did not fully react”. It did fully react. The ethanoate ion left behind is a base, and it is that ion pulling protons off water afterwards that raises the pH.

The half-equivalence point

Halfway to equivalence — at 12.5 cm3 in our curves — exactly half of the weak acid has been converted to its conjugate base. So the flask contains equal amounts of the two.

Put that into the Ka expression and something very tidy drops out.

At half equivalence Ka = [H+][A−] ÷ [HA]   and since [A−] = [HA],
Ka = [H+]    so    pH = pKa

Our ethanoic acid curve reads exactly 4.76 at 12.5 cm3, and pKa for ethanoic acid is 4.76. That is not a coincidence — it is a standard way of measuring a Ka in the lab. For a weak base the same argument gives pOH = pKb at half equivalence.

FeatureStrong acid + strong baseWeak acid + strong baseWeak base + strong acid
Starting pH1.002.8811.12
Buffer regionnoneyes, before equivalenceyes, before equivalence
pH at half equivalenceno special meaning= pKa = 4.76pOH = pKb, so pH 9.25
pH at equivalence7.008.735.28
Salt left behindNaCl, neutralCH3COONa, alkalineNH4Cl, acidic
Range of steep sectionabout pH 4 to 10about pH 7 to 11about pH 3 to 7

🧩 Identifying a curve you have never seen

  1. Read the starting pH. Low and near 1 means a strong acid; near 3 means a weak acid; near 11 means a weak base; near 13 means a strong base.
  2. Look for a shoulder shortly after the start. A shoulder means a weak species is in the flask.
  3. Find the middle of the vertical section and read its pH. Above 7 means the acid was weak; below 7 means the base was weak.
  4. If there is no clear vertical section, both were weak.

Worked examples

WORKED EXAMPLE

A curve starts at pH 2.9, flattens into a shoulder, then rises steeply to an equivalence point at pH 8.7. Identify the acid and the base, and explain the equivalence pH.

Step 1: use the starting pH A 0.1 mol dm−3 strong acid would start near 1.0, so 2.9 means the acid is weak. Step 2: use the finish and the shoulder The curve ends high and there is one buffer region, so the base is strong. Step 3: explain the equivalence pH The salt formed contains the conjugate base of a weak acid, e.g. CH3COO−. CH3COO− + H2O ↔ CH3COOH + OH− That reaction makes hydroxide ions, pushing the pH above 7. Weak acid titrated with a strong base; the salt hydrolyses to give OH− quote the hydrolysis equation — “because the acid is weak” alone rarely gets full marks
WORKED EXAMPLE

25.0 cm3 of a weak acid needs 22.0 cm3 of NaOH to reach equivalence. The pH at 11.0 cm3 is 4.20. Find Ka.

Step 1: check what 11.0 cm3 means 22.0 ÷ 2 = 11.0, so this is the half-equivalence point. Step 2: apply the half-equivalence rule Half the acid has become its conjugate base, so [HA] = [A−] and pH = pKa. pKa = 4.20 Step 3: convert to Ka Ka = 10−4.20 = 6.31 × 10−5 Ka = 6.31 × 10−5 spotting that 11.0 is half of 22.0 is the whole question
WORKED EXAMPLE

Explain why ammonia is not titrated against ethanoic acid in school laboratories.

Step 1: classify both reagents Ammonia is a weak base and ethanoic acid is a weak acid. Step 2: describe the curve shape The pH drifts gradually from about 2.9 to about 9.2 with no vertical section. Step 3: link the shape to the practical problem An indicator needs a sharp pH jump to change colour at one point. There is no steep section, so no indicator can show the end point accurately a pH meter can still find it from the steepest part, but colour indicators cannot

💡 Exam tip

⚠ Common mix-up

Up next: The pOH Scale. We keep flipping between hydrogen ions and hydroxide ions, so it is time to give the hydroxide side its own scale and learn the shortcut that connects the two.

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