Lesson 2 of 3
pH and endpoints
Use units and supplied titration data.
A-Level 8873, revised syllabus (2026-2027)
I can explain pH and equilibrium terms.Syllabus
Syllabus 8873, 3(d). Explain pH and equilibrium terms.
I can calculate strong-solution hydrogen concentration and pH.Syllabus
Syllabus 8873, 3(e). Calculate strong-solution hydrogen concentration and pH.
I can choose titration indicators from data.Syllabus
Syllabus 8873, 3(f). Choose titration indicators from data.
Count ions, then take the logarithm
Concentration units and stoichiometric factors come before the calculator.
Worked exampleA strong monoprotic acid
Find the pH of 3.20 × 10-3 mol dm-3 HCl at 25 degrees C.
Try it first. Then show the worked stepsHide the worked steps
- Assume complete dissociation; one mole HCl provides one mole H+.
- [H+] = 3.20 × 10-3 mol dm-3. Water contributes negligibly at this concentration.
- pH = -log10(3.20 × 10-3) = 2.49485.
pH = 2.495. The pH itself has no unit. Using three decimal places preserves the three significant figures of the concentration.
Worked exampleA base with two hydroxide ions
A fully dissolved Ca(OH)2 solution has concentration 2.50 × 10-3 mol dm-3. Find its pH at 25 degrees C.
Try it first. Then show the worked stepsHide the worked steps
- Ca(OH)2(aq) supplies two OH- per formula unit: [OH-] = 5.00 × 10-3 mol dm-3.
- Use Kw: [H+] = (1.00 × 10-14)/(5.00 × 10-3) = 2.00 × 10-12 mol dm-3.
- pH = -log10(2.00 × 10-12) = 11.69897.
pH = 11.699. This uses the stated dissolved concentration; it does not assume every mass of solid calcium hydroxide will dissolve.
Worked exampleNeutralise before finding pH
Mix 25.0 cm3 of 0.0400 mol dm-3 HCl with 15.0 cm3 of 0.0400 mol dm-3 NaOH. Assume additive volumes and 25 degrees C.
Try it first. Then show the worked stepsHide the worked steps
- Initial H+ amount = 0.0400 × 0.0250 = 1.00 × 10-3 mol.
- Initial OH- amount = 0.0400 × 0.0150 = 6.00 × 10-4 mol.
- The 1:1 neutralisation leaves 4.00 × 10-4 mol H+.
- Total volume = 0.0400 dm3, so [H+] = 0.0100 mol dm-3.
pH = 2.000. Averaging the two starting pH values would ignore both stoichiometry and the logarithmic scale.
Try it step by step
Work it out in steps: [H+] or [OH-], then pH.
Show the answer
Problem 1.
[H+] or [OH-]: HNO3 ionises completely and gives one H+ each, so [H+] = 0.0250 mol dm-3.
pH: pH = -log(0.0250) = 1.60.
Problem 2.
[H+] or [OH-]: Each Ba(OH)2 gives two OH-, so [OH-] = 2 x 0.00400 = 0.00800 mol dm-3.
pH: [H+] = 1.00 x 10-14 / 0.00800 = 1.25 x 10-12 mol dm-3, so pH = 11.90.
Problem 3.
[H+] or [OH-]: NaOH gives one OH- each, so [OH-] = 0.00500 mol dm-3.
pH: [H+] = 1.00 x 10-14 / 0.00500 = 2.00 x 10-12 mol dm-3, so pH = 11.70.
Check your understanding
A strong acid, not extremely dilute, is diluted tenfold. How does its pH change?
Show the answer
It rises by 1, because [H+] falls by a factor of ten.
pH = -log[H+], so a tenfold fall in [H+] raises pH by one. In extremely dilute acid, H+ from water matters and this rule breaks down.
Choose a colour change inside the sharp rise
The equivalence point and the observed endpoint are related but not identical.
At equivalence, the acid and base have reacted in the stoichiometric proportion. An indicator changes colour over a pH range; its endpoint should occur within the steep portion of the titration curve, where a very small added volume causes the whole indicator transition. Choose using the supplied curve or transition data, not merely a memorised colour.
| Indicator | Approximate transition pH | Acid-side / base-side colour |
|---|---|---|
| Methyl orange | 3.1-4.4 | Red / yellow |
| Bromothymol blue | 6.0-7.6 | Yellow / blue |
| Phenolphthalein | 8.2-10.0 | Colourless / pink |
| Titration | Region near equivalence | Usual interpretation |
|---|---|---|
| Strong acid / strong base | Steep change spans acid and alkaline regions at suitable concentrations. | Several indicators may work; inspect the given curve. |
| Weak acid / strong base | Equivalence is alkaline because the conjugate base reacts with water. | Phenolphthalein is commonly suitable. |
| Strong acid / weak base | Equivalence is acidic because the conjugate acid reacts with water. | Methyl orange is commonly suitable. |
| Weak acid / weak base | Often no sufficiently sharp pH change. | A visual indicator is generally unsuitable for a precise endpoint. |
Worked exampleUse given data rather than the name alone
A weak-acid/strong-base titration changes sharply from pH 7.8 to 11.0 near equivalence. Which listed indicator is suitable?
Try it first. Then show the worked stepsHide the worked steps
- Methyl orange changes well before the steep interval.
- Bromothymol blue also finishes its transition below the stated interval.
- Phenolphthalein changes from 8.2 to 10.0, wholly inside the steep interval.
Choose phenolphthalein. The endpoint then needs very little extra titrant to complete the colour change.
For reliable observations, add titrant dropwise near the endpoint, swirl to mix, and use a consistent faint permanent colour. A darker final colour after extra titrant is not "more complete": it means the endpoint has been overshot.
Check your understanding
Ethanoic acid is titrated with sodium hydroxide. The steep part of the curve runs from about pH 7 to pH 11. Which indicator suits it?
Show the answer
Phenolphthalein, which changes over pH 8.2-10.0
Its whole colour-change range sits inside the steep rise, so it changes sharply at the end-point. The equivalence pH is above 7 because ethanoate is a weak base.