Part 5: Titrations, indicators and Core Practical 3

Choose a meaningful end point, obtain reliable titres and trace a dilution calculation back to the original acid concentration.

Equivalence and end point are different ideas

At equivalence the reactants have been mixed in the stoichiometric ratio in the balanced equation. The end point is the experimental signal, usually an indicator colour change. A useful indicator changes within the steep pH change near equivalence, so the two volumes are close. An indicator is added in small quantity because it is itself an acid–base system and too much can affect the titre.

Phenolphthalein is colourless on the acidic side and pink on the alkaline side; methyl orange is red in sufficiently acidic solution and yellow on the alkaline side, with an orange transition. State both the starting and finishing appearance for the actual direction of addition. 'Turns pink' is wrong if acid is being added to alkali with phenolphthalein.

Good technique protects the measured amount

Rinse the burette with the titrant, fill its jet and remove the filling funnel before taking the initial reading. A bubble that fills during the titration makes the apparent delivered volume too large: some measured liquid filled the jet instead of reaching the flask. A funnel left in place may drip extra titrant into the burette after the initial reading.

Use a volumetric pipette with a filler to transfer the aliquot. Allow the calibrated pipette to drain as instructed; do not blow out the residual drop in a standard transfer pipette. Swirl the conical flask so each addition reacts promptly. A white tile makes a faint colour change easier to judge, and washing down the flask walls with deionised water brings splashes into the reacting mixture without adding reagent moles.

Read a colourless meniscus at eye level to avoid parallax, record both readings, and calculate titre = final − initial. Record a burette with 0.1 cm³ graduations to the nearest 0.05 cm³ when required. Do a rough titration to locate the end point, then add titrant dropwise near it in accurate runs. Concordant results show consistency; a systematic error can still make all of them inaccurate.

Core Practical 3: dilute and titrate hydrochloric acid

In Pearson's CP3 example, pipette 25.0 cm³ of the original HCl into a 250 cm³ volumetric flask and make up to the mark, then mix. Pipette a 25.0 cm³ aliquot of the diluted acid into a conical flask. Titrate against previously standardised NaOH from the burette with a small amount of phenolphthalein. Use the first persistent pale pink end point under the stated procedure, not a deep pink excess of alkali.

Dilution puts the expected titre in a useful measurable range. It does not change the moles of acid in the whole volumetric flask, but the titrated aliquot contains only one tenth of them. Use the standardised NaOH concentration from CP2 or a supplied verified value, not an approximate preparation label.

Keep standardised alkali appropriately stoppered because it reacts with atmospheric CO₂. This alters its chemical composition and can alter indicator-dependent titration behaviour; do not assume a contaminated solution remains a known pure NaOH standard. Phenolphthalein solutions can contain flammable solvent, so keep away from ignition sources. Wear eye protection and fill burettes below eye level.

A complete calculation back to the original bottle

Original illustrative titres are: rough 23.10 cm³; accurate runs 22.40, 22.45 and 22.40 cm³. The accurate results span 0.05 cm³ and give mean (22.40 + 22.45 + 22.40)/3 = 22.4167 cm³, reported as 22.42 cm³. Keep extra digits during subsequent calculation. The rough titre is not part of the mean.

With c(NaOH) = 0.0984 mol dm⁻³, n(NaOH) in the mean titre = 0.0984 × 0.0224167 = 0.0022058 mol. The reaction is 1:1, so this is n(HCl) in the 25.0 cm³ diluted aliquot. c(diluted HCl) = 0.0022058/0.0250 = 0.088232 mol dm⁻³. The original acid was diluted by 250/25.0 = 10, giving c(original HCl) = 0.882 mol dm⁻³ to 3 s.f.

An alternative is to multiply the aliquot amount by ten to get 0.022058 mol in the whole flask, then divide by 0.0250 dm³, the original acid volume used. Both routes agree. Dividing the whole-flask amount by 0.250 dm³ gives the diluted concentration and stops one step too soon.

Acid–base does not always mean 1:1

For H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, one acid molecule supplies two acidic protons. If 20.00 cm³ acid neutralises 25.0 cm³ of 0.120 mol dm⁻³ NaOH, n(NaOH) = 0.00300 mol, n(H₂SO₄) = 0.00150 mol and c(H₂SO₄) = 0.00150/0.02000 = 0.0750 mol dm⁻³. Applying c₁V₁ = c₂V₂ without the reaction ratio would give twice the correct acid concentration.

Carbonate requires two protons for complete conversion to CO₂ and water: CO₃²⁻ + 2H⁺ → CO₂ + H₂O. Methyl orange can detect this complete acidification; phenolphthalein instead detects an earlier carbonate-to-hydrogencarbonate stage. This is why choosing an indicator by the name of one reagent alone is unsafe.

Transfer application: titrate the excess reagent

A back titration is useful when a solid reacts slowly or its direct end point is unclear. Add a known excess of acid, let the reaction finish, then titrate the acid remaining. The difference between acid added and acid left is the amount consumed by the sample. Do not treat the leftover amount as the amount that reacted.

An original 0.600 g impure CaCO₃ sample receives 25.0 cm³ of 0.500 mol dm⁻³ HCl: 0.01250 mol HCl. The remaining acid needs 15.00 cm³ of 0.200 mol dm⁻³ NaOH, so 0.00300 mol HCl remains. Consumed HCl = 0.00950 mol; n(CaCO₃) = 0.00950/2 = 0.00475 mol. Mass CaCO₃ = 0.00475 × 100.1 = 0.475475 g; percentage purity = 79.2%. This assumes impurities do not also consume acid and all remaining acid is titrated.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources