Part 2: Standard solutions and titration calculations

Prepare a known concentration, find a reliable end point, and carry the aliquot factor through an unstructured calculation.

Make a known solution quantitatively

A standard solution has an accurately known concentration. Choose a suitable pure, stable solute. Calculate m = cVM; for 250.0 cm³ of 0.100 mol dm⁻³ Na₂CO₃, m = 0.100 × 0.2500 × 106 = 2.65 g.

Weigh by difference, dissolve fully in a beaker, transfer through a funnel into a volumetric flask, rinse the beaker, rod and funnel into it, cool to calibration temperature, and add water to the mark dropwise. Stopper and invert repeatedly. Rinsing transfers all solute; mixing gives uniform concentration. Do not merely add 250 cm³ water to the solid.

Explain each preparation step through n/V

Concentration equals solute moles divided by final solution volume. Leaving solid in the weighing bottle or failing to rinse the beaker makes the delivered amount smaller than assumed. Adding water beyond the flask mark makes the volume too large. Both produce a solution less concentrated than its calculated value, but by different mechanisms.

A flask still wet with distilled water is usually acceptable because you make the solution up to its final calibrated volume. A wet weighing vessel is different: if water contributes to the recorded “solid” mass, fewer solute moles are present than assumed. Unmixed solution can give aliquots that do not represent the intended bulk concentration.

Weighing by difference measures the mass actually removed from a container; it does not by itself prove that all removed solid reached the flask. Quantitative transfer and rinsing are still needed. Choose the balance resolution and flask size so their uncertainties are appropriate for the target concentration.

Titrate with a consistent end point

Rinse the pipette with the solution it will deliver and use a filler to transfer a fixed aliquot to a conical flask. Rinse the burette with its reagent, fill the jet without air bubbles, remove the funnel and record the initial reading at eye level. Use a few drops of suitable indicator and a white tile.

Make a rough titration, then add reagent dropwise near the end point while swirling. Record both readings, subtract to obtain each titre, and repeat until concordant. Typical school burette readings are recorded to 0.05 cm³, written to two decimal places. Follow the stated instrument precision.

For strong acid–strong alkali either methyl orange or phenolphthalein can be suitable. Methyl orange is red in acid and yellow in alkali, with an orange end point; phenolphthalein is colourless in acid and pink in alkali. A strong alkali with weak acid normally uses phenolphthalein, a strong acid with weak base methyl orange. Do not choose an indicator merely because it changes colour somewhere.

Process a complete results table

Use initial and final burette readings to calculate every titre. A burette does not have to start at zero. In the constructed table, use the two closely agreeing accurate titres and exclude the rough result from the mean. If the question specifies a concordance criterion, apply that criterion rather than inventing one.

Mean = (23.40 + 23.45)/2 = 23.425 cm³, reported as 23.43 cm³ to two decimal places. Keep 23.425 in subsequent calculations if the question does not direct otherwise. Do not round the raw readings to whole numbers or include the rough result just to have more data.

Worked example with an aliquot

A sample containing Na₂CO₃ is made up to 250.0 cm³. A 25.0 cm³ aliquot requires 20.00 cm³ of 0.100 mol dm⁻³ HCl for complete neutralisation. n(HCl) = 0.100 × 0.02000 = 0.00200 mol. From Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O, carbonate in the aliquot = 0.00100 mol.

The flask contains ten aliquots, so total carbonate = 0.0100 mol. Mass = 0.0100 × 106 = 1.06 g. If the original sample weighed 1.25 g and other components did not consume acid, purity = 1.06/1.25 × 100 = 84.8%. Label “in aliquot” and “in full flask” to prevent a factor-of-ten error.

A direct calculation with a non-1:1 ratio

25.0 cm³ of 0.120 mol dm⁻³ NaOH neutralises 18.75 cm³ H₂SO₄. Start with H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Known NaOH amount = 0.120 × 0.0250 = 0.00300 mol. Acid amount is half: 0.00150 mol. Acid concentration = 0.00150/0.01875 = 0.0800 mol dm⁻³.

The final division uses the volume belonging to the substance whose concentration is requested. The 25.0 cm³ volume has already served its purpose in finding NaOH moles. Reusing it for acid concentration would answer a different question.

For a diluted unknown, first find its diluted concentration and then undo the dilution factor. If 10.0 cm³ original vinegar is diluted to 100.0 cm³, a measured diluted concentration of 0.0750 mol dm⁻³ corresponds to 0.750 mol dm⁻³ originally. This factor of ten is separate from any acid:base coefficient ratio.

Back titration: subtract the excess before finding the sample

Back titration is an application of the same AS mole principles when an excess reagent and subsequent titration are described. An original example: a 0.600 g impure MgO sample reacts completely with 50.0 cm³ of 0.500 mol dm⁻³ HCl. The remaining acid requires 10.00 cm³ of 0.500 mol dm⁻³ NaOH. Assume impurities do not react.

Initial HCl = 0.0500 × 0.500 = 0.0250 mol. Leftover HCl equals the NaOH amount = 0.01000 × 0.500 = 0.00500 mol. Therefore acid consumed by MgO = 0.0200 mol. From MgO + 2HCl → MgCl₂ + H₂O, n(MgO) = 0.0100 mol.

With M(MgO) = 40.3 g mol⁻¹, pure MgO mass = 0.403 g and purity = 0.403/0.600 × 100 = 67.2%. The second titration measures acid left over, not acid consumed. If only an aliquot of the remaining acid were titrated, scale that remainder to the whole flask before subtracting.

Explain direction of error

Illustrative titres 20.00, 20.10 and 20.80 cm³: average the concordant first two to obtain 20.05 cm³; investigate the last. Close agreement is precision, not proof of accuracy. For ±0.05 cm³ per reading, titre uncertainty is ±0.10 cm³, or about 0.50% of 20.05 cm³.

Residual water in the burette dilutes its reagent, so more volume is needed for the same analyte moles. Distilled water in the conical flask changes concentration but not analyte moles, so it does not change the ideal titre. An air bubble initially in the burette jet can make apparent delivered volume too large.

Link the apparatus error to the reported result

For an unknown acid in the flask and standard NaOH in the burette, water left in the burette dilutes the NaOH. More volume is needed. If the calculation uses the labelled undiluted concentration, it overestimates NaOH moles and therefore overestimates acid concentration. The direction follows from this arrangement; swapping the unknown and standard requires fresh reasoning.

Overshooting the end point also makes this calculated acid concentration too high. Residual distilled water in the conical flask does not add acid moles, so it does not change the ideal titre. Rinsing that flask with extra unknown acid would add unmeasured acid moles and increase the titre.

Distinguish an end point, observed through the indicator, from the equivalence point defined by stoichiometric amounts. A suitable indicator changes over the steep pH change near equivalence. Adding excessive indicator can itself perturb the titration because indicators are acid–base substances.

Quick checks

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

Sources