Standard solution and an acid titration: which error actually matters?
A candidate must prepare 250.0 cm³ of a standard solution of anhydrous sodium carbonate of concentration 0.0500 mol dm⁻³ (M = 106 g mol⁻¹) and use it to find the concentration of a hydrochloric acid solution. She weighs 1.33 g of the solid on a balance reading to two decimal places, dissolves it in a beaker and transfers the solution to a 250 cm³ volumetric flask, rinsing the beaker twice into the flask before making up to the mark and inverting several times. For each titration she rinses the conical flask with distilled water and then with about 10 cm³ of the acid, shaking out the excess, before pipetting in 25.0 cm³ of the acid and adding two drops of methyl orange. Her burette readings give titres of 24.10 cm³, 23.75 cm³ and 23.70 cm³. She uses the mean of the last two titres to calculate the acid concentration and reports it to four significant figures.
1. Show by calculation whether 1.33 g is an acceptable mass to weigh for this standard solution, and state the effect of any difference on the acid concentration she calculates.
Read the worked answer
Required mass = c × V × M = 0.0500 × 0.2500 × 106 = 1.325 g.
She weighed 1.33 g, which is 0.005 g high, i.e. 0.005/1.325 × 100 ≈ 0.38% high. The true concentration of her standard solution is therefore 1.33/(106 × 0.2500) = 0.0502 mol dm⁻³, so an answer calculated assuming exactly 0.0500 mol dm⁻³ will be about 0.4% too high.
Judgement: this is acceptable. A two-decimal balance has an uncertainty of roughly ±0.005 g on a single weighing, so the difference is at the level of the instrument and cannot be removed by 'weighing more carefully'. It is also much smaller than the uncertainties from reading a burette to ±0.05 cm³ (about 0.2% per titre, doubled because two readings are taken) and from judging the indicator end-point (typically 0.05–0.1 cm³, about 0.3%). What does matter is that she records the mass actually weighed to two decimal places and uses it, rather than silently rounding it back to the theoretical value.
2. Evaluate the effect of rinsing the conical flask with the acid before each titration. What should she have done instead?
Read the worked answer
Rinsing with the acid leaves a few drops of acid clinging to the flask. Those extra moles are not delivered from the burette, so the carbonate solution has to neutralise them as well as the pipetted 25.0 cm³. The titre is therefore too large.
Since she calculates [HCl] = 2 × [Na₂CO₃] × V(Na₂CO₃)/V(acid), an inflated V(Na₂CO₃) gives an acid concentration that is too high - a systematic error affecting every titre, not a random one.
Correct practice: rinse the conical flask with distilled water only. Water dilutes the pipetted acid slightly but does not change the number of moles of acid in it, and the titration measures moles, not concentration, in the flask.
3. Explain why the first titre is discarded, and calculate the acid concentration from the two concordant titres.
Read the worked answer
The first reading is usually high because the burette has just been filled: solution wets the walls above the tap and drains down during the titration, and any air bubble in the jet is displaced, so the volume delivered is not exactly the reading difference. It is a conditioning run, not a measurement.
Mean titre = (23.75 + 23.70)/2 = 23.725 cm³
n(Na₂CO₃) = 0.0500 × 23.725/1000 = 1.186 × 10⁻³ mol Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂, so n(HCl) = 2 × 1.186 × 10⁻³ = 2.373 × 10⁻³ mol
[HCl] = 2.373 × 10⁻³/0.0250 = 0.0949 mol dm⁻³
Note the reporting: 0.0949 (three significant figures) is honest for this apparatus; quoting 0.09487 overstates the precision of a burette read to 0.05 cm³.