A discordant titration reading caused by an air bubble
A student prepares a standard solution by dissolving 0.630 g of oxalic acid dihydrate (H₂C₂O₄·2H₂O, M = 126 g mol⁻¹) in distilled water and making the volume up to 250.0 cm³. Portions of 25.0 cm³ are titrated with sodium hydroxide solution of unknown concentration, using two drops of phenolphthalein. The burette readings (final minus initial) are 19.8, 19.9, 21.6 and 19.9 cm³. The 21.6 cm³ reading was taken first, immediately after filling the burette, and the student recalls a small air bubble in the tip that disappeared during that titration. The student proposes to average all four readings and report the result. The teacher insists on at least three concordant readings within 0.10 cm³ before any calculation is accepted, and asks for the concentration with and without the rejected value.
1. Calculate the oxalic acid concentration, then the NaOH concentration both with all four readings and with the rejected reading removed. Decide which value should be reported and justify the rejection.
Read the worked answer
Oxalic acid: 0.630/126 = 5.00 × 10⁻³ mol in 250.0 cm³, so 0.0200 M. Each 25.0 cm³ portion therefore contains 5.00 × 10⁻⁴ mol of acid.
H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O, so NaOH required = 2 × 5.00 × 10⁻⁴ = 1.00 × 10⁻³ mol.
All four readings: mean = (19.8 + 19.9 + 21.6 + 19.9)/4 = 20.3 cm³, giving [NaOH] = 1.00 × 10⁻³/0.0203 = 0.049 M.
Concordant three: mean = 19.87 cm³, giving [NaOH] = 1.00 × 10⁻³/0.01987 = 0.0503 M.
Report 0.050 mol dm⁻³. The 21.6 cm³ reading lies 1.7 cm³ outside the spread of the others, roughly seventeen times the permitted 0.10 cm³, and has an identified physical cause: the air bubble occupied part of the burette tip at the start, so the recorded volume includes gas that was never delivered as liquid. The true volume delivered must have been close to 19.9 cm³. This is a determinate error with a known cause, not random scatter, so rejecting it is justified - and the reading should be repeated rather than merely deleted.
2. The burette was rinsed with distilled water but not with the NaOH solution. State the direction of the resulting error and explain, with the same reasoning for a pipette rinsed only with water.
Read the worked answer
Burette: residual water dilutes the NaOH as it is poured in. More titrant volume is then needed to deliver the same number of moles, so the titre is too large and the calculated [NaOH] is too low. Because the dilution affects the whole burette filling, this is a systematic error, not scatter.
Pipette: residual water dilutes the 25.0 cm³ oxalic acid portion that is transferred, so fewer moles of acid reach the flask. Less NaOH is required, the titre is too small, and the calculated [NaOH] is too high.
Note that the two errors push the answer in opposite directions, so they partially cancel - which is exactly why a candidate should never assume 'small rinsing errors do not matter'. The correct practice is to rinse the burette with the NaOH solution and the pipette with the oxalic acid solution, then discard those rinsings.
A sensible candidate also flags a separate systematic problem: NaOH absorbs CO₂ from the air to form carbonate, which with phenolphthalein is titrated only to hydrogen carbonate, so the solution behaves as a slightly weaker base than its label claims.
3. Predict the effect on the calculated [NaOH] if methyl orange were used instead of phenolphthalein, and justify your answer.
Read the worked answer
The calculated [NaOH] would come out too low, because the endpoint would be reached early.
Methyl orange changes colour over roughly pH 3.1–4.4, whereas the equivalence point of this titration lies in the basic region, around pH 8, because the product is sodium oxalate, the salt of a weak acid. The solution would therefore be judged 'endpoint reached' while it is still acidic and unreacted NaOH remains in the flask - that is, after less titrant has been delivered. A smaller titre gives a smaller calculated concentration.
Phenolphthalein, changing over about pH 8.2–10.0, brackets the equivalence point correctly for a weak acid–strong base titration and gives a pale pink endpoint that fades on standing.
The wider lesson is that indicator choice is part of the design of the experiment, not decoration. The same reasoning explains why a strong acid–strong base titration tolerates either indicator, while a weak base–strong acid titration requires methyl orange.