Standardising a Vinegar Titration When One Run Disagrees
A student determines the ethanoic acid content of a commercial vinegar sample. She pipettes 25.0 cm3 of vinegar into a 250.0 cm3 volumetric flask and makes it up to the mark with distilled water. She titrates 25.0 cm3 portions of this diluted solution with 0.100 mol dm-3 sodium hydroxide, using phenolphthalein. Her three titre values are 18.60 cm3, 18.75 cm3 and 19.55 cm3. In the last titration the pale pink colour fades after about 15 s, before she records the volume; in the first two it persists for more than 30 s. Before filling, she rinsed the burette with distilled water but not with the sodium hydroxide solution. The label states that the vinegar contains 5.0 g of ethanoic acid per 100 cm3. (M(CH3COOH) = 60.0 g mol-1.)
1. Decide whether the 19.55 cm3 titre should be included in the mean. Justify your decision using both the endpoint observation and the numerical spread, and state the mean titre you would use.
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The 19.55 cm3 value is 0.80–0.95 cm3 above the other two titres, while those two differ by only 0.15 cm3. That is a large discrepancy compared with the repeatability of the other runs. More importantly, the endpoint in that run faded after about 15 s. A valid phenolphthalein endpoint should be a pale pink that persists for at least 30 s; a colour that fades indicates that the run was stopped before a proper endpoint or that the recorded volume is not a reliable endpoint volume. The run should therefore be rejected rather than averaged. Using 18.60 cm3 and 18.75 cm3 gives a mean titre of (18.60 + 18.75)/2 = 18.675 cm3, quoted as 18.68 cm3. Including 19.55 cm3 would give 18.97 cm3 and raise the calculated acid content by about 1.6%, but that apparent improvement does not make the faulty run valid.
2. Using your mean titre, calculate the ethanoic acid content of the original vinegar in g per 100 cm3 and compare it with the label value.
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Moles of NaOH used = 0.100 mol dm-3 × (18.675/1000) dm3 = 1.8675 × 10-3 mol. The acid and base react in a 1:1 mole ratio, so the 25.0 cm3 aliquot of diluted vinegar contains 1.8675 × 10-3 mol of ethanoic acid. The dilution factor is 250.0/25.0 = 10, so 25.0 cm3 of the original vinegar contains 1.8675 × 10-2 mol. Concentration of the original vinegar = 1.8675 × 10-2 mol / 0.0250 dm3 = 0.747 mol dm-3. Mass concentration = 0.747 × 60.0 = 44.8 g dm-3, which is 4.48 g per 100 cm3. The label claims 5.0 g per 100 cm3, so the calculated value is about 10% lower. Moreover, because the burette was rinsed with water, the sodium hydroxide was diluted and the calculated value is an upper estimate; the true value would be even lower. The result therefore does not support the label claim.
3. Explain the effect of rinsing the burette with distilled water rather than with the sodium hydroxide solution on the calculated ethanoic acid content.
Read the worked answer
Rinsing with distilled water leaves water in the burette, so the sodium hydroxide solution is diluted when the burette is filled. Its actual concentration is therefore less than the labelled 0.100 mol dm-3. To deliver the same amount of NaOH, a larger volume is needed, so the measured titre is larger than it should be. In the calculation, the student uses moles = 0.100 × titre volume. Because the true concentration is lower, the true moles delivered are less than 0.100 × titre volume; the calculated amount of NaOH, and hence the calculated amount of ethanoic acid, is too high. This is a systematic error that raises the apparent acid content. Rinsing the burette with a little of the sodium hydroxide solution before filling would remove this error.