Vinegar Titration: A Label Claim and a Persistent Bias
A student analyses a local brand of vinegar labelled '5.0 g ethanoic acid per 100 cm³'. She pipettes 25.0 cm³ of the vinegar into a 250 cm³ volumetric flask and makes the mark with distilled water. She rinses and fills a burette with 0.100 mol dm⁻³ sodium hydroxide solution. She titrates 25.0 cm³ portions of the diluted vinegar with phenolphthalein, using a white tile because the vinegar is dark. Her burette readings are: 0.00 → 22.40 cm³; 0.00 → 22.35 cm³; 0.10 → 22.55 cm³. The label claim is a legal minimum, and the acceptable analytical tolerance is ±3% of the labelled value. (Mᵣ CH₃COOH = 60.0; the density of the vinegar is 1.01 g cm⁻³.)
1. Calculate the ethanoic acid content in g per 100 cm³ and decide whether the bottle meets the label claim and the ±3% tolerance. Show your reasoning.
Read the worked answer
Use the mean of the concordant titres. The three titres are 22.40, 22.35 and 22.45 cm³; mean = 22.40 cm³.
n(NaOH) = 0.100 × 22.40/1000 = 2.240 × 10⁻³ mol.
CH₃COOH + NaOH → CH₃COONa + H₂O, so n(CH₃COOH) in the 25.0 cm³ aliquot = 2.240 × 10⁻³ mol.
The aliquot came from 250 cm³, so n in the whole flask = 2.240 × 10⁻³ × 10 = 2.240 × 10⁻² mol. This is the acid in the original 25.0 cm³ of vinegar.
mass = 2.240 × 10⁻² × 60.0 = 1.344 g per 25.0 cm³.
Per 100 cm³: 1.344 × 4 = 5.38 g.
So the vinegar contains 5.38 g ethanoic acid per 100 cm³.
The legal minimum is 5.0 g per 100 cm³, so 5.38 g passes that minimum. However, the ±3% tolerance is 4.85–5.15 g per 100 cm³, and 5.38 g is about 7.6% above the label. The result therefore does not support the label within the stated tolerance; either the batch is over-strength or there is a systematic error in the analysis. A repeat with a fresh bottle and a check against a standard would be needed before reporting a failure.
2. Explain why phenolphthalein is appropriate for this titration and why methyl orange is not.
Read the worked answer
The titration is weak acid (ethanoic acid) against strong base (NaOH). At the equivalence point the solution contains sodium ethanoate, which is the salt of a weak acid and a strong base; it hydrolyses to give an alkaline solution, typically about pH 8–9.
Phenolphthalein changes over about pH 8.2–10.0 (colourless to pink), so its endpoint falls close to the equivalence pH. The first permanent faint pink is the correct endpoint.
Methyl orange changes over about pH 3.1–4.4 (red to yellow). That is in the acidic region, well before the equivalence point. The student would stop adding NaOH while unreacted ethanoic acid is still present, so the titre would be too small and the calculated acid content too low. Indicator choice must match the pH jump, not personal preference.
3. The pipette was rinsed with distilled water but not with the diluted vinegar before delivering the 25.0 cm³ portion. State and explain the effect on the calculated acid concentration.
Read the worked answer
Residual distilled water in the pipette dilutes the portion that is delivered. The 25.0 cm³ nominal volume therefore contains slightly less ethanoic acid than it should. A smaller amount of acid requires a smaller volume of NaOH, so the titre is lower.
The calculation assumes n(acid) is proportional to the titre; a low titre gives a low calculated acid content. This is a systematic error (it affects every run in the same direction), not a random one, so averaging repeats will not remove it.
The pipette should be rinsed with a little of the diluted vinegar before use. The conical flask, by contrast, should not be rinsed with vinegar; a little water in the flask is harmless because it does not change the amount of acid being titrated.