Vinegar Titration: Choosing Titres and Explaining an Overestimate
A student is determining the ethanoic acid content of a commercial vinegar. She pipettes 25.0 cm³ of the vinegar into a 250.0 cm³ volumetric flask and makes it up to the mark with distilled water. She rinses a burette with distilled water only, then fills it with sodium hydroxide solution of concentration 0.0987 mol dm⁻³. Using phenolphthalein, she titrates 25.0 cm³ portions of the diluted vinegar and records burette readings (final minus initial) of 24.10 cm³ (rough), 23.90 cm³, 23.95 cm³ and 23.90 cm³. The end-point is a permanent pale pink that persists for 30 seconds. She reports the vinegar as 5.9% (w/v) ethanoic acid. Her teacher says the manufacturer's labelled value for this brand is 5.0% (w/v), and asks her to check her work rather than simply change the answer.
1. Select the appropriate titres, calculate the mean titre, and calculate the ethanoic acid concentration in % (w/v). Show all working.
Read the worked answer
Selecting titres. The rough titration is used only to estimate the end-point and is discarded. The three accurate titres are 23.90, 23.95 and 23.90 cm³. The spread is 23.95 − 23.90 = 0.05 cm³, well within the usual 0.10 cm³ tolerance, so all three are concordant and all are averaged.
Mean titre = (23.90 + 23.95 + 23.90) / 3 = 23.92 cm³
Moles of NaOH. n = cV = 0.0987 mol dm⁻³ × 0.02392 dm³ = 2.361 × 10⁻³ mol
Moles of acid. CH₃COOH + NaOH → CH₃COONa + H₂O, so the ratio is 1:1. n(CH₃COOH) in the 25.0 cm³ aliquot = 2.361 × 10⁻³ mol
Scale to the original vinegar. The aliquot came from a 250.0 cm³ flask, so total moles in the flask = 2.361 × 10⁻³ × 10 = 2.361 × 10⁻² mol. This is the amount in the original 25.0 cm³ of undiluted vinegar.
Mass = 2.361 × 10⁻² mol × 60 g mol⁻¹ = 1.417 g
% (w/v) = (1.417 g / 25.0 cm³) × 100 = 5.67% w/v, i.e. about 5.7% w/v.
2. The student's reported 5.9% is higher than both your calculated value and the label. Explain how the water-rinsed burette affects the result, and assess whether it accounts for the discrepancy.
Read the worked answer
Rinsing with distilled water leaves water clinging to the burette, which dilutes the sodium hydroxide slightly below 0.0987 mol dm⁻³. A more dilute base needs a larger volume to neutralise the same amount of acid, so the recorded titre is too big. Because the calculation treats that inflated volume as if it were 0.0987 mol dm⁻³, the calculated moles of acid, and hence the percentage, are both too high - exactly the direction of the student's error.
So the fault is real and it does explain part of the gap. But the corrected value (5.7%) is still above the label (5.0%), so the rinse is not the whole story. Other plausible contributors are: reading the top rather than the bottom of the meniscus, a parallax error, running past the first permanent pink, a contaminated pipette, or simply that the label is a minimum content. The correct procedure is to rinse the burette with a little of the sodium hydroxide solution before filling it.
3. The student repeats the titration using methyl orange instead of phenolphthalein. Predict whether her calculated percentage would rise, fall or stay the same, and justify your prediction.
Read the worked answer
It would fall.
This is a weak acid (ethanoic acid) titrated with a strong base (NaOH). At the equivalence point the solution contains ethanoate ions, which hydrolyse (CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻) and make the solution basic - the equivalence pH is roughly 8–9, not 7.
Methyl orange changes over about pH 3.1–4.4. In this titration that range is reached while there is still unneutralised acid in the flask, so the indicator changes before the equivalence point. The recorded titre is therefore too small, giving fewer moles of acid and a lower percentage.
Phenolphthalein (pH 8.3–10) straddles the equivalence pH of this titration and is the correct choice; the indicator must be matched to the salt formed, not chosen arbitrarily.