Checking a '5% Acidity' Vinegar Label
A small vinegar bottling workshop near Battambang prints '5% acidity' on its labels. A Grade 12 student volunteers to check one bottle for her practical investigation. With a 10.00 mL pipette she titrates three 10.00 mL samples against sodium hydroxide solution of concentration 0.500 mol/L, using phenolphthalein, and obtains titres of 16.80 mL, 16.85 mL and 16.75 mL. She measures the density of the vinegar as 1.012 g/mL. The label does not say whether '5%' means mass per volume (w/v) or mass per mass (w/w). M(CH₃COOH) = 60.05 g/mol and Ka = 1.8 × 10⁻⁵. The workshop owner insists that one titration is enough, and that adding distilled water to the conical flask before titrating 'dilutes the acid and spoils the result'. Her teacher asks her to report a verdict the owner can act on, not just a number.
1. Calculate the mean titre, the amount of acetic acid in the sample and its concentration. Express the result as % w/v and % w/w, and decide whether the label is accurate.
Read the worked answer
Mean titre = (16.80 + 16.85 + 16.75)/3 = 16.80 mL. The spread is 0.10 mL, about 0.6%, which is acceptable precision.
n(NaOH) = 0.500 × 0.01680 = 8.40 × 10⁻³ mol. The reaction is CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O, 1:1, so the 10.00 mL sample contains 8.40 × 10⁻³ mol of acid.
c = 8.40 × 10⁻³ / 0.01000 = 0.840 mol/L Mass per litre = 0.840 × 60.05 = 50.4 g/L → 5.04% w/v
For w/w, 1.000 L of vinegar has mass 1012 g, so 50.4/1012 × 100 = 4.99% w/w.
Verdict: both readings of the ambiguous label pass. The true value is 5.04% w/v, which is the same solution as 4.99% w/w; the difference of 0.04 percentage points is well inside a ±5% tolerance and inside the uncertainty from the burette reading alone (±0.05 mL on 16.80 mL is ±0.3%). The label is accurate and the ambiguity happens not to matter here - though it would matter for a stronger product.
2. Assess the owner's two claims, and quantify what happens if the student overshoots the end point to 17.10 mL.
Read the worked answer
Adding distilled water to the flask is harmless. The titration measures moles of acid, and moles do not change when water is added; only the concentration in the flask falls, which slightly sharpens the visual end point. The owner has confused concentration with amount - the same error as thinking a dilute sample needs less alkali.
The single-titration claim is weaker but not absurd. One carefully run titration with a clear end point gives a result within about 1%, which is enough to confirm or reject a 5% claim. What repetition buys is detection of a systematic blunder: a dirty pipette, an air bubble under the burette tap, or a misread meniscus would be invisible in a single run. The correct advice is to titre roughly first, then do at least two concordant runs.
Overshoot: at 17.10 mL, c = (0.500 × 0.01710)/0.0100 = 0.855 mol/L, giving 0.855 × 60.05 = 51.3 g/L = 5.13% w/v. That is +1.8% relative to the true value - still inside a ±5% labelling tolerance, but it is larger than the real difference being tested. Reporting three titres with the mean is the defence.
3. Calculate the pH of the vinegar, and explain why pH = −log(0.840) = 0.08 would be badly wrong. Then predict the pH after a tenfold dilution.
Read the worked answer
Acetic acid is weak, so use Ka = [H⁺]²/(c − [H⁺]) ≈ [H⁺]²/c:
[H⁺] = √(1.8 × 10⁻⁵ × 0.840) = √(1.512 × 10⁻⁵) = 3.89 × 10⁻³ mol/L pH = −log(3.89 × 10⁻³) = 2.41
pH = 0.08 would assume complete ionisation, like a strong acid of the same concentration. That overstates [H⁺] by about 216 times. Only 3.89 × 10⁻³/0.840 = 0.46% of the molecules are ionised, so the vast majority contribute to the concentration but not to the acidity. The approximation is justified because [H⁺] is far below 5% of c.
Tenfold dilution to 0.0840 mol/L: [H⁺] = √(1.8 × 10⁻⁵ × 0.0840) = 1.23 × 10⁻³, pH = 2.91. For a weak acid the pH rises by only about 0.5 per tenfold dilution, not by 1 as it would for a strong acid. This is worth telling the workshop owner: watering the product down changes the titrated concentration in direct proportion, but changes the taste far less than the pH suggests.