Vinegar Titration: Outlier, Claim, and Procedural Error
A student determines the acetic acid content of household vinegar by titration. She pipettes 10.0 mL of vinegar (density 1.01 g/mL) into a flask, adds phenolphthalein, and titrates with 0.500 M NaOH. Three trials give 24.10 mL, 24.30 mL, and 26.80 mL. The third endpoint is a vivid pink that persists; the first two are pale pink. The teacher says the label claims 7% acetic acid and asks whether the data support that claim. The student must decide whether to average all three trials, discard one, or repeat the titration, and must explain how each choice affects the reported percentage. She also has to explain one procedural error that would make the result too high.
1. Compute the acetic acid mass percent using the two consistent trials; show whether the 7% claim is supported. Use molar mass 60.05 g/mol.
Read the worked answer
Use the two consistent trials: mean volume = (24.10 + 24.30)/2 = 24.20 mL = 0.02420 L. Moles NaOH = 0.500 × 0.02420 = 0.01210 mol. The reaction is 1:1, so moles acetic acid = 0.01210 mol. Mass acid = 0.01210 × 60.05 = 0.7266 g. Sample mass = 10.0 mL × 1.01 g/mL = 10.1 g. Mass percent = 0.7266/10.1 × 100 ≈ 7.19%. This supports the 7% claim within ordinary titration error, about 0.2 percentage points above the label.
2. Justify whether the 26.80 mL trial should be included. What evidence in the setup supports your decision?
Read the worked answer
The 26.80 mL trial should not be averaged without investigation. It is 2.5 mL above the other two, about 10% higher, and the endpoint is described as vivid pink rather than pale pink. That suggests the student overshot the endpoint, adding excess NaOH. Including it would raise the mean to (24.10 + 24.30 + 26.80)/3 = 25.07 mL and give about 7.45% acetic acid, still plausible but less accurate. The better choice is to discard the trial as an outlier, report the 7.19% from the consistent trials, and repeat a third titration to confirm. The decision should be justified by the endpoint evidence, not merely by the number.
3. Identify one procedural error that would make the calculated percentage too high, and explain the direction of the error.
Read the worked answer
One error: rinsing the burette with distilled water instead of with the NaOH solution. Residual water dilutes the titrant, so a larger volume is needed to reach the endpoint. The calculated moles of NaOH, and therefore of acetic acid, would be too high, making the reported percentage too high. Another error: an air bubble in the burette tip that disappears during titration; the recorded volume includes the bubble, so the volume reading is too high and the percentage is overestimated. Either error affects the result in the same direction.