Vinegar Titration: A Decimal Error That Looks Like a Confirmed Label
A student titrates 25.0 mL of white vinegar with 0.100 M NaOH. With phenolphthalein, a persistent pale pink appears after 22.4 mL of NaOH. The student computes the acetic acid concentration as (0.100 M)(0.0224 L)/(0.0250 L) = 0.0896 M. Using a density of 1.00 g/mL and a molar mass of 60.05 g/mol, the student reports 5.38% acetic acid by mass. The bottle label says 5.0% acidity, so the student claims the result is confirmed. A second student uses bromothymol blue and obtains a persistent blue endpoint at 21.8 mL. The teacher asks the class to check the percent-by-mass conversion, evaluate the indicator choice, and decide whether the two titrations agree within reasonable error. A classmate points out that 25.0 mL of 5.0% vinegar would need much more than 22.4 mL of 0.100 M NaOH.
1. Check the percent-by-mass calculation. Is 5.38% correct for this sample? Show the conversion.
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The concentration calculation is correct: (0.100 M)(0.0224 L)/(0.0250 L) = 0.0896 M. But the mass percent is wrong. 0.0896 mol/L × 60.05 g/mol = 5.38 g/L. A 25.0 mL sample has mass 25.0 g because the density is 1.00 g/mL, and it contains 5.38 g/L × 0.0250 L = 0.1345 g acetic acid. Percent by mass = 0.1345 g / 25.0 g × 100 = 0.538%, not 5.38%. The student's reported value is ten times too high, so the label's 5.0% is not confirmed by this calculation.
2. Evaluate the indicator choices. Can the difference between 22.4 mL and 21.8 mL explain the discrepancy with the label?
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Phenolphthalein is appropriate for a weak acid–strong base titration because the equivalence point is basic, roughly pH 8–9, and phenolphthalein changes in that region. Bromothymol blue changes at lower pH, about 6–7.6, so it will signal an endpoint before the equivalence point and give a smaller volume, which is consistent with 21.8 mL versus 22.4 mL. That difference is 0.6 mL, only about 2.7% of the volume. It cannot explain a factor-of-ten discrepancy with the label. Indicator choice is a minor systematic error here; the major problem is the percent conversion or the sample volume.
3. What volume of 0.100 M NaOH should 25.0 mL of 5.0% vinegar require? Use that to judge whether the titration procedure is plausible.
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25.0 mL of 5.0% vinegar contains 25.0 g × 0.050 = 1.25 g acetic acid. Moles = 1.25 g / 60.05 g/mol = 0.0208 mol. NaOH volume = 0.0208 mol / 0.100 mol/L = 0.208 L = 208 mL. The observed 22.4 mL is almost exactly one-tenth of that, so either the vinegar is about 0.54% acetic acid, the sample was actually 2.50 mL, the NaOH was about 1.00 M, or a decimal or unit error was made. The group should repeat with a fresh sample and a buret, record the actual sample volume, and check the NaOH concentration by titration with a primary standard.