Choosing an Indicator in a Weak Acid–Strong Base Titration
A student titrates 25.0 cm³ samples of ethanoic acid solution of unknown concentration with 0.100 mol/dm³ sodium hydroxide. Using methyl orange, the indicator turns from red to yellow after 22.90 cm³ of alkali. The student then repeats the titration using phenolphthalein and obtains concordant results of 24.20, 24.15 and 24.20 cm³. The student is unsure which value to use. They know that the equivalence point of a weak acid–strong base titration is not at pH 7. A classmate suggests averaging all five results to improve accuracy. The teacher says the methyl orange result is not an outlier to be discarded but a systematic error caused by indicator choice. The student must decide which data to report and whether the classmate's suggestion is valid.
1. Explain why methyl orange gives a smaller titre than phenolphthalein for this titration.
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At the equivalence point, the solution contains sodium ethanoate. The ethanoate ion is the conjugate base of a weak acid and hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. This makes the equivalence pH alkaline, roughly pH 8–9, not pH 7. Phenolphthalein changes colour over about pH 8.3–10.0, close to the equivalence pH, so its endpoint is a good approximation. Methyl orange changes over about pH 3.1–4.4, which is reached while the acid is still only partly neutralised. The methyl orange endpoint therefore occurs before the equivalence point, so less sodium hydroxide has been added and the titre is smaller. This is a systematic error, not a random outlier.
2. Calculate the concentration of the ethanoic acid using the valid data.
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The valid phenolphthalein titres are 24.20, 24.15 and 24.20 cm³, which are concordant. Mean titre = (24.20 + 24.15 + 24.20) / 3 = 24.183 cm³ ≈ 24.18 cm³. n(NaOH) = c × V = 0.100 × 0.02418 = 2.418 × 10⁻³ mol. CH₃COOH + NaOH → CH₃COONa + H₂O, so the acid and alkali react in a 1:1 ratio. n(CH₃COOH) in 25.0 cm³ = 2.418 × 10⁻³ mol. c(CH₃COOH) = n / V = 2.418 × 10⁻³ / 0.0250 = 0.0967 mol/dm³ (approximately 0.0967 M). Using 24.20 cm³ alone gives 0.0968 mol/dm³, which is within normal rounding.
3. Evaluate the classmate's suggestion to average all five results.
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Averaging all five results would be wrong. The 22.90 cm³ value comes from a different indicator and has a systematic error: methyl orange changes before the equivalence point, so including it would bias the mean downwards. The classmate is confusing random error, which can be reduced by averaging, with systematic error, which cannot be removed that way. The phenolphthalein results are the valid set because phenolphthalein's colour-change interval matches the alkaline equivalence point of this weak acid–strong base titration. The student should report the mean of the concordant phenolphthalein titres, not an average of all five values.