Overshot Endpoint and an Outlier in a Titration
In an honors chemistry lab, students standardize a sodium hydroxide solution using potassium hydrogen phthalate (KHP) and then use it to determine the molarity of an unknown hydrochloric acid solution. Each student massed about 0.500 g of KHP and titrated to a phenolphthalein endpoint. One student, Jordan, reports NaOH volumes of 24.85 mL, 24.92 mL, 25.10 mL, and 27.60 mL. The fourth trial turned dark magenta after Jordan added an extra 2 mL past the first pink flash. The class then titrated 25.00 mL of the unknown acid; the average of the first three NaOH trials gives 22.40 mL for the unknown. Jordan proposes to drop the 27.60 mL trial 'because it is bad' and average the rest.
1. Should Jordan's fourth trial be excluded? Justify the decision with evidence and describe how to document it.
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The fourth trial should be excluded, but not merely because Jordan calls it bad. There is a documented procedural reason: the endpoint was overshot by about 2 mL, and the solution turned dark magenta. In a phenolphthalein titration, the first persistent pale pink is the endpoint; excess NaOH raises the pH well above the endpoint and makes the recorded volume too large. The 27.60 mL value is also separated from the other three by about 2.5 mL, far more than the roughly 0.25 mL spread among 24.85, 24.92, and 25.10 mL.
The exclusion should be recorded in the lab notebook with the observation 'overshot endpoint, dark magenta, approximately 2 mL excess,' not simply 'outlier.' The student should report the mean of the three valid trials and state that the fourth trial was rejected for a known systematic procedural error. If a statistical outlier test is used, it should be secondary to the chemical reason; a value should not be discarded only because it raises or lowers the average.
2. Calculate the NaOH molarity and the unknown HCl molarity from the valid data, stating assumptions and uncertainty.
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Using 0.500 g KHP and a molar mass of 204.22 g/mol, moles of KHP = 0.500/204.22 = 2.448 × 10−3 mol. The valid NaOH volumes average to (24.85 + 24.92 + 25.10)/3 = 24.9567 mL = 0.0249567 L. Because KHP reacts 1:1 with NaOH, molarity of NaOH = 2.448 × 10−3 mol / 0.0249567 L = 0.0981 M.
For the unknown acid, moles of NaOH used = 0.0981 M × 0.02240 L = 2.197 × 10−3 mol. The reaction HCl + NaOH → NaCl + H2O is 1:1, so moles HCl = 2.197 × 10−3 mol in 25.00 mL. Molarity HCl = 2.197 × 10−3 / 0.02500 = 0.0879 M. Assumptions include a pure KHP sample, quantitative reaction, and equivalence of the endpoint to the stoichiometric point; the visible endpoint adds uncertainty of roughly one drop, about 0.05 mL, plus the uncertainty in the KHP mass and the unknown volume. The result should be reported as approximately 0.088 M, not with excessive significant figures.
3. What conceptual misunderstanding might the dark endpoint reveal, and how should the teacher respond?
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The dark magenta endpoint suggests Jordan may be treating 'more color' as 'more complete' or may not understand that the indicator changes over a pH range and that the first permanent pale pink is the target. It may also reflect a procedural habit of adding titrant rapidly near the endpoint instead of dropwise. The color is not a measure of how much acid was present; it is a signal that the equivalence point has been passed.
The teacher should respond with a short post-lab conference or mini-lesson: show a pH titration curve with the phenolphthalein range marked, demonstrate the pale-pink endpoint with a known sample, and have Jordan practice approaching the endpoint dropwise with swirling. The student should then recalculate using the valid trials and write an explanation of why the fourth trial was high. This addresses the reasoning rather than simply telling Jordan to be more careful, and it connects the procedural error to the underlying acid-base chemistry.