Standardizing NaOH: When to Reject a Titration Trial
A student standardizes a NaOH solution using potassium hydrogen phthalate (KHP, molar mass 204.22 g/mol) and phenolphthalein. Four KHP samples are weighed: 0.4082 g, 0.4110 g, 0.4095 g, and 0.4078 g. Each is dissolved in about 50 mL of deionized water and titrated to a persistent pink endpoint. The corresponding NaOH volumes are 19.85 mL, 20.02 mL, 19.95 mL, and 18.10 mL. The student notices that the last trial required noticeably less titrant and asks whether it should be discarded. The burette was rinsed with deionized water, not with the NaOH solution, before the trials began. The student needs a defensible final concentration and an explanation of how the outlier and the burette preparation affect the result. She has no additional KHP.
1. Calculate the NaOH molarity for each trial, apply a reasonable outlier test, and state whether the 18.10 mL trial should be retained. Show the final concentration you would report.
Read the worked answer
Use n(KHP) = m / 204.22 g mol−1 and M(NaOH) = n / V(L). The four trial molarities are:
- (0.4082/204.22)/0.01985 = 0.1007 M
- (0.4110/204.22)/0.02002 = 0.1005 M
- (0.4095/204.22)/0.01995 = 0.1005 M
- (0.4078/204.22)/0.01810 = 0.1103 M
The last value is high because its titrant volume is unusually low. For n = 4, Dixon's Q = (0.1103 − 0.1007)/(0.1103 − 0.1005) ≈ 0.98, which exceeds the 95% critical value (about 0.93). Reject the 18.10 mL trial. The mean of the first three is 0.1006 M; report 0.1006 M with four significant figures. Retaining all four would give 0.1031 M and would bias the result from a likely premature endpoint.
2. Identify the most likely cause of the low-volume trial and explain how the burette rinsing affects all four calculated molarities. Recommend two concrete procedural changes.
Read the worked answer
The low-volume trial most likely reached the endpoint prematurely: incomplete swirling, an air bubble in the burette tip, or reading the meniscus before the pink color persisted. The water rinse is a separate systematic error. Residual water dilutes the NaOH in the burette, so more titrant volume is delivered for the same moles of KHP; because calculated molarity is n/V, all calculated molarities are then slightly too low. That systematic effect would not produce one high value, so it does not explain the 18.10 mL trial. Procedural changes: condition the burette with a small portion of the NaOH solution, and dissolve each KHP sample completely while swirling continuously. Also check for and expel air bubbles and read the bottom of the meniscus against a white background.
3. The student reports 0.1006 M. A second student repeats with a properly conditioned burette and obtains 0.1019 M. Is the difference consistent with the known error? Explain.
Read the worked answer
The second result is about 1.3% higher than 0.1006 M. A water-rinsed burette predicts a low calculated molarity because the titrant is diluted, so the direction matches the known error. The magnitude is also plausible for a small amount of residual water in a burette. The difference therefore supports the explanation that the first result is systematically low. It does not rescue the rejected 18.10 mL trial, which is an isolated endpoint error rather than the same systematic dilution effect.