Chemistry - Titration with an Unstandardized Base and a Fading Endpoint
A student standardizes a sodium hydroxide solution and uses it to titrate 25.00 mL of hydrochloric acid of unknown concentration. The burette readings for three trials are: initial 0.40 mL, final 24.90 mL; initial 1.20 mL, final 25.80 mL; initial 0.10 mL, final 26.10 mL. The NaOH concentration is nominally 0.100 M, but the student prepared it by dissolving 2.00 g of NaOH pellets in 500.0 mL of water and did not standardize against a primary standard. Phenolphthalein is used, and the third trial turns faint pink that persists for 20 seconds, then fades after 2 minutes. The molar mass of NaOH is 40.00 g/mol, and the student reports the average of all three trials as the acid concentration. The instructor requires the result to be reported to three significant figures with an uncertainty judgement.
1. Calculate the volume of NaOH used in each trial and the mean volume. Should all three trials be averaged? Justify your decision.
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Titration volumes:
- Trial 1: 24.90 − 0.40 = 24.50 mL
- Trial 2: 25.80 − 1.20 = 24.60 mL
- Trial 3: 26.10 − 0.10 = 26.00 mL
The first two trials differ by only 0.10 mL, which is consistent with normal reading variation. Trial 3 is 1.40 mL higher than Trial 2 and 1.50 mL higher than Trial 1. That is far outside the spread of the first two and is consistent with passing the endpoint. The fading pink after 20 seconds supports that interpretation: a stable endpoint should persist for at least 30 seconds, and fading suggests excess base has been neutralized by atmospheric CO2 or that the endpoint was overshot.
Do not average all three. A defensible accepted mean is (24.50 + 24.60)/2 = 24.55 mL. Averaging all three gives 25.03 mL, which happens to move the calculated acid concentration toward the nominal 0.100 M; that coincidence is not evidence that Trial 3 is valid.
2. Using your accepted mean, calculate the HCl concentration. Then assess whether the nominal 0.100 M NaOH is trustworthy given the preparation details.
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Using 24.55 mL of 0.100 M NaOH:
moles NaOH = 0.100 mol/L × 0.02455 L = 2.455 × 10⁻³ mol.
The reaction is HCl + NaOH → NaCl + H2O, so moles HCl = 2.455 × 10⁻³ mol.
[HCl] = 2.455 × 10⁻³ mol / 0.02500 L = 0.0982 M.
The nominal NaOH concentration comes from 2.00 g / 40.00 g/mol = 0.0500 mol in 0.5000 L, which is 0.100 M. But that is only a nominal value. NaOH pellets absorb water and CO2 from the air, so the mass of 'NaOH' includes water and carbonate; the true concentration is likely lower than 0.100 M. Without standardization against potassium hydrogen phthalate or another primary standard, the calculated HCl concentration is an apparent value, not an accurate one. The correct judgement is to report 0.0982 M with a caveat that the base was not standardized, not to present it as a definitive concentration.
3. The student says Trial 3 should be kept because the pink persisted for 20 seconds. Explain the error and state what should be done.
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A 20-second persistence is too short to count as a stable endpoint. The standard visual criterion is a faint pink that lasts for at least 30 seconds; if it fades, the solution is still taking up base or CO2, and the recorded volume is not the equivalence-point volume. Keeping Trial 3 because it 'turned pink' mistakes a transient colour for an endpoint.
The practical correction is to discard Trial 3 and run a fresh trial, adding NaOH dropwise near the endpoint and swirling after each drop. The student should also standardize the NaOH against a primary standard, because a fading endpoint and an unstandardized base are separate problems: one is a technique error, the other is a calibration error. Reporting the mean of all three trials hides both problems.