Two titrations, two indicators: which alkali molarity is acceptable?
Two candidates in the same HSC Chemistry batch titrate 25.0 mL of 0.050 M oxalic acid against the sodium hydroxide solution from a bottle labelled “NaOH ≈ 0.25 M”. Candidate 1 uses phenolphthalein and records burette readings 0.0 → 9.9, 9.9 → 19.8, 19.8 → 29.6 and 29.6 → 39.9 mL. She takes the mean of all four titres as 10.0 mL, calculates the molarity of the alkali as 0.250 M and writes that the label is exactly correct. Candidate 2 uses methyl orange, obtains a sharp colour change after 4.2 mL and reports the alkali as 0.60 M, arguing that his value is more reliable because his end point was sharper. Both candidates used the same burette, the same pipette and the same stock solutions. Before awarding practical marks the examiner asks you to decide which report is acceptable, and to state what each candidate should have done differently.
1. Identify the concordant titres, compute the correct molarity, and judge the label claim.
Read the worked answer
Titres: 9.9, 9.9, 9.8 and 10.3 mL. The first three lie within 0.1 mL of one another, so they are the concordant set; 10.3 mL differs from their mean (9.87 mL) by 0.4 mL and is an outlier, to be rejected and replaced by a fresh reading.
Mean titre = (9.9 + 9.9 + 9.8)/3 = 9.87 mL.
Moles of oxalic acid = 0.0250 × 0.050 = 1.25 × 10⁻³ mol. Oxalic acid is dibasic, so moles NaOH = 2.50 × 10⁻³.
Molarity = 2.50 × 10⁻³/0.00987 = 0.253 M.
The difference from the label is about 1.3%, within the combined burette, pipette and indicator uncertainties, so the label is consistent with the experiment. “Exactly correct” is not a claim one titration can support; the correct statement is “agrees with the label within experimental error”. Averaging all four readings (10.0 mL, 0.250 M) happens to look tidier but conceals an unacceptable reading.
2. Explain why methyl orange is unsuitable here and deduce the direction of Candidate 2's error.
Read the worked answer
Oxalic acid is a weak acid and NaOH a strong base, so the equivalence point lies in the basic region, around pH 8–9. Phenolphthalein changes over about pH 8.3–10.0 and therefore brackets the equivalence point. Methyl orange changes over about pH 3.1–4.4, so it signals the end of the titration well before equivalence - hence the small titre of 4.2 mL.
Molarity is inversely proportional to titre volume, so an end point that comes too early gives too small a volume and therefore too large a calculated concentration. That is exactly what happened: 0.60 M against a true value near 0.25 M.
A sharp colour change is not the same as an accurate end point; sharpness only reflects a steep pH change near that indicator's range. Candidate 2 must repeat with phenolphthalein and obtain at least three concordant readings.
3. Account for Candidate 1's 10.3 mL reading and state what she should do about it.
Read the worked answer
The reading is one-sided and 0.4 mL above the other three, which is far beyond reading error on a burette graduated to 0.1 mL. Likely causes are: the last drops were run in too fast, so the end point was overshot; the flask was not swirled enough near the end point; a bubble was dislodged from the jet; the stopcock leaked; or the meniscus was read at an angle.
She should reject it, record it with the reason, and take a fourth reading until she has three concordant titres. Averaging a non-concordant reading shifts the result and hides a technique fault - here the shift is small, but the same habit would be damaging with a larger outlier.
Good practice: read the lower meniscus at eye level, add dropwise (and then half-dropwise) near the end point, rinse the flask walls with distilled water, and keep the jet filled.