Indicator Choice and a 3.6% Low Titre
A student titrates 25.0 mL of a monobasic weak acid HA against sodium hydroxide. The acid was prepared as 0.100 M from a stock bottle; the NaOH solution is labelled 0.100 M but was made from pellets that had stood open to air. Ka of HA = 1.8 × 10⁻⁵. Using methyl orange (colour change over pH 3.1–4.4) the student sees a persistent colour at 24.1 mL and records this as the endpoint. Two repeat runs give 24.1 mL and 24.2 mL.
A demonstrator asks for a re-check, points to a shelf holding phenolphthalein (8.2–10.0), bromothymol blue (6.0–7.6) and a recently calibrated pH meter, and asks the student to justify the indicator from the shape of the titration curve rather than from habit. The student must also explain the 3.6% gap between the observed and the expected titre before repeating anything, and state what extra measurement would settle the question.
1. Compute the pH at half-equivalence and at the equivalence point, then quantify the indicator error for methyl orange versus phenolphthalein. Which indicator should be used?
Read the worked answer
Half-equivalence (12.5 mL). [HA] = [A⁻], so pH = pKa = −log(1.8 × 10⁻⁵) = 4.74. This is the point to read Ka from, not to stop at.
Equivalence (25.0 mL). Total volume 50.0 mL, n(A⁻) = 2.50 mmol, so [A⁻] = 0.0500 M. The conjugate base hydrolyses with Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰.
[OH⁻] = √(Kb·C) = √(5.56 × 10⁻¹⁰ × 0.0500) = 5.27 × 10⁻⁶ M → pOH = 5.28 → pH = 8.72.
Phenolphthalein error. At the acid end of its range, pH 8.2, [H⁺] = 6.31 × 10⁻⁹, so [A⁻]/[HA] = Ka/[H⁺] = 2853, leaving a fraction 1/2854 = 3.5 × 10⁻⁴ unneutralised - an error of 0.00035 × 25.0 ≈ 0.009 mL, which is invisible.
Methyl orange error. At pH 4.4, [A⁻]/[HA] = 1.8 × 10⁻⁵/3.98 × 10⁻⁵ = 0.452, so the neutralised fraction is 0.452/1.452 = 0.311, i.e. the colour changes at 0.311 × 25.0 ≈ 7.8 mL - less than a third of the true equivalence volume.
Verdict. The equivalence pH of 8.72 lies on the steep part of the curve and is bracketed by phenolphthalein's 8.2–10.0 window; methyl orange finishes on the buffer plateau, where pH barely moves with added base, so the endpoint is both early and insensitive. Phenolphthalein is the defensible choice.
2. The student's replicate titres are 24.1 and 24.2 mL against an expected 25.0 mL. Diagnose the discrepancy and state what should be done next.
Read the worked answer
First classify the error. The two replicates agree to 0.4%, so the scatter is small while the offset is large and one-signed. That signature means a systematic error, not technique noise; averaging more runs will simply reproduce 24.15 mL with false confidence.
Then convert it to a concentration. 25.0 mL of 0.100 M HA requires 2.50 mmol of OH⁻. The observed titre implies c(NaOH) = 2.50/24.1 = 0.1037 M, or equivalently the acid is only 0.0964 M. Either the base is about 3.7% stronger than its label or the acid about 3.6% weaker; the titration alone cannot distinguish these.
Candidate causes, and their directions. Sodium hydroxide pellets exposed to air absorb CO₂ and water. Water uptake dilutes the base, which would make the titre larger than 25.0 mL; carbonate formation also pushes the phenolphthalein endpoint to a larger volume. Both point the wrong way, so the loose-lidded NaOH bottle is probably not the culprit here - which is exactly the reasoning that stops the student blaming the obvious suspect.
What to do. Standardise the NaOH against dried potassium hydrogen phthalate (a primary standard, weighed accurately, phenolphthalein endpoint) to fix c(NaOH) independently; then re-titrate the acid. Check the 25.0 mL pipette for drainage time and cleanliness, and confirm the acid is genuinely monobasic - a hidden second proton would shift results the other way. Only after the base is standardised does the 24.1 mL number become interpretable.
3. Would the calibrated pH meter (±0.05 pH) improve matters? Quantify what it can and cannot deliver here.
Read the worked answer
At the endpoint the gain is real but small. The indicator error was already about 0.009 mL; the meter removes even that and, more usefully, removes any personal colour-judgement bias. The limit is that near equivalence the curve is almost vertical, so ±0.05 pH corresponds to a tiny increment of titrant - negligible against a 25 mL burette read to 0.05 mL. So the meter sharpens an endpoint that was already sharp.
For Ka the meter is the wrong tool at this precision. At half-equivalence, pH = pKa = 4.74 with ±0.05, so Ka = 10^(−4.74 ± 0.05) is uncertain by a factor 10^0.05 = 1.12, i.e. about ±11%. Quoting Ka to three significant figures from a ±0.05 meter would be unjustified; the value supports only an order-of-magnitude or two-figure statement.
Judgement. The 3.6% discrepancy is a standardisation problem, not a detection problem. Buying a better detector cannot fix a wrong concentration. The correct sequence is: standardise the base, then choose the indicator from the computed equivalence pH (8.72 → phenolphthalein), and use the meter only as a cross-check on the endpoint and to sketch the curve. A student who reaches for the meter first has misdiagnosed which link in the chain is weak.