Choosing an Indicator and Auditing a Titration Stopped at pH 7
A student titrates 25.0 mL of a 0.100 M solution of a weak monoprotic acid HA (Ka = 1.8 × 10⁻⁵) with standardised 0.100 M NaOH from a burette, using a calibrated pH meter alongside the visual indicator. The stockroom supplies phenolphthalein (colour change 8.2–10.0) and methyl orange (3.1–4.4). The student's reasoning is that 'the endpoint of an acid–base titration is the neutral point, pH 7', so the run is stopped the moment the meter reads 7.00, and the acid concentration is then computed from that burette reading. A demonstrator disputes this, arguing the equivalence point of this particular system is nowhere near 7 and that the student's number is biased in a predictable direction. Before the result is entered in the practical record, the demonstrator wants the equivalence pH, the correct indicator choice with justification, and a quantified estimate of the error the student has introduced.
1. Compute the pH at half-neutralisation and at the equivalence point for this titration, showing the approximations used.
Read the worked answer
Equivalence requires n(NaOH) = n(HA): 25.0 mL × 0.100 M = 2.50 mmol, so 25.0 mL of NaOH, total volume 50.0 mL.
Half-neutralisation (12.5 mL added): [HA] = [A⁻], so the Henderson–Hasselbalch equation gives pH = pKa = −log(1.8 × 10⁻⁵) = 4.74.
Equivalence: all HA has become A⁻, with [A⁻] = 2.50 mmol/50.0 mL = 0.0500 M. A⁻ is a weak base: Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. [OH⁻] = √(Kb·C) = √(5.6 × 10⁻¹⁰ × 0.0500) = √(2.8 × 10⁻¹¹) = 5.3 × 10⁻⁶ M pOH = 5.28 ⇒ pH = 8.72.
The approximation [OH⁻] ≪ C is valid here (5 × 10⁻⁶ against 0.05, about 0.01%), so no quadratic is needed. The key structural point is that the equivalence pH is set by the conjugate base, not by neutrality; a weak acid always lands on the basic side, and the weaker the acid the further above 7 it sits.
2. Which indicator should be used, and how should the student's 'pH 7 is the endpoint' rule be corrected?
Read the worked answer
Phenolphthalein. Its transition range (8.2–10.0) brackets the equivalence pH of 8.72, so the colour change and the stoichiometric point nearly coincide; the indicator error is a fraction of a drop. Methyl orange changes at 3.1–4.4, which on this curve corresponds to roughly half-neutralisation - using it would stop the titration far short and under-report the acid concentration drastically.
The rule the student applied confuses three different things. The equivalence point is defined by stoichiometry (moles base = moles acid) and its pH depends on the species present. The endpoint is the observable signal, chosen to fall as close to the equivalence pH as possible. Neutrality (pH 7 at 25 °C) is a property of pure water and is irrelevant unless both acid and base are strong, where the equivalence pH happens to be 7. Correct rule: compute or estimate the equivalence pH from the nature of the salt formed, then select an indicator whose range straddles it; confirm with a pH meter or a titration curve when accuracy matters.
3. The run was stopped at pH 7.00. Estimate the volume of NaOH delivered and the resulting error in the reported acid concentration.
Read the worked answer
At pH 7.00 the solution is still a buffer, so use Henderson–Hasselbalch:
7.00 = 4.74 + log([A⁻]/[HA]) ⇒ log([A⁻]/[HA]) = 2.26 ⇒ [A⁻]/[HA] = 10^2.26 ≈ 182.
Fraction neutralised = 182/183 = 0.9945, i.e. 99.45% of the 2.50 mmol, so volume delivered = 0.9945 × 25.0 = 24.86 mL instead of 25.00 mL.
Reported concentration: c = (0.100 × 24.86)/25.0 = 0.0994 M against the true 0.1000 M, an error of −0.55%.
Judgement: 0.55% happens to sit inside this student's likely random error, so the number is not disastrous - but the error is systematic and always negative, because the indicator always fires before equivalence for a weak acid, so averaging repeats will not remove it. The size scales with how far the equivalence pH sits above 7: for a much weaker acid the fraction unneutralised at pH 7 grows sharply and the same mistake becomes a several-per-cent bias. That is why 'stop at 7' is unacceptable practice even when it looks harmless on one trial.