Weak-acid titration: indicator choice and a wet burette
Two students each titrate 25.0 cm³ portions of the same weak monoprotic acid HA with 0.100 mol/L NaOH. Student 1 obtains three concordant titres and reports a mean of 22.40 cm³, using phenolphthalein. Student 2 rinses the burette with distilled water immediately before filling it with the NaOH solution and reports a mean titre of 22.95 cm³. For HA, Ka = 1.8 × 10⁻⁵ and Kw = 1.0 × 10⁻¹⁴. Available indicators: methyl orange (pH 3.1–4.4), bromothymol blue (pH 6.0–7.6) and phenolphthalein (pH 8.2–10.0). A third student suggests that methyl orange would be "safer, because it changes colour earlier and so less base is wasted", and proposes stopping the titration at the first hint of colour change.
1. Calculate the concentration of HA from Student 1's data and the pH at the equivalence point, then justify the indicator used.
Read the worked answer
Concentration. At the end point the moles of base added equal the moles of acid in the portion (1:1 stoichiometry, monoprotic acid):
n(NaOH) = 0.100 × 22.40 × 10⁻³ = 2.240 × 10⁻³ mol
c(HA) = 2.240 × 10⁻³ / (25.0 × 10⁻³) = 0.0896 mol/L
Equivalence-point pH. At equivalence all HA has become A⁻, in a total volume of 25.0 + 22.40 = 47.40 cm³:
c(A⁻) = 2.240 × 10⁻³ / 47.40 × 10⁻³ = 0.0473 mol/L
A⁻ is the conjugate base of a weak acid, so it hydrolyses. Kb = Kw/Ka = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰.
[OH⁻] = √(Kb × c) = √(5.6 × 10⁻¹⁰ × 0.0473) = √(2.6 × 10⁻¹¹) ≈ 5.1 × 10⁻⁶ mol/L
pOH ≈ 5.29, so pH ≈ 8.7.
Indicator. The indicator must change colour over a pH range containing 8.7. Phenolphthalein (8.2–10.0) does; methyl orange (3.1–4.4) and bromothymol blue (6.0–7.6) do not. The equivalence point of a weak acid–strong base titration is basic, not neutral, precisely because the conjugate base is present.
2. Quantify the effect of Student 2's burette handling and state the direction of the resulting error.
Read the worked answer
Rinsing with distilled water leaves the burette wet inside, so the NaOH solution is diluted as it is added. The delivered solution is therefore at a concentration below 0.100 mol/L, and more volume is needed to deliver the same number of moles of OH⁻ - the titre is too large.
Using Student 2's titre as if the concentration were exactly 0.100 mol/L:
c(HA) = 0.100 × 22.95 / 25.0 = 0.0918 mol/L
error = (22.95 − 22.40)/22.40 × 100 % ≈ +2.5 % high
The error is systematic and one-directional: every repeat by Student 2 is biased the same way, so averaging more runs will not help, and the runs will look beautifully concordant while all being wrong. Note also that the dilution is greatest at the start (the burette is wettest then) and decreases as the burette fills with solution, which slightly distorts the shape of the whole run rather than applying a single clean factor.
The correct procedure is to rinse the burette with a little of the NaOH solution (discarded) before filling, so any residual liquid is at the same concentration as the titrant.
3. Evaluate the third student's proposal to use methyl orange and stop at the first colour change.
Read the worked answer
The proposal confuses "less titrant used" with "correct measurement", and it must be rejected on two counts.
First, an indicator is chosen so that its colour change brackets the equivalence pH (here 8.7), not so that the titration ends sooner. Methyl orange changes at pH 3.1–4.4, where the solution is still strongly acidic; at that point only a fraction of the HA has been neutralised, so the volume recorded is not the equivalence volume. Substituting it into c(HA) = c(NaOH) × V/25.0 cm³ would underestimate the acid concentration by a large margin, by an amount that depends on Ka and on the student's colour judgement.
Second, "stopping at the first hint of colour change" abandons the concordant-titre method that makes titrations precise. The accepted end point is a faint but persistent colour lasting about 30 s, obtained by adding dropwise near the end point; a premature stop from a whole-drop addition is a random error of roughly 0.05 cm³, which is comparable to the drop size itself.
The valid comparison is: choose the indicator from the equivalence pH, and improve precision through concordant repeats, not by shortening the titration.