Reading a Titration: An Outlier, a Wet Burette and the Wrong End-point Colour
Three students standardise the same sodium hydroxide solution. Each pipettes 25.0 cm³ into a conical flask and titrates with hydrochloric acid labelled 0.100 mol/dm³.
Their burette readings give titres of 24.30 cm³, 23.80 cm³ and 23.75 cm³.
Two procedural details are noted:
- Student A rinsed the burette with distilled water and then filled it with the acid.
- Student B rinsed the 25.0 cm³ pipette with distilled water only before use.
A third student, C, uses phenolphthalein and stops at the first drop that leaves a permanent pink colour in the flask.
The demonstrator says all three will report a concentration that differs from the true value, but not by the same amount, and that B's error is the most serious. She also says the reported answer should be quoted to three significant figures. Assume the acid concentration on the label is correct and the reaction is 1:1.
1. Decide which titres may be averaged, calculate the concentration of the sodium hydroxide, and justify the significant figures you quote.
Read the worked answer
Selecting titres. Concordant titres agree within about 0.10 cm³. Here 23.80 and 23.75 differ by 0.05 cm³ and are accepted. 24.30 differs from the nearer value by 0.50 cm³ - far outside the tolerance - so it is rejected as an outlier and takes no part in the average.
Mean titre = (23.80 + 23.75)/2 = 23.775 cm³ → 23.78 cm³.
Moles of HCl = c × V = 0.100 × 23.78 × 10⁻³ = 2.378 × 10⁻³ mol.
Ratio 1:1, so moles NaOH in the flask = 2.378 × 10⁻³ mol.
[NaOH] = n/V = 2.378 × 10⁻³ / 25.0 × 10⁻³ = 0.09512 mol/dm³ → 0.0951 mol/dm³.
Significant figures. Three significant figures is right: the 0.100 mol/dm³ acid has three, the 25.0 cm³ pipette volume has three, and the burette is read to 0.05 cm³ (three to four figures). Quoting 0.09512 would claim a precision the apparatus cannot deliver. Including the rejected 24.30 would have pulled the average to 23.95 cm³ and given 0.0958 mol/dm³ - a difference in the third figure, which is exactly the figure the exam expects you to control.
2. For Student A, state the direction of the error and explain its cause. Then do the same for Student B, and explain why the demonstrator rates B's error as the more serious of the two.
Read the worked answer
Student A - residual water in the burette. The water left in the burette dilutes the acid as it is poured in, so each cubic centimetre delivers fewer moles than the label implies. More acid must be run in to reach the end-point, so the recorded titre is too large. The student then computes moles = 0.100 × V_titre, which is also too large, and divides by the correct 25.0 cm³. Reported [NaOH] comes out too high. The size of the error is the ratio of residual water to burette volume - a few drops, so of the order of 0.1–1%: real, but small, and it shrinks as the titration proceeds because the burette gradually fills with undiluted acid.
Student B - wet pipette. Water in the pipette dilutes the alkali before it is transferred, so the 25.0 cm³ actually delivered contains fewer moles of NaOH than a clean pipette would give. Less acid is needed, the titre is too small, and the reported concentration is too low. The size of the error here is the residual volume against 25.0 cm³ - around 0.5 cm³ of water is 2%, an order of magnitude worse than A's, and it repeats in every run.
Why B is worse. A's error is confined to the acid inside the burette and is diluted out; B's error changes the amount of the very substance being measured, on every trial, and no concordance check will reveal it because all of B's titres are consistently biased. Precision looks perfect while accuracy is lost - the more dangerous failure.
3. Explain what Student C did wrong, what volume error it produces, and whether methyl orange or phenolphthalein is the better indicator for this titration.
Read the worked answer
C's error. With alkali in the flask and acid in the burette, phenolphthalein starts pink (the solution is alkaline) and goes colourless at the end-point. C waited for a permanent pink to appear, which is the colour change seen when acid is in the flask and alkali in the burette - the titration read from the wrong direction. To reach a persistent pink starting from an alkaline flask, C must have overshot the end-point by at least a drop or more, adding excess acid. The recorded volume is too large, so the calculated [NaOH] is too high.
Which indicator? This is a strong acid–strong base titration, and the pH at the equivalence point is 7, with a very steep section of the curve roughly from pH 3 to pH 11. Both indicators sit inside that steep section:
- Phenolphthalein, pH 8.3–10.0 (colourless → pink), end-point on the alkaline side.
- Methyl orange, pH 3.1–4.4 (red → yellow), end-point on the acid side.
Either is acceptable because one drop changes pH across the steep region, so neither introduces a significant volume error. The choice matters only when one of the two species is weak: for a weak base titrated with a strong acid, the equivalence pH is acidic and methyl orange is the correct choice, while phenolphthalein would change colour far too late. For this strong/strong system, C's mistake was not the indicator but the colour it was read at.