Standardising NaOH and Judging a Vinegar Label
A student determines the ethanoic acid content of a bottle of rice vinegar (label: 4.5 g ethanoic acid per 100 mL; M = 60.05 g mol-1). She first standardises a sodium hydroxide solution against 0.1000 mol/L hydrochloric acid: 25.00 mL portions of NaOH are titrated to the phenolphthalein endpoint, giving 24.10 mL, 24.05 mL, 25.55 mL and 24.15 mL of HCl. She then dilutes 25.00 mL of the vinegar to 250.0 mL and titrates 25.00 mL aliquots of the diluted solution with her NaOH, obtaining 18.45 mL, 18.50 mL and 18.40 mL. All glassware is class B: burette readings to ±0.05 mL, bulb pipettes to ±0.03 mL. During the vinegar titrations she noticed an air bubble in the burette tip at the start, which had disappeared by the last reading.
1. Which hydrochloric acid volumes should be used to calculate the NaOH concentration, and why? Calculate c(NaOH).
Read the worked answer
Use 24.10, 24.05 and 24.15 mL. Those three are concordant: their spread is 0.10 mL, which is exactly the resolution of two burette readings of ±0.05 mL, so they plausibly estimate the same endpoint. The 25.55 mL reading sits 1.4 mL above the group - about fourteen times that spread. A difference that large is not repeatable random scatter; it is the signature of running past the endpoint (or mis-reading the meniscus), so it estimates something different and must be rejected, not averaged in.
Mean of the three = (24.10 + 24.05 + 24.15) / 3 = 24.10 mL.
NaOH + HCl react 1:1, so n(NaOH) = n(HCl) in each flask: c(NaOH) = (0.1000 mol/L × 24.10 mL) / 25.00 mL = 0.09640 mol/L.
Including the rejected value would give a mean of 24.46 mL and c(NaOH) = 0.0978 mol/L - a 1.5 % high bias that would then propagate into every vinegar result. Selecting concordant data is a decision about which numbers estimate the same quantity, not a way of getting the answer you prefer.
2. Carry the result through to grams of ethanoic acid per 100 mL of vinegar and judge the label claim. Is the difference bigger than her random uncertainty?
Read the worked answer
Vinegar titrations: mean of 18.45, 18.50 and 18.40 mL = 18.45 mL.
n(NaOH) = 0.09640 × 0.01845 = 1.779 × 10^-3 mol. Acid and base react 1:1, so this is also the acid in the 25.00 mL aliquot.
The aliquot is one tenth of the 250.0 mL made from 25.00 mL of vinegar, so n(acid) in 25.00 mL of vinegar = 1.779 × 10^-2 mol. c(vinegar) = 1.779 × 10^-2 / 0.02500 = 0.7114 mol/L. Mass concentration = 0.7114 × 60.05 = 42.7 g/L = 4.27 g/100 mL.
Uncertainty: burette readings contribute about 0.3 % on an 18.5 mL titre and about 0.2 % on the 24.1 mL standardisation; two pipette deliveries add about 0.2 %. Combined in quadrature this is under 0.5 %, and even a generous treatment stays below 1 %.
Judgement: 4.27 against a stated 4.5 is a 5 % shortfall, roughly ten times her random uncertainty. It cannot be explained by reading error, so a systematic cause is present - a dirty pipette, an imperfectly standardised base, or a nominal label figure. She should report 4.3 g/100 mL with that caveat rather than simply writing the label value.
3. She later realises the air bubble was present at the start of the vinegar titrations and gone by the end. State the direction of the resulting error in the acid content, and whether it rescues the label claim.
Read the worked answer
The recorded titre is (final reading − initial reading). While the bubble occupies the tip, that space holds air instead of liquid. As she titrates, the bubble is flushed out and replaced by solution drawn from the graduated column, so the level falls by the delivered liquid plus the bubble volume. The recorded volume change is therefore larger than the volume actually delivered into the flask.
Since n(NaOH) is calculated directly from the recorded volume, n is overestimated, and so is the ethanoic acid content. The true value would be slightly below 4.27 g/100 mL.
So the bubble pushes in the wrong direction to rescue the label: it makes the measured figure too high, yet the measurement already falls 5 % short of 4.5 g/100 mL. The discrepancy is therefore real, or caused by a separate systematic effect. The correct procedure is to fill the tip, expel all bubbles, and only then take the initial reading - a procedural control, not a correction applied afterwards.