The Air Bubble and the Discarded Titre
A titration is carried out to find the concentration of a hydrochloric acid solution. Anhydrous sodium carbonate (1.325 g, M = 106 g mol⁻¹) is dissolved in distilled water and made up to 250 cm³ in a volumetric flask. 25.0 cm³ portions are titrated with the acid using methyl orange. The average of three concordant titres is 24.20 cm³. Later the student notices that the burette had an air bubble at the tip at the start, and that the bubble disappeared during the first titration. The first titre was discarded and not used in the average. The bubble volume is estimated at 0.30 cm³. The student's report gives the acid concentration as 0.103 mol dm⁻³. A second student, working nearby, rinses the pipette with the sodium carbonate solution instead of distilled water before taking a 25.0 cm³ portion.
1. Calculate the concentration of the hydrochloric acid from the student's average titre, showing the equation and each step.
Read the worked answer
Sodium carbonate in the flask:
Total n(Na₂CO₃) = 1.325 g ÷ 106 g mol⁻¹ = 0.0125 mol in 250 cm³.
In 25.0 cm³: n = 0.0125 × (25.0 ÷ 250) = 0.00125 mol.
Equation:
Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O
n(HCl) = 2 × 0.00125 = 0.00250 mol.
Concentration from the average titre:
V(HCl) = 24.20 cm³ = 0.02420 dm³
c(HCl) = 0.00250 mol ÷ 0.02420 dm³ = 0.1033 mol dm⁻³ ≈ 0.103 mol dm⁻³ (3 s.f.)
The reported value agrees with the calculation.
2. The student is worried that the air bubble invalidated the result. Decide whether the reported concentration is affected, explaining your reasoning.
Read the worked answer
The bubble affected only the first titration, which was discarded. If the burette was refilled and no new bubble formed, the three concordant titres used in the average were not affected by the bubble. The reported value of 0.103 mol dm⁻³ is therefore valid at 3 s.f.
For comparison, if the bubble had remained throughout every titration, the recorded volume would be larger than the volume actually delivered by about 0.30 cm³. The true volume would be 24.20 − 0.30 = 23.90 cm³, giving c = 0.00250 ÷ 0.02390 = 0.105 mol dm⁻³. In that case the reported concentration would be slightly too low. But because the bubble disappeared in the discarded first titration, this correction should not be applied.
3. Predict and explain the effect on the titre and on the calculated acid concentration if the second student rinses the pipette with sodium carbonate solution instead of distilled water.
Read the worked answer
Rinsing the pipette with sodium carbonate solution leaves some extra carbonate solution in the pipette after it has been filled to the mark. This extra carbonate is delivered into the conical flask along with the measured 25.0 cm³, so the flask contains slightly more Na₂CO₃ than the student assumes from the mass of solid.
More carbonate requires more hydrochloric acid, so the titre increases. The student still calculates n(HCl) from the mass-based amount of Na₂CO₃, which has not changed. Using c = n ÷ V with a larger measured V gives a lower calculated acid concentration than the true value. Rinsing with distilled water avoids this systematic error.