Two-Endpoint Titration of a Carbonate–Bicarbonate Mixture
A first-year chemistry laboratory receives a 1.000 g sample labelled “sodium carbonate/sodium bicarbonate mixture”. The sample is dissolved and made up to 250.0 mL. A 25.00 mL aliquot is titrated with standardized 0.1000 M HCl. With phenolphthalein, the colour just disappears after 4.80 mL of acid. The analyst then adds methyl orange and continues the same titration; a further 10.65 mL is required to reach the methyl-orange endpoint. The balance has a readability of ±0.001 g, and each burette reading is estimated to ±0.05 mL. The laboratory needs the mass percentage of each component and a judgement on whether the result is fit to report. The analyst also wants to know what would happen to the two calculated masses if the phenolphthalein endpoint had been overrun by 0.10 mL.
1. Assign the two endpoints and calculate the mass and mass percentage of Na₂CO₃ and NaHCO₃ in the sample.
Read the worked answer
Phenolphthalein changes colour when CO₃²⁻ has been converted to HCO₃⁻. Methyl orange changes when all HCO₃⁻ has been converted to CO₂.
For the aliquot: n(Na₂CO₃) = 0.1000 × 0.00480 = 4.80 × 10⁻⁴ mol. In the 250.0 mL flask (10 aliquots): 4.80 × 10⁻³ mol. Mass = 4.80 × 10⁻³ × 105.99 = 0.5088 g.
The acid added after the phenolphthalein endpoint is 10.65 mL. Of this, 4.80 mL converts the HCO₃⁻ formed from carbonate, so the original HCO₃⁻ uses 10.65 − 4.80 = 5.85 mL. n(NaHCO₃) = 0.1000 × 0.00585 = 5.85 × 10⁻⁴ mol in the aliquot, ×10 = 5.85 × 10⁻³ mol. Mass = 5.85 × 10⁻³ × 84.01 = 0.4914 g.
Total = 0.5088 + 0.4914 = 1.0002 g, consistent with the 1.000 g sample. Mass %: Na₂CO₃ = 50.9%, NaHCO₃ = 49.1%.
2. What happens to the two calculated masses if the phenolphthalein endpoint is overrun by 0.10 mL?
Read the worked answer
If V₁ becomes 4.90 mL: Carbonate mass = 0.1000 × 0.00490 × 10 × 105.99 = 0.5194 g, an increase of 0.0106 g (+2.1%).
The original bicarbonate volume becomes 10.65 − 4.90 = 5.75 mL: Mass = 0.1000 × 0.00575 × 10 × 84.01 = 0.4831 g, a decrease of 0.0083 g (−1.7%).
The total becomes 1.0025 g, still close to 1.000 g. The endpoint error shifts acid volume from one component to the other, so the total is partly self-correcting while the individual percentages are more affected.
3. Is the result fit to report? Comment on the uncertainty and on how the two component masses are correlated.
Read the worked answer
A burette uncertainty of ±0.05 mL corresponds to about 5 × 10⁻⁶ mol per aliquot, or about 0.005 g in the 250 mL total for each component-roughly 0.5–1% relative. The balance contributes ±0.001 g, which is negligible here.
The sample is fit to report as approximately 50.9% Na₂CO₃ and 49.1% NaHCO₃. However, the two masses are correlated: a positive volume error in the first endpoint raises the carbonate result and lowers the original bicarbonate result. The analyst should report the individual percentages with their uncertainty and note that the total mass is more robust than either component alone.