Back-titration of a Mixed Carbonate Antacid Tablet
An analyst at a contract laboratory receives a batch of antacid tablets labelled 'calcium carbonate 250 mg and magnesium carbonate 250 mg per tablet'. One tablet weighs 0.6500 g. The analyst grinds it, treats it with 50.00 cm³ of 0.5000 mol dm⁻³ hydrochloric acid, warms gently, filters off 0.1500 g of insoluble binder, and dilutes the filtrate and washings to 250.0 cm³. A 25.00 cm³ aliquot is titrated with 0.1000 mol dm⁻³ sodium hydroxide to a methyl-orange endpoint, requiring 14.00 cm³. A blank titration of 25.00 cm³ of the diluted acid requires 25.00 cm³ of the same sodium hydroxide. The analyst reports the result directly, without boiling the solution before titration. The specification is ±5% of each labelled mass. The laboratory's quality officer asks for an independent assessment of the calculation, the method bias, and whether the batch passes.
1. Calculate the amount of carbonate in one tablet, the mean molar mass of the carbonate mixture, and the mass of calcium carbonate and magnesium carbonate. Hence decide whether the batch meets the ±5% specification. State every assumption.
Read the worked answer
Total acid added
n(HCl) = 0.05000 dm³ × 0.5000 mol dm⁻³ = 0.02500 mol.
The 250.0 cm³ flask contains this acid, and each 25.00 cm³ aliquot is one-tenth of the flask. The blank confirms this: 0.002500 mol HCl in the aliquot needs 25.00 cm³ of 0.1000 mol dm⁻³ NaOH.
Excess acid in the sample aliquot
n(NaOH) = 0.01400 dm³ × 0.1000 mol dm⁻³ = 0.001400 mol.
This equals the excess HCl in the aliquot. Scaling to the whole flask gives 0.01400 mol excess HCl.
Acid consumed by carbonate
n(HCl consumed) = 0.02500 − 0.01400 = 0.01100 mol.
Each carbonate ion consumes two protons:
CO₃²⁻ + 2H⁺ → CO₂ + H₂O
so n(carbonate) = 0.01100/2 = 0.005500 mol.
Mass and composition
The carbonate mass is the tablet mass minus the binder:
0.6500 − 0.1500 = 0.5000 g.
Mean molar mass = 0.5000 g / 0.005500 mol = 90.91 g mol⁻¹.
Let x = mol CaCO₃ and y = mol MgCO₃.
x + y = 0.005500
100.09x + 84.31y = 0.5000
Substituting y = 0.005500 − x gives:
15.78x + 0.4637 = 0.5000
x = 0.002300 mol; y = 0.003200 mol.
Mass CaCO₃ = 0.002300 × 100.09 = 0.2302 g = 230 mg.
Mass MgCO₃ = 0.003200 × 84.31 = 0.2698 g = 270 mg.
The label claims 250 mg of each. The deviations are about −8% for CaCO₃ and +8% for MgCO₃. Both exceed a ±5% tolerance, so the batch fails the specification even though the total carbonate mass is correct. Assumptions: the binder is acid-insoluble and contains no carbonate; all carbonate is present as CaCO₃ and MgCO₃; the endpoint corresponds to complete neutralisation of excess strong acid; and no carbonate is lost during filtration.
2. The analyst omitted the boiling step. Predict the direction of the error in the reported calcium carbonate content and explain the chemistry behind it. Would a blank titration correct this error?
Read the worked answer
The reported carbonate content would be too low, not too high.
When the tablet is dissolved in excess HCl, the carbonate is converted to dissolved CO₂:
CO₃²⁻ + 2H⁺ → CO₂(aq) + H₂O.
Some CO₂ remains dissolved in the filtrate. During the back-titration with NaOH, the hydroxide first neutralises the excess HCl, but near the methyl-orange endpoint the solution is still acidic enough for some CO₂ to react:
CO₂ + OH⁻ → HCO₃⁻.
That reaction consumes NaOH that should have been attributed to excess HCl. A larger NaOH titre means a larger calculated excess of HCl, so the acid consumed by carbonate is underestimated and the carbonate result is low.
Boiling before titration removes dissolved CO₂ and prevents this bias. The solution should be cooled before titration so that the endpoint and the indicator colour change are not affected.
A blank does not correct this error. The blank contains diluted acid but no dissolved CO₂ from the sample, so it cannot reproduce the sample-specific CO₂ consumption. A blank corrects reagent impurities and acid concentration, not an error caused by retained sample CO₂. The correct fix is to boil or sparge the sample solution and then titrate.
3. Critically evaluate the choice of methyl orange and the practice of titrating a 25.00 cm³ aliquot. Suggest two improvements that would make the result fit for a regulatory report.
Read the worked answer
Methyl orange is broadly appropriate for this back-titration because the solution contains excess strong acid and the endpoint should be in the acidic range. Methyl orange changes over about pH 3.1–4.4, where the excess HCl has been neutralised but carbonate is still mainly present as CO₂/H₂CO₃ rather than being converted further to HCO₃⁻. Phenolphthalein would be a poor choice: its endpoint near pH 8.3 would require additional NaOH to convert H₂CO₃ to HCO₃⁻, so the titre would no longer measure only the excess HCl and the carbonate result would be badly biased.
The aliquot procedure is acceptable in principle because the 250.0 cm³ flask is homogeneous and a 25.00 cm³ aliquot is exactly one-tenth of the flask. The main risks are volumetric errors, incomplete transfer during filtration, and a non-sharp endpoint in the presence of CO₂. A visual methyl-orange endpoint is also subjective.
Two improvements for a regulatory report would be:
- Boil the acidified sample to remove CO₂, cool it, and then titrate; or use a CO₂-free procedure with an autotitrator and a pH electrode to determine the endpoint objectively.
- Run duplicate or triplicate aliquots, include a certified reference carbonate and a full procedural blank, and calibrate the pH electrode or indicator against a standard buffer. The report should also state the temperature, the exact endpoint criterion, and the correction for the acid-insoluble binder.