Back titration of an antacid tablet: is the label claim really failed?
A pharmacist supplies a student with a batch of antacid tablets, each labelled as containing 500 mg of calcium carbonate. The student is asked to check the label by back titration. One tablet is crushed and transferred to a 250 cm³ conical flask with about 50 cm³ of distilled water. A 25.0 cm³ portion of 0.500 mol dm⁻³ hydrochloric acid is added from a pipette and the mixture is warmed gently until effervescence stops. The student cools the flask and titrates the excess acid with 0.250 mol dm⁻³ sodium hydroxide, using phenolphthalein. Concordant titres of 18.55, 18.60 and 18.65 cm³ are obtained (mean 18.60 cm³). The student calculates the mass of calcium carbonate in the tablet as 393 mg and writes in the report that the tablets are substandard and should be withdrawn. The technician points out that the solution was not boiled before titration, and that the tablet also contains a magnesium stearate lubricant and a starch binder.
1. Show, from the titration data, that the tablet appears to contain 393 mg of calcium carbonate, and express this as a percentage of the stated 500 mg.
Read the worked answer
Acid added: n(HCl) = (25.0/1000) × 0.500 = 1.250 × 10⁻² mol
Alkali used: n(NaOH) = (18.60/1000) × 0.250 = 4.650 × 10⁻³ mol
NaOH and HCl react 1:1, so the acid left over is 4.650 × 10⁻³ mol.
Acid that reacted with the tablet: (12.50 − 4.65) × 10⁻³ = 7.850 × 10⁻³ mol
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, so n(CaCO₃) = 7.850 × 10⁻³ / 2 = 3.925 × 10⁻³ mol
m = 3.925 × 10⁻³ × 100.1 = 0.393 g = 393 mg
Percentage of label = (393/500) × 100 = 78.6 %
The arithmetic is correct; the question is whether the number means what the student thinks it means.
2. Assess the reliability of the student's conclusion, paying particular attention to the failure to boil the solution before the phenolphthalein titration, and to the presence of the lubricant and binder.
Read the worked answer
Phenolphthalein changes colour near pH 8.3–9.0, and at that pH dissolved carbon dioxide is neutralised as H₂CO₃ → HCO₃⁻. Any CO₂ still in solution therefore consumes hydroxide in addition to the excess acid, so the apparent excess acid is too large and the apparent carbonate too small.
Quantify the worst case: the tablet liberated 3.925 × 10⁻³ mol CO₂. If none had escaped, NaOH would have been 4.650 × 10⁻³ + 3.925 × 10⁻³ = 8.575 × 10⁻³ mol; acid reacted would be 12.50 × 10⁻³ − 8.575 × 10⁻³ = 3.925 × 10⁻³ mol; CaCO₃ would be 1.96 × 10⁻³ mol = 196 mg (39 %). The error can therefore be as large as ~40 percentage points downwards, so 78.6 % is a lower limit, not a measurement. CO₂ also causes a drifting, fading endpoint, so even the titre is unreliable.
Starch is insoluble and does not consume acid. Magnesium stearate is a salt of a weak acid and does consume acid, which would raise the apparent CaCO₃ - it cannot explain a low result; if anything the true carbonate content is lower still.
Verdict: the conclusion is not justified by this evidence. The tablet cannot be condemned on a method that has an uncontrolled systematic error of this size.
3. Describe how you would modify the procedure so that the result could be reported, and state one further check you would carry out before publishing a verdict on the batch.
Read the worked answer
Improved procedure
- Grind the tablet to a fine powder and stir it with the acid so that no lump survives; confirm the acid is in at least 100 % excess.
- Warm to about 60 °C until effervescence ceases, then boil gently for 2–3 minutes to strip out CO₂; cool, and stopper the flask to prevent re-absorption of atmospheric CO₂.
- Titrate to a pink colour that persists for at least 30 s. A pH meter (endpoint near pH 8.3, or a potentiometric end-point) is preferable to an indicator with a CO₂-sensitive colour change.
- Run a blank with the same volume of acid and no tablet, and subtract it.
Further check: repeat on at least two more tablets from different parts of the batch, and confirm the method on a standard sample of pure CaCO₃ of known mass. Only if the corrected, replicated results still fall below the label claim - allowing for the stated tolerance - can the batch be failed.