Verifying a '98% Pure' Sodium Hydroxide Label: Adulteration or a Bottle Left Open?
Animo's chemistry group is given a bottle labelled 'sodium hydroxide pellets, 98% pure' and told to verify the label. They weigh 2.50 g of the pellets, dissolve them in distilled water and make the solution up to 250 cm³ in a volumetric flask. Using a 25.0 cm³ pipette they titrate portions against standard sulphuric acid of concentration 0.100 mol dm⁻³, with methyl orange as indicator. Their burette readings are 24.80 cm³ for a rough titration, then 25.00, 24.95 and 25.05 cm³. Their calculation gives a purity near 80%, far below the label. Amaka says the pellets must have been adulterated by the supplier. Bola says the fault may lie with the bottle, which was left open in a humid store for several weeks, and not with the supplier at all. The teacher will not accept a conclusion until the group has tested both explanations against the chemistry, and she also asks whether methyl orange was the right indicator for a sample that may contain carbonate.
1. Decide which burette readings may be used, calculate the mean titre, and from it obtain the concentration of the sodium hydroxide solution.
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Which readings? The 24.80 cm³ run is the rough titration, used only to find the approximate end-point, so it is discarded. The remaining three readings (25.00, 24.95, 25.05) span 0.10 cm³, which is within normal concordance limits, so all three are used.
Mean titre. (25.00 + 24.95 + 25.05) / 3 = 25.00 cm³.
Mole ratio. H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so 1 mol acid neutralises 2 mol hydroxide.
Moles of acid. n(H₂SO₄) = 0.100 mol dm⁻³ × 0.02500 dm³ = 2.50 × 10⁻³ mol.
Moles of hydroxide in the 25.0 cm³ portion. n(NaOH) = 2 × 2.50 × 10⁻³ = 5.00 × 10⁻³ mol.
Concentration. c = 5.00 × 10⁻³ mol ÷ 0.0250 dm³ = 0.200 mol dm⁻³.
Check. A 0.200 mol dm⁻³ solution of a strong base with a strong acid gives an end-point in the acidic range, where methyl orange changes, so the indicator choice is at least self-consistent for a pure hydroxide.
2. Calculate the percentage purity by mass and state clearly what the calculation assumes about the sample.
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Scale up from the portion to the flask. The 250 cm³ flask holds ten 25.0 cm³ portions, so n(NaOH) = 0.200 × 0.250 = 0.0500 mol.
Mass of sodium hydroxide actually present. m = 0.0500 × 40 = 2.00 g.
Percentage purity. (2.00 ÷ 2.50) × 100 = 80.0%.
Size of the discrepancy. The label claims 98%, which would require 2.45 g of NaOH in the 2.50 g weighed; the titration accounts for only 2.00 g, a shortfall of 0.45 g, or 18 percentage points.
Assumptions that must be stated.
- Every mole of acid consumed is neutralised by hydroxide that behaves as NaOH - no other base and no acidic impurity is present.
- The 2.50 g weighed is representative of the whole bottle.
- The pellets were weighed dry; if they were damp, part of the 2.50 g was water, which consumes no acid at all.
Judgement. The titration can only report a 'NaOH equivalent'. It cannot by itself distinguish between a supplier who sold weak material and a sample that has changed in the store, so the number alone does not settle Amaka's claim.
3. Evaluate Bola's hypothesis chemically, explain what methyl orange does to a carbonate impurity, and propose tests that would settle the argument.
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The chemistry of an open bottle. NaOH is deliquescent and absorbs both water vapour and carbon dioxide: 2NaOH + CO₂ → Na₂CO₃ + H₂O.
Why this lowers the apparent purity. Under methyl orange the titration runs to the acidic end-point, so carbonate is neutralised completely: CO₃²⁻ + 2H⁺ → CO₂ + H₂O. Per gram, Na₂CO₃ (M = 106) consumes far less acid than NaOH (M = 40): 1 mol of carbonate (106 g) takes the same 2 mol of H⁺ as 80 g of NaOH. Absorbed water consumes no acid at all. Either form of contamination therefore adds mass that contributes little or nothing, so the calculated NaOH content per gram of sample falls. Bola's explanation is chemically sound.
Was methyl orange right? Yes for total alkali - it counts hydroxide and carbonate together. But because it hides the split, a second, phenolphthalein stage would be more informative: phenolphthalein titrates OH⁻ → H₂O and CO₃²⁻ → HCO₃⁻; the extra acid needed to go on to the methyl orange end-point measures the carbonate alone.
Tests to settle it. (i) Add BaCl₂ solution to a portion: a white precipitate of BaCO₃ that fizzes with dilute acid confirms carbonate. (ii) Run the two-indicator titration and calculate the carbonate content. (iii) Repeat the whole procedure on a freshly opened, sealed bottle and on a sample dried in a desiccator - if the fresh sample titrates near 98%, the supplier is cleared and the store is at fault.
Conclusion to write. Not 'adulterated', but 'consistent with carbonate and water absorption during storage; confirm by barium chloride test and repeat with a fresh sample'.