Assay and Clean-Up of a Propanoic Acid Consignment
A small chemical supplier receives a 1.50 g consignment labelled 'propanoic acid, 99% pure'. The quality officer dissolves it in water and titrates with 0.100 M NaOH to a phenolphthalein endpoint, consuming 185.0 mL. The supplier's own route is oxidation of propan-1-ol, so the likely impurity is propanal or propanone carried through the work-up; molar masses are 74.0 g mol⁻¹ (propanoic acid) and 58.0 g mol⁻¹ (either carbonyl). The customer uses the acid in an esterification for a food flavouring and has rejected an earlier batch because the product discoloured on storage, a known sign of aldehydic impurity. Available in the stockroom: Tollens' reagent, Fehling's solution, 2,4-dinitrophenylhydrazine, iodine in NaOH, saturated NaHCO₃, dilute HCl and a distillation kit. The acid is worth ₹4,200 per kilogram; the batch must be sold, purified or rejected within a week.
1. Calculate the purity from the titration and decide what to do with the consignment.
Read the worked answer
Moles of base: n = 0.100 × 0.1850 = 1.850 × 10⁻² mol.
Moles of acid: the acid is monobasic, so n(acid) = 1.850 × 10⁻² mol (Phenolphthalein turns at the carboxylate equivalence point; neither propanal nor propanone has an acidic hydrogen, so the carbonyl impurity consumes no base - the whole titre belongs to the acid).
Mass and purity: m = 1.850 × 10⁻² × 74.0 = 1.37 g. % purity = 1.37 / 1.50 × 100 = 91.3%, i.e. about 0.13 g (8.7%) of impurity.
Decision: the 99% claim is false by a wide margin, so the certificate must not be passed on. Note the economics before choosing between purification and rejection: 1.50 g at ₹4,200 kg⁻¹ is worth about ₹6, so the sample itself is trivial - the real risk is the customer relationship and the reject already on record for discolouration. The consignment should be held, the impurity identified, and either re-purified with a documented assay or rejected; selling it as 99% is not defensible.
2. Using the stockroom reagents, describe a decisive sequence to establish whether the impurity is propanal or propanone. Justify the order in which you use the tests.
Read the worked answer
Step 1 - confirm a carbonyl is present: 2,4-dinitrophenylhydrazine. Both propanal and propanone give an orange/red 2,4-dinitrophenylhydrazone precipitate; propanoic acid does not react. A positive result confirms the impurity is carbonyl, not unreacted propan-1-ol or a solvent, so it converts the suspicion into evidence.
Step 2 - split aldehyde from ketone: Tollens' reagent. Warm with the ammoniacal silver nitrate.
- Propanal is oxidised to propanoate and reduces Ag⁺ to Ag: a silver mirror forms. Aldehyde.
- Propanone is not oxidised by Tollens' reagent; no mirror. Ketone.
Why Tollens' before Fehling's: Tollens' discriminates cleanly between aldehyde and ketone and is not confused by the aromatic/α-hydroxy cases that make Fehling's the more selective and narrower test.
Step 3 - only if Tollens' is negative (ketone): iodine in NaOH. A methyl ketone such as propanone gives the yellow iodoform precipitate CHI₃ with its characteristic smell; this confirms the ketone and rules out a non-methyl ketone.
Conclusion logic: positive 2,4-DNP plus positive Tollens' = propanal, which matches the customer's discolouration complaint and identifies the batch as the aldehyde-contaminated type. Do not use the iodoform test as the primary test, because aldehydes can also be oxidised under those strongly basic, oxidising conditions and give misleading results.
3. Recommend a chemical method to remove the impurity without losing the acid, and explain why this is preferred to simply distilling the sample.
Read the worked answer
Use the acid–base difference. Dissolve the sample in a little water/ether and shake with saturated NaHCO₃:
CH₃CH₂COOH + NaHCO₃ → CH₃CH₂COONa + CO₂ + H₂O
The acid becomes the water-soluble sodium salt (visible effervescence confirms the reaction); propanal, which has no acidic hydrogen, stays in the organic phase and is washed away.
Separate the layers, then acidify the aqueous layer with dilute HCl:
CH₃CH₂COONa + HCl → CH₃CH₂COOH + NaCl
The acid is liberated, isolated, dried and distilled to give material that can be re-assayed by the same titration before release.
Why not just distil? The boiling points are far apart (propanoic acid ≈141 °C, propanal ≈48 °C), so distillation would separate them in principle - but a 1.5 g sample at 141 °C suffers charring and heavy mechanical loss in the column, propanal slowly self-condenses/oxidises in the flask, and the acid is corrosive on a small glassware scale. More importantly, distillation separates on a physical difference that gives no proof of what was removed, whereas the bicarbonate route is selective for the acid by chemistry, keeps the acid as a salt that cannot be lost with the volatile aldehyde, and leaves a product whose identity can be confirmed. Re-assay the purified acid by titration before issuing a corrected certificate.