Chemistry • Year 12 • Module 7 • Lesson 14

Carboxylic Acids: Structure, Properties & Reactions

Build HSC Band 5–6 extended-response technique: evaluate data-driven scenarios, critique a source, and reach evidence-based judgements about carboxylic acid chemistry.

Master · Band 5–6 · Extended Response

1. Data scenario — identifying unknowns in the Australian sourdough industry

8 marks Band 5–6

Scenario. An Australian artisan bakery tests three unidentified organic compounds (X, Y, Z) isolated from a sourdough starter culture. All three have molecular formula C₃H₆O₂. The results of four chemical tests are given in the table below.

Test	Compound X	Compound Y	Compound Z
NaHCO₃ solution	Effervescence	No reaction	No reaction
NaOH solution	Reacts	Reacts	No reaction
Fehling’s reagent	No reaction	No reaction	No reaction
Br₂(aq)	No decolourisation	No decolourisation	No decolourisation

C₃H₆O₂ isomers include: propanoic acid, methyl ethanoate, ethyl methanoate, and 1,3-dioxolane (cyclic). Additional Ka data: propanoic acid Ka = 1.34 × 10⁻⁵ mol/L.

Q1. Evaluate the test data to identify compounds X, Y, and Z, and justify the boiling point ranking for the three compounds. In your response you must:

Identify each compound, naming the functional group class and IUPAC name.
Explain the chemical reasoning behind each positive and negative test result for compound X, using at least two tests.
Use the concept of resonance to explain why compound X is more acidic than compound Z.
Predict and rank the boiling points of all three compounds at 1 atm, giving a structural reason for each ranking decision.
Reach an evidence-based judgement about which compound is most likely the sourdough acid detected by the bakery, using the Ka data and the test results.

Tackle methodically: (1) use the NaHCO₃ test to lock in compound X; (2) use the NaOH/NaHCO₃ combination to narrow Y vs Z; (3) consult Br₂ and Fehling’s to confirm saturation and exclude aldehyde; (4) apply resonance for acid strength; (5) rank BPs by IMF type (dimer > single H-bond > dipole only).

2. Source critique — a student’s claim about formic acid and bull ant venom

7 marks Band 5–6

“When an Australian bull ant (Myrmecia sp.) stings you, it injects formic acid (methanoic acid, HCOOH) which is a strong acid. This is why the sting burns — the acid fully ionises in tissue fluids and releases a large amount of H⁺ ions all at once, creating a pH close to zero at the sting site. Because it is a strong acid, formic acid reacts more vigorously with sodium carbonate than does ethanoic acid (vinegar), and adding baking soda to the sting immediately neutralises all of the acid.”

— Student biology-chemistry essay, Year 12.

Q2. This student’s claim contains multiple scientific errors. Identify the errors, explain the correct chemistry, and discuss how the correct behaviour of formic acid in tissue fluids could be demonstrated experimentally. In your response you must:

Identify and correct at least three specific factual errors in the passage.
For each error, provide the correct chemical explanation with reference to Ka, ionisation equilibrium, or reaction equations where appropriate.
Explain whether adding baking soda (NaHCO₃) would in fact fully neutralise all the formic acid at a sting site, and why.
Describe one experimental procedure (method, expected observation, and conclusion) that would demonstrate the weak acid nature of formic acid in aqueous solution.

Identify errors systematically: (1) is formic acid a strong acid? Check Ka = 1.77 × 10⁻⁴; (2) does it fully ionise? (3) would pH reach near zero? (4) does it react “more vigorously” with Na₂CO₃ than ethanoic acid? (5) does NaHCO₃ neutralise all the acid? — think about the equilibrium position.

Answers — Do not peek before attempting

Q1 — Marking criteria (8 marks)

Identification of compounds (2 marks):

X: NaHCO₃ + effervescence → carboxylic acid. No Fehling’s reaction → not an aldehyde. No Br₂ decolourisation → saturated. C₃H₆O₂ carboxylic acid = propanoic acid (CH₃CH₂COOH). (1 mark)
Y: reacts with NaOH but not NaHCO₃ → not acidic enough for NaHCO₃; no Fehling’s → not aldehyde; not carboxylic acid. C₃H₆O₂ with no C=C, NaOH reactive = ester that hydrolyses under NaOH. Methyl ethanoate (CH₃COOCH₃) or ethyl methanoate (HCOOC₂H₅); either accepted. (1 mark)
Z: no reaction with NaHCO₃ or NaOH; no Fehling’s; no Br₂. Saturated, non-acidic, non-ester-hydrolysable = most likely a cyclic ether or the remaining structural isomer. Accept any valid C₃H₆O₂ isomer that fits all negatives (e.g. 1,3-dioxolane). (1 mark — accept any chemically valid assignment consistent with data)

Test reasoning for X — 2 tests explained (2 marks):

NaHCO₃ test: propanoic acid (pKa 4.87) is acidic enough to donate H⁺ to HCO₃⁻ (pKa 6.35), forming H₂CO₃ → H₂O + CO₂(g) = effervescence. Esters and most other functional groups cannot do this. (1 mark)
Fehling’s negative: carboxylic acids cannot be further oxidised at the carbonyl carbon — the carbon is already in its highest oxidation state in −COOH. No Cu₂O precipitate. (1 mark)

Resonance argument for X > Z (1 mark): After propanoic acid (X) loses a proton, the propanoate anion (CH₃CH₂COO⁻) is resonance-stabilised — the negative charge is delocalised equally across both oxygen atoms. Compound Z (a cyclic ether) has no acidic proton at all and no resonance mechanism, so it does not ionise. Resonance stabilisation of the conjugate base lowers the energy of the products, shifting equilibrium right → stronger acid.

Boiling point ranking (2 marks): Propanoic acid (X) > methyl ethanoate/ethyl methanoate (Y) > Z (cyclic ether). X forms H-bonded dimers (two simultaneous H-bonds per pair) — highest IMF energy; Y is a polar ester with dipole–dipole interactions but no O−H bond → no H-bonding; Z has only weak dipole or dispersion forces. (1 mark for correct ranking; 1 mark for structural reasoning for at least two positions.)

Sourdough acid judgement (1 mark): Compound X (propanoic acid) is the most likely sourdough acid because: (a) it is the only compound that reacts with NaHCO₃, confirming a carboxylic acid functional group consistent with organic acid production by bacteria; (b) its Ka = 1.34 × 10⁻⁵ gives a pH of ~2.9 at typical sourdough concentrations, consistent with the pH 3.5–4.5 measured in sourdough; (c) propanoic acid is a known product of heterofermentative bacteria in sourdough cultures.

Q2 — Marking criteria (7 marks)

Error 1 — “Formic acid is a strong acid” (2 marks: 1 identify, 1 correct): Formic acid (methanoic acid) is a weak acid — Ka = 1.77 × 10⁻⁴ mol/L, which is much less than 1. Strong acids have Ka ≫ 1 (or are defined as fully dissociating). The ionisation equilibrium is: HCOOH + H₂O ↔ HCOO⁻ + H₃O⁺ (double arrow, not single arrow). At 0.1 mol/L, only about 4.1% of HCOOH molecules are ionised.

Error 2 — “Fully ionises / pH near zero” (1 mark): A 0.1 mol/L solution of formic acid has pH ≈ 2.4 (not near 0). pH 0 corresponds to [H₃O⁺] = 1 mol/L, which would require a fully ionised 1 mol/L strong acid. Formic acid only partially ionises, giving [H₃O⁺] ≈ 0.004 mol/L at 0.1 mol/L concentration → pH ≈ 2.4.

Error 3 — “Reacts more vigorously with Na₂CO₃ than ethanoic acid” (1 mark): Both formic acid and ethanoic acid react with Na₂CO₃ to produce CO₂ gas. Formic acid is somewhat stronger (Ka ~10× larger) and would react slightly faster, but this is not a dramatic difference. More importantly, the reaction proceeds for both: the claim that it reacts “more vigorously” is only partially true (both react; the rate difference is modest). The student may be confusing acid strength with concentration — at the same concentration, formic acid gives a higher [H₃O⁺], but both acids are weak and the Na₂CO₃ reaction goes to completion for both.

NaHCO₃ neutralisation at the sting (1 mark): Applying NaHCO₃ would partially neutralise the formic acid (HCOOH + NaHCO₃ → HCOONa + H₂O + CO₂). However, it would not neutralise all of it — the acid is continuously being released from the venom, and the equilibrium between HCOOH and HCOO⁻ means a portion of the acid always remains in the undissociated form. Adding a limited amount of NaHCO₃ will raise pH and reduce the [H₃O⁺], but complete neutralisation requires stoichiometric excess.

Experimental demonstration (1 mark — must include method, observation, conclusion): Example: Measure the pH of a 0.1 mol/L formic acid solution using a pH meter and compare to a 0.1 mol/L HCl solution. Expected observation: formic acid pH ≈ 2.4; HCl pH = 1.0. Conclusion: formic acid is a weak acid because its pH is higher than the strong acid of the same concentration, confirming partial ionisation (Ka « 1). Alternative: electrical conductivity measurement — weak acid solution has lower conductivity than the equivalent strong acid.