Chemistry • Year 12 • Module 7 • Lesson 14
Carboxylic Acids: Structure, Properties & Reactions
Apply Ka data, reaction equations and boiling-point reasoning to real scenarios drawn from Australian industry and everyday chemistry.
1. Interpret Ka data for common carboxylic acids
The table below gives acid dissociation constants (Ka) and pKa values for five carboxylic acids at 25°C. Use the data to answer the questions that follow. 10 marks
| Acid | IUPAC name | Formula | Ka (mol/L) | pKa | Australian context |
|---|---|---|---|---|---|
| Methanoic acid | Methanoic acid | HCOOH | 1.77 × 10−4 | 3.75 | Formic acid — venom of Australian bull ant (Myrmecia sp.) |
| Ethanoic acid | Ethanoic acid | CH3COOH | 1.76 × 10−5 | 4.75 | Acetic acid — vinegar, AWRI Australian wine fermentation |
| Propanoic acid | Propanoic acid | CH3CH2COOH | 1.34 × 10−5 | 4.87 | Produced in artisan sourdough fermentation (Australian bakeries) |
| Lactic acid | 2-hydroxypropanoic acid | CH3CH(OH)COOH | 1.38 × 10−4 | 3.86 | Primary acid in Australian sourdough bread; produced by lactic acid bacteria |
| Citric acid | 2-hydroxypropane-1,2,3-tricarboxylic acid | C6H8O7 | 7.40 × 10−4 (Ka1) | 3.13 | Dominant acid in Australian citrus industry (lemons, limes, oranges) |
Table 1. Acid dissociation constants for selected carboxylic acids at 25°C. Sources: CRC Handbook of Chemistry and Physics (104th ed.); AWRI Technical Review.
1.1 Rank the five acids from strongest to weakest using the Ka data. State the chemical principle that links Ka value to acid strength. 2 marks
1.2 Methanoic acid has a Ka approximately 10 times greater than ethanoic acid, despite both being simple carboxylic acids. Suggest a structural reason for this difference. (Hint: consider the “R group” in each acid.) 2 marks
1.3 Lactic acid has a Ka of 1.38 × 10−4 mol/L, making it stronger than ethanoic acid. Account for this using the concept of the inductive effect. 2 marks
1.4 A 0.1 mol/L solution of ethanoic acid has a pH of approximately 2.9. A 0.1 mol/L solution of hydrochloric acid (HCl) has a pH of 1.0. Explain this difference using the Ka data in the table. 2 marks
1.5 Citric acid is the dominant acid in Australian-grown lemons and contributes to the sharp flavour of lemon juice. A food scientist measures the pH of lemon juice as 2.4. Using the Ka data, predict whether undissociated citric acid molecules or citrate ions are in greater concentration in lemon juice at this pH. Justify your prediction. 2 marks
2. Graph interpretation — boiling points of homologous series
The graph below shows the boiling points of the first six members of three homologous series: n-alkanes, primary alcohols, and carboxylic acids, all plotted against carbon chain length. 8 marks
2.1 Describe the general trend in boiling point as carbon chain length increases for all three series. 1 mark
2.2 At C2, carboxylic acids have a boiling point approximately 40°C higher than alcohols of the same chain length. Explain why, using the concept of dimerisation. 3 marks
2.3 The gap between alcohols and carboxylic acids remains relatively constant across C1–C6. Predict whether this gap would increase or decrease for C10 acids versus C10 alcohols, and justify your prediction. 2 marks
2.4 Identify the type(s) of intermolecular forces responsible for the boiling point trend in n-alkanes, and explain why alkane BPs are the lowest at each carbon number. 2 marks
3. Cause-and-effect chain — wine turns to vinegar (Australian winemaking)
The Australian Wine Research Institute (AWRI) warns that wines stored at high temperature and exposed to air can spoil: ethanol is oxidised to ethanoic acid (acetic acid) by Acetobacter bacteria. Trace the chemistry using the cause–effect chain below. 6 marks
| Cause (given) | → | Effect (your answer) |
| Ethanol is oxidised by Acetobacter in the presence of O2 | → | |
| Ethanoic acid dissolves in water (remaining wine) | → | |
| Ethanoate ions (CH3COO−) are formed at equilibrium | → | |
| Baking soda (NaHCO3) is added to the spoiled wine | → |
Overall outcome (so…) — the wine has spoiled because…
4. Compare and contrast — carboxylic acids vs alcohols
Complete the comparison table for ethanoic acid (CH3COOH) and ethanol (CH3CH2OH). 8 marks (1 per row)
| Feature | Ethanoic acid (CH3COOH) | Ethanol (CH3CH2OH) |
|---|---|---|
| Functional group name | ||
| Boiling point (°C) | ||
| Reason for BP difference | ||
| pKa (approx.) | ||
| Reaction with NaHCO3 | ||
| Conjugate base formed | ||
| Stability of conjugate base | ||
| Reaction with NaOH (products) | ||
Q1.1 — Acid strength ranking
Strongest to weakest (using Ka): citric acid (Ka1 = 7.40×10−4) > methanoic acid (1.77×10−4) > lactic acid (1.38×10−4) > ethanoic acid (1.76×10−5) > propanoic acid (1.34×10−5). Principle: the larger the Ka (or smaller the pKa), the more the equilibrium lies toward ionisation, the greater the [H3O+], and the stronger the acid. (1 mark for correct ranking; 1 mark for principle.)
Q1.2 — Methanoic vs ethanoic acid Ka difference
In methanoic acid (HCOOH), the “R group” is simply H — there is no electron-donating alkyl group. In ethanoic acid (CH3COOH), the CH3 group donates electron density inductively toward the carboxyl carbon, slightly destabilising the carboxylate anion and raising the pKa (weakening the acid). Methanoic acid's H has no electron-donating effect, so the carboxylate is relatively more stable → lower pKa → stronger acid. (1 mark for identifying the electron-donating effect of CH3; 1 mark for linking this to reduced conjugate base stability.)
Q1.3 — Lactic acid and inductive effect
Lactic acid has an −OH group on the carbon adjacent to −COOH. The electronegative oxygen of the −OH group withdraws electron density inductively through the C−C bond, partially stabilising the negative charge on the carboxylate ion beyond what resonance alone provides. This additional stabilisation of the conjugate base shifts the equilibrium further toward ionisation, giving lactic acid a higher Ka than simple ethanoic acid. (1 mark for identifying the −OH as the electronegative group; 1 mark for linking inductive electron withdrawal to conjugate base stabilisation.)
Q1.4 — pH comparison with HCl
HCl is a strong acid — it fully dissociates in water, so [H3O+] = 0.1 mol/L → pH 1.0. Ethanoic acid is a weak acid (Ka = 1.76×10−5) — at 0.1 mol/L only about 1.3% of molecules are ionised, giving [H3O+] ≈ 0.0013 mol/L → pH ≈ 2.9. The much smaller degree of ionisation (dictated by the small Ka) produces far fewer H3O+ ions at the same initial concentration. (1 mark for HCl full dissociation; 1 mark for ethanoic acid partial ionisation linked to Ka.)
Q1.5 — Citric acid speciation in lemon juice
At pH 2.4 (which is below the pKa1 of 3.13), the solution is more acidic than the half-dissociation point. Below the pKa, undissociated acid molecules predominate over the anion form. Therefore, undissociated citric acid molecules are in greater concentration than citrate ions in lemon juice at pH 2.4. (1 mark for correct identification; 1 mark for justification using pKa vs pH comparison.)
Q2.1 — BP trend with chain length
Boiling point increases with increasing carbon chain length for all three series. This is because longer chains have greater surface area and more electrons, which increases the strength of London/dispersion forces between molecules.
Q2.2 — C2 acid vs alcohol BP (dimerisation)
Both ethanoic acid and ethanol possess one O−H bond (1 mark). However, two ethanoic acid molecules form a cyclic dimer via two simultaneous H-bonds: O−H (mol 1) ··· O=C (mol 2) and O=C (mol 1) ··· H−O (mol 2) (1 mark). To vaporise, both H-bonds of the dimer must be broken at once — this requires significantly more energy than breaking the single H-bond between two ethanol molecules (1 mark). Hence carboxylic acids have higher BPs despite similar or lower molar mass than the corresponding alcohol.
Q2.3 — C10 gap prediction
The gap would likely decrease slightly at C10. As chain length increases, dispersion forces (proportional to chain length/surface area) grow for both series, while the contribution of dimerisation (a fixed “bonus” from the −COOH end group) stays approximately constant. Dispersion forces dominate at high carbon numbers, narrowing the relative advantage of dimerisation. (1 mark for correct direction; 1 mark for reasoning.)
Q2.4 — Alkane IMF
London (dispersion) forces only. Alkanes are non-polar — no permanent dipole and no capacity for H-bonding — so only weak transient dipole interactions operate. These are the weakest type of IMF, giving alkanes the lowest boiling points at every carbon number compared to polar or H-bonding compounds.
Q3 — Wine / vinegar cause–effect chain
Row 1: Ethanoic acid (acetic acid, CH3COOH) is produced.
Row 2: A partial ionisation equilibrium is established: CH3COOH + H2O ↔ CH3COO− + H3O+; pH falls (wine becomes acidic/sour).
Row 3: [H3O+] increases; the solution is weakly acidic with pH ≈ 2.9–3.5 depending on concentration.
Row 4: Effervescence (CO2 bubbles) observed; sodium ethanoate, water and carbon dioxide are produced: CH3COOH + NaHCO3 → CH3COONa + H2O + CO2.
Overall: The wine has spoiled because bacterial oxidation converted ethanol (the desired product) to ethanoic acid, making the solution taste vinegary, dropping the pH, and making it reactive with carbonate-containing substances.
Q4 — Comparison table
Functional group: carboxyl (−COOH) | hydroxyl (−OH). BP: 118°C | 78°C. BP difference: Ethanoic acid forms H-bonded dimers (2 simultaneous H-bonds per pair); ethanol forms only 1 H-bond per molecular interaction. pKa: ~4.75 | ~16. NaHCO3: effervescence (CO2) | no reaction. Conjugate base: ethanoate CH3COO− | ethoxide CH3CH2O−. Stability: resonance-stabilised (charge on both O) | unstabilised (charge on one O). Reaction with NaOH: CH3COOH + NaOH → CH3COONa + H2O | CH3CH2OH does not react with NaOH under standard conditions.