The difference between vinegar (pH ~3), a phenol solution (pH ~5), and an alcohol solution (pH ~7) reflects three orders of magnitude in Ka — understanding why requires applying the resonance and equilibrium concepts you built in Module 6 to the molecular structures you now know from Module 7.
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A pharmacist has three unlabelled colourless aqueous solutions. She adds blue litmus paper to each — all three turn red. She then adds sodium hydrogen carbonate: Solution A bubbles vigorously; Solutions B and C show no bubbling. To distinguish B and C, she adds sodium carbonate: Solution B produces faint bubbling; Solution C produces none.
Before reading: What compound class do you think is in B and in C? Explain your reasoning using what you know about acid strength from Module 6.
pKa is not just a number — it is a ranking of how willing a molecule is to give up a proton. Understanding why different organic functional groups fall at different positions on the scale requires the same resonance reasoning that explains carboxylate stability in Module 6.
The Ka and pKa Framework (from Module 6):
Ka is the equilibrium constant for HA + H₂O ⇌ A⁻ + H₃O⁺. Large Ka → equilibrium lies right → strong acid. pKa = −log₁₀(Ka): lower pKa = stronger acid. Each unit decrease in pKa represents a 10-fold increase in acid strength.
The pKa difference between carboxylic acid, phenol, and alcohol is not arbitrary — it is a direct consequence of how stable the conjugate base is after the proton is lost. Stability is determined by whether the negative charge can be delocalised by resonance.
Principle: More stable A⁻ → weaker base → equilibrium lies further right → stronger acid (lower pKa).
The reactions of NaOH, Na₂CO₃, and NaHCO₃ with organic acid classes are thermodynamic probes — each tells you whether the organic acid is stronger than the reference acid produced in the reaction.
Thermodynamic logic: A reaction proceeds left → right when the acid on the left is stronger (lower pKa) than the acid on the right.
| Test Reagent | Carboxylic Acid (pKa ~5) | Phenol (pKa ~10) | Alcohol (pKa ~16) |
|---|---|---|---|
| Litmus (acid test) | Turns red (acidic) | Turns red (weakly acidic) | No change (neutral) |
| NaHCO₃ solution | CO₂ gas evolved ✓ | No reaction ✗ | No reaction ✗ |
| Na₂CO₃ solution | CO₂ gas evolved ✓ | Borderline/faint ✓ | No reaction ✗ |
| NaOH solution | Salt + H₂O ✓ | Sodium phenoxide ✓ | No reaction ✗ |
| Na metal | H₂ gas evolved ✓ | H₂ gas evolved ✓ | H₂ gas evolved ✓ |
The pKb scale ranks organic bases by the same principle as pKa ranks acids — the more available the lone pair for proton acceptance, the stronger the base.
Base equilibrium: B + H₂O ⇌ BH⁺ + OH⁻ | Kb = [BH⁺][OH⁻] / [B]
Lower pKb = stronger base = more OH⁻ produced = higher pH.
Predicting reaction direction using pKa:
If pKa(acid on left) < pKa(conjugate acid of base on left) → reaction proceeds right.
If pKa(acid on left) > pKa(conjugate acid of base) → no significant reaction.
Problem: Predict whether each reaction proceeds in aqueous solution. Justify using pKa values.
(a) Ethanoic acid + Na₂CO₃ (b) Phenol + NaHCO₃ (c) Propan-1-amine + HCl (d) Ethanamide + HCl
Problem: Four compounds (A, B, C, D) are tested. Results:
| Test | A | B | C | D |
|---|---|---|---|---|
| Litmus | Red | No change | Red | No change |
| NaHCO₃ | CO₂ gas | No reaction | No reaction | No reaction |
| NaOH | Reacts | No reaction | Reacts | No reaction |
| Na metal | H₂ gas | No reaction | H₂ gas | H₂ gas |
Compounds are: butanoic acid, butan-1-ol, phenol, butylamine. Match A–D.
Problem: Three 0.1 mol/L solutions: X = ethanoic acid, Y = phenol, Z = ethanol. (a) Arrange X, Y, Z in order of increasing pH with Ka justification. (b) Two chemical tests to conclusively distinguish all three. (c) Add Na₂CO₃ to X — write the ionic equation; calculate the pH of the resulting 0.1 mol/L sodium ethanoate solution. Ka(ethanoic acid) = 1.8 × 10⁻⁵.
Organic Acid Strength (strongest → weakest):
Diagnostic reactions: NaHCO₃ → CO₂ = carboxylic acid only | NaOH reacts = carboxylic acid or phenol | Na metal H₂ = any O-H compound
Organic base strength: Alkylamine (Kb ~10⁻⁴) > NH₃ > Arylamine (~10⁻¹⁰) > Amide (~10⁻¹⁵)
Key relationship: Ka × Kb = Kw = 10⁻¹⁴ (for conjugate acid-base pair)
Using the formula-panel pKa values, predict which of the following pairs of reactions will proceed. For each, identify the acid on the left and the product acid, compare pKa values, and justify your prediction (1–2 sentences each):
Without looking at your notes, rank the following compounds in order of increasing acid strength, and for each, write one sentence explaining what happens to the negative charge in the conjugate base and why this determines position in the ranking:
Compounds: propan-1-ol | propanoic acid | phenol | water
Q1. A student tests an unknown compound with NaHCO₃ solution — no CO₂ observed. She then tests with NaOH solution — a reaction occurs. Which compound is most consistent with these results?
Q2. Which correctly ranks ethylamine, aniline, and ethanamide in order of increasing base strength, with the correct structural reason?
Q3. A student adds excess NaHCO₃ to a mixture of propanoic acid and phenol. What does the student observe, and what is the composition of the resulting solution?
Q4. Methylamine has Kb = 4.4 × 10⁻⁴. Its conjugate acid (methylammonium) therefore has pKa ≈ 10.6. Which of the following acids would react significantly with methylamine in aqueous solution?
Q5. A 0.1 mol/L solution of sodium ethanoate (CH₃COO⁻Na⁺) in water has a pH greater than 7. Which statement best explains this?
Q6. (4 marks) Explain why propanoic acid (pKa 4.87) reacts with sodium hydrogen carbonate solution to produce CO₂, but phenol (pKa 10.0) does not. In your answer, refer to pKa values, the identity of the product acid, and the thermodynamic direction of each reaction.
Q7. (5 marks) A student has three unlabelled bottles containing 0.1 mol/L solutions of pentanoic acid, phenol, and pentan-1-ol. Describe two chemical tests that would identify all three compounds. For each test, state: (i) the reagent and observation for each compound, and (ii) the pKa reasoning that explains each result.
Q8. (6 marks) Arrange ethylamine, aniline, and ethanamide in order of increasing base strength. For each compound: (a) state the approximate Kb value, (b) identify the structural feature of the nitrogen lone pair, and (c) explain how this determines the compound's position in the ranking. Use the concept of lone pair availability throughout your response.
The pharmacist's three solutions: A (CO₂ with NaHCO₃) = carboxylic acid. B (faint reaction with Na₂CO₃ only) = phenol (pKa ~10 — barely reacts with CO₃²⁻ since HCO₃⁻ pKa 10.3 is only slightly weaker). C (no reaction with either) = alcohol (pKa ~16 — far too weak to react with either carbonate reagent). How well did your reasoning match this pKa logic?
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