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 resonance and equilibrium concepts to the molecular structures you know from Module 7.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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: 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.
pKa ladder: carboxylic acids (pKa ~4–5) are strong enough to react with NaHCO₃; phenol (pKa ~10) is too weak for NaHCO₃ but reacts with NaOH; alcohols (pKa ~16) do not react with either.
Which organic compound has the LOWEST pKa (strongest acid)?
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).
Conjugate base stability: carboxylate (RCOO⁻) has high resonance delocalisation across two equivalent O atoms → most stable → weakest base → strongest acid (pKa ~5). Phenoxide has partial resonance into ring → intermediate stability → pKa ~10. Alkoxide has no resonance → least stable → strongest base → weakest acid (pKa ~16).
Why does ethanoic acid (pKa ~5) have a much lower pKa than ethanol (pKa ~16)?
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 | Turns red ✓ | Turns red ✓ | No change |
| NaHCO₃ solution | CO₂ gas ✓ | No reaction ✗ | No reaction ✗ |
| Na₂CO₃ solution | CO₂ gas ✓ | Borderline/faint | No reaction ✗ |
| NaOH solution | Salt + H₂O ✓ | Sodium phenoxide ✓ | No reaction ✗ |
| Na metal | H₂ gas ✓ | H₂ gas ✓ | H₂ gas ✓ |
Diagnostic flowchart: NaHCO₃ → CO₂ identifies carboxylic acid; NaOH reaction identifies phenol; Na metal H₂ identifies alcohol. Na metal reacts with ALL three — it cannot distinguish between classes.
True or False: Phenol reacts with sodium hydrogen carbonate (NaHCO₃) solution to produce CO₂ gas, because phenol is an acid.
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.
Kb ~4 × 10⁻⁴Kb = 1.8 × 10⁻⁵Kb ~4 × 10⁻¹⁰Kb ~10⁻¹⁵Predicting reaction direction using pKa: If pKa(acid on left) < pKa(conjugate acid of base on left) → reaction proceeds right.
Arrange ethylamine, aniline, and ethanamide in order of INCREASING base strength:
(a) Ethanoic acid + Na₂CO₃ (b) Phenol + NaHCO₃ (c) Propan-1-amine + HCl (d) Ethanamide + HCl
pKa(ethanoic acid) = 4.74; pKa(H₂CO₃) = 6.4; pKa(HCO₃⁻) = 10.3; pKa(phenol) = 10; Kb(alkylamines) ~4 × 10⁻⁴; Kb(ethanamide) ~10⁻¹⁵
Ethanoic acid pKa = 4.74 < H₂CO₃ pKa = 6.4 → ethanoic acid stronger → PROCEEDS. CO₂ evolved.
2CH₃COOH + Na₂CO₃ → 2CH₃COO⁻Na⁺ + H₂O + CO₂↑
Phenol pKa = 10 > H₂CO₃ pKa = 6.4 → phenol weaker than H₂CO₃ → DOES NOT PROCEED. No CO₂ evolved.
HCl is a strong acid (pKa << 0). Conjugate acid of propan-1-amine pKa ≈ 10.6. Since pKa(HCl) << 10.6 → PROCEEDS completely. Propylammonium chloride formed.
Kb(ethanamide) ~10⁻¹⁵ → Ka(protonated ethanamide) = 10⁻¹⁴/10⁻¹⁵ = 10 (pKa ≈ −1). The protonated form would immediately lose H⁺ back to solution → DOES NOT PROCEED. Amide lone pair is fully delocalised into C=O — not available for proton acceptance.
Four compounds (A–D) tested. Compounds are butanoic acid, butan-1-ol, phenol, butylamine.
| 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 |
A: Red litmus + CO₂ with NaHCO₃ + reacts NaOH + H₂ with Na → only carboxylic acids react with NaHCO₃ to give CO₂. A = butanoic acid.
C: Red litmus + NO CO₂ with NaHCO₃ + reacts NaOH + H₂ with Na → acidic but NaHCO₃ negative, NaOH positive. C = phenol (pKa ~10 — too weak for NaHCO₃, but NaOH deprotonates it).
B: No litmus change + no reactions with acidic reagents + no Na reaction. No O-H bond. Is a base. B = butylamine.
D: By elimination: D = butan-1-ol. No litmus change, no NaHCO₃/NaOH reaction, but H₂ with Na metal (O-H present).
Three 0.1 mol/L solutions: X = ethanoic acid, Y = phenol, Z = ethanol. (a) Arrange 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 pH of the resulting 0.1 mol/L sodium ethanoate solution. Ka(ethanoic acid) = 1.8 × 10⁻⁵.
Ethanoic acid: [H₃O⁺] = √(Ka × C) = √(1.8 × 10⁻⁵ × 0.1) = 1.34 × 10⁻³ → pH ≈ 2.87
Phenol: Ka ~10⁻¹⁰ → [H₃O⁺] = 3.16 × 10⁻⁶ → pH ≈ 5.5
Ethanol: essentially no dissociation → pH ≈ 7.0
Increasing pH: X (2.87) < Y (5.5) < Z (7.0)
Test 1 — NaHCO₃: X (ethanoic acid, pKa 4.74 < 6.4): CO₂ evolved. Y and Z: no CO₂. → Distinguishes X from Y and Z.
Test 2 — NaOH: X and Y react (both acids); Z (ethanol, pKa ~16): no significant reaction. But since X was already identified, this distinguishes Y (phenol, reacts with NaOH) from Z (alcohol, no reaction).
2CH₃COOH + CO₃²⁻ → 2CH₃COO⁻ + H₂O + CO₂(g)
pH of 0.1 mol/L CH₃COO⁻Na⁺: Ethanoate is the conjugate base of a weak acid → basic solution.
Kb(CH₃COO⁻) = Kw / Ka = 10⁻¹⁴ / (1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰
[OH⁻] = √(Kb × C) = √(5.56 × 10⁻¹¹) = 7.46 × 10⁻⁶ mol/L
pOH = −log(7.46 × 10⁻⁶) = 5.13 → pH = 14 − 5.13 = 8.87 ≈ 8.9 (basic)
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For each of the following, classify the compound and predict its behaviour with NaHCO₃ and NaOH. Justify using pKa values.
1. 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?
2. Which correctly ranks ethylamine, aniline, and ethanamide in order of increasing base strength, with the correct structural reason?
3. 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?
4. 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?
5. A 0.1 mol/L solution of sodium ethanoate (CH₃COO⁻Na⁺) in water has a pH greater than 7. Which statement best explains this?
Question 6 (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.
Question 7 (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.
Question 8 (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.
Q1 — B. Phenol (pKa ~10) is weaker than H₂CO₃ (pKa ~6.4) → no CO₂ with NaHCO₃. NaOH is a strong base → deprotonates phenol → sodium phenoxide + water. Option A wrong: propanoic acid IS strong enough to produce CO₂ with NaHCO₃. Option C wrong: propan-1-ol does not react with NaOH under standard aqueous conditions.
Q2 — A. Ethanamide (Kb ~10⁻¹⁵, lone pair fully delocalised into C=O) < aniline (Kb ~10⁻¹⁰, lone pair partially delocalised into benzene ring) < ethylamine (Kb ~4 × 10⁻⁴, free lone pair on N, enhanced by ethyl inductive donation).
Q3 — B. Propanoic acid (pKa ~4.9 < 6.4) → reacts with NaHCO₃ → CO₂ + sodium propanoate. Phenol (pKa ~10 > 6.4) → thermodynamically cannot react with NaHCO₃ → remains as un-ionised phenol.
Q4 — B. For a reaction to proceed: pKa(acid) must be < pKa(conjugate acid of amine) = 10.6. Phenol pKa ~10 < 10.6 → phenol is slightly stronger acid → reaction proceeds. Ethanol pKa ~16 and butan-1-ol pKa ~16 are both weaker than methylammonium (pKa 10.6) → do not react.
Q5 — B. Ethanoate (CH₃COO⁻) is the conjugate base of ethanoic acid, a weak acid. It partially accepts protons from water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Kb = Kw/Ka = 5.56 × 10⁻¹⁰. More OH⁻ than H₃O⁺ → pH > 7. Na⁺ ions are spectator ions.
Q6 (4 marks): Propanoic acid (pKa 4.87) reacts with NaHCO₃ because propanoic acid is stronger than the product acid, H₂CO₃ (pKa 6.4). Since pKa(propanoic acid) < pKa(H₂CO₃), the reaction proceeds left → right: CH₃CH₂COOH + HCO₃⁻ → CH₃CH₂COO⁻ + H₂O + CO₂(g). CO₂ evolution drives the equilibrium further right. Phenol (pKa 10.0) does not react because phenol is a weaker acid than H₂CO₃ (pKa 6.4). The reaction would need to proceed right → left — thermodynamically unfavourable. Therefore no CO₂ is produced.
Q7 (5 marks): Test 1 — NaHCO₃: Pentanoic acid (pKa ~4.8 < 6.4): effervescence — CO₂ produced → identified. Phenol and pentan-1-ol: no CO₂ (both pKa > 6.4). Test 2 — NaOH: Phenol (pKa ~10): reacts → sodium phenoxide. Pentan-1-ol (pKa ~16): no significant reaction. → Phenol identified by NaOH reaction; pentan-1-ol identified by neither NaHCO₃ nor NaOH reaction.
Q8 (6 marks): Increasing base strength: ethanamide < aniline < ethylamine. Ethanamide (Kb ~10⁻¹⁵): N lone pair fully delocalised into adjacent C=O via resonance — lone pair participates in pi system, not available to accept H⁺ → essentially not basic. Aniline (Kb ~4 × 10⁻¹⁰): N lone pair partially delocalised into benzene ring's pi system via three resonance structures — partial delocalisation reduces lone pair availability → weakly basic. Ethylamine (Kb ~4 × 10⁻⁴): N lone pair freely available (not involved in any adjacent pi bond) — no adjacent pi system. Ethyl group donates electron density to N inductively, making lone pair MORE electron-rich → most readily accepts H⁺ → strongest base.
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).
1. What are the approximate pKa values for carboxylic acids, phenol, and alcohols?
2. Why is phenol more acidic than ethanol?
3. Which test distinguishes carboxylic acid from phenol?
4. Use Ka × Kb = Kw to find the pKa of the conjugate acid of ethylamine (Kb = 4 × 10⁻⁴).
5. Why does sodium ethanoate solution have pH > 7?
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