Year 12 Chemistry Module 7, Organic Chemistry ⏱ ~45 min Lesson 18 of 23 Core

Organic Acids & Bases, pKa, Strength & Reactions

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.

⚖️
Today's hook: In 2019, a pharmaceutical company had to recall a batch of medication because an amine impurity with a pKa difference of only 0.3 units wasn't neutralised efficiently during purification. pKa puts a precise number on acid/base strength, by the end of this lesson you will be able to use pKa values to predict which direction any organic acid-base equilibrium lies, and why some amines are a thousand times more basic than others.
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Connect: diagnose three unknown solutions

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions
Before You Read

The Pharmacist's Three Unlabelled Solutions

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.

Learning Intentions
goals

Know

  • pKa values for carboxylic acids (~5), phenols (~10), alcohols (~16)
  • Kb values for alkylamines, arylamines, amides
  • Diagnostic reactions: NaHCO₃, Na₂CO₃, NaOH, Na metal

Understand

  • How resonance stabilisation of conjugate base determines pKa
  • Why lone pair availability determines amine base strength
  • Thermodynamic logic of acid-base reactions (pKa comparison)

Can Do

  • Predict whether a reaction will proceed using pKa values
  • Design a test scheme to distinguish organic acid classes
  • Calculate pH of salt solutions using Ka × Kb = Kw
Scan these before reading
vocab
Organic acidAn organic compound that donates a proton; carboxylic acids (RCOOH) are the most common class.
pKapKa = −log₁₀(Ka); lower pKa = stronger acid; ethanoic acid pKa ≈ 4.76; electron-withdrawing groups lower pKa.
Inductive effectElectron-withdrawing groups (e.g., Cl, F) near the –COOH group stabilise the carboxylate anion (RCOO⁻), increasing acid strength.
Organic baseAn organic compound that accepts a proton; amines (RNH₂) are the most common organic bases.
Basicity of aminesElectron-donating alkyl groups increase Kb (more basic than NH₃); electron-withdrawing aryl groups decrease Kb.
ZwitterionA species with both positive and negative charges; amino acids exist as zwitterions at their isoelectric point. Amino acids are the monomers that proteins hydrolyse into, each peptide (amide) bond cleaved by water, the biomolecule hydrolysis dot point developed in Lesson 23.
📚 Core Content
2

Explain: interpret Ka and pKa

01
The pKa Scale: Quantifying Organic Acid Strength

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.

Compares strong/carboxylic acids, phenol, and water/alcohols by pKa and their reactivity with NaHCO₃ and NaOH.

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.

Beyond the syllabus (extension): Phenol and other aromatic (benzene-ring) compounds are not part of NSW HSC Chemistry, so every phenol reference in this lesson is enrichment only. The in-scope comparison you are assessed on is carboxylic acid vs alcohol; treat the phenol "middle rung" as optional extra context.
Exam tip (in scope): When comparing the acid strengths of a carboxylic acid and an alcohol, include the Ka/pKa values. "Ethanoic acid (pKa 4.74, Ka 1.8 × 10⁻⁵) is far stronger than ethanol (pKa ~16)" earns more marks than "ethanoic acid is stronger than ethanol."
Common Error: "Carboxylic acids are strong acids." Wrong, carboxylic acids are WEAK acids (Ka << 1, only ~1.3% ionised at 0.1 mol/L). They are the strongest organic acid class in Module 7, but still weak by definition.

Which organic compound has the LOWEST pKa (strongest acid)?

Precision: pKa compares equilibrium acid strength, not how quickly a reaction occurs. Reaction rate depends on kinetics and conditions.
3

Represent: stabilise the conjugate base

02
Why the Three Classes Have Different pKa Values: Resonance Revisited

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 comparison across carboxylate, phenoxide, and alkoxide

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).

HSC Must-Do: For any question asking WHY carboxylic acids are stronger acids than alcohols, give three components: (1) identify the conjugate bases; (2) describe resonance stabilisation difference (carboxylate = delocalised across 2 O; alkoxide = localised on 1 O); (3) connect to Ka/pKa (more stable A⁻ → weaker base → equilibrium lies right → stronger acid).
Common Error: "Carboxylic acids are stronger because the O-H bond is weaker." Bond strength is not the primary HSC explanation, thermodynamic stability of the conjugate base (resonance) is expected.

Why does ethanoic acid (pKa ~5) have a much lower pKa than ethanol (pKa ~16)?

4

Respond: predict diagnostic observations

03
Diagnostic Reactions: Using Reagents to Distinguish Acid Strength

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.

✓ Ethanoic acid (pKa 4.74) + NaHCO₃ → CH₃COO⁻Na⁺ + H₂O + CO₂↑
Because ethanoic acid (pKa 4.74) < H₂CO₃ (pKa 6.4), stronger acid on left ✓
✗ Phenol (pKa ~10) + NaHCO₃ → NO REACTION
Because phenol (pKa 10) > H₂CO₃ (pKa 6.4), product acid would be stronger → reaction reversed ✗
✓ Phenol (pKa ~10) + NaOH → C₆H₅O⁻Na⁺ + H₂O
NaOH is a strong base, no pKa constraint; deprotonates phenol ✓
Test Reagent Carboxylic Acid (pKa ~5) Phenol (pKa ~10) Alcohol (pKa ~16)
LitmusTurns red ✓Turns red ✓No change
NaHCO₃ solutionCO₂ gas ✓No reaction ✗No reaction ✗
Na₂CO₃ solutionCO₂ gas ✓Borderline/faintNo reaction ✗
NaOH solutionSalt + H₂O ✓Sodium phenoxide ✓No reaction ✗
Na metalH₂ gas ✓H₂ gas ✓H₂ gas ✓
Diagnostic flowchart for identifying organic acid class as a sequence

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.

Exam tip (in scope): The NaHCO₃ test is the key discriminating test: only a carboxylic acid produces CO₂ with NaHCO₃, so it separates the acid from an alcohol in NSW identification questions. Extension: the further NaOH step to separate phenol from alcohol involves aromatic phenol, which is beyond NSW HSC Chemistry.
Common Error: "Phenol reacts with NaHCO₃." This is wrong. Phenol (pKa ~10) is weaker than H₂CO₃ (pKa ~6.4), the reaction is thermodynamically unfavourable. Na metal reacts with ALL three classes (all have O-H), so it confirms O-H presence but cannot distinguish between classes.

True or False: Phenol reacts with sodium hydrogen carbonate (NaHCO₃) solution to produce CO₂ gas, because phenol is an acid.

5

Feedback: compare amines and amides

04
Organic Bases: Amines, Amides, and the Basicity Scale

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.

Alkylamines
e.g. ethylamine
Kb ~4 × 10⁻⁴
free lone pair on N (not involved in adjacent C=O) + alkyl electron donation → stronger than NH₃
Ammonia (ref)
NH₃
Kb = 1.8 × 10⁻⁵
Reference point for comparison
Aryl amines
e.g. aniline
Kb ~4 × 10⁻¹⁰
Lone pair delocalised into benzene ring → much less available
Amides
e.g. ethanamide
Kb ~10⁻¹⁵
Lone pair fully delocalised into C=O → essentially not basic

Predicting reaction direction using pKa: If pKa(acid on left) < pKa(conjugate acid of base on left) → reaction proceeds right.

✓ Ethanoic acid (pKa 4.74) + ethylamine → proceeds
Conjugate acid of ethylamine pKa ~10.6 → ethanoic acid (4.74) << 10.6 → forward spontaneous

✗ Ethanol (pKa ~16) + methylamine → no significant reaction
Ethanol (pKa 16) > conjugate acid of methylamine (pKa ~10.6) → wrong direction ✗
HSC Must-Do: Use Ka × Kb = Kw = 10⁻¹⁴ to find unknown Ka or Kb. If Kb(ethylamine) = 4 × 10⁻⁴, then Ka(ethylammonium) = 10⁻¹⁴ / (4 × 10⁻⁴) = 2.5 × 10⁻¹¹ → pKa ≈ 10.6. This pKa comparison then governs reaction prediction.
Common Error: "Any amine reacts with any O-H compound." Only true if the acid is stronger than the conjugate acid of the amine (pKa of acid < pKa of amine's conjugate acid). Ethylamine reacts with ethanoic acid (pKa 4.74 < 10.6 ✓) but NOT with ethanol (pKa ~16 > 10.6 ✗).

Arrange ethylamine, aniline, and ethanamide in order of INCREASING base strength:

6

Apply: identify and rank unknowns

📋 Worked Examples
WE
Predicting Reactions Using pKa
GIVEN

(a) Ethanoic acid + Na₂CO₃  (b) Phenol + NaHCO₃  (c) Propan-1-amine + HCl  (d) Ethanamide + HCl

GIVEN DATA

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

1, (a)

Ethanoic acid pKa = 4.74 < H₂CO₃ pKa = 6.4 → ethanoic acid stronger → PROCEEDS. CO₂ evolved.
2CH₃COOH + Na₂CO₃ → 2CH₃COO⁻Na⁺ + H₂O + CO₂↑

2, (b)

Phenol pKa = 10 > H₂CO₃ pKa = 6.4 → phenol weaker than H₂CO₃ → DOES NOT PROCEED. No CO₂ evolved.

3, (c)

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.

4, (d)

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.

(a) Proceeds ✓, CO₂ evolved. (b) Does not proceed ✗. (c) Proceeds completely ✓, salt formed. (d) Does not proceed ✗, amide is not basic.
Step 1 of 6
WE
Identifying Unknown Compounds from Test Results
GIVEN

Four compounds (A–D) tested. Compounds are butanoic acid, butan-1-ol, phenol, butylamine.

TestABCD
LitmusRedNo changeRedNo change
NaHCO₃CO₂ gasNo reactionNo reactionNo reaction
NaOHReactsNo reactionReactsNo reaction
Na metalH₂ gasNo reactionH₂ gasH₂ gas
1

A: Red litmus + CO₂ with NaHCO₃ + reacts NaOH + H₂ with Na → only carboxylic acids react with NaHCO₃ to give CO₂. A = butanoic acid.

2

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).

3

B: No litmus change + no reactions with acidic reagents + no Na reaction. No O-H bond. Is a base. B = butylamine.

4

D: By elimination: D = butan-1-ol. No litmus change, no NaHCO₃/NaOH reaction, but H₂ with Na metal (O-H present).

A = butanoic acid | B = butylamine | C = phenol | D = butan-1-ol
WE
Comparing Acid Strengths and pH Calculation
GIVEN

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.

1, pH order

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)

2, Two tests

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).

3, Ionic equation + pH

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)

Increasing pH: X (2.87) < Y (5.5) < Z (7.0). Tests: NaHCO₃ identifies X; NaOH separates Y from Z. pH of sodium ethanoate solution ≈ 8.9.
→
Key Formulas & pKa Values
Acid dissociation constant
Ka = [A⁻][H₃O⁺] / [HA]  |  pKa = −log₁₀(Ka)
Lower pKa = stronger acid; each unit = 10× change in acid strength
Base dissociation constant
Kb = [BH⁺][OH⁻] / [B]  |  pKb = −log₁₀(Kb)
Ka × Kb = Kw = 10⁻¹⁴
pKa reference values
Carboxylic acids ~4–5 | Phenol ~10 | Alcohols ~16
NaHCO₃ test: only acids with pKa < 6.4 produce CO₂
pKb reference values
Alkylamine Kb ~10⁻⁴ | Arylamine ~10⁻¹⁰ | Amide ~10⁻¹⁵
Lone pair availability determines base strength
Interactive Tool, Organic Acid-Base Strength Explorer Open fullscreen ↗
Use the Organic Acids tool. Carboxylic acids are classified as WEAK acids because they…
🔀 Sort the Steps +7 XP
Arrange these steps for comparing the acid strength of two organic acids using pKa values:
Determine which conjugate base (A⁻) is more stabilised by inductive or resonance effects
The more stabilised conjugate base corresponds to the stronger acid (greater Ka, lower pKa)
State which compound is the stronger acid and justify using structural evidence
Write the ionisation equation for each acid: HA ⇌ H⁺ + A⁻
Identify any electron-withdrawing groups (e.g. halogens) near the carboxyl group
7

Practice: classify and justify acid-base behaviour

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Practice Activity
A
Acid-Base Classification and Prediction

For each of the following, classify the compound and predict its behaviour with NaHCO₃ and NaOH. Justify using pKa values.

  1. Pentanoic acid (pKa ~4.8), does it react with NaHCO₃? With NaOH?
  2. Phenol (pKa ~10), does it react with NaHCO₃? With NaOH?
  3. Butan-1-ol (pKa ~16), does it react with NaHCO₃? With NaOH?
❓ Check Your Understanding
SA
Short Answer

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.

Reveal Answers

Multiple Choice Answers

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.

Short Answer Sample Answers

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.

Revisit: The Pharmacist's Three Solutions

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).

8

Review: retrieve acid-base predictions

Check what actually stuck
🏆 Review, Key Ideas
Quick Recall, Organic Acids & Bases

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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