Chemistry • Year 12 • Module 6 • Lesson 7
Conjugate Pairs, Amphiprotic Substances & Water's Role
Lock in the one-proton rule, the core amphiprotic species, and the meaning of Kw before moving to calculations.
1. Term–definition match
Write the matching term from this list in the right-hand column. Each term is used once. conjugate acid • conjugate base • conjugate acid-base pair • amphiprotic • autoionisation • Kw • amphoteric • pOH • neutral solution • amphiprotic intermediate 10 marks
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The species formed when a Brønsted-Lowry base accepts exactly one proton. | |
| 1.2 | The species formed when a Brønsted-Lowry acid donates exactly one proton. | |
| 1.3 | Two species on opposite sides of an equilibrium that differ by exactly one H+. | |
| 1.4 | A substance that can both donate and accept a proton depending on its reaction partner (Brønsted-Lowry definition). | |
| 1.5 | The broader term for a substance that can react with both acids and bases; includes Lewis acid-base behaviour. | |
| 1.6 | The spontaneous proton transfer between two water molecules: 2H2O ⇌ H3O+ + OH−. | |
| 1.7 | The equilibrium constant = [H3O+][OH−]; equals 1.0 × 10−14 at 25°C. | |
| 1.8 | −log[OH−]; at 25°C, pH + this quantity = 14. | |
| 1.9 | A solution in which [H3O+] = [OH−]; occurs at pH 7.00 only at 25°C. | |
| 1.10 | An ion that has already lost at least one proton from a polyprotic acid but still retains at least one more ionisable proton; e.g. HCO3−, H2PO4−. |
2. True or false — with correction
Circle T or F. If false, write the corrected statement on the line below. 10 marks (1 T/F + 1 correction each)
2.1 The conjugate base of H2SO4 is SO42− because sulfuric acid loses two protons. T / F
2.2 Water is amphiprotic because in the autoionisation 2H2O ⇌ H3O+ + OH−, one molecule donates and one molecule accepts a proton. T / F
2.3 A pure water solution at 37°C has pH 7.00 because it is neutral. T / F
2.4 The NESA-named salt Na2HPO4 provides the amphiprotic ion HPO42− in solution. T / F
2.5 A stronger acid has a stronger conjugate base because both can easily transfer protons. T / F
3. Cloze — fill the blanks
Complete the paragraph using the word bank below. Each word is used once. 8 marks
Word bank: amphiprotic • one • 14 • 10−14 • weaker • opposite • autoionisation • neutral
A conjugate acid-base pair consists of two species that differ by exactly __________ proton, appear on __________ sides of the equilibrium equation, and have charges differing by exactly one unit. The inverse strength rule states that a stronger acid has a __________ conjugate base. The process by which water reacts with itself is called __________. This produces H3O+ and OH− in equal amounts, making pure water __________ at 25°C. The equilibrium constant for this process, Kw, equals __________ at 25°C. Taking −log of both sides gives the relationship pH + pOH = __________. Because HCO3− can both donate and accept a proton, it is described as __________.
4. Concept map — connect the key ideas
Draw labelled arrows between the six terms below. Each arrow must carry a linking phrase. Aim for at least 6 labelled arrows. 6 marks
Supplied terms: amphiprotic substance • conjugate acid-base pair • autoionisation of water • Kw • HCO3− • pH + pOH = 14.
5. Function recall
Answer each question in 1–2 sentences using precise terms. 8 marks (2 each)
5.1 What is the role of Kw in linking [H3O+] and [OH−] in any aqueous solution?
5.2 What does the inverse strength relationship tell us about the conjugate base of a strong acid?
5.3 Why is Kw larger at 60°C than at 25°C? Use Le Chatelier’s Principle.
5.4 Explain why HCO3− added to a NaOH solution acts as an acid rather than a base.
Q1 — Term–definition match
1.1 conjugate acid • 1.2 conjugate base • 1.3 conjugate acid-base pair • 1.4 amphiprotic • 1.5 amphoteric • 1.6 autoionisation • 1.7 Kw • 1.8 pOH • 1.9 neutral solution • 1.10 amphiprotic intermediate.
Q2 — True / false with corrections
2.1 False. The conjugate base of H2SO4 is HSO4−; removing exactly one proton gives HSO4−. Only then does a second proton removal produce SO42− (conjugate base of HSO4−). These are two separate conjugate pairs: H2SO4/HSO4− and HSO4−/SO42−.
2.2 True.
2.3 False. Pure water at 37°C is neutral (by definition, [H3O+] = [OH−]) but its pH is approximately 6.81, not 7.00. Neutral pH = 7.00 only at 25°C.
2.4 True.
2.5 False. The inverse strength rule states a stronger acid has a weaker conjugate base (less tendency to accept H+ back). A strong acid fully donates its proton; its conjugate base has almost no tendency to re-accept it.
Q3 — Cloze answers
one / opposite / weaker / autoionisation / neutral / 10−14 / 14 / amphiprotic.
Q4 — Sample concept map
Acceptable arrows include: amphiprotic substance → is an example of → HCO3−; HCO3− → forms two → conjugate acid-base pairs; autoionisation of water → defines → Kw; Kw → leads to → pH + pOH = 14; autoionisation of water → proves water is → amphiprotic substance; conjugate acid-base pair → governs → Kw. Award 1 mark per valid labelled arrow, maximum 6.
Q5.1 — Role of Kw
Kw = [H3O+][OH−] is a fixed constraint at a given temperature. If [H3O+] is known, [OH−] = Kw / [H3O+] can always be calculated, and vice versa. It applies to all aqueous solutions, not only pure water.
Q5.2 — Inverse strength rule
The conjugate base of a strong acid is an extremely weak base with almost no tendency to re-accept a proton. For example, the conjugate base of HCl (a strong acid) is Cl−, which is negligible as a base.
Q5.3 — Kw and temperature
Water self-ionisation is endothermic (heat is absorbed). By Le Chatelier’s Principle, increasing temperature shifts the equilibrium to the right, increasing [H3O+] and [OH−], so Kw = [H3O+][OH−] increases.
Q5.4 — HCO3− in NaOH
In NaOH solution, OH− ions are present as a strong base. OH− pulls a proton from HCO3−, causing HCO3− to act as an acid (proton donor), producing CO32− + H2O. The reaction partner determines which mode HCO3− adopts.