Year 12 Chemistry Module 6, Acid/Base Reactions ⏱ ~45 min Lesson 7 of 19 IQ2

Conjugate Pairs, Amphiprotic Substances & Water's Role

In 1893, Friedrich Kohlrausch measured the electrical conductivity of the purest water he could produce at 25°C and found it was not zero, it was 5.5 × 10⁻⁸ S cm⁻¹, meaning water conducted electricity ever so slightly. He concluded that water itself must ionise into H⁺ and OH⁻ ions. That tiny conductance established Kw = 10⁻¹⁴, the number that sets the pH of every aqueous solution ever measured.

Today's hook, In 1893, Friedrich Kohlrausch measured ultra-pure water and found it conducted a tiny electrical current, proving water spontaneously ionises. That experiment established Kw = 10⁻¹⁴ and is why every aqueous solution has both H⁺ and OH⁻ in it. How can water be simultaneously an acid and a base in its own ionisation?

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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

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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

One Compound, Two Behaviours

A chemist prepares three solutions: one containing sodium hydrogen carbonate (NaHCO₃, baking soda), one containing hydrochloric acid, and one containing sodium hydroxide. She adds NaHCO₃ to the HCl solution, bubbles form vigorously and the solution becomes less acidic. She then adds NaHCO₃ to the NaOH solution, the solution becomes less basic, and no bubbles form.

The same compound, in the same amount, made one solution less acidic and another solution less basic. Before you read on, write down your explanation: how can one substance act as a base in one reaction and an acid in another? Is this a contradiction, or does it reveal something fundamental about how acid-base reactions work?

Learning Intentions

Know

A conjugate acid-base pair differs by one H⁺
Amphiprotic substances can act as both acids and bases (e.g. HSO₄⁻, HCO₃⁻, H₂O)
Water self-ionises with Kw = [H₃O⁺][OH⁻] = 10⁻¹⁴ at 25°C

Understand

The inverse strength relationship: stronger acid → weaker conjugate base
Why intermediate ions of polyprotic acids are amphiprotic
How Kw allows calculation of [H₃O⁺] or [OH⁻] in any aqueous solution

Can Do

Identify conjugate acids and bases for any given species
Write balanced equations showing amphiprotic substances acting as acids and bases
Calculate [H₃O⁺] and [OH⁻] using Kw in pure water and mixed solutions

Key Terms

Conjugate acid

The species formed when a Brønsted-Lowry base accepts a proton (H⁺).

Conjugate base

The species formed when a Brønsted-Lowry acid donates a proton (H⁺).

Conjugate acid-base pair

Two species differing by exactly one H⁺ that interconvert in equilibrium.

Amphiprotic substance

A substance that can donate or accept a proton depending on the reaction partner; e.g., HCO₃⁻, H₂O, HSO₄⁻.

Autoionisation of water

H₂O + H₂O ⇌ H₃O⁺ + OH⁻; occurs to a very small extent; Kw = [H₃O⁺][OH⁻] = 10⁻¹⁴.

Amphoteric vs amphiprotic

Amphiprotic: can donate/accept H⁺ (Brønsted-Lowry). Amphoteric: can react with both acids and bases (includes Lewis definition).

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Conjugate Pairs, Amphiprotic Species & Kw

Deep dive, Extension. Conjugate pairs, amphiprotic behaviour and water’s role are core. Full phosphate-chain constants, temperature-dependent Kw calculations, and the Lewis/amphoteric distinction are Extension.

Conjugate base: HA → A⁻ + H⁺ (A⁻ is the conjugate base of HA) A⁻ has one fewer H and one more negative charge than HA

Conjugate acid: B + H⁺ → BH⁺ (BH⁺ is the conjugate acid of B) BH⁺ has one more H and one less negative charge than B

Amphiprotic (as acid): HX ⇌ H⁺ + X⁻ | Amphiprotic (as base): HX + H⁺ ⇌ H₂X⁺ Can both donate AND accept H⁺ depending on reaction partner

Water self-ionisation: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C | At 25°C: [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L → pH = 7.00 (neutral)

pH + pOH = pKw = 14 at 25°C | [OH⁻] = Kw / [H₃O⁺] pKw = −log(Kw) | Kw increases with temperature (endothermic self-ionisation)

Content

Cross-lesson links: Conjugate pairs introduced in L01 are deepened here; the inverse strength rule (Ka × Kb = Kw) connects to calculations in L09. Amphiprotic species, especially HCO₃⁻ and H₂PO₄⁻, reappear in blood pH buffering (L08, L13). Kw = 10⁻¹⁴ is the constant used in every pH/pOH calculation from L08 onward.

Conjugate acid-base pairs

Conjugate Acid-Base Pairs: Deepening the Relationship

Every Brønsted-Lowry acid-base reaction produces a new acid and a new base on the product side, and the relationship between each parent species and its conjugate is always and only the transfer of a single proton, nothing more and nothing less.

The conjugate acid-base pair relationship was introduced in L01, but Module 6 requires deeper fluency: the ability to identify conjugate pairs in multi-step and polyprotic acid equilibria, to write both equations showing a species acting as acid and as base, and to use the inverse strength relationship to predict equilibrium direction.

Recall the rules: a conjugate base is formed by removing exactly one H⁺ from the acid, the conjugate base has one fewer H and one more negative charge. A conjugate acid is formed by adding exactly one H⁺ to the base, the conjugate acid has one more H and one less negative charge.

The inverse strength relationship is critical: a strong acid has an extremely weak conjugate base (essentially no tendency to accept H⁺ back). A weak acid has a relatively stronger conjugate base (meaningful tendency to accept H⁺ back, which is why the weak acid equilibrium does not go to completion). This inverse relationship is captured quantitatively as Ka × Kb = Kw for a conjugate pair.

The extended conjugate pair chain for phosphoric acid illustrates successive proton loss:

H₃PO₄ ⇌ H⁺ + H₂PO₄⁻ (Ka1 = 7.5 × 10⁻³)

H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻ (Ka2 = 6.2 × 10⁻⁸)

HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ (Ka3 = 4.8 × 10⁻¹³)

Each successive conjugate base is a weaker acid than its predecessor, removing a proton from an already-negative ion is increasingly difficult.

As acid → conjugate base

Donates H⁺ → H₂PO₄⁻

Donates H⁺ → HPO₄²⁻

Donates H⁺ → PO₄³⁻

Donates H⁺ → CO₃²⁻

As base → conjugate acid

Cannot meaningfully accept H⁺ (H₄PO₄⁺ negligible)

Accepts H⁺ → H₃PO₄

Accepts H⁺ → H₂PO₄⁻

Accepts H⁺ → H₂CO₃

Must Do: When asked to identify conjugate pairs, always verify three things: (1) the two species are on opposite sides of the equation; (2) they differ by exactly one H⁺; (3) the charge difference is exactly 1. Failing any one check means you have not identified a genuine conjugate pair.

Common Error: Students identify SO₄²⁻ as the conjugate base of H₂SO₄, removing two protons at once. The conjugate base of H₂SO₄ is HSO₄⁻ (one proton removed). The conjugate base of HSO₄⁻ is SO₄²⁻. These are two separate conjugate pair relationships, H₂SO₄/HSO₄⁻ and HSO₄⁻/SO₄²⁻, and they must not be conflated.

Exam Tip: In acid-base calculations, always write the balanced equation first, identify the conjugate pair, and state any assumptions before substituting into the Ka expression.

A conjugate base is formed by removing exactly one H⁺ (−1 charge change); a conjugate acid by adding one H⁺. Inverse strength rule: strong acid → negligible conjugate base; weak acid → stronger conjugate base; Ka × Kb = Kw for any pair. Conjugate base of H₂SO₄ = HSO₄⁻ (not SO₄²⁻, that removes two protons).

Pause, copy the highlighted definition into your book before moving on.

Which species is the conjugate base of HSO₄⁻?

Amphiprotic substances

Amphiprotic Substances: Definition and Key Examples

We just saw that conjugate pairs differ by exactly one H⁺ and that conjugate base strength is inversely related to the parent acid's Ka. That raises a question: What happens when a species is a conjugate base of one acid and the conjugate acid of another, can it behave as either? This card answers it → amphiprotic substances have both an ionisable H⁺ and an H⁺-accepting site, so they act as acid or base depending on the reaction partner.

An amphiprotic substance is one that can either donate or accept a proton depending entirely on its reaction partner, and this dual capability is a direct consequence of the substance being an intermediate in a polyprotic acid-base system.

A substance is amphiprotic if it contains at least one ionisable H⁺ (so it can act as an acid, proton donor) AND at least one lone pair or basic site capable of accepting a proton (so it can act as a base, proton acceptor).

In practice, the most important amphiprotic substances in HSC chemistry are the intermediate ions of polyprotic acids species that have already lost one proton from a polyprotic acid but still retain at least one more ionisable proton.

Water is the archetypal amphiprotic substance: it can donate H⁺ (acting as an acid, producing OH⁻) or accept H⁺ (acting as a base, producing H₃O⁺). The hydrogen carbonate ion (HCO₃⁻) is amphiprotic: it can donate its remaining H⁺ to a stronger base (weak acid: HCO₃⁻ ⇌ H⁺ + CO₃²⁻, Ka = 4.7 × 10⁻¹¹) or accept H⁺ from a stronger acid (as a base: HCO₃⁻ + H⁺ ⇌ H₂CO₃).

Acting as acid (proton donor)

H₂O ⇌ H⁺ + OH⁻
HCO₃⁻ ⇌ H⁺ + CO₃²⁻
H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻
HPO₄²⁻ ⇌ H⁺ + PO₄³⁻

Acting as base (proton acceptor)

H₂O + H⁺ → H₃O⁺
HCO₃⁻ + H⁺ ⇌ H₂CO₃
H₂PO₄⁻ + H⁺ ⇌ H₃PO₄
HPO₄²⁻ + H⁺ ⇌ H₂PO₄⁻

Must Do: For any HSC question involving an amphiprotic salt such as NaHCO₃, KH₂PO₄ or Na₂HPO₄, you must: (1) identify the amphiprotic ion; (2) write the equation showing it acting as an acid; (3) write the equation showing it acting as a base. Both equations are required, writing only one mode earns partial marks. These are explicitly NESA-named examples.

Common Error: Students confuse "amphiprotic" with "amphoteric." Amphoteric is broader, it includes Lewis acid-base behaviour. Amphiprotic specifically means a substance can both donate AND accept a proton (Brønsted-Lowry). In HSC Module 6, use "amphiprotic" for syllabus dot points. All amphiprotic substances are amphoteric, but not all amphoteric substances are amphiprotic.

Insight: Whether an amphiprotic substance acts as acid or base depends on its reaction partner. HCO₃⁻ acts as a base with HCl (HCl forces HCO₃⁻ to accept H⁺): HCO₃⁻ + HCl → H₂CO₃ + Cl⁻. HCO₃⁻ acts as an acid with NaOH (OH⁻ pulls H⁺ from HCO₃⁻): HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O. This is exactly the chemist's observation in the Think First.

An amphiprotic substance has at least one ionisable H⁺ AND can accept a proton, it acts as acid or base depending on the partner. Key examples: H₂O, HCO₃⁻, H₂PO₄⁻, HPO₄²⁻, HSO₄⁻. All are intermediate ions in polyprotic systems. "Amphiprotic" (Brønsted-Lowry, H⁺ transfer only) ≠ "amphoteric" (broader Lewis definition); use "amphiprotic" in HSC Module 6 answers.

Add the highlighted point to your notes before the check below.

True or false: All amphoteric substances are also amphiprotic.

Syllabus-named amphiprotic salts

The Syllabus Examples: NaHCO₃ and KH₂PO₄ in Full Detail

We just saw that amphiprotic substances contain both a donatable H⁺ and an H⁺-accepting site. That raises a question: Which specific amphiprotic salts does the syllabus give as examples? This card answers it → NaHCO₃ (amphiprotic ion HCO₃⁻) and KH₂PO₄ (amphiprotic ion H₂PO₄⁻); both equations must be written for each. Na₂HPO₄ is worked through as well, because the same chain rule covers it and it appears often in practice questions.

The NSW Chemistry Stage 6 syllabus names sodium hydrogen carbonate and potassium dihydrogen phosphate as its worked examples of “the amphiprotic nature of some salts”. They are examples rather than a closed list, so an exam may use a different amphiprotic salt, being able to write both the acid and the base equation for any of them is the competency that matters.

NaHCO₃, Sodium Hydrogen Carbonate

Dissolves completely to give Na⁺ and HCO₃⁻ ions. Na⁺ is the conjugate of NaOH (strong base), a neutral spectator. HCO₃⁻ is the amphiprotic ion, this is the baking soda from the opening scenario, and it is why the same white powder can calm both an acid and a base.

HCO₃⁻ acting as an acid (proton donor): HCO₃⁻(aq) ⇌ H⁺(aq) + CO₃²⁻(aq) Ka2 = 4.7 × 10⁻¹¹
HCO₃⁻ acting as a base (proton acceptor): HCO₃⁻(aq) + H⁺(aq) ⇌ H₂CO₃(aq) (reverse of Ka1)

Na₂HPO₄, Sodium Hydrogen Phosphate (additional example, not syllabus-named)

Dissolves completely to give Na⁺ and HPO₄²⁻ ions. Na⁺ is the conjugate of NaOH (strong base), it is a neutral spectator. HPO₄²⁻ is the amphiprotic ion. It sits in the middle of the phosphate system, it has already lost two protons from H₃PO₄ and retains one more ionisable proton.

HPO₄²⁻ acting as an acid (proton donor): HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq) Ka3 = 4.8 × 10⁻¹³
HPO₄²⁻ acting as a base (proton acceptor): HPO₄²⁻(aq) + H⁺(aq) ⇌ H₂PO₄⁻(aq) (reverse of Ka2)

KH₂PO₄, Potassium Dihydrogen Phosphate

Dissolves completely to give K⁺ and H₂PO₄⁻ ions. K⁺ is the conjugate of KOH (strong base), neutral spectator. H₂PO₄⁻ is the amphiprotic ion. It has lost one proton (Ka1 step) and retains two more ionisable protons.

H₂PO₄⁻ acting as an acid: H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq) Ka2 = 6.2 × 10⁻⁸
H₂PO₄⁻ acting as a base: H₂PO₄⁻(aq) + H⁺(aq) ⇌ H₃PO₄(aq) (reverse of Ka1)

NaHCO₃ (syllabus)

Amphiprotic ion: HCO₃⁻
As acid: HCO₃⁻ ⇌ H⁺ + CO₃²⁻
As base: HCO₃⁻ + H⁺ ⇌ H₂CO₃

Na₂HPO₄ (extra)

Amphiprotic ion: HPO₄²⁻
As acid: HPO₄²⁻ ⇌ H⁺ + PO₄³⁻
As base: HPO₄²⁻ + H⁺ ⇌ H₂PO₄⁻

KH₂PO₄ (syllabus)

Amphiprotic ion: H₂PO₄⁻
As acid: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻
As base: H₂PO₄⁻ + H⁺ ⇌ H₃PO₄

Chain-navigation method: For each ion, identify its position in the phosphate chain (H₃PO₄ → H₂PO₄⁻ → HPO₄²⁻ → PO₄³⁻). Acting as acid = move one step right (lose one H⁺). Acting as base = move one step left (gain one H⁺). This produces the correct equation every time without memorising four separate equations.

Common Error: Students write HPO₄²⁻ losing two protons to form PO₄³⁻ in one step, or H₂PO₄⁻ losing two protons to form PO₄³⁻ in one step. Amphiprotic behaviour always involves the transfer of exactly one proton per equation one step in the chain. Two-proton transfers in a single equation are incorrect.

Syllabus-named amphiprotic salts: NaHCO₃ → HCO₃⁻ (acid: HCO₃⁻ ⇌ H⁺ + CO₃²⁻; base: HCO₃⁻ + H⁺ ⇌ H₂CO₃); KH₂PO₄ → H₂PO₄⁻ (acid: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻; base: H₂PO₄⁻ + H⁺ ⇌ H₃PO₄). Also know Na₂HPO₄ → HPO₄²⁻ (acid: HPO₄²⁻ ⇌ H⁺ + PO₄³⁻; base: HPO₄²⁻ + H⁺ ⇌ H₂PO₄⁻). Chain rule: acting as acid = one step right; acting as base = one step left. One proton per equation only.

Pause, write the highlighted definition into your book.

Complete: The amphiprotic ion in Na₂HPO₄ is . When acting as an acid it forms . When acting as a base it forms .

The phosphate proton chain

The Phosphate Proton Chain, Visual Summary

We just saw the four equations for Na₂HPO₄ and KH₂PO₄, each amphiprotic ion moves one step in the phosphate chain. That raises a question: What does the full chain look like at a glance, and which species can act only as acid, only as base, or as both? This card answers it → the chain diagram maps all four phosphate species with their Ka values and direction rules.

H₃PO₄ loses protons stepwise through two amphiprotic intermediates to PO₄³⁻.

Red arrows = acting as acid (−H⁺, move right). Green arrows = acting as base (+H⁺, move left). H₂PO₄⁻ and HPO₄²⁻ are amphiprotic, they can move either direction.

Phosphate chain: H₃PO₄ → H₂PO₄⁻ (Ka1 7.5 × 10⁻³) → HPO₄²⁻ (Ka2 6.2 × 10⁻⁸) → PO₄³⁻ (Ka3 4.8 × 10⁻¹³). H₃PO₄ = acid only (no further H⁺ to accept back). PO₄³⁻ = base only (no ionisable H⁺). H₂PO₄⁻ and HPO₄²⁻ = amphiprotic. Each Ka decreases because removing H⁺ from an increasingly negative ion is harder.

Add the highlighted point to your notes before the check below.

In the phosphate chain, which species can ONLY act as a base (cannot donate H⁺)?

Water's self-ionisation and Kw

Water's Self-Ionisation and Kw

We just saw the full phosphate chain showing H₂O, HCO₃⁻, H₂PO₄⁻ and HPO₄²⁻ as amphiprotic. That raises a question: Water ionises itself, what is that equilibrium constant, and why does it mean pH 7 is not universally "neutral"? This card answers it → Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C; neutral means [H₃O⁺] = [OH⁻], not pH = 7.

In 1893, Friedrich Kohlrausch purified water to an unprecedented degree and measured its conductivity: 5.5 × 10⁻⁸ S cm⁻¹, tiny, but not zero. Ions had to be present. The only source was water itself, ionising spontaneously. The tiny equilibrium constant he derived, Kw = 1.0 × 10⁻¹⁴ at 25°C, is the number that sets the neutral point at pH 7 and underpins every pH calculation that follows in this module.

Pure water undergoes self-ionisation (autoprotolysis), a small but thermodynamically significant fraction of water molecules transfer a proton from one molecule to another:

2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)

In this reaction, one water molecule acts as the Brønsted-Lowry acid (proton donor → OH⁻) and the other acts as the Brønsted-Lowry base (proton acceptor → H₃O⁺). This confirms water as amphiprotic, in the same reaction it simultaneously acts as both acid and base.

The equilibrium constant for this reaction is the water dissociation constant: Kw = [H₃O⁺][OH⁻]. At 25°C, Kw = 1.0 × 10⁻¹⁴. The concentration of pure water is omitted from the expression (pure liquid has activity = 1).

Because self-ionisation produces equal amounts: [H₃O⁺] = [OH⁻] = √(1.0 × 10⁻¹⁴) = 1.0 × 10⁻⁷ mol/L at 25°C → pH = 7.00. Kw is temperature-dependent: it increases with temperature because the self-ionisation is endothermic (Le Chatelier: increasing T shifts right, increasing both [H₃O⁺] and [OH⁻]).

Mini-rule

Neutral means [H₃O⁺] = [OH⁻], not automatically pH 7. Pure water is always neutral, but its pH equals 7.00 only at 25°C. Because self-ionisation is endothermic, heating water raises Kw, so neutral pH falls below 7 (e.g. about 6.63 at 50°C) while cooling raises it above 7, in every case [H₃O⁺] still equals [OH⁻]. Judge acidity/basicity by comparing [H₃O⁺] with [OH⁻] at the stated temperature, not by comparing pH with 7.

10°C

Kw: 2.9 × 10⁻¹⁵
Neutral pH: 7.27
pH > 7 but water is neutral

25°C

Kw: 1.0 × 10⁻¹⁴
Neutral pH: 7.00
Standard reference temperature

37°C

Kw: 2.4 × 10⁻¹⁴
Neutral pH: 6.81
Blood temperature, neutral < 7

60°C

Kw: 9.6 × 10⁻¹⁴
Neutral pH: 6.51
Hot water, neutral well below 7

Must Do: Neutral pH = 7 ONLY at 25°C. At any other temperature, neutral pH = −log(√Kw at that temperature). HSC questions may specify a non-standard temperature and ask whether a solution with pH 7 is acidic, basic, or neutral. At 37°C, pH 7 is slightly basic (neutral pH ≈ 6.81 and pH 7 > 6.81 → [H₃O⁺] < [OH⁻]).

Common Error: "Pure water has pH 7 because it is neutral." The causal relationship is backwards. Pure water is neutral because [H₃O⁺] = [OH⁻], that is the definition. At 25°C, this happens to give pH 7. At 60°C, pure water is still neutral but pH ≈ 6.51, not 7. Neutrality is defined by the equality of [H₃O⁺] and [OH⁻], not by the number 7.

Insight: Self-ionisation occurs even in strongly acidic or basic solutions, Kw always applies. In 1.0 mol/L HCl: [H₃O⁺] = 1.0 mol/L and [OH⁻] = Kw/[H₃O⁺] = 1.0 × 10⁻¹⁴ mol/L. OH⁻ ions are still present, just at extraordinarily low concentration. This is important for calculating pH of mixtures and understanding why Kw applies in all aqueous solutions, not just pure water.

Water self-ionises: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq); Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C. Neutral means [H₃O⁺] = [OH⁻], not pH = 7. pH 7 = neutral only at 25°C; at 37°C neutral pH ≈ 6.81. Kw increases with temperature (endothermic equilibrium); Kw applies to every aqueous solution, not just pure water.

Pause, copy the highlighted definition into your book before moving on.

At 37°C (Kw = 2.4 × 10⁻¹⁴), a solution has pH 7.0. This solution is:

Using Kw in any solution

Using Kw: [H₃O⁺] and [OH⁻] in Any Aqueous Solution

We just saw that Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C and that neutral means [H₃O⁺] = [OH⁻]. That raises a question: How do you use Kw as a calculation tool in acid and base problems, not just in pure water? This card answers it → [OH⁻] = Kw/[H₃O⁺]; pH + pOH = 14 at 25°C; gives the standard route to base pH calculations.

Kw is not just a curiosity about pure water, it is a universal constraint on every aqueous solution that links [H₃O⁺] and [OH⁻] at all times, allowing you to calculate one from the other regardless of what else is dissolved.

In any aqueous solution at 25°C, [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴. This is a fixed constraint, if [H₃O⁺] increases (acid added), [OH⁻] must decrease proportionally; if [OH⁻] increases (base added), [H₃O⁺] must decrease.

Three important consequences for HSC calculations:

Given [H₃O⁺], you can always find [OH⁻] = Kw/[H₃O⁺], and vice versa.
Taking −log of both sides: −log[H₃O⁺] + (−log[OH⁻]) = −log(Kw) → pH + pOH = pKw = 14 at 25°C. Essential for base pH calculations in L08.
Classification at 25°C: acidic if [H₃O⁺] > [OH⁻] → pH < 7; basic if [OH⁻] > [H₃O⁺] → pH > 7; neutral if [H₃O⁺] = [OH⁻] → pH = 7. These use 7 as reference only at 25°C.

Acidic

[H₃O⁺] > [OH⁻]
pH < 7 at 25°C
[OH⁻] = 1.0 × 10⁻¹⁴ / [H₃O⁺]

Neutral

[H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷
pH = 7 at 25°C
[OH⁻] = 1.0 × 10⁻⁷

Basic

[OH⁻] > [H₃O⁺]
pH > 7 at 25°C
[OH⁻] = 1.0 × 10⁻¹⁴ / [H₃O⁺]

Must Do: Memorise and be able to derive: pH + pOH = 14 at 25°C. This is used in every base pH calculation in L08. Standard method for strong base: calculate [OH⁻] → pOH = −log[OH⁻] → pH = 14 − pOH. Derive it once from Kw = [H₃O⁺][OH⁻] so you understand where it comes from.

Common Error: Students apply pH + pOH = 14 at temperatures other than 25°C. At 37°C, Kw = 2.4 × 10⁻¹⁴ → pKw = −log(2.4 × 10⁻¹⁴) = 13.62. At 37°C, pH + pOH = 13.62, not 14. Using 14 at non-standard temperatures gives wrong pH values. Unless a question specifies otherwise, assume 25°C and pH + pOH = 14.

Kw links [H₃O⁺] and [OH⁻] in every aqueous solution: [OH⁻] = Kw/[H₃O⁺]. Taking −log: pH + pOH = pKw = 14 at 25°C ONLY (at 37°C pKw = 13.62). Base pH route: [OH⁻] → pOH = −log[OH⁻] → pH = 14 − pOH (25°C). Acidic: [H₃O⁺] > [OH⁻]; basic: [OH⁻] > [H₃O⁺], these definitions hold at any temperature.

Pause, copy the highlighted definition into your book before moving on.

At 25°C, a solution has [OH⁻] = 2.0 × 10⁻³ mol/L. What is [H₃O⁺]?

Misconceptions, worked examples and activities

Common Misconceptions

Conjugate Pairs & Amphiprotic Substances

"SO₄²⁻ is the conjugate base of H₂SO₄.", Incorrect. A conjugate base is formed by removing exactly ONE proton. The conjugate base of H₂SO₄ is HSO₄⁻. Only HSO₄⁻ losing a second proton gives SO₄²⁻, making HSO₄⁻/SO₄²⁻ a separate conjugate pair.

"Amphiprotic and amphoteric mean the same thing.", Incorrect. Amphoteric is broader (includes Lewis acid-base). Amphiprotic specifically means proton donor AND acceptor (Brønsted-Lowry). In Module 6, always use "amphiprotic."

"Neutral pH is always 7.", Incorrect. Neutral pH = −log(√Kw) and depends on temperature. At 37°C, neutral pH ≈ 6.81. pH 7 at 37°C is slightly basic. Only at 25°C does neutral pH equal 7.00.

"HPO₄²⁻ can't donate a proton, it's already negatively charged.", Incorrect. Negative charge does not prevent proton donation. HPO₄²⁻ retains one O–H bond (Ka3 = 4.8 × 10⁻¹³). It can donate this H⁺ to a sufficiently strong base, forming PO₄³⁻.

Conjugate Pairs and Amphiprotic Equations

For the reaction H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O: (a) identify the Brønsted-Lowry acid and base and write both conjugate pairs. (b) Write equations showing H₂PO₄⁻ acting as (i) an acid and (ii) a base, and state which NESA-named salt provides H₂PO₄⁻ ions.

GIVEN: Reaction H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O. Need to track H⁺ transfer.
FIND: BL acid, BL base, both conjugate pairs; both equations for amphiprotic behaviour; NESA salt.

METHOD (a, identify acid/base): H₂PO₄⁻ → HPO₄²⁻: H₂PO₄⁻ has lost one H⁺ → H₂PO₄⁻ is the BL acid; HPO₄²⁻ is its conjugate base. OH⁻ → H₂O: OH⁻ has gained one H⁺ → OH⁻ is the BL base; H₂O is its conjugate acid.

Conjugate pairs: Pair 1: H₂PO₄⁻ (acid) / HPO₄²⁻ (conjugate base), differ by one H⁺, opposite sides ✓. Pair 2: OH⁻ (base) / H₂O (conjugate acid), differ by one H⁺, opposite sides ✓.

METHOD (b, H₂PO₄⁻ as acid): H₂PO₄⁻ donates H⁺ → moves one step right in phosphate chain. Equation: H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq). Use ⇌, this is weak acid ionisation (Ka2 = 6.2 × 10⁻⁸).
H₂PO₄⁻ as base: Accepts H⁺ → moves one step left in chain. Equation: H₂PO₄⁻(aq) + H⁺(aq) ⇌ H₃PO₄(aq). Use ⇌, this is the reverse of Ka1.

ANSWER: (a) Acid = H₂PO₄⁻; conjugate base = HPO₄²⁻. Base = OH⁻; conjugate acid = H₂O. Conjugate pairs: H₂PO₄⁻/HPO₄²⁻ and OH⁻/H₂O. (b)(i) H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq). (ii) H₂PO₄⁻(aq) + H⁺(aq) ⇌ H₃PO₄(aq). NESA salt: KH₂PO₄ (potassium dihydrogen phosphate).

Kw Calculations at Non-Standard Temperature

At 50°C, Kw = 5.5 × 10⁻¹⁴. (a) Calculate the pH of pure water at 50°C. (b) A solution at 50°C has pH 6.8. Is it acidic, basic, or neutral? Justify. (c) Calculate [OH⁻] in a solution at 50°C with [H₃O⁺] = 2.5 × 10⁻³ mol/L. Is this solution acidic or basic?

GIVEN: Kw = 5.5 × 10⁻¹⁴ at 50°C.
FIND: (a) pH of pure water at 50°C; (b) classification of pH 6.8; (c) [OH⁻] and classification for [H₃O⁺] = 2.5 × 10⁻³.

METHOD (a): In pure water, [H₃O⁺] = [OH⁻] = √Kw = √(5.5 × 10⁻¹⁴) = 2.35 × 10⁻⁷ mol/L. pH = −log(2.35 × 10⁻⁷) = 6.63. Pure water at 50°C is neutral (by definition, [H₃O⁺] = [OH⁻]) but has pH 6.63, not 7.

METHOD (b): Neutral pH at 50°C = 6.63. A solution with pH 6.8 has [H₃O⁺] = 10⁻⁶·⁸ = 1.58 × 10⁻⁷ mol/L. Since pH 6.8 > 6.63 (neutral pH at 50°C), [H₃O⁺] < [OH⁻] → the solution is basic even though its pH is below 7.

METHOD (c): [OH⁻] = Kw / [H₃O⁺] = (5.5 × 10⁻¹⁴) / (2.5 × 10⁻³) = 2.2 × 10⁻¹¹ mol/L. Compare: [H₃O⁺] = 2.5 × 10⁻³ >> [OH⁻] = 2.2 × 10⁻¹¹ → acidic. pH = −log(2.5 × 10⁻³) = 2.60, well below neutral pH 6.63 at 50°C.

ANSWER: (a) pH of pure water at 50°C = 6.63 (neutral but not pH 7). (b) pH 6.8 at 50°C is basic. Neutral pH is 6.63 at 50°C, so pH 6.8 > 6.63 means [H₃O⁺] < [OH⁻]. (c) [OH⁻] = 2.2 × 10⁻¹¹ mol/L; solution is acidic ([H₃O⁺] >> [OH⁻]).

Amphiprotic Behaviour and Kw Synthesis

Sodium hydrogen phosphate (Na₂HPO₄) is used in phosphate buffer solutions for biological research. (a) Explain why HPO₄²⁻ is amphiprotic, writing equations to support your answer. (b) In a buffer at pH 7.4 and 25°C, calculate [OH⁻] and determine whether [H₃O⁺] or [OH⁻] is greater. (c) A researcher claims that because the buffer is used at 37°C, the neutral pH at that temperature is 7.0. Evaluate this claim (Kw = 2.4 × 10⁻¹⁴ at 37°C). (d) Explain why HPO₄²⁻ in this buffer predominantly acts as a base at pH 7.4, using Ka values.

GIVEN: Na₂HPO₄; pH 7.4 at 25°C; Kw = 2.4 × 10⁻¹⁴ at 37°C; Ka3(HPO₄²⁻) = 4.8 × 10⁻¹³; Ka2 = 6.2 × 10⁻⁸.

METHOD (a, amphiprotic definition): Amphiprotic means a substance can both donate and accept a proton depending on its reaction partner. HPO₄²⁻ is amphiprotic because it retains one ionisable H⁺ (can act as acid) and has lone pairs on oxygen atoms (can act as base).
Acting as acid: HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq) Ka3 = 4.8 × 10⁻¹³
Acting as base: HPO₄²⁻(aq) + H⁺(aq) ⇌ H₂PO₄⁻(aq) (reverse of Ka2)

METHOD (b, [OH⁻] at pH 7.4): [H₃O⁺] = 10⁻⁷·⁴ = 3.98 × 10⁻⁸ mol/L. [OH⁻] = Kw/[H₃O⁺] = (1.0 × 10⁻¹⁴)/(3.98 × 10⁻⁸) = 2.51 × 10⁻⁷ mol/L. [OH⁻] > [H₃O⁺] → the buffer is basic at 25°C (pH 7.4 > 7.0).

METHOD (c, evaluate claim at 37°C): Neutral pH at 37°C = −log(√(2.4 × 10⁻¹⁴)) = −log(1.549 × 10⁻⁷) = 6.81. The researcher's claim is incorrect neutral pH at 37°C is 6.81, not 7.0. pKw = −log(2.4 × 10⁻¹⁴) = 13.62, so pH + pOH = 13.62 at 37°C (not 14). A solution with pH 7.0 at 37°C is actually slightly basic (pH 7.0 > 6.81).

METHOD (d, why HPO₄²⁻ acts as base at pH 7.4): At pH 7.4, [H₃O⁺] = 3.98 × 10⁻⁸ mol/L. HPO₄²⁻ as acid: Ka3 = 4.8 × 10⁻¹³, extremely small, meaning HPO₄²⁻ has very little tendency to donate H⁺. The reverse of Ka2 gives Kb = Kw/Ka2 = 1.0 × 10⁻¹⁴/6.2 × 10⁻⁸ = 1.6 × 10⁻⁷, significantly larger than Ka3, so HPO₄²⁻ is a stronger base than it is an acid.

ANSWER: (a) HPO₄²⁻ is amphiprotic: as acid HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ (Ka3 = 4.8 × 10⁻¹³); as base HPO₄²⁻ + H⁺ ⇌ H₂PO₄⁻. (b) [OH⁻] = 2.51 × 10⁻⁷ mol/L; [OH⁻] > [H₃O⁺], basic. (c) Claim incorrect, neutral pH at 37°C = 6.81; pH + pOH = 13.62 at 37°C. (d) Ka3 = 4.8 × 10⁻¹³ is extremely small, HPO₄²⁻ has minimal acid tendency; its Kb = 1.6 × 10⁻⁷ >> Ka3, so it predominantly accepts protons at pH 7.4.

Activities

Activity

Activity A, Sort & Classify: Conjugate Pairs and Amphiprotic Species

For each species listed, (i) state whether it is amphiprotic (yes/no) and give a reason, and (ii) write its equation acting as an acid AND as a base where amphiprotic, or explain why it cannot act in one mode. One row contains a deliberate error, identify and correct it.

Species	Amphiprotic?	As acid (if applicable)	As base (if applicable)
HCO₃⁻
HPO₄²⁻
H₂O
Cl⁻
OH⁻
H₂PO₄⁻
NH₄⁺
H₂SO₄ → SO₄²⁻ + 2H⁺ (conjugate base is SO₄²⁻)	⚠ This row contains a deliberate error, identify and correct it

Activity B, Analyse & Connect: Kw at Non-Standard Temperatures

Use the Kw data provided to answer the questions below.

Temperature	Kw	Neutral pH	Is pH 7.0 acidic, basic, or neutral at this temperature?
10°C	2.9 × 10⁻¹⁵
25°C	1.0 × 10⁻¹⁴
37°C	2.4 × 10⁻¹⁴
60°C	9.6 × 10⁻¹⁴

Calculate the neutral pH at each temperature and complete column 3.
Determine whether a solution with pH 7.0 is acidic, basic, or neutral at each temperature. Complete column 4.
Explain why neutral pH decreases as temperature increases, using Le Chatelier's principle and the self-ionisation equation: 2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq).
A student states: "At 60°C, pure water must be acidic because its pH is only 6.51, below 7." Explain the error in this reasoning.

✍️ Fill in the Blanks +4 XP

Complete the conjugate pair relationships:

In the reaction HCl + H₂O → H₃O⁺ + Cl⁻, the conjugate base of HCl is . An amphiprotic substance can act as both a proton and a proton . A strong acid has a conjugate base.

Interactive · Conjugate Pair Matcher

Use the matcher. Water is amphiprotic. What is the conjugate base of H₂O?

Complete the Learn phase to unlock Practice.

✓

Practice questions

Multiple Choice

+5 XP

A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct

Short Answer

Apply Band 3–4 4 marks

Question 6. For each of the following species, state whether it is amphiprotic and write equations showing its amphiprotic behaviour where applicable. If not amphiprotic, explain why not.

(a) H₂PO₄⁻ (b) HPO₄²⁻ (c) PO₄³⁻ (d) Cl⁻

Apply Band 4 4 marks

Question 7. At 60°C, Kw = 9.6 × 10⁻¹⁴.

(a) Calculate the pH of pure water at 60°C. (b) State whether a solution with pH 6.8 at 60°C is acidic, basic, or neutral. Justify with a calculation. (c) Calculate [OH⁻] in a solution at 60°C with pH 5.2. (d) Does pH + pOH = 14 apply at 60°C? Explain your reasoning.

Evaluate Band 6 7 marks

Question 8. Potassium dihydrogen phosphate (KH₂PO₄) is used as a pH buffer in food preservation. (a) Identify the amphiprotic ion in KH₂PO₄ and write equations showing its behaviour as both a Brønsted-Lowry acid and base. (2 marks) (b) When KH₂PO₄ is dissolved in water at 25°C, the pH of the solution is 4.4. Calculate [OH⁻] at this pH and state whether [H₃O⁺] or [OH⁻] is greater. (2 marks) (c) A food scientist claims: "Adding KH₂PO₄ to a strongly acidic food preservative will make it less acidic, because H₂PO₄⁻ will act as a base and accept the excess H⁺." Evaluate this claim using Brønsted-Lowry theory and relevant Ka values. (3 marks)

Answers

Key ideas the question bank tests

Identifying an amphiprotic species. HCO₃⁻ can donate H⁺ (HCO₃⁻ ⇌ H⁺ + CO₃²⁻, acid) and accept H⁺ from an acid (HCO₃⁻ + H⁺ ⇌ H₂CO₃, base), so it is amphiprotic. Cl⁻ (conjugate of strong acid HCl) cannot donate H⁺ and has negligible base tendency; OH⁻ can only accept H⁺; SO₄²⁻ has no ionisable H⁺.

Using Kw to find [H₃O⁺] and classify a solution. [H₃O⁺] = Kw/[OH⁻] = (1.0 × 10⁻¹⁴)/(2.0 × 10⁻³) = 5.0 × 10⁻¹² mol/L. Since [OH⁻] = 2.0 × 10⁻³ >> [H₃O⁺] = 5.0 × 10⁻¹², the solution is basic. Equal concentrations of H₃O⁺ and OH⁻ would be neutral.

HPO₄²⁻ acting as both acid and base. HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ shows it as a Brønsted-Lowry acid (Ka3 step); HPO₄²⁻ + H⁺ ⇌ H₂PO₄⁻ shows it as a base (reverse Ka2). Together they demonstrate amphiprotic character. Having lost two protons does not prevent further amphiprotic behaviour: one ionisable H⁺ remains, and PO₄³⁻ is a stable, well-characterised ion.

Classifying pH at a non-standard temperature. Neutral pH at 37°C = −log(√(2.4 × 10⁻¹⁴)) = −log(1.549 × 10⁻⁷) = 6.81. A solution at pH 6.8 has pH < 6.81 → [H₃O⁺] > [OH⁻] → acidic. The number 7 is the neutral reference only at 25°C.

Naming the two conjugate pairs in a reaction. For NH₃ + H₂O ⇌ NH₄⁺ + OH⁻: NH₃ (base) accepts H⁺ from H₂O → NH₄⁺ (conjugate acid); H₂O (acid) donates H⁺ → OH⁻ (conjugate base). Pair 1 NH₃/NH₄⁺ and Pair 2 H₂O/OH⁻ each sit on opposite sides and differ by exactly one H⁺.

Q6 Sample Answer

(a) H₂PO₄⁻, YES, amphiprotic. It retains ionisable H⁺ (can act as acid) and has lone pairs on oxygen (can act as base).
As acid: H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq) (Ka2 = 6.2 × 10⁻⁸)
As base: H₂PO₄⁻(aq) + H⁺(aq) ⇌ H₃PO₄(aq) (reverse of Ka1)

(b) HPO₄²⁻, YES, amphiprotic.
As acid: HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq) (Ka3 = 4.8 × 10⁻¹³)
As base: HPO₄²⁻(aq) + H⁺(aq) ⇌ H₂PO₄⁻(aq) (reverse of Ka2)

(c) PO₄³⁻, NOT amphiprotic. PO₄³⁻ has no ionisable H⁺ remaining, it cannot act as a Brønsted-Lowry acid (no proton to donate). It can only act as a base. It is the final conjugate base of the phosphate system.

(d) Cl⁻, NOT amphiprotic. Cl⁻ is the conjugate base of HCl (a strong acid). It has no ionisable H⁺ (cannot act as acid) and its tendency to accept H⁺ is essentially zero. Cl⁻ is a neutral spectator ion in aqueous solution.

Q7 Sample Answer

(a) pH of pure water at 60°C:
[H₃O⁺] = [OH⁻] = √Kw = √(9.6 × 10⁻¹⁴) = 3.10 × 10⁻⁷ mol/L
pH = −log(3.10 × 10⁻⁷) = 6.51
Pure water at 60°C is neutral (by definition, [H₃O⁺] = [OH⁻]) but has pH 6.51, not 7.

(b) Classification of pH 6.8 at 60°C:
Neutral pH at 60°C = 6.51. pH 6.8 > 6.51 → [H₃O⁺] < [OH⁻].
Therefore pH 6.8 at 60°C is basic. Even though pH 6.8 < 7, the relevant reference at 60°C is 6.51 (not 7).

(c) [OH⁻] at pH 5.2:
[H₃O⁺] = 10⁻⁵·² = 6.31 × 10⁻⁶ mol/L
[OH⁻] = Kw/[H₃O⁺] = (9.6 × 10⁻¹⁴)/(6.31 × 10⁻⁶) = 1.52 × 10⁻⁸ mol/L

(d) pH + pOH at 60°C:
pKw = −log(9.6 × 10⁻¹⁴) = 13.02
pH + pOH = pKw = 13.02 at 60°C, NOT 14. The relationship pH + pOH = 14 applies only at 25°C because that is the temperature at which Kw = 1.0 × 10⁻¹⁴ exactly.

Q8 Sample Answer (Band 6)

(a) Amphiprotic ion in KH₂PO₄ = H₂PO₄⁻
KH₂PO₄ dissolves: K⁺ (neutral spectator) + H₂PO₄⁻ (amphiprotic ion).
As acid (proton donor): H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq) Ka2 = 6.2 × 10⁻⁸
As base (proton acceptor): H₂PO₄⁻(aq) + H⁺(aq) ⇌ H₃PO₄(aq) (reverse of Ka1)

(b) [OH⁻] at pH 4.4:
[H₃O⁺] = 10⁻⁴·⁴ = 3.98 × 10⁻⁵ mol/L
[OH⁻] = (1.0 × 10⁻¹⁴)/(3.98 × 10⁻⁵) = 2.51 × 10⁻¹⁰ mol/L
[H₃O⁺] = 3.98 × 10⁻⁵ >> [OH⁻] = 2.51 × 10⁻¹⁰ → [H₃O⁺] is greater. The solution is acidic (pH 4.4 < 7).

(c) Evaluate the food scientist's claim:
The claim is partially correct but oversimplified. H₂PO₄⁻ can indeed act as a Brønsted-Lowry base, accepting H⁺ from a stronger acid to form H₃PO₄ (reverse of Ka1 = 7.5 × 10⁻³). If the "strongly acidic preservative" has [H⁺] significantly greater than Ka1, then H₂PO₄⁻ would accept the H⁺ and reduce acidity, the claim would be correct in principle.
However, the claim is limited because: (1) If [H⁺] is very high, the buffer capacity of H₂PO₄⁻ is finite, excess strong acid will eventually overwhelm it; (2) At very low pH, H₂PO₄⁻ may already be fully protonated to H₃PO₄ and unable to accept more H⁺; (3) Whether H₂PO₄⁻ acts as acid or base depends on the relative strength of the reaction partner.
Conclusion: The scientist correctly identifies the base behaviour of H₂PO₄⁻, but overstates its effectiveness in strongly acidic conditions without specifying concentrations and Ka values.

✓

Retrieve and reflect

Check what actually stuck

Band 6 Synthesis Challenge

A student states: "The presence of HPO₄²⁻ ions in a solution means the solution must be at pH 7, because HPO₄²⁻ is the amphiprotic ion that buffers blood at neutral pH."

Evaluate this claim fully. Address: (a) whether the claim about pH is correct; (b) whether it is correct to call HPO₄²⁻ the buffer at blood pH (use Ka values); (c) what determines whether HPO₄²⁻ acts as acid or base in a given solution; and (d) the correct role of HPO₄²⁻ in physiological buffering.

Return to Your Initial Thinking

Return to your Think First response. Recall Kohlrausch's 1893 measurement: even ultra-pure water ionises slightly, Kw = 10⁻¹⁴, confirming water is amphiprotic. The chemist observed NaHCO₃ acting as a base with HCl and as an acid with NaOH. You now have the framework to explain this precisely:

NaHCO₃ + HCl: HCl is a strong acid, Cl⁻ is an extremely weak base. HCO₃⁻ acts as a base, accepting H⁺ from HCl: HCO₃⁻ + HCl → H₂CO₃ + Cl⁻. H₂CO₃ then decomposes to CO₂ + H₂O (the bubbles observed). The solution becomes less acidic.
NaHCO₃ + NaOH: OH⁻ is a strong base. HCO₃⁻ acts as an acid, donating H⁺ to OH⁻: HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O. No CO₂ is produced because H₂CO₃ is not formed. The solution becomes less basic.
The key insight: HCO₃⁻ is amphiprotic. Whether it acts as acid or base depends entirely on which is stronger, its reaction partner as a base (relative to CO₃²⁻) or as an acid (relative to H₂CO₃). With a stronger acid (HCl), it acts as a base. With a stronger base (OH⁻), it acts as an acid. This is not a contradiction, it is the defining feature of amphiprotic behaviour.

Quick Recall Drill

Rapid-Fire Questions

What is the conjugate base of H₂SO₄?

Answer

HSO₄⁻ (one proton removed, charge increases by 1). NOT SO₄²⁻.

Write the equation for HPO₄²⁻ acting as a base.

Answer

HPO₄²⁻(aq) + H⁺(aq) ⇌ H₂PO₄⁻(aq) (one step left in phosphate chain)

At 37°C (Kw = 2.4 × 10⁻¹⁴), what is the neutral pH?

Answer

pH = −log(√(2.4 × 10⁻¹⁴)) = −log(1.55 × 10⁻⁷) = 6.81

Why does Kw increase with temperature?

Answer

Self-ionisation (2H₂O ⇌ H₃O⁺ + OH⁻) is endothermic, increasing temperature shifts equilibrium right by Le Chatelier's principle, increasing both [H₃O⁺] and [OH⁻], so Kw increases.

What determines whether an amphiprotic substance acts as acid or base?

Answer

Its reaction partner, if the partner is a stronger acid (provides H⁺), the amphiprotic substance acts as a base; if the partner is a stronger base (accepts H⁺), it acts as an acid.

Distinguish amphiprotic from amphoteric.

Answer

Amphiprotic (Brønsted-Lowry): can both donate AND accept H⁺. Amphoteric (broader): can react with both acids and bases, includes Lewis acid-base behaviour. In Module 6, use "amphiprotic."

Lesson 7 Complete, I can identify conjugate pairs, write both equations for any amphiprotic salt, and classify solutions using Kw at any temperature.

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