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.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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?
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.
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⁻?
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).
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.
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 NESA require you to write equations for in the HSC? This card answers it → Na₂HPO₄ (amphiprotic ion HPO₄²⁻) and KH₂PO₄ (amphiprotic ion H₂PO₄⁻); both equations must be written for each.
These two salts are named explicitly in the NSW Chemistry Stage 6 syllabus — which means they will appear in HSC exams, and the ability to write both their acid and base equations from memory is a non-negotiable competency for this lesson.
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.
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.
NESA-named amphiprotic salts: Na₂HPO₄ → HPO₄²⁻ (acid: HPO₄²⁻ ⇌ H⁺ + PO₄³⁻; base: HPO₄²⁻ + H⁺ ⇌ H₂PO₄⁻); KH₂PO₄ → H₂PO₄⁻ (acid: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻; base: H₂PO₄⁻ + H⁺ ⇌ H₃PO₄). 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 ______.
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.
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⁺)?
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⁻]).
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:
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:
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⁺]?
"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⁻.
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.
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?
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.
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 | ||
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 |
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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.
Select the best answer for each question.
1. Which of the following species is amphiprotic?
2. At 25°C, Kw = 1.0 × 10⁻¹⁴. A solution has [OH⁻] = 2.0 × 10⁻³ mol/L. Which correctly identifies [H₃O⁺] and the nature of the solution?
3. Na₂HPO₄ is dissolved in water. A student writes: Eq 1: HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq). Eq 2: HPO₄²⁻(aq) + H⁺(aq) ⇌ H₂PO₄⁻(aq). Which statement correctly evaluates these equations?
4. At 37°C, Kw = 2.4 × 10⁻¹⁴. A student measures the pH of a physiological buffer and obtains pH 6.8. Which statement correctly describes the classification of this solution at 37°C?
5. A student identifies the two conjugate pairs in NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) as: Pair 1: NH₃/NH₄⁺. Pair 2: H₂O/OH⁻. Which evaluation is correct?
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⁻
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.
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)
Q1: C — HCO₃⁻
HCO₃⁻ can donate H⁺ (HCO₃⁻ ⇌ H⁺ + CO₃²⁻, acid) and accept H⁺ from an acid (HCO₃⁻ + H⁺ ⇌ H₂CO₃, base) — 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⁺.
Q2: B
[H₃O⁺] = Kw/[OH⁻] = (1.0 × 10⁻¹⁴)/(2.0 × 10⁻³) = 5.0 × 10⁻¹² mol/L. [OH⁻] = 2.0 × 10⁻³ >> [H₃O⁺] = 5.0 × 10⁻¹² → basic. Option A wrong — equal concentrations would be neutral. Option C has correct [H₃O⁺] but wrong classification. Option D uses incorrect calculation.
Q3: C
Both equations are chemically correct. Eq 1: HPO₄²⁻ acts as BL acid (Ka3 step). Eq 2: HPO₄²⁻ acts as BL base (reverse Ka2). Together they demonstrate amphiprotic character. Option A wrong — having lost two protons does not prevent further amphiprotic behaviour; one H⁺ remains. Option D wrong — PO₄³⁻ is a stable, well-characterised ion.
Q4: D
Neutral pH at 37°C = −log(√(2.4 × 10⁻¹⁴)) = −log(1.549 × 10⁻⁷) = 6.81. pH 6.8 < 6.81 → [H₃O⁺] > [OH⁻] → acidic. Option A correct reasoning but incomplete. Option B correct numbers but wrong direction. Option C invents an incorrect neutral pH value.
Q5: A
Both pairs are correctly identified. NH₃ (base) accepts H⁺ from H₂O → NH₄⁺ (conjugate acid). H₂O (acid) donates H⁺ → OH⁻ (conjugate base). Pair 1: NH₃/NH₄⁺ on opposite sides, differ by one H⁺ ✓. Pair 2: H₂O/OH⁻ on opposite sides, differ by one H⁺ ✓.
(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.
(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.
(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.
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 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:
What is the conjugate base of H₂SO₄?
HSO₄⁻ (one proton removed, charge increases by 1). NOT SO₄²⁻.
Write the equation for HPO₄²⁻ acting as a base.
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?
pH = −log(√(2.4 × 10⁻¹⁴)) = −log(1.55 × 10⁻⁷) = 6.81
Why does Kw increase with temperature?
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?
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.
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."
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