Pure water is neither just an acid nor just a base — it is both simultaneously, and the tiny fraction of water molecules that react with each other at any given instant is the foundation of every pH calculation in this module.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
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? Write your reasoning before the lesson gives you the framework to test it.
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.
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).
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.
HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq) Ka3 = 4.8 × 10⁻¹³ — very weak acidHPO₄²⁻(aq) + H⁺(aq) ⇌ H₂PO₄⁻(aq) (reverse of Ka2)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⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq) Ka2 = 6.2 × 10⁻⁸H₂PO₄⁻(aq) + H⁺(aq) ⇌ H₃PO₄(aq) (reverse of Ka1)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.
The fraction of water molecules that react with each other at any instant is extraordinarily small — but it is precisely this self-ionisation that establishes the H₃O⁺ and OH⁻ present in every aqueous solution, making Kw the foundation of every pH calculation in the 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⁻]).
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:
"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⁻.
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 (H₃PO₄ → H₂PO₄⁻ → HPO₄²⁻ → PO₄³⁻). 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).
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. At temperatures above 25°C, neutral pH < 7, so solutions with pH between neutral pH and 7 are actually basic.
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⁻]).
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⁺. [H₃O⁺] = 3.98 × 10⁻⁸ >> Ka3 = 4.8 × 10⁻¹³ — far more H₃O⁺ available than HPO₄²⁻ can generate, so HPO₄²⁻ preferentially accepts 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.
These are the minimum facts and equations to have in your notes for Lesson 7.
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 | At this temperature, is a solution with pH 7.0 acidic, basic, or neutral? |
|---|---|---|---|
| 10°C | 2.9 × 10⁻¹⁵ | ||
| 25°C | 1.0 × 10⁻¹⁴ | ||
| 37°C | 2.4 × 10⁻¹⁴ | ||
| 60°C | 9.6 × 10⁻¹⁴ |
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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 of the following correctly identifies [H₃O⁺] and the nature of the solution?
3. The salt Na₂HPO₄ is dissolved in water. A student writes: Equation 1: HPO₄²⁻(aq) ⇌ H⁺(aq) + PO₄³⁻(aq). Equation 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 is asked to identify the two conjugate pairs in the reaction: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). The student writes: "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 it is 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 your answer 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 a Brønsted-Lowry 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)
Return to your Think First response. 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:
Review your original response. Did you identify the role of the reaction partner? Did you use the terms amphiprotic or conjugate pair? Update your notes with any missing reasoning.
Q1: C — HCO₃⁻
HCO₃⁻ can donate H⁺ to a base (HCO₃⁻ ⇌ H⁺ + CO₃²⁻, acting as acid) and accept H⁺ from an acid (HCO₃⁻ + H⁺ ⇌ H₂CO₃, acting as base) — amphiprotic. Cl⁻ (conjugate of strong acid HCl) cannot donate H⁺ and has negligible base tendency. OH⁻ can only accept H⁺ (base only). SO₄²⁻ has no ionisable H⁺ and is a very weak base — not meaningfully amphiprotic at HSC level.
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. pH = −log(5.0 × 10⁻¹²) = 11.3 > 7, confirms basic. Option A wrong — equal concentrations would be neutral, but [OH⁻] = 2.0 × 10⁻³ ≠ 1.0 × 10⁻⁷. Option C correct [H₃O⁺] but wrong classification. Option D uses incorrect calculation.
Q3: C
Both equations are chemically correct. Equation 1: HPO₄²⁻ acts as BL acid (proton donor → PO₄³⁻, Ka3 step). Equation 2: HPO₄²⁻ acts as BL base (proton acceptor → H₂PO₄⁻, reverse Ka2). Together they demonstrate amphiprotic character. Option A wrong — having lost two protons does not prevent further amphiprotic behaviour; one H⁺ remains. Option B wrong — negative charge does not prevent proton donation. 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 is correct reasoning but incomplete (doesn't use Kw). Option B uses same numbers but classifies as basic — incorrect 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 of NH₃). H₂O (acid) donates H⁺ → OH⁻ (conjugate base of H₂O). Pair 1: NH₃/NH₄⁺ on opposite sides, differ by one H⁺ ✓. Pair 2: H₂O/OH⁻ on opposite sides, differ by one H⁺ ✓. Option B confuses direction — NH₄⁺ is the conjugate acid (NH₃ + H⁺, not NH₃ − H⁺). Option C wrong — H₂O and OH⁻ are on opposite sides (reactant and product).
(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 (accepting H⁺). 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 (the inverse strength rule means the conjugate of a very strong acid is a very weak base). 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. At other temperatures, Kw changes and so does pKw.
(a) Amphiprotic ion in KH₂PO₄ = H₂PO₄⁻
KH₂PO₄ dissolves: K⁺ (neutral spectator, conjugate of KOH strong base) + 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 (i.e., the acid is stronger than H₃PO₄), 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 (e.g., from a strong acid like HCl at low pH), the buffer capacity of H₂PO₄⁻ is finite — adding 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 — with a very strong acid, it acts as base (correct), but this does not mean it will always neutralise all excess H⁺ in a "strongly acidic" system.
Conclusion: The scientist correctly identifies the base behaviour of H₂PO₄⁻, but overstates its effectiveness in strongly acidic conditions without specifying concentrations and Ka values.
Answer questions on conjugate acid-base pairs, amphiprotic substances and Kw before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–7.