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Inquiry Question 5: How are acids and bases defined and how do they behave in aqueous solution?

Investigate the Brønsted-Lowry theory of acids and bases, including conjugate acid/base pairs and the behaviour of amphiprotic species

A focused answer to the HSC Chemistry Module 5 dot point on Brønsted-Lowry acid-base theory. Definitions, conjugate acid-base pairs, amphiprotic species (water and bicarbonate), how the theory extends Arrhenius, and the worked HSC past exam questions.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-22

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In plain English

Think of a proton ( $H^+$ ) as a parcel being handed between two people. Whoever hands the parcel over is the Brønsted-Lowry acid; whoever receives it is the Brønsted-Lowry base. Once the handover happens, the roles are set up to reverse: the person who just gave away the parcel could take it back (so they are now capable of being a base), and the person who just received it could hand it onward (so they are now capable of being an acid). That "before and after" pair, differing by exactly one parcel, is a conjugate acid-base pair. Some people are flexible couriers who can either give or receive the parcel depending on who they are standing next to, water and bicarbonate both do this in the body and in water treatment, and chemists call that flexibility "amphiprotic". The old Arrhenius theory only tracked whether $H^+$ or $OH^-$ ions turned up dissolved in water, so it could not describe couriers working outside a water-based room; Brønsted-Lowry tracks the handover itself, so it works anywhere a proton can be passed.

What this dot point is asking

NESA wants you to define Brønsted-Lowry acids and bases, identify conjugate acid-base pairs in a chemical equation, explain how a species can be amphiprotic, and compare Brønsted-Lowry to the earlier Arrhenius model. This is the conceptual foundation for every acid-base calculation in HSC Chemistry, including pH and pOH, titration analysis, and buffer systems.

The answer

Definitions

Brønsted-Lowry acid: a species that donates a proton ( $H^+$ ).
Brønsted-Lowry base: a species that accepts a proton ( $H^+$ ).

The definition focuses on the proton transfer itself, not on whether the reaction occurs in water.

Conjugate acid-base pairs

When an acid donates a proton, it becomes a base (because it can now accept the proton back). When a base accepts a proton, it becomes an acid. The acid and base that differ by a single $H^+$ form a conjugate acid-base pair.

For the reaction:

HCl_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + Cl^-_{(aq)}

HCl donates $H^+$ , so $HCl$ is the acid. Its conjugate base is $Cl^-$ .
Water accepts $H^+$ , so $H_2O$ is the base. Its conjugate acid is $H_3O^+$ .

Two conjugate pairs: $HCl / Cl^-$ and $H_3O^+ / H_2O$ .

An owned reaction diagram makes the proton handover explicit: the same $H^+$ leaves the acid and lands on the base, creating the two conjugate species in a single step.

Amphiprotic species

An amphiprotic species can act as either a Brønsted-Lowry acid or a Brønsted-Lowry base depending on what it reacts with. The most important examples:

Water.

Acts as a base: $H_2O + HCl \rightleftharpoons H_3O^+ + Cl^-$ .
Acts as an acid: $H_2O + NH_3 \rightleftharpoons OH^- + NH_4^+$ .

Bicarbonate ion ( $HCO_3^-$ ).

Acts as a base: $HCO_3^- + H_3O^+ \rightleftharpoons H_2CO_3 + H_2O$ .
Acts as an acid: $HCO_3^- + OH^- \rightleftharpoons CO_3^{2-} + H_2O$ .

Hydrogen sulfate ion ( $HSO_4^-$ ). Similarly acts as both an acid and a base.

Amino acids (like glycine, $H_2NCH_2COOH$ ) are amphiprotic because they contain both an acidic $-COOH$ group and a basic $-NH_2$ group.

A useful term to distinguish: amphoteric is the broader concept (can react with both acids and bases), which includes species like $Al_2O_3$ that are not necessarily proton donors. Amphiprotic specifically means proton donor and acceptor.

Comparison with Arrhenius theory

Arrhenius (1887): an acid produces $H^+$ in water, a base produces $OH^-$ in water.

Brønsted-Lowry (1923) extends this in three ways:

Defines acid-base behaviour by proton transfer, not by what ions form in water.
Works in non-aqueous solvents.
Explains the basicity of species like $NH_3$ , $CO_3^{2-}$ , $HCO_3^-$ that contain no hydroxide.

Every Arrhenius acid is also a Brønsted-Lowry acid, but the reverse is not true.

Worked example

Identify the Brønsted-Lowry acids and bases and the conjugate pairs in:

NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH_4^+{}_{(aq)} + OH^-_{(aq)}

Forward direction.

Water loses a proton (becomes $OH^-$ ). So $H_2O$ is the acid.
Ammonia gains a proton (becomes $NH_4^+$ ). So $NH_3$ is the base.

Reverse direction.

The $NH_4^+$ ion donates a proton, so it is the conjugate acid of $NH_3$ .
The $OH^-$ ion accepts a proton, so it is the conjugate base of $H_2O$ .

Conjugate pairs:

First pair: $NH_4^+ / NH_3$ (acid / base).
Second pair: $H_2O / OH^-$ (acid / base).

Note that water acts as an acid here but as a base in $H_2O + HCl \rightleftharpoons H_3O^+ + Cl^-$ . Water is amphiprotic.

Examples in context

Example 1. Sydney Water Prospect treatment plant pH adjustment. The Prospect water filtration plant adjusts incoming Warragamba water to a pH near 7.8 by dosing carbon dioxide or hydrated lime. The chemistry is a textbook Brønsted-Lowry transfer: hydrated lime deprotonates water pulled from a moderately acidic supply to give $Ca^{2+}_{(aq)} + 2OH^-_{(aq)}$ , with the hydroxide accepting a proton from any free $H_3O^+$ in the source water. Operators see the conjugate-base $HCO_3^-$ formed from the dissolved $CO_2$ acting amphiprotically, buffering the network against tiny upstream pH fluctuations as water moves through the supply tunnel to Sydney homes.

Example 2. Bicarbonate as the body's amphiprotic ion. Bicarbonate is the most important amphiprotic species in human physiology. In stomach acid it accepts a proton: $HCO_3^- + H_3O^+ \rightarrow H_2CO_3 + H_2O$ , neutralising acidity in antacids such as Mylanta. In the bloodstream the same ion donates a proton to keep plasma pH near 7.4: $HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+$ . A single ion species playing both roles, depending on its partner, is exactly what Brønsted-Lowry theory predicts but the older Arrhenius framework cannot describe.

Try this

Q1. Define a Brønsted-Lowry acid and identify the conjugate base of $H_2SO_4$ , $NH_4^+$ and $HCO_3^-$ . [3 marks]

Cue. Acid as proton donor; conjugate bases are $HSO_4^-$ , $NH_3$ and $CO_3^{2-}$ respectively, each formed by removing one $H^+$ .

Q2. Calculate the $[H_3O^+]$ produced when 0.0500 mol of HCl is dissolved in 250 mL of water, assuming full dissociation. [2 marks]

Cue. Strong acid donates one proton per molecule, $[H_3O^+] = 0.0500 / 0.250 = 0.200 \text{ mol L}^{-1}$ .

Q3. Hydrogen carbonate is described as amphiprotic. (a) Write equations showing $HCO_3^-$ acting as an acid and as a base in water. (b) Identify the conjugate pairs in each equation. (c) Explain why Arrhenius theory cannot accommodate this behaviour. [2+2+1 marks]

Cue. (a) Acid: $HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+$ ; base: $HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-$ . (b) Label by one-proton difference. (c) Arrhenius limits acids to producers of $H^+$ in water and cannot describe a species that both donates and accepts protons.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC4 marksUsing the equation HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻, identify each species as a Brønsted-Lowry acid or base, and explain the term amphiprotic with reference to HCO₃⁻.

Show worked answer →

A 4 mark answer needs the acid/base assignment, the conjugate pairs, and a clear demonstration that $HCO_3^-$ is amphiprotic.

In the forward direction:

Water donates a proton to $HCO_3^-$ , so $H_2O$ is the Brønsted-Lowry acid.
The $HCO_3^-$ ion accepts a proton, so $HCO_3^-$ is the Brønsted-Lowry base.

In the reverse direction:

The $H_2CO_3$ molecule donates a proton, so it is the conjugate acid of $HCO_3^-$ .
The $OH^-$ ion accepts a proton, so it is the conjugate base of $H_2O$ .

Conjugate pairs: $HCO_3^- / H_2CO_3$ and $H_2O / OH^-$ .

Amphiprotic means a species can act either as a Brønsted-Lowry acid (donating $H^+$ ) or as a Brønsted-Lowry base (accepting $H^+$ ). In the equation above, $HCO_3^-$ acts as a base. But $HCO_3^-$ can also donate a proton, for example $HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+$ , where it acts as an acid. Because it can do both, $HCO_3^-$ is amphiprotic.

Markers reward (1) correct assignment in the forward and reverse directions, (2) explicit naming of the two conjugate pairs, (3) the definition of amphiprotic with two equations showing $HCO_3^-$ in both roles.

2017 HSC2 marksExplain why the Brønsted-Lowry theory of acids is considered an improvement on the Arrhenius theory.

Show worked answer →

Arrhenius defined an acid as a substance that produces $H^+$ in aqueous solution and a base as a substance that produces $OH^-$ . This definition is limited to aqueous solutions and cannot explain basic behaviour without hydroxide ions.

Brønsted-Lowry defines an acid as a proton donor and a base as a proton acceptor. This extends the theory in two ways:

It works in non-aqueous solvents (for example, $NH_3$ acting as a base toward $HCl$ in liquid ammonia).
It explains the basic behaviour of species like $NH_3$ , $CO_3^{2-}$ and $HCO_3^-$ that contain no $OH^-$ but still accept protons in water.

Markers reward (1) clearly stating both definitions, (2) at least one specific extension Brønsted-Lowry accounts for that Arrhenius cannot.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFor the reaction

H_2S + H_2O \rightleftharpoons HS^- + H_3O^+

, identify the Brønsted-Lowry acid and base in the forward direction, and state the two conjugate acid-base pairs.

Show worked solution →

A 3-mark identify needs the correct acid/base labels plus both conjugate pairs.

Forward direction. $H_2S$ donates a proton to become $HS^-$ , so $H_2S$ is the Brønsted-Lowry acid. Water accepts that proton to become $H_3O^+$ , so $H_2O$ is the Brønsted-Lowry base.

Conjugate pairs. $H_2S / HS^-$ (acid / conjugate base) and $H_3O^+ / H_2O$ (conjugate acid / base).

Marking criteria: 1 mark for correctly identifying the acid, 1 mark for correctly identifying the base, 1 mark for both conjugate pairs named correctly (differing by one $H^+$ each).

foundation3 marksWrite two equations showing

H_2PO_4^-

(dihydrogen phosphate ion) acting first as a Brønsted-Lowry acid and then as a Brønsted-Lowry base in water, and explain what property this demonstrates.

Show worked solution →

Acting as an acid (donates $H^+$ to water):

H_2PO_4^-{}_{(aq)} + H_2O_{(l)} \rightleftharpoons HPO_4^{2-}{}_{(aq)} + H_3O^+_{(aq)}

Acting as a base (accepts $H^+$ from water):

H_2PO_4^-{}_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3PO_4{}_{(aq)} + OH^-_{(aq)}

Because $H_2PO_4^-$ can both donate and accept a proton depending on its partner, it is amphiprotic, in the same way as $HCO_3^-$ and water itself.

Marking criteria: 1 mark for the correct acid-role equation, 1 mark for the correct base-role equation (both with state symbols), 1 mark for explicitly naming and explaining "amphiprotic".

core4 marksA 0.150 mol sample of a monoprotic Brønsted-Lowry acid, HA, is fully dissociated in enough water to make 500.0 mL of solution. Calculate the resulting

[H_3O^+]

and hence the pH of the solution, to 2 decimal places. (

pH = -\log_{10}[H_3O^+]

Show worked solution →

Step 1: write the dissociation equation. A monoprotic acid donates one proton per molecule:

HA_{(aq)} + H_2O_{(l)} \rightarrow A^-_{(aq)} + H_3O^+_{(aq)}

One mole of HA gives one mole of $H_3O^+$ when fully dissociated (a strong acid assumption).

Step 2: concentration of HA.

c = \frac{n}{V} = \frac{0.150\ \text{mol}}{0.5000\ \text{L}} = 0.300\ \text{mol L}^{-1}

Step 3: $[H_3O^+]$ . Full dissociation with a 1:1 mole ratio means:

[H_3O^+] = 0.300\ \text{mol L}^{-1}

Step 4: pH.

pH = -\log_{10}(0.300) = 0.5229

Step 5: round to 2 decimal places (matching the precision requested).

pH = 0.52

Marking criteria: 1 mark for the correct 1:1 dissociation relationship, 1 mark for the correct concentration of HA, 1 mark for correctly equating $[H_3O^+]$ to that concentration (full dissociation), 1 mark for the correct pH to 2 decimal places. Note this method assumes HA behaves as a strong (fully dissociating) Brønsted-Lowry acid, as stated in the question.

core5 marksThe titration curve below is an owned illustrative curve for 25.0 mL of a monoprotic Brønsted-Lowry acid solution titrated with 0.100 mol L⁻¹ NaOH, reaching the equivalence point at 20.0 mL of titrant. (a) Identify whether the acid is strong or weak, justifying your answer using a feature of the curve. (b) Explain, in Brønsted-Lowry terms, the proton-transfer reaction occurring at the equivalence point.

Show worked solution →

(a) Strong or weak. The curve shows a low, flat initial pH (around pH 1) before the equivalence point, and a very steep, near-vertical jump in pH exactly at the equivalence point (20.0 mL). A weak acid would start at a higher initial pH and show a much more gradual rise (a buffering region) before a smaller, less abrupt jump. The low starting pH plus the large steep jump indicate this is a strong acid.

(b) Proton-transfer reaction at equivalence. At the equivalence point, moles of $OH^-$ added exactly equal the initial moles of $H_3O^+$ present. In Brønsted-Lowry terms, hydroxide ion (behaving as a Brønsted-Lowry base, accepting a proton) reacts with the hydronium ion (behaving as a Brønsted-Lowry acid, donating a proton):

H_3O^+_{(aq)} + OH^-_{(aq)} \rightarrow 2H_2O_{(l)}

Because the conjugate base of a strong acid is negligibly basic, the resulting solution at equivalence is neutral, pH ≈ 7, matching the curve.

Marking criteria: (a) 1 mark for correctly identifying "strong", 1 mark for justification using BOTH the low starting pH and the steep/large jump. (b) 1 mark for identifying $H_3O^+$ as the Brønsted-Lowry acid and $OH^-$ as the base, 1 mark for the correct neutralisation equation, 1 mark for linking this to the neutral equivalence pH shown on the curve.

exam6 marksAssess the claim that 'the Brønsted-Lowry theory made the Arrhenius theory obsolete' with reference to the behaviour of ammonia and bicarbonate ion in aqueous and non-aqueous systems.

Show worked solution →

This is a 6-mark ASSESS: markers reward a judgement supported by specific evidence, not a simple description of both theories.

Band 6 PLAN.

Thesis: Brønsted-Lowry did not make Arrhenius theory obsolete so much as generalise it; Arrhenius remains correct and useful within its aqueous domain, but Brønsted-Lowry is required to fully explain species like $NH_3$ and $HCO_3^-$ .
Evidence 1 (ammonia): Arrhenius cannot classify $NH_3$ as a base directly, since $NH_3$ contains no $OH^-$ to "produce"; only after reacting with water ( $NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH_4^+{}_{(aq)} + OH^-_{(aq)}$ ) does Arrhenius indirectly explain the resulting basicity. Brønsted-Lowry explains it directly: $NH_3$ accepts a proton from water, so it is a base regardless of solvent.
Evidence 2 (bicarbonate, amphiprotic behaviour): $HCO_3^-$ acts as an acid toward $OH^-$ and as a base toward $H_3O^+$ . Arrhenius theory has no mechanism at all for a single species switching roles; Brønsted-Lowry explains both roles with one consistent proton-transfer definition.
Evidence 3 (non-aqueous systems): Brønsted-Lowry acid-base chemistry occurs in solvents such as liquid ammonia, where Arrhenius theory (defined only for aqueous solution) cannot apply at all.
Judgement: for everyday aqueous strong-acid/strong-base chemistry, Arrhenius's simpler definition is still an accurate special case; therefore Brønsted-Lowry theory extends and subsumes Arrhenius rather than making it "obsolete" in the sense of incorrect.

Model paragraph (excerpt). The Brønsted-Lowry theory is best understood as a generalisation of Arrhenius theory rather than a replacement that renders it wrong. Arrhenius correctly predicts that $HCl$ is acidic and $NaOH$ is basic in water, and Brønsted-Lowry agrees, classifying $HCl$ as a proton donor and $NaOH$ 's hydroxide as a proton acceptor. The real gap Arrhenius leaves is species like ammonia and bicarbonate: $NH_3$ has no hydroxide to "produce" yet is unambiguously basic, and $HCO_3^-$ behaves as an acid in one reaction and a base in another, a role-switching behaviour ("amphiprotic") that a fixed-ion-production model cannot describe at all. Because the proton-transfer definition covers every case the ion-production definition covers, plus these extra cases, and even extends beyond water to non-aqueous solvents like liquid ammonia, Brønsted-Lowry theory is more accurately described as generalising Arrhenius theory than as making it obsolete.

Marker's note: top-band answers (1) state an explicit judgement (not "both theories are useful" as a non-answer), (2) use at least two specific chemical examples with correct equations, (3) explicitly address the non-aqueous extension, and (4) conclude by characterising the RELATIONSHIP between the theories (generalisation, not obsolescence) rather than simply listing differences.