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Conjugate acid-base pairs and amphiprotic species: WACE Year 12 Chemistry

Syllabus dot point

“Identify conjugate acid-base pairs in proton-transfer reactions and explain amphiprotic behaviour”

7 min read

Quick answer

In the Bronsted-Lowry model, every proton transfer involves two conjugate acid-base pairs. A conjugate pair differs by exactly one proton (H+\text{H}^+): the acid has it, the base does not. The stronger an acid, the weaker its conjugate base. An amphiprotic species, such as the hydrogen carbonate ion, can either donate or accept a proton depending on what it reacts with.

What this dot point is asking

The Bronsted-Lowry theory defines an acid as a proton donor and a base as a proton acceptor. Because a proton transfer involves giving from one species and accepting by another, acids and bases always come in linked conjugate pairs.

Conjugate acid-base pair

Two species that differ by one proton. When an acid donates a proton it becomes its conjugate base; when a base accepts a proton it becomes its conjugate acid.

Identifying the pairs

Consider ethanoic acid reacting with water:

CH3COOH+H2O⇌CH3COO−+H3O+\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+

Ethanoic acid donates a proton to become ethanoate, so CH3COOH\text{CH}_3\text{COOH} and CH3COO−\text{CH}_3\text{COO}^- are one conjugate pair (acid and its conjugate base). Water accepts a proton to become hydronium, so H2O\text{H}_2\text{O} and H3O+\text{H}_3\text{O}^+ are the second pair (base and its conjugate acid). Every Bronsted-Lowry equation contains exactly two such pairs.

Two pairs, differing by one proton

In any proton-transfer equation, find the species that differs by a single H+\text{H}^+ on the same side relationship: the larger member is the acid, the smaller is its conjugate base. There are always two pairs, one on each side of the proton being passed.

The inverse strength relationship

For a conjugate pair, acid strength and conjugate base strength are inversely related: Ka×Kb=KwK_a \times K_b = K_w. A strong acid such as HCl is essentially fully ionised, so its conjugate base Cl−\text{Cl}^- is negligibly basic and does not affect pH. Conversely, the conjugate base of a weak acid (such as ethanoate from ethanoic acid) is itself a meaningful weak base. This is exactly why a salt of a weak acid, such as sodium ethanoate, gives a slightly basic solution.

Amphiprotic species

Some species can either donate or accept a proton. These are called amphiprotic (a subset of amphoteric behaviour involving proton transfer).

Amphiprotic species

A species that can act as either a Bronsted-Lowry acid (donating a proton) or a base (accepting a proton), depending on the other reactant.

The classic example is the hydrogen carbonate ion, HCO3−\text{HCO}_3^-:

  • As an acid (donating a proton): HCO3−⇌CO32−+H+\text{HCO}_3^- \rightleftharpoons \text{CO}_3^{2-} + \text{H}^+
  • As a base (accepting a proton): HCO3−+H+⇌H2CO3\text{HCO}_3^- + \text{H}^+ \rightleftharpoons \text{H}_2\text{CO}_3

Water itself is amphiprotic, which is why it self-ionises. The dihydrogen phosphate ion H2PO4−\text{H}_2\text{PO}_4^- is another common example. Amphiprotic ions are central to buffer systems such as the carbonate buffer in blood.

Identifying conjugate pairs

Question

In the reaction NH3+H2O⇌NH4++OH−\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-, identify the two conjugate acid-base pairs.

Solution

Ammonia accepts a proton from water to become the ammonium ion. So NH3\text{NH}_3 is a base and NH4+\text{NH}_4^+ is its conjugate acid: that is pair one (NH4+/NH3\text{NH}_4^+ / \text{NH}_3, differing by one proton).

Water donates the proton to become hydroxide, so H2O\text{H}_2\text{O} is the acid and OH−\text{OH}^- is its conjugate base: that is pair two (H2O/OH−\text{H}_2\text{O} / \text{OH}^-). Here water behaves as an acid, whereas with ethanoic acid it behaved as a base, which shows water is amphiprotic.

Why this matters

Recognising conjugate pairs lets you predict whether a salt solution is acidic, basic or neutral, explain buffer action, and write balanced proton-transfer equations confidently. Amphiprotic species are the working components of the buffers that keep biological and environmental systems stable.

Pairing across the wrong species

A common error is pairing an acid with the base on the other pair (for example matching CH3COOH\text{CH}_3\text{COOH} with OH−\text{OH}^-) when conjugates must differ by exactly one proton. Another is forgetting that water can be either the acid or the base in different equations, so its conjugate changes accordingly. Always check that your two members differ by a single H+\text{H}^+.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style5 marks
(a) For the reaction HSO4−(aq)+H2O(l)⇌SO42−(aq)+H3O+(aq)\text{HSO}_4^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{SO}_4^{2-}(aq) + \text{H}_3\text{O}^+(aq), identify the two conjugate acid-base pairs. (b) The hydrogen sulfate ion is amphiprotic. Write an equation showing it acting as a base. (c) Given Ka(HSO4−)=1.2×10−2K_a(\text{HSO}_4^-) = 1.2 \times 10^{-2}, calculate KbK_b for its conjugate base SO42−\text{SO}_4^{2-} at 25 ∘C25\ ^\circ\text{C}.
Show worked answer →

A 5 mark question rewards the pairs, the base equation, and the KbK_b calculation.

(a) HSO4−\text{HSO}_4^- donates a proton to become SO42−\text{SO}_4^{2-}, so the pair is HSO4−/SO42−\text{HSO}_4^-/\text{SO}_4^{2-} (acid/conjugate base). Water accepts the proton to become H3O+\text{H}_3\text{O}^+, so the second pair is H3O+/H2O\text{H}_3\text{O}^+/\text{H}_2\text{O} (acid/conjugate base).

(b) Acting as a base it accepts a proton:

HSO4−(aq)+H+(aq)⇌H2SO4(aq).\text{HSO}_4^-(aq) + \text{H}^+(aq) \rightleftharpoons \text{H}_2\text{SO}_4(aq).

(c) For a conjugate pair Ka×Kb=Kw=1.0×10−14K_a \times K_b = K_w = 1.0 \times 10^{-14}, so

Kb=KwKa=1.0×10−141.2×10−2=8.3×10−13.K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-2}} = 8.3 \times 10^{-13}.

The very small KbK_b confirms SO42−\text{SO}_4^{2-} is an extremely weak base, consistent with HSO4−\text{HSO}_4^- being a moderately strong acid.

Markers reward both correct pairs, the proton-accepting equation, and Kb=8.3×10−13K_b = 8.3 \times 10^{-13}.

WACE 2023-style4 marks
Explain what is meant by an amphiprotic species and, using the dihydrogen phosphate ion H2PO4−\text{H}_2\text{PO}_4^-, write equations showing it behaving as both a Bronsted-Lowry acid and a Bronsted-Lowry base.
Show worked answer →

A 4 mark answer needs the definition plus a correct equation for each role.

An amphiprotic species can either donate a proton (acting as a Bronsted-Lowry acid) or accept a proton (acting as a base), depending on the other reactant.

As an acid (donating a proton):

H2PO4−(aq)⇌HPO42−(aq)+H+(aq).\text{H}_2\text{PO}_4^-(aq) \rightleftharpoons \text{HPO}_4^{2-}(aq) + \text{H}^+(aq).

As a base (accepting a proton):

H2PO4−(aq)+H+(aq)⇌H3PO4(aq).\text{H}_2\text{PO}_4^-(aq) + \text{H}^+(aq) \rightleftharpoons \text{H}_3\text{PO}_4(aq).

Markers reward the donate-or-accept definition and one correct equation each, with the species differing by exactly one proton.

Practise this

Sources & how we know this

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