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How are acids and bases linked in conjugate pairs, and how can one species act as both an acid and a base?

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

A focused answer to the WACE Year 12 Chemistry dot point on conjugate acid-base pairs, how they differ by one proton, the inverse strength relationship, and amphiprotic species, with a worked example and common exam mistakes.

Reviewed by: AI editorial process; not yet individually human-reviewed

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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.

Identifying the pairs

Consider ethanoic acid reacting with water:

CH3COOH+H2Oβ‡ŒCH3COOβˆ’+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 CH3COOβˆ’\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.

The inverse strength relationship

For a conjugate pair, acid strength and conjugate base strength are inversely related: KaΓ—Kb=KwK_a \times K_b = K_w. A strong acid such as HCl is essentially fully ionised, so its conjugate base Clβˆ’\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).

The classic example is the hydrogen carbonate ion, HCO3βˆ’\text{HCO}_3^-:

  • As an acid (donating a proton): HCO3βˆ’β‡ŒCO32βˆ’+H+\text{HCO}_3^- \rightleftharpoons \text{CO}_3^{2-} + \text{H}^+
  • As a base (accepting a proton): HCO3βˆ’+H+β‡ŒH2CO3\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 H2PO4βˆ’\text{H}_2\text{PO}_4^- is another common example. Amphiprotic ions are central to buffer systems such as the carbonate buffer in blood.

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.

Exam-style practice questions

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

WACE 20215 marks(a) For the reaction HSO4βˆ’(aq)+H2O(l)β‡ŒSO42βˆ’(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(HSO4βˆ’)=1.2Γ—10βˆ’2K_a(\text{HSO}_4^-) = 1.2 \times 10^{-2}, calculate KbK_b for its conjugate base SO42βˆ’\text{SO}_4^{2-} at 25 ∘C25\ ^\circ\text{C}.
Show worked answer β†’

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

(a) HSO4βˆ’\text{HSO}_4^- donates a proton to become SO42βˆ’\text{SO}_4^{2-}, so the pair is HSO4βˆ’/SO42βˆ’\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:

HSO4βˆ’(aq)+H+(aq)β‡ŒH2SO4(aq).\text{HSO}_4^-(aq) + \text{H}^+(aq) \rightleftharpoons \text{H}_2\text{SO}_4(aq).

(c) For a conjugate pair KaΓ—Kb=Kw=1.0Γ—10βˆ’14K_a \times K_b = K_w = 1.0 \times 10^{-14}, so

Kb=KwKa=1.0Γ—10βˆ’141.2Γ—10βˆ’2=8.3Γ—10βˆ’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 SO42βˆ’\text{SO}_4^{2-} is an extremely weak base, consistent with HSO4βˆ’\text{HSO}_4^- being a moderately strong acid.

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

WACE 20234 marksExplain what is meant by an amphiprotic species and, using the dihydrogen phosphate ion H2PO4βˆ’\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 β†’

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):

H2PO4βˆ’(aq)β‡ŒHPO42βˆ’(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):

H2PO4βˆ’(aq)+H+(aq)β‡ŒH3PO4(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.

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