Brønsted–Lowry acid–base equilibria in aqueous solution (A-level only) — AQA A-Level Chemistry
Test yourself on Brønsted–Lowry acid–base equilibria in aqueous solution (A-level only) with AQA A-Level practice questions.
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Brønsted–Lowry acid–base equilibria in aqueous solution (A-level only) explained
Brønsted and Lowry defined an acid as a proton donor.
Read the full explanation
A proton here means H⁺, a hydrogen atom that has lost its only electron, so a bare nucleus. In water that proton is never free: it bonds to a lone pair on a water molecule, which is why HCl + H₂O → H₃O⁺ + Cl⁻ is the honest equation and [H⁺] is a shorthand for [H₃O⁺]. What remains after the proton has gone is the conjugate base, Cl⁻ here. The definition says nothing about how much of the acid donates. A strong acid such as HNO₃ gives up essentially every proton; a weak acid such as CH₃COOH gives up only a small fraction and sits at an equilibrium. Strength is about that fraction; concentration is about how many moles are in a cubic decimetre.
A base is a proton acceptor.
A base is a proton acceptor, and to accept a proton it must have a lone pair of electrons available to form the new bond. Ammonia has a lone pair on nitrogen, so NH₃ + H₂O ⇌ NH₄⁺ + OH⁻: the ammonia takes a proton from water, and NH₄⁺ is its conjugate acid. This definition is wider than 'a substance that produces hydroxide ions', which would exclude ammonia and the carbonate ion altogether. An alkali is the narrower idea: a base that dissolves in water and so releases OH⁻(aq). Water itself is amphoteric, acting as a base towards HCl and as an acid towards NH₃, which is why it turns up on both sides of this topic and why its own slight dissociation matters later.
Acid–base equilibria involve the transfer of protons.
Every acid and base reaction is a proton changing owner, so every one contains two conjugate pairs whose members differ by exactly one H⁺. In CH₃COOH + NH₃ ⇌ CH₃COO⁻ + NH₄⁺ the pairs are CH₃COOH with CH₃COO⁻ and NH₄⁺ with NH₃. Each acid is paired with the base that came from it by losing the proton, not with the base it reacted with. Because these are equilibria, the reaction need not go to completion, and the position tells you about relative strengths: the equilibrium always lies away from the stronger acid and towards the weaker one. That is why the conjugate base of a strong acid, such as Cl⁻, is an extremely poor base; it has almost no tendency to take the proton back.
Your focus
- Identify the proton donor in the equilibrium CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻.
- Write the conjugate base of HNO₃, of H₂SO₄ and of NH₄⁺.
- Distinguish a strong acid from a concentrated acid using the extent of dissociation.
Show all 9 objectives
- Identify the proton acceptor in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ and mark the lone pair involved.
- Write the conjugate acid of OH⁻, of NH₃ and of CH₃COO⁻.
- Explain why water is described as amphoteric, using one equation in which it donates a proton and one in which it accepts one.
- Label both conjugate acid and base pairs in CH₃COOH + NH₃ ⇌ CH₃COO⁻ + NH₄⁺.
- Predict which side the equilibrium HCl + CH₃COO⁻ ⇌ Cl⁻ + CH₃COOH lies on and justify it from acid strength.
- Explain, using an equation, why the conjugate base of a strong acid is a very weak base.
Brønsted–Lowry acid–base equilibria in aqueous solution (A-level only) exam tips
Marking Points
- one mark for defining an acid as a proton donor or H⁺ donor
- one mark for identifying the species that loses the proton in a given equation
- one mark for naming the conjugate base as the particle remaining after the proton has been donated
- one mark for writing H₃O⁺ rather than H⁺ when the acid reacts with water
- one mark for defining a base as a proton acceptor or H⁺ acceptor
- one mark for identifying the lone pair that accepts the proton
- one mark for naming the conjugate acid formed when the base gains the proton
- one mark for distinguishing an alkali as a base that dissolves to give OH⁻(aq)
- one mark for identifying both conjugate acid and base pairs in a given equation
- one mark for showing the proton being transferred from the acid to the base
- one mark for stating that the members of a pair differ by one H⁺
- one mark for deducing which side the equilibrium lies on from the relative strengths of the two acids
Examiner Tips
- 💡Write the word 'proton' in the definition; 'hydrogen ion' is usually accepted but 'hydrogen' alone is not.
- 💡To find a conjugate base, remove exactly one H⁺ from the acid and reduce the charge by one.
- 💡If asked why a species can act as a base, point to the lone pair, not to a negative charge alone.
- 💡To write a conjugate acid, add one H⁺ and increase the charge by one.
- 💡Label the pairs as acid 1 / base 1 and acid 2 / base 2 across the equation; the numbering makes the pairing obvious.
- 💡The stronger acid is always on the side the equilibrium lies away from.
Common Mistakes
- defining an acid as a substance that produces hydrogen ions in solution, which is the Arrhenius idea rather than Brønsted–Lowry
- calling an acid a 'hydrogen donor', which is refused because a hydrogen atom carries an electron
- confusing strength with concentration, so concentrated ethanoic acid is described as a strong acid
- giving the conjugate base of H₂SO₄ as SO₄²⁻ rather than HSO₄⁻
- defining a base as a substance that produces hydroxide ions, which excludes ammonia
- using base and alkali as interchangeable words
- giving the conjugate acid of NH₃ as NH₂⁻, which is its conjugate base
- forgetting that water can act as a base towards HCl and as an acid towards NH₃
- pairing an acid with the base on the opposite side of the equation rather than with its own conjugate
- saying the members of a conjugate pair differ by a hydrogen atom instead of a proton
- using a single arrow, so the reaction is treated as going to completion
- treating water as a spectator when it is in fact one member of a conjugate pair