Brønsted-Lowry Acids and Bases Revision Notes | A Level AQA Chemistry

1. Why Brønsted-Lowry Changes Everything

You probably met acids at GCSE as substances that produce $H^+$ ions in water. That definition works for hydrochloric acid in a beaker — but it falls apart the moment you ask: what happens when there is no water? The Brønsted-Lowry model fixes this by shifting the focus from water to the proton itself.

At A-level, the key question is not where the proton comes from, but who gives it and who takes it. That single shift in perspective unlocks buffer chemistry, $K_a$ calculations, and pH equilibria — all of which build directly on what you are about to learn here.

2. The Brønsted-Lowry Definitions

Picture a reaction between hydrogen chloride gas and ammonia gas — no water involved at all. A white solid, ammonium chloride, forms instantly. Something has clearly been transferred between the two molecules. That something is a proton, $H^+$.

The ammonia molecule, $NH_3$, has a lone pair of electrons on its nitrogen atom. That lone pair attracts the proton released by $HCl$ and forms a new $N-H$ bond, producing the ammonium ion $NH_4^+$.

The equation for this reaction is: $HCl(g) + NH_3(g) \rightarrow NH_4^+(s) + Cl^-(s)$. Notice that $HCl$ loses a proton and $NH_3$ gains one — proton transfer is the defining event.

3. Proton Transfer and Acid-Base Equilibria

Most acid-base reactions in aqueous solution are reversible. The forward reaction involves proton donation from the acid to the base; the reverse reaction involves the products donating and accepting the same proton back. This is why we write acid-base equilibria with a reversible arrow $\rightleftharpoons$.

Consider ethanoic acid dissolving in water: $CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)$. In the forward direction, $CH_3COOH$ donates a proton to $H_2O$. In the reverse direction, $H_3O^+$ donates a proton back to $CH_3COO^-$. Both directions involve proton transfer — the equilibrium describes which direction is favoured.

For a weak acid like ethanoic acid, the equilibrium lies well to the left: most molecules remain undissociated. For a strong acid like hydrochloric acid, the equilibrium lies so far to the right that we treat the reaction as essentially complete and use a one-way arrow.

4. Conjugate Acid-Base Pairs

Here is where the model becomes particularly powerful. When an acid donates a proton, the species left behind can accept a proton in the reverse reaction — it has become a base. This pair of species, differing by exactly one proton, is called a conjugate acid-base pair.

Return to the ethanoic acid equilibrium. $CH_3COOH$ donates $H^+$ to form $CH_3COO^-$: these two are a conjugate pair. $H_2O$ accepts $H^+$ to form $H_3O^+$: these two are the second conjugate pair. Every Brønsted-Lowry equilibrium contains exactly two conjugate acid-base pairs.

The rule for identifying conjugate pairs is straightforward: the conjugate base of an acid has one fewer proton and one more negative charge. The conjugate acid of a base has one more proton and one fewer negative charge. Check this against the table above — $CH_3COO^-$ has one fewer $H$ and one more negative charge than $CH_3COOH$.

Worked example: Identifying Conjugate Acid-Base Pairs

Identify both conjugate acid-base pairs in the equilibrium: $NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$

1. $H_2O$ donates $H^+$ to $NH_3$ — Track which species loses a proton going left to right. $H_2O$ becomes $OH^-$ — it has lost one proton, so $H_2O$ is the acid.
2. $NH_3$ accepts $H^+$ to become $NH_4^+$ — $NH_3$ gains one proton to form $NH_4^+$, so $NH_3$ is the Brønsted-Lowry base in the forward reaction.
3. Acid: $H_2O$ / Conjugate base: $OH^-$ — $H_2O$ and $OH^-$ differ by exactly one proton. $OH^-$ is the conjugate base of $H_2O$.
4. Base: $NH_3$ / Conjugate acid: $NH_4^+$ — $NH_3$ and $NH_4^+$ differ by exactly one proton. $NH_4^+$ is the conjugate acid of $NH_3$.
5. In the reverse reaction: $NH_4^+$ donates $H^+$ to $OH^-$ — This confirms the pairs: $NH_4^+$ acts as an acid and $OH^-$ acts as a base in the reverse direction — consistent with the conjugate pairs identified.

Answer: Conjugate pair 1: $H_2O$ (acid) / $OH^-$ (conjugate base). Conjugate pair 2: $NH_3$ (base) / $NH_4^+$ (conjugate acid).

5. Amphoteric Species and the Role of Water

In the ammonia equilibrium above, water appears as the acid in one conjugate pair ($H_2O$/$OH^-$), while $NH_3$/$NH_4^+$ forms the second pair. Across different reactions, water can also act as a base — for example, accepting a proton from a stronger acid such as $HCl$ to form $H_3O^+$. A species that can act as both a Brønsted-Lowry acid and a Brønsted-Lowry base depending on its reaction partner is described as amphoteric.

Water is the classic amphoteric species. With a stronger acid like $HCl$, water acts as the base and accepts a proton to form $H_3O^+$. With a stronger base like $NH_3$, water acts as the acid and donates a proton to form $OH^-$. The same molecule, two different roles — determined entirely by what it is reacting with.

The hydrogencarbonate ion, $HCO_3^-$, is another important amphoteric species. It can donate a proton (acting as an acid) to form $CO_3^{2-}$, or accept a proton (acting as a base) to form $H_2CO_3$. Recognising amphoteric species is a common higher-order demand in AQA questions.

6. Exam Technique and Mark-Winning Precision

AQA questions on this topic tend to follow a predictable pattern: you are given an equation and asked to identify the acid, the base, and the two conjugate pairs. The marks are awarded for precision of language, not just for naming the right species.

When a question uses the command word identify, name the species and state its role explicitly. When the command word is explain, you must include a reason — for example: '$HCl$ is the Brønsted-Lowry acid because it donates a proton to $H_2O$.' The word 'because' is doing real work here and earns the explanation mark.

For conjugate pair questions, always present both members of each pair together and label which is the acid and which is the base. Writing '$CH_3COOH / CH_3COO^-$' without labels risks losing a mark if the examiner cannot determine which direction you mean.