The big idea: The Brønsted–Lowry theory defines acids and bases by what they do with a proton — a hydrogen ion, H⁺.
- A Brønsted–Lowry acid is a proton (H⁺) donor. - A Brønsted–Lowry base is a proton (H⁺) acceptor.
A 'proton' here just means a hydrogen ion — a hydrogen atom that has lost its single electron, leaving a bare nucleus (one proton).
Brønsted–Lowry acid
- A proton (H⁺) donor.
- Loses an H⁺ in the reaction.
- Examples: HCl, H2SO4, CH3COOH, H3O+.
Brønsted–Lowry base
- A proton (H⁺) acceptor.
- Gains an H⁺ in the reaction.
- Examples: OH⁻, NH3, CO32-, H2O.
It takes two: An acid can only donate a proton if a base is there to accept it. So every proton-transfer reaction has both an acid and a base — one passes the H⁺, the other catches it.
Example: HCl + H2O → H3O+ + Cl⁻. Here HCl donates (acid) and water accepts (base).
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When an acid donates its proton, what is left behind is a base — it could accept that proton back. The two are linked as a conjugate acid–base pair: they differ by exactly one H⁺.
Differ by one proton: - Conjugate base = the acid minus one H⁺. - Conjugate acid = the base plus one H⁺.
Losing an H⁺ also removes a + charge (or adds a −). So Cl⁻ is the conjugate base of HCl, and NH4+ is the conjugate acid of NH3.
| Acid | loses H⁺ → | Conjugate base |
|---|---|---|
| HCl | −H⁺ | Cl⁻ |
| H2SO4 | −H⁺ | HSO4⁻ |
| CH3COOH | −H⁺ | CH3COO⁻ |
| NH4+ | −H⁺ | NH3 |
| H3O+ | −H⁺ | H2O |
| H2O | −H⁺ | OH⁻ |
Two pairs in one equation: In CH3COOH + H2O ⇌ CH3COO⁻ + H3O+:
- Pair 1: CH3COOH (acid) / CH3COO⁻ (its conjugate base). - Pair 2: H2O (base) / H3O+ (its conjugate acid).
The acid on the left and its conjugate base on the right always differ by just one H⁺.
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Some species can do both: An amphiprotic species can act as either a Brønsted–Lowry acid or a base — it can donate a proton in one reaction and accept one in another.
To be amphiprotic a species must have both a hydrogen it can donate and a lone pair (or charge) that lets it accept one.
| Species | As an acid (donates H⁺) | As a base (accepts H⁺) |
|---|---|---|
| H2O | → OH⁻ | → H3O+ |
| HCO3⁻ | → CO32- | → H2CO3 |
| HSO4⁻ | → SO42- | → H2SO4 |
| H2PO4⁻ | → HPO42- | → H3PO4 |
Worked example — water acting both ways
Show, with equations, how water can behave as both a Brønsted–Lowry acid and a Brønsted–Lowry base.
Solution
- As a base (accepts H⁺) — reacting with the acid HCl, water gains a proton to become H3O+:
- As an acid (donates H⁺) — reacting with the base NH3, water loses a proton to become OH⁻:
- Because water can both donate and accept a proton, it is amphiprotic.
Final answer
Water is amphiprotic: it accepts H⁺ (→ H3O+) with an acid, and donates H⁺ (→ OH⁻) with a base.
Spot an amphiprotic ion: The give-away is a hydrogen-containing anion like HCO3⁻, HSO4⁻ or H2PO4⁻ — it still has an H to donate, and a negative charge that lets it accept one.
How this is tested: Brønsted–Lowry shows up as a quick Paper 1A MCQ — 'pick the genuine conjugate pair (differing by one H⁺)', 'what is added to a base to form its conjugate acid?', or 'which species is amphiprotic?'.
On Paper 2 you are often asked to write an equation showing a species behaving as a proton donor (acid) or acceptor (base), and to identify the conjugate pairs in a given equation.
How to score it: A conjugate pair differs by exactly one H⁺ and one unit of charge. To form a conjugate acid, add an H⁺; to form a conjugate base, remove an H⁺. When you write an equation, make sure it balances for atoms and charge.
IB-style question — hydrogencarbonate (a)
(a) Hydrogencarbonate, HCO3⁻, is amphiprotic. Write an equation to show it acting as a Brønsted–Lowry acid in water. [1]
How to score the marks
- Acid = proton donor, so HCO3⁻ must lose an H⁺ to the water molecule (which accepts it).
- Losing one H⁺ turns HCO3⁻ into CO32- (charge goes from −1 to −2); water becomes H3O+:
- Check it balances for atoms and charge (left −1, right −2 + 1 = −1).
Final answer
HCO3⁻ + H2O ⇌ CO32- + H3O+ — HCO3⁻ donates a proton, so it acts as the acid.
IB-style question — hydrogencarbonate (b)
(b) Identify the conjugate base of HCO3⁻ and the conjugate acid of HCO3⁻. [2]
How to score the marks
- Conjugate base — remove one H⁺ from HCO3⁻: this gives CO32-.
- Conjugate acid — add one H⁺ to HCO3⁻: this gives H2CO3 (carbonic acid).
- Each conjugate species differs from HCO3⁻ by exactly one H⁺ and one unit of charge.
Final answer
Conjugate base = CO32-; conjugate acid = H2CO3.