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Conjugate pairs and amphiprotic species
Using the Brønsted–Lowry model to identify conjugate acid–base pairs, write ionisation equations for acids and bases in water (including polyprotic acids), explain the amphiprotic behaviour of water, hydrogen carbonate and dihydrogen phosphate, predict which way a proton transfer goes, and explain why some salts are acidic or basic.
Bicarbonate of soda neutralises acid in an antacid. It also neutralises the base in a cake recipe when it reacts with cream of tartar. The same ion, HCO₃⁻, can take a proton or give one away, depending on what it meets. Once you see every acid–base reaction as a single proton changing hands, you can predict what HCO₃⁻ will do, why soap feels slippery, and why a solution of a "neutral" salt can turn litmus red.
Conjugate acid–base pairs
When an acid donates a proton, what's left behind can accept a proton back. That leftover particle is the acid's conjugate base. Likewise, when a base accepts a proton it becomes its conjugate acid. A conjugate pair is two species that differ by exactly one H⁺.
$$\underset{\text{acid}}{\mathrm{HF}} + \underset{\text{base}}{\mathrm{H_2O}} \rightleftharpoons \underset{\text{conj. acid}}{\mathrm{H_3O^+}} + \underset{\text{conj. base}}{\mathrm{F^-}}$$
The pairs here are HF/F⁻ and H₃O⁺/H₂O. Always write a pair with the acid first, and check that the two differ by one H⁺ and one unit of charge. HSO₄⁻ and SO₄²⁻ are a conjugate pair; H₂SO₄ and SO₄²⁻ are not.
Identify the conjugate acid–base pairs in CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq).
Ionisation equations in water
When an acid dissolves, it donates a proton to water, forming hydronium ions. When a base dissolves, it accepts a proton from water, forming hydroxide ions. Use a single arrow (→) for strong acids and bases, which react completely, and an equilibrium arrow (⇌) for weak ones.
| Species | Equation in water |
|---|---|
| hydrochloric acid (strong) | HCl(g) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq) |
| ethanoic acid (weak) | CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq) |
| ammonia (weak base) | NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) |
| sodium hydroxide (strong base) | NaOH(s) → Na⁺(aq) + OH⁻(aq) (dissociation: the base is the OH⁻ ion) |
| ethanoate ion (weak base) | CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq) |
Polyprotic acids can donate more than one proton, one at a time. Each step has its own equation, and each step is weaker than the one before, because it is harder to pull a positive proton away from an ion that is already negative.
H₂SO₄(aq) + H₂O(l) → H₃O⁺(aq) + HSO₄⁻(aq) (complete)
HSO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + SO₄²⁻(aq) ($K_a$ = 1.0 × 10⁻²)
H₃PO₄ ⇌ H₂PO₄⁻ ⇌ HPO₄²⁻ ⇌ PO₄³⁻ (each step loses one H⁺ to water; $K_a$ = 7.1 × 10⁻³, 6.3 × 10⁻⁸, 4.5 × 10⁻¹³)
Strong acids have weak conjugate bases
The stronger an acid is at giving away its proton, the weaker its conjugate base is at taking one back. HCl ionises completely, so Cl⁻ has essentially no tendency to accept a proton: it is a negligibly weak base. Ethanoic acid holds its proton fairly tightly, so ethanoate is a noticeably basic ion.
This lets you predict which way a proton transfer goes: the equilibrium favours the side with the weaker acid and weaker base. The proton ends up on whichever base holds it more tightly. For a reaction HA + B ⇌ A⁻ + HB⁺, the equilibrium constant is $K = K_a(\mathrm{HA}) \div K_a(\mathrm{HB^+})$.
Predict whether ethanoic acid reacts substantially with the carbonate ion. ($K_a$(CH₃COOH) = 1.8 × 10⁻⁵, $K_a$(HCO₃⁻) = 4.7 × 10⁻¹¹.)
- Start with CH₃COOH and H₂O and press "Transfer the proton". What is K? Where have you seen that number before?
- Change the acceptor to OH⁻. How big is K now, and which way does the equilibrium lie?
- Choose HCl and H₂O. Then choose NH₄⁺ and H₂O. Compare the two values of K: which is the strong acid?
- Choose HCO₃⁻ as the donor and OH⁻ as the acceptor. Then choose H₃O⁺ as the donor and HCO₃⁻ as the acceptor. What does the sim say about HCO₃⁻ each time?
- Choose H₂O as both the donor and the acceptor. What value of K do you get, and what is its usual name?
Amphiprotic species
An amphiprotic species can both donate and accept a proton: it can act as a Brønsted–Lowry acid or base. To be amphiprotic it must have a proton it can lose and a lone pair (usually a negative charge) to take one on. Water is the most important example: it is a base towards HCl and an acid towards NH₃. In pure water, it does both at once (6F):
$$\mathrm{H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)}$$
The syllabus names two salts. For each, write one equation where it acts as an acid and one where it acts as a base.
Sodium hydrogen carbonate
as an acid: HCO₃⁻(aq) + OH⁻(aq) → CO₃²⁻(aq) + H₂O(l)
as a base: HCO₃⁻(aq) + H₃O⁺(aq) → H₂CO₃(aq) + H₂O(l) (then H₂CO₃ → CO₂ + H₂O)
Potassium dihydrogen phosphate
as an acid: H₂PO₄⁻(aq) + H₂O(l) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq)
as a base: H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃PO₄(aq) + OH⁻(aq)
The Na⁺ and K⁺ ions are spectators, so you can write the equations with just the anion. Other amphiprotic species include HSO₄⁻, HPO₄²⁻ and HS⁻.
Amphiprotic is not the same as amphoteric. Amphoteric means able to react as an acid or a base in any sense, and it is mostly used for oxides such as ZnO and Al₂O₃ (6A), which have no protons to donate. All amphiprotic species are amphoteric, but not all amphoteric substances are amphiprotic.
Whether an amphiprotic salt solution ends up acidic or basic depends on which tendency is stronger. HCO₃⁻ is a better base than acid, so NaHCO₃ solution has a pH of about 8.3. H₂PO₄⁻ is a better acid than base, so KH₂PO₄ solution has a pH of about 4.7.
Acidic, basic and neutral salts
A salt dissolves into a cation and an anion. Ask of each ion: is it the conjugate of a weak acid or base? If so, it reacts with water. If it comes from a strong acid or base, it doesn't.
| Salt made from | Example | Ion that reacts with water | pH of solution |
|---|---|---|---|
| strong acid + strong base | NaCl, KNO₃ | neither | 7 (neutral) |
| weak acid + strong base | CH₃COONa, Na₂CO₃ | anion: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ | above 7 (basic) |
| strong acid + weak base | NH₄Cl, NH₄NO₃ | cation: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ | below 7 (acidic) |
| weak acid + weak base | CH₃COONH₄ | both | depends on which is stronger |
This matters for titrations: at the equivalence point of a weak acid–strong base titration, the flask contains a basic salt, so the pH is above 7 (6I).
Test yourself
What is the conjugate base of H₂PO₄⁻?
Remove one H⁺: H₂PO₄⁻ → HPO₄²⁻.
In NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), the conjugate acid–base pairs are
Each pair differs by one H⁺ and is written acid first.
Which 0.1 mol/L solution has a pH below 7?
NH₄⁺ is the conjugate acid of the weak base NH₃, so it donates protons to water.
Which species is amphiprotic?
HSO₄⁻ can lose a proton (to SO₄²⁻) or gain one (to H₂SO₄). Al₂O₃ is amphoteric but has no protons.
Given $K_a$(HF) = 6.8 × 10⁻⁴ and $K_a$(NH₄⁺) = 5.6 × 10⁻¹⁰, the reaction HF + NH₃ ⇌ F⁻ + NH₄⁺
K = 6.8 × 10⁻⁴ ÷ 5.6 × 10⁻¹⁰ = 1.2 × 10⁶: HF is a much stronger acid than NH₄⁺.
Label the species in the reaction.
Sort these 0.1 mol/L salt solutions by pH.
The conjugate base of a strong acid is a very base. A species that can donate or accept a proton is . When ammonia accepts a proton it forms its conjugate acid, the ion. An acid that can donate more than one proton is . A proton transfer favours the formation of the acid and base.
Potassium dihydrogen phosphate (KH₂PO₄) and sodium hydrogen carbonate (NaHCO₃) are both described as amphiprotic. (a) Explain what amphiprotic means. (b) Write equations showing H₂PO₄⁻ acting as an acid and as a base in water. (c) A 0.1 mol/L solution of NaHCO₃ has pH 8.3. Explain this using the Brønsted–Lowry model.
(a) An amphiprotic species can act as either a Brønsted–Lowry acid (proton donor) or base (proton acceptor), depending on what it reacts with.
(b) As an acid: H₂PO₄⁻(aq) + H₂O(l) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq). As a base: H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃PO₄(aq) + OH⁻(aq).
(c) In water, Na⁺ is a spectator ion. HCO₃⁻ can donate a proton to water, HCO₃⁻(aq) + H₂O(l) ⇌ CO₃²⁻(aq) + H₃O⁺(aq), or accept one from water, HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq). HCO₃⁻ is a better proton acceptor than donor, so the second reaction occurs to a greater extent. More OH⁻ than H₃O⁺ is produced, so the solution is slightly basic, pH 8.3.