Acid–base reactions are proton transfers
Lesson 2.12.1 introduced the Brønsted–Lowry definitions: an acid donates a proton (H⁺), and a base accepts one. AP Unit 4 uses them to analyze every acid–base reaction:
HCl donates H⁺, so it’s the acid; NH₃ accepts H⁺, so it’s the base. Afterward, NH₄⁺ can donate H⁺ again, so it’s the conjugate acid共轭酸 of NH₃; Cl⁻ can accept H⁺, so it’s the conjugate base共轭碱 of HCl. The two bases in this reaction are therefore NH₃ and Cl⁻, and the two acids are HCl and NH₄⁺.
The members of a conjugate pair differ by one H⁺: the conjugate acid of NH₃ is NH₄⁺, and the conjugate acid of NH₂⁻ is NH₃.
Amphiprotic species
Some particles can both donate and accept H⁺. They are amphiprotic两性物质:
- H₂O accepts H⁺ from HCl (acting as a base) and donates H⁺ to NH₃ (acting as an acid).
- HCO₃⁻ donates H⁺ to OH⁻, forming CO₃²⁻, and accepts H⁺ from H₃O⁺, forming H₂CO₃.
- HC₂O₄⁻: oxalic acid, H₂C₂O₄, donates its protons in two steps, and HC₂O₄⁻ is both the conjugate base of the first step and the acid of the second:
Click each particle to label it acid, base, conjugate acid, or conjugate base:
A weak acid in water, particle by particle
A strong acid such as HCl gives up essentially all its protons to water, so the solution contains H₃O⁺ and Cl⁻ and no HCl molecules. A weak acid such as HF gives up only a small fraction of its protons, so the solution is mostly HF molecules with a few H₃O⁺ and F⁻ ions, equal in number.
HF is a weak acid. Which diagram best represents the solute particles in an HF solution? (Water molecules are left out.)
Only a small fraction of a weak acid ionizes: mostly HF molecules, with a few H₃O⁺ and F⁻ ions in equal numbers. A shows complete ionization (a strong acid); C has unequal positive and negative charges; D shows no ionization at all, which is also wrong.
Comparing strengths with K
A reaction in which two weak acids compete for a proton can be written:
Equilibrium favors the side with the weaker acid and the weaker base. So:
- K > 1: the right side is favored; HA is a stronger acid than HB, and B⁻ is a stronger base than A⁻.
- K < 1: the left side is favored; HB is the stronger acid, and A⁻ is the stronger base.
For example, has K = 3.7 × 10⁴ (the ratio of the two acids’ Kₐ values), so CH₃COOH is a stronger acid than HCN, and CN⁻ is a stronger base than CH₃COO⁻. Likewise, goes essentially to completion, so CO₃²⁻ is a much stronger base than SO₄²⁻.
This comparison is qualitative; pH calculations come in Unit 8. Choose two acids to see K and the direction of the equilibrium:
Kₐ values (25 °C) from OpenStax, Chemistry 2e, Appendix H. This comparison is qualitative; pH calculations come in Unit 8.
Example 1: HCO₃⁻ plays both roles
Label the acid, base, conjugate acid, and conjugate base in each reaction: (1) ; (2) .
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Identify: the test. In the Brønsted–Lowry model, an acid donates and a base accepts it; a conjugate pair differs by exactly one . For each reaction, follow the .
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Reaction (1): follow the . becomes , losing an ; becomes , gaining it:
- Acid: ; conjugate base:
- Base: ; conjugate acid:
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Reaction (2): follow the . becomes , losing an ; becomes , gaining it:
- Acid: ; conjugate base:
- Base: ; conjugate acid:
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Compare the role of . It donates in (1) (acid) and accepts in (2) (base): is amphiprotic, and its role depends on what it reacts with.
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Check. All four conjugate pairs (/, /, /, /) differ by exactly one and by a charge of 1.
Example 2: Strength from K
For HF + CH₃COO⁻ ⇌ F⁻ + CH₃COOH, K = Kₐ(HF) ÷ Kₐ(CH₃COOH) = (6.4 × 10⁻⁴) ÷ (1.8 × 10⁻⁵) = 36. Compare the strengths of the two acids and of the two bases.
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Identify: what K says. First confirm K:
K > 1, so the right side (products) is favored.
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Compare the acids. A proton-transfer equilibrium favors the side with the weaker acid. The acid on the right is and the acid on the left is HF, so is weaker and HF is the stronger acid (it also has the larger Kₐ).
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Compare the bases. The equilibrium likewise favors the weaker base. The base on the right is and the base on the left is , so is weaker and is the stronger base.
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Connect to the pattern. The stronger the acid, the weaker its conjugate base:
The stronger acid and stronger base (HF and ) are on the same side, and the equilibrium favors the other side.
- H₂O in HCl + H₂O → H₃O⁺ + Cl⁻
H₂O accepts H⁺.
- H₂O in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
H₂O donates H⁺.
- NH₄⁺ in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
NH₄⁺ is the conjugate acid; it can donate H⁺.
- OH⁻ in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
OH⁻ is the conjugate base of H₂O.
- Cl⁻ in NH₃ + HCl → NH₄⁺ + Cl⁻
Cl⁻ is the conjugate base of HCl.
- HCO₃⁻ in HCO₃⁻ + OH⁻ → CO₃²⁻ + H₂O
HCO₃⁻ donates H⁺.
- HCO₃⁻ in HCO₃⁻ + H₃O⁺ → H₂CO₃ + H₂O
HCO₃⁻ accepts H⁺.
- HCN in CH₃COOH + CN⁻ ⇌ CH₃COO⁻ + HCN
HCN is the conjugate acid of CN⁻.
- HC₂O₄⁻ in HC₂O₄⁻ + H₂O ⇌ C₂O₄²⁻ + H₃O⁺
HC₂O₄⁻ donates H⁺.
- NH₂⁻ in NH₂⁻ + H₂O → NH₃ + OH⁻
NH₂⁻ accepts H⁺, forming its conjugate acid, NH₃.
Common mistakes
Mistake 1: “The conjugate base of a strong acid is a strong base.” It’s the reverse: the stronger an acid, the more readily it gives up H⁺, and the less its conjugate base wants H⁺ back. Cl⁻, the conjugate base of HCl, is almost not basic at all.
Mistake 2: “Water is just the solvent; it doesn’t take part in acid–base reactions.” Water is amphiprotic: it acts as a base toward HCl and as an acid toward NH₃.
Mistake 3: “K > 1 means the reactant acid is weak, so it reacts easily.” K > 1 means the products are favored, and equilibrium favors the weaker acid and weaker base, so the reactant acid is the stronger one.
Quick check
Which two particles act as bases in NH₃ + HCl → NH₄⁺ + Cl⁻?
NH₃ accepts H⁺, and Cl⁻, the conjugate base of HCl, can also accept H⁺. B and C each pair an acid with its conjugate base.
What is the conjugate acid of NH₂⁻?
A conjugate acid has one more H⁺ than its base: NH₂⁻ + H⁺ → NH₃.
HF + CN⁻ ⇌ F⁻ + HCN has K ≈ 1.3 × 10⁶. Which statement is correct?
K > 1, so equilibrium favors the weaker acid, HCN, and the weaker base, F⁻. HF is therefore the stronger acid and CN⁻ the stronger base.
HClO + CH₃COO⁻ ⇌ ClO⁻ + CH₃COOH has K < 1. What can be concluded?
K < 1, so the left side, with the weaker acid HClO and the weaker base CH₃COO⁻, is favored. CH₃COOH is therefore the stronger acid and ClO⁻ the stronger base.
Which particle is amphiprotic?
HCO₃⁻ can donate H⁺ (forming CO₃²⁻) and accept H⁺ (forming H₂CO₃). Cl⁻ and CO₃²⁻ can only accept H⁺; NH₄⁺ can only donate it.
The reaction HSO₄⁻ + CO₃²⁻ → SO₄²⁻ + HCO₃⁻ goes essentially to completion. Of CO₃²⁻, SO₄²⁻, HCO₃⁻, and HSO₄⁻, which is the strongest base?
Equilibrium favors the weaker base, SO₄²⁻, so CO₃²⁻ is a stronger base than SO₄²⁻; HCO₃⁻, the conjugate acid of CO₃²⁻, is a weaker base still; and HSO₄⁻ acts as the acid here.
AP-style free response
Oxalic acid (H₂C₂O₄) is a diprotic weak acid.
(a) Write equations for the two steps in which oxalic acid donates protons to water, and identify the conjugate acid–base pairs in each step.
(b) Which of H₂C₂O₄, HC₂O₄⁻, and C₂O₄²⁻ is amphiprotic? Justify your answer.
(c) HC₂O₄⁻ + CH₃COO⁻ ⇌ C₂O₄²⁻ + CH₃COOH is found to have K > 1. Compare the acid strengths of HC₂O₄⁻ and CH₃COOH, and the base strengths of CH₃COO⁻ and C₂O₄²⁻.
(d) NaOH solution is added to NaHC₂O₄ solution. Write the net ionic equation for the reaction, and identify the acid and the base.
Model answer and scoring
(a) Step 1: , with conjugate pairs H₂C₂O₄/HC₂O₄⁻ and H₃O⁺/H₂O. Step 2: , with conjugate pairs HC₂O₄⁻/C₂O₄²⁻ and H₃O⁺/H₂O.
(b) HC₂O₄⁻. It can donate H⁺ to become C₂O₄²⁻ (acting as an acid) and accept H⁺ to become H₂C₂O₄ (acting as a base).
(c) K > 1, so equilibrium favors the weaker acid, CH₃COOH, and the weaker base, C₂O₄²⁻. HC₂O₄⁻ is therefore a stronger acid than CH₃COOH, and CH₃COO⁻ is a stronger base than C₂O₄²⁻.
(d) . HC₂O₄⁻ is the acid and OH⁻ the base.
Scoring: (a) both equations and their conjugate pairs; (b) HC₂O₄⁻, with a reason for each direction; (c) 1 point each for the acid and base comparisons; (d) net ionic equation (no Na⁺), with the acid and base.
Summary
- Acid–base reactions are proton transfers: the acid donates H⁺ and the base accepts it; the members of a conjugate pair differ by one H⁺.
- Amphiprotic species (H₂O, HCO₃⁻, HC₂O₄⁻) can act as either acid or base.
- A weak acid solution is mostly acid molecules, with a few H₃O⁺ ions and conjugate-base ions; a strong acid ionizes almost completely.
- Equilibrium favors the weaker acid and weaker base: with K > 1 the reactant acid and base are stronger, and with K < 1 the product acid and base are.
AP tip: When comparing strengths from K, first state which side is favored, then say that side has the weaker acid and the weaker base. Refer to the direction of proton transfer; “K is large, so it’s strong” isn’t a reason.