Acids, bases and salts: three definitions
Electrolytes are substances that give ions in water. Strong electrolytes (like NaCl, HCl) ionise almost fully; weak electrolytes (like acetic acid, NH₃) ionise only a little.
1. Arrhenius: an acid gives H⁺ (really H₃O⁺) in water; a base gives OH⁻ in water. Limit: works only in water, and cannot explain why NH₃ is a base.
2. Brønsted–Lowry: an acid is a proton (H⁺) donor; a base is a proton acceptor. When an acid gives H⁺, what is left is its conjugate base. When a base takes H⁺, it becomes its conjugate acid. A conjugate pair differs by one H⁺. Example: in NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, water is the acid and OH⁻ its conjugate base; NH₃ is the base and NH₄⁺ its conjugate acid. Water can act as both (amphoteric). Strong acid → very weak conjugate base.
3. Lewis: an acid is an electron-pair acceptor; a base is an electron-pair donor. Examples: BF₃, AlCl₃, H⁺, Co³⁺ are Lewis acids; NH₃, H₂O, OH⁻ are Lewis bases. This works even without H⁺.
Salts form when an acid reacts with a base. The acid and base they come from decide whether their solution is acidic, basic or neutral (see hydrolysis below).
Ionisation of acids and bases: Ka, Kb and degree of ionisation
A weak acid HA sets up an equilibrium: HA + H₂O ⇌ H₃O⁺ + A⁻, with Ka = [H₃O⁺][A⁻]/[HA]. A weak base: B + H₂O ⇌ BH⁺ + OH⁻, with Kb = [BH⁺][OH⁻]/[B]. Bigger Ka means stronger acid. We also use pKa = −log Ka: smaller pKa means stronger acid.
Degree of ionisation α = fraction of molecules that split. For concentration C: Ka = Cα²/(1 − α) ≈ Cα² (when α is small). So α ≈ √(Ka/C) and [H⁺] = Cα = √(Ka·C).
Ostwald's dilution law: α grows as C falls. More water = more room for ions to stay apart.
Conjugate pair: Ka × Kb = Kw, so pKa + pKb = 14 at 25 °C.
Polybasic acids (H₂SO₄, H₂CO₃, H₃PO₄) lose H⁺ one at a time: Ka₁ > Ka₂ > Ka₃, because it is harder to pull H⁺ from an already negative ion. Usually only Ka₁ matters for pH.
What makes an acid strong? Across a period, a more electronegative atom holds the electrons, so H–A breaks easily (CH₄ < NH₃ < H₂O < HF). Down a group, the H–A bond gets weaker as A gets bigger (HF < HCl < HBr < HI).
Common ion effect: adding an ion that is already in the equilibrium (like acetate from sodium acetate to acetic acid) pushes it backward, so the weak acid ionises even less.
Ionisation of water and the pH scale
Water ionises slightly: H₂O + H₂O ⇌ H₃O⁺ + OH⁻. Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K (ionic product of water). In pure water [H₃O⁺] = [OH⁻] = 10⁻⁷ M. Kw rises with temperature, because ionisation takes in heat.
pH = −log[H₃O⁺], pOH = −log[OH⁻], and pH + pOH = pKw = 14 at 25 °C.
- Acidic: [H⁺] > 10⁻⁷ M, pH < 7.
- Neutral: pH = 7.
- Basic: [H⁺] < 10⁻⁷ M, pH > 7.
A change of 1 pH unit means a 10-times change in [H⁺]. For very dilute acids (like 10⁻⁸ M HCl) you must add water's own 10⁻⁷ M, so the pH is just below 7, not 8.
Try it at home: boil red cabbage leaves in water and cool. Add drops to lemon juice, water and soap water. Red means acidic, purple neutral, green-yellow basic.
Hydrolysis of salts and the pH of their solutions
Salt ions from a weak parent react with water; this is salt hydrolysis.
- Strong acid + strong base (NaCl): no hydrolysis, pH = 7.
- Weak acid + strong base (CH₃COONa): CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, basic. pH = 7 + ½(pKa + log C).
- Strong acid + weak base (NH₄Cl): NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, acidic. pH = 7 − ½(pKb + log C).
- Weak acid + weak base (CH₃COONH₄): both ions react. pH = 7 + ½(pKa − pKb), independent of C.
Exam corner
Expect: identify conjugate pairs, classify Lewis acids/bases, pH of strong and weak acids/bases, Ka from pH, α from Ka, Ka × Kb = Kw, and pH of salt solutions. Always check if the answer is sensible: an acid's pH must be below 7.
Key formulas and definitions
- Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ (25 °C); pH + pOH = 14
- pH = −log[H⁺]; pOH = −log[OH⁻]; pKa = −log Ka
- Weak acid: Ka = Cα²/(1 − α) ≈ Cα²; α ≈ √(Ka/C); [H⁺] = √(Ka·C)
- Conjugate pair: Ka × Kb = Kw; pKa + pKb = 14
- Salt of weak acid + strong base: pH = 7 + ½(pKa + log C)
- Salt of strong acid + weak base: pH = 7 − ½(pKb + log C); weak + weak: pH = 7 + ½(pKa − pKb)
Worked examples
1. Find the conjugate base of H₂SO₄ and HCO₃⁻, and the conjugate acid of NH₃ and H₂O.
Remove one H⁺ for the conjugate base: HSO₄⁻ and CO₃²⁻. Add one H⁺ for the conjugate acid: NH₄⁺ and H₃O⁺.
2. Find the pH of 0.001 M HCl.
HCl is strong, so [H⁺] = 10⁻³ M. pH = −log 10⁻³ = 3.
3. Find the pH of 0.01 M NaOH.
[OH⁻] = 10⁻² M, so pOH = 2. pH = 14 − 2 = 12.
4. Find the pH of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵).
α = √(Ka/C) = √(1.8 × 10⁻⁴) = 0.0134. [H⁺] = 0.1 × 0.0134 = 1.34 × 10⁻³ M. pH = 3 − log 1.34 = 3 − 0.13 = 2.87.
5. The pH of 0.1 M HCOOH is 2.39. Find Ka and α.
[H⁺] = 10⁻²·³⁹ = 4.07 × 10⁻³ M. α = 4.07 × 10⁻³/0.1 = 0.0407. Ka ≈ [H⁺]²/C = (4.07 × 10⁻³)²/0.1 = 1.66 × 10⁻⁴.
6. For NH₃, Kb = 1.8 × 10⁻⁵. Find Ka of NH₄⁺ and say which is stronger: NH₃ as a base or NH₄⁺ as an acid.
Ka = Kw/Kb = 10⁻¹⁴/1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. NH₃ (Kb 10⁻⁵) is a much stronger base than NH₄⁺ is an acid.
7. Find the pH of 10⁻⁸ M HCl.
Water also gives H⁺. Let [H⁺] = 10⁻⁸ + x, [OH⁻] = x. (10⁻⁸ + x)x = 10⁻¹⁴ → x ≈ 9.5 × 10⁻⁸. [H⁺] = 1.05 × 10⁻⁷ M. pH ≈ 6.98 (slightly acidic, not 8).
8. Find the pH of 0.1 M sodium acetate (pKa of acetic acid = 4.74).
Salt of weak acid + strong base: pH = 7 + ½(pKa + log C) = 7 + ½(4.74 − 1) = 7 + 1.87 = 8.87. Basic, as expected.
Common mistakes
- Using Ka for a strong acid. Strong acids ionise fully: [H⁺] = C directly.
- Thinking dilution makes a weak acid stronger. α increases, but [H⁺] still falls and pH rises.
- Forgetting water's 10⁻⁷ M for very dilute acids, and getting a pH above 7 for an acid.
- Mixing up the conjugate pair: the conjugate base of an acid has one H⁺ less, not one OH⁻ more.