All revision notes topics

R3.1 Proton transfer reactionsIB Chemistry HL: Revision notes

Section 1

Brønsted–Lowry acids and bases, pH and Kw

A Brønsted–Lowry acid donates a proton; a base accepts one. A conjugate acid–base pair differs by one proton (CH₃COOH/CH₃COO⁻). Amphiprotic species (H₂O, HCO₃⁻, H₂PO₄⁻) can do both.

pH = −log₁₀[H⁺]; Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K. Strong acids and bases ionise fully; weak ones partially, and equilibria lie towards the weaker conjugate. Neutralisation equations: oxides, hydroxides, carbonates and hydrogencarbonates all give a salt and water (plus CO₂ for carbonates). A strong acid–strong base pH curve has a steep section from about pH 3 to 11 and equivalence at pH 7.

Key termsconjugate acid–base pairamphiproticKw

Section 2

pOH and interconversion (HL)

pOH = −log₁₀[OH⁻] and [OH⁻] = 10⁻ᵖᴼᴴ. From Kw: pH + pOH = 14.00 at 298 K. Route map: [H⁺] ↔ pH, [OH⁻] ↔ pOH, and [H⁺][OH⁻] = Kw links the two sides. Example: [OH⁻] = 1.63 × 10⁻³ gives pOH = 2.79 and pH = 11.21.

Key termspOH

Section 3

Ka, Kb, pKa and pKb (HL)

For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻] ÷ [HA]; for B + H₂O ⇌ BH⁺ + OH⁻, Kb = [BH⁺][OH⁻] ÷ [B]. pKa = −log₁₀Ka. A larger Ka (smaller pKa) means a stronger acid; a larger Kb (smaller pKb) a stronger base.

Multiplying the expressions for a conjugate pair cancels [HA] and [A⁻]: Ka × Kb = Kw, so pKa + pKb = 14.00 at 298 K. The stronger the acid, the weaker its conjugate base.

For a weak acid, assume [HA] ≈ initial concentration: [H⁺] = √(Ka × c). For a weak base, [OH⁻] = √(Kb × c).

Key termsKaKbpKa
Exam tip

State the assumption: ionisation is small, so [HA]eqm ≈ [HA]initial.

Section 4

pH of salt solutions (HL)

The pH of a salt solution depends on the relative strengths of the parent acid and base. Ions from strong acids and bases (Na⁺, K⁺, Cl⁻, NO₃⁻) do not hydrolyse.

  • NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺ → NH₄Cl is acidic.
  • RCOO⁻ + H₂O ⇌ RCOOH + OH⁻ → sodium ethanoate is basic.
  • CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ → sodium carbonate is basic.
  • HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻ → sodium hydrogencarbonate is weakly basic. A salt of a weak acid and a weak base has a pH set by comparing Ka of the cation with Kb of the anion.
Key termshydrolysis

Section 5

pH curves for the four combinations (HL)

For 0.100 mol dm⁻³ solutions:

  • Strong acid + strong base: starts at pH 1, steep section about 3–11, equivalence pH 7.
  • Weak acid + strong base: starts higher (about 3), a buffer region, half-equivalence pH = pKa, steep section about 7–11, equivalence pH > 7.
  • Strong acid + weak base (acid added to base): starts about 11, half-equivalence pH = pKa of BH⁺, steep section about 7–3, equivalence pH < 7.
  • Weak acid + weak base: no steep section; the pH changes gradually through equivalence.
Key termshalf-equivalence pointbuffer region

Section 6

Indicators, end point and equivalence point (HL)

An indicator is a weak acid, HIn ⇌ H⁺ + In⁻, where HIn and In⁻ have different colours. Adding acid shifts the equilibrium left (HIn colour); adding alkali shifts it right (In⁻ colour). The colour changes when [HIn] ≈ [In⁻], i.e. at pH ≈ pKa(HIn).

The equivalence point is where stoichiometric amounts have reacted; the end point is where the indicator changes colour. Choose an indicator whose range lies within the steep section: phenolphthalein (8.2–10.0) for weak acid–strong base; methyl orange or methyl red for strong acid–weak base; any for strong–strong; none for weak–weak.

Key termsend pointequivalence pointindicator
Common mistake

End point and equivalence point are not the same thing; a good indicator makes them coincide.

Section 7

Buffer solutions (HL)

A buffer resists changes in pH when small amounts of acid or alkali are added. An acidic buffer contains a weak acid and its conjugate base (CH₃COOH/CH₃COO⁻); a basic buffer contains a weak base and its conjugate acid (NH₃/NH₄⁺). Added H⁺ is removed by the base (CH₃COO⁻ + H⁺ → CH₃COOH); added OH⁻ by the acid (CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O). Both components are present in large amounts, so their ratio barely changes.

The pH depends on the pKa and the ratio of the concentrations: pH = pKa + log₁₀([A⁻] ÷ [HA]). Diluting a buffer changes both concentrations equally, so the pH hardly changes (though its capacity falls).

Key termsbufferacidic bufferbasic buffer

That's the notes covered.

Carry on to the next subtopic.