Acids, bases and pH
Acids, bases and pH covers proton transfer, equilibrium in aqueous solution and titration behaviour. The exam tests calculations involving strong and weak acids, buffers, salt hydrolysis and indicator choice.
Core rules
- Brønsted-Lowry: an acid is a proton donor, a base a proton acceptor; conjugate pairs differ by H+.
- Strong acids and bases dissociate completely in water; weak acids and bases establish an equilibrium with Ka or Kb.
- Kw = [H+][OH−] = 1.0 × 10−14 at 298 K; pH = −log10[H+], pOH = −log10[OH−], and pH + pOH = 14 at 298 K only.
- For a weak acid HA, Ka = [H+][A−]/[HA]; pKa = −log10Ka. If dissociation is small, [H+] ≈ √(Ka × c).
- A buffer is a mixture of a weak acid and its conjugate base (or weak base and conjugate acid); it resists pH change. Henderson-Hasselbalch: pH = pKa + log10([A−]/[HA]).
- During a titration, the equivalence point is where moles of acid and base react exactly. Strong acid-strong base gives pH 7; weak acid-strong base gives pH > 7 because the conjugate base hydrolyses. Choose an indicator whose range brackets the vertical section.
Worked example
Calculate the pH of 0.100 mol dm−3 ethanoic acid, Ka = 1.8 × 10−5. Assume x = [H+] and x is small compared with 0.100. Ka ≈ x2/0.100, so x2 = 1.8 × 10−6 and x = 1.34 × 10−3 mol dm−3. pH = −log10(1.34 × 10−3) = 2.87. The assumption is valid because x is about 1.3% of 0.100.
Traps that cost marks
- Using the strong-acid formula for a weak acid without checking Ka.
- Forgetting that pH + pOH = 14 is temperature-dependent.
- Calculating buffer pH with concentrations of strong acid and salt rather than the weak acid/conjugate base pair.
- Assuming the equivalence point is always pH 7.
- Choosing an indicator by colour rather than by its pH range.