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LearnCambridge97012025–2027Paper 4 · A Level Structured QuestionsEquilibria

Paper 4 · Module

Equilibria

2 subtopics≈ 105 min in total

Start the module: pH, Ka and buffers Open graphing calculator

pH, Ka and buffers
$\mathrm{pH}=-\log[\mathrm{H^+}]$ . Strong acids: $[\mathrm{H^+}]=$ acid concentration (× number of H⁺). Water: $K_w=[\mathrm{H^+}][\mathrm{OH^-}]=1.00\times10^{-14}$ at 298 K; for strong bases, $[\mathrm{H^+}]=K_w/[\mathrm{OH^-}]$ . Weak acids: $K_a=\frac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$ ; assume $[\mathrm{H^+}]=[\mathrm{A^-}]$ , so $[\mathrm{H^+}]=\sqrt{K_a[\mathrm{HA}]}$ ; $\mathrm{p}K_a=-\log K_a$ . A buffer (weak acid + its salt) resists pH change: $[\mathrm{H^+}]=K_a\frac{[\mathrm{HA}]}{[\mathrm{A^-}]}$ , or $\mathrm{pH}=\mathrm{p}K_a+\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$ . Titration curves show a steep section at equivalence; choose an indicator whose colour change lies within it. At half-neutralisation, pH = pKa.
≈ 55 min
Kp, Ksp and partition
For gas equilibria, partial pressure $p_A=x_A\times P_\text{total}$ , where mole fraction $x_A=n_A/n_\text{total}$ . $K_p$ uses partial pressures: for $\mathrm{N_2O_4\rightleftharpoons2NO_2}$ , $K_p=\frac{p_{\mathrm{NO_2}}^2}{p_{\mathrm{N_2O_4}}}$ . Only temperature changes $K_p$ . The solubility product $K_{sp}$ for a sparingly soluble salt, e.g. $K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}]$ ; a precipitate forms if the ionic product exceeds $K_{sp}$ . Adding a common ion lowers solubility. The partition coefficient $K_{pc}=\frac{[X]_\text{solvent 1}}{[X]_\text{solvent 2}}$ describes how a solute distributes between two immiscible solvents.
≈ 50 min