Chemical Equilibrium: Acids, Bases and Buffers
What this domain tests
pH calculations for strong and weak acids, buffers, salt hydrolysis and titration curves.
Core rules
- Brønsted–Lowry: acids donate protons, bases accept them; conjugates differ by H+.
- Kw = [H+][OH−] = 1.0 × 10−14 at 298 K; pH = −log[H+]; pKa = −log Ka.
- Strong acids dissociate completely; for a weak acid of concentration c, [H+] ≈ √(Ka c), provided dissociation is small.
- A buffer is a weak acid plus its conjugate base, or a weak base plus its conjugate acid. Henderson–Hasselbalch: pH = pKa + log([A−]/[HA]).
- At the equivalence point of a weak acid–strong base titration the salt hydrolyses and pH > 7; weak base–strong acid gives pH < 7.
Worked example
Find the pH of 0.100 mol dm−3 ethanoic acid with Ka = 1.8 × 10−5. [H+] = √(1.8 × 10−5 × 0.100) = √(1.8 × 10−6) = 1.34 × 10−3 mol dm−3. pH = −log(1.34 × 10−3) = 2.87. The approximation is justified because x is only about 1.3% of c.
Traps
Applying [H+] = c to a weak acid; forgetting to convert pH back to concentration in buffer problems; using initial amounts instead of the new ratio after adding strong base to a buffer; assuming pH 7 at every equivalence point; ignoring dilution when solutions are mixed.