ZIMSEC A Level · Chemistry · Form 5, Form 6
Dynamic equilibrium
In a closed system, forward and reverse rates are equal and concentrations remain constant.
- Le Chatelier's principle: the position of equilibrium shifts to oppose a change in concentration, pressure or temperature. Catalysts do not change the position.
- Kc = product of [products] ÷ product of [reactants], each raised to the power of its coefficient in the equation; Kp uses partial pressures. Only temperature changes K.
Worked example: For N₂ + 3H₂ ⇌ 2NH₃, write the expression for Kc.
Kc = [NH₃]² ÷ ([N₂][H₂]³)
Brønsted–Lowry acids and bases
Acids are proton donors; bases are proton acceptors. Conjugate pairs differ by one H⁺.
pH
pH = −log[H⁺]. Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K.
- Strong acid: [H⁺] = concentration. 0.010 mol dm⁻³ HCl has pH 2.
- Weak acid: Ka = [H⁺][A⁻]/[HA]; [H⁺] ≈ √(Ka × c).
Worked example: Find the pH of 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³).
[H⁺] = √(1.74 × 10⁻⁵ × 0.100) = 1.32 × 10⁻³ → pH = 2.88
Buffers
A buffer resists pH change on adding small amounts of acid or alkali: a weak acid with its conjugate base (e.g. ethanoic acid/sodium ethanoate). Blood is buffered by carbonic acid/hydrogencarbonate. pH = pKa + log([A⁻]/[HA]).
Indicators and titration curves
Choose an indicator whose colour change range lies on the steep part of the curve: methyl orange for strong acid–weak base; phenolphthalein for weak acid–strong base.
Check yourself
- What is the effect of increasing pressure on N₂ + 3H₂ ⇌ 2NH₃?
- Find the pH of 0.0050 mol dm⁻³ NaOH.
- Name a buffer system in blood.
Show answers
- Equilibrium shifts right (fewer gas molecules), increasing the yield of NH₃.
- [OH⁻] = 0.0050; [H⁺] = 1.0 × 10⁻¹⁴ ÷ 0.0050 = 2.0 × 10⁻¹²; pH = 11.7
- Carbonic acid/hydrogencarbonate (H₂CO₃/HCO₃⁻).
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