Equilibria (A Level)

Revise Equilibria (A Level) for Chemistry 9701 (AS Level) — revision notes and instant AI marking.

📖 Revision notes · preview

Cambridge International AS & A Level Chemistry (9701)

Equilibria (A Level)

The big idea: This builds on the AS chapter Equilibria. The same equilibrium-constant idea now handles acids (K_a, pH), water (K_w), buffers, sparingly soluble salts (K_sp) and a solute shared between two solvents (K_pc). Most marks come from writing the right expression and handling the logs.

Summary — What This Chapter Covers

Conjugate acid–base pairs.
pH, K_a, pK_a and K_w; pH of strong acids, strong alkalis and weak acids.
Buffers: what they are, how they work, how to make them, the HCO₃⁻ buffer in blood, and their pH.
Solubility product K_sp and the common ion effect.
Partition coefficient K_pc and how polarity affects it.

1. Acids and Bases

25.1

Conjugate acid–base pairs

A Brønsted–Lowry acid is a proton (H⁺) donor and a base is a proton acceptor. When an acid donates a proton, what remains is its conjugate base; when a base accepts a proton, it becomes its conjugate acid. A conjugate acid–base pair is two species that differ by one H⁺.

Pairs CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
acid 1 base 2 base 1 acid 2

The pairs are CH₃COOH/CH₃COO⁻ and H₃O⁺/H₂O. In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, the pairs are NH₄⁺/NH₃ and H₂O/OH⁻; water acts as an acid here.

Definitions

Key expressions pH = −log₁₀[H⁺(aq)] K_a = [H⁺][A⁻]/[HA] pK_a = −log₁₀K_a
K_w = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K

A larger K_a (smaller pK_a) means a stronger acid. To go back from pH: [H⁺] = 10^−pH.

Solution	Method
strong acid	fully dissociated: [H⁺] = concentration × number of H⁺ released
strong alkali	[OH⁻] from concentration; [H⁺] = K_w/[OH⁻]
weak acid	[H⁺] = √(K_a × [HA]), assuming [H⁺] = [A⁻] and that [HA] at equilibrium ≈ starting concentration

Worked Example

Calculate the pH of (a) 0.050 mol dm⁻³ HCl, (b) 0.020 mol dm⁻³ NaOH, (c) 0.10 mol dm⁻³ ethanoic acid, K_a = 1.74 × 10⁻⁵ mol dm⁻³.

Practice Question

A 0.050 mol dm⁻³ solution of a weak acid HA has pH 3.00. Calculate K_a and pK_a.

25.1

Buffer solutions

A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali are added.

Making a buffer

A weak acid and its conjugate base, e.g. ethanoic acid with sodium ethanoate.
A weak acid partly neutralised by a strong base (the salt forms in the flask).
A weak base and its conjugate acid, e.g. NH₃ with NH₄Cl.

How it works (ethanoic acid / ethanoate)

CH₃COOH ⇌ CH₃COO⁻ + H⁺. The solution contains large amounts of both CH₃COOH and CH₃COO⁻.

Add H⁺: it is removed by the conjugate base, CH₃COO⁻ + H⁺ → CH₃COOH.
Add OH⁻: it is removed by the acid, CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.
Because both reservoirs are large, the ratio [CH₃COOH]/[CH₃COO⁻] hardly changes, so [H⁺] and pH hardly change.

Uses, including blood

Blood is held at pH 7.35–7.45 mainly by the carbonic acid / hydrogencarbonate buffer: H₂CO₃ ⇌ H⁺ + HCO₃⁻. Extra H⁺ is removed by HCO₃⁻ + H⁺ → H₂CO₃; extra OH⁻ by H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O. H₂CO₃ is linked to dissolved CO₂ (CO₂ + H₂O ⇌ H₂CO₃), which the lungs can remove. Buffers are also used in shampoos, food, fermentation, enzyme experiments and calibrating pH meters.

Buffer pH ⁺ _a ⁻ _a ⁻

Worked Example

⁻³ ₃ ⁻³ ₃ _a ⁻⁵ ⁻³

25.1

🔓 Read the full Equilibria (A Level) note → You're seeing the preview · sign in to read it all

Also in the full note

Solubility product, Ksp
A solute shared between two solvents

What's inside

📖 Revision notes 🎯 Learn mode ✦ AI flashcards ✓ Instant AI marking 🧊 3D explorers 🧪 Experiments & simulations 📈 Progress tracking

📄 Practise Equilibria (A Level) with Chemistry 9701 past papers Every paper with its mark scheme — answer online, marked instantly. Open →

Read the full Equilibria (A Level) notes free

That's the preview — create a free account to read the rest, plus flashcards and practice questions with instant AI marking. No credit card.

Unlock the full notes free →

Equilibria (A Level)