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Flashcards · 9701 · A Level Topic 25

Equilibria (A Level) flashcards

14 cards on A Level Topic 25, subtopics 25.1 to 25.2. Pick a subtopic or study them all.

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25.1 Acids and bases

Definitions of pH, Ka, pKa and Kw
pH = −log₁₀[H⁺] Ka = [H⁺][A⁻] ÷ [HA] (mol dm⁻³) pKa = −log₁₀Ka Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K
pH of 0.020 mol dm⁻³ NaOH
[OH⁻] = 0.020; [H⁺] = Kw ÷ [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 0.020 = 5.0 × 10⁻¹³ mol dm⁻³; pH = 12.30.
pH of a weak acid, and the two assumptions
[H⁺] = √(Ka × [HA]). Assume [H⁺] = [A⁻] (ignore water) and [HA] at equilibrium ≈ starting concentration. e.g. 0.10 mol dm⁻³, Ka 1.7 × 10⁻⁵: [H⁺] = 1.30 × 10⁻³, pH = 2.88.
Buffer solution
A solution that resists changes in pH when small amounts of acid or alkali are added. Made from a weak acid and its conjugate base (e.g. ethanoic acid + sodium ethanoate) or a weak base and its conjugate acid.
How does an ethanoic acid / ethanoate buffer work?
Large reservoirs of CH₃COOH and CH₃COO⁻. Added H⁺: CH₃COO⁻ + H⁺ → CH₃COOH. Added OH⁻: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The ratio hardly changes, so the pH hardly changes.
pH of a buffer
[H⁺] = Ka × [HA] ÷ [A⁻], or pH = pKa + log([A⁻] ÷ [HA]). When [HA] = [A⁻], pH = pKa.
How is the pH of blood controlled?
Carbonic acid / hydrogencarbonate buffer, near pH 7.4: H₂CO₃ ⇌ H⁺ + HCO₃⁻. HCO₃⁻ removes added H⁺; H₂CO₃ removes added OH⁻.
Solubility product, Ksp
The equilibrium constant for a sparingly soluble ionic compound in a saturated solution: the product of its ion concentrations, each raised to its coefficient (solid not included). e.g. Ksp(CaF₂) = [Ca²⁺][F⁻]², mol³ dm⁻⁹.
Ksp of CaF₂ is 3.2 × 10⁻¹¹ mol³ dm⁻⁹ (illustrative). Calculate its solubility.
Ksp = s × (2s)² = 4s³ = 3.2 × 10⁻¹¹, so s³ = 8.0 × 10⁻¹² and s = 2.0 × 10⁻⁴ mol dm⁻³.
The common ion effect
A sparingly soluble salt is less soluble in a solution already containing one of its ions: the equilibrium shifts to the left. Ksp itself is unchanged.

25.2 Partition coefficients

Partition coefficient, Kpc
The ratio of the concentrations of a solute in two immiscible solvents at equilibrium, at a stated temperature, with the solute in the same physical state in both. No units; state which solvent is on top.
Effect of polarity on Kpc(organic/water)
A non-polar solute (e.g. I₂) dissolves better in the non-polar solvent: large Kpc. A polar or hydrogen-bonding solute (e.g. NH₃) dissolves better in water: small Kpc.
1.00 g X in 100 cm³ water is shaken with 50 cm³ organic solvent; Kpc(organic/water) = 4.0. Mass extracted?
(x ÷ 50) ÷ ((1.00 − x) ÷ 100) = 4.0 → 2x ÷ (1.00 − x) = 4.0 → 6x = 4.00 → x = 0.67 g.
Is one large extraction or several smaller ones more effective?
Several smaller ones. In the example above, two 25 cm³ portions extract 0.50 g + 0.25 g = 0.75 g, compared with 0.67 g from one 50 cm³ portion.

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