Equilibria

📚The chemistry

25.1 Acids and bases

Conjugate pairs

In the Brønsted–Lowry theory (topic 7) an acid donates a proton and a base accepts one. When an acid loses a proton, what is left can accept it back, so it is a base: its conjugate base. Every acid–base reaction has two conjugate acid–base pairs, which differ by exactly one H⁺:

CH₃COOH/CH₃COO⁻ is one pair; H₃O⁺/H₂O is the other.

pH, K_w, K_a and pK_a

K_a is the acid dissociation constant: the equilibrium constant for HA ⇌ H⁺ + A⁻. A larger K_a (smaller pK_a) means a stronger acid. Ethanoic acid has K_a = 1.74 × 10⁻⁵ mol dm⁻³, pK_a 4.76. (K_b and K_w = K_a × K_b are not tested.)

Calculating pH

• Strong acid — fully dissociated, so [H⁺] = concentration of the acid (for a monoprotic acid). 0.050 mol dm⁻³ HCl: pH = −log 0.050 = 1.30.
• Strong alkali — find [OH⁻], then [H⁺] = K_w/[OH⁻]. 0.0500 mol dm⁻³ NaOH: [H⁺] = 1.00 × 10⁻¹⁴/0.0500 = 2.00 × 10⁻¹³, pH = 12.70. Watch for Ba(OH)₂ and similar, which give two OH⁻ per formula unit.
• Weak acid — assume that [H⁺] = [A⁻] (each molecule that dissociates gives one of each) and that so little dissociates that [HA] at equilibrium ≈ the concentration put in, c. Then \[ K_\mathrm{a} = \frac{[\mathrm{H^{+}}]^2}{c} \qquad [\mathrm{H^{+}}] = \sqrt{K_\mathrm{a}\,c} \]

Buffer solutions

A buffer solution is one that resists changes in pH when small amounts of acid or alkali are added. (It does not keep the pH perfectly constant, and a large amount of acid or alkali overwhelms it.)

Making one: mix a weak acid and its conjugate base — usually a weak acid with one of its salts, e.g. ethanoic acid and sodium ethanoate. Or partly neutralise a weak acid with a strong alkali, so that some of the acid is converted into its salt. (A weak base with its salt, such as ammonia and ammonium chloride, also works.)

How it works: the mixture contains large reservoirs of both CH₃COOH and CH₃COO⁻:

• Add acid (H⁺): it is removed by the large supply of conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH. The position shifts left, and [H⁺] hardly changes.
• Add alkali (OH⁻): it is removed by the weak acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. More acid dissociates to replace the H⁺, and pH hardly changes.

Both halves are needed: the salt supplies the large [A⁻] that a weak acid on its own does not have.

Calculating the pH: rearrange K_a, using the concentrations of acid and salt as made up:

When [HA] = [A⁻], [H⁺] = K_a and pH = pK_a — a useful check, and the point of maximum buffering.

Uses: anywhere pH must be held steady — calibrating pH meters, shampoos and biological or enzyme experiments, fermentation and food processing. The most important is blood, held at pH 7.35–7.45 mainly by the carbonic acid / hydrogencarbonate system:

Extra H⁺ (for example from lactic acid during exercise) combines with HCO₃⁻ to form H₂CO₃; extra OH⁻ is neutralised by H₂CO₃, forming HCO₃⁻ and water. Carbonic acid is itself in equilibrium with dissolved CO₂, which the lungs control by breathing.

Solubility product, K_sp

A sparingly soluble ionic solid in contact with its saturated solution is at equilibrium. For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), and for a general salt M_xA_y:

The solubility product is the product of the concentrations of the ions in a saturated solution, each raised to the power of its number in the formula. The solid does not appear. Units follow from the expression: AgCl mol² dm⁻⁶; PbI₂ mol³ dm⁻⁹.

From solubility to K_sp: if s mol dm⁻³ dissolves, AgCl gives [Ag⁺] = [Cl⁻] = s, so K_sp = s²; PbI₂ gives [Pb²⁺] = s and [I⁻] = 2s, so K_sp = s(2s)² = 4s³. Solubility may first need converting from g dm⁻³ to mol dm⁻³.

A precipitate forms when mixing two solutions if the ionic product (the same expression using the concentrations after mixing) exceeds K_sp.

The common ion effect

A salt is less soluble in a solution that already contains one of its ions than in pure water. Adding Cl⁻ to saturated AgCl shifts AgCl(s) ⇌ Ag⁺ + Cl⁻ to the left (Le Chatelier), and AgCl precipitates. Calculating it: K_sp(AgCl) = 1.8 × 10⁻¹⁰. In water, s = √K_sp = 1.3 × 10⁻⁵ mol dm⁻³. In 0.100 mol dm⁻³ NaCl, [Cl⁻] ≈ 0.100 (the small amount from AgCl is negligible), so s = [Ag⁺] = 1.8 × 10⁻¹⁰/0.100 = 1.8 × 10⁻⁹ mol dm⁻³ — about 7000 times less.

25.2 Partition coefficients

When a solute is shaken with two immiscible solvents (such as water and hexane), it distributes itself between them until equilibrium is reached. The partition coefficient, K_pc, is the ratio of its concentrations in the two solvents at equilibrium, at a given temperature, for a solute in the same physical state in both:

Always say which solvent is on top: K_pc(hexane/water) is the reciprocal of K_pc(water/hexane). Concentrations can be in any consistent unit, g dm⁻³ or mol dm⁻³, so K_pc has no units.

The value depends on polarity — “like dissolves like”. A non-polar solute such as iodine is far more soluble in a non-polar solvent such as hexane than in water, so K_pc(hexane/water) is large. A polar solute, especially one that can hydrogen bond with water (small alcohols, carboxylic acids, amines), favours the water, so K_pc(organic/water) is small. This is the basis of solvent extraction, and of chromatography (topic 37).