Conjugate acid–base pairs

· Brønsted–Lowry acid = proton donor; Brønsted–Lowry base = proton acceptor.
· Conjugate acid = species formed when a base gains H⁺.
· Conjugate base = species formed when an acid loses H⁺.
· Conjugate acid–base pair = two species differing by one H⁺ only.
· Example: CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺
· Pairs: CH₃COOH / CH₃COO⁻ and H₂O / H₃O⁺.
· In equations, identify the species that donates H⁺ and what it becomes; then identify the species that accepts H⁺ and what it becomes.

pH, Ka, pKa and Kw definitions

· pH = −log₁₀[H⁺].
· [H⁺] = 10⁻ᵖᴴ.
· For weak acid HA ⇌ H⁺ + A⁻:
· Ka = [H⁺][A⁻] / [HA].
· pKa = −log₁₀Ka.
· Ka = 10⁻ᵖᴷᵃ.
· Kw = [H⁺][OH⁻].
· At 298 K, usually use Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ unless data gives a different value.
· Therefore, at 298 K: pH + pOH = 14.00.
· Smaller pKa = larger Ka = stronger acid.

The pH scale summarises how [H⁺] changes with acidity. Remember that pH is logarithmic, so a decrease of 1 pH unit means 10 times greater [H⁺]. Source

Strong acid and strong alkali pH calculations

· Strong acids fully dissociate in water.
· For monoprotic strong acids: [H⁺] = acid concentration.
· Example: 0.100 mol dm⁻³ HCl → [H⁺] = 0.100 mol dm⁻³ → pH = 1.00.
· For diprotic strong acids such as H₂SO₄, use the information given in the question; often [H⁺] ≈ 2 × [H₂SO₄] if complete dissociation is assumed.
· Strong alkalis fully dissociate to give OH⁻.
· For NaOH/KOH: [OH⁻] = alkali concentration.
· Use Kw = [H⁺][OH⁻] to find [H⁺], then pH = −log₁₀[H⁺].
· At 298 K, quicker method: pOH = −log₁₀[OH⁻], then pH = 14.00 − pOH.

Weak acid pH calculations

· Weak acid equilibrium: HA ⇌ H⁺ + A⁻.
· Expression: Ka = [H⁺][A⁻] / [HA].
· For weak monoprotic acid of concentration c, usually assume:
· [H⁺] = [A⁻] = x and [HA] ≈ c.
· Therefore: Ka ≈ x² / c.
· So: [H⁺] ≈ √(Ka × c).
· Then: pH = −log₁₀[H⁺].
· If approximation is not valid, use the full equilibrium expression and solve using the method expected from the data.
· Exam tip: weak acid has lower [H⁺] and therefore higher pH than a strong acid of the same concentration.

Buffer solutions

· Buffer solution = solution that resists changes in pH when small amounts of acid or alkali are added.
· Acidic buffer contains weak acid HA and its conjugate base A⁻.
· Made by mixing: weak acid + salt of its conjugate base, e.g. CH₃COOH + CH₃COONa.
· Can also be made by partial neutralisation of a weak acid with a strong alkali.
· Key equilibrium: HA ⇌ H⁺ + A⁻.
· When acid is added: A⁻ + H⁺ → HA, removing added H⁺.
· When alkali is added: HA + OH⁻ → A⁻ + H₂O, removing added OH⁻.
· For buffer calculations: [H⁺] = Ka × [HA] / [A⁻].
· Also: pH = pKa + log₁₀([A⁻] / [HA]).
· Use moles after reaction first if acid/alkali is added before calculating buffer pH.
· In buffers, the ratio [A⁻] : [HA] controls pH; the total concentration affects buffer capacity.

This graph shows that more concentrated buffers resist pH change more effectively. It links directly to the idea of buffer capacity and why buffers need significant amounts of both acid and conjugate base. Source

HCO₃⁻ buffer in blood

· Blood pH is controlled partly by the hydrogencarbonate / carbonic acid buffer system.
· Main equilibrium: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻.
· HCO₃⁻ acts as a base by removing excess H⁺: H⁺ + HCO₃⁻ ⇌ H₂CO₃.
· H₂CO₃ can release H⁺ when pH rises.
· This buffer is important because enzymes and metabolic reactions require blood pH to remain within a narrow range.

This diagram supports the syllabus example of HCO₃⁻ in blood pH control. Use it to connect buffer chemistry with biological pH regulation. Source

Solubility product, Ksp

· Solubility product, Ksp = equilibrium constant for dissolving a sparingly soluble ionic solid.
· For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq):
· Ksp = [Ag⁺][Cl⁻].
· For Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq):
· Ksp = [Mg²⁺][OH⁻]².
· Do not include solids in the Ksp expression.
· Powers in the Ksp expression come from the balanced equation coefficients.
· Units of Ksp vary depending on the expression, so derive units from the concentration terms if required.

Ksp calculations

· From solubility to Ksp: write the dissolving equation, express ion concentrations in terms of solubility s, then substitute.
· Example: AgCl(s) ⇌ Ag⁺ + Cl⁻
· If molar solubility = s, then [Ag⁺] = s, [Cl⁻] = s, so Ksp = s².
· Example: Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻
· If molar solubility = s, then [Mg²⁺] = s, [OH⁻] = 2s, so Ksp = s(2s)² = 4s³.
· From Ksp to solubility: set up ion concentrations in terms of s, substitute into Ksp, then solve for s.
· Always check whether the question asks for molar solubility, ion concentration, or mass dissolved.

Common ion effect

· Common ion effect = reduced solubility of a sparingly soluble salt when a solution already contains one of its ions.
· Example: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).
· Adding NaCl increases [Cl⁻], so equilibrium shifts left, decreasing AgCl solubility.
· This is an application of Le Chatelier’s principle.
· In calculations, use the existing common ion concentration as the initial concentration.
· If common ion concentration is large, the extra amount from dissolving may be negligible.
· Common exam trap: Ksp does not change when a common ion is added; only the solubility changes.

Exam calculation checklist

· pH from [H⁺]: use pH = −log₁₀[H⁺].
· [H⁺] from pH: use [H⁺] = 10⁻ᵖᴴ.
· Strong acid: assume complete dissociation.
· Strong alkali: calculate [OH⁻], then use Kw or pOH.
· Weak acid: use Ka, equilibrium setup and approximation where valid.
· Buffer: adjust moles of HA and A⁻ first, then calculate pH.
· Ksp: write the balanced dissolving equation before writing the expression.
· Common ion: include the starting concentration of the common ion in the equilibrium setup.