Acids, Bases, Buffers and Partition Coefficients

This guide covers subtopics 25.1, Acids and bases, and 25.2, Partition coefficients, from Topic 25, Equilibria, of Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. Both are A Level content, grouped together as the syllabus’s two applications of equilibrium constants beyond Kc and Kp.

Before studying this

This resource assumes the Brønsted-Lowry theory, strong vs weak acids, and qualitative titration-curve/indicator work from Acids and Bases: The Brønsted-Lowry Theory — this page adds the quantitative layer (actual pH and Ka calculations) that AS Level explicitly excludes. It also assumes the equilibrium-constant concept from Chemical Equilibria.

Syllabus coverage

CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY 9701 — A Level, Topic 25

25.1 Acids and bases — understanding conjugate acid-base pairs; defining pH, Ka, pKa and Kw mathematically and using them in calculations; calculating [H⁺(aq)] and pH for strong acids, strong alkalis and weak acids; defining a buffer solution, explaining how one is made and how it controls pH, including the role of HCO₃⁻ in blood; calculating the pH of buffer solutions; understanding and using solubility product, Ksp; writing a Ksp expression and calculating Ksp from concentrations and vice versa; understanding the common ion effect and performing calculations using it.

25.2 Partition coefficients — stating what a partition coefficient, Kpc, is; calculating and using Kpc for a solute in the same physical state in two solvents; understanding the factors affecting Kpc in terms of the polarities of solute and solvents.

Conjugate acid-base pairs

In the Brønsted-Lowry sense, an acid is a proton (H⁺) donor and a base is a proton acceptor. When an acid HA donates a proton, what remains, A⁻, is called its conjugate base — it is capable of accepting a proton back to re-form HA. Equally, when a base B accepts a proton to form BH⁺, that BH⁺ is the conjugate acid of B. Every acid-base equilibrium contains two such pairs, related by the loss or gain of a single H⁺:

HA + B ⇌ A⁻ + BH⁺

(acid₁) (base₂) (base₁, conjugate base of HA) (acid₂, conjugate acid of B)

Worked example. For CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺, identify both conjugate pairs.

CH₃COOH is the acid; losing H⁺ gives CH₃COO⁻, its conjugate base. H₂O is the base; gaining H⁺ gives H₃O⁺, its conjugate acid. The two conjugate pairs are CH₃COOH/CH₃COO⁻ and H₃O⁺/H₂O.

This is exactly the relationship exploited in a buffer solution, below: CH₃COOH and CH₃COO⁻ (from CH₃COONa) are a conjugate acid-base pair present together in solution.

pH, Ka and pKa

pH, Ka and pKa are all defined mathematically:

pH = −log₁₀[H⁺(aq)]

Ka = [H⁺][A⁻] / [HA] (for a weak acid HA ⇌ H⁺ + A⁻)

pKa = −log₁₀ Ka

A larger Ka (smaller pKa) means a stronger (more dissociated) weak acid. Kw, the ionic product of water (Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K), links [H⁺] and [OH⁻] in any aqueous solution. (Kb and the equation Kw = Ka × Kb are not required.)

Strong acid. A strong acid dissociates essentially completely, so [H⁺] equals the acid’s concentration directly (for a monoprotic acid).

Worked example. Calculate the pH of 0.100 mol dm⁻³ HCl.

[H⁺] = 0.100 mol dm⁻³ (complete dissociation)

pH = −log₁₀(0.100) = 1.00

Strong alkali. Use Kw to find [H⁺] from the alkali’s [OH⁻].

Worked example. Calculate the pH of 0.0500 mol dm⁻³ NaOH.

[OH⁻] = 0.0500 mol dm⁻³

[H⁺] = Kw / [OH⁻] = 1.00 × 10⁻¹⁴ / 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³

pH = −log₁₀(2.00 × 10⁻¹³) = 12.70

Weak acid. A weak acid only partially dissociates. Since dissociation is small, the equilibrium concentration of HA is approximated as equal to its initial concentration, C, and [H⁺] ≈ [A⁻] (both come from the same dissociation):

Ka ≈ [H⁺]² / C, so [H⁺] = √(Ka × C)

Worked example. Calculate the pH of 0.100 mol dm⁻³ CH₃COOH (Ka = 1.80 × 10⁻⁵ mol dm⁻³).

[H⁺] = √(Ka × C) = √(1.80 × 10⁻⁵ × 0.100) = √(1.80 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³

pH = −log₁₀(1.34 × 10⁻³) = 2.87

Note this is far higher (less acidic) than a strong acid of the same concentration would give (pH 1.00), because only a small fraction of the weak acid molecules have actually dissociated.

Buffer solutions

A buffer solution resists changes in pH when small amounts of acid or base are added. An acidic buffer is made from a weak acid and a salt of its conjugate base (e.g. CH₃COOH and CH₃COONa) — both present in significant, comparable concentrations.

How a buffer controls pH. The weak acid (CH₃COOH) is available to neutralise any added OH⁻; the conjugate base (CH₃COO⁻, from the salt) is available to neutralise any added H⁺:

CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

CH₃COO⁻ + H⁺ → CH₃COOH

Because both a reservoir of acid and a reservoir of base are present together, small additions shift the position of the CH₃COOH ⇌ H⁺ + CH₃COO⁻ equilibrium only slightly, and pH barely changes.

Calculating buffer pH. Since a buffer contains significant, deliberately chosen concentrations of both HA and A⁻ (not related by simple dissociation), the full Ka expression is rearranged directly:

Ka = [H⁺][A⁻] / [HA], so [H⁺] = Ka × [HA] / [A⁻]

Worked example. Calculate the pH of a buffer containing 0.100 mol dm⁻³ CH₃COOH and 0.200 mol dm⁻³ CH₃COONa (Ka = 1.80 × 10⁻⁵ mol dm⁻³).

[H⁺] = Ka × [HA]/[A⁻] = 1.80 × 10⁻⁵ × (0.100/0.200) = 9.00 × 10⁻⁶ mol dm⁻³

pH = −log₁₀(9.00 × 10⁻⁶) = 5.05

Buffers in blood. The HCO₃⁻/H₂CO₃ (hydrogencarbonate/carbonic acid) system buffers blood pH close to 7.4. Excess H⁺ is removed by HCO₃⁻ (HCO₃⁻ + H⁺ → H₂CO₃), and excess OH⁻ is neutralised by the H₂CO₃ reservoir — the same principle as the acetic acid/acetate buffer, using a biologically available conjugate pair.

Solubility product, Ksp

For a sparingly soluble ionic solid in equilibrium with its saturated solution, the solubility product, Ksp, is the equilibrium constant for that dissolving equilibrium, with the solid’s “concentration” (constant, as a pure solid) omitted.

Worked example. Write the Ksp expression for AgCl, and calculate its solubility in pure water given Ksp(AgCl) = 1.80 × 10⁻¹⁰ mol² dm⁻⁶.

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Ksp = [Ag⁺][Cl⁻]

In pure water, [Ag⁺] = [Cl⁻] = s (the molar solubility), so Ksp = s²:

s = √Ksp = √(1.80 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol dm⁻³

The power rule for unequal stoichiometry. Each ion’s concentration in the Ksp expression is raised to the power of its coefficient in the dissolving equation — this only shows up when the two ions aren’t produced in a 1:1 ratio.

Worked example. Write the Ksp expression for PbI₂, and calculate its molar solubility given Ksp(PbI₂) = 7.10 × 10⁻⁹ mol³ dm⁻⁹.

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)

Ksp = [Pb²⁺][I⁻]²

If the molar solubility is s, then [Pb²⁺] = s but [I⁻] = 2s (two iodide ions are released per formula unit dissolved), so:

Ksp = s × (2s)² = 4s³

s = ³√(Ksp / 4) = ³√(7.10 × 10⁻⁹ / 4) = ³√(1.775 × 10⁻⁹) = 1.21 × 10⁻³ mol dm⁻³

The common ion effect. Solubility is suppressed when a solution already contains one of the ions from another source, since Ksp still has to be satisfied — a higher concentration of one ion forces the other lower.

Worked example. Calculate the solubility of AgCl in 0.100 mol dm⁻³ NaCl(aq) (Ksp(AgCl) = 1.80 × 10⁻¹⁰ mol² dm⁻⁶).

[Cl⁻] ≈ 0.100 mol dm⁻³ (almost entirely from the NaCl, since so little AgCl dissolves in comparison)

[Ag⁺] = Ksp / [Cl⁻] = 1.80 × 10⁻¹⁰ / 0.100 = 1.80 × 10⁻⁹ mol dm⁻³

This is over 7000 times less soluble than in pure water — the added Cl⁻ suppresses AgCl’s solubility, exactly as Le Chatelier’s principle predicts for the equilibrium AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).

Partition coefficients

When a solute is shaken with two immiscible solvents (e.g. water and an organic solvent like ether), it distributes between them in a fixed ratio at equilibrium, described by the partition coefficient, Kpc:

Kpc = [solute in solvent A] / [solute in solvent B]

(defined only where the solute exists as the same species in both solvents — no dissociation or association in either layer).

Worked example. Kpc (ether/water) = 4.00 for solute X. 5.00 g of X is shaken with 50 cm³ of ether and 50 cm³ of water until equilibrium is reached. Calculate the mass of X in each layer.

Let the mass in ether = m; mass in water = (5.00 − m). Since the volumes are equal, the concentration ratio equals the mass ratio directly:

Kpc = m / (5.00 − m) = 4.00

m = 4.00(5.00 − m) = 20.0 − 4.00m

5.00m = 20.0, so m = 4.00 g in ether, and 1.00 g in water

What affects Kpc. The value reflects how the solute’s polarity matches each solvent — a solute favours (partitions more into) whichever solvent is closer to its own polarity, so a non-polar organic solute typically favours a non-polar organic solvent over water.

Common mistakes

Using the weak-acid approximation for a strong acid, or vice versa. Strong acids dissociate completely ([H⁺] = concentration directly); weak acids need the Ka-based approximation — mixing these methods up is the most common error in this topic.

Forgetting the buffer pH formula uses the ratio of concentrations, not their absolute values. Doubling both [HA] and [A⁻] together doesn’t change the buffer’s pH at all, since the ratio stays the same.

Treating a common-ion solubility calculation like a pure-water one. In pure water, [Ag⁺] = [Cl⁻] = s; with a common ion already present at a much higher concentration, that assumption breaks — use the given/dominant concentration for the common ion instead.

Forgetting Kpc has no units when solute concentrations are expressed consistently, and applies only when the solute doesn’t dissociate or dimerise differently in the two solvents.

Quick revision checklist

• pH = −log₁₀[H⁺]; Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K
• Strong acid/alkali: [H⁺] or [OH⁻] = concentration directly
• Weak acid: [H⁺] = √(Ka × C)
• Buffer: [H⁺] = Ka × [HA]/[A⁻]; works because both HA and A⁻ are present in reservoir amounts
• Ksp = product of ion concentrations (powers = stoichiometric coefficients)
• Common ion effect: solubility falls when one ion is already present
• Kpc = [solute in A] / [solute in B], same species, no dissociation

Related resources

• Acids and Bases: The Brønsted-Lowry Theory — the qualitative AS foundation this topic quantifies
• Chemical Equilibria — Kc and Kp, the equilibrium-constant concept this topic extends
• Chemical Energetics: Lattice Energy, Entropy and Gibbs Free Energy — feasibility reasoning that complements equilibrium-constant reasoning
• Cambridge AS & A Level Chemistry hub

Written against Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. Always check the current syllabus for your examination year.