A Level Chemistry: Acids, Bases, Buffers and Partition Coefficients — Revision Notes

Condensed for revision. For the full explanation, use the Acids, Bases, Buffers and Partition Coefficients study guide, then test yourself with the practice questions. For Brønsted–Lowry acids, strong and weak acids, and neutralisation at AS Level, see the AS Acids and Bases revision notes.

Syllabus: Cambridge International AS & A Level Chemistry 9701, 2025–2027, A Level content: subtopics 25.1 Acids and bases and 25.2 Partition coefficients.

Acids and bases (25.1)

Conjugate acid–base pairs

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

Reaction	Acid / conjugate base	Base / conjugate acid
CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺	CH₃COOH / CH₃COO⁻	H₂O / H₃O⁺
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻	H₂O / OH⁻	NH₃ / NH₄⁺
HCl + NH₃ → NH₄⁺ + Cl⁻	HCl / Cl⁻	NH₃ / NH₄⁺

Water can act as an acid or as a base, depending on what it reacts with.

Definitions: pH, Ka, pKa and Kw

pH  = –log₁₀[H⁺]              so  [H⁺] = 10^(–pH)
Ka  = [H⁺][A⁻] / [HA]          for HA ⇌ H⁺ + A⁻        units: mol dm⁻³
pKa = –log₁₀ Ka               so  Ka = 10^(–pKa)
Kw  = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K

• Ka is the acid dissociation constant: a larger Ka (a smaller pKa) means a stronger acid.
• In pure water at 298 K, [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³, so pH = 7.00.
• Kb and the relationship Kw = Ka × Kb are not tested.

Calculating [H⁺] and pH

Type	Method
Strong acid (fully dissociated), e.g. HCl → H⁺ + Cl⁻	[H⁺] = concentration of the acid
Strong alkali (fully dissociated), e.g. NaOH → Na⁺ + OH⁻	find [OH⁻], then [H⁺] = Kw / [OH⁻]
Weak acid (partly dissociated)	[H⁺] = √(Ka × [HA])

Strong acid: 0.050 mol dm⁻³ HCl.

[H⁺] = 0.050 mol dm⁻³      pH = –log(0.050) = 1.30

Strong alkali: 0.020 mol dm⁻³ NaOH. (0.010 mol dm⁻³ Ba(OH)₂ gives the same [OH⁻], because each formula unit releases two OH⁻.)

[OH⁻] = 0.020 mol dm⁻³
[H⁺]  = 1.00 × 10⁻¹⁴ / 0.020 = 5.0 × 10⁻¹³ mol dm⁻³      pH = 12.30

Weak acid: 0.10 mol dm⁻³ ethanoic acid, Ka ≈ 1.7 × 10⁻⁵ mol dm⁻³ (approximate value at 298 K). Two assumptions: [H⁺] = [CH₃COO⁻] (ionisation of water ignored), and [CH₃COOH] at equilibrium ≈ the starting concentration (dissociation is very small).

Ka = [H⁺]² / [CH₃COOH]
[H⁺] = √(1.7 × 10⁻⁵ × 0.10) = √(1.7 × 10⁻⁶) = 1.304 × 10⁻³ mol dm⁻³
pH = –log(1.304 × 10⁻³) = 2.88      (keep the unrounded [H⁺] for the log)
pKa = –log(1.7 × 10⁻⁵) = 4.77

Ka from pH (illustrative values): a 0.050 mol dm⁻³ solution of a weak acid HA has pH 3.00.

[H⁺] = 10⁻³·⁰⁰ = 1.0 × 10⁻³ mol dm⁻³
Ka = (1.0 × 10⁻³)² / 0.050 = 2.0 × 10⁻⁵ mol dm⁻³

Buffer solutions

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

How a buffer is made:

• a weak acid and its conjugate base, e.g. ethanoic acid with sodium ethanoate; or by adding less than the neutralising amount of NaOH to excess ethanoic acid (partial neutralisation);
• a weak base and its conjugate acid, e.g. ammonia with ammonium chloride. Added H⁺ is removed by NH₃ + H⁺ → NH₄⁺; added OH⁻ is removed by NH₄⁺ + OH⁻ → NH₃ + H₂O.

How the ethanoic acid / ethanoate buffer controls pH: the buffer contains large reservoirs of both CH₃COOH and CH₃COO⁻.

CH₃COOH ⇌ CH₃COO⁻ + H⁺
added H⁺:   CH₃COO⁻ + H⁺  → CH₃COOH
added OH⁻:  CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

Added H⁺ or OH⁻ is removed, and because the reservoirs are large the ratio [CH₃COOH] : [CH₃COO⁻] hardly changes, so [H⁺] and the pH hardly change.

Uses of buffers:

• Blood is kept close to pH 7.4 by the carbonic acid / hydrogencarbonate buffer:

  CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻
  added H⁺:   HCO₃⁻ + H⁺  → H₂CO₃
  added OH⁻:  H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O

  HCO₃⁻ removes excess H⁺; H₂CO₃ removes excess OH⁻. Enzymes and other proteins in the body work only in a narrow pH range.

• Other uses: calibrating pH meters; keeping the pH constant for enzyme-catalysed reactions and cell cultures; shampoos and some foods and medicines.

Calculating the pH of a buffer

[H⁺] = Ka × [HA] / [A⁻]         or   pH = pKa + log([A⁻] / [HA])

Example: 0.10 mol dm⁻³ ethanoic acid with 0.20 mol dm⁻³ sodium ethanoate; Ka ≈ 1.7 × 10⁻⁵ mol dm⁻³ (approximate).

[H⁺] = 1.7 × 10⁻⁵ × 0.10 / 0.20 = 8.5 × 10⁻⁶ mol dm⁻³
pH = –log(8.5 × 10⁻⁶) = 5.07

For a partial-neutralisation buffer, first work out the moles of HA left and A⁻ formed; the total volume is the same for both, so you can use the mole ratio directly. When [HA] = [A⁻], pH = pKa.

Solubility product, Ksp

Definition: Ksp is the equilibrium constant for a sparingly soluble ionic compound in a saturated solution: the product of the concentrations of its ions, each raised to the power of its coefficient in the equilibrium equation. The solid is not included.

Equilibrium	Ksp expression	Units
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)	[Ag⁺][Cl⁻]	mol² dm⁻⁶
CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)	[Ca²⁺][F⁻]²	mol³ dm⁻⁹
PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)	[Pb²⁺][I⁻]²	mol³ dm⁻⁹

Ksp from solubility (illustrative values): the solubility of AgCl is 1.34 × 10⁻⁵ mol dm⁻³.

[Ag⁺] = [Cl⁻] = 1.34 × 10⁻⁵ mol dm⁻³
Ksp = (1.34 × 10⁻⁵)² = 1.8 × 10⁻¹⁰ mol² dm⁻⁶

Solubility from Ksp (illustrative values): Ksp of CaF₂ = 3.2 × 10⁻¹¹ mol³ dm⁻⁹. Let the solubility be s, so [Ca²⁺] = s and [F⁻] = 2s.

Ksp = s × (2s)² = 4s³ = 3.2 × 10⁻¹¹
s³ = 8.0 × 10⁻¹²        s = 2.0 × 10⁻⁴ mol dm⁻³

Will a precipitate form? Calculate the ionic product using the concentrations after mixing. If it is greater than Ksp, a precipitate forms until the product falls to Ksp.

The common ion effect

A sparingly soluble salt is less soluble in a solution that already contains one of its ions (a common ion). The added ion shifts the position of equilibrium, e.g. AgCl(s) ⇌ Ag⁺ + Cl⁻, to the left, so more solid remains undissolved. Ksp itself does not change (at constant temperature).

Calculation (illustrative values): solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰ mol² dm⁻⁶) in 0.10 mol dm⁻³ NaCl. The Cl⁻ from the dissolving AgCl is negligible compared with 0.10 mol dm⁻³.

[Ag⁺] = Ksp / [Cl⁻] = 1.8 × 10⁻¹⁰ / 0.10 = 1.8 × 10⁻⁹ mol dm⁻³

Compare the solubility in pure water: √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ mol dm⁻³, several thousand times greater.

Partition coefficients (25.2)

Definition: the partition coefficient, Kpc, is the ratio of the concentrations of a solute in two immiscible solvents when an equilibrium has been established, at a stated temperature. The solute must be in the same physical state in both solvents.

Kpc = [X in solvent 1] / [X in solvent 2]          (no units)

State which solvent is on top of the ratio; Kpc(organic/water) is the reciprocal of Kpc(water/organic).

Using Kpc (illustrative values): 1.00 g of X in 100 cm³ of water is shaken with 50 cm³ of an organic solvent. Kpc(organic/water) = 4.0. Let x g of X move into the organic layer.

Kpc = (x / 50) / ((1.00 – x) / 100) = 4.0
2x / (1.00 – x) = 4.0     →   6x = 4.00     →   x = 0.67 g extracted; 0.33 g stays in the water

Using the same 50 cm³ as two 25 cm³ portions: each time (x / 25) / ((m – x) / 100) = 4.0 gives x = m / 2, so 0.50 g then 0.25 g is extracted, a total of 0.75 g. Several small extractions remove more solute than one large one.

Factors affecting the value of Kpc (polarity):

Solute	Dissolves better in	Kpc(organic/water)
non-polar, or weakly polar (e.g. I₂, hydrocarbons)	the non-polar organic solvent (“like dissolves like”)	large (much greater than 1)
polar, or able to form hydrogen bonds with water (e.g. NH₃, small alcohols)	water	small

The more similar the polarity of the solute is to that of a solvent, the more of the solute is found in that solvent. A more polar organic solvent will also take up a polar solute better, changing the value.

Exam traps

• A conjugate pair differs by exactly one H⁺: H₂SO₄ and SO₄²⁻ are not a conjugate pair.
• For a strong alkali, never write pH = –log[OH⁻]; use Kw to find [H⁺] first.
• For a weak acid, [H⁺] is not equal to the acid concentration; use √(Ka × [HA]) and state the assumptions.
• In a buffer calculation, [A⁻] comes from the salt (or the NaOH added), not from the acid’s own dissociation.
• Always write the units of Ka, Kw and Ksp; Ksp units depend on the formula (mol² dm⁻⁶ for AgCl, mol³ dm⁻⁹ for CaF₂).
• In Ksp for CaF₂, square [F⁻] and remember [F⁻] = 2s, so Ksp = 4s³.
• The common ion lowers solubility; it does not change Ksp.
• Kpc has no units, and you must say which solvent is on top of the ratio.

Self-test

1. Identify the two conjugate acid–base pairs in NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.
2. Calculate the pH of 0.010 mol dm⁻³ nitric acid.
3. Calculate the pH of 0.050 mol dm⁻³ KOH (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶).
4. (Illustrative values.) A weak acid HA has Ka = 1.0 × 10⁻⁴ mol dm⁻³. Calculate the pH of a 0.040 mol dm⁻³ solution.
5. Define a buffer solution.
6. Using an equation, explain how HCO₃⁻ helps to control the pH of blood when H⁺ ions are added.
7. 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid is mixed with 25.0 cm³ of 0.200 mol dm⁻³ NaOH. Using Ka ≈ 1.7 × 10⁻⁵ mol dm⁻³, calculate the pH of the buffer formed.
8. Write the Ksp expression, with units, for lead(II) iodide, PbI₂.
9. (Illustrative values.) Ksp of CaF₂ = 3.2 × 10⁻¹¹ mol³ dm⁻⁹. Calculate the solubility of CaF₂ in 0.10 mol dm⁻³ NaF.
10. (Illustrative values.) At equilibrium, a solute has concentration 0.24 mol dm⁻³ in hexane and 0.030 mol dm⁻³ in water. Calculate Kpc(hexane/water) and state what it suggests about the solute.

Answers:

1. NH₄⁺ / NH₃ (acid / conjugate base) and H₃O⁺ / H₂O (conjugate acid / base).
2. [H⁺] = 0.010 mol dm⁻³; pH = 2.00.
3. [H⁺] = 1.00 × 10⁻¹⁴ / 0.050 = 2.0 × 10⁻¹³ mol dm⁻³; pH = 12.70.
4. [H⁺] = √(1.0 × 10⁻⁴ × 0.040) = √(4.0 × 10⁻⁶) = 2.0 × 10⁻³ mol dm⁻³; pH = 2.70.
5. A solution that resists changes in pH when small amounts of acid or alkali are added.
6. HCO₃⁻ + H⁺ → H₂CO₃. The added H⁺ is removed by the large reservoir of HCO₃⁻, so the pH hardly changes.
7. Acid: 0.0100 mol; NaOH: 0.00500 mol. After reaction: 0.00500 mol CH₃COOH and 0.00500 mol CH₃COO⁻, so [HA] = [A⁻] and pH = pKa = –log(1.7 × 10⁻⁵) = 4.77.
8. Ksp = [Pb²⁺][I⁻]²; units mol³ dm⁻⁹.
9. [F⁻] ≈ 0.10 mol dm⁻³; solubility = [Ca²⁺] = 3.2 × 10⁻¹¹ / (0.10)² = 3.2 × 10⁻⁹ mol dm⁻³.
10. Kpc = 0.24 / 0.030 = 8.0 (no units). The solute is much more soluble in hexane, so it is non-polar or only weakly polar.

These are original notes written for revision. Ka, Ksp and Kpc values marked illustrative or approximate are not data-book values. Check the full syllabus wording in the official 9701 syllabus.