Equilibria (A Level)

0. What Examiners Want

This guide covers Topic 25 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught.

1. Conjugate Pairs, K_w and pH

At AS you used the Brønsted–Lowry definitions: an acid is a proton (H⁺) donor and a base is a proton acceptor. When an acid gives away its proton, what is left can take a proton back, so it is a base. The two species are linked by the transfer of one proton.

Conjugate acid–base pair. An acid and the base formed when the acid loses a proton (or, equivalently, a base and the acid formed when it gains a proton). The acid has one more H⁺ than its conjugate base.

In every Brønsted–Lowry equilibrium there are two conjugate pairs. Label them acid 1 / base 1 and acid 2 / base 2, and join each acid to the base with one fewer H⁺.

Water can act as an acid or a base. Some ions, such as HCO₃⁻ and HC₂O₄⁻, can do both, so they belong to two different pairs. The conjugate acid of a strong base is very weak, and the conjugate base of a strong acid (such as Cl⁻) is too weak to accept a proton in water.

The ionic product of water, K_w. Water ionises very slightly: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). Because [H₂O] is effectively constant it is included in the constant:

K_w = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K

In pure water [H⁺] = [OH⁻] = √(1.00 × 10⁻¹⁴) = 1.00 × 10⁻⁷ mol dm⁻³. This is the neutral solution at 298 K.

pH. pH is the negative logarithm (base 10) of the hydrogen ion concentration:

pH = −log₁₀[H⁺]    and    [H⁺] = 10^−pH

The same ‘p’ notation applies to the other constants: pK_a = −log₁₀ K_a, and so K_a = 10^−pK_a. Each pH unit is a factor of 10 in [H⁺].

The pH scale. Because K_w = [H⁺][OH⁻] is constant, any rise in [H⁺] means a fall in [OH⁻] and vice versa. At 298 K pH + pOH = 14.

2. pH of Strong Acids and Alkalis

A strong acid or strong alkali is completely ionised in water, so [H⁺] or [OH⁻] is found directly from the concentration. The H⁺ from the ionisation of water is negligible next to the H⁺ from the acid and is ignored.

Strong acid (monobasic). [H⁺] = concentration of the acid.

Worked example 1. Calculate the pH of 0.0250 mol dm⁻³ HCl. [H⁺] = 0.0250 mol dm⁻³; pH = −log(0.0250) = 1.60.

Worked example 2. A 10.0 cm³ sample of 0.100 mol dm⁻³ HCl is diluted to 1.00 dm³. [H⁺] = 0.100 × 10.0 / 1000 = 1.00 × 10⁻³ mol dm⁻³, so pH = 3.00. A 100-fold dilution raises the pH by 2.

Strong alkali. [OH⁻] = concentration × (number of OH⁻ in the formula). Then [H⁺] = K_w / [OH⁻].

Worked example 3. Calculate the pH of 0.0500 mol dm⁻³ NaOH (K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶).

1. [OH⁻] = 0.0500 mol dm⁻³
2. [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³
3. pH = −log(2.00 × 10⁻¹³) = 12.70

Short cut. pOH = −log[OH⁻] = 1.30, and pH = 14.00 − 1.30 = 12.70, because pH + pOH = 14 at 298 K. Use this only when the question says 298 K or gives K_w = 1.00 × 10⁻¹⁴.

Worked example 4 (a strong base with two OH⁻). 0.0100 mol dm⁻³ Ba(OH)₂ gives [OH⁻] = 2 × 0.0100 = 0.0200 mol dm⁻³, so pOH = 1.70 and pH = 12.30.

Worked example 5 (from pH to mass). What mass of Ba(OH)₂ is needed to make 100.0 cm³ of solution of pH 12.80? [M_r = 171.3]

1. [H⁺] = 10^−12.80 = 1.585 × 10⁻¹³ mol dm⁻³
2. [OH⁻] = 1.00 × 10⁻¹⁴ ÷ 1.585 × 10⁻¹³ = 0.0631 mol dm⁻³
3. [Ba(OH)₂] = 0.0631 ÷ 2 = 0.03155 mol dm⁻³
4. moles = 0.03155 × 0.1000 = 3.155 × 10⁻³ mol; mass = 3.155 × 10⁻³ × 171.3 = 0.540 g

3. Weak Acids: K_a and pK_a

A weak acid is only partly ionised. For a monobasic acid HA the equilibrium is HA(aq) ⇌ H⁺(aq) + A⁻(aq), and the equilibrium law gives the acid dissociation constant:

K_a = [H⁺][A⁻] / [HA]    units: mol dm⁻³     pK_a = −log₁₀ K_a

A large K_a (small pK_a) means a stronger acid. Write the expression from the balanced equation for the specific acid, for example for butanoic acid K_a = [CH₃CH₂CH₂COO⁻][H⁺] / [CH₃CH₂CH₂COOH].

At the same concentration a strong acid has far more H⁺ than a weak acid, so a much lower pH. The weak acid exists mainly as un-ionised HA.

A ladder of pK_a values. Values quoted in different data sources differ slightly, so always use the numbers in your question.

The two assumptions. For a solution of a weak acid HA of concentration c:

1. The H⁺ from the ionisation of water is negligible, so [H⁺] = [A⁻].
2. The acid is so slightly ionised that [HA]_equilibrium ≈ c, the starting concentration.

Then K_a = [H⁺]² / c, which gives three calculation types.

Worked example 1: pH from K_a. Calculate the pH of 0.100 mol dm⁻³ ethanoic acid (K_a = 1.74 × 10⁻⁵ mol dm⁻³).

1. [H⁺]² = K_a × c = 1.74 × 10⁻⁵ × 0.100 = 1.74 × 10⁻⁶
2. [H⁺] = 1.32 × 10⁻³ mol dm⁻³
3. pH = 2.88

Worked example 2: K_a from pH. A 0.0100 mol dm⁻³ solution of methanoic acid has pH 2.90.

1. [H⁺] = 10^−2.90 = 1.26 × 10⁻³ mol dm⁻³
2. K_a = (1.26 × 10⁻³)² / 0.0100 = 1.59 × 10⁻⁴ mol dm⁻³; pK_a = 3.80

Worked example 3: concentration from pH and pK_a. A solution of benzoic acid (pK_a = 4.20) has pH 2.85. Find its concentration.

1. K_a = 10^−4.20 = 6.31 × 10⁻⁵ mol dm⁻³; [H⁺] = 10^−2.85 = 1.41 × 10⁻³ mol dm⁻³
2. [HA] = [H⁺]² / K_a = (1.41 × 10⁻³)² / 6.31 × 10⁻⁵ = 0.0316 mol dm⁻³
3. Check: the acid is about 4.5 % ionised (1.41 × 10⁻³ / 0.0316), so the approximation [HA] ≈ c is reasonable.

4. Buffer Solutions

Buffer solution. A solution that resists changes in pH when small amounts of acid or alkali are added. It contains a weak acid and its conjugate base (a weak base and its conjugate acid also works).

Ways to make an acidic buffer (the same ideas apply to basic buffers):

1. Mix a weak acid with a salt of the weak acid, for example CH₃COOH with CH₃COONa. The salt supplies the conjugate base.
2. Add a limited amount of strong alkali to an excess of the weak acid. This converts part of the acid to its conjugate base: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The acid must be in excess.
3. For an alkaline buffer, mix a weak base with its conjugate acid (NH₃ with NH₄Cl), or add a limited amount of strong acid to excess weak base.

How it works. In a mixture of ethanoic acid and sodium ethanoate there are two reserve supplies. The acid is mostly un-ionised and the salt is fully ionised, so the solution has a large reserve of CH₃COOH and a large reserve of CH₃COO⁻.

The weak acid removes added alkali and the conjugate base removes added acid. Because both reserves are large, their ratio hardly changes, so [H⁺] and pH hardly change.

• Add H⁺: CH₃COO⁻ + H⁺ → CH₃COOH. The equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifts left, removing the added H⁺. The concentration of the conjugate base falls slightly and the acid rises slightly, but both are large, so the ratio changes little.
• Add OH⁻: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. The added alkali removes H⁺, so the equilibrium shifts right and more acid ionises to replace it.
• For an ammonia buffer: NH₄⁺ + OH⁻ → NH₃ + H₂O and NH₃ + H⁺ → NH₄⁺.
• For a solution containing HC₂O₄⁻ (an amphiprotic ion): HC₂O₄⁻ + H⁺ → H₂C₂O₄ and HC₂O₄⁻ + OH⁻ → C₂O₄²⁻ + H₂O.

The same 0.0100 mol of strong acid or alkali changes the pH of water by 5 units but the pH of the buffer by less than 0.1.

Limits. A buffer works best when [acid] = [conjugate base] (then pH = pK_a) and within about 1 pH unit of the pK_a. A large amount of acid or alkali uses up one of the reserves and the pH then changes sharply. The more concentrated the buffer, the more acid or alkali it can absorb.

Uses of buffers. Buffers keep the pH constant in many industrial processes (electroplating, dye manufacture, leather treatment), in culture media for bacteria, in shampoos and eye drops, and in the calibration of pH meters. Living things also depend on them.

The CO₂ / HCO₃⁻ buffer in blood. Human blood is held between pH 7.35 and 7.45. Respiration produces CO₂, which dissolves in blood plasma (catalysed by the enzyme carbonic anhydrase):

CO₂(aq) + H₂O(l) ⇌ H⁺(aq) + HCO₃⁻(aq)

The hydrogencarbonate buffer. HCO₃⁻ removes extra H⁺ to form CO₂, which is breathed out; dissolved CO₂ replaces H⁺ if it falls.

• If [H⁺] rises, the equilibrium shifts left: HCO₃⁻ + H⁺ → CO₂ + H₂O. The CO₂ is lost through the lungs, so the extra H⁺ is removed and the pH stays almost constant.
• If [H⁺] falls, the equilibrium shifts right: more CO₂ reacts with water to give H⁺ and HCO₃⁻.
• Other buffers in blood include haemoglobin and plasma proteins (HPr ⇌ H⁺ + Pr⁻) and H₂PO₄⁻ / HPO₄²⁻.
• If the pH of blood falls too far (acidosis) body functions are disturbed. Rain water is similarly buffered by H₂CO₃ / HCO₃⁻ but acidic pollution overwhelms the small reserves.

5. Buffer Calculations

For a buffer of a weak acid HA and its conjugate base A⁻, the equilibrium expression is rearranged for [H⁺]:

[H⁺] = K_a × [HA] / [A⁻]    or    pH = pK_a + log([A⁻] / [HA])

[HA] and [A⁻] are the concentrations in the buffer, taken as equal to the amounts added (the acid is only slightly ionised and the salt supplies the base). Because both are in the same volume, the ratio of the amounts (in mol) can replace the ratio of concentrations.

Choose the method from the type of solution before you start the arithmetic.

Worked example 1. Calculate the pH of a buffer containing 0.600 mol dm⁻³ propanoic acid and 0.800 mol dm⁻³ sodium propanoate (K_a = 1.35 × 10⁻⁵ mol dm⁻³).

• [H⁺] = 1.35 × 10⁻⁵ × 0.600 / 0.800 = 1.01 × 10⁻⁵ mol dm⁻³; pH = 4.99

Worked example 2 (amount of salt needed). How many moles of sodium ethanoate must be added to 1.00 dm³ of 0.100 mol dm⁻³ ethanoic acid to make a buffer of pH 4.90? (K_a = 1.74 × 10⁻⁵)

• [H⁺] = 10^−4.90 = 1.26 × 10⁻⁵ mol dm⁻³
• [CH₃COO⁻] = K_a [CH₃COOH] / [H⁺] = 1.74 × 10⁻⁵ × 0.100 / 1.26 × 10⁻⁵ = 0.138 mol dm⁻³, so 0.138 mol in 1.00 dm³

Worked example 3 (mixing two solutions). 25.0 cm³ of 0.200 mol dm⁻³ CH₃COOH is mixed with 15.0 cm³ of 0.100 mol dm⁻³ CH₃COONa.

• moles of acid = 0.0250 × 0.200 = 5.00 × 10⁻³ mol; moles of salt = 0.0150 × 0.100 = 1.50 × 10⁻³ mol
• [H⁺] = 1.74 × 10⁻⁵ × (5.00 × 10⁻³) / (1.50 × 10⁻³) = 5.80 × 10⁻⁵; pH = 4.24 (the total volume cancels)

Worked example 4 (acid partly neutralised by alkali). 50.0 cm³ of 0.200 mol dm⁻³ CH₃COOH is mixed with 20.0 cm³ of 0.100 mol dm⁻³ NaOH.

• moles of acid at the start = 0.0100 mol; moles of OH⁻ = 0.00200 mol. The alkali reacts completely: acid left = 0.00800 mol; ethanoate formed = 0.00200 mol
• [H⁺] = 1.74 × 10⁻⁵ × 0.00800 / 0.00200 = 6.96 × 10⁻⁵; pH = 4.16

Worked example 5 (finding the starting concentration). 250 cm³ of a methanoic acid solution is mixed with 150 cm³ of 0.200 mol dm⁻³ sodium methanoate. The buffer has pH 3.50 (K_a of HCOOH = 1.60 × 10⁻⁴). Find the concentration of the methanoic acid.

• moles of salt = 0.150 × 0.200 = 0.0300 mol; [H⁺] = 10^−3.50 = 3.16 × 10⁻⁴
• [HA]/[A⁻] = [H⁺]/K_a = 3.16 × 10⁻⁴ / 1.60 × 10⁻⁴ = 1.98, so moles of acid = 1.98 × 0.0300 = 0.0593 mol
• concentration in the 250 cm³ used = 0.0593 / 0.250 = 0.237 mol dm⁻³

Worked example 6 (the effect of adding acid or alkali). A 1.00 dm³ buffer contains 0.100 mol CH₃COOH and 0.100 mol CH₃COONa. The pH is pK_a = −log(1.74 × 10⁻⁵) = 4.76. Add 0.0100 mol HCl (volume unchanged).

• The HCl reacts with the base: ethanoate = 0.100 − 0.0100 = 0.090 mol; acid = 0.100 + 0.0100 = 0.110 mol
• [H⁺] = 1.74 × 10⁻⁵ × 0.110 / 0.090 = 2.13 × 10⁻⁵; pH = 4.67 (a fall of 0.09; the same HCl in water gives pH 2.00)
• Adding 0.0100 mol NaOH instead gives acid 0.090 mol, ethanoate 0.110 mol and pH 4.85.

6. Solubility Product, K_sp

Ionic compounds such as AgCl and BaSO₄ are called ‘insoluble’ but dissolve to a very small extent. In contact with its saturated solution a sparingly soluble salt is in dynamic equilibrium. For silver chloride, AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).

Solubility product, K_sp. For a sparingly soluble salt, the product of the concentrations of its ions in a saturated solution, each raised to the power of its coefficient in the dissolving equation, at a stated temperature. The solid does not appear in the expression.

Write the equation first, then the expression, then the units. The last column links K_sp to the solubility s (mol dm⁻³).

For Fe(OH)₂: Fe(OH)₂(s) ⇌ Fe²⁺(aq) + 2OH⁻(aq), so K_sp = [Fe²⁺][OH⁻]². The units come from the powers: (mol dm⁻³) × (mol dm⁻³)² = mol³ dm⁻⁹.

Worked example 1: K_sp from solubility. The solubility of magnesium hydroxide in water is 1.65 × 10⁻⁴ mol dm⁻³ at 298 K. Calculate K_sp.

1. Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq)
2. [Mg²⁺] = 1.65 × 10⁻⁴; [OH⁻] = 2 × 1.65 × 10⁻⁴ = 3.30 × 10⁻⁴ mol dm⁻³
3. K_sp = (1.65 × 10⁻⁴)(3.30 × 10⁻⁴)² = 1.80 × 10⁻¹¹ mol³ dm⁻⁹

Worked example 2: K_sp from solubility in g dm⁻³. A saturated solution of lead(II) iodide, PbI₂, contains 0.680 g dm⁻³ (M_r = 461.0).

1. solubility s = 0.680 / 461.0 = 1.475 × 10⁻³ mol dm⁻³
2. [Pb²⁺] = s, [I⁻] = 2s; K_sp = 4s³ = 4 × (1.475 × 10⁻³)³ = 1.28 × 10⁻⁸ mol³ dm⁻⁹

Worked example 3: solubility from K_sp. Calculate the solubility of lead(II) chloride in water (K_sp = 1.6 × 10⁻⁵ mol³ dm⁻⁹).

1. PbCl₂(s) ⇌ Pb²⁺ + 2Cl⁻, so K_sp = [Pb²⁺][Cl⁻]² = s(2s)² = 4s³
2. s = (K_sp / 4)^1/3 = (4.0 × 10⁻⁶)^1/3 = 1.59 × 10⁻² mol dm⁻³

Worked example 4: solubility in g dm⁻³. For BaSO₄ (K_sp = 1.0 × 10⁻¹⁰ mol² dm⁻⁶): s = √(1.0 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol dm⁻³; with M_r = 233.4 the solubility is 1.0 × 10⁻⁵ × 233.4 = 2.3 × 10⁻³ g dm⁻³.

Predicting precipitation. Mix two solutions and find the concentrations of the ions immediately after mixing (each is diluted). The ion product Q is the K_sp expression evaluated with these concentrations.

• If Q > K_sp, a precipitate forms.
• If Q ≤ K_sp, no precipitate forms (the solution is unsaturated or just saturated).

Worked example 5. 50.0 cm³ of 2.0 × 10⁻⁴ mol dm⁻³ AgNO₃ is mixed with 50.0 cm³ of 2.0 × 10⁻⁴ mol dm⁻³ NaCl. [Ag⁺] = [Cl⁻] = 1.0 × 10⁻⁴ mol dm⁻³ (doubling the volume halves each); Q = 1.0 × 10⁻⁸, which is greater than K_sp(AgCl) = 1.8 × 10⁻¹⁰, so AgCl precipitates. With 1.0 × 10⁻⁵ mol dm⁻³ solutions each ion would be 5.0 × 10⁻⁶ and Q = 2.5 × 10⁻¹¹ < K_sp: no precipitate.

7. Common Ion Effect and Precipitation

Common ion effect. The solubility of an ionic compound is lower in a solution that already contains one of its ions (a common ion) than it is in pure water. The added common ion shifts the dissolving equilibrium to the left, so more solid is formed.

Adding NaCl to a saturated solution of AgCl raises [Cl⁻], so [Ag⁺][Cl⁻] exceeds K_sp and AgCl precipitates. K_sp itself does not change; the solubility does.

To calculate solubility in a solution with a common ion, take [common ion] from the solution (the contribution from the sparingly soluble salt is so small that it can be ignored) and find the concentration of the other ion from K_sp. The solubility is equal to that concentration, divided by the number of that ion in the formula.

Worked example 1. Calculate the solubility of AgCl (K_sp = 1.8 × 10⁻¹⁰ mol² dm⁻⁶) (a) in water and (b) in 0.100 mol dm⁻³ NaCl.

• (a) s = √K_sp = 1.34 × 10⁻⁵ mol dm⁻³
• (b) [Cl⁻] = 0.100 mol dm⁻³, so [Ag⁺] = 1.8 × 10⁻¹⁰ / 0.100 = 1.8 × 10⁻⁹ mol dm⁻³, which is the solubility of AgCl, about 7000 times smaller than in water.

Worked example 2. Calculate the solubility of PbCl₂ in 0.100 mol dm⁻³ HCl (K_sp = 1.6 × 10⁻⁵ mol³ dm⁻⁹). [Cl⁻] = 0.100, so [Pb²⁺] = 1.6 × 10⁻⁵ / (0.100)² = 1.6 × 10⁻³ mol dm⁻³, ten times lower than the 1.59 × 10⁻² mol dm⁻³ in water.

Worked example 3 (hydroxide as common ion; pH of a saturated solution). Using K_sp[Mg(OH)₂] = 1.8 × 10⁻¹¹ mol³ dm⁻⁹:

• In water: s = (1.8 × 10⁻¹¹ / 4)^1/3 = 1.65 × 10⁻⁴ mol dm⁻³; [OH⁻] = 3.30 × 10⁻⁴ mol dm⁻³; pOH = 3.48 and pH = 10.52
• In NaOH of pH 12.0 ([OH⁻] = 0.0100 mol dm⁻³): [Mg²⁺] = 1.8 × 10⁻¹¹ / (0.0100)² = 1.8 × 10⁻⁷ mol dm⁻³, roughly 1000 times less than in water.

A comparison to explain. BaSO₄ dissolves to 1.0 × 10⁻⁵ mol dm⁻³ in water but only about 1.0 × 10⁻⁹ mol dm⁻³ in 0.100 mol dm⁻³ H₂SO₄: [Ba²⁺] = 1.0 × 10⁻¹⁰ / 0.100 = 1.0 × 10⁻⁹ mol dm⁻³, because the sulfate from the acid is a common ion.

8. Partition Coefficients

When a solute X is shaken with two immiscible solvents (that is, solvents that do not mix and form two layers), it distributes itself between them until a dynamic equilibrium is reached. X(solvent 1) ⇌ X(solvent 2).

Partition coefficient, K_pc. The equilibrium constant that relates the concentrations of a solute in two immiscible solvents at a stated temperature:

K_pc = [X in solvent 1] / [X in solvent 2]    (at constant temperature)

The value applies when the solute is in the same physical and molecular state in both solvents (for example NH₃ molecules in water and in the organic solvent). There are no units, because the concentrations are in the same units. Always state which solvent is in the numerator.

A separating funnel after shaking. At equilibrium the ratio of the two concentrations is a constant at a given temperature.

What controls the value? The size of K_pc depends on how soluble the solute is in each solvent, which depends on the intermolecular forces it can form with each solvent:

• Polar or hydrogen-bonding solutes (such as NH₃) dissolve well in water because they form hydrogen bonds with it. They are less soluble in an organic solvent such as CHCl₃, which can offer only weaker permanent dipole forces. K_pc (organic / water) is less than 1.
• Non-polar solutes (such as I₂) cannot break the strong hydrogen bonds between water molecules, so they are much less soluble in water. They dissolve in non-polar solvents such as cyclohexane by instantaneous dipole–induced dipole forces. K_pc (cyclohexane / water) is much greater than 1.
• The more similar the polarity of the solute and the solvent, the more soluble the solute is in that solvent (‘like dissolves like’).

Worked example 1: mass extracted. A 50.0 cm³ aqueous solution contains 1.20 g of X. It is shaken with 25.0 cm³ of ether, K_pc = [X]_ether / [X]_water = 6.00. Find the mass of X extracted into the ether.

1. Let m g be in the ether. Then 1.20 − m g remains in the water.
2. K_pc = (m / 25.0) / ((1.20 − m) / 50.0) = 2m / (1.20 − m) = 6.00
3. 2m = 7.20 − 6.00m, so m = 0.900 g extracted, 0.300 g left in the water

(Volumes may be left in cm³, because they appear in the ratio. Concentrations can be in g per cm³ or mol per dm³, provided they are the same in both layers.)

Worked example 2: calculating K_pc. 0.0500 mol of a compound Y dissolved in 40.0 cm³ of ether is shaken with 100.0 cm³ of water. At equilibrium 0.0110 mol of Y is in the water. Find K_pc (ether / water).

• in water: 0.0110 mol / 0.1000 dm³ = 0.110 mol dm⁻³; in ether: (0.0500 − 0.0110) = 0.0390 mol / 0.0400 dm³ = 0.975 mol dm⁻³
• K_pc = 0.975 / 0.110 = 8.86

Worked example 3: one extraction or two? Using the data of example 1, split the 25.0 cm³ of ether into two portions of 12.5 cm³. After each portion the fraction of X left in the water is 1 / (1 + K_pc × V_org / V_aq) = 1 / (1 + 6.00 × 12.5 / 50.0) = 0.400. After two portions the fraction left is 0.400² = 0.160, so 0.192 g remains and 1.008 g is extracted, more than the 0.900 g removed by one 25.0 cm³ portion.

Several small portions of solvent extract more than one large portion of the same total volume.

Link to chromatography. In paper chromatography a solute partitions between the mobile phase (the moving solvent) and the stationary phase (water held in the paper). A solute that is more soluble in the mobile phase moves further and faster. The solute must be in the same state in both phases for a single K_pc to apply, so associations such as dimerisation of a carboxylic acid in an organic solvent are outside the calculations asked.

9. Quick Sheet and Checklist

Checklist. Before the exam you should be able to:

• [ ] identify conjugate acid–base pairs in any equation and complete a Brønsted–Lowry equilibrium for an acid and water
• [ ] write the expressions for pH, K_w, K_a and pK_a and convert between K_a and pK_a
• [ ] calculate the pH of a strong acid and a strong alkali (including Ba(OH)₂), and find a mass or concentration from a pH
• [ ] calculate the pH of a weak acid, K_a from a pH, and a concentration from a pH and a pK_a, stating the two assumptions
• [ ] define a buffer, say how to make one, and explain its action with two equations
• [ ] describe the role of HCO₃⁻ in blood and give two uses of buffers
• [ ] calculate the pH of a buffer, including from mixed volumes, after partial neutralisation and after adding strong acid or alkali
• [ ] write K_sp expressions with units and calculate K_sp from solubility and solubility from K_sp
• [ ] predict precipitation by comparing Q with K_sp after dilution
• [ ] explain and calculate the effect of a common ion on solubility
• [ ] define K_pc, calculate a mass extracted or a K_pc, and explain its size using polarity and intermolecular forces