Cambridge International AS & A Level Chemistry 9701 · A Level topic 25

Equilibria

What this chapter covers25.1–25.2

Lemon juice, stomach acid, seawater and blood are all aqueous solutions, and in every one of them the concentration of hydrogen ions matters. A change in blood pH of a few tenths of a unit is a medical emergency; a small fall in the pH of seawater slows the growth of shells and coral. Yet the hydrogen-ion concentrations involved are tiny and span an enormous range. This chapter builds the quantitative tools that make such solutions manageable: the pH scale, the ionic product of water, K_w, and the acid dissociation constant, K_a.

Those tools are then applied to three kinds of equilibrium that control what happens in real solutions. A buffer resists changes in pH when small amounts of acid or alkali are added. A solubility product, K_sp, decides how much of a sparingly soluble salt can dissolve and when a precipitate forms. A partition coefficient, K_pc, describes how a solute shares itself between two solvents that do not mix — the basis of solvent extraction and of how drugs move between water and fatty tissue. Each is an equilibrium constant, and each is used in exactly the way K_c was used at AS: write the expression, substitute equilibrium concentrations, solve.

What topic 25 asks you to do

25.1 Acids and bases — use the terms conjugate acid and conjugate base and identify conjugate acid–base pairs; define pH, K_a, pK_a and K_w mathematically and use them in calculations; calculate [H⁺] and pH for strong acids, strong alkalis and weak acids; define a buffer solution, explain how one is made and how it controls pH, using equations; describe the uses of buffers, including the role of HCO₃⁻ in controlling the pH of blood; calculate the pH of buffer solutions; understand, write and calculate with the solubility product, K_sp; use the common ion effect and calculate with it.

25.2 Partition coefficients — state what is meant by a partition coefficient, K_pc; calculate and use K_pc when the solute is in the same physical state in both solvents; explain how the polarities of solute and solvents affect its value.

The syllabus states that K_b and the relationship K_w = K_a × K_b will not be tested; they are not used in this chapter.

What you are assumed to know already

Brønsted–Lowry acids as proton donors and bases as proton acceptors; strong and weak acids and bases; the pH scale in outline (topic 7.2).
Dynamic equilibrium, Le Chatelier's principle, K_c expressions and their units (topic 7.1).
Amount of substance, concentration in mol dm⁻³, and titration calculations (topic 2).
Ionic equations and state symbols; intermolecular forces and polarity (topics 2 and 3).

Conjugate acids and conjugate bases25.1.1, 25.1.2

In the Brønsted–Lowry model an acid is a proton donor and a base is a proton acceptor. A proton cannot be donated unless something accepts it, so every Brønsted–Lowry reaction involves one acid and one base on each side of the equation. Consider hypobromous acid dissolving in water:

HBrO(aq) + H₂O(l) ⇌ BrO⁻(aq) + H₃O⁺(aq)

In the forward reaction HBrO donates a proton to a water molecule: HBrO is the acid and H₂O is the base. The reaction is reversible, so look at the reverse reaction too: H₃O⁺ donates a proton to BrO⁻. On the right-hand side, therefore, H₃O⁺ is an acid and BrO⁻ is a base.

HBrO and BrO⁻ are linked: remove one proton from HBrO and you have BrO⁻; add one to BrO⁻ and you have HBrO back. Such a pair is a conjugate acid–base pair. BrO⁻ is the conjugate base of HBrO, and HBrO is the conjugate acid of BrO⁻. In the same way H₃O⁺ is the conjugate acid of H₂O.

Definition

A conjugate acid–base pair is two species that differ by one proton (H⁺). The member with the extra proton is the conjugate acid; the member without it is the conjugate base.

Figure 25.1 The two conjugate acid–base pairs in the equilibrium HBrO + H₂O ⇌ BrO⁻ + H₃O⁺. Each pair is linked across the equilibrium sign, never across the plus sign: the acid on the left is paired with the base on the right.

Three habits prevent almost every error with conjugate pairs. First, pair species across the ⇌ sign, never across a + sign — the two reactants are an acid and a base, not a pair. Second, check that the two members differ by exactly one H⁺: the formula changes by one H and the charge by one unit, and the conjugate base always carries the lower (more negative) charge. Third, remember that one species can play both roles. Water is the base in the HBrO equilibrium but the acid when ammonia dissolves:

NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) pairs: NH₄⁺/NH₃ and H₂O/OH⁻

The strength of an acid and the strength of its conjugate base are linked. A strong acid such as HCl gives up its proton almost completely, so Cl⁻ has very little tendency to take one back: Cl⁻ is a very weak base. A weak acid such as ethanoic acid holds its proton more firmly, so its conjugate base, the ethanoate ion, is a noticeably stronger base — which is exactly why ethanoate ions can mop up added H⁺ in a buffer later in this chapter.

Definitions examiners did not credit

Asked to define a conjugate acid–base pair, many candidates wrote a definition that would apply to any acid and any base — "an acid donates a proton and a base accepts one". That describes an acid and a base, not a conjugate pair. The creditworthy idea is the relationship between the two members: they differ by a single proton. When labelling species in an equation, some used OH⁻ instead of water as the base when ethanoic acid dissolves; others gave a conjugate base with a higher positive charge than its acid, which is impossible if a proton has been removed.

AnimationDefinitions of acids and bases

A short tour of the Brønsted–Lowry and Lewis definitions. Only the Brønsted–Lowry model is needed for the calculations in this chapter.

AnimationInterpreting equations: which species is the acid?

For each equation choose the species acting as the acid and the species acting as the base, then name the conjugate of each.

Loading the model…

pH: a logarithmic scale for [H⁺]25.1.3

The hydrogen-ion concentration of aqueous solutions runs from more than 1 mol dm⁻³ in concentrated acid to about 10⁻¹⁴ mol dm⁻³ in concentrated alkali. Numbers that differ by fourteen powers of ten are awkward to compare, so chemists use a logarithmic scale:

pH = −log₁₀[H⁺(aq)] and so [H⁺(aq)] = 10^−pH

Here [H⁺(aq)] is the concentration of hydrogen ions in mol dm⁻³; pH itself has no units. [H⁺] and [H₃O⁺] mean the same thing in these calculations. The minus sign makes the pH of most solutions a positive number, and it means that a higher [H⁺] gives a lower pH. Because the scale is logarithmic, a change of one pH unit is a tenfold change in [H⁺]: a solution of pH 3 has ten times the hydrogen-ion concentration of one of pH 4, and a thousand times that of one of pH 6.

Figure 25.2 The pH scale set against [H⁺(aq)]. Each step of one pH unit corresponds to a factor of ten in hydrogen-ion concentration. The four solutions marked are calculated in this chapter.

Worked example 25.1 · Moving between pH and [H⁺]

Question. (a) Calculate the pH of a solution with [H⁺] = 0.0250 mol dm⁻³. (b) Calculate [H⁺] in a solution of pH 3.40.

(a) Relationship	pH = −log₁₀[H⁺]
Calculation	pH = −log₁₀(0.0250) = 1.60
(b) Relationship	[H⁺] = 10^−pH
Calculation	[H⁺] = 10^−3.40 = 3.98 × 10⁻⁴ mol dm⁻³
Check	(a) 0.0250 lies between 10⁻¹ and 10⁻², so the pH must lie between 1 and 2. (b) pH 3.40 lies between 3 and 4, so [H⁺] must lie between 10⁻³ and 10⁻⁴. Estimating the order of magnitude first catches most calculator slips.

Significant figures in pH

The digits after the decimal point of a pH carry the precision of [H⁺]; the digit before it only fixes the power of ten. A concentration known to three significant figures therefore gives a pH to two decimal places (0.0250 → 1.60). Mark schemes usually accept a pH to a stated minimum of significant figures, so give at least two decimal places unless the question asks otherwise, and never round an intermediate [H⁺] before taking the logarithm.

AnimationCalculations of [H+] from pH

Practise the reverse calculation, [H+] = 10^−pH. Estimate the power of ten before pressing the calculator keys.

The ionic product of water, K_w25.1.3

Even the purest water conducts electricity very slightly, because a tiny fraction of its molecules transfer protons to one another:

H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) or simply H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)

Water is present in such a large, effectively constant, concentration that it is left out of the equilibrium expression. The resulting constant is the ionic product of water:

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

K_w applies to every aqueous solution at that temperature, not just to pure water. If an acid raises [H⁺], the equilibrium shifts to the left and [OH⁻] falls so that the product stays at 1.00 × 10⁻¹⁴. This is what lets you find [H⁺] — and so the pH — of an alkaline solution from its hydroxide-ion concentration.

Pure water and neutrality

In pure water every H⁺ ion is formed together with one OH⁻ ion, so [H⁺] = [OH⁻] and K_w = [H⁺]². At 298 K, [H⁺] = √(1.00 × 10⁻¹⁴) = 1.00 × 10⁻⁷ mol dm⁻³ and the pH is 7.00.

The ionisation of water is endothermic, so raising the temperature shifts the equilibrium to the right and K_w increases. Both [H⁺] and [OH⁻] rise equally, so the pH of pure water falls below 7 when it is heated — yet the water is still neutral, because [H⁺] still equals [OH⁻]. "Neutral" means [H⁺] = [OH⁻]; "pH 7" means neutral only at 298 K.

A reported misconception about K_w

Examiners report two recurring errors. Many candidates think that increasing the temperature of water has no effect on its pH; in fact the pH decreases while the ratio [H⁺] : [OH⁻] stays at 1 : 1. Weaker responses argued the opposite extreme — that because K_w has increased the water must have become acidic, or alkaline. Neither follows: K_w rising raises both ion concentrations equally. Stronger answers calculated [H⁺] and [OH⁻] at the new temperature and compared them.

AnimationKw and the pH of water

How Kw is used to calculate the pH of pure water, and why that pH is not always 7.00. Note that the water stays neutral at every temperature.

The pH of a strong acid25.1.4(a)

A strong acid is one that is fully dissociated in aqueous solution. Hydrochloric acid, nitric acid and (for its first proton) sulfuric acid behave in this way:

HCl(aq) → H⁺(aq) + Cl⁻(aq)

Because the reaction goes to completion, one mole of a monoprotic strong acid gives one mole of H⁺. For a strong monoprotic acid of concentration c, therefore, [H⁺] = c and pH = −log₁₀c. The H⁺ from water (at most 10⁻⁷ mol dm⁻³, and suppressed further by the acid) is negligible unless the acid is extremely dilute.

Worked example 25.2 · A strong acid

Question. Calculate the pH of 0.100 mol dm⁻³ nitric acid, and of the solution made by diluting 10.0 cm³ of it to 250 cm³ with water.

Given	HNO₃ is a strong monoprotic acid, fully dissociated
Relationship	[H⁺] = c(acid); pH = −log₁₀[H⁺]
Before dilution	[H⁺] = 0.100 mol dm⁻³; pH = 1.00
After dilution	c = 0.100 × 10.0/250 = 4.00 × 10⁻³ mol dm⁻³; pH = −log₁₀(4.00 × 10⁻³) = 2.40
Check	A 25-fold dilution should raise the pH by log₁₀25 = 1.40 units. 1.00 + 1.40 = 2.40 ✔

A diprotic strong acid needs care. If a question states or implies that both protons of sulfuric acid are fully released, then [H⁺] = 2c. Read what the question says about the acid rather than assuming.

AnimationCalculating the pH of a strong acid

A step-by-step method for a strong acid: concentration of acid → [H+] → pH.

AnimationCalculations of strong acid pH

Four short questions. Give each answer to two decimal places, as the exercise asks.

The pH of a strong alkali25.1.4(b)

A strong alkali such as sodium hydroxide is fully dissociated, so its solution contains a known concentration of hydroxide ions, not hydrogen ions:

NaOH(aq) → Na⁺(aq) + OH⁻(aq) [OH⁻] = c

The pH, however, is defined through [H⁺]. K_w is the bridge between the two:

[H⁺] = K_w / [OH⁻] then pH = −log₁₀[H⁺]

For a Group 2 hydroxide such as Ba(OH)₂, each formula unit releases two OH⁻ ions, so [OH⁻] = 2c.

Worked example 25.3 · Strong alkalis

Question. Calculate the pH at 298 K of (a) 0.0500 mol dm⁻³ NaOH(aq); (b) 0.0100 mol dm⁻³ Ba(OH)₂(aq). K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

(a) [OH⁻]	0.0500 mol dm⁻³
[H⁺]	1.00 × 10⁻¹⁴ ÷ 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³
pH	−log₁₀(2.00 × 10⁻¹³) = 12.70
(b) [OH⁻]	2 × 0.0100 = 0.0200 mol dm⁻³
[H⁺]; pH	1.00 × 10⁻¹⁴ ÷ 0.0200 = 5.00 × 10⁻¹³; pH = 12.30
Check	Both are strongly alkaline (pH well above 7). The shortcut pH = 14.00 − pOH, where pOH = −log₁₀[OH⁻], gives the same answers at 298 K: 14.00 − 1.30 = 12.70.

The reverse calculation — finding the concentration of an alkali from its pH — runs the same steps backwards: [H⁺] = 10^−pH, then [OH⁻] = K_w/[H⁺].

Mixing a strong acid with a strong alkali

When a strong acid and a strong alkali are mixed in unequal amounts, H⁺ + OH⁻ → H₂O removes the smaller amount completely. Find which is in excess, calculate the moles left over, divide by the total volume, and then use the strong acid or strong alkali method. For example, 25.0 cm³ of 0.100 mol dm⁻³ HCl mixed with 20.0 cm³ of 0.100 mol dm⁻³ NaOH leaves 5.0 × 10⁻⁴ mol of H⁺ in 45.0 cm³: [H⁺] = 0.0111 mol dm⁻³ and pH = 1.95.

AnimationCalculating the pH of a strong base

The route from [OH−] to [H+] through Kw, worked for 1 mol dm−3 sodium hydroxide.

AnimationCalculations of strong base pH

Five practice questions on strong bases. For Ba(OH)2 remember the factor of two.

AnimationStrong acid–strong base calculations

pH after partial neutralisation: find the reagent in excess, divide by the total volume, then calculate the pH.

Weak acids: K_a and pK_a25.1.3

A weak acid is only partly dissociated in water. In 0.100 mol dm⁻³ ethanoic acid only about one molecule in a hundred has given up its proton at any moment; the rest remain as CH₃COOH molecules. The dissociation is a dynamic equilibrium:

HA(aq) ⇌ H⁺(aq) + A⁻(aq) e.g. CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq)

Its equilibrium constant is the acid dissociation constant, K_a:

K_a = [H⁺(aq)][A⁻(aq)] / [HA(aq)] units: mol dm⁻³

K_a measures the strength of the acid — the extent of its dissociation — and, like every equilibrium constant, it changes only with temperature. The larger K_a, the further the equilibrium lies to the right and the stronger the acid. Because K_a values are small and span many powers of ten, they are often converted to a logarithmic scale in the same way as [H⁺]:

pK_a = −log₁₀K_a K_a = 10^−pK_a

The minus sign reverses the order: the smaller the pK_a, the stronger the acid. Ethanoic acid (K_a = 1.74 × 10⁻⁵ mol dm⁻³) has pK_a = 4.76; hypobromous acid (K_a = 2.00 × 10⁻⁹) has pK_a = 8.70 and is much weaker.

K_a is not [H⁺]²/[HA], and pK_a is not inversely proportional to K_a

Asked for the general K_a expression, candidates often wrote [H⁺]²/[HA]. That form is an approximation valid only for a solution of the weak acid alone, where [H⁺] = [A⁻]; it is wrong as a definition and fails completely in a buffer. Examiners have also reported answers stating that pK_a is inversely proportional to K_a. The relationship is logarithmic: pK_a = −log₁₀K_a. Finally, a K_a expression uses square brackets for equilibrium concentrations of species in solution; water is not included.

**Table 25.1** Strength and concentration are different things. Values quoted are those used in this chapter's examples.
acid	K_a / mol dm⁻³	pK_a	relative strength
4-chlorobutanoic acid, Cl(CH₂)₃CO₂H	3.02 × 10⁻⁵	4.52	strongest at the top; weakest at the bottom
ethanoic acid, CH₃COOH	1.74 × 10⁻⁵	4.76
propanoic acid, CH₃CH₂COOH	1.35 × 10⁻⁵	4.87
chloric(I) acid, HClO	3.72 × 10⁻⁸	7.43
hypobromous acid, HBrO	2.00 × 10⁻⁹	8.70

Notice what K_a does not depend on: concentration. "Strong" and "weak" describe how far an acid dissociates; "concentrated" and "dilute" describe how much acid is dissolved per dm³. A concentrated solution of a weak acid can have a lower pH than a very dilute solution of a strong acid.

The pH of a weak acid25.1.4(c)

For a weak acid, [H⁺] is not equal to the acid concentration, because most of the acid is undissociated. K_a supplies the missing link. For a solution of a weak acid HA of concentration c on its own, two approximations make the calculation straightforward:

[H⁺] = [A⁻]. Each molecule that dissociates gives one H⁺ and one A⁻; the H⁺ from water is negligible by comparison.
[HA] at equilibrium ≈ c. So little of the acid dissociates that its concentration is essentially unchanged.

Substituting both into the K_a expression gives:

K_a = [H⁺]² / c so [H⁺] = √(K_a × c)

Worked example 25.4 · pH of a weak acid

Question. Calculate the pH of 0.100 mol dm⁻³ ethanoic acid. K_a = 1.74 × 10⁻⁵ mol dm⁻³.

Given	c = 0.100 mol dm⁻³; K_a = 1.74 × 10⁻⁵ mol dm⁻³
Relationship	[H⁺] = √(K_ac), assuming [H⁺] = [A⁻] and [HA] ≈ c
Substitution	[H⁺] = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶)
Calculation	[H⁺] = 1.32 × 10⁻³ mol dm⁻³; pH = −log₁₀(1.32 × 10⁻³) = 2.88
Check	Only 1.32 × 10⁻³ ÷ 0.100 = 1.3% of the acid is dissociated, so assumption 2 was reasonable. The pH is higher than that of 0.100 mol dm⁻³ HCl (1.00), as it must be for a weak acid.

The same relationship can be run backwards: if the pH of a weak acid solution of known concentration is measured, K_a can be found. This is how K_a values are determined in practice.

Worked example 25.5 · K_a and pK_a from a measured pH

Question. A 0.0500 mol dm⁻³ solution of a weak monoprotic acid HA has pH 3.12. Calculate K_a and pK_a.

[H⁺]	10^−3.12 = 7.59 × 10⁻⁴ mol dm⁻³ = [A⁻]
K_a	[H⁺]²/[HA] = (7.59 × 10⁻⁴)² ÷ 0.0500 = 1.15 × 10⁻⁵ mol dm⁻³
pK_a	−log₁₀(1.15 × 10⁻⁵) = 4.94
Check	Dissociation is 1.5%, so [HA] ≈ c is still fair. Not dividing by c at all, or forgetting to square [H⁺], are the usual slips; a K_a larger than c would be a clear warning sign.

Figure 25.3 Calculated pH of hydrochloric acid and ethanoic acid across four powers of ten of concentration. The strong-acid line has a gradient of −1 (tenfold dilution raises pH by 1); the weak-acid line has a gradient of −½, because [H⁺] depends on √c. The dashed line is the √(K_ac) approximation; it departs from the full solution only when the acid is so dilute that a large fraction dissociates.

Figure 25.3 shows two consequences of the weak-acid model. At any given concentration the weak acid has a much higher pH than the strong acid. And diluting a weak acid tenfold raises its pH by only about half a unit, because diluting the acid also shifts its equilibrium to the right and releases more H⁺ — Le Chatelier's principle at work.

Think like a chemist · when the approximation fails

The assumption [HA] ≈ c fails when the acid is not very weak or very dilute, because then a significant fraction dissociates. The more complete treatment uses [HA] = c − [H⁺], which leads to a quadratic equation. Cambridge questions at this level expect the approximation unless told otherwise; the point to understand is why it works (little dissociation) so that you can recognise when it would not.

Using the strong-acid method for a weak acid

A common error on a question about 0.0400 mol dm⁻³ hypobromous acid was to take [H⁺] = 0.0400 and obtain pH 1.40 — the strong-acid method — instead of using K_a to get pH 5.05. Examiners also noted candidates who could not convert "4.00 × 10⁻³ mol in 100 cm³" into a concentration of 4.00 × 10⁻² mol dm⁻³ before starting. Always ask first: is this acid strong or weak, and do I have a concentration in mol dm⁻³?

AnimationFinding the pH of a weak acid: method

Why [H+] ≠ [CH3COOH] for ethanoic acid, and how Ka is used instead. Note the two assumptions the method makes.

AnimationWeak acid pH: a worked example

Step through the calculation for 0.1 mol dm−3 ethanoic acid and compare your answer with Worked example 25.4.

AnimationCalculations of weak acid pH

Practice questions on weak acids. Write the Ka expression before substituting.

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Choosing the right method

Every pH question in this topic begins by deciding which kind of solution you have. Table 25.2 summarises the choice.

**Table 25.2** Choosing the pH method.
solution	find [H⁺] from	key point
strong monoprotic acid, c	[H⁺] = c	fully dissociated
strong alkali, [OH⁻] known	[H⁺] = K_w/[OH⁻]	[OH⁻] = 2c for M(OH)₂
weak acid alone, c and K_a	[H⁺] = √(K_ac)	[H⁺] = [A⁻]; [HA] ≈ c
strong acid + strong alkali mixed	excess H⁺ or OH⁻ ÷ total volume	neutralisation goes to completion
weak acid + its conjugate base (buffer)	[H⁺] = K_a[HA]/[A⁻]	buffers: see Worked example 25.6; [H⁺] ≠ [A⁻]

AnimationpH calculations summary

Follow the decision chain to see which calculation each kind of solution needs.

AnimationpH calculations: mixed practice

Five mixed questions. Identify the type of solution before choosing the method.

AnimationCalculations summary: chains of questions

Chains of linked questions on pH, Ka and Kw. Use this after the whole of this part.

Quick check 25.1

Identify the two conjugate acid–base pairs in CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺.
answer
CH₃COOH (acid) / CH₃COO⁻ (conjugate base); H₃O⁺ (acid) / H₂O (conjugate base).
Calculate the pH of 2.5 × 10⁻³ mol dm⁻³ KOH(aq) at 298 K.
answer
[H⁺] = 1.00 × 10⁻¹⁴/2.5 × 10⁻³ = 4.0 × 10⁻¹²; pH = 11.40.
The pH of pure water at 50 °C is 6.63. Is the water acidic, neutral or alkaline? Explain.
answer
Neutral: in pure water [H⁺] = [OH⁻] at every temperature. K_w is larger at 50 °C, so both concentrations are higher and the pH is lower than 7.
Acid P has pK_a 3.2 and acid Q has pK_a 4.8. Which is stronger, and by what factor is its K_a larger?
answer
P is stronger; its K_a is 10^1.6 ≈ 40 times larger.

Examiner's overall observation · Acids, pH, K_w and K_a

Answered well: writing the mathematical expressions for pH and K_w; converting a pH into [H⁺]; routine pH calculations for a strong alkali and for a weak acid from K_a; finding K_a from a measured pH, where candidates showed good mathematical skills with very few calculator or rounding errors.

Found difficult: defining a conjugate acid–base pair precisely; comparing the concentrations of a strong and a weak acid that have the same pH, which needs the K_a expression rearranged for [HA]; predicting how temperature affects the pH of water; going on from K_a to pK_a when both are asked for.

Recurring errors: K_a written as [H⁺]²/[HA]; the strong-acid method used for a weak acid; moles not converted to a concentration; ratios inverted; OH⁻ used in place of water as the base; conjugate bases carrying a more positive charge than their acids; "pH stays the same" for heated water.

What successful answers did: stated which species are the conjugate pair and that they differ by one H⁺; decided first whether an acid was strong or weak; wrote the K_a expression before substituting; carried unrounded values to the final logarithm.

What a buffer solution is, and how one is made25.1.5(a), 25.1.5(b)

Add one drop of concentrated hydrochloric acid to a litre of pure water and the pH falls from 7 to about 3. Add the same drop to a litre of blood, or of seawater, or of the solution in a shampoo bottle labelled "pH balanced", and the pH barely moves. These solutions are buffered.

Definition

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

Every word of that definition earns its place. A buffer resists change: the pH does alter slightly, it is not held constant. And it does so only for small additions: add enough acid and any buffer is overwhelmed.

The most frequently reported buffer error

Across many sessions examiners report the same imprecise definition: "a buffer keeps the pH constant", "maintains the pH" or "the pH remains unchanged". None of these is credited. The creditworthy answer says the solution resists (or minimises) change in pH when small amounts of acid or alkali are added. Answers that omitted the word pH, or omitted "small", also lost the mark.

What a buffer contains

The buffers in this syllabus are acidic buffers: a weak acid, HA, together with a large amount of its conjugate base, A⁻. Both are needed. The weak acid can react with added OH⁻; the conjugate base can react with added H⁺. Both must be present in large concentrations compared with the amounts of acid or alkali to be added — that is what gives the buffer its capacity.

There are two ways to make such a mixture:

Weak acid + a soluble salt of that acid. Ethanoic acid mixed with sodium ethanoate. The salt is fully ionic, so it supplies CH₃COO⁻ directly.
Excess weak acid + a strong alkali. Ethanoic acid partly neutralised by sodium hydroxide. The alkali converts some of the acid into its conjugate base, CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O, leaving a mixture of the two. The acid must be in excess — if all of it were neutralised there would be no HA left to react with added alkali.

A solution of a weak acid alone is not an effective buffer: it contains plenty of HA but only a tiny concentration of A⁻, so it cannot absorb added H⁺. Nor is a strong acid with its salt (HCl + NaCl) a buffer: Cl⁻ is far too weak a base to remove added H⁺.

Naming the components

Asked to suggest one organic and one inorganic compound that could each be added to aqueous butanoic acid to give a buffer, many candidates managed only the inorganic one (sodium hydroxide). The organic answer is a salt of the same acid — sodium butanoate, CH₃CH₂CH₂COONa. A correct organic compound was rare, even though it is simply route 1 above.

AnimationHow do buffers work?

See what happens to an acidic buffer when acid and then alkali are added. Watch which species is used up each time.

How a buffer controls pH25.1.5(c)

Take the ethanoic acid – sodium ethanoate buffer. It contains:

a large concentration of CH₃COOH, because the weak acid is only slightly dissociated and its dissociation is suppressed further by the added ethanoate ions;
a large concentration of CH₃COO⁻, from the fully ionised salt;
a small concentration of H⁺, fixed by the equilibrium CH₃COOH ⇌ H⁺ + CH₃COO⁻.

When a small amount of acid is added, the added H⁺ ions are removed by the large reservoir of conjugate base:

CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq)

In equilibrium terms, the position of CH₃COOH ⇌ H⁺ + CH₃COO⁻ shifts to the left. [CH₃COOH] rises a little and [CH₃COO⁻] falls a little, but both are so large that their ratio hardly changes.

When a small amount of alkali is added, the added OH⁻ ions are removed by the large reservoir of weak acid:

CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l)

(Equivalently: OH⁻ removes H⁺ to form water, and more CH₃COOH dissociates to replace it — the equilibrium shifts to the right.) Again the ratio [CH₃COOH]/[CH₃COO⁻] barely changes.

Figure 25.4 How an acidic buffer resists change in pH. Added H⁺ is removed by the reservoir of A⁻; added OH⁻ is removed by the reservoir of HA. Because both reservoirs are large, the ratio [HA]/[A⁻] — and therefore [H⁺] — changes very little.

Why does the ratio matter? Rearranging the K_a expression shows that in any mixture of HA and A⁻,

[H⁺] = K_a × [HA] / [A⁻]

K_a is constant at constant temperature, so [H⁺], and hence the pH, depends only on the ratio of acid to conjugate base. The reactions above change that ratio only slightly, so the pH changes only slightly. This single relationship explains how a buffer works and how to calculate its pH.

Equations that do not show buffering

Examiners regularly see equations that are correct chemistry but do not answer the question. OH⁻ + H₃O⁺ → 2H₂O does not show how the buffer components remove alkali. Nor does H₂CO₃ + H₂O → HCO₃⁻ + H₃O⁺, which is simply the acid dissociating. The two equations required show the conjugate base reacting with H⁺ and the weak acid reacting with OH⁻.

Buffers that are not a named acid and salt

The same logic applies to any mixture of a weak acid and its conjugate base. Hydrogenphosphate buffers, used in biochemistry, contain H₂PO₄⁻ (the acid) and HPO₄²⁻ (its conjugate base): HPO₄²⁻ + H⁺ → H₂PO₄⁻ removes acid, and H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O removes alkali. Amino acids in solution can also act as buffers, because the zwitterion carries both a group that can accept a proton (–COO⁻) and one that can donate a proton (–NH₃⁺).

AnimationThe half-equivalence point

When exactly half of a weak acid has been neutralised, [HA] = [A−], so [H+] = Ka and pH = pKa. This is the fastest route through many buffer questions.

Calculating the pH of a buffer25.1.6

A buffer calculation always uses the full K_a expression, never the weak-acid shortcut. In a buffer [H⁺] is not equal to [A⁻], because nearly all the A⁻ comes from the salt. Instead, two different approximations are made:

[HA] at equilibrium ≈ the concentration (or amount) of weak acid put in, because its slight dissociation is suppressed by the A⁻ already present;
[A⁻] at equilibrium ≈ the concentration of conjugate base put in (from the salt, or formed by the alkali).

[H⁺] = K_a × [HA] / [A⁻] pH = −log₁₀[H⁺]

Because HA and A⁻ are in the same solution, the volume cancels in the ratio: you may use moles of HA and A⁻ directly. Taking logarithms gives an equivalent form that many find convenient:

pH = pK_a + log₁₀([A⁻] / [HA])

Two consequences are worth remembering. When [HA] = [A⁻], pH = pK_a — the half-neutralisation result. And a buffer's pH is set mainly by the pK_a of its acid; the ratio then fine-tunes it by about ±1 unit.

Getting the log form upside down

In a question on seawater, a common error was pH 5.20, produced by the incorrect equation pH = pK_a − log([A⁻]/[HA]). A quick sense-check prevents it: adding more conjugate base (the proton acceptor) must make the solution less acidic, so a ratio [A⁻]/[HA] greater than 1 must give a pH above pK_a.

Worked example 25.6 · Buffer from a weak acid and its salt

Question. Equal volumes of 0.100 mol dm⁻³ ethanoic acid and 0.150 mol dm⁻³ sodium ethanoate are mixed. Calculate the pH of the buffer. K_a = 1.74 × 10⁻⁵ mol dm⁻³.

Given	after mixing, [HA] = 0.0500 and [A⁻] = 0.0750 mol dm⁻³ (both halved; the ratio is 0.100 : 0.150 either way)
Relationship	[H⁺] = K_a[HA]/[A⁻]
Substitution	[H⁺] = 1.74 × 10⁻⁵ × 0.0500/0.0750 = 1.16 × 10⁻⁵ mol dm⁻³
Answer	pH = −log₁₀(1.16 × 10⁻⁵) = 4.94
Check	pK_a = 4.76; there is more A⁻ than HA, so the pH should be a little above 4.76 ✔

Worked example 25.7 · Buffer made by partly neutralising a weak acid

Question. 10.0 cm³ of 0.100 mol dm⁻³ NaOH is added to 30.0 cm³ of 0.100 mol dm⁻³ ethanoic acid. Calculate the pH of the resulting solution. K_a = 1.74 × 10⁻⁵ mol dm⁻³.

Initial moles	CH₃COOH: 0.100 × 30.0/1000 = 3.00 × 10⁻³; NaOH: 0.100 × 10.0/1000 = 1.00 × 10⁻³
Reaction	CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O goes to completion; OH⁻ is limiting
Moles after	CH₃COOH: 3.00 − 1.00 = 2.00 × 10⁻³; CH₃COO⁻: 1.00 × 10⁻³
[H⁺]	1.74 × 10⁻⁵ × (2.00 × 10⁻³)/(1.00 × 10⁻³) = 3.48 × 10⁻⁵ mol dm⁻³
Answer	pH = 4.46
Check	More acid than conjugate base remains, so pH is below pK_a (4.76) ✔. The total volume (40.0 cm³) cancels, so it need not be used.

Examiners describe this second type as highly discriminating. Most candidates found the initial moles and could take a logarithm; the step that separated the strongest answers was working out the moles of acid and conjugate base after the neutralisation. Writing the equation and a before/after line, as above, makes that step visible.

Designing a buffer of a chosen pH

The same relationship can be rearranged to find how much salt must be added to give a buffer of a required pH. For example, to make pH 4.50 from 0.200 mol of ethanoic acid: [H⁺] = 10^−4.50 = 3.16 × 10⁻⁵; moles of A⁻ = K_a × n(HA)/[H⁺] = 1.74 × 10⁻⁵ × 0.200 ÷ 3.16 × 10⁻⁵ = 0.110 mol, which is 0.110 × 82.0 = 9.02 g of sodium ethanoate. Examiners found that candidates could convert the pH to [H⁺] and could convert moles of salt to mass, but many could not use the K_a expression to find the moles of conjugate base in between.

AnimationCalculating the pH of a buffer solution

A worked example: ethanoic acid mixed with sodium ethanoate. Compare the method with Worked example 25.6.

AnimationCalculations: buffer pH

Three practice questions on buffers made from a weak acid and its salt.

AnimationCalculating the buffer pH from acid and alkali

A buffer made by partly neutralising ethanoic acid with sodium hydroxide. Find the moles of each species left before using Ka.

AnimationWeak acid–strong base calculation

A step-by-step multiple-choice route through a partial-neutralisation buffer. Work with the Ka printed in the exercise, which is not the usual value for ethanoic acid.

How much does the pH change — and when does a buffer fail?25.1.5(c), 25.1.6

To calculate the pH after an addition, treat the added acid or alkali as reacting completely with the appropriate buffer component, then recalculate the ratio.

Worked example 25.8 · The pH change on adding acid

Question. 1.00 dm³ of buffer contains 0.100 mol CH₃COOH and 0.100 mol CH₃COO⁻. Calculate the pH before and after adding 0.0100 mol of HCl. K_a = 1.74 × 10⁻⁵ mol dm⁻³.

Before	ratio 1 : 1, so pH = pK_a = −log₁₀(1.74 × 10⁻⁵) = 4.76
Reaction	CH₃COO⁻ + H⁺ → CH₃COOH: 0.0100 mol A⁻ converted to HA
After	HA = 0.110 mol; A⁻ = 0.090 mol; [H⁺] = 1.74 × 10⁻⁵ × 0.110/0.090 = 2.13 × 10⁻⁵
Answer	pH = 4.67, a fall of only 0.09 units
Compare	0.0100 mol HCl in 1.00 dm³ of pure water gives [H⁺] = 0.0100 and pH 2.00 — a fall of 5 units.

Figure 25.5 Calculated pH of 1.00 dm³ of pure water and of 1.00 dm³ of an ethanoic acid – ethanoate buffer as hydrochloric acid (right) or sodium hydroxide (left) is added. The buffer's pH drifts slowly; water's pH changes by several units with the first additions.

Buffer capacity: when the buffer is overwhelmed

The ratio argument works only while both reservoirs remain large. If more acid is added than there is conjugate base to remove it, all the A⁻ is converted to HA and the excess H⁺ stays in solution; the pH then falls towards the value set by the excess strong acid. The same happens with alkali once all the HA is used up. The more concentrated the buffer components, the greater its capacity.

Explaining a buffer that has failed

In one question, dilute sulfuric acid was mixed with a small sample of a propanoate buffer and the final pH was close to 1. Full credit needed two ideas: all the propanoate ions had been converted to propanoic acid (the buffer's conjugate base was used up), and H⁺ from the sulfuric acid was now in excess, so the pH was set by the strong acid. Examiners reported that the award of both marks was rare: many recognised the first idea but did not explain why the pH was close to 1.

AnimationChanges in the pH of a buffer

Calculate the pH change when hydrochloric acid is added to a buffer of known composition.

AnimationCalculations: changes in pH

Four questions on buffers after an addition. Convert the added acid or alkali to moles first.

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Uses of buffers, and the control of pH in blood25.1.5(d)

Wherever a process works only within a narrow pH range, a buffer is used to hold it there:

biological systems — enzymes lose their activity outside a narrow pH range, so body fluids are buffered;
laboratory work — calibrating pH meters; controlling the pH in electrophoresis, in some precipitation reactions and in biochemical experiments (hydrogenphosphate buffers);
consumer products — shampoos, cosmetics, food and drinks, where pH affects stability, taste or effect on the skin;
the environment — the carbonic acid / hydrogencarbonate system buffers the pH of seawater.

The HCO₃⁻ buffer in blood

Human blood is held at a pH of about 7.4; even small deviations are dangerous. The most important buffer in blood is the carbonic acid – hydrogencarbonate system. Carbon dioxide produced by respiration dissolves in the blood and forms carbonic acid, which dissociates into hydrogencarbonate ions:

CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq)

H₂CO₃ is the weak acid; HCO₃⁻ is its conjugate base.

When acid enters the blood — for example lactic acid produced in exercising muscles — the added H⁺ is removed by hydrogencarbonate ions: HCO₃⁻ + H⁺ → H₂CO₃ (which can also be written HCO₃⁻ + H⁺ → CO₂ + H₂O).
When alkali is added, OH⁻ is removed by carbonic acid: H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O.

In normal blood the concentration of HCO₃⁻ is roughly twenty times that of H₂CO₃. The large reservoir of hydrogencarbonate is well suited to the body's main challenge, which is the continual production of acids by metabolism. The carbonic acid formed can leave the system as carbon dioxide, which is breathed out.

Figure 25.6 The carbonic acid – hydrogencarbonate buffer. Hydrogencarbonate ions remove added acid; carbonic acid removes added alkali.

Worked example 25.9 · A carbonate buffer

Question. In a sample of seawater [HCO₃⁻]/[H₂CO₃] = 14.1. Calculate the pH. pK_a(H₂CO₃) = 6.35.

K_a	10^−6.35 = 4.47 × 10⁻⁷ mol dm⁻³
[H⁺]	K_a × [H₂CO₃]/[HCO₃⁻] = 4.47 × 10⁻⁷ ÷ 14.1 = 3.17 × 10⁻⁸ mol dm⁻³
Answer	pH = 7.50 (or pH = 6.35 + log₁₀14.1 = 7.50)
Check	More base than acid, so pH above pK_a ✔ — the check that catches the 5.20 error described earlier.

Chemistry connection

The same species appear in thermal-stability questions: hydrogencarbonates decompose on gentle heating, 2HCO₃⁻ → CO₃²⁻ + CO₂ + H₂O, and to topic 7, where it is a textbook example of Le Chatelier's principle. In the oceans, extra dissolved CO₂ pushes the equilibrium to the right, raising [H⁺] and lowering the pH.

Past-paper practice · Set 25B · Buffer solutions

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 25B.1[2]

Answer and marking guidance

(i) A solution whose pH resists change when small amounts of acid or base are added ✔. (ii) Organic: sodium butanoate (a salt of butanoic acid, CH₃CH₂CH₂COO⁻) AND inorganic: NaOH ✔. Examiner insight: most knew the syllabus definition, but some were imprecise, stating that the pH would remain unchanged or stay constant. Many found (ii) challenging: NaOH was often given, but a correct organic compound was rare.

Question 25B.2[7]

Answer and marking guidance

(i) A solution that resists (opposes) changes in pH ✔ when small amounts of acid (H⁺) or alkali (OH⁻) are added ✔. (ii) With acid: HCO₃⁻ + H⁺ → H₂CO₃ ✔ (or HCO₃⁻ + H₃O⁺ → H₂CO₃ + H₂O); with alkali: H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O ✔. (iii) K_a = 10^−6.35 = 4.47 × 10⁻⁷ ✔; [H⁺] = 4.47 × 10⁻⁷ ÷ 14.1 = 3.17 × 10⁻⁸ ✔; pH = 7.5 ✔ (7.50). Examiner insight: most knew the definition, but some stated that the pH would remain unchanged or stay constant. (ii) was generally well answered; a common error was a reaction with water, such as H₂CO₃ + H₂O → HCO₃⁻ + H₃O⁺. In (iii) the correct answer was often given; a common error was 5.20, from the incorrect equation pH = pK_a − log([A⁻]/[HA]).

Question 25B.3[8]

Answer and marking guidance

(d) HCO₃⁻ + H⁺ → H₂CO₃ (or → CO₂ + H₂O) ✔; H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O ✔. (e)(i) CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺ ✔; labels acid, base, base, acid ✔. (ii) moles NaOH = 0.15 × 20/1000 = 0.0030 AND initial moles CH₃COOH = 0.25 × 30/1000 = 0.0075 ✔; after reaction CH₃COOH = 0.0045 mol AND CH₃COO⁻ = 0.0030 mol ✔; [CH₃COOH] = 0.090, [CH₃COO⁻] = 0.060 mol dm⁻³, [H⁺] = 1.75 × 10⁻⁵ × 0.090/0.060 = 2.625 × 10⁻⁵ ✔; pH = 4.6 ✔. Examiner insight: (d) was well known. In (e)(i) some used OH⁻ instead of water as the base. (e)(ii) was very discriminating: most calculated the initial moles and a pH from their [H⁺]; better performing candidates went on to find the equilibrium moles of acid and salt.

Question 25B.4[5]

This question continues from 25A.4 (the same paper and question).

Answer and marking guidance

(b) [H⁺] = 10^−5.00 = 1.0 × 10⁻⁵ AND use of K_a = [H⁺][A⁻]/[HA] (or pH = pK_a + log([A⁻]/[HA])) ✔; moles of A⁻ = 1.35 × 10⁻⁵ × (5.00/74.0) ÷ 1.0 × 10⁻⁵ = 0.0912 ✔; mass of sodium propanoate = 0.0912 × 96.0 = 8.76 g ✔. (c) All of the propanoate ions have been protonated (converted to propanoic acid) ✔; H⁺ from the sulfuric acid is now in excess, so the pH is set by the strong acid ✔. Examiner insight: in (b) most could find [H⁺] from the pH and convert moles of propanoate to a mass, but many could not calculate the moles of propanoate using the K_a expression. In (c) the award of both marks was rare: some recognised that the propanoate had been neutralised, but most did not explain why the pH was close to 1.

Quick check 25.2

Explain why a mixture of hydrochloric acid and sodium chloride is not a buffer.
answer
Cl⁻ is the conjugate base of a strong acid and has almost no tendency to accept H⁺, so added acid is not removed; there is also no weak acid to remove added OH⁻.
Write the equation for the reaction that removes added OH⁻ in a propanoic acid / sodium propanoate buffer.
answer
CH₃CH₂COOH + OH⁻ → CH₃CH₂COO⁻ + H₂O
A buffer contains 0.020 mol HA and 0.040 mol A⁻; pK_a = 4.20. Calculate its pH.
answer
pH = 4.20 + log₁₀(0.040/0.020) = 4.50.
Half of a sample of a weak acid is neutralised by NaOH; the pH is 3.86. State the pK_a of the acid.
answer
3.86 — at half-neutralisation [HA] = [A⁻], so pH = pK_a.

Examiner's overall observation · Buffer solutions

Answered well: recalling the definition in its syllabus form by many candidates; writing the two blood-buffer equations with HCO₃⁻ and H₂CO₃; calculating a pH once a value of [H⁺] had been found; calculating the initial moles of acid and alkali in a partial-neutralisation buffer.

Found difficult: finding the moles of weak acid and conjugate base remaining after partial neutralisation; using the K_a expression to find how much salt is needed for a given pH; recognising that at half-neutralisation pH = pK_a; explaining both parts of what happens when a buffer is overwhelmed; naming an organic salt that would make a buffer with a given acid.

Recurring errors: "keeps the pH constant" or "maintains pH"; omitting "pH" or "small amounts" from the definition; equations showing water or H₃O⁺ rather than the buffer components reacting; using K_a = [H⁺]²/[HA] in a buffer, where [H⁺] and [A⁻] are not equal; the log ratio inverted.

What successful answers did: wrote the neutralisation equation with a before/after line of moles; used the full K_a expression with moles of HA and A⁻; checked whether the pH should be above or below pK_a.

Solubility equilibria and the solubility product, K_sp25.1.7, 25.1.8

Barium sulfate is swallowed as a "barium meal" before an X-ray of the digestive system, even though barium ions are toxic. It is safe because so little barium sulfate dissolves. "Insoluble" in everyday chemistry means sparingly soluble: a small, definite amount dissolves, and the rest stays as solid.

Shake a sparingly soluble ionic solid with water and ions leave the lattice and enter solution. As their concentration builds up, ions also collide with the solid and rejoin the lattice. Eventually the two rates are equal: the solution is saturated, and a dynamic equilibrium exists between the solid and its aqueous ions.

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)

Figure 25.7 A saturated solution of lead(II) iodide. Ions leave and rejoin the lattice at equal rates; the concentrations of the aqueous ions are fixed by K_sp. The solid, whose concentration cannot change, does not appear in the expression.

The equilibrium constant for this heterogeneous equilibrium is the solubility product. As with all heterogeneous equilibria, the solid is left out of the expression — its "concentration" is constant — so K_sp contains only the aqueous ions, each raised to the power of its coefficient in the equation:

K_sp = [Pb²⁺(aq)][I⁻(aq)]² units: mol³ dm⁻⁹

Definition

The solubility product, K_sp, of a sparingly soluble ionic compound is the product of the concentrations of its ions in a saturated solution, each concentration raised to the power of the number of those ions in the formula, at a stated temperature.

For M_xA_y(s) ⇌ xM^y+(aq) + yA^x−(aq): K_sp = [M^y+]^x[A^x−]^y

Like every equilibrium constant, K_sp depends only on temperature. It applies only to sparingly soluble salts: for very soluble salts the solutions are so concentrated that ions interact strongly and the simple expression no longer holds.

**Table 25.3** Writing K_sp expressions and their units.
salt	equilibrium	K_sp	units
BaSO₄	BaSO₄(s) ⇌ Ba²⁺ + SO₄²⁻	[Ba²⁺][SO₄²⁻]	mol² dm⁻⁶
CaF₂	CaF₂(s) ⇌ Ca²⁺ + 2F⁻	[Ca²⁺][F⁻]²	mol³ dm⁻⁹
Ag₂SO₃	Ag₂SO₃(s) ⇌ 2Ag⁺ + SO₃²⁻	[Ag⁺]²[SO₃²⁻]	mol³ dm⁻⁹
Mn(OH)₂	Mn(OH)₂(s) ⇌ Mn²⁺ + 2OH⁻	[Mn²⁺][OH⁻]²	mol³ dm⁻⁹
Al(OH)₃	Al(OH)₃(s) ⇌ Al³⁺ + 3OH⁻	[Al³⁺][OH⁻]³	mol⁴ dm⁻¹²

Errors in K_sp expressions

Examiners have reported: omitting the square brackets that mean "concentration"; including the solid, e.g. [Sr(OH)₂(s)], in the expression; and writing the coefficient inside the bracket, as in [Fe²⁺][2OH⁻]². The coefficient in the equation becomes a power, not a multiplier inside the bracket: K_sp = [Fe²⁺][OH⁻]². The units follow from the total power: two ions → mol² dm⁻⁶, three → mol³ dm⁻⁹, four → mol⁴ dm⁻¹².

Calculating K_sp from solubility, and solubility from K_sp25.1.9

The solubility, s, of a salt is the amount that dissolves to form a saturated solution, usually in mol dm⁻³ (sometimes given in g dm⁻³, which must first be divided by M_r). The stoichiometry of the dissolving equation links s to the ion concentrations: if s mol dm⁻³ of PbI₂ dissolves, then [Pb²⁺] = s and [I⁻] = 2s.

**Table 25.4** The link between solubility s and K_sp for common formula types (dissolving in pure water).
formula type	example	ion concentrations	K_sp in terms of s	s in terms of K_sp
MA	BaSO₄	s, s	s²	√K_sp
MA₂ or M₂A	PbI₂, Ag₂SO₃	s, 2s	s × (2s)² = 4s³	∛(K_sp/4)
MA₃	Al(OH)₃	s, 3s	s × (3s)³ = 27s⁴	∜(K_sp/27)

The factors 4 and 27 are where most marks are lost: (2s)² is 4s², not 2s², and (3s)³ is 27s³. Writing the concentration of each ion explicitly before substituting avoids the error.

Worked example 25.10 · K_sp from a solubility in g dm⁻³

Question. The solubility of calcium fluoride, CaF₂, at 298 K is 0.0160 g dm⁻³. Calculate K_sp, with units. (M_r CaF₂ = 78.1)

Solubility	s = 0.0160 ÷ 78.1 = 2.05 × 10⁻⁴ mol dm⁻³
Ion concentrations	[Ca²⁺] = s = 2.05 × 10⁻⁴; [F⁻] = 2s = 4.10 × 10⁻⁴ mol dm⁻³
K_sp	[Ca²⁺][F⁻]² = 2.05 × 10⁻⁴ × (4.10 × 10⁻⁴)² = 3.44 × 10⁻¹¹ mol³ dm⁻⁹
Check	Equivalent to 4s³ = 4 × (2.05 × 10⁻⁴)³ ✔. Forgetting to square [F⁻] gives 8.4 × 10⁻⁸; forgetting the factor 2 gives 8.6 × 10⁻¹².

Worked example 25.11 · Solubility from K_sp

Question. K_sp of barium sulfate at 298 K is 1.08 × 10⁻¹⁰ mol² dm⁻⁶. Calculate its solubility in water in mol dm⁻³.

Relationship	BaSO₄ ⇌ Ba²⁺ + SO₄²⁻; K_sp = s × s = s²
Answer	s = √(1.08 × 10⁻¹⁰) = 1.04 × 10⁻⁵ mol dm⁻³
Check	Units: √(mol² dm⁻⁶) = mol dm⁻³ ✔. A tiny solubility, as expected for a salt used in a barium meal.

Treating K_sp as the solubility

When asked to calculate the solubility of manganese(II) hydroxide from its K_sp, some candidates approached the question as if 1.1 × 10⁻¹¹ was the solubility, and the fact that [OH⁻] is twice [Mn²⁺] was not well appreciated. In a question on silver sulfite, the most common error was to give s (1.55 × 10⁻⁵) when the question asked for [Ag⁺], which is 2s. Read exactly which concentration is requested.

Comparing solubilities

K_sp values can be compared directly to rank solubilities only for salts of the same formula type (both MA, or both MA₂), because only then is the relationship between s and K_sp the same. To compare salts of different types, calculate s for each.

The common ion effect25.1.10

What happens to a saturated solution of lead(II) iodide if a few drops of concentrated potassium iodide are added? KI is very soluble and fully ionised, so it raises [I⁻]. The ionic product [Pb²⁺][I⁻]² now exceeds K_sp, so by Le Chatelier's principle the equilibrium PbI₂(s) ⇌ Pb²⁺ + 2I⁻ shifts to the left: Pb²⁺ and I⁻ ions combine and a yellow precipitate of PbI₂ forms until the product of concentrations has fallen back to K_sp.

The common ion effect

A sparingly soluble salt is less soluble in a solution that already contains one of its ions (a common ion) than it is in pure water. The added ion shifts the solubility equilibrium towards the solid; K_sp itself is unchanged.

Quantitatively, the common ion's concentration is fixed by the added salt, so only the other ion's concentration is unknown. Because the salt is sparingly soluble, the amount it adds to the common ion is negligible, which simplifies the algebra.

Worked example 25.12 · Solubility in a solution containing a common ion

Question. Calculate the solubility of BaSO₄ in 0.0100 mol dm⁻³ sodium sulfate. K_sp = 1.08 × 10⁻¹⁰ mol² dm⁻⁶.

Given	[SO₄²⁻] = 0.0100 + s ≈ 0.0100 mol dm⁻³ (s is tiny)
Relationship	K_sp = [Ba²⁺][SO₄²⁻] = s × 0.0100
Answer	s = 1.08 × 10⁻¹⁰ ÷ 0.0100 = 1.08 × 10⁻⁸ mol dm⁻³
Check	About a thousand times less than in pure water (1.04 × 10⁻⁵) ✔. The approximation is safe: 1.08 × 10⁻⁸ is negligible beside 0.0100.

Take care with the stoichiometry of the common ion. For Fe(OH)₃ in 0.010 mol dm⁻³ Ba(OH)₂, [OH⁻] = 0.020 mol dm⁻³, and it enters K_sp = [Fe³⁺][OH⁻]³ cubed: s = 2.0 × 10⁻³⁹ ÷ (0.020)³ = 2.5 × 10⁻³⁴ mol dm⁻³. Examiners described this calculation as very difficult for candidates.

Figure 25.8 Calculated solubility of PbI₂ (K_sp = 1.0 × 10⁻⁹ mol³ dm⁻⁹) as the concentration of iodide from added KI increases (both axes logarithmic). Once [I⁻] from KI is much larger than the iodide from PbI₂ itself, the approximation s = K_sp/[I⁻]² holds and each tenfold increase in [I⁻] lowers the solubility a hundredfold.

Uses of the common ion effect

Adding an excess of a common ion drives precipitation further towards completion — useful in gravimetric analysis, where as much of an ion as possible must be precipitated and weighed, and in removing unwanted metal ions from solution. Different K_sp values also allow ions to be precipitated one at a time: the ion forming the salt with the lowest K_sp (for the same formula type) precipitates first as the reagent is added.

A familiar situation made unfamiliar

When asked what is seen when saturated KI solution is added to a filtrate still containing Pb²⁺ and I⁻, many candidates described redox reactions producing iodine. The expected answer was simple: a yellow precipitate (of PbI₂) because the common ion I⁻ raises [Pb²⁺][I⁻]² above K_sp. In another session, candidates who wrote only "common ion effect" lost the explanation mark because they omitted the key word concentration: the increased concentration of the common ion shifts the equilibrium to the left.

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Will a precipitate form?25.1.9, 25.1.10

K_sp is the maximum value that the product of the ion concentrations can have at equilibrium. If two solutions are mixed, calculate the product of the ion concentrations in the mixture — using the same expression as K_sp but with the concentrations actually present, sometimes called the ionic product — and compare it with K_sp:

product greater than K_sp: the solution would be supersaturated, so solid precipitates until the product falls to K_sp;
product equal to K_sp: the solution is just saturated;
product less than K_sp: no precipitate; more solid could dissolve.

Worked example 25.13 · Predicting precipitation

Question. Equal volumes of 2.0 × 10⁻⁴ mol dm⁻³ BaCl₂(aq) and 2.0 × 10⁻⁴ mol dm⁻³ Na₂SO₄(aq) are mixed. Does barium sulfate precipitate? K_sp = 1.08 × 10⁻¹⁰ mol² dm⁻⁶.

Concentrations after mixing	each is halved by the doubling of volume: [Ba²⁺] = [SO₄²⁻] = 1.0 × 10⁻⁴ mol dm⁻³
Product	1.0 × 10⁻⁴ × 1.0 × 10⁻⁴ = 1.0 × 10⁻⁸ mol² dm⁻⁶
Answer	1.0 × 10⁻⁸ > 1.08 × 10⁻¹⁰, so yes, a precipitate forms
Check	Forgetting the dilution on mixing is the usual error; here it does not change the conclusion, but near the boundary it can.

Past-paper practice · Set 25C · Solubility product and the common ion effect

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 25C.1[4]

Answer and marking guidance

(i) K_sp = [Al³⁺][OH⁻]³ ✔. (ii) [OH⁻] = 3 × 2.47 × 10⁻⁹ = 7.41 × 10⁻⁹ ✔; K_sp = 2.47 × 10⁻⁹ × (7.41 × 10⁻⁹)³ = 1.01 × 10⁻³³ ✔; units mol⁴ dm⁻¹² ✔. Examiner insight: most gave a correct expression, and (ii) was answered well by many. The commonest error was 3.72 × 10⁻³⁵ — from not multiplying the solubility by 3 before cubing.

Question 25C.2[2]

Answer and marking guidance

Mn(OH)₂ ⇌ Mn²⁺ + 2OH⁻, so K_sp = s(2s)² = 4s³ = 1.1 × 10⁻¹¹ ✔; s = 1.4 × 10⁻⁴ mol dm⁻³ ✔. Examiner insight: this was found more difficult. Some treated 1.1 × 10⁻¹¹ as the solubility, and the fact that [OH⁻] is twice [Mn²⁺] was not well appreciated.

Question 25C.3[3]

Answer and marking guidance

[Hg²⁺] = s = 1.00 × 10⁻⁷ ÷ 454.4 = 2.20 × 10⁻¹⁰ mol dm⁻³ ✔; K_sp = [Hg²⁺][I⁻]² = 4s³ = 4.26 × 10⁻²⁹ ✔; units mol³ dm⁻⁹ ✔. Examiner insight: this discriminated well. Most gave correct units; some did not convert to mol dm⁻³, and could gain partial credit (error carried forward) for 4.00 × 10⁻²¹.

Question 25C.4[4]

Answer and marking guidance

(i) K_sp = x(3x)³ = 27x⁴ = 2.0 × 10⁻³⁹; x = 9.28 × 10⁻¹¹ mol dm⁻³ ✔. (ii) [OH⁻] = 2 × 0.010 = 0.020; [Fe³⁺] × (0.020)³ = 2.0 × 10⁻³⁹ ✔; solubility = 2.5 × 10⁻³⁴ mol dm⁻³ ✔. (iii) the common ion effect ✔. Examiner insight: (i) was found difficult; a significant number lost the mark after writing 3³ = 9. (ii) was found very difficult. Most scored the mark in (iii).

Question 25C.5[4]

Answer and marking guidance

(i) K_sp = [Pb²⁺][I⁻]² = 5.68 × 10⁻³ × (4.20 × 10⁻⁴)² = 1.0 × 10⁻⁹ ✔; mol³ dm⁻⁹ ✔. (ii) A yellow precipitate (of PbI₂) forms ✔; because of the common ion effect — the added I⁻ makes [Pb²⁺][I⁻]² exceed K_sp ✔. Examiner insight: (i) was generally answered well; 4.0 × 10⁹ was a commonly seen wrong answer. In (ii) many did not recognise this familiar situation — the addition of a high concentration of a common ion to a saturated solution — and described redox reactions producing iodine instead.

Quick check 25.3

Write the K_sp expression for silver chromate, Ag₂CrO₄, and give its units.
answer
K_sp = [Ag⁺]²[CrO₄²⁻]; mol³ dm⁻⁹.
The solubility of a salt MX₂ is 1.0 × 10⁻³ mol dm⁻³. Calculate K_sp.
answer
4s³ = 4 × (1.0 × 10⁻³)³ = 4.0 × 10⁻⁹ mol³ dm⁻⁹.
Explain why CaF₂ is less soluble in NaF(aq) than in water.
answer
F⁻ is a common ion. Its increased concentration shifts CaF₂(s) ⇌ Ca²⁺ + 2F⁻ to the left, so less CaF₂ dissolves; K_sp is unchanged.
Does adding more solid PbI₂ to a saturated solution change [Pb²⁺]?
answer
No — the solid does not appear in K_sp; the solution is already saturated, so the extra solid simply remains undissolved.

Examiner's overall observation · Solubility product and the common ion effect

Answered well: writing K_sp expressions for a given salt; calculating K_sp from a solubility and including correct units — this discriminated well, with most candidates giving correct units; recognising and naming the common ion effect.

Found difficult: calculating a solubility from K_sp when the ions are in a 1 : 2 or 1 : 3 ratio; using a common ion whose concentration enters K_sp squared or cubed; describing what is actually seen when a common ion is added to a saturated solution; converting a solubility in g dm⁻³ to mol dm⁻³ before using it.

Recurring errors: K_sp taken to be the solubility; the stoichiometric factor put inside the bracket ([2OH⁻]²) or omitted; the solid included in the expression; the solubility s given when the concentration of one ion was asked for; wrong powers in the units; "common ion effect" stated without saying that the concentration of the common ion has increased.

What successful answers did: wrote the dissolving equation, then the concentration of each ion in terms of s, before substituting; stated the equilibrium shift and its cause; checked the units from the total power.

Partition between two immiscible solvents25.2.1

Shake an aqueous solution of iodine with cyclohexane in a separating funnel and let the layers settle. The pale brown water layer loses most of its colour, and the cyclohexane layer turns violet: iodine has moved into the organic solvent. Yet not quite all of it moves. However long you wait, a small, fixed fraction remains in the water.

The iodine molecules are crossing the boundary between the two liquids in both directions. When the rate at which they leave the water equals the rate at which they return from the cyclohexane, a dynamic equilibrium is reached, and the ratio of the two concentrations is constant at a given temperature:

I₂(aq) ⇌ I₂(cyclohexane) K_pc = [I₂(cyclohexane)] / [I₂(aq)]

Figure 25.9 A separating funnel. The two immiscible liquids form layers, the less dense on top. After shaking (with the pressure released) and settling, the solute X is distributed between the layers in the ratio fixed by K_pc, and the lower layer can be run off through the tap.

Definition

The partition coefficient, K_pc, is the ratio of the concentrations of a solute in two immiscible solvents when the system is at equilibrium, at a stated temperature.

K_pc = [X(solvent 1)] / [X(solvent 2)]

Four points follow from the definition.

It is an equilibrium constant. K_pc depends only on temperature (and on the solute and the two solvents). It is not affected by how much solute is used or by the volumes of the solvents — those change the amounts in each layer, not the ratio of concentrations.
It needs equilibrium. If the layers are analysed before equilibrium is reached, the ratio calculated will not equal K_pc.
The order must be stated. K_pc for iodine between cyclohexane and water is the reciprocal of that between water and cyclohexane. Always check which solvent is on top of the fraction in the question.
It has no units, because it is a ratio of two concentrations measured in the same units.

The syllabus restricts calculations to a solute that is in the same physical state in both solvents — that is, a solute that does not ionise, associate or react in either solvent — so that the same particle is being counted on each side.

Imprecise definitions

Examiners repeatedly report that "many candidates understood the idea of a partition coefficient but definitions often lacked precision". Answers lost credit by not referring to concentrations, by referring to "two solutions" rather than one solute in two immiscible solvents, or — the commonest omission among weaker candidates — by leaving out equilibrium. The statement "K_pc is an equilibrium constant" was increasingly seen and is correct, but on its own it does not say what is being compared.

Calculating with a partition coefficient25.2.2

All partition calculations use one relationship. Because concentration = amount ÷ volume, and the solute is the same substance in both layers, masses can be used in place of moles:

K_pc = (m₁ / V₁) ÷ (m₂ / V₂)

where m₁ and m₂ are the masses (or amounts) of solute in solvents 1 and 2, and V₁, V₂ are their volumes (in the same units). The other fact you almost always need is conservation of the solute: the masses in the two layers add up to the mass you started with.

Worked example 25.14 · Calculating K_pc from experimental data

Question. 0.500 g of a solute X is dissolved in 50.0 cm³ of water and shaken with 25.0 cm³ of hexane until equilibrium is reached. The hexane layer then contains 0.380 g of X. Calculate K_pc for X between hexane and water.

Mass in water	0.500 − 0.380 = 0.120 g
Relationship	K_pc = [X(hexane)] / [X(water)]
Substitution	K_pc = (0.380/25.0) ÷ (0.120/50.0) = 0.01520 ÷ 0.00240
Answer	K_pc = 6.33 (no units)
Check	More than three-quarters of X went into only half the volume of hexane, so the concentration in hexane must be several times that in water ✔

Worked example 25.15 · How much is extracted?

Question. K_pc for a solute between an organic solvent and water is 4.00. 1.00 g of the solute is dissolved in 100 cm³ of water. Calculate the mass extracted by (a) one portion of 50.0 cm³ of organic solvent; (b) two successive portions of 25.0 cm³.

(a) Set up	let x g move into the organic layer; (1.00 − x) g stays in the water
Substitution	4.00 = (x/50.0) ÷ ((1.00 − x)/100) → 4.00 = 2x/(1.00 − x)
Solve	4.00 − 4.00x = 2x → x = 0.667 g extracted
(b) First 25.0 cm³	4.00 = (x/25.0) ÷ ((1.00 − x)/100) → 4.00 = 4x/(1.00 − x) → x = 0.500 g; 0.500 g left in water
Second 25.0 cm³	same ratio: half of the remaining 0.500 g = 0.250 g
Answer	total extracted = 0.500 + 0.250 = 0.750 g
Check	The same total volume of solvent extracts more when used in two portions (0.750 g against 0.667 g).

Figure 25.10 Calculated mass extracted from 1.00 g of solute in 100 cm³ of water by 50 cm³ of organic solvent (K_pc = 4.00) used in one, two or five portions. Several small extractions remove more solute than one large one, although each extra portion gains less.

The result in Figure 25.10 is general: because every extraction leaves the same fraction of what remains, repeated extraction with small portions is more efficient than a single extraction with the same total volume. This is why a chemist extracting a product from an aqueous mixture shakes it with two or three small portions of solvent and then combines the organic layers.

Setting up the equation: the errors examiners list

Most lost marks in partition calculations come from the set-up, not the arithmetic. Reported errors include: using the original mass in one layer instead of the mass remaining there (e.g. 1.77 = (x/50)/(0.5/75) instead of (x/50)/((0.5 − x)/75)); subtracting inside the wrong bracket; pairing a mass with the wrong volume (so that a volume of water came out as 121.8 or 66.5 cm³ instead of 82 cm³); putting the ratio upside down relative to the K_pc given; and, in a second extraction, ignoring the new volume of solvent and assuming the concentration is unchanged. Label each concentration with its solvent before substituting.

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What controls the size of K_pc25.2.3

A partition coefficient compares how well each solvent holds on to the solute. The general rule is "like dissolves like": a solute is more soluble in a solvent whose intermolecular forces resemble its own, because the solute–solvent interactions formed can compensate for the solute–solute and solvent–solvent interactions broken.

Non-polar solutes (I₂, hydrocarbons, most halogenoalkanes) interact with other molecules mainly through instantaneous dipole–induced dipole forces. They are far more soluble in non-polar solvents such as hexane or cyclohexane than in water, whose molecules are held together by strong hydrogen bonds. K_pc(non-polar solvent/water) is large — 93.8 for iodine between cyclohexane and water.
Polar solutes that can hydrogen bond (small alcohols, amides such as ethanamide, carboxylic acids with short chains) form hydrogen bonds with water. They favour the aqueous layer, so K_pc(organic/water) is small or less than 1.
Changing the solvent changes K_pc. Replacing cyclohexane (non-polar) with hexan-2-one (polar, because of its C=O group) makes the organic layer more like water; iodine is less soluble in it than in cyclohexane, so K_pc between the new solvent and water is lower.
The balance within a molecule matters. Lengthening the hydrocarbon chain of a solute increases its non-polar character and shifts the partition towards the organic solvent; adding –OH, –NH₂ or –COOH groups shifts it towards water.

Chemistry connection · octan-1-ol and water

The partition coefficient between octan-1-ol and water is widely used as a measure of how a molecule is shared between fatty (non-polar) and watery (polar) environments, which affects how a drug is absorbed and distributed in the body. Octan-1-ol has a polar –OH group but a long non-polar chain, so it behaves as a model for the membranes of cells. Several examination questions use this pair of solvents with drug-like solutes.

Reasoning about polarity

Asked how K_pc for iodine would change on replacing cyclohexane by hexan-2-one, most candidates found the question difficult, and some thought iodine is polar and so more soluble in hexan-2-one. Full credit needed a chain of three ideas: K_pc would be lower; because hexan-2-one is more polar than cyclohexane (I₂ and cyclohexane are non-polar); so I₂ is less soluble in hexan-2-one. In another question, a calculated K_pc lower than the true value was explained by the system not having reached equilibrium — concentration in water still too high — a statement examiners saw much more rarely than vague references to experimental error.

Past-paper practice · Set 25D · Partition coefficients

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 25D.1[3]

Answer and marking guidance

(i) The ratio of the concentrations of a solute in two (immiscible) solvents at equilibrium ✔. (ii) 3.50 = (1.62/100) ÷ (0.38/V) ✔; V = 82 cm³ ✔ (82.1). Examiner insight: (i) was answered well by most; weaker candidates omitted a reference to equilibrium. Many correct answers were seen in (ii); errors included 121.8 cm³, from 3.50 = (1.62/V) ÷ (0.38/100), and 66.5 cm³, from 3.50 = (2.00/100) ÷ (0.38/V).

Question 25D.2[3]

Answer and marking guidance

(i) K_pc = (0.935/50) ÷ (0.065/50) = 14.4 ✔ (14.38). (ii) 14.4 = ((0.935 − x)/50) ÷ (x/100) ✔, where x is the mass in octan-1-ol; x = 0.114 g ✔. Examiner insight: (i) was usually answered well. Many found (ii) challenging; common errors were 0.82 g, from 14.4 = (x/50) ÷ ((0.935 − x)/100), and 0.13 g, from 14.4 = (0.935/50) ÷ (x/100).

Question 25D.3[4]

Only the parts on partition are reproduced; part (a)(i), on the role of H₂O₂, is omitted.

Answer and marking guidance

(a)(ii) K_pc = [I₂(cyclohexane)] ÷ [I₂(aq)]: 93.8 = (0.390/15) ÷ (x/20) ✔; mass of I₂ in the aqueous layer, x = 5.54 × 10⁻³ g ✔. (iii) K_pc would be lower; hexan-2-one is more polar than cyclohexane (hexan-2-one is polar, cyclohexane non-polar); so I₂ is less soluble in hexan-2-one — all three for two marks ✔✔. Examiner insight: only a minority calculated the correct mass in (ii); errors included 0.396 g and 0.384 g from subtracting 0.390 inside the wrong bracket. Most found (iii) difficult, and some thought iodine is polar and more soluble in hexan-2-one.

Question 25D.4[6]

Answer and marking guidance

(a) The system had not reached equilibrium ✔; so the concentration in water was still too high (or that in octan-1-ol too low) ✔. (b)(i) 7.62 ÷ 6.760 = 1.13 g dm⁻³ ✔. (ii) 100 cm³ of the octan-1-ol layer contains 0.762 g; 6.760 = ((0.762 − x)/100) ÷ (x/100) ✔; 7.76x = 0.762, x = 0.0982 g extracted into the water ✔; 0.0982 ÷ 74 = 1.3 × 10⁻³ mol ✔. Examiner insight: in (a) many suggested that the concentration in water was too high, but the statement that the system was not at equilibrium was much rarer; no credit was given for unspecified experimental errors. (b)(i) was generally answered well. (b)(ii) was found very difficult and three marks were rare: many ignored the "100 cm³ of water" and assumed the aqueous concentration was the same as in (b)(i).

Question 25D.5[4]

Answer and marking guidance

(i) The ratio of the concentrations of a solute in two solvents ✔; at equilibrium ✔. (ii) K_pc = [procaine]_oct ÷ [procaine]_water: 1.77 = (x/50) ÷ ((0.500 − x)/75) ✔; 1.77 = 1.5x/(0.500 − x), so x = 0.271 g ✔. Examiner insight: in (i) some did not refer to concentrations, or referred to "two solutions" rather than one solute in two immiscible solvents; more candidates stated that K_pc is an equilibrium constant. In (ii) most good answers started with an expression such as 1.77 = (x/50)/(0.5 − x/75); a common incorrect expression was 1.77 = (x/50)/(0.5/75).

Quick check 25.4

State two things that must be specified alongside a value of K_pc.
answer
The temperature, and which solvent's concentration is on top of the ratio (the order of the solvents).
K_pc(ether/water) = 5.0 for a solute. After equilibrium with equal volumes of the two solvents, 0.60 g is in the water. What mass is in the ether?
answer
Equal volumes, so masses are in the ratio 5 : 1: 3.0 g.
Predict whether K_pc(hexane/water) is larger for ethanol or for hexan-1-ol. Explain.
answer
Hexan-1-ol: its long non-polar chain favours hexane, whereas ethanol's small chain lets hydrogen bonding with water dominate.
Why must a solute be in the same physical state in both solvents for the simple K_pc expression to apply?
answer
Because the ratio compares the same particle in each solvent; if the solute ionised, associated or reacted in one solvent, the species counted would differ.

Examiner's overall observation · Partition coefficients

Answered well: calculating K_pc directly from masses and volumes; many correct answers to straightforward single-step partition calculations, and good recall that K_pc is an equilibrium constant.

Found difficult: calculations where an unknown mass must be shared between the layers — only a minority reached the correct mass of iodine left in water; second extractions with a different volume of solvent, where the award of all three marks was rare; explaining the effect of solvent polarity on K_pc; explaining why a measured ratio differs from K_pc.

Recurring errors: definitions without "concentration", "immiscible solvents" or "equilibrium"; the original mass used in place of the mass remaining; the ratio set up upside down; masses paired with the wrong volumes; iodine described as polar; experimental errors invented rather than the lack of equilibrium identified.

What successful answers did: began with an expression such as K_pc = (x/V_org) ÷ ((m − x)/V_aq), written with the solvents labelled, then solved it; linked every polarity argument to solubility and then to the direction of the change in K_pc.

Misconceptions and how the topic is assessed25.1–25.2

The misconceptions below are the ones that recur in examiner reports on this topic. Each is set out as the incorrect idea, why it fails, the correct model and what it costs in an examination.

**Table 25.5** Recurring misconceptions.
misconception	why it is wrong	correct model	examination consequence
"A buffer keeps the pH constant."	Added H⁺ or OH⁻ changes the ratio [HA]/[A⁻] slightly, so [H⁺] changes slightly.	A buffer resists changes in pH when small amounts of acid or alkali are added.	Definition mark lost; the same idea leads to wrong answers when a buffer is overwhelmed.
K_a = [H⁺]²/[HA] is the definition of K_a.	It assumes [H⁺] = [A⁻], true only for a weak acid on its own.	K_a = [H⁺][A⁻]/[HA] always; in a buffer [A⁻] comes mainly from the salt.	Expression mark lost; buffer pH calculations collapse.
Conjugate pair = "an acid and a base".	Any acid and base react; that does not make them conjugate.	Two species that differ by one H⁺, linked across the ⇌ sign.	Definition not credited; pairs identified across the + sign.
Heating water leaves its pH at 7, or makes it acidic.	K_w increases, raising [H⁺] and [OH⁻] equally.	pH falls; water stays neutral because [H⁺] = [OH⁻].	Tick-box and explanation marks lost.
K_sp is the solubility.	K_sp is a product of ion concentrations with powers; solubility is the amount of salt dissolved.	Relate them through stoichiometry: K_sp = s², 4s³ or 27s⁴.	Whole calculation lost.
A coefficient goes inside the bracket: [2OH⁻]².	Square brackets mean the actual concentration of the species.	The coefficient becomes the power: [OH⁻]²; the concentration is found separately (2s).	Expression mark and later numerical marks lost.
"Common ion effect" is a full explanation.	It names the effect but does not explain it.	The increased concentration of the common ion shifts the solubility equilibrium to the left, so less solid dissolves or a precipitate forms.	Explanation mark lost.
K_pc changes with the volumes of the solvents.	Volumes change the amounts in each layer, not the ratio of concentrations at equilibrium.	K_pc depends only on the solute, the two solvents and temperature.	Extraction calculations set up wrongly.
Iodine is polar, so it dissolves in polar organic solvents.	I₂ is a symmetrical non-polar molecule.	Non-polar solutes favour non-polar solvents; a more polar organic solvent lowers K_pc(organic/water) for I₂.	Prediction and explanation reversed.

How the topic is assessed

**Table 25.6** Question families seen in the structured papers reviewed for this chapter.
question family	typical demand	chemistry needed
Definitions	conjugate acid–base pair; buffer; K_sp; partition coefficient; expressions for pH, K_w, K_a	precise wording: one H⁺; resists, small amounts; concentrations at equilibrium; immiscible solvents
pH calculations	strong alkali; weak acid from K_a; K_a and pK_a from pH; comparing a strong and a weak acid of equal pH	K_w; √(K_ac); logarithms both ways
Buffers	define; name components; two equations; pH of a mixture or after partial neutralisation; mass of salt for a target pH; explain a buffer that has failed	[H⁺] = K_a[HA]/[A⁻]; moles before and after reaction; HCO₃⁻/H₂CO₃
Solubility product	write K_sp with units; K_sp from solubility (mol or g dm⁻³) or from ion concentrations; solubility from K_sp; solubility with a common ion; observation on adding a common ion	stoichiometric powers; Le Chatelier
Partition	define; calculate K_pc; mass extracted or remaining; volume of solvent; effect of solvent polarity; why a measured value differs	ratio of concentrations; conservation of solute; intermolecular forces

Questions on this topic are frequently combined with others in one structured question — a K_sp part following a lattice-energy part, a partition part inside an organic synthesis question, a buffer part inside an amino-acid question. Recognise the equilibrium being tested, whatever the context.

Self-test

Twelve questions across the whole unit, each with the reasoning behind the answer.

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Definitions to learn

**Table 25.7** The definitions and relationships this unit examines.
term	definition or relationship
conjugate acid–base pair	two species that differ by one proton (H⁺)
pH	pH = −log₁₀[H⁺(aq)]
ionic product of water, K_w	K_w = [H⁺(aq)][OH⁻(aq)] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K
acid dissociation constant, K_a	for HA ⇌ H⁺ + A⁻: K_a = [H⁺][A⁻]/[HA]; units mol dm⁻³
pK_a	pK_a = −log₁₀K_a; smaller pK_a, stronger acid
buffer solution	a solution that resists changes in pH when small amounts of acid or alkali are added
solubility product, K_sp	the product of the concentrations of the ions in a saturated solution of a sparingly soluble salt, each raised to the power of its coefficient in the dissolving equation
common ion effect	the reduction in solubility of a sparingly soluble salt in a solution that already contains one of its ions
partition coefficient, K_pc	the ratio of the concentrations of a solute in two immiscible solvents at equilibrium

Data used in this chapter

K_a, pK_a, K_sp and K_pc values in the worked examples and figures are those printed in examination questions where one was available (for example K_a(HBrO) = 2.00 × 10⁻⁹, K_a(propanoic acid) = 1.35 × 10⁻⁵ mol dm⁻³, pK_a(H₂CO₃) = 6.35, K_sp(BaSO₄) = 1.08 × 10⁻¹⁰ mol² dm⁻⁶, K_pc(I₂, cyclohexane/water) = 93.8); K_a(ethanoic acid) is taken as 1.74 × 10⁻⁵ mol dm⁻³. K_sp(PbI₂) in Figure 25.8 is the value, 1.0 × 10⁻⁹ mol³ dm⁻⁹, obtained from the ion concentrations in one of the practice questions. The K_pc of 4.00 and the solute data in Worked examples 25.14 and 25.15 are illustrative values chosen for the calculation, not measurements. These are not the official data booklet; always use the value a question gives. K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K, the value printed with the examination papers.

The curves in Figures 25.3, 25.5, 25.8 and 25.10 are calculated from these constants, not measured. The weak-acid and buffer curves solve the equilibria in full; the dashed lines show the approximations used in examination answers.

Summary

25.1 Acids, bases and pH

A conjugate acid–base pair differs by one H⁺; pairs are linked across the ⇌ sign.
pH = −log₁₀[H⁺]; [H⁺] = 10^−pH; one pH unit = tenfold change in [H⁺].
K_w = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K; K_w rises with temperature; neutral means [H⁺] = [OH⁻].
Strong acid: [H⁺] = c. Strong alkali: [H⁺] = K_w/[OH⁻]. Weak acid: [H⁺] = √(K_ac).
K_a = [H⁺][A⁻]/[HA]; pK_a = −log₁₀K_a; larger K_a or smaller pK_a means a stronger acid.

25.1 Buffers

A buffer resists changes in pH when small amounts of acid or alkali are added; it contains large amounts of a weak acid and its conjugate base.
Made from a weak acid + its salt, or from excess weak acid + a strong alkali.
A⁻ + H⁺ → HA removes acid; HA + OH⁻ → A⁻ + H₂O removes alkali; the ratio [HA]/[A⁻] changes little.
[H⁺] = K_a[HA]/[A⁻], or pH = pK_a + log([A⁻]/[HA]); moles may replace concentrations; pH = pK_a when [HA] = [A⁻].
Blood: H₂CO₃/HCO₃⁻; HCO₃⁻ + H⁺ → H₂CO₃; H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O.

25.1 Solubility product

K_sp = [M^y+]^x[A^x−]^y for a saturated solution; solid omitted; depends only on temperature.
MA: K_sp = s²; MA₂: 4s³; MA₃: 27s⁴. Convert g dm⁻³ to mol dm⁻³ first.
A common ion lowers solubility; with a common ion of concentration c, s = K_sp/cⁿ.
If the product of ion concentrations exceeds K_sp, a precipitate forms.

25.2 Partition coefficients

K_pc = ratio of concentrations of a solute in two immiscible solvents at equilibrium; no units; depends on temperature.
K_pc = (m₁/V₁) ÷ (m₂/V₂); total solute is conserved.
Several small extractions remove more than one large one of the same total volume.
Like dissolves like: non-polar solutes favour non-polar solvents; hydrogen-bonding solutes favour water.

Examination checklist

Can I define a conjugate acid–base pair and pick out both pairs in an equation?
Can I write expressions for pH, K_w, K_a and pK_a and move between pH and [H⁺], K_a and pK_a?
Do I decide first whether an acid is strong or weak, and convert moles to mol dm⁻³?
Can I calculate the pH of a strong alkali, including M(OH)₂, using K_w?
Can I explain why heated water has pH below 7 but is still neutral?
Does my buffer definition say "resists", "pH" and "small amounts of acid or alkali"?
Can I write both buffer equations with the buffer components, including for HCO₃⁻/H₂CO₃ in blood?
For a partial-neutralisation buffer, do I write the moles of HA and A⁻ after reaction?
Do I check whether a buffer pH should be above or below pK_a?
Can I explain what happens when a buffer is overwhelmed, with both ideas?
Do my K_sp expressions use powers (not multipliers inside brackets), omit the solid and carry correct units?
Can I use K_sp = s², 4s³, 27s⁴, and give the ion concentration that is actually asked for?
Can I calculate a solubility in the presence of a common ion, and describe what is seen?
Does my K_pc definition mention concentrations, immiscible solvents and equilibrium?
Do I label each concentration with its solvent and use the mass remaining in each layer?
Can I predict and explain how solvent or solute polarity changes K_pc?

Knowledge organiser

idea	key facts and relationships	must-remember distinctions and common errors
Conjugate pairs	differ by one H⁺; acid ↔ its conjugate base across ⇌	not "an acid and a base"; conjugate base has the lower charge
pH	pH = −log[H⁺]; [H⁺] = 10^−pH	1 unit = ×10; pH to 2 d.p.
K_w	[H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ (298 K)	rises with T; neutral = [H⁺] = [OH⁻], not pH 7
Strong acid / alkali	[H⁺] = c; [H⁺] = K_w/[OH⁻]	[OH⁻] = 2c for Ba(OH)₂; excess ÷ total volume
K_a, pK_a	K_a = [H⁺][A⁻]/[HA]; pK_a = −log K_a	not [H⁺]²/[HA] as a definition; smaller pK_a = stronger
Weak acid pH	[H⁺] = √(K_ac)	assumes [H⁺] = [A⁻], [HA] ≈ c
Buffer	resists pH change for small additions; HA + A⁻ in large amounts	not "keeps pH constant"; HCl/NaCl is not a buffer
Buffer pH	[H⁺] = K_a[HA]/[A⁻]; pH = pK_a + log([A⁻]/[HA])	moles after neutralisation; ratio not inverted; half-neutralised → pH = pK_a
Blood	HCO₃⁻ + H⁺ → H₂CO₃; H₂CO₃ + OH⁻ → HCO₃⁻ + H₂O	equations with buffer components, not water
K_sp	[M]^x[A]^y in saturated solution; s², 4s³, 27s⁴	no solid; powers not multipliers; units from total power
Common ion	solubility falls; s = K_sp/[common ion]ⁿ	explain via concentration and equilibrium shift; precipitate seen
Precipitation	product of ion concentrations > K_sp → precipitate	dilute on mixing first
K_pc	[X(1)]/[X(2)] at equilibrium; (m₁/V₁)/(m₂/V₂)	state solvent order; mass remaining, not original; no units
Factors on K_pc	like dissolves like; T	I₂ non-polar; polar solvent lowers K_pc(org/water) for I₂

Equilibria

What this chapter covers25.1–25.2

Conjugate acids and conjugate bases25.1.1, 25.1.2

pH: a logarithmic scale for [H+]25.1.3

The ionic product of water, Kw25.1.3

Pure water and neutrality

The pH of a strong acid25.1.4(a)

The pH of a strong alkali25.1.4(b)

Mixing a strong acid with a strong alkali

Weak acids: Ka and pKa25.1.3

The pH of a weak acid25.1.4(c)

Choosing the right method

What a buffer solution is, and how one is made25.1.5(a), 25.1.5(b)

What a buffer contains

How a buffer controls pH25.1.5(c)

Buffers that are not a named acid and salt

Calculating the pH of a buffer25.1.6

Designing a buffer of a chosen pH

How much does the pH change — and when does a buffer fail?25.1.5(c), 25.1.6

Buffer capacity: when the buffer is overwhelmed

Uses of buffers, and the control of pH in blood25.1.5(d)

The HCO3− buffer in blood

Solubility equilibria and the solubility product, Ksp25.1.7, 25.1.8

Calculating Ksp from solubility, and solubility from Ksp25.1.9

Comparing solubilities

The common ion effect25.1.10

Uses of the common ion effect

Will a precipitate form?25.1.9, 25.1.10

Partition between two immiscible solvents25.2.1

Calculating with a partition coefficient25.2.2

What controls the size of Kpc25.2.3

Misconceptions and how the topic is assessed25.1–25.2

How the topic is assessed

Self-test

Definitions to learn

Data used in this chapter

Summary

25.1 Acids, bases and pH

25.1 Buffers

25.1 Solubility product

25.2 Partition coefficients

Examination checklist

Knowledge organiser

pH: a logarithmic scale for [H⁺]25.1.3

The ionic product of water, K_w25.1.3

Weak acids: K_a and pK_a25.1.3

The HCO₃⁻ buffer in blood

Solubility equilibria and the solubility product, K_sp25.1.7, 25.1.8

Calculating K_sp from solubility, and solubility from K_sp25.1.9

What controls the size of K_pc25.2.3