What is a partition coefficient in the context of A-Level Physical Chemistry?

In A-Level Physical Chemistry, a partition coefficient is a ratio that describes how a solute distributes itself between two immiscible solvents at equilibrium. It is a measure of the solubility of the solute in each solvent and is crucial for understanding how substances separate in different phases.

How is the partition coefficient calculated in Chemistry experiments?

The partition coefficient is calculated by dividing the concentration of the solute in one solvent by its concentration in the other solvent, at equilibrium. Mathematically, it is expressed as K = [solute]_solvent1 / [solute]_solvent2, where K is the partition coefficient.

Why is the partition coefficient important in the study of equilibria?

The partition coefficient is important in equilibria studies because it helps predict how a solute will distribute itself between two phases. This is essential for processes like extraction, purification, and drug delivery, where understanding solute behavior in different environments is crucial.

What factors can affect the partition coefficient of a substance?

Several factors can affect the partition coefficient, including temperature, the nature of the solvents, and the chemical properties of the solute, such as polarity and molecular size. Changes in these factors can alter the solubility of the solute in each solvent, thus affecting the partition coefficient.

How does the concept of partition coefficient apply to real-world scenarios?

In real-world scenarios, partition coefficients are used in pharmaceuticals to understand drug distribution in the body, in environmental science to predict pollutant distribution between water and soil, and in chemical engineering for designing separation processes. Understanding partition coefficients allows for better control and optimization of these processes.

Why is "Forgetting to divide by the (different) solvent volumes." a common mistake in Partition coefficients?

Why it happens: Treating mass as concentration. How to avoid it: Kpc uses CONCENTRATIONS (mass ÷ volume); volumes only cancel when they are equal. Source: 9701 Examiner Reports 2022-2024

Why is "Inverting the concentration ratio." a common mistake in Partition coefficients?

Why it happens: Not tracking which solvent is on top. How to avoid it: Keep the solvents in the order given (e.g. ether/water).

Why is "Thinking one large extraction beats several small ones." a common mistake in Partition coefficients?

Why it happens: Assuming total volume is all that matters. How to avoid it: Multiple smaller portions extract MORE solute overall.

Why is "Applying Kpc when the solute changes form in a solvent." a common mistake in Partition coefficients?

Why it happens: Forgetting the 'same molecular state' assumption. How to avoid it: Kpc only holds if the solute does not associate or dissociate in either solvent.

Cambridge International A Levels Chemistry (9701)

Partition coefficients

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Short Notes - Partition coefficients

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Detailed Study Notes

Detailed notes on Equilibria for Cambridge International A Levels Chemistry, covering key concepts, explanations, examples, and exam-focused revision points.

Partition Coefficients — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

How a solute distributes between two immiscible solvents: the partition coefficient, extraction calculations, why several small extractions beat one large one, and the assumptions involved.

At a glance

Partition coefficient: $K_{pc} = \dfrac{[\text{X in solvent 1}]}{[\text{X in solvent 2}]}$ — a constant ratio at fixed temperature.
Applies to one solute shared between two immiscible solvents.
Use it to find how much solute moves into an extracting solvent (solvent extraction).
Several small extractions remove more solute than one large extraction.
Assumes the solute is in the same molecular form in both solvents (no association/dissociation).
Concentration = mass ÷ volume — keep the volumes in unless they cancel.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

25.2 — State what is meant by a partition coefficient.
25.2 — Calculate and use a partition coefficient for a solute between two immiscible solvents.
25.2 — Understand the assumptions on which the partition coefficient depends.

What the partition coefficient is

The fixed ratio of a solute's concentrations in two immiscible solvents at equilibrium.

When a solute is shaken with two immiscible solvents (solvents that do not mix, e.g. ether and water) it distributes itself between them. At a given temperature, the ratio of its concentrations reaches a constant value — the partition (distribution) coefficient:

$K_{pc} = \frac{[\text{solute in solvent 1}]}{[\text{solute in solvent 2}]}$

Worked. If [X] in ether = 0.80 mol dm⁻³ and [X] in water = 0.20 mol dm⁻³: $K_{pc}\,(\text{ether/water}) = \frac{0.80}{0.20} = 4.0$

Kpc is usually quoted without units (the concentration units cancel). Keep the solvents in the order stated, or you will quote the reciprocal value.

Kpc = [X]₁ / [X]₂ at constant temperature.
Two immiscible solvents, one solute.
Keep the solvent order consistent.

See the full worked example for partition coefficients →

Solvent extraction calculations

Use Kpc with concentrations (mass ÷ volume); several small extractions are most efficient.

Solvent extraction uses Kpc to predict how much solute moves into an extracting solvent. Set up concentrations as mass ÷ volume in each layer.

Single extraction. 1.00 g of Z in 100 cm³ water, shaken with 100 cm³ ether (Kpc ether/water = 3.0). Let x g enter the ether: $K_{pc} = \frac{x/100}{(1-x)/100} = \frac{x}{1-x} = 3.0 \Rightarrow x = 0.75\,\text{g}$

Two smaller extractions (2 × 50 cm³). First portion gives: $\frac{y/50}{(1-y)/100} = \frac{2y}{1-y} = 3.0 \Rightarrow y = 0.60\,\text{g}$ Then 0.40 g remains; the second 50 cm³ extracts: $\frac{2z}{0.40-z} = 3.0 \Rightarrow z = 0.24\,\text{g}$ Total extracted = 0.60 + 0.24 = 0.84 g — more than the single extraction's 0.75 g.

This is the key result: dividing the solvent into several smaller portions extracts more solute than using it all at once.

Concentration = mass ÷ volume in each layer.
Volumes cancel only when they are equal.
Several small extractions > one large extraction.

See the full worked example for partition coefficients →

Assumptions and limitations

Kpc is constant only if the solute keeps the same molecular form in both solvents.

The partition coefficient is constant only if certain assumptions hold:

The temperature is constant (Kpc varies with temperature).
The solvents are immiscible.
The solute is in the same molecular state in both solvents — it must not associate (e.g. dimerise) or dissociate (e.g. ionise) in one of them.

If the solute changes its molecular form in one solvent, the simple concentration ratio no longer holds. For example, a carboxylic acid that dimerises in an organic solvent, or ionises in water, will not give a constant Kpc unless the calculation accounts for the species actually present.

Constant T, immiscible solvents.
Same molecular form in both solvents.
Fails if the solute dimerises or ionises in one solvent.

See the full worked example for partition coefficients →

Quick recap

Kpc = [X in solvent 1]/[X in solvent 2] at constant temperature.
Use concentrations (mass ÷ volume); volumes cancel only if equal.
Several small extractions remove more solute than one large one.
Assumes the same molecular form in both solvents.
Fails if the solute associates or dissociates in a solvent.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Kpc = [solute in solvent 1] / [solute in solvent 2].
Constant only at fixed temperature, immiscible solvents.
Same molecular form in both solvents (no dimerising/ionising).
Multiple small extractions > one large extraction.

How it’s examined

Usually a short Paper 4 calculation: find Kpc, or the mass extracted, and often a single-vs-multiple-extraction comparison. A definition or assumptions part is common. Examiner reports flag treating mass as concentration (forgetting to divide by volume) and inverting the concentration ratio.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 25.2; 9701 Examiner Reports 2022-2024; 9701/41 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-20.

Take this whole topic with you

Step-by-step worked examples — Partition coefficients

Step-by-step solutions to past-paper-style questions on partition coefficients, written exactly the way a tutor would explain them at the board.

Question type:

1Defining the partition coefficient (2 marks)
Getting startedWord problem• definition
Question
Define the partition coefficient for a solute X distributed between two immiscible solvents.
Step-by-step solution
1. Step 1
  The fixed ratio of concentrations of X in the two solvents at equilibrium (constant temperature).
  $K_{pc} = \frac{[\text{X in solvent 1}]}{[\text{X in solvent 2}]}$
Answer
Kpc = [X]₁ / [X]₂ — the constant ratio of concentrations between the two immiscible solvents.
Examiner tip
State 'at constant temperature' and that the solvents are immiscible.
2Calculating Kpc (2 marks)
Getting startedDirect calculation• calculation
Question
At equilibrium, the concentration of X in ether is 0.80 mol dm⁻³ and in water is 0.20 mol dm⁻³. Calculate Kpc (ether/water).
Step-by-step solution
1. Step 1
  Divide the two concentrations.
  $K_{pc} = \frac{0.80}{0.20} = 4.0$
Answer
Kpc = 4.0 (no units here, as the concentration units cancel).
Examiner tip
Keep the solvents in the order stated (ether/water) so Kpc has the right value.
3Finding a concentration from Kpc (3 marks)
Building confidenceDirect calculation• calculation
Question
Kpc (ether/water) for Y is 3.0. If [Y] in water is 0.050 mol dm⁻³ at equilibrium, find [Y] in ether.
Step-by-step solution
1. Step 1
  Rearrange Kpc = [ether]/[water].
  $[\text{ether}] = K_{pc} \times [\text{water}]$
2. Step 2
  Substitute.
  $[\text{ether}] = 3.0 \times 0.050 = 0.15\,\text{mol dm}^{-3}$
Answer
[Y] in ether = 0.15 mol dm⁻³.
4Single solvent extraction (4 marks)
Building confidenceMulti-step problem• extraction
Question
1.00 g of Z is dissolved in 100 cm³ of water and shaken with 100 cm³ of ether. Kpc (ether/water) = 3.0. Find the mass of Z extracted into the ether.
Step-by-step solution
1. Step 1
  Let x g go into ether; (1 − x) g stays in water. Equal volumes (100 cm³ each), so the volumes cancel.
  $K_{pc} = \frac{x/100}{(1-x)/100} = \frac{x}{1-x} = 3.0$
2. Step 2
  Solve for x.
  $x = 3(1-x) \Rightarrow 4x = 3 \Rightarrow x = 0.75$
Answer
0.75 g of Z is extracted into the ether (0.25 g remains in water).
Examiner tip
With equal volumes the volume terms cancel; otherwise keep them in (concentration = mass/volume).
5Single vs double extraction (4 marks)
StretchMulti-step problem• extraction
Question
Repeat the extraction of 1.00 g of Z (in 100 cm³ water, Kpc = 3.0) using TWO separate 50 cm³ portions of ether. Compare with the single 100 cm³ extraction.
Step-by-step solution
1. Step 1
  First 50 cm³: let y extracted. Kpc = (y/50)/((1−y)/100) = 2y/(1−y) = 3.
  $2y = 3(1-y) \Rightarrow 5y = 3 \Rightarrow y = 0.60$
2. Step 2
  0.40 g remains. Second 50 cm³: Kpc = (z/50)/((0.40−z)/100) = 2z/(0.40−z) = 3.
  $2z = 3(0.40 - z) \Rightarrow 5z = 1.20 \Rightarrow z = 0.24$
3. Step 3
  Total extracted = 0.60 + 0.24.
  $0.84\,\text{g} \;>\; 0.75\,\text{g (single)}$
Answer
0.84 g extracted in two portions vs 0.75 g in one — multiple smaller extractions are more efficient.
Examiner tip
The standard result: several small extractions remove more solute than one large one.
6Assumptions and limitations (3 marks)
StretchWord problem• assumptions
Question
State the assumptions required for the partition coefficient to be constant, and one situation where it fails. (3)
Step-by-step solution
1. Step 1
  The solute is in the SAME molecular state in both solvents (no association or dissociation).
2. Step 2
  Temperature is constant and the solvents are immiscible.
3. Step 3
  It fails if the solute dimerises/dissociates in one solvent (e.g. an acid that ionises in water), so the simple ratio no longer holds.
Answer
Same molecular form in both solvents, constant T, immiscible solvents; fails if the solute associates/dissociates in one solvent.
Examiner tip
The key assumption is that the solute does not change its molecular state between the two solvents.

Key Formulae — Partition coefficients

The formulae you need to memorise for partition coefficients on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Partition coefficient
$K_{pc} = \frac{[\text{solute in solvent 1}]}{[\text{solute in solvent 2}]}$
$[\,]$
equilibrium concentration of the solute in each solvent
When to use
Relating the concentrations of one solute distributed between two immiscible solvents at constant T.

Key Definitions and Keywords — Partition coefficients

Definitions to memorise and the exact keywords mark schemes credit for partition coefficients answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Partition (distribution) coefficient
Examiner keyword
The constant ratio of the concentrations of a solute distributed between two immiscible solvents at a given temperature.
Immiscible solvents
Examiner keyword
Solvents that do not mix and form two separate layers (e.g. ether and water).
Solvent extraction
Examiner keyword
Using an immiscible solvent to remove a solute from a solution; several small extractions remove more than one large one.

Common Mistakes and Misconceptions — Partition coefficients

The traps other students keep falling into on partition coefficients questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to divide by the (different) solvent volumes.
9701 Examiner Reports 2022-2024
Why it happens
Treating mass as concentration.
How to avoid it
Kpc uses CONCENTRATIONS (mass ÷ volume); volumes only cancel when they are equal.
✕Inverting the concentration ratio.
Why it happens
Not tracking which solvent is on top.
How to avoid it
Keep the solvents in the order given (e.g. ether/water).
✕Thinking one large extraction beats several small ones.
Why it happens
Assuming total volume is all that matters.
How to avoid it
Multiple smaller portions extract MORE solute overall.
✕Applying Kpc when the solute changes form in a solvent.
Why it happens
Forgetting the 'same molecular state' assumption.
How to avoid it
Kpc only holds if the solute does not associate or dissociate in either solvent.

Partition coefficients — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Partition coefficients

Short Notes - Partition coefficients

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Defining the partition coefficient (2 marks)

2Calculating Kpc (2 marks)

3Finding a concentration from Kpc (3 marks)

4Single solvent extraction (4 marks)

5Single vs double extraction (4 marks)

6Assumptions and limitations (3 marks)

Partition coefficient

Partition (distribution) coefficient

Immiscible solvents

Solvent extraction

✕Forgetting to divide by the (different) solvent volumes.

✕Inverting the concentration ratio.

✕Thinking one large extraction beats several small ones.

✕Applying Kpc when the solute changes form in a solvent.

Partition coefficients — frequently asked questions