The shape of the lesson
By the end of the lesson, learners can
- Defining Kpc: Partition is an equilibrium like any other: molecules move across the boundary in both directions until the rates are equal, and the ratio of concentrations is then fixed.
- How much is extracted?: Let x be the amount that moves into the organic layer.
- Kpc = [X]organic ÷ [X]aqueous, no units.
- Divide by volume: each layer's amount over its own volume.
- Polarity: non-polar solute, Kpc > 1; polar solute, Kpc < 1.
- Several small portions extract more than one large one.
One picture every idea returns to
When a solute X is shaken with two immiscible solvents, such as hexane and water, it distributes itself between them and reaches a dynamic equilibrium, X(aq) ⇌ X(organic). The partition coefficient is the ratio of the equilibrium concentrations, Kpc = [X]organic ÷ [X]aqueous, which is constant at a fixed temperature and has no units. A non-polar solute such as iodine is more soluble in a non-polar solvent, so Kpc (organic over water) is greater than 1; a polar solute such as ammonia, which hydrogen bonds to water, has Kpc less than 1. Because the amount extracted depends on the volumes as well as Kpc, several small portions of solvent extract more than one large portion.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, One solute, two solvents.. Shake a solute with two immiscible solvents and it shares itself between them in a fixed ratio. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Defining Kpc | Partition is an equilibrium like any other: molecules move across the boundary in both directions until the rates are equal, and the ratio of concentrations is then fixed. Project tab 6 of the lab (Acid-base equilibria lab, partition coefficient) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: How much is extracted? | Let x be the amount that moves into the organic layer. Project tab 6 of the lab (Acid-base equilibria lab, partition coefficient) and let a student drive it while the class predicts. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in How much is extracted?, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| K pc is a ratio of concentrations, not masses: divide each amount by its own layer's volume. | Keep the order as stated (organic over aqueous here); inverting it gives 1 ÷ K pc. The amount left in water after extraction is the starting amount minus x, and each new portion starts from what is left. A large K pc means the solute prefers the organic layer, which needs a non-polar solute. |
Partition coefficients: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Define the partition coefficient, Kpc, of a solute X between an organic solvent and water.
After shaking, the hexane layer contains 0.0500 mol dm−3 of a solute and the water layer contains 6.00 × 10−4 mol dm−3. Calculate Kpc (hexane / water) and comment on the polarity of the solute.
2.00 g of Z is dissolved in 50.0 cm3 of water and shaken with 50.0 cm3 of an organic solvent. Kpc (organic / water) = 5.00. Calculate the mass of Z extracted.
The same 2.00 g of Z in 50.0 cm3 of water (Kpc = 5.00) is instead extracted with two 25.0 cm3 portions. Calculate the total mass extracted and compare with the previous answer.
Predict whether Kpc (hexane / water) is greater or less than 1 for (a) iodine and (b) ammonia. Explain.
Explain why Kpc stays the same when more solute is added to the two-layer mixture at constant temperature.
Answer key and marking notes
- The ratio of the concentration of X in the organic solvent to its concentration in water, [X]organic ÷ [X]aqueous.
- Measured at equilibrium at a stated temperature, with the two solvents immiscible.
- Kpc = 0.0500 ÷ 6.00 × 10−4 = 83.3.
- It is much greater than 1, so the solute prefers hexane and is likely to be non-polar.
- Let x g be extracted: (x ÷ 50.0) ÷ ((2.00 − x) ÷ 50.0) = 5.00.
- x = 5.00(2.00 − x), so 6.00x = 10.0.
- x = 1.67 g extracted (0.333 g remains in the water).
- First portion: (x ÷ 25.0) ÷ ((2.00 − x) ÷ 50.0) = 5.00, so 2x = 5.00(2.00 − x), x = 1.43 g.
- Second portion, from 0.571 g left: 2y = 5.00(0.571 − y), y = 0.408 g.
- Total 1.84 g, more than 1.67 g with a single 50.0 cm3 portion.
- (a) Greater than 1: I2 is non-polar and its id-id forces match those of hexane.
- (b) Less than 1: NH3 is polar and forms hydrogen bonds with water.
- Partition is a dynamic equilibrium, X(aq) ⇌ X(organic).
- Adding solute raises both concentrations in the same proportion, so the ratio, like any equilibrium constant, only changes with temperature.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which statement correctly defines the partition coefficient, Kpc, of a solute X between hexane and water?
- Athe ratio [X]hexane ÷ [X]water when X has reached equilibrium between the two immiscible solvents at a stated temperature
- Bthe ratio of the mass of X in hexane to the mass of X in water, whatever the volumes
- Cthe product [X]hexane × [X]water at equilibrium
- Dthe ratio [X]hexane ÷ [X]water at any moment during shaking
Iodine is shaken with a mixture of hexane and water until equilibrium is reached. Which statement about Kpc = [I2]hexane ÷ [I2]water is correct?
- Ait is much greater than 1, because non-polar iodine dissolves better in non-polar hexane
- Bit is less than 1, because iodine forms hydrogen bonds with water
- Cit equals 1, because iodine dissolves equally in any solvent
- Dit is less than 1, because water is denser than hexane
Kpc for X between hexane and water is 4.00. At equilibrium, the water layer contains X at 0.0500 mol dm−3. What is the concentration of X in the hexane layer?
- A0.200 mol dm−3
- B0.0125 mol dm−3
- C0.0500 mol dm−3
- D4.05 mol dm−3
The diagram shows the concentrations of a solute X in the two layers after shaking. What is Kpc (hexane / water), and what does it suggest about X?
- A83.3; X is probably non-polar
- B0.0120; X is probably polar
- C83.3; X is probably polar and hydrogen bonds with water
- D3.00 × 10−5; X is probably non-polar
A chemist has 100 cm3 of ether to extract an organic product from an aqueous solution. Which method extracts the most product?
- Ashake with four separate 25 cm3 portions of ether, one after another
- Bshake once with all 100 cm3 of ether
- Cshake once with 100 cm3 of ether for twice as long
- Dit makes no difference, because Kpc is constant
1.00 g of X is dissolved in 100 cm3 of water and shaken with 50.0 cm3 of ether. Kpc (ether / water) = 4.00. What mass of X is extracted into the ether?
- A0.667 g
- B0.800 g
- C0.889 g
- D0.333 g
1.00 g of X in 100 cm3 of water (Kpc ether / water = 4.00) is extracted with two successive 25.0 cm3 portions of ether. What total mass of X is extracted?
- A0.750 g
- B0.667 g
- C0.500 g
- D1.00 g
0.0100 mol of Y is dissolved in 50.0 cm3 of water and shaken with 10.0 cm3 of hexane. Kpc (hexane / water) = 20.0. What percentage of Y is extracted into the hexane?
- A80.0%
- B95.2%
- C99.0%
- D20.0%
Quiz key
Kpc is the ratio of the equilibrium concentrations of X in the two immiscible solvents, constant at a given temperature. A ratio of masses only equals the ratio of concentrations when the two volumes are equal, so masses must be divided by their volumes. The concentrations are divided, not multiplied. Before equilibrium is reached the ratio is still changing, so it is only the equilibrium ratio that is constant.
I2 is non-polar, with only instantaneous dipole-induced dipole forces, which match those in hexane; it is only slightly soluble in water. So far more iodine ends up in the hexane layer and Kpc is much greater than 1 (the hexane layer turns purple). Iodine has no hydrogen atoms and is non-polar, so it cannot hydrogen bond with water. Solubility depends on matching intermolecular forces, so it is not equal in all solvents. Density only decides which layer is on top, not where the solute goes.
[X]hexane = Kpc × [X]water = 4.00 × 0.0500 = 0.200 mol dm−3. 0.0125 divides by 4.00, inverting the ratio. 0.0500 assumes equal concentrations, which would need Kpc = 1. 4.05 adds Kpc to the concentration, which has no meaning.
Kpc = 0.0500 ÷ 6.00 × 10−4 = 83.3, with no units. A value much greater than 1 shows that X strongly prefers the non-polar hexane, so X is probably non-polar. 0.0120 is the inverse ratio (water over hexane). 83.3 is right, but a solute that hydrogen bonds with water would stay in the water and give Kpc below 1. 3.00 × 10−5 multiplies the concentrations instead of dividing them.
Each extraction removes a fixed fraction of the solute still in the water, so several small portions remove more in total than one large portion of the same total volume. For example, with Kpc = 4.00 and 1.00 g in 100 cm3 water, one 50.0 cm3 portion extracts 0.667 g but two 25.0 cm3 portions extract 0.750 g. A single portion reaches only one equilibrium. Shaking longer does not help once equilibrium is reached. Kpc is constant, but the amount extracted also depends on the volumes and the number of extractions.
(x ÷ 50.0) ÷ ((1.00 − x) ÷ 100) = 2x ÷ (1.00 − x) = 4.00, so 2x = 4.00 − 4.00x and x = 0.667 g. 0.800 g ignores the volumes, using x ÷ (1.00 − x) = 4.00 as if the layers were equal in size. 0.889 g swaps the volumes, putting 100 cm3 under the ether. 0.333 g is the mass left in the water, not the mass extracted.
First portion: (x ÷ 25.0) ÷ ((1.00 − x) ÷ 100) = 4x ÷ (1.00 − x) = 4.00, so x = 0.500 g. The water now holds 0.500 g; the second portion again takes half of it, 0.250 g. Total = 0.750 g. 0.667 g is the answer for one 50.0 cm3 portion, which is less effective. 0.500 g counts only the first portion. 1.00 g assumes every last trace is removed, but each extraction leaves some X in the water at equilibrium.
(x ÷ 10.0) ÷ ((0.0100 − x) ÷ 50.0) = 5x ÷ (0.0100 − x) = 20.0, so 5x = 0.200 − 20.0x, x = 0.00800 mol: 80.0% extracted. 95.2% (20 ÷ 21) ignores the volumes; the small hexane volume reduces the amount it can hold. 99.0% swaps the two volumes. 20.0% is the fraction left in the water, not the fraction extracted.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.