Partition Coefficient and Solvent Extraction – Solvent Extraction and Colligative Properties (Senior 6 Chemistry) | Ganzaa

Solvent Extraction and Colligative Properties

Lessons

Senior 6 · Chemistry

Solvent Extraction and Colligative Properties

5 lessons

Lesson 1 of 5

Partition Coefficient and Solvent Extraction

25 min

Getting an insecticide out of a flower

Pyrethrum daisies grown on the volcanic soils around Musanze contain pyrethrins, natural insecticides used in household sprays. At the Horizon SOPYRWA factory in Musanze the dried flowers are milled and extracted to give a crude extract, which is then refined. A 2012 sector report describes how: the powder is mixed with hexane, which dissolves the pyrethrins, and steam then separates the hexane from the oily extract; about 60 kg of dried flowers give 1 kg of extract. Why hexane? Because pyrethrins dissolve far better in it than in water. This lesson turns that idea into a law you can calculate with.

Definition

Immiscible liquids: liquids that do not mix but form two separate layers. Example: hexane and water.

Partition coefficient (distribution coefficient), $K_{pc}$ or $K_{D}$ : the ratio of the concentrations of a solute in two immiscible solvents in contact, at equilibrium, at a stated temperature:

$K_{pc} = \frac{[\mathrm{X}]_{\text{solvent 1}}}{[\mathrm{X}]_{\text{solvent 2}}}$

It has no units, because both concentrations are in the same units. Always state which solvent is on top of the fraction. Example: if a solute's concentration is 0.60 mol dm⁻³ in hexane and 0.030 mol dm⁻³ in water, $K_{pc}(\text{hexane/water}) = 20$ .

Solvent extraction: removing a solute from one solvent by shaking it with a second, immiscible solvent in which the solute is more soluble.

The partition law

Shake iodine solution in water with hexane and let the layers settle. Iodine molecules keep crossing the interface in both directions. When the rate at which they cross from water into hexane equals the rate back, the system is in dynamic equilibrium, and the ratio of the two concentrations stays fixed. Adding more iodine raises both concentrations, but not their ratio. This is the partition law (Nernst's distribution law, 1891):

$\mathrm{I_2(aq) \rightleftharpoons I_2(hexane)} \qquad K_{pc} = \frac{[\mathrm{I_2}]_{\text{hexane}}}{[\mathrm{I_2}]_{\text{aq}}}$

It is the equilibrium law of Unit 7 applied to one substance and two phases. It holds only when:

the temperature is constant ( $K_{pc}$ changes with temperature);
the two solvents are immiscible and do not react with the solute;
the solute is in the same molecular form in both solvents: it must not dissociate in one of them or associate in the other. Ethanoic acid, for example, forms dimers, $\mathrm{(CH_3COOH)_2}$ , in hexane, so the simple ratio is not constant;
the solutions are dilute and not saturated.

A non-polar solute such as iodine prefers a non-polar solvent ("like dissolves like"), so $K_{pc}(\text{hexane/water})$ is large and the violet hexane layer takes most of the iodine (Figure 1.1).

Figure 1.1 Solvent extraction in a separating funnel, held in a ring clamp on a stand. Hexane (density 0.66 g cm⁻³) floats on water; a solvent denser than water, such as dichloromethane (1.33 g cm⁻³), would form the LOWER layer. The funnel is stoppered, inverted and shaken, with the tap opened now and then to release vapour, then left to settle.

Several small extractions beat one large one

Suppose a mass $a$ of solute is dissolved in a volume $V_{aq}$ of water, and it is shaken with a volume $V_{org}$ of organic solvent. Let $x$ be the mass left in the water at equilibrium. Then

$K_{pc} = \frac{(a - x) \div V_{org}}{x \div V_{aq}} = \frac{(a - x)\,V_{aq}}{x\,V_{org}} \quad\Rightarrow\quad x = a\,\frac{V_{aq}}{V_{aq} + K_{pc}V_{org}}$

If the same total volume of solvent is split into $n$ equal portions, each extraction leaves the same fraction behind, so after $n$ extractions

$x_n = a\left(\frac{V_{aq}}{V_{aq} + K_{pc}\dfrac{V_{org}}{n}}\right)^{n}$

This fraction gets smaller as $n$ increases: for the same volume of solvent, several small extractions remove more solute than one large extraction. That is why a chemist "washes" a solution two or three times, and why an industrial extractor runs the solvent through the solid again and again.

Choosing a solvent. A good extracting solvent is immiscible with the first solvent, dissolves the solute much better (large $K_{pc}$ ), does not react with it, and is volatile, so it can be boiled off afterwards; it should also be as safe and cheap as possible.

Table 1.1: Uses of partition

Use	What partitions between what
extracting natural products (pyrethrins, essential oils, plant medicines)	from the plant material into an organic solvent such as hexane
decaffeinating coffee	caffeine from moist beans into a solvent such as ethyl ethanoate or supercritical $\mathrm{CO_2}$
purifying a product in the laboratory	impurities washed from an organic layer into water, or the reverse
paper chromatography	solutes between water held in the paper and the moving solvent
designing medicines	octan-1-ol/water partition coefficient predicts how easily a drug crosses cell membranes

Worked example

Problem. Butanedioic acid (a diprotic acid) was shaken with water and ethoxyethane at 25 °C until equilibrium. 25.0 cm³ of the aqueous layer needed 20.0 cm³ of 0.100 mol dm⁻³ NaOH for neutralisation; 25.0 cm³ of the ethoxyethane layer needed 3.00 cm³ of the same NaOH. Calculate $K_{pc}$ (ethoxyethane/water).

Solution. Each mole of acid reacts with 2 mol NaOH.

$[\text{acid}]_{aq} = \frac{\frac{1}{2}(0.100 \times 0.0200)}{0.0250} = 0.0400\,\mathrm{mol\,dm^{-3}}$

$[\text{acid}]_{ether} = \frac{\frac{1}{2}(0.100 \times 0.00300)}{0.0250} = 0.00600\,\mathrm{mol\,dm^{-3}}$

$K_{pc} = \frac{0.00600}{0.0400} = 0.150$

Answer: 0.150. The acid is more soluble in water, because its two –COOH groups form hydrogen bonds with water. (Since the same volumes and the same NaOH were used, $K_{pc}$ is simply the ratio of the titres, $\frac{3.00}{20.0}$ .)

Worked example

Problem. 100 cm³ of water contains 2.00 g of compound X. $K_{pc}$ (dichloromethane/water) for X is 5.0. Calculate the mass of X extracted by (a) one 50 cm³ portion of dichloromethane, (b) two successive 25 cm³ portions.

Solution. (a) Let $x$ g remain in the water:

$5.0 = \frac{(2.00 - x) \times 100}{50x} \;\Rightarrow\; 100(2.00 - x) = 250x \;\Rightarrow\; x = \frac{200}{350} = 0.571\,\mathrm{g}$

Extracted: $2.00 - 0.571 = 1.43\,\mathrm{g}$ (71 %).

(b) Each 25 cm³ portion leaves the fraction $\frac{100}{100 + 5.0 \times 25} = \frac{100}{225} = 0.444$ in the water. After two portions:

$x_2 = 2.00 \times (0.444)^2 = 0.395\,\mathrm{g} \qquad \text{extracted} = 2.00 - 0.395 = 1.60\,\mathrm{g}\ (80\,\%)$

Answer: (a) 1.43 g; (b) 1.60 g. The same 50 cm³ of solvent does more work in two portions. (Three portions of 16.7 cm³ would extract 1.68 g.)

Warning

Common mistakes. (1) Putting the solvents the wrong way up in the ratio: always write which is on top, e.g. $K_{pc}(\text{hexane/water})$ . (2) Using masses instead of concentrations when the volumes differ: divide each mass by its own volume. (3) Assuming the organic layer is always on top: it depends on density.

Safety. Hexane, ethoxyethane and other extracting solvents are highly flammable, and their vapour builds up pressure in a shaken funnel. Keep flames away, work in a fume cupboard, hold the stopper in, and vent through the open tap with the stem pointing up and away from people.

Tip

For any extraction problem, write one line: $K_{pc} = \frac{\text{mass in org} \div V_{org}}{\text{mass in aq} \div V_{aq}}$ , put the unknown mass $x$ in the right place, and solve. For repeated extractions, reuse the same fraction left behind each time.

FranklyChemistry: Partition equilibria and solvent extraction, principles and questions. Watch how the partition law is set up as an equilibrium and how single and repeated extractions are calculated.

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Practice

Define partition coefficient and state two conditions under which the partition law holds.
After shaking, 50.0 cm³ of hexane contained 0.950 g of Z and 50.0 cm³ of water contained 0.0500 g. Calculate $K_{pc}$ (hexane/water).
100 cm³ of water contains 1.00 g of Z. Using your value from question 2, calculate the mass extracted by one 20.0 cm³ portion of hexane.
Repeat question 3 with two 10.0 cm³ portions, and comment on the result.
Why can ethanol not be used to extract iodine from water, although iodine dissolves well in ethanol?
Dichloromethane (density 1.33 g cm⁻³) is used to extract caffeine from tea. Which layer in the funnel contains the caffeine? Explain.
The ratio [ethanoic acid] in hexane ÷ [ethanoic acid] in water is not constant when the total amount of acid changes. Suggest why.
A student must remove as much of a costly product as possible from 200 cm³ of water, using 60 cm³ of solvent. Evaluate two plans: one 60 cm³ extraction, or three 20 cm³ extractions.

Show answersHide answers

The ratio of the concentrations of a solute in two immiscible solvents at equilibrium at constant temperature. Conditions (any two): constant temperature; solvents immiscible and unreactive; solute in the same molecular form in both (no dissociation or association); dilute, unsaturated solutions.
$K_{pc} = \frac{0.950 \div 50.0}{0.0500 \div 50.0} = 19.0$ .
Mass left $x = 1.00 \times \frac{100}{100 + 19.0 \times 20.0} = 0.208\,\mathrm{g}$ ; extracted $0.792\,\mathrm{g}$ .
Left: $1.00 \times \left(\frac{100}{100 + 190}\right)^2 = 0.119\,\mathrm{g}$ ; extracted $0.881\,\mathrm{g}$ , more than with one 20 cm³ portion, though the same volume of hexane was used.
Ethanol is miscible with water, so no second layer forms and nothing can be separated.
The lower layer: dichloromethane is denser than water, so it sinks, and caffeine is more soluble in it.
Ethanoic acid forms hydrogen-bonded dimers in hexane (and ionises slightly in water), so it is not in the same molecular form in both solvents.
Three 20 cm³ extractions remove more product than one 60 cm³ extraction, because each leaves the same fraction behind and the fractions multiply. They take longer and each step loses a little liquid, but for a costly product the repeated extraction is better.

Sources and credits

Pyrethrum processing: Pyrethrum, Horizon SOPYRWA; and Sector Document for Pyrethrum, Agri Knowledge Centre, Kigali, 3 October 2012, via the FinDev Gateway.
Image: Flowers of Tanacetum cinerariifolium 20180430.jpg by そらみみ (Soramimi), via Wikimedia Commons, licensed under CC BY-SA 4.0.
Image: Separatory funnel with oil and colored water.jpg by PRHaney, via Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Partition Equilibria and Solvent Extraction, Principles and Questions, FranklyChemistry, embedded from YouTube.
The text, diagram, table, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1Getting an insecticide out of a flower

2The partition law

3Several small extractions beat one large one

Getting an insecticide out of a flower

The partition law

Several small extractions beat one large one