This statement is about partition coefficents. Before you go on, you should find and read the statement in your copy of the syllabus.
If you have two immiscible liquids like ether and water, and shake them up in a separating funnel, they obviously form two layers. The ether is less dense than water, and so forms the top layer. Now suppose you shake up a mixture of ether and water containing a substance which is soluble in both of them. Let's suppose that the substance, X, is more soluble in ether than it is in water. Particles of X will cross the boundary between the two liquid layers, and you will soon get a dynamic equilibrium set up. For every particle which moves into the top layer, one will move back down into the bottom one. You could write an equation for this: . . . and like any other equilibrium, you can find an equilibrium constant: This equilibrium constant is called the _{pc}.Like other equilibrium constants, partition coefficients are constant at a constant temperature, but they have some other restrictions as well. They only work properly for fairy dilute solutions, and the solute must be in the same chemical form in both solvents. It mustn't react, or ionise or associate (join together in dimers, for example).
Notice that the partition coefficient is a simple ratio of two concentrations. It doesn't matter what concentration units you use - as long as you use the same ones top and bottom. You could use mol dm Technically, the square brackets can only be used for concentration in mol dm I'm not prepared to do that, and so shall use the term "concentration of X" rather than [X] where non-standard concentration units are used.
When a solution of 1.00 g of X in 100 cm If you are asked to calculate a partition coefficient between two solvents, the concentration of the first solvent mentioned goes on top of the K You have enough information to calculate both concentrations in g cm concentration of X in ether = 0.80/10 g cm If 0.80 g were transferred to the ether, 1.00 - 0.80 g = 0.20 g were left in the water. concentration of X in water = 0.20/100 g cm So: Obviously, you could work out the concentrations in ether and in water as actual numbers before you put them into the expression. Do it however you feel most comfortable. Partition coefficients like this don't have units - the units cancel out because they are the same top and bottom.
We will use the same case as before - the same solvents, the same X and the same partition coefficient we have just calculated. This time we will work out how much would have been extracted into the ether layer if we had shaken the original solution of 1.00 g of X in 100 cm We are trying to work out the mass of X extracted. Let's call that m. Now work out an expression for the concentration of the solution of X in ether. concentration of X in ether = m/5 g cm What about the water? There will be (1.00 - m) g of X left in the water. So: concentration of X in water = (1.00 - m)/100 g cm Now you can put all this into the partition coefficient expression. Remember that we have already calculated the partition coefficient of X between ether and water as 40. You are then just faced with a simple, but slightly tedious, bit of algebra:
In the original calculation to find the partition coefficient, you were told that if you shook the original solution of 1.00 g of X in 100 cm Shaking it with 5 cm That would leave 0.33 g of X behind in the 100 cm How much of X would you extract Let's call the mass of X extracted by the second lot of ether n - so that we don't get confused. Work out an expression for the concentration of the solution of X in ether. concentration of X in ether = n/5 g cm What about the water? There will be (0.33 - n) g of X left in the water after the second extraction. So: concentration of X in water = (0.33 - n)/100 g cm Now you can put all this into the partition coefficient expression for X between ether and water as before. That means that if you were to combine the two 5 cm You were originally told that if you had only done this once, using the ether as a single lot of 10 cm You get a more efficient extraction by splitting your solvent up into smaller volumes as above. You use this sort of technique during the preparation of some organic compounds. You extract what you are trying to make from some messy solution in water so that it ends up in an organic solvent. You then remove the solvent by careful distillation. CIE asked a question similar to this in June 2009 paper 4 Q8. You had to calculate a value for the partition coefficient, and then use it in a two step extraction exactly as above. The calculation was worth 4 marks. This may be tedious, but you can't afford not to be able to do it. | |

Note: If you have a copy of my calculations book, I would suggest that you don't read about these multi-step calculations in the book. I use a much more efficient, and less tedious, way of doing them, but CIE doesn't use it. You risk confusing yourself. | |

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© Jim Clark 2011 (last modified April 2014) |