1 / 11

Equilibria between different phases

Equilibria between different phases. Advanced Higher Unit 2b(ii). Separating Funnel. Separates immiscible liquids with different densities Usually water, plus an organic solvent (ether, chloroform). [solute](organic). K =. [solute](aq). Partition Coefficient.

tim
Télécharger la présentation

Equilibria between different phases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Equilibria between different phases Advanced Higher Unit 2b(ii)

  2. Separating Funnel • Separates immiscible liquids with different densities • Usually water, plus an organic solvent (ether, chloroform)

  3. [solute](organic) K = [solute](aq) Partition Coefficient • When a solute is added to two such immiscible liquids, some will dissolve in each layer • The distribution will not be even • Ratio of distributed called PARTITION COEFFICIENT

  4. Solvent extraction • Partition coefficient is used in solvent extraction and for purification • For example, with an impure carboxylic acid, the acid will dissolve in ether but impurities will not

  5. Calculation Consider the following partition coefficient of an organic acid: [acid](ether) K = = 5 [acid](aq) If 10g of acid is dissolved in 100ml of water and 100ml ether, the quantity of acid which can be extracted can be calculated...

  6. x / 100 = 5 (10 – x) / 100 x = 5 (10 – x) x = 50 – 5x 6x = 50 x = 8.3 g [acid](ether) K = = 5 [acid](aq) Concentration organic acid (ether) = x / 100 Concentration organic acid (aq) = (10 – x) / 100

  7. An alternative method • Repeat the previous calculation, but this time for using 100 ml water and 50 ml ether for a first extraction, then repeating with another 50 ml ether for a second extraction

  8. [acid](ether) K = = 5 [acid](aq) x / 50 = 5 (10 – x) / 100 Concentration organic acid (ether) = x / 50 Concentration organic acid (aq) = (10 – x) / 100 x = 5 (10 – x) / 100 50 x = 50 – 5x / 100 50 100x = 50 – 5x 50 2x = 50 – 5x 7x = 50 x = 7.14 g

  9. [acid](ether) K = = 5 [acid](aq) x / 50 = 5 (2.86 – x) / 100 Concentration organic acid (ether) = x / 50 Concentration organic acid (aq) = (2.86 – x) / 100 x = 5 (2.86 – x) / 100 50 x = 14.3 – 5x / 100 50 100x = 14.3 – 5x 50 2x = 14.3 – 5x Total extracted: 7.14 + 2.04 = 9.18g 7x = 14.3 x = 2.04 g

  10. Chromatography • Method of separation which depends on the partition of substances between two phases • One is mobile, the other is stationary • Stationary phase – depends on type of chromatography • Paper • Thin layer • Column • Gas/liquid • High performance liquid

  11. Task • Each group take one of the chromatography types and prepare a one page notes sheet for the class and for the Think-chemistry website

More Related