Refer to the \"Appendix - Multiple Extractions\" then answer the following quest
ID: 531159 • Letter: R
Question
Refer to the "Appendix - Multiple Extractions" then answer the following questions.
1. Exactly 2.0g of Kryptonite is dissolved in 100mL of Solvent A and then 100mL is immiscible Solvent B is added. The distribution coefficient is K=SB/SA= 0.42. If a series of extractions is done using 100mL of fresh B each time, what quantity of Kryptonite will be left in Solvent A after 1 extraction? After 2 extractions? Show your calculations.
APPENDIX-MULTIPLE ENTRACTIONS one solvent and a second, immiscible, solvent is added. We will assume a somewhat ideal situation that the impurities X, Y and z are completely luble in the new solvent and therefore remain in the original liquid phase. However, it rarely happens that 100 of M will be transferred to the second solvent in a single extraction. More often, M will be divided between the two liquid layers in proportions described by the Distribution Coefficient, K,: Concentration of M in one solvent g of M per ml. of one solvent Concentration of M in second so g of M per ml. of second solvent This K value is analogous to the equilibrium constant for a reversible chemical reaction. In both cases, the relative concentrations are constant, and the situation is dynamic; in the two-phase liquid system, molecules of M are continually passing back and forth through the liquid interface, but do so in equal numbers so that the concentrations litre or grams/litre in each of the two layers remains the same. Even when K is small, repeated extractions make it possible to transfer most of the desired material out of the original solvent. Suppose you had 1.0 g of M in 100 ml of water and that the distribution of M between ether and water was given by: 1.5 SE 1.5 which is equivalent to SW [(1.5)/(1.5 +1.0) 0.6, or 60 similarly, 1.0/2.5 40%l First Extraction, with 100 ml of ether: 60% of the 1.0 g, -0.6 g of M will transfer into the ether layer which the student separates and sets aside; the remaining 40 0.4 g remains in the water layer and is subjected to the next Second Extraction, with 100 ml of fresh ether 60 of the 0.4 g, or 0.6 x .4 0.24 g goes into the organic layer which is separated and saved; the rest of the M, 0.4 x 0.4 0.16 g, stays in the aqueous layer and another extraction is carried out Third Extraction, with more ether At this stage, the M will be divided into 0.6 x 0.16 0.096 g, and 0.064 g; separate the organic layer. Combine the three 100 ml portions of ether to get (0.6 0.24 0.096) 0.936 g, or about of the original 1.0 g of M, separated from the X, Y & Z impurities which stayed in the water yor. You can recover even more M by doing more extractions with additional portions of ether.Explanation / Answer
2.0g of Kryptonite is dissolved in 100mL of Solvent A and then 100mL is immiscible Solvent B is added.
The distribution coefficient is K=SB/SA= 0.42/1.0 which is equivalent to 29% / 71%.
If a series of extractions is done using 100mL of fresh B each time,
The quantity of Kryptonite will be left after Ist extraction with 100 ml of solvent B:
29% of the 2.0g = 0.59 g of Kryptonite will transfer into the Solvent B layer which then separates, the remaining 71% = 1.41 g remains in the solvent A layer and is subjected to the next extraction.
The quantity of Kryptonite will be left after IInd extraction with 100 ml of solvent B:
29% of the 1.41 g = 0.41 g of Kryptonite will transfer into the Solvent B layer which then separates, the remaining 71% = 0.99 g remains in the solvent A layer and is subjected to the next extraction.