1. Exactly 2.0g of Kryptonite is dissolved in 100mL of Solvent A and then 100mL

Question

1. Exactly 2.0g of Kryptonite is dissolved in 100mL of Solvent A and then 100mL is immisclible 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.
2. A commercial lubricant, Dubdeefordy, has K=SN/SM=12.7 for immiscible solvents M and N. If 14.7g of Dubdeefordy is shaken with 115mL if solvent M and 45mL of solvent N, what weight if Dubdeefordy will be found in each layer at equilibrium? Show your calculations.
The picture provided is the "Appendix-Multiple Extractions" DIX-MULTIPLE IONS 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 the original 1.0 g of M, separated from the X, Y & Z impurities which stayed in the water layer. You can recover even more M by doing more extractions with additional portions of ether.

Explanation / Answer

K = SN/SM = 12.7

starting = 14.7 g

M = 115 ml

N = 45 ml

let x amount of Duberfordy is extracted into solvent N

then,

12.7 = (x/45)/[(14.7 - x)/115]

1.62 - 0.11x = 0.02x

x = 1.62/0.13 = 12.46 g

So at equilibrium,

concentration in N = 12.46 g

concentration in M = 12.47 - 12.46 = 0.01 g

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