Question

2. What volume of an organic solvent must be used to effect 90% separation in one extraction when only 2.7g of a certain compound dissolves in 100ml of water? (K=15) 3. What percentage of the organic compound could be recovered if two extractions were made, each time using half of the volume calculated 2?

Answer 1

Ans 2:

Expression for partition coefficient can be written as follows:

K = (Wo/Vo) / (Waq/Vaq)

Where,

K = partition coefficient = 15

Wo = Weight of solute in organic phase = (2.7 x 0.90)= 2.43g

Vo = Volume of organic solvent

Waq = Weight of solute in aqueous phase =(2.7 - Wo) = 2.7 – 2.43 = 0.27g

Vaq = volume of aqueous phase = 100mL

Thus, partition coefficient can be written as follows:

K = 15=(2.43/Vo) / (0.27/100)

15 = 2.43 / (Vo x 0.0027)

Vo = 2.43 / (15 x 0.0027)

Vo = 60mL

Thus, volume of organic solvent required to effect 90% separation is 60 mL

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Ans 3:

The percentage of organic compound that can be recovered after two extractions of 30mL each can be calculated as follows:

K = (Wo/Vo) / (Waq/Vaq)

For first extraction:

Vo = 30mL

Vaq = 100mL

Wo + Waq = 2.7

K = 15

Therefore, 15 = (Wo/30) / (Waq/100)

15 = Wo x 100 / Waq x 30

(15 x 30) /100= Wo / Waq

Wo/Waq = 4.5

Wo = 4.5 Waq

Wo = 4.5 (2.7 – Wo)

Wo = 12.15 – 4.5Wo

Wo + 4.5Wo = 12.15

5.5Wo = 12.15

Wo = 12.15/5.5

Wo = 2.20

Hence, Waq = 2.7-2.2 = 0.5 g

Thus, after first extraction, amount of solute in organic phase is 2.2g and that in aqueous phase is 0.5g

For second extraction:

Vo = 30mL

Vaq = 100mL

Wo + Waq = 0.5

K = 15

Therefore, 15 = (Wo/30) / (Waq/100)

15 = Wo x 100 / Waq x 30

(15 x 30) /100= Wo / Waq

Wo/Waq = 4.5

Wo = 4.5 Waq

Wo = 4.5 (0.5 – Wo)

Wo = 2.25 – 4.5Wo

Wo + 4.5Wo = 2.25

5.5Wo = 2.25

Wo = 2.25/5.5

Wo = 0.409

Hence, Waq = 0.5 – 0.409 = 0.091 g

Thus, in second extraction, amount of solute in organic phase is 0.409g and that in aqueous phase is 0.091g

Thus, after two extractions of 30mL each, total solute in organic phase is 2.20 + 0.409 = 2.609g

Therefore, % of solute recovered = (2.609 / 2.7) x 100 = 96.63%

Hence, percentage of organic compound that can be recovered after two extractions is 96.63%