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CHEM 2030/ Chapter 23 Extra QuestionA solute with a partition coefficient of 4 is extracted from 10 mL of phase 1 into phase 2.a) What volume of phase 2 is needed to extract 99% of the solute in one extraction? q is the amount remaining in phase 1, so if we are extracting 99% we need to divide that ...

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CHEM 2030/2130 Chapter 23 Extra Question

A solute with a partition coefficient of 4.0 is extracted from 10.0 mL of phase 1 into phase 2. a) What volume of phase 2 is needed to extract 99% of the solute in one extraction?

-

q is the amount remaining in phase 1, so if we are extracting 99% we need to divide that by 100 and subtract it from 1 to get q. o q = 1 – (99/100) = 0.01

q=

V1 K D V 2 +V 1

0.01=

10.0 4.0 (V 2)+ 10.0

0.04 V 2 +0.1=10.0

0.04 V 2=9.9

V 2=247.5 mL

CHEM2030/2130|SMavilla|F20

CHEM 2030/2130 Chapter 23 Extra Question

b) What is the total volume of solvent 2 needed to remove 99% of the solute in three equal extractions instead? -

q is the amount remaining in phase 1, so if we are extracting 99% we need to divide that by 100 and subtract it from 1 to get q. o q = 1 – (99/100) = 0.01

(

q3 =

V1 K D V 2+V 1

(

0.01=

)

3

10.0 4.0 (V 2 )+ 10.0

)

√ 0.01=

10.0 4.0 (V 2 )+ 10.0

0.215=

10.0 4.0 (V 2)+ 10.0

3

3

0.86 V 2 +2.15=10.0

0.86 V 2=7.85

V 2=9.13 ml

** It’s always useful to think logically about these questions. Is it expected that the volume for 3 extractions to extract 99% of the solute would be so small? We know that, given equal total volumes, the more extraction we do the better the extraction. So if we want to extract 99% and do either 1 or 3 extractions, it would make sense that the total volume used would be less for the multiple extractions.

CHEM2030/2130|SMavilla|F20...