An organic analyte is extracted from the aqueous phase into dichloromethane, whe

Question

An organic analyte is extracted from the aqueous phase into dichloromethane, where D_e = 4.0 and 150 mL of each phase is employed. What will the extraction efficiency be if ten repetitive extractions are employed? How many extractions would be needed for this system in (a) in order to achieve a minimum of 99% extraction efficiency? How could 100% extraction efficiency be obtained for the system in (a)? Why does dimerisation of an organic acid (HA) assist in extraction thereof into an organic solvent? Include a chemical reaction equation in your answer.

Explanation / Answer

Fraction in aqueous

q=(Vaq/DVorg+Vaq)n

n= no of extraction =1

D=distribution coefficient=4

Vorg=Vaq =150 ml

So q=(150/(4*150+150))10=1.024*10-7

The fraction of solute in the organic phase is 1 – 1.024*10-7, or 1. Extraction efficiency is the percentage of solute moving into the extracting phase. The extraction efficiency, therefore, is 100%.

Means it is 100 % efficiency

(b)

For minimum 99%

q=0.01

0.01=(150/(4*150+150))n

n=1.17=2

(c)

100% efficiency means in aqueous phase amount of solute is 0.

When q=0.then efficiency 100%.

So

using equation and put all values.

q=0.2n

For n=3 q=0.008

n=4 q=0.00016=0

So 2 extraction required.

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