Solvent Extraction: Definition & Process

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  • 0:00 What Is Solvent Extraction?
  • 1:21 The Solvent Extraction Process
  • 5:15 Sample Problem
  • 6:37 Lesson Summary
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Lesson Transcript
Instructor: Saranya Chatterjee

Saranya has a masters degree in Chemistry and in Secondary Education. She has taught high school, AP chemistry for 2 years and is teaching undergraduate college chemistry for 3 years.

This lesson will define solvent extraction and discuss and explain the process. It will derive the expression of distribution coefficient and work out some related mathematical problems.

What Is Solvent Extraction?

Solvent extraction, also called liquid-liquid extraction (LLE) and partitioning, is a method to separate compounds based on their relative solubilities in two different immiscible liquids. Immiscible liquids are ones that cannot get mixed up together and separate into layers when shaken together. These liquids are usually water and an organic solvent. LLE is an extraction of a substance from one liquid into another liquid phase. The most common use of the distribution principle is in the extraction of substances by solvents, which are often employed in a laboratory or in large scale manufacturing. Organic compounds are generally much more soluble in organic solvents, like benzene, chloroform, and ether, than in water and these solvents are immiscible with water. Organic compounds are then quite easily separated from the mixture with inorganic compounds in aqueous medium by adding benzene, chloroform, etc. Upon shaking, these separate into two layers. Since organic compounds have their distribution ratio largely in favor of the benzene phase, more of them would pass into a non-aqueous layer. Finally this non-aqueous layer is removed and distilled to obtain the purified compound.

The Solvent Extraction Process

In such solvent extraction, it's advantageous to do extraction in successive stages using smaller lots of solvents rather doing extraction once using the entire lot. Let's suppose that a solute A is present in 100 cc of water, and 100 cc of ether will be used for its extraction. Now suppose that the distribution coefficient of A between ether and water is 4, which means:

K = Concentration of A in ether / Concentration of A in water = 4

(i) When the whole of 100 cc of ether is used at a time for extraction, suppose w1 grams of solute pass into ether layer and w2 grams are left in aqueous layer, so that:

w1/100 divided by w2/100 = 4

That is: w1/w2 = 4

or, w1 / w1 + w2 = 4/5

This means that 100 cc of ether has separated 4/5 (or 80%) of the solute originally present.

(ii) Now let's use 100 cc of ether in two successive extractions, using 50 cc each time. In the first stage:

w1 / 50 divided by w2 / 100 = 4

or, w1 / w2 = 2

or, w1 / w1 + w2 = 2/3

In the first extraction 2/3 (that is, 66.7%) is extracted. So 1/3 of the original amount is still retained in aqueous medium. In the second extraction, again using 50 cc of ether, we will further extract 2/3 of 1/3 (that is, 2/9) of the original amount. In other words, in two extractions using the same 100 cc ether we can separate (2/3 + 2/9), or 88.9%, of the original amount of the compound. Thus, a two-stage extraction is more efficient. If the same 100 cc of solution is used in four or five lots, a still greater proportion could be extracted.

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