The theoretical limit for extracting solute S from phase 1 (volume V 1 ) into phase 2 (volume V 2 ) is attained by dividing V 2 into an infinite number of infinitesimally small portions and conducting an infinite number of extractions With a partition coefficient K S 2 S 1 , the limiting fraction of solute remaining in phase 1 is 2 qlimit e (V2 V1)K Let k V 1 V 2 50 mL and let K 2 Let volume V2 be divided into n equal portions to conduct n extractions Find the fraction of S extracted into phase 2 for n 1, 2, 10 extractions How many portions are required to attain 95 of the theoretical limit

Question: The theoretical limit for extracting solute S from phase 1 (volume V 1 ) into phase 2 (volume V 2 ) is attained by dividing

The theoretical limit for extracting solute S from phase 1 (volume V₁) into phase 2 (volume V₂) is attained by dividing V₂ into an infinite number of infinitesimally small portions and conducting an infinite number of extractions. With a partition coefficient K = [S]₂/[S]₁, the limiting fraction of solute remaining in phase 1 is 2 qlimit = e^{- (V2/V1)K}. Let k = V₁ = V₂ = 50 mL and let K = 2. Let volume V2 be divided into n equal portions to conduct n extractions. Find the fraction of S extracted into phase 2 for n = 1, 2, 10 extractions. How many portions are required to attain 95% of the theoretical limit?

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