Q16P The theoretical limit for extrac... [FREE SOLUTION]

Chapter 23: Q16P (page 629)

The theoretical limit for extracting solute Sfrom ${Phase}_{1}$ $(volume V_{1})$ into phase2 $(volume V_{2})$ 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 $1 i s^{2 q} l i m i t = e - (V_{2} / V_{1}) K (V_{1} = V_{2} = 50 mL and K = 2)$ . Let volume $V_{2}$ be divided intoequal portions to conduct extractions. Find the fraction of S extracted into phase 2 for n = 1,2, and 10extractions. How many portions are required to attain 95%of the theoretical limit?

Expert verified

To conclude the number of portions required in order to attain the 95% of the limit would be and the extracted fraction would be

$0.95 脳 0.864655 = 0.82143$

For Partition coefficient $K = {[S]}_{2} / {[S]}_{1}$ ,the limited fraction of solute in is

$1 i s^{2 q} limit = e^{- (V_{2} / V_{1}) K} (V_{1} = V_{2} = 50 mL and K = 2)$ .The fraction of is extracted for $for n = 1, 2, and 10$ extractions.

Limited fractions of solute remaining in first phase

$q_{limit} = e^{(V_{2} / V_{1}) k} q_{limit} = e^{- 50 mL / 50 mL 脳 2} q_{limit} = 0.135335$

The maximum fraction extracted would be

$1 - q_{limit} = 1 - 0.135335 1 - q_{limit} = 0.864665$

For column C calculations as $q = {(\frac{V_{1}}{V_{1} + D 脳 V_{2}})}^{n}$

For column D calculations as $1 - q$

For column E calculations as $\frac{1 - q}{1 - q_{limit}}$

To conclude the number of portions required in order to attain the of the limit would be and the extracted fraction would be

$0.95 脳 0.864655 = 0.82143$

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