Question

A chemist wants to extract a solute from 100 mL of water using only 300 mL of ether. The partition coefficient between ether and water is 3.34. Calculate ????, the fraction of solute that would remain in the water under each of the extraction conditions.
a.The chemist performs a single extraction with 300 mL of ether.
b.The chemist performs three extractions with 100 mL of ether.
c.The chemist performs six extractions with 50 mL of ether.

Answer 1

Answer:

a. X = 0,909

b. X = 0,965

c X = 0,997

Explanation:

The partition coefficient (k) is defined as:

k = Solute in ether / solute in water

a. $3,34 = \frac{\frac{X}{300mL} }{\frac{1-X}{100mL} }$

Where X is the fraction of solute extracted

3,34 = X / 3-3X

10,02-10,02X = X

10,02 = 11,02X

X = 0,909

b. First extraction:

$3,34 = \frac{\frac{X}{100mL} }{\frac{1-X}{100mL} }$

3,34 = X / 1-X

3,34 - 3,34X = X

3,34 = 4,34X

X = 0,770

That means solute in water will be: 1-0,770 = 0,23

Second extraction:

The second extraction will extract the same fraction of solute, as now you have 0,230 of solute in water you will extract:

0,230×0,770 = 0,177

Third extraction:

In the same way, the third extraction will extract:

(0,230-0,177)×0,770 = 0,018

Fraction in water×Fraction extracted

That means total solute extracted is:

0,770 + 0,177 + 0,018 = 0,965

c. Extracting with 50mL  of ether:

First extraction

$3,34 = \frac{\frac{X}{50mL} }{\frac{1-X}{100mL} }$

3,34 = 2X / 1-X

3,34 - 3,34X = 2X

3,34 = 5,34X

X = 0,625

Second extraction:

(1-0,625)×0,625= 0,234

Third Extraction:

(1-0,625-0,234)×0,625= 0,088

Fourth extraction:

(1-0,625-0,234-0,088)×0,625= 0,033

Fifth extraction:

(1-0,625-0,234-0,088-0,033)×0,625= 0,013

Sixth extraction:

(1-0,625-0,234-0,088-0,033-0,013)×0,625= 0,004

Total extractions gives:

0,625+0,234+0,088+0,033+0,013+0,004 = 0,997

I hope it helps!