A solute with partition co-efficient of 4.0 is extracted from 10 mL of phase 1 into phase 2.

(a) What volume of phase 2 is needed to extract 99% of the solute in one extraction?

(b) What is the total volume of solvent 2 needed to remove 99% of the solute in three equal extractions?

Extraction is the physical transfer of a solute from one phase to another.

Partition Co-efficient, $K=\frac{\left[S_2\right]}{\left[S_1\right]}$; where solute, S is partitioned between phase 1 and 2.

Fraction remaining in phase 1 after one extraction = $q=\frac{V_1}{V_1+K{V}_2}$

Fraction remaining in phase 1 after n extractions = $q^n={\left(\frac{V_1}{V_1+K{V}_2}\right)}^n$

Where $V_1$and $V_2$ are volume of solvent in phase 1 and 2 respectively.

$q=1-\frac{99}{100}=0.01$

Volume of phase 2 in extraction,$q=\frac{V_1}{V_1+K{V}_2}$

Or, $0.01=\frac{10}{4.0\times V_2+V_1}$

Or, $0.04V_2+0.1=10$

Or, $0.04V_2=10-0.1$

Or, $0.04V_2=9.9$

Or, $V_2=247.5ml$ml

Fraction remaining in phase 1 after n extractions = $q^n={\left(\frac{V_1}{V_1+K{V}_2}\right)}^3$

When n= 3 , $q={\left(\frac{V_1}{V_1+K{V}_2}\right)}^3$

$q=1-\frac{99}{100}-0.01$

∴$0.01={\left[\frac{10}{4.0\times V_2+10}\right]}^3$

Or, $\sqrt[3]{0.01}=\left[\frac{10}{4.0\times V_2+10}\right]$

Or, $0.215=\left[\frac{10}{4.0\times V_2+10}\right]$

Or, $0.86V_2+2.15=10$

Or, $0.86V_2=7.85$

Or, $V_2=9.13ml$

So, Volume of solvent 2 is needed 247.5 ml in one extraction and 9.13 ml is needed in three equal extraction.