Chapter 23, Problem 23.AE

(a)

Interpretation Introduction

Interpretation:

The volume of phase 2 is needed to extract $\text{99}\text{\%}$ of the solute has to be calculated.

Concept introduction:

Partition coefficient:

Partition of solutes is depends on “like dissolves like” concept.  This means the solubility of solute is more in a solvent whose polarity is similar to that of the solute.

The partition coefficient of solute $\left(\text{S}\right)$ is given by

$\text{Solute}\left(\text{in}\text{phase}\text{1}\right)\stackrel{}{\rightleftharpoons }\text{Solute}\left(\text{in}\text{phase}\text{2}\right)$

$\begin{array}{l}\text{K=}\frac{{\text{A}}_{{\text{S}}_{\text{2}}}}{{\text{A}}_{{\text{S}}_{\text{1}}}}\approx \frac{{\left[\text{S}\right]}_{\text{2}}}{{\left[\text{S}\right]}_{\text{1}}}\\ \text{where,}\\ \text{K}\text{-}\text{partition}\text{coefficient}\\ {\text{A}}_{{\text{S}}_{\text{1}}}\text{-}\text{Acitivity}\text{of}\text{solute}\text{in}\text{phase}\text{1}\\ {\text{A}}_{{\text{S}}_{\text{2}}}\text{-}\text{Acitivity}\text{of}\text{solute}\text{in}\text{phase}\text{2}\\ {\left[\text{S}\right]}_{\text{1}}\text{-}\text{Concentration}\text{of}\text{solute}\text{in}\text{phase}\text{1}\\ {\left[\text{S}\right]}_{\text{2}}\text{-}\text{Concentration}\text{of}\text{solute}\text{in}\text{phase}\text{2}\end{array}$

Answer to Problem 23.AE

• The volume of phase 2 is needed to extract $\text{99}\text{\%}$ of the solute is $\text{248}\text{mL}\text{.}$

Explanation of Solution

Given information,

Partition coefficient value is 4.0.

Extracted volume of phase 1 is $\text{10}\text{mL}\text{.}$

Calculate the volume of phase 2

$\begin{array}{l}\text{Fraction}\text{remaining}\text{=}\text{q}\text{=}\frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{1}}\text{+}{\text{KV}}_{\text{2}}}\\ \text{where,}\\ {\text{V}}_{\text{1}}\text{-}\text{volume}\text{of}\text{phase}\text{1}\\ {\text{V}}_{\text{2}}\text{-}\text{volume}\text{of}\text{phase}\text{2}\\ {\text{K}}_{\text{1}}\text{-}\text{partition}\text{coefficient}\end{array}$

Substitute the given values into the above expression

$\begin{array}{l}\text{q}\text{=}\frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{1}}\text{+}\text{}{\text{KV}}_{\text{2}}}=\text{0}\text{.01}\text{=}\frac{\text{10}}{\text{10}\text{+}\left(\text{4}\text{.0}\right){\text{V}}_{\text{2}}}\\ \\ \text{10}\text{=}\text{0}\text{.01}\left(\text{10}\text{+}\text{4}{\text{.0V}}_{\text{2}}\right)\text{=}\text{0}\text{.1}\text{+}\text{0}{\text{.04V}}_{\text{2}}\\ \\ {\text{V}}_{\text{2}}\text{=}\frac{\text{10}\text{-}\text{0}\text{.1}}{\text{0}\text{.04}}\text{=}\text{247}\text{.5}\text{=}\text{248}\text{mL}\text{.}\end{array}$

The volume of phase 2 is needed to extract $\text{99}\text{\%}$ of the solute is $\text{248}\text{mL}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The total volume of solvent 2 is needed to remove $\text{99}\text{\%}$ of the solute in three equal extractions has to be calculated.

Concept introduction:

Partition coefficient:

Partition of solutes is depends on “like dissolves like” concept.  This means the solubility of solute is more in a solvent whose polarity is similar to that of the solute.

The partition coefficient of solute $\left(\text{S}\right)$ is given by

$\text{Solute}\left(\text{in}\text{phase}\text{1}\right)\stackrel{}{\rightleftharpoons }\text{Solute}\left(\text{in}\text{phase}\text{2}\right)$

$\begin{array}{l}\text{K=}\frac{{\text{A}}_{{\text{S}}_{\text{2}}}}{{\text{A}}_{{\text{S}}_{\text{1}}}}\approx \frac{{\left[\text{S}\right]}_{\text{2}}}{{\left[\text{S}\right]}_{\text{1}}}\\ \text{where,}\\ \text{K}\text{-}\text{partition}\text{coefficient}\\ {\text{A}}_{{\text{S}}_{\text{1}}}\text{-}\text{Acitivity}\text{of}\text{solute}\text{in}\text{phase}\text{1}\\ {\text{A}}_{{\text{S}}_{\text{2}}}\text{-}\text{Acitivity}\text{of}\text{solute}\text{in}\text{phase}\text{2}\\ {\left[\text{S}\right]}_{\text{1}}\text{-}\text{Concentration}\text{of}\text{solute}\text{in}\text{phase}\text{1}\\ {\left[\text{S}\right]}_{\text{2}}\text{-}\text{Concentration}\text{of}\text{solute}\text{in}\text{phase}\text{2}\end{array}$

Answer to Problem 23.AE

• The total volume of solvent 2 is needed to remove $\text{99}\text{\%}$ of the solute in three equal extractions is $\text{27}\text{.3}\text{mL}\text{.}$

Explanation of Solution

Given information,

Partition coefficient value is 4.0.

Extracted volume of phase 1 is $\text{10}\text{mL}\text{.}$

Fraction remaining in a given phase after $\text{n}$ extractions is ${\text{q}}^{\text{n}}$

${\text{q}}^{\text{n}}\text{=}{\left(\frac{{\text{V}}_{\text{1}}}{{\text{V}}_{\text{1}}\text{+}{\text{KV}}_{\text{2}}}\right)}^{\text{n}}$

Calculate the volume of phase 2 for first extraction

$\begin{array}{l}{\text{q}}^{\text{3}}\text{=}\text{0}\text{.01}\text{=}{\left(\frac{\text{10}}{\text{10}\text{+}\text{4}{\text{.0V}}_{\text{2}}}\right)}^{\text{3}}\\ \sqrt[\text{3}]{\text{0}\text{.01}}\text{=}\frac{\text{10}}{\text{10}\text{+}{\text{4V}}_{\text{2}}}\\ \\ \text{0}\text{.21544}\left(\text{10}\text{+}{\text{4V}}_{\text{2}}\right)\text{=}\text{10}\\ \\ \text{2}\text{.1544}\text{+}\text{0}{\text{.8618V}}_{\text{2}}\text{=}\text{10}\\ \\ {\text{V}}_{\text{2}}\text{=}\frac{\text{10}\text{-}\text{2}\text{.1544}}{\text{0}\text{.8618}}\text{=}\text{9}\text{.10}\text{mL}\text{.}\end{array}$

Calculate the total volume of solvent 2 needed, the number of extraction is three.

$\text{9}\text{.1}\text{mL}\text{×}\text{3}\text{=}\text{27}\text{.3}\text{mL}\text{.}$

The total volume of solvent 2 is needed to remove $\text{99}\text{\%}$ of the solute in three equal extractions is $\text{27}\text{.3}\text{mL}\text{.}$