Chapter 23, Problem 23.21P

(a)

Interpretation Introduction

Interpretation:

The value of linear velocity and volume flow rate has to be calculated.

Answer to Problem 23.21P

• The value of linear velocity is $\text{13}\text{.9}\text{m/min}\text{.}$
• The value of volume flow rate is $\text{3}\text{mL/min}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Column}\text{length}\text{is}\text{30}\text{.1}\text{m}\text{.}\\ \text{Thickness}\text{of}\text{the}\text{stationary}\text{phase}\text{is}\text{3}\text{.1}\text{µm}\text{.}\\ \text{Column}\text{inner}\text{diameter}\text{is}\text{0}\text{.530}\text{mm}\text{.}\\ \text{Unretained}\text{solute}\text{passes}\text{through}\text{in}\text{2}\text{.16}\text{min}\text{.}\\ \text{Particular}\text{solute}\text{passes}\text{through}\text{in}\text{17}\text{.32}\text{min}\text{.}\end{array}$

Calculate the linear flow rate

$\begin{array}{l}\text{Linear}\text{flow}\text{rate}\text{=}\frac{\text{30}\text{.1}\text{m}}{\text{2}\text{.16}\text{min}}\\ \\ \text{=}\text{13}\text{.9351}\text{=}\text{13}\text{.9}\text{m/min}\text{.}\end{array}$

The value of linear velocity is $\text{13}\text{.9}\text{m/min}\text{.}$

Calculate the radius of the tube

$\begin{array}{l}\text{Inner}\text{diameter}\text{of}\text{the}\text{tube}\text{=}\text{530µm}\text{-}\text{2}\left(\text{3}\text{.1}\text{µm}\right)\text{=}\text{523}\text{.8}\text{µm}\\ \Rightarrow \text{radius}\text{=}\text{261}\text{.9}\text{µm}\end{array}$

Calculate the volume

$\begin{array}{l}\text{Volume}\text{=}{\text{πr}}^{\text{2}}\text{×length}\text{=}\text{π}{\left(\text{261}\text{.9}\text{×}{\text{10}}^{\text{-4}}\text{cm}\right)}^{\text{2}}\left(\text{30}\text{.1}{\text{×10}}^{\text{2}}\text{cm}\right)\\ \text{=}\left(\text{3}\text{.14}\right){\left(\text{261}\text{.9}\text{×}{\text{10}}^{\text{-4}}\text{cm}\right)}^{\text{2}}\left(\text{30}\text{.1}{\text{×10}}^{\text{2}}\text{cm}\right)\\ \text{=}\text{6}\text{.49}\text{mL}\text{.}\end{array}$

Calculate the volume flow rate

$\left(\frac{\text{6}\text{.49}\text{mL}}{\text{2}\text{.16}\text{min}}\right)\text{=}\text{3}\text{.0046}\text{=}\text{3}\text{.00}\text{mL/min}\text{.}$

The volume flow rate is $\text{3}\text{mL/min}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The retention factor of solute and the fraction of time spent in the stationary phase has to be calculated.

Concept introduction:

Retention factor:

The retention factor $\left(\text{k}\right)$ is given by

$\begin{array}{l}\text{k}\text{=}\frac{{\text{t}}_{\text{r}}\text{-}{\text{t}}_{\text{m}}}{{\text{t}}_{\text{m}}}\\ \text{Where,}\\ \text{k}\text{-}\text{retention}\text{factor}\\ {\text{t}}_{\text{r}}\text{-}\text{time}\text{required}\text{to}\text{elute}\text{the}\text{peak}\\ {\text{t}}_{\text{m}}\text{-}\text{time}\text{required}\text{for}\text{mobile}\text{phase}\text{to}\text{pass}\text{through}\text{the}\text{column}\text{.}\end{array}$

Answer to Problem 23.21P

• The retention factor of solute is $\text{7}\text{.02}\text{.}$
• The fraction of time spent by solute in the stationary phase is $\text{0}\text{.875}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Value}\text{of}{\text{t}}_{\text{r}}\text{is}\text{17}\text{.32}\text{min}\text{.}\\ \text{Value}\text{of}{\text{t}}_{\text{m}}\text{is}\text{2}\text{.16}\text{min}\text{.}\end{array}$

Calculate the retention factor of solute

$\begin{array}{l}\text{k}\text{=}\frac{{\text{t}}_{\text{r}}\text{-}{\text{t}}_{\text{m}}}{{\text{t}}_{\text{m}}}\text{=}\frac{\text{17}\text{.32}\text{-}\text{2}\text{.16}}{\text{2}\text{.16}}\\ \\ \text{=}\frac{\text{15}\text{.16}}{\text{2}\text{.16}}\text{=}\text{7}\text{.01852}\text{=}\text{7}\text{.02}\text{.}\end{array}$

The retention factor of solute is $\text{7}\text{.02}\text{.}$

Calculate the fraction of time spent by solute in stationary phase

$\begin{array}{l}\text{k}\text{=}\frac{{\text{t}}_{\text{s}}}{{\text{t}}_{\text{m}}}\left(\text{where}{\text{t}}_{\text{s}}\text{is}\text{the}\text{time}\text{in}\text{stationary}\text{phase}\right)\\ \text{Fraction}\text{of}\text{time}\text{=}\frac{{\text{t}}_{\text{s}}}{{\text{t}}_{\text{s}}\text{+}\text{}{\text{t}}_{\text{m}}}\text{=}\frac{{\text{kt}}_{\text{m}}}{{\text{kt}}_{\text{m}}\text{+}{\text{t}}_{\text{m}}}\text{=}\frac{\text{k}}{\text{k}\text{+}\text{}\text{1}}\\ \\ \text{=}\frac{\text{7}\text{.02}}{\text{7}\text{.02}\text{+}\text{1}}\text{=}\text{0}\text{.87531}\text{=}\text{0}\text{.875}\text{.}\end{array}$

The fraction of time spent by solute in the stationary phase is $\text{0}\text{.875}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The value of partition coefficient $\left(\text{K}\right)$ has to be calculated.

Concept introduction:

Partition coefficient:

Partition of a solute 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

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

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

Answer to Problem 23.21P

• The value of partition coefficient $\left(\text{K}\right)$ is $\text{295}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}\text{Column}\text{length}\text{is}\text{30}\text{.1}\text{m}\text{.}\\ \text{Thickness}\text{of}\text{the}\text{stationary}\text{phase}\text{is}\text{3}\text{.1}\text{µm}\text{.}\\ \text{Column}\text{inner}\text{diameter}\text{is}\text{0}\text{.530}\text{mm}\text{.}\\ \text{Unretained}\text{solute}\text{passes}\text{through}\text{in}\text{2}\text{.16}\text{min}\text{.}\\ \text{Particular}\text{solute}\text{passes}\text{through}\text{in}\text{17}\text{.32}\text{min}\text{.}\end{array}$

Calculate volume of stationary phase

$\begin{array}{l}\text{Volume}\text{of}\text{coating}\text{=}\text{2πr}\text{×}\text{thickness}\text{×}\text{length}\\ \text{=}\text{2}\left(\text{3}\text{.14}\right)\left[\left(\text{261}\text{.9}\text{+}\text{}\text{1}\text{.55}\right)\text{×}{\text{10}}^{\text{-4}}\right]\left(\text{3}\text{.1}\text{×}{\text{10}}^{\text{-4}}\text{cm}\right)\left(\text{30}\text{.1}\text{×}{\text{10}}^{\text{2}}\text{cm}\right)\\ \text{=}\text{0}\text{.15437}\text{=}\text{0}\text{.154}\text{mL}\text{.}\end{array}$

Calculate the value of partition coefficient

$\begin{array}{l}\text{k}\text{=}\text{K}\frac{{\text{V}}_{\text{s}}}{{\text{V}}_{\text{m}}}\\ \\ \text{K}\text{=}\left(\text{7}\text{.02}\right)\frac{\text{6}\text{.49}\text{mL}}{\text{0}\text{.154}\text{mL}}\text{=}\frac{\text{45}\text{.5598}}{\text{0}\text{.154}}\text{=}\frac{{\text{C}}_{\text{s}}}{{\text{C}}_{\text{m}}}\\ \\ \text{=}\text{295}\text{.8428}\text{=}\text{295}\text{.}\end{array}$

The value of partition coefficient $\left(\text{K}\right)$ is $\text{295}\text{.}$