Chapter 26, Problem 26.16QAP

Interpretation Introduction

(a)

Interpretation:

The resolution for species B and C from the given data should be determined.

Concept introduction:

The resolution of the column is defined as the separation of two species of the column.

Answer to Problem 26.16QAP

The resolution is $2.01$.

Explanation of Solution

Given:

The expression of resolution of the column is:

$R_S=\frac{2\left[{\left(t_R\right)}_C-{\left(t_R\right)}_B\right]}{W_C+W_B}$ ....... (I)

Here, the retention time of species $C$ is ${\left(t_R\right)}_C$, the retention time of species $B$ is ${\left(t_R\right)}_B$, the width time of the peak base of the species $C$ is $W_C$ and the width time of the peak base of the species $B$ is $W_B$.

Substitute $15.7\min$ for ${\left(t_R\right)}_B$, $13.3\min$ for ${\left(t_R\right)}_C$, $1.07\min$ for $W_C$ and $1.32\min$ for $W_D$ in Equation (I).

$\begin{array}{c}{R}_S=\frac{2\left[\left(15.7\min \right)-\left(13.3\min \right)\right]}{\left(1.07\min \right)+\left(1.32\min \right)}\\ =\frac{2\left(2.4\min \right)}{\left(2.39\min \right)}\\ =2.01\end{array}$

Thus, the resolution is $2.01$.

Interpretation Introduction

(b)

Interpretation:

The selectivity factor for species B and C from the given data should be determined.

Concept introduction:

The resolution of the column is defined as the separation of two species of the column.

Answer to Problem 26.16QAP

The selectivity factor is $1.24$.

Explanation of Solution

The expression of selectivity factor is:

$a_{C,B}=\frac{{\left(t_R\right)}_C-t_M}{{\left(t_R\right)}_B-t_M}$ ........ (II)

Here, the non-retained retention time is $t_M$.

Substitute $15.7\min$ for ${\left(t_R\right)}_C$, $13.3\min$ for ${\left(t_R\right)}_B$ and $3.1\min$ for $t_M$ in Equation (II).

$\begin{array}{c}{a}_{C,B}=\frac{\left(15.7\min \right)-\left(3.1\min \right)}{\left(13.3\min \right)-\left(3.1\min \right)}\\ =\frac{12.6\min }{10.2\min }\\ =1.24\end{array}$

Thus, the selectivity factor is $1.24$.

Interpretation Introduction

(c)

Interpretation:

The length of column necessary to separate B and C species with a resolution of $2.5$ should be determined.

Concept introduction:

The resolution of the column is defined as the separation of two species of the column.

Answer to Problem 26.16QAP

The length of column necessary to separate the two species with a resolution of $2.5$ is $37.98\text{cm}$.

Explanation of Solution

The expression of length of the column is:

$L=N_{2}H$ ....... (III)

Here, the number of plates needed to separate the two species is $N$ and the height of the plate is $H$.

The expression of relation of the resolution and number of plates is:

$\begin{array}{l}\frac{\left(R_S\right)}{{\left(R_S\right)}_2}=\frac{\sqrt{N_1}}{\sqrt{N_2}}\\ \sqrt{N_2}=\frac{{\left(R_S\right)}_2\sqrt{N_1}}{\left(R_S\right)}\\ N_2={\left\{\frac{{\left(R_S\right)}_2\sqrt{N_1}}{\left(R_S\right)}\right\}}^2\\ N_2=\frac{N_1{\left(R_S\right)}_2^2}{{\left(R_S\right)}^2}\end{array}$

Here, the number of plates needed is $N_1$ and the given resolution is ${\left(R_S\right)}_2$.

Substitute $\frac{N_1{\left(R_S\right)}_2^2}{{\left(R_S\right)}^2}$ for $N_2$ in Equation (III).

$L=\left\{\frac{N_1{\left(R_S\right)}_2^2}{{\left(R_S\right)}^2}\right\}H$ ......... (IV)

Substitute $2.01$ for $\left(R_S\right)$, $2.5$ for ${\left(R_S\right)}_2$, $2455$ for $N_1$ and $0.01\text{cm}$ for $H$ in Equation (IV).

$\begin{array}{c}L=\left\{\frac{\left(2455\right){\left(2.5\right)}^2}{{\left(2.01\right)}^2}\right\}\left(0.01\text{cm}\right)\\ =\left(3797.86\right)\left(0.01\text{cm}\right)\\ =37.98\text{cm}\end{array}$

Thus, the length of column necessary to separate the two species with a resolution of $2.5$ is $37.98\text{cm}$.

Interpretation Introduction

(d)

Interpretation:

The time required to separate B and C species on the column of part c should be determined.

Concept introduction:

The resolution of the column is defined as the separation of two species of the column.

Answer to Problem 26.16QAP

The time required to separate the two species on the column of part c is $24.29\min$.

Explanation of Solution

The expression of the relation of time required to separate the two species on the column is:

$\frac{{\left(t_R\right)}_C}{\left(t_R\right)}=\frac{{\left(R_s\right)}^2}{{\left(R_S\right)}_1{}^2}$ ........ (V)

Here, the given resolution is ${\left(R_S\right)}_1$ and time required to separate the two species on the column is $t_R$.

Substitute $15.7\min$ for ${\left(t_R\right)}_C$, $2.01$ for $R_S$ and $2.5$ for ${\left(R_S\right)}_1$ in Equation (V).

$\begin{array}{l}\frac{\left(15.7\min \right)}{\left(t_R\right)}=\frac{{\left(2.01\right)}^2}{{\left(2.5\right)}^2}\\ t_R=\frac{\left(15.7\min \right){\left(2.5\right)}^2}{{\left(2.01\right)}^2}\\ t_R=24.29\min \end{array}$

Thus, the time required to separate the two species on the column of part c is $24.29\min$.