Chapter 26, Problem 26.23QAP

Interpretation Introduction

(a)

Interpretation:

The time spends by the component A and B in the stationary phase is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The time spends by the component A and B in the stationary phase are $25\min$ and $45\min$ respectively.

Explanation of Solution

Draw the diagram for the chromatogram of a two component mixture with the time period.

Figure-(1)

The Figure (1) shows that the time spend by the component A in the stationary phase is $25\min$ and the time spend by the component in the stationary phase is $45\min$.

Thus, the time spends by the component A and B in the stationary phase are $25\min$ and $45\min$ respectively.

Interpretation Introduction

(b)

Interpretation:

The retention time for A and B are to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The retention time for the component A and B are $30\min$ and $50\min$ respectively.

Explanation of Solution

Write the expression for the retention time of component A.

$t_{\text{R}}=t_{\text{SA}}+t_{\text{M}}$   ...... (I)

Here, the time spend by the component A in stationary phase is $t_{\text{SA}}$ and the time spend in the mobile phase is $t_{\text{M}}$.

Write the expression for the retention time of component B.

$t_{\text{R}}=t_{\text{SB}}+t_{\text{M}}$   ...... (II)

Here, the time spend by the component B in stationary phase is $t_{\text{SB}}$

Draw the diagram for the mixture of two components to show the time spent by the components in the stationary phase and the mobile phase.

Figure-(2)

The Figure (2) shows that the time spend by the component A in the stationary phase is $25\min$, the time spend by the component in the stationary phase is $45\min$ and the time spent in the mobile phase is $5\min$.

Substitute $25\min$ for $t_{\text{SA}}$ and $5\min$ for $t_{\text{M}}$ in Equation (I).

$\begin{array}{c}{t}_{\text{R}}=25\min +5\min \\ =30\min \end{array}$

Thus, the retention time of the component A is $30\min$.

Substitute $45\min$ for $t_{\text{SB}}$ and $5\min$ for $t_{\text{M}}$ in Equation (II).

$\begin{array}{c}{t}_{\text{R}}=45\min +5\min \\ =50\min \end{array}$

Thus, the retention time of the component B is $50\min$.

Interpretation Introduction

(c)

Interpretation:

The retention factors for the two components are to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The retention factors for the component A and B are $5$ and $9$.

Explanation of Solution

Write the expression for the retention factor for the component A.

$k_{\text{A}}=\frac{t_{\text{SA}}}{t_{\text{M}}}$   ...... (III)

Write the expression for the retention factor for the component B.

$k_{\text{B}}=\frac{t_{\text{SB}}}{t_{\text{M}}}$   ...... (IV)

Substitute $25\min$ for $t_{\text{SA}}$ and $5\min$ for $t_{\text{M}}$ in Equation (III).

$\begin{array}{c}{k}_{\text{A}}=\frac{25\min }{5\min }\\ =5\end{array}$

Thus, the retention factor for the component A is $5$.

Substitute $45\min$ for $t_{\text{SB}}$ and $5\min$ for $t_{\text{M}}$ in Equation (IV).

$\begin{array}{c}{k}_{\text{B}}=\frac{45\min }{5\min }\\ =9\end{array}$

Thus, the retention factor for the component B is $9$.

Interpretation Introduction

(d)

Interpretation:

The full width of the each peak and full width at half maximum values are to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The full width for the component A and B are $18\min$ and $22\min$ respectively.

The half width for the component A and B are $9\min$ and $11\min$ respectively.

Explanation of Solution

Draw the diagram to show the full width and half width of both the components.

Figure-(3)

The Figure-(3) shows that the full width of the component A is $W_{\text{A}}=18\min$ and the full width for the component B is $W_{\text{B}}=22\min$.

Thus, the full width for the component A and B are $18\min$ and $22\min$ respectively.

Write the expression for the half width of the component A.

$W_{\text{A}1/2}=\frac{W_{\text{A}}}{2}$   ...... (V)

Write the expression for the half width of the component B.

$W_{\text{B}1/2}=\frac{W_{\text{B}}}{2}$   ...... (VI)

Substitute $18\min$ for $W_{\text{A}}$ in Equation (V).

$\begin{array}{c}{W}_{\text{A}1/2}=\frac{18\min }{2}\\ =9\min \end{array}$

Thus, the half width for the component A is $9\min$.

Substitute $22\min$ for $W_{\text{B}}$ in Equation (VI).

$\begin{array}{c}{W}_{\text{B}1/2}=\frac{22\min }{2}\\ =11\min \end{array}$

Thus, the half width for the component B is $11\min$.

Interpretation Introduction

(e)

Interpretation:

The resolution of two peaks is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The resolution of two peaks is $1$.

Explanation of Solution

Write the expression for the resolution of two peaks.

$R_{\text{S}}=\frac{2\left(t_{\text{RB}}-t_{\text{RA}}\right)}{W_{\text{A}}+W_{\text{B}}}$   ...... (VII)

Substitute $18\min$ for $W_{\text{A}}$, $22\min$ for $W_{\text{B}}$, $30\min$ for $t_{\text{RA}}$ and $50\min$ for $t_{\text{RB}}$ in Equation (VII).

$\begin{array}{c}{R}_{\text{S}}=\frac{2\left(50\min -30\min \right)}{18\min +22\min }\\ =\frac{2\left(20\min \right)}{40\min }\\ =\frac{40}{40}\\ =1\end{array}$

Thus, the resolution of two peaks is $1$.

Interpretation Introduction

(f)

Interpretation:

The average number of plates is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The average number of plates is $64$.

Explanation of Solution

Write the expression for the number of the plate for the component A.

$N_{\text{A}}=16{\left(\frac{t_{\text{RA}}}{W_{\text{A}}}\right)}^2$   ...... (VIII)

Write the expression for the number of the plate for the component B.

$N_{\text{B}}=16{\left(\frac{t_{\text{RB}}}{W_{\text{B}}}\right)}^2$   ...... (IX)

Write the expression for the average number of plates.

$N_{\text{avg}}=\frac{N_{\text{A}}+N_{\text{B}}}{2}$   ...... (X)

Substitute $30\min$ for $t_{\text{RA}}$ and $18\min$ for $W_{\text{A}}$ in Equation (VIII).

$\begin{array}{c}{N}_{\text{A}}=16{\left(\frac{30\min }{18\min }\right)}^2\\ =16\left(\frac{900}{324}\right)\\ =16\left(2.777\right)\\ =44.44\end{array}$

Substitute $50\min$ for $t_{\text{RB}}$ and $22\min$ for $W_{\text{B}}$ in Equation (IX).

$\begin{array}{c}{N}_{\text{B}}=16{\left(\frac{50\min }{22\min }\right)}^2\\ =16\left(\frac{2500}{484}\right)\\ =16\left(5.165\right)\\ =82.64\end{array}$

Substitute $82.64$ for $N_{\text{B}}$ and $44.44$ for $N_{\text{A}}$ in Equation (X).

$\begin{array}{c}{N}_{\text{avg}}=\frac{44.44+82.64}{2}\\ =\frac{127.08}{2}\\ =63.54\\ \approx 64\end{array}$

Thus, the average number of plates is $64$.

Interpretation Introduction

(g)

Interpretation:

The average height of the plate is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The average height of plate is $\text{0}\text{.40}\text{cm}$.

Explanation of Solution

Write the expression for the average height of the plate for the column.

$H_{\text{avg}}=\frac{L}{N_{\text{avg}}}$   ...... (XI)

Here, the length of the column is $L$.

Substitute $25\text{cm}$ for $L$ and $64$ for $N_{\text{avg}}$ in Equation (XI).

$\begin{array}{c}{H}_{\text{avg}}=\frac{25\text{cm}}{64}\\ =0.3906\text{cm}\\ \approx 0.40\text{cm}\end{array}$

Thus, the average height of plate is $\text{0}\text{.40}\text{cm}$.

Interpretation Introduction

(h)

Interpretation:

The length of the column to achieve resolution of $1.75$ is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The length of the column to achieve resolution of $1.75$ is $78.4\text{cm}$.

Explanation of Solution

Write the expression for the average number of plate at $1.75$ resolution.

$\begin{array}{l}\frac{{\left(R_{\text{S}}\right)}_1}{{\left(R_{\text{S}}\right)}_2}=\sqrt{\frac{N_1}{N_2}}\\ N_2={\left(\sqrt{N_1}\frac{{\left(R_{\text{S}}\right)}_2}{{\left(R_{\text{S}}\right)}_1}\right)}^2\\ N_2=N_1{\left(\frac{{\left(R_{\text{S}}\right)}_2}{{\left(R_{\text{S}}\right)}_1}\right)}^2\end{array}$   ...... (XII)

Here, the average number of plate at resolution 1 is $N_1$, the average number of plate at resolution 1.75 is $N_2$, the initial resolution is ${\left(R_{\text{S}}\right)}_1$ and the final given resolution is ${\left(R_{\text{S}}\right)}_2$.

Write the expression for the column length to achieve resolution of $1.75$.

$L=N_2{H}_{\text{avg}}$   ...... (XIII)

Substitute $1$ for ${\left(R_{\text{S}}\right)}_1$, $1.75$ for ${\left(R_{\text{S}}\right)}_2$ and $64$ for $N_1$ in Equation (XII).

$\begin{array}{c}{N}_2=64{\left(\frac{1.75}{1}\right)}^2\\ =64\left(3.0625\right)\\ =196\end{array}$

Substitute $196$ for $N_2$ and $0.40\text{cm}$ for $H_{\text{avg}}$ in Equation (XIII).

$\begin{array}{c}L=196\left(0.40\text{cm}\right)\\ =78.4\text{cm}\end{array}$

Thus, the length of the column to achieve resolution of $1.75$ is $78.4\text{cm}$.

Interpretation Introduction

(i)

Interpretation:

The time required to achieve resolution of $1.75$ is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The time required to achieve resolution of $1.75$ is $153\min$.

Explanation of Solution

Write the expression for the time required at $1.75$ resolution.

$\frac{{\left(R_{\text{S}}\right)}_1}{{\left(R_{\text{S}}\right)}_2}=\frac{{\left(t_{\text{R}}\right)}_1}{{\left(t_{\text{R}}\right)}_2}$   ...... (XIV)

Here, the time required to achieve the resolution 1 is ${\left(t_{\text{R}}\right)}_1$, the time required to achieve resolution 1.75 is ${\left(t_{\text{R}}\right)}_2$, the initial resolution is ${\left(R_{\text{S}}\right)}_1$ and the final given resolution is ${\left(R_{\text{S}}\right)}_2$.

Substitute $1$ for ${\left(R_{\text{S}}\right)}_1$, $1.75$ for ${\left(R_{\text{S}}\right)}_2$ and $50\min$ for ${\left(t_{\text{R}}\right)}_1$ in Equation (XIV).

$\begin{array}{l}{\left(\frac{1}{1.75}\right)}^2=\frac{50\min }{{\left(t_{\text{R}}\right)}_2}\\ {\left(t_{\text{R}}\right)}_2=\frac{50\min }{0.3264}\\ {\left(t_{\text{R}}\right)}_2=153.14\min \\ {\left(t_{\text{R}}\right)}_2\approx 153\min \end{array}$

Thus, the time required to achieve resolution of $1.75$ is $153\min$.

Interpretation Introduction

(j)

Interpretation:

The measures required to increase the resolution to achieve the baseline separation are to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The resolution of the peak is increased by increasing the retention time of the species.

Explanation of Solution

Write the expression for the resolution of the peak.

$R_{\text{S}}=\frac{\left[{\left(t_r\right)}_{\text{B}}-{\left(t_r\right)}_{\text{A}}\right]}{W_{\text{A}}+W_{\text{B}}}$   ...... (XV)

Here, the retention time for the species A is ${\left(t_r\right)}_{\text{A}}$, the retention time for the species B is ${\left(t_r\right)}_{\text{B}}$, the width of the peak A is $W_{\text{A}}$ and the width of the peak B is $W_{\text{B}}$.

The Equation (XV) shows that the retention time is directly proportional to the resolution of the peak. Hence, the retention time have to increase to increase the resolution of the peak.

Thus, the resolution of the peak is increased by increasing the retention time of the species.

Interpretation Introduction

(k)

Interpretation:

The measures to achieve a better separation in a shorter time with the same column as in part j is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Answer to Problem 26.23QAP

The on increasing the resolution keeping the length of the column constant the better separation in a shorter time with the same column as in part j is achieved.

Explanation of Solution

Write the expression for the average number of the plates.

$N_{\text{avg}}=16R_{\text{S}}^2{\left(\frac{\alpha }{\alpha -1}\right)}^2\left(\frac{1+K_{\text{B}}}{K_{\text{B}}}\right)$   ...... (XVI)

Here, the resolution of the peaks is $R_{\text{S}}$, the average number of the plates is $N_{\text{avg}}$, the selectively factor is $\alpha$ and the retention factor for the solute B is $k_{\text{B}}$.

The Equation (XVI) shows that the average number of the plates is directly proportional to the square of the resolution. Hence, the average number of the plates increases on increasing the resolution keeping the length of the column constant. Due to this the better separation in a shorter time with the same column as in part j may be achieved.

Thus, the on increasing the resolution keeping the length of the column constant the better separation in a shorter time with the same column as in part j is achieved.