This problem reviews concepts from Chapter 23. An unretained solute passes through a chromatography column in 3.7 min and analyte requires 8.4 min.

(a) Find the adjusted retention time and retention factor for the analyte.

(b) Find the phase ratio b for a 0.32-mm-diameter column with a 1.0-mm-thick film of stationary phase.

(c) Find the partition coefficient for the analyte.

(d) Determine the retention time on a similar length of 0.32-mm diameter column with a 0.5-mm-thick film of the same stationary phase at the same temperature.

Compute the adjusted retention time$({t_r}^{\text{'}})$by applying the formula below:

${{\mathrm{t}}_{\mathrm{r}}}^{\text{'}}={\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}$

8.4min-3.7 min

4.7 min

The retention factor is the time required to elute that peak minusthe time from mobile phase to pass through the column, expressed in multiples of${\mathrm{t}}_{\mathrm{m}}$. Calculate the retention factor using the formula below:

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ =\frac{8.4\min -3.7\min }{3.7\min } \end{gathered}$$

GIVEN:

- Diameter of column =$0.32\mathrm{mm}$

- Thickness of stationary phase =$1.0\mathrm{\mu m}$

Find: phase ratiorole="math" localid="1654785803260" $\beta$

The phase ratio role="math" localid="1654785811025" $\beta$is the volume of the mobile phase divided by the volume of stationary phase. It can be deduced to the following expression:

$\mathrm{\beta }=\frac{\mathrm{r}}{2{\mathrm{d}}_{\mathrm{f}}}$

where:

$$\begin{gathered} -\mathrm{r}=\mathrm{column}\mathrm{ratio} \\ -{\mathrm{d}}_{\mathrm{f}}=\mathrm{stationary}\mathrm{fase}\mathrm{film}\mathrm{thickness} \end{gathered}$$

Solve for the phase ratio:

$\mathrm{\beta }=\frac{0.32/2\mathrm{mm}}{2\left(1.0\mathrm{\mu m}\right)\left(\begin{array}{c}1\mathrm{mm}\\ 1000\mathrm{\mu m}\end{array}\right)}=80$

Find: partition coefficient$\left(\mathrm{K}\right)$of the analyte

The retention factor $\left(\mathrm{k}\right)$is related to the partition coefficient role="math" localid="1654783983446" $\left(\mathrm{K}\right)$and phase ratio $\left(\mathrm{\beta }\right)$, based on the following expression:

$k=\frac{K}{\beta }$

Thus, the partition coefficient $\left(K\right)$can be calculated by rearranging the expression:

$\mathrm{K}=\mathrm{k}\times \mathrm{\beta }$

$$\begin{gathered} =1.27\times 80 \\ =101.6 \end{gathered}$$

Given:

- Diameter of column $=0.32\mathrm{mm}$

- Thickness of stationary phase$=0.5\mathrm{\mu m}$

Find: retention time$({\mathrm{t}}_{\mathrm{r}})$of the same stationary phase and at the same temperature

Based on the formula of phase ratio role="math" localid="1654785312894" $\left(\beta \right)$, decreasing the thickness of the stationary phase, increases the role="math" localid="1654785292964" $\beta$.

Calculating the new phase ratio role="math" localid="1654784786511" $\left({\beta }^{·}\right)$:

role="math" localid="1654785571991" ${\beta }^{\text{'}}=\frac{0.32/2mm}{2\left[0.5\mu m\right]\left(\begin{array}{c}1mm\\ 1000\mu m\end{array}\right)}=160$

$\mathrm{The}\mathrm{new}\mathrm{phase}\mathrm{ratio}\left({\beta }^{\text{'}}\right)\mathrm{is}2\mathrm{times}\mathrm{the}\mathrm{original}\mathrm{value}.$

${\beta }^{\text{'}}=2\beta$

$$\begin{gathered} \mathrm{Thus},\mathrm{the}\mathrm{retention}\mathrm{time}\left({\mathrm{t}}_{\mathrm{m}}\right)\mathrm{is}\mathrm{expected}\mathrm{to}\mathrm{decrease}. \\ \mathrm{Then},\mathrm{compute}\mathrm{for}\mathrm{the}\mathrm{new}\mathrm{retention}\mathrm{factor}\left({\mathrm{k}}^{\text{'}}\right): \end{gathered}$$

${\mathrm{k}}^{\text{'}}=\frac{\mathrm{K}}{{\beta }^{\mathit{\text{'}}}}$

thus,

${\mathrm{k}}^{\text{'}}=\frac{\mathrm{K}}{2\mathrm{\beta }}=\frac{1}{2}\left(1.27\right)=0.635$

Lastly, compute for the retention time $\left(t_r\right)$:

$$\begin{gathered} {\mathrm{k}}^{\text{'}}=0.635=\frac{{\mathrm{t}}_{\mathrm{r}}-3.7}{3.7} \\ {\mathrm{t}}_{\mathrm{r}}=6.05\min \end{gathered}$$

$$\begin{gathered} a)k=1.27 \\ b)\beta =80 \\ c)K=101.6 \\ D)t_r=6.05min \end{gathered}$$