1. Use equation ${}^{25-1}$to estimate the length of a column required to achieve${}^{1.0\times {10}^4}$plates if the stationary phase particles size is${}^{10.5,5.0,3.0,\mathrm{or}1.5\mathrm{\mu m}}$

2. If the retention time was ${}^{20}$mins on the ${}^{10.0\mathrm{\mu m}}$ particle size column, what is the retention time on the ${}^{5.0,3.0,\mathrm{or}1.5\mathrm{\mu m}}$columns from part (a)? Assume that flow rate is constant for all columns.

3. Use equation${}^{25-2}$to estimate the pressure of the column in (a) given that the pressure of the${}^{10.0\mathrm{\mu m}}$column was${}^{4.4\mathrm{Mpa}}$

4. If the flow rate was${}^{{}^{2.0\mathrm{mL}/\min }}$ , what is the baseline width for the peaks on ${}^{10.5,5.0,3.0,\mathrm{or}1.5\mathrm{\mu m}}$columns form part (a)?

5. Which of these column configurations would require a UHPLC instrument?

The equation ${}^{\mathbf{25}\mathbf{-}\mathbf{1}}$gives us the below,

$\mathbf{N}\mathbf{=}\frac{\mathbf{3000}\mathbf{\times }\mathbf{L}}{{\mathbf{d}}_{\mathbf{p}}}$

We have express as below,

$\mathbf{L}\mathbf{=}\frac{\mathbf{N}\mathbf{\times }{\mathbf{d}}_{\mathbf{p}}}{\mathbf{3000}}$

When the stationary phase particle size is${}^{\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$ , the length of the column is,

role="math" localid="1655028224449" $$\begin{gathered} \mathbf{L}\mathbf{=}\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\times }\mathbf{10}}{\mathbf{3000}} \\ \mathbf{L}\mathbf{=}\mathbf{33}\mathbf{cm} \end{gathered}$$

When the stationary phase particle size is${}^{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$ , the length of the column is,

$$\begin{gathered} \mathbf{L}\mathbf{=}\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\times }\mathbf{5}}{\mathbf{3000}} \\ \mathbf{L}\mathbf{=}\mathbf{17}\mathbf{cm} \end{gathered}$$

When the stationary phase particle size is${}^{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$ , the length of the column is,

$$\begin{gathered} \mathbf{L}\mathbf{=}\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\times }\mathbf{3}}{\mathbf{3000}} \\ \mathbf{L}\mathbf{=}\mathbf{10}\mathbf{cm} \end{gathered}$$

When the stationary phase particle size is${}^{\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\mu m}}$ , the length of the column is,

$$\begin{gathered} \mathbf{L}\mathbf{=}\frac{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{5}}{\mathbf{3000}} \\ \mathbf{L}\mathbf{=}\mathbf{5}\mathbf{cm} \end{gathered}$$

Since the flow rate is constant for all columns, the retention time decreases proportionally with the size of the particle.

The given is${}^{\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$particle has the retention time of${}^{\mathbf{20}}$ mins. So the ${}^{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$particle has${}^{\mathbf{50}\mathbf{\%}}$ less retention time.

${\mathbf{t}}_{\mathbf{r}}\mathbf{=}\mathbf{20}\mathbf{min}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{=}\mathbf{10}\mathbf{min}$

The${}^{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$ particle would have ${}^{\mathbf{70}\mathbf{\%}}$less retention time ${}^{\mathbf{30}\mathbf{\%}}$

${\mathbf{t}}_{\mathbf{r}}\mathbf{=}\mathbf{20}\mathbf{min}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{=}\mathbf{3}\mathbf{min}$

The ${}^{\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\mu m}}$particle would have ${}^{\mathbf{85}\mathbf{\%}}$less retention time ${}^{\mathbf{15}\mathbf{\%}}$

${\mathbf{t}}_{\mathbf{r}}\mathbf{=}\mathbf{20}\mathbf{min}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{=}\mathbf{3}\mathbf{min}$

The equation ${}^{\mathbf{25}\mathbf{-}\mathbf{2}}$gives us the below equation,

$\mathbf{p}\mathbf{=}\mathbf{f}\mathbf{\times }\frac{{\mathbf{u}}_{\mathbf{v}}\mathbf{\times }\mathbf{\eta }\mathbf{\times }\mathbf{L}}{\mathbf{\pi }\mathbf{\times }{\mathbf{r}}^{\mathbf{2}}\mathbf{\times }{{\mathbf{d}}_{\mathbf{p}}}^{\mathbf{2}}}$

From the above formula, we can know that the pressure is inversely proportional to the square of the size of the particle.

The given is, for the${}^{\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$particle${}^{\left(d_1\right)}$has the pressure${}^{\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{MPa}\left(p_1\right)}$

So the particle ${}^{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\mu m}\mathbf{}\left(d_2\right)}$will have the pressure ${}^{{\mathbf{p}}_{\mathbf{2}}}$.

role="math" localid="1655029638610" $$\begin{gathered} \frac{{\mathbf{p}}_{\mathbf{1}}}{{\mathbf{p}}_{\mathbf{2}}}\mathbf{=}\frac{{{\mathbf{d}}_{\mathbf{2}}}^{\mathbf{2}}}{{{\mathbf{d}}_{\mathbf{1}}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{p}}_{\mathbf{1}}\mathbf{\times }{{\mathbf{d}}_{\mathbf{1}}}^{\mathbf{2}}}{{{\mathbf{d}}_{\mathbf{2}}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{\left(}\mathbf{10}\mathbf{\mu m}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\times }\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{MPa}}{{\mathbf{\left(}\mathbf{5}\mathbf{\mu m}\mathbf{\right)}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{18}\mathbf{MPa} \end{gathered}$$

So the particle ${}^{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\mu m}\mathbf{}\mathbf{\left(}{\mathbf{d}}_{\mathbf{2}}\mathbf{\right)}}$will have the pressure ${}^{{\mathbf{p}}_{\mathbf{2}}}$.

$$\begin{gathered} \frac{{\mathbf{p}}_{\mathbf{1}}}{{\mathbf{p}}_{\mathbf{2}}}\mathbf{=}\frac{{{\mathbf{d}}_{\mathbf{2}}}^{\mathbf{2}}}{{{\mathbf{d}}_{\mathbf{1}}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{p}}_{\mathbf{1}}\mathbf{\times }{{\mathbf{d}}_{\mathbf{1}}}^{\mathbf{2}}}{{{\mathbf{d}}_{\mathbf{2}}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{\left(}\mathbf{10}\mathbf{\mu m}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\times }\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{MPa}}{{\mathbf{\left(}\mathbf{3}\mathbf{\mu m}\mathbf{\right)}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{49}\mathbf{MPa} \end{gathered}$$

So the particle ${}^{\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\mu m}\mathbf{}\mathbf{\left(}{\mathbf{d}}_{\mathbf{2}}\mathbf{\right)}}$will have the pressure ${}^{{\mathbf{p}}_{\mathbf{2}}}$.

$$\begin{gathered} \frac{{\mathbf{p}}_{\mathbf{1}}}{{\mathbf{p}}_{\mathbf{2}}}\mathbf{=}\frac{{{\mathbf{d}}_{\mathbf{2}}}^{\mathbf{2}}}{{{\mathbf{d}}_{\mathbf{1}}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{p}}_{\mathbf{1}}\mathbf{\times }{{\mathbf{d}}_{\mathbf{1}}}^{\mathbf{2}}}{{{\mathbf{d}}_{\mathbf{2}}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{\left(}\mathbf{10}\mathbf{\mu m}\mathbf{\right)}}^{\mathbf{2}}\mathbf{\times }\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{MPa}}{{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\mu m}\mathbf{)}}^{\mathbf{2}}} \\ \mathbf{}{\mathbf{p}}_{\mathbf{2}}\mathbf{=}\mathbf{196}\mathbf{MPa} \end{gathered}$$

The baseline width can be calculated by using the below formula,

$\mathbf{N}\mathbf{=}\frac{\mathbf{16}\mathbf{\times }{{\mathbf{t}}_{\mathbf{r}}}^{\mathbf{2}}}{{\mathbf{\omega }}^{\mathbf{2}}}$

The baseline width in time unit is,

$\mathbf{\omega }\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{N}}}\mathbf{\times }\mathbf{4}\mathbf{\times }{\mathbf{t}}_{\mathbf{r}}$

To calculate the baseline width in volume unit, we need to multiply${}^{\mathbf{\omega }}$ with flow rate.

For a${}^{\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$ particle, the baseline width in time unit is,

role="math" localid="1655030126576" $$\begin{gathered} \mathbf{\omega }\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}}}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{20}\mathbf{min} \\ \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{min} \end{gathered}$$

The same baseline width in volume unit is,

role="math" localid="1655030532545" $$\begin{gathered} \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{min}\mathbf{\times }\mathbf{2}\mathbf{mL}\mathbf{/}\mathbf{min} \\ \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{16}\mathbf{mL} \end{gathered}$$

For a${}^{\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$ particle, the baseline width in time unit is,

$$\begin{gathered} \mathbf{\omega }\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}}}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{10}\mathbf{min} \\ \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{min} \end{gathered}$$

The same baseline width in volume unit is,

$$\begin{gathered} \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{min}\mathbf{\times }\mathbf{2}\mathbf{mL}\mathbf{/}\mathbf{min} \\ \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{mL} \end{gathered}$$

For a${}^{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\mu m}}$ particle, the baseline width in time unit is,

$$\begin{gathered} \mathbf{\omega }\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}}}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{6}\mathbf{min} \\ \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{min} \end{gathered}$$

The same baseline width in volume unit is,

$$\begin{gathered} \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{min}\mathbf{\times }\mathbf{2}\mathbf{mL}\mathbf{/}\mathbf{min} \\ \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{48}\mathbf{mL} \end{gathered}$$

For a${}^{\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\mu m}}$ particle, the baseline width in time unit is,

$$\begin{gathered} \mathbf{\omega }\mathbf{=}\sqrt{\frac{\mathbf{1}}{\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}}}\mathbf{\times }\mathbf{4}\mathbf{\times }\mathbf{3}\mathbf{min} \\ \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{min} \end{gathered}$$

The same baseline width in volume unit is,

$$\begin{gathered} \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{12}\mathbf{min}\mathbf{\times }\mathbf{2}\mathbf{mL}\mathbf{/}\mathbf{min} \\ \mathbf{\omega }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{mL} \end{gathered}$$

For the part${}^{\left(e\right)}$,${}^{\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{\mu m}}$ particle would require UHPLC instrument.