Question

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?

Short answer

For the part${}^{\left(a\right)}$, the length of the columns are${}^{\mathbf{L}\mathbf{=}\mathbf{33}\mathbf{cm}\mathbf{,}\mathbf{}\mathbf{17}\mathbf{cm}\mathbf{,}\mathbf{}\mathbf{10}\mathbf{cm}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{5}\mathbf{cm}\mathbf{}}$ respectively.

For the part${}^{\left(b\right)}$,${}^{{\mathbf{t}}_{\mathbf{r}}\mathbf{=}\mathbf{10}\mathbf{min}\mathbf{,}\mathbf{}\mathbf{6}\mathbf{}\mathbf{min}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{3}\mathbf{}\mathbf{min}}$ respectively.

For the part ${}^{\left(c\right)}$, the pressure${}^{\mathbf{p}\mathbf{=}\mathbf{}\mathbf{18}\mathbf{MPa}\mathbf{,}\mathbf{49}\mathbf{MPa}\mathbf{,}\mathbf{}\mathbf{196}\mathbf{MPa}}$ respectively.

For the part ${}^{\left(d\right)}$, the baseline width is

$\mathbf{w}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{mL}\mathbf{;}\mathbf{}\mathbf{w}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{24}\mathbf{min}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{48}\mathbf{mL}\mathbf{;}\mathbf{}\mathbf{w}\mathbf{=}\mathbf{o}\mathbf{.}\mathbf{12}\mathbf{min}\mathbf{,}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{mL}$respectively.

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

Step-by-step solution

Calculating the column length of the stationary phase particles.

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,

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

Calculating the retention time of the stationary phase particles.

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 ${}^{\mathbf{15}\mathbf{\%}}$

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

Calculating the pressure of the stationary phase particles.

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

Calculating the pressure of the stationary phase particles.

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 ${}^{{\mathbf{p}}_{\mathbf{2}}}$.

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

Calculating the baseline width in time unit and volume unit of the particles.

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,

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

Identifying the size of the particle that requires the UHPLC instrument.

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