Question

Two compounds with partition coefficients of 15 and 18 are to be

separated on a column with ${\mathbf{V}}_{\mathbf{mi}}\mathbf{/}{\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}$and localid="1654950413291" ${\mathbf{t}}_{\mathbf{m}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{min}$. Calculate the

number of theoretical plates needed to produce a resolution of 1.5 .

Short answer

The solution is

To produce a resolution of 1.5,N is 1296 plates

Step-by-step solution

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We use the supplied partition coefficients K , $\frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}$, and ${\mathrm{t}}_{\mathrm{m}}$to compute for the retention

factor k to find the number of theoretical plates N required to achieve a resolution of 1.5 . ${\mathrm{t}}_{\mathrm{r}1}$and${\mathrm{t}}_{\mathrm{r}2}$are calculated using the retention factor. The unadjusted relative retention durations would be computed for N using the ${\mathrm{t}}_{\mathrm{r}}$values that have been solved.

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It is given

$$\begin{gathered} {\mathrm{K}}_1:15 \\ {\mathrm{K}}_2:18 \\ \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}:3.0 \\ {\mathrm{t}}_{\mathrm{m}:1.0\min } \\ \mathrm{Resolution}:1.5 \\ \mathrm{Resolution}=\frac{\sqrt{\mathrm{N}}}{4}\left(\mathrm{Y}-1\right) \end{gathered}$$

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To solve for the at a given resolution, we use equation : 1

$\mathrm{Resolution}=\frac{\sqrt{\mathrm{N}}}{4}\left(\mathrm{Y}-1\right)$

Equation l is used to find the N at a certain resolution, where N is the number of theoretical plates and $\mathrm{\gamma }$is the uncorrected relative retention.

To apply equation 1, we must first solve equation 2 for $\mathrm{\gamma }$:

$\mathrm{Y}=\frac{{\mathrm{t}}_{\mathrm{r}2}}{{\mathrm{t}}_{\mathrm{r}2}}........\left(2\right)$

where $\mathrm{\gamma }$ is the uncorrected relative retention and ${\mathrm{t}}_{\mathrm{r}}1$is the compound's retention duration. ${\mathrm{t}}_{\mathrm{r}}1$is the compound's retention time.

Because the values of ${\mathrm{t}}_{\mathrm{r}}$ and $\mathrm{Y}$ are unknown, we must first obtain k from k using equation3:

$\mathrm{k}=\mathrm{K}\frac{{\mathrm{V}}_{\mathrm{S}}}{{\mathrm{V}}_{\mathrm{m}}}$

Where k is the retention factor,

k is the partition coefficient,

${\mathrm{V}}_{\mathrm{m}}$is the volume of the mobile phase, and ${\mathrm{V}}_{\mathrm{s}}$is the volume of the stationary phase.

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When we plug in the known values of k and $\frac{{\mathrm{V}}_{\mathrm{S}}}{{\mathrm{V}}_{\mathrm{m}}}$into equation 3, we get:

$$\begin{gathered} {\mathrm{k}}_1={\mathrm{K}}_1\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{m}}} \\ =\left(15\right)\left(3.0^{-1}\right) \\ =5.0 \\ {\mathrm{k}}_2={\mathrm{K}}_2\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{m}}} \\ =\left(18\right)\left(3.0^{-1}\right) \\ =6.0 \end{gathered}$$

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We get: by rearranging equation 4 and inserting in the known values:

$\mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}}$

Where k is the retention ${\mathrm{factor}}_{\mathrm{r}}$ is the retention time of the compound, and

${\mathrm{t}}_{\mathrm{m}}$is the retention time of the mobile phase.

Rearrange equation 4, and plugging the known values, we get:

$$\begin{gathered} {\mathrm{k}}_1=\frac{{\mathrm{t}}_{\mathrm{r}1}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ {\mathrm{k}}_1=\frac{{\mathrm{t}}_{\mathrm{r}1}}{{\mathrm{t}}_{\mathrm{m}}}-1 \\ {\mathrm{t}}_{\mathrm{r}1}=\left({\mathrm{k}}_1+1\right)\left({\mathrm{t}}_{\mathrm{m}}\right) \\ =\left(5.0+1\right)\left(1.0\min \right) \\ =6.0\min \\ {\mathrm{k}}_2=\frac{{\mathrm{t}}_{\mathrm{r}2}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ {\mathrm{k}}_2=\frac{{\mathrm{t}}_{\mathrm{r}2}}{{\mathrm{t}}_{\mathrm{m}}}-2 \\ {\mathrm{t}}_{\mathrm{r}2}=\left({\mathrm{k}}_2+1\right)\left({\mathrm{t}}_{\mathrm{m}}\right) \\ =\left(6.0+1\right)\left(1.0\min \right) \\ =7.0\min \end{gathered}$$

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Now that we know both ${\mathrm{t}}_{\mathrm{r}}$numbers, we can insert them into equation 2 to get Y :

$$\begin{gathered} \mathrm{Y}=\frac{{\mathrm{t}}_{\mathrm{r}2}}{{\mathrm{t}}_{\mathrm{r}1}} \\ =\frac{7.0}{6.0} \\ =1.1667 \end{gathered}$$

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We get N : Y by rearranging equation l and plugging Y into the equation.

$$\begin{gathered} \mathrm{N}={\left[\frac{4\left(\mathrm{Resolution}\right)}{\mathrm{Y}-1}\right]}^2 \\ ={\left[\frac{\left(4\right)\left(1.5\right)}{1.1667-1}\right]}^2=1296 \end{gathered}$$

To produce a resolution of 1.5,N is 1296 plates