Question

An open tubular column has an inner diameter of $\mathbf{207}\mathbf{\mu m}$and the thickness of the stationary phase on the inner wall is $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{\mu m}$. Unretained solute passes through in 63s and a particular solute emerges in 433s . Find the partition coefficient for this solute and find the fraction of time spent in the stationary phase.

Short answer

The solution is

Fraction of time spent in the stationary phase is 0.854

Step-by-step solution

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In this task, we have a tubular column with an inner diameter of $207\mathrm{\mu m}$and a stationary phase thickness of $0.50\mathrm{\mu m}$on the inner wall. We'll calculate the partition coefficient for this solute and the fraction of time spent in the stationary phase because the unretained liquid passes through in 63s and a specific solute emerges in 433s .

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It is given

Mobile phase

$$\begin{gathered} {\mathrm{d}}_{\mathrm{inner}}=207\mathrm{\mu m}-2\mathrm{thickness}=206\mathrm{\mu m} \\ {\mathrm{r}}_{\mathrm{inner}}={\mathrm{d}}_{\mathrm{inner}}/2=103\mathrm{\mu m} \end{gathered}$$

Stationary phase

$$\begin{gathered} {\text{d}}_{\mathrm{outer}}={\mathrm{d}}_{\mathrm{inner}}+\mathrm{thickness}=206.5\mathrm{\mu m} \\ {r}_{outer}=d_{outer}/2=103.25\mu m \\ {t}_m=63s \\ {t}_r=433s \end{gathered}$$

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First we will calculate the retention for this solute:

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{t}}_{\mathrm{r}}-{\mathrm{t}}_{\mathrm{m}}}{{\mathrm{t}}_{\mathrm{m}}} \\ \mathrm{k}=\frac{433\mathrm{s}-63\mathrm{s}}{63\mathrm{s}} \\ \mathrm{k}=5.87 \end{gathered}$$

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We know the partition coefficient equation is as follows:

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

we know the value of , thus we'll figure out the value of $\frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}$

$\frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{{\mathrm{\pi r}}^2\times \mathrm{I}}{2\mathrm{\pi r}\times \mathrm{thickness}\times \mathrm{I}}$

Remove the values that are repeated presently.

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{\cancel{\mathrm{\pi }}{\mathrm{r}}^2\times \cancel{\mathrm{i}}}{2\cancel{\mathrm{\pi }}\mathrm{r}\times \mathrm{thickness}\times \cancel{\mathrm{i}}} \\ \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{{\mathrm{r}}_{\mathrm{inner}}^2}{2{\mathrm{r}}_{\mathrm{outer}}\times \mathrm{thickness}} \\ \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{\left(103\mathrm{\mu m}\right)}{2\times 103.25\mathrm{\mu m}\times 0.5\mathrm{\mu m}} \\ \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}}=102.8 \end{gathered}$$

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The partition coefficient can then be calculated as follows:

$$\begin{gathered} \mathrm{K}=\mathrm{k}\times \frac{{\mathrm{V}}_{\mathrm{m}}}{{\mathrm{V}}_{\mathrm{s}}} \\ \mathrm{K}=5.87\times 102.8 \\ \mathrm{K}=603 \end{gathered}$$

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We'll also figure out how much time we spent in the stationary phase:

$$\begin{gathered} \mathrm{Fraction}=\frac{{\mathrm{t}}_{\mathrm{s}}}{{\mathrm{t}}_{\mathrm{s}}+{\mathrm{t}}_{\mathrm{m}}} \\ \mathrm{Fraction}=\frac{\mathrm{k}\times {\mathrm{t}}_{\mathrm{m}}}{\mathrm{k}\times {\mathrm{t}}_{\mathrm{m}}+{\mathrm{t}}_{\mathrm{m}}} \\ \mathrm{Fraction}=\frac{\mathrm{k}}{\mathrm{k}+1} \\ \mathrm{Fraction}=\frac{5.87}{5.87+1} \\ \mathrm{Fraction}=0.854 \end{gathered}$$