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Chapter 23: QBE (page 628)

Consider a chromatography experiment in which two components with retention factors $k_{1} = 4.00$ and $k_{2} = 5.00$ are injected into column with $N = 1.00 脳 10^{3}$ theoretical plates. The retention time for the less-retained component is $t r_{1} = 10.0 m i n$ .
(a)Calculate $t_{m}$ and $t_{12}$ .Find $w_{\frac{1}{2}}$ (width at half height) and w (at the base) for each peak.
(b)Using graph paper , sketch the chromatogram analogous to figure 23-7,supposing that two peaks have same amplitude(height). Draw the half widths accurately.
(c)Calculate the resolution of the two peaks and compare this value with those drawn in Figure 23-10.

Short Answer

Expert verified

(a) $t_{m} = 2.00 m i n, t_{r 2} = 12.00 m i n$

are 1.265 min and 1.518 min.

$w_{\frac{1}{2}}$ (for peak1 ) and $w_{\frac{1}{2}}$ (for peak2) are 0.745 min and 0.894 min

(b) Resolution =1.437

Step by step solution

01

Step1:

Finding the value of, $t_{m}, t_{r 2}$ w and $W_{\frac{1}{2}}$

We know that $k_{1} = \frac{t_{r 1} - t_{m}}{t_{m}}$ or, $t_{m} = \frac{t_{r 1}}{k_{1} + 1}$

Or, $t_{m} = \frac{10.00}{4 + 1} = 2.00 m i n$

We know that , $t_{r 2} = t_{m} (K_{2} + 1)$ Or, $t_{r 2} = 2.00 (5.00 + 1) = 12 m i n$

Now $N = \frac{5.55 脳 t_{r}^{2}}{{w^{2}}_{\frac{1}{2}}}$ Or, $w_{\frac{1}{2}} = \sqrt{\frac{5.55 脳 t_{r}^{2}}{N}}$

For $w_{\frac{1}{2}} = \sqrt{\frac{5.55 脳 {(10)}^{2}}{1.00 脳 10^{3}}} = 0.745 m i n$

$w_{\frac{1}{2}} = \sqrt{\frac{5.55 脳 {(12)}^{2}}{1.00 脳 10^{3}}} = 0.894 m i n$

$N = \frac{16 t_{r}^{2}}{w}$ Or, $w = \sqrt{\frac{16 脳 t_{r}^{2}}{N}}$

$w_{1} = \sqrt{\frac{16 脳 {(10)}^{2}}{1.00 脳 10^{3}}} = 1.265 m i n \ w_{2} = \sqrt{\frac{16 脳 {(12)}^{2}}{1.00 脳 10^{3}}} = 1.518 m i n$

02

(b)Step 2 : Sketching the graph :

03

finding Resolution:

(c) Resolution = $\frac{鈭� t}{w_{a v}} = \frac{t''_{r} - t'_{r}}{w_{a v}}$

Or, Resolution= $\frac{(12 - 10)}{\frac{1}{2} (1.265 + 1.518)}$ =1.437

For quantitive analysis resolution >1.5 is highly desirable.

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Most popular questions from this chapter

An open tubular column is 30.01 mlong and has an inner diameter of 0.530mm. It is coated on the inside wall with a layer of stationary phase that is $3.1 渭尘$ thick. Unretained solute passes through in 2.16min, whereas a particular solute has a retention time of 17.32min. (a) Find the linear velocity and volume flow rate. (b) Find the retention factor for the solute and the fraction of time spent in the stationary phase.(c) Find the partition coefficient, K 5 cs/cm, for this solute.

The solute in Problem 23 - 8 is initially dissolved in 80.0 mL of water. It is then extracted six times with 10.0mL portions of chloroform. Find the fraction of solute remaining in the aqueous phase.

(a) Why is it difficult to extract the complex of aluminum into an organic solvent but easy to extract the complex?

(b) If you need to bring the $EDTA$ complex into the organic solvent, should you add a phase transfer agent with a hydrophobic cation or a hydrophobic anion?

23-47. A chromatographic peak has a width,W,of $4.0 mL$ and a retention volume of role="math" localid="1654842133798" $49 mL$ . What width is expected for a band with a retention volume of $127 mL$ ? Assume that the only band spreading occurs on the column itself.

Match statements 1鈥�5 with the band broadening terms in the second list.

1. Depends on radius of open tubular column.

2. Not present in an open tubular column.

3. Depends on length and radius of connecting tubing.

4. Increases with diffusion coefficient of solute.

5. Increases with thickness of stationary phase film.

A Multiple paths

B Longitudinal diffusion

$C_{m}$ Equilibration time in mobile phase

$C_{s}$ Equilibration time in stationary phase

EC Extra column band broadening

See all solutions

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