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23-41. Describe how nonlinear partition isotherms lead to non-Gaussian band shapes. Draw the chromatographic peak shapes produced by an overloaded gas chromatography column and an overloaded liquid chromatography column.

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We can now depict the chromatographic peak morphologies that an overloaded gas chromatography column and liquid chromatography column produce:

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In this job, we will sketch the chromatographic peak forms produced by an overloaded gas and liquid chromatography column and explain how nonlinear partition isotherms lead to non-Gaussian band shapes.

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Peak with a Gaussian form.

When the partition coefficient K does not rely on the solute concentration on a column, it is said to be "t .

- If we had real columns, K would alter as the concentration of solute increased.

- An isotherm is a graph of CSvsCm at a certain temperature (an ideal isotherm would produce a symmetric peak with a retention time ( tr) that is independent of the solute concentration).

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Overload.

- Fronting is a phenomenon that occurs in gas chromatography.

- In liquid chromatography, tailing occurs.

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We'll start by drawing a graph of Csvs Cm.

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The upper and lower isotherms are due to an overloaded column (too much solute in the column).

Fronting peaks (asymmetry factorBA<1 ) cause the upper isotherm; the intensity of fronting rises with the amount of extra solute injected.

- Tailing peaks cause a lower isotherm (asymmetry factorBA>1 ).

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We can now depict the chromatographic peak morphologies that an overloaded gas chromatography column and liquid chromatography column produce:

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Most popular questions from this chapter

The theoretical limit for extracting solute Sfrom Phase1(volumeV1)into phase2 (volumeV2)is attained by dividing V2into an infinite number of infinitesimally small portions and conducting an infinite number of extractions. With a partition coefficient, K=[S]2/[S]1the limiting fraction of solute remaining in phase 1is2qlimit=e-(V2/V1)K(V1=V2=50mLandK=2). Let volume V2be divided intoequal portions to conduct extractions. Find the fraction of S extracted into phase 2 for n = 1,2, and 10extractions. How many portions are required to attain 95%of the theoretical limit?

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 layer with negligible thickness containing 10.0nmol of methano(D=1.6ร—10-9m2/s)was placed in a tube of water 5.00 cm in diameter and allowed to spread by diffusion. Using Equation 23-27, prepare a graph showing the Gaussian concentration profile of the methanol zone after 1.00,10.0, and 100 min. Prepare a second graph showing the same experiment with the enzyme rib nuclease(D=1.12ร—10-9m2/s).

23-1.If you are extracting a substance from water into ether, is it more effective to do one extraction with of ether or three extractions with ?

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(a) Define the distribution coefficient, D for this system.

(b) Explain the difference between D and K the partition coefficient.

(c) Calculate D at pH8.00ifk=50.0

(d) Will D be greater or less at pH10thanatpH8? Explain why.

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