In chromatography, why is the optimal flow rate greater if the stationary phase particle size is smaller?

Short Answer

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The solution is

the smaller the particle, the faster the equilibrium between the stationary and mobile phases is achieved.

Step by step solution

01

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In this job, we will explain why the ideal flow rate in chromatography is higher when the stationary phase particle size is lower.

02

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Representative diffusion coefficients at 289K are shown in Table 23-1:

- As the size of the diffusing particle and the viscosity of the fluid both rise, the friction coefficient increases.

- Some larger molecules diffuse slower than small ones due to their vast radius.

03

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As a result, the smaller the particle, the faster the equilibrium between the stationary and mobile phases is achieved.

- As seen in Table 23-1, tiny molecules have a greater diffusion coefficient.

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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 weak base B(Kb=1.0×10-5)equilibrates between water (phase 1) and benzene (phase 2)

(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.

The three chromatograms shown here obtained with 2.5, 1.0 and 0.4 μLof ethyl acetate injected on the same column under the same conditions. Explain why the peak becomes less symmetrical with increasing sample size.

2.5μL 1.0L 0.4μL

(a) For the asymmetric chromatogram in Figure 23-14, calculate the asymmetry factor, BA.(b) The asymmetric chromatogram in Figure 23-14 has a retention time equal to 15.0 min and a w0.1of 44s . Find the number of theoretical plates. (c) The width of a Gaussian peak at a height equal to110 of the peak height is 4.297σ. Suppose that the peak in (b) is symmetric with A=B=22s. Use Equations 23-30 and 23-32 to find the plate number.

A Gas chromatogram of a mixture of toluene and ethyl acetate is shown here:

(a)Use the width of each peak (measured at the base)to calculate the number of theoretical plates in the column. Estimate all lengths to the nearest 0.1 mm.

(b)Using the width of the toluene peak at its base, calculate the width expected at half height. Compare the measured and calculated values. When the thickness of the the line is significant relative to the length being measured, it is important to take the pen line width into account. You can measure from the edge of one line to the corresponding edge of the other line,as shown here.

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