(a). Sketch a graph of the van Deemnter equation (plate height versus linear flow rate).what would the curve look like if the multiple path term were 0? If the longtitundinal diffusinal diffusion term were 0?

(b). Explain why the van Deemter curve for 1.8μmparticles in figure 25-3is nearly flat at high flow rate.what can you say about each of the terms in the van Deemeter equation for 1.8μmparticles.

(c). Explain why the 2.7μmsuperficially porous particle enables separations similar to those achieved by 1.8μmtotally porous particles,but the superficially porous particle requires lower pressure.

Short Answer

Expert verified

a) Sketch a graph of the van Deemnter.

b) Comparing the van Deemter curve of the particles in Figure to the graph

c) Superficially porous particles have a diameter non porous silica core.

Step by step solution

01

Sketch a graph of the van Deemnter 

The van Deemter equation was graphed to help visualise each term in the equation. Figure 24-3 was then examined using the van Deemter equation and its implications.

The figure below shows the graph of the van Deemter equation when the multiple paths term ( A) is , when the longitudinal diffusion term ( B) is , when the equilibration time term (C ) is constant, and when all terms are .

02

Comparing the van Deemter

b) why the van Deemter curve

Comparing the van Deemter curve of the 1.8μmparticles in Figure 24-3to the graph in (a), we can see that the graph that it most resembles is the graph of c=0. Although not obvious from the graphs, the equation of the curve of 1.8μmparticles has a small multiple paths term (A ) relative to that of the other larger particles in the graph.

Based on the graph, the curve of the particles plateaus at the smallest plate height ( H) compared to the other graphs. The reduction in plate heights can be related to the decrease in the band spreading due to multiple flow paths.

Given this relationship, the decrease in the H plateau value of the particles means that there is a decrease in A as the particle size decreases.

c) superficially porous particle

Superficially porous particles have a 5μmdiameter non porous silica core. Compared to other fully porous particles with a larger radii, superficially porous particles can perform at the same level or better.

This is due to the structure of superficially porous particles, having a 0.25 thick porous silica layer on the outside of its 5μmdiameter non porous silica core

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Most popular questions from this chapter

a. Why is high pressure needed in HPLC?

b. For a given column length , why do smaller particles give a higher plate number?

c. What is bonded phase in liquid chromatography?

What does it mean for a separation procedure to be ”rugged” and why is it desirable?

The figure shows the separation of two enantiomers on a chiral stationary phase.


Sepation of enantiomers of Ritalin by HPLC with a chiral stationary phase.[Data from R.Bakhitar,L.Ramos,and F.L.S.Tse, “Quantification of methlylphenidate in plasma using chiral Liquid-chromatography/Tandem mass spectrometry: Application to Toxicokinetric studies,”Anal Chim Acta 2002,469,261.]

(a)From trandw1/2 find N for each peak.

(b) Fromtrandw1/2find the resolution.

(c)Giventm=1.62min, use Equation23-23with the average N to predict the resolution.

Chromatography–mass spectrometry. HPLC separation of

enantiomers of the drug Ritalin on a chiral stationary phase was

shown in Problem 25-13.

(a) Detection is by atmospheric pressure chemical ionization with

selected reaction monitoring of the m/z 23484 transitions. Explain

how this detection works and propose structures for m/z 234 and m/z 84.

(b) For quantitative analysis, the internal standardH32-Ritalin with

a deuterated methyl group was added. Deuterated enantiomers have

the same retention times as unlabelled enantiomers. Which selected

reaction monitoring transition should be monitored to produce a

chromatogram of the internal standard in which unlabelled Ritalin

will be invisible?

The rate at which heat is generated inside a chromatographycolumn from friction of flowing liquid is power (watts, W=J/s)= volume flow raterole="math" localid="1656474760665" (m3/s)×pressure drop (pascals, role="math" localid="1656474828044" Pa=kg/[m?s2]).

(a) Explain the analogy between heat generated in a chromatography column and heat generated in an electric circuit (power = current×voltage).

(b) At what rate (watts = J/s) is heat generated for a flow of 1 mL/minwith a pressure difference of3500 bar between the inlet and outlet?

You will need to convert mL/min tom3/s.Also 1 bar=105Pa.

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