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Chapter 25: Q2P (page 707)

Why are the relative eluent strengths of solvents in adsorption chromatography fairly independent of solute?

Short Answer

Expert verified

Adsorption chromatography uses a solid stationary phase and a liquid or gas mobile phase. As a result, it is independent of the solute because it is based on the difference in adsorption affinity between the adsorbent and the solvent. If the solvent's affinity is higher. A solute will be eluted from the absorbent.

Step by step solution

01

Step 1:Explaining the relative eluent chromatography of solvents.

  • Because adsorption chromatography employs a mobile phase in the liquid or solid state and a stationary phase in the solid state, the relative eluent strength of the solvent in adsorption chromatography is relatively independent of solute.
  • As a result, the solvent is competing with the solute for adsorption sites.
02

Explaining the independence of relative eluent strength.

  • Each solvent strikes a balance between adsorption on the solid surface and solvent solubility.
  • As a result, the solvent will move with the mobile phase until the equilibrium state is reached. The solvent is absorbed by the stationary phase

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Most popular questions from this chapter

Morphine and morphine 3-b-d-glucuronide were separated on two different 50 3 4.6 mm columns with 3-mm particles.61 Column A was C18-silica run at 1.4 mL/min and column B was bare silica run at 2.0 mL/min.

(a) Estimate the volume,Vm, and time,tm, at which unretained solute would emerge from each column. The observed times are 0.65 min for column A and 0.50 min for column B.

(b) Column A was eluted with 2 vol% acetonitrile in water containing 10 mM ammonium formate at pH 3. Morphine 3-β-d-glucuro-nide emerged at 1.5 min and morphine at 2.8 min. Explain the order of elution.

(c) Find the retention factor k for each solute on column A, usingtm5 0.65 min.

(d) Column B was eluted with a 5.0-min gradient beginning at 90 vol% acetonitrile in water and ending at 50 vol% acetonitrile in water. Both solvents contained 10 mM ammonium formate, pH 3. Morphine emerged at 1.3 min and morphine 3-b-d-glucuronide emerged at 2.7 min. Explain the order of elution. Why does the gradient go from high to low acetonitrile volume fraction?

(e) From Equation 25-12 in Box 25-4, estimate k* on Column B assuming S = 4 and withtm5 0.50 min.

Question: Explain how to use a gradient for the first run to decide whether isocratic or gradient elution would be more appropriate.

Question: what are the general steps in developing an isocratic separation for reversed-phase chromatography?

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. Cocaine metabolism in rats can be studied by injecting the drug and periodically with drawing blood to measure levels of metabolites by HPLC–mass spectrometry. For quantitative analysis, isotopically labelled internal standards are mixed with the blood sample. Blood was analysed by reversed-phase chromatography with an acidic eluent and atmospheric pressure chemical ionization mass spectrometry for detection. The mass spectrum of the collisionally activated dissociation products from the m/z 304 positive ion is shown in the figure on the next page. Selected reaction monitoring (m/z 304 from mass filter Q1 and m/z 182 from Q3 in Figure 22-33) gave a single chromatographic peak at 9.22 min for cocaine. The internal standard H52-cocaine gave a single peak at 9.19 min for m/z 309 (Q1) 182(Q3).

(a) Draw the structure of the ion at m/z 304.

(b) Suggest a structure for the ion at m/z 182.

(c) The intense peaks at m/z 182 and 304 do not have C2isotopic partners at m/z 183 and 305. Explain why.

(d) Rat plasma is exceedingly complex. Why does the chromatogram show just one clean peak?

(e) Given that H52-cocaine has only two major mass spectral peaks at m/z 309 and 182, which atoms are labelled with deuterium?

(f) Explain how you would use H52-cocaine for measuring cocaine in blood.

Spectrum for Problem 25-25.

Left: Mass spectrum of collisionally activated dissociation products from m/z 304 positive ion from atmospheric pressure chemical ionization mass spectrum of cocaine.

Right: Chromatograms obtained by selected reaction monitoring. [Data from G. Singh, V. Arora, P. T. Fenn, B. Mets, and I. A. Blair, “Isotope Dilution Liquid Chromatography Tandem Mass Spectrometry Assay for Trace Analysis of Cocaine and Its Metabolites in Plasma,” Anal. Chem. 1999, 71, 2021.]
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