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Why is it advantageous to use large particles \(\left( {{\bf{50}}{\rm{ }}\mu {\bf{m}}} \right)\) for solid phase extraction, but small particles \(\left( {{\bf{5}}{\rm{ }}\mu {\bf{m}}} \right)\) for chromatography?

Short Answer

Expert verified
  • If you're working with an aqueous base sample and want to inject it into a GC instrument to convert it to a volatile organic, solid phase extraction could help you achieve your aim.
  • Chromatography helps in purification,precise separation and analysis. It necessitates very small sample volumes.

Step by step solution

01

Definition ofChromatography.

  • Chromatography is a fashion for separating the factors, or solutes, of a mixture grounded on the relative quantities of each solute distributed between a flowing fluid sluice, known as the mobile phase, and a stationary phase that's conterminous.
02

Determine the advantages for solid particles and small particles for chromatography.

Solid phase extraction:

  • As per the physical and chemical properties the solid phase extraction helps in separating compounds dissolved or suspended in a liquid solution from other compounds in the mixture
  • Without application of high-pressure large particles can be allowed for the sample to drain through the column.

Chromatography

  • The separation of mixtures whilst using a mobile phase (moving solvent) on stationary phase\( \to \)mostly due to different polarity of each compound in the mixture.
  • Small particles - cause an increase in the efficiency of the separation. Using of these molecules requirehigh-pressure application in order to "push" the solvent through the column.
  • It can analyse pharmaceuticals, food particles, plastics, pesticides, air and water samples, and tissue extracts, among other things.

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Most popular questions from this chapter

When you flip a coin, the probability of its landing on each side is \(p = q = \frac{1}{2}\)in Equations 28-2 and 28-3. If you flip it \(n\)times, the expected number of heads equals the expected number of tails \( = np = nq = \frac{1}{2}n.\)The expected standard deviation for \(n\)flips is\({\sigma _n} = \sqrt {npq} \). From Table 4-1, we expect that \(68.3\% \)of the results will lie within \( \pm 1{\sigma _n}\) and \(95.5\% \)of the results will lie within\( \pm 2{\sigma _n}\).

(a) Find the expected standard deviation for the number of heads in \({\bf{1000}}\) coin flips.

(b) By interpolation in Table 4-1, find the value of \(z\)that includes \(90\% \)of the area of the Gaussian curve. We expect that \(90\% \)of the results will lie within this number of standard deviations from the mean.

(c) If you repeat the\({\bf{1000}}\)coin flips many times, what is the expected range for the number of heads that includes\(90\% \) of the results? (For example, your answer might be, "The range \({\bf{490}}\) to \({\bf{510}}\) will be observed \(90\% \)of the time.")

Barium titanate, a ceramic used in electronics, was analyzed by the following procedure: Into a Pt crucible was placed \(1.2\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}\) and \(0.8\;{\rm{g}}\)of \({\rm{N}}{{\rm{a}}_2}\;{{\rm{B}}_4}{{\rm{O}}_7}\)plus \(0.3146\;{\rm{g}}\)of unknown. After fusion at \({1000^\circ }{\rm{C}}\)in a furnace for\(30\;{\rm{min}}\), the cooled solid was extracted with \(50\;{\rm{mL}}\)of\(6{\rm{MHCl}}\), transferred to a \(100 - {\rm{mL}}\) volumetric flask, and diluted to the mark. A \(25.00 - {\rm{mL}}\)aliquot was treated with \(5\;{\rm{mL}}\)of \(15\% \)tartaric acid (which complexes \({\rm{T}}{{\rm{i}}^{4 + }}\)and keeps it in aqueous solution) and \(25\;{\rm{mL}}\)of ammonia buffer,\({\rm{pH}}9.5\). The solution was treated with organic reagents that complex\({\rm{B}}{{\rm{a}}^{2 + }}\), and the \({\rm{Ba}}\)complex was extracted into \({\rm{CC}}{{\rm{l}}_4}.\)After acidification (to release the \({\rm{B}}{{\rm{a}}^{2 + }}\) from its organic complex), the \({\rm{B}}{{\rm{a}}^{2 + }}\)was backextracted into\(0.1{\rm{MHCl}}\). The final aqueous sample was treated with ammonia buffer and methylthymol blue (a metal ion indicator) and titrated with \(32.49\;{\rm{mL}}\) of \(0.01144{\rm{M}}\)EDTA. Find the weight per cent of Ba in the ceramic.

If 105 particles are taken, what is the relative standard deviation of each measurement?

To pre-concentrate cocaine and benzoylecgonine from river water described at the opening of this chapter, solid-phase extraction was carried out at \({\rm{pH}}\,\,2\) using the mixed-mode cation-exchange resin in Figure 28-19. After passing \(500\;{\rm{mL}}\)of river water through \(60{\rm{mg}}\)of resin, the retained analytes were eluted first with \(2\;{\rm{mL}}\)of \({\rm{C}}{{\rm{H}}_3}{\rm{OH}}\)and then with \(2\;\,\,{\rm{mL }}of\,\,\,2\% \) ammonia solution in\({\rm{C}}{{\rm{H}}_3}{\rm{OH}}\). Explain the purpose of using \({\rm{pH}}2\) for retention and dilute ammonia for elution.

Referring to Table 28-7, explain how an anion-exchange resin can be used for absorption and analysis of \({\bf{S}}{{\bf{O}}_2}\)released by combustion.

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