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Efficiency of solid-phase microextraction. Equation24-9gives the mass of analyte extracted into a solid-phase microextraction fiber as a function of the partition coefficient between the fiber coating and the solution.

(a) A commercial fiber with a100-μ³¾-³Ù³ó¾±³¦°ìcoating has a film volume of6.9×10-4mL. Suppose that the initial concentration of analyte in solution is

c0=0.10μ²µ/mL(100ppb).Use a spreadsheet to prepare a graph showing the mass of analyte extracted into the fiber as a function of solution volume for partition coefficients of 10000,5000,1000and100and. Let the solution volume vary from 0to100mL.

(b) Evaluate the limit of Equation24-9asVr gets big relative to KVf. Does the extracted mass in your graph approach this limit?

(c) What percentage of the analyte fromof solution is extracted into the fiber when and whenK=100andwhenk=10000?

Short Answer

Expert verified

a)The graph:

b) This is stated to be in agreement with the graph for K=100,m=6.9ng.

For K=10000,m=690ng,this number goes ahead of the graph, and reaching the limiting concentration in the fibre would need at least one litre of solution.

c) The fraction extracted when K=100=0.69%.

ThefractionextractedwhenK=10000=41%.

Step by step solution

01

Concept used. 

Solid-phase microextraction:

Solid-phase microextraction is the extraction of chemicals from liquids, air, or sludge without the need of a solvent. The major component is the fused-silica fibre coated with 10-100μmthickness in film of stationary phase exactly to those used in gas chromatography.

m=KV1c0VaKV1+VaWhereVf=volumeoffilmonthefibreVs=volumeofsolutionthatisextractedco=initialconcentrationofanalyteinthesolutionthatisextractedK=partitioncoefficientforsolutebetweenfilmandsolution.

02

Step 2: The graph showing the mass of analyte extracted into the fiber.

a)

The mass of analyte and volume of solution, as well as partition coefficient, volume of the film on fibre, and starting concentration of solution, are entered into a spreadsheet.

The volume of the solution is displayed on the axis, and the extracted mass is plotted on the xaxis, along with the partition coefficients.

03

Evaluate the limit.

b)

It is known that

m=KV1ccVoKVf+VsIfVs>>KVfthenm=KVFcoForVf=6.9×10-4mlco=0.1μ²µ/mlm=(6.9×10-5)(K)μ²µThisisstatedtobeinagreementwiththegraphforK=100,m=6.9ng.ForK=10000,m=690ng,thisnumbergoesaheadofthegraph,andreachingthelimitingconcentrationinthefibrewouldneedatleastonelitreofsolution.

04

Step 4:The percentage of the analyte from  of solution is extracted into the fiber.

c)whenKisthesetto100,06.85ngisextractedintothefibre,accordingtothespeadsheet.whenthevalueofKreaches10000.408ngisextractedintothefibre.in10.0ml,thetotalanalyteis(0.10μ²µ/ml)(10.0ml)=1.0μ²µWhenK=100,thefractionextractediscomputedasFractionextractedforK=100=6.86ng1.0μ²µ=0.0069(0.69%)Whenk=10000,thethefractionextractediscomputedasFractionextractedforK=10000=408ng1.0Hg=0.41(41%)ThefractionextractedwhenK=100=0.69%.ThefractionextractedwhenK=10000=41%.

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Most popular questions from this chapter

(a) Explain how solid phase microextraction works. Why is cold trapping necessary during injection with this technique? Is all the analyte in an unknown extracted into the fiber in solid phase microextraction?

(b) Explain the differences between stir-bar sportive extraction and solid-phase microextraction. Which is more sensitive and why?

(a) If retention times are 1.0 min for CH4, 12.0 min for octane, 13.0 min for unknown, and 15.0 min for nonane, find the Kovats retention index for the unknown. (b) What would be the Kovats index for the unknown if the phase ratio of the column were doubled? (c) What would be the Kovats index for the unknown if the length of the column were halved

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