Question

Efficiency of solid-phase microextraction. Equation$\mathbf{24}\mathbf{-}\mathbf{9}$gives 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 a$\mathbf{\text{100-μm-thick}}$coating has a film volume of$\mathbf{6}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{mL}$. Suppose that the initial concentration of analyte in solution is

${\mathbf{c}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{\mu g}\mathbf{/}\mathbf{mL}\mathbf{\left(}\mathbf{100}\mathbf{ppb}\mathbf{\right)}$.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 $\mathbf{10000}\mathbf{,}\mathbf{5000}\mathbf{,}\mathbf{1000}\mathbf{}\mathbf{and}\mathbf{}\mathbf{100}$and. Let the solution volume vary from $\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{100}\mathbf{mL}$.

(b) Evaluate the limit of Equation$\mathbf{24}\mathbf{-}\mathbf{9}\mathbf{}\mathit{a}\mathit{s}\mathbf{}{\mathit{V}}_{\mathbf{r}}$ gets big relative to $\mathit{K}{\mathit{V}}_{\mathit{f}}$. 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 when$\mathit{K}\mathbf{=}\mathbf{100}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathbf{}\mathit{k}\mathbf{=}\mathbf{10000}\mathbf{}\mathbf{?}$

Short answer

a)The graph:

b) This is stated to be in agreement with the graph for $K\mathit{=}100\mathit{,}\mathrm{m}=6.9\mathrm{n}\mathrm{g}$.

For $\mathrm{K}=10000,\mathrm{m}=690\mathrm{ng}$,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 $\mathrm{K}=100=0.69\%.$

$\mathrm{The}\mathrm{fraction}\mathrm{extracted}\mathrm{when}\mathrm{K}=10000=41\%.$

Step-by-step solution

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 $10-100\mathrm{\mu }m$thickness in film of stationary phase exactly to those used in gas chromatography.

$$\begin{gathered} m\mathit{=}\frac{K{V}_{\mathit{1}}{c}_{\mathit{0}}{V}_a}{K{V}_{\mathit{1}}\mathit{+}{V}_a} \\ Where\mathit{}{V}_f\mathit{=}\mathrm{volume}\mathrm{of}\mathrm{film}\mathrm{on}\mathrm{the}\mathrm{fibre} \\ {\mathrm{V}}_{\mathrm{s}}=\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{that}\mathrm{is}\mathrm{extracted} \\ {\mathrm{c}}_{\mathrm{o}}=\mathrm{initial}\mathrm{concentration}\mathrm{of}\mathrm{analyte}\mathrm{in}\mathrm{the}\mathrm{solution}\mathrm{that}\mathrm{is}\mathrm{extracted} \\ \mathrm{K}=\mathrm{partition}\mathrm{coefficient}\mathrm{for}\mathrm{solute}\mathrm{between}\mathrm{film}\mathrm{and}\mathrm{solution}. \end{gathered}$$

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 $x$axis, along with the partition coefficients.

Evaluate the limit.

b)

It is known that

$$\begin{gathered} \mathrm{m}=\frac{K{\mathrm{V}}_1{\mathrm{c}}_{\mathrm{c}}{\mathrm{V}}_{\mathrm{o}}}{{\mathrm{KV}}_{\mathrm{f}}+{\mathrm{V}}_{\mathrm{s}}} \\ \mathrm{If}{\mathrm{V}}_{\mathrm{s}}>>{\mathrm{KV}}_{\mathrm{f}}\mathrm{then}\mathrm{m}={\mathrm{KV}}_{\mathrm{F}}{\mathrm{c}}_{\mathrm{o}} \\ \mathrm{For} \\ {\mathrm{V}}_{\mathrm{f}}=6.9\times {10}^{-4}\mathrm{ml} \\ {\mathrm{c}}_{\mathrm{o}}=0.1\mathrm{\mu g}/\mathrm{ml} \\ \mathrm{m}=(6.9\times {10}^{-5})\left(\mathrm{K}\right)\mathrm{\mu g} \\ \mathrm{This}\mathrm{is}\mathrm{stated}\mathrm{to}\mathrm{be}\mathrm{in}\mathrm{agreement}\mathrm{with}\mathrm{the}\mathrm{graph}\mathrm{for}K=100,\mathrm{m}=6.9\mathrm{n}\mathrm{g}. \\ \mathrm{For}\mathrm{K}=10000,\mathrm{m}=690\mathrm{ng},\mathrm{this}\mathrm{number}\mathrm{goes}\mathrm{ahead}\mathrm{of}\mathrm{the}\mathrm{graph},\mathrm{and}\mathrm{reaching}\mathrm{the} \\ \mathrm{limiting}\mathrm{concentration}\mathrm{in}\mathrm{the}\mathrm{fibre}\mathrm{would}\mathrm{need}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{litre}\mathrm{of}\mathrm{solution}. \end{gathered}$$

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

$$\begin{gathered} c) \\ \mathrm{when}\mathrm{K}\mathrm{is}\mathrm{the}\mathrm{set}\mathrm{to}100,06.85\mathrm{ng}\mathrm{is}\mathrm{extracted}\mathrm{into}\mathrm{the}\mathrm{fibre},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{speadsheet}. \\ \mathrm{when}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{K}\mathrm{reaches}10000.408\mathrm{ng}\mathrm{is}\mathrm{extracted}\mathrm{into}\mathrm{the}\mathrm{fibre}. \\ \mathrm{in}10.0\mathit{}ml,\mathrm{the}\mathrm{total}\mathrm{analyte}\mathrm{is}(0.10\mathrm{\mu g}/\mathrm{ml}\left)\right(10.0\mathrm{ml})=1.0\mathrm{\mu g} \\ \mathrm{When}\mathrm{K}=100,\mathrm{the}\mathrm{fraction}\mathrm{extracted}\mathrm{is}\mathrm{computed}\mathrm{as}\mathrm{Fraction}\mathrm{extracted}\mathrm{for} \\ \mathrm{K}=100=\frac{6.86\mathrm{ng}}{1.0\mathrm{\mu g}}=0.0069(0.69\%) \\ \mathrm{When}\mathrm{k}=10000,\mathrm{the}\mathrm{the}\mathrm{fraction}\mathrm{extracted}\mathrm{is}\mathrm{computed}\mathrm{as}\mathrm{Fraction}\mathrm{extracted}\mathrm{for} \\ \mathrm{K}=10000=\frac{408\mathrm{ng}}{1.0\mathrm{Hg}}=0.41(41\%) \\ \mathrm{The}\mathrm{fraction}\mathrm{extracted}\mathrm{when}K\mathit{=}100=0.69\%. \\ \mathrm{The}\mathrm{fraction}\mathrm{extracted}\mathrm{when}\mathrm{K}=10000=41\%. \end{gathered}$$