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$\mathit{K}\mathbf{=}\mathbf{100}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathbf{}\mathit{k}\mathbf{=}\mathbf{10000}\mathbf{}\mathbf{?}$

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.

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

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.

b)

It is known that

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

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