Question

Fluorescence quenching in micelles. Consider an aqueous solution with a high concentration of micelles and relatively low concentrations of the fluorescent molecule pyrene and a quencher (cetylpyridinium chloride, designated Q), both of which dissolve in the micelles.

Quenching occurs if pyrene and Q are in the same micelle. Let the total concentration of quencher be [Q] and the concentration of micelles be [M]. The average number of quenchers per micelle is$\overline{\mathbf{Q}}\mathbf{=}\left[Q\right]\mathbf{/}\left[M\right]$. If Q is randomly distributed among the micelles, then the probability that a particular micelle has n molecules of Q is given by the Poisson distribution:

Probability of n molecules of Q in micelle =${\mathbf{P}}_{\mathbf{n}}\mathbf{=}\frac{{\overline{\mathbf{Q}}}_{\mathbf{n}}}{\mathbf{n}\mathbf{!}}\mathbf{e}\mathbf{-}\overline{\mathbf{Q}}$

whereis n factorial$\mathbf{(}\mathbf{=}\mathbf{n}\mathbf{[}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{\left]}\mathbf{\right[}\mathbf{n}\mathbf{-}\mathbf{2}\mathbf{]}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{[}\mathbf{1}\mathbf{\left]}\mathbf{\right)}$. The probability that there are no molecules of Q in a micelle is

Probability ofmolecules of Q in micelle = ${\mathbf{P}}_{\mathbf{n}}\mathbf{=}\frac{\overline{{\mathbf{Q}}^{\mathbf{0}}}}{\mathbf{0}\mathbf{!}}\mathbf{e}\mathbf{-}\overline{\mathbf{Q}}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\overline{\mathbf{Q}}}$

because 0!=1

Let ${\mathit{l}}_{\mathbf{0}}$be the fluorescence intensity of pyrene in the absence of Q and let I_Qbe the intensity in the presence of Q (both measured at the same concentration of micelles). The quotient ${\mathit{l}}_{\mathit{Q}}\mathbf{/}{\mathit{l}}_{\mathbf{0}}$must be ${\mathit{e}}^{\mathbf{-}\overline{\mathbf{Q}}}$which is the probability that a micelle does not possess a quencher molecule. Substituting $\overline{\mathbf{Q}}\mathbf{=}\mathbf{\left[}\mathbf{Q}\mathbf{\right]}\mathbf{/}\mathbf{\left[}\mathbf{M}\mathbf{\right]}\mathbf{}$gives

${\mathit{l}}_{\mathbf{Q}}\mathbf{/}{\mathit{l}}_{\mathbf{0}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\overline{\mathbf{Q}}}\mathbf{=}{\mathit{e}}^{\mathbf{-}\left[Q\right]\mathbf{/}\left[M\right]}$

Micelles are made of the surfactant molecule, sodium dodecyl sulfate. When surfactant is added to a solution, no micelles form until a minimum concentration called the critical micelle concentration (CMC) is attained. When the total concentration of surfactant, [S], exceeds the critical concentration, then the surfactant found in micelles is$\mathbf{\left[}\mathbf{S}\mathbf{\right]}\mathbf{-}\mathbf{\left[}\mathbf{CMC}\mathbf{\right]}$. The molar concentration of micelles is

$\left[M\right]\mathbf{=}\frac{\left[S\right]\mathbf{-}\left[\mathrm{CMS}\right]}{{\mathbf{N}}_{\mathbf{av}}}$

where N_av is the average number of molecules of surfactant in each micelle.

Combining Equationsandgives an expression for fluorescence as a function of total quencher concentration, [Q]:

${\mathbf{l}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{l}}_{\mathbf{0}}}{{\mathbf{l}}_{\mathbf{Q}}}\mathbf{=}\frac{\left[Q\right]{\mathbf{N}}_{\mathbf{av}}}{\left[S\right]\mathbf{-}\left[\mathrm{CMS}\right]}$

By measuring fluorescence intensity as a function of [Q] at fixed [S], we can find the average number of molecules of S per micelle if we know the critical micelle concentration (which is independently measured in solutions of S). The table gives data for $\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{\mu M}$

pyrene in a micellar solution with a total concentration of sodium dodecyl sulfate [S]=20.8mM

(a) If micelles were not present, quenching would be expected to follow the Stern-Volmer equation. Show that the graph of ${\mathbf{l}}_{\mathbf{0}}\mathbf{/}{\mathbf{l}}_{\mathbf{Q}}$versus [Q] is not linear.

(b) The critical micelle concentration is 8.1mM.Prepare a graph of${\mathbf{l}}_{\mathbf{n}}\left(l_0/l_Q\right)$versus [Q]. Use Equation 5 to find N_av, the average number of sodium dodecyl sulfate molecules per micelle.

(c) Find the concentration of micelles, [M], and the average number of molecules of Q per micelle,$\overline{\mathbf{Q}}$, when$\left[Q\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{mM}$

(d) Compute the fractions of micelles containing,, andmolecules of Q when$\left[Q\right]\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{mM}$

Short answer

(a) See graph (b) ${\mathrm{N}}_{\mathrm{sv}}=55.88$(c) $\left[\mathrm{M}\right]=0.227\mathrm{mM}$, $\tilde{\mathrm{Q}}=0.881\mathrm{molecules}\mathrm{per}\mathrm{micelle}$(d) $P_0=0.414,P_1=0.365$,$P_2=0.161$

(a)Solution of part (a)

The graph$l_0/l_Q$ versus is not linear.

(b) Solution of part (b)

Step-by-step solution

Graph of l0/lQ versus [Q]

Calculation of Nsv

The slope of graph is

$\frac{N_{sv}}{\left[\mathrm{S}\right]-\left[\mathrm{CMS}\right]}$

The slope is4.4mM, so we can write

$4.4=\frac{{\mathrm{N}}_{\mathrm{sv}}}{\left[\mathrm{S}\right]-\left[\mathrm{CMS}\right]}$

The N_svis

_{$$\begin{gathered} N_{sv}=4.4.\left(\left[S\right]-\left[CMS\right]\right) \\ {N}_{sv}=4.4.\left(20.8-8.1\right) \\ {N}_{sv}=55.88 \end{gathered}$$}

(c) Solution of part (c)

Concentration of micelles

$$\begin{gathered} \left[\mathrm{M}\right]=\frac{\left[\mathrm{S}\right]-\left[\mathrm{CMS}\right]}{{\mathrm{N}}_{\mathrm{sv}}} \\ \left[\mathrm{M}\right]=\frac{20.8-8.1}{55.880} \\ \left[\mathrm{M}\right]=0.227\mathrm{mM} \end{gathered}$$

The average number of molecule per micelle

$$\begin{gathered} \tilde{\mathrm{Q}}=\frac{\left[\mathrm{Q}\right]}{\left[\mathrm{M}\right]} \\ \tilde{\mathrm{Q}}=\frac{0.200\mathrm{mM}}{0.227\mathrm{mM}} \\ \tilde{\mathrm{Q}}=0.881\mathrm{molecules}\mathrm{per}\mathrm{micelle} \end{gathered}$$

(d) Solution of part (d)

The fraction of micelle is

$P_n=\frac{Q}{n!}.e$

For n =0

$$\begin{gathered} P_0=\frac{{\left(0.881\right)}^0}{0!}.e^{-0.881} \\ {P}_0=0.414 \end{gathered}$$

Forn=1

$$\begin{gathered} P_1=\frac{{\left(0.881\right)}^1}{1!}.e^{-0.881} \\ {P}_1=0.365 \end{gathered}$$

For n=2

$$\begin{gathered} P_2=\frac{{\left(0.881\right)}^2}{2!}.e^{-0.881} \\ {P}_2=0.161 \end{gathered}$$