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 isQ=[Q]/[M]. 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 =Pn=Qnn!e-Q

whereis n factorial(=n[n-1][n-2]....[1]). The probability that there are no molecules of Q in a micelle is

Probability ofmolecules of Q in micelle = Pn=Q00!e-Q=e-Q

because 0!=1

Let l0be the fluorescence intensity of pyrene in the absence of Q and let IQbe the intensity in the presence of Q (both measured at the same concentration of micelles). The quotient lQ/l0must be e-Qwhich is the probability that a micelle does not possess a quencher molecule. Substituting Q=[Q]/[M]gives

lQ/l0=e-Q=e-[Q]/[M]

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[S]-[CMC]. The molar concentration of micelles is

[M]=[S]-[CMS]Nav

where Nav 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]:

ln=l0lQ=[Q]Nav[S]-[CMS]

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 3.8μM

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 l0/lQversus [Q] is not linear.

(b) The critical micelle concentration is 8.1mM.Prepare a graph ofln(l0/lQ)versus [Q]. Use Equation 5 to find Nav, 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,Q, when[Q]=0.200mM

(d) Compute the fractions of micelles containing,, andmolecules of Q when[Q]=0.200mM

Short Answer

Expert verified

(a) See graph (b) Nsv=55.88(c) [M]=0.227mM, Q~=0.881moleculespermicelle(d) P0=0.414,P1=0.365,P2=0.161

(a)Solution of part (a)

The graphl0/lQ versus is not linear.

(b) Solution of part (b)

Step by step solution

01

Graph of l0/lQ versus [Q]

02

Calculation of Nsv

The slope of graph is

NsvS-CMS

The slope is4.4mM, so we can write

4.4=NsvS-CMS

The Nsvis

Nsv=4.4.S-CMSNsv=4.4.20.8-8.1Nsv=55.88

(c) Solution of part (c)

03

Concentration of micelles

M=S-CMSNsvM=20.8-8.155.880M=0.227mM

04

The average number of molecule per micelle

Q~=QMQ~=0.200mM0.227mMQ~=0.881moleculespermicelle

(d) Solution of part (d)

The fraction of micelle is

Pn=Qn!.e

For n =0

P0=0.88100!.e-0.881P0=0.414

Forn=1

P1=0.88111!.e-0.881P1=0.365

For n=2

P2=0.88122!.e-0.881P2=0.161

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Most popular questions from this chapter

The end of Box 19-2 states that “the advantage of up conversion for biomedical probes is that low energy near-infrared (800 to 1000 nm) incident radiation stimulates little background emission from the complex biological matrix that can be highly fluorescent under visible radiation.” Suggest why near-infrared radiation stimulates less emission than visible radiation and why this behavior is useful.

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The sensor compartment is separated from seawater by a silicone membrane through which CO2, but not dissolved ions, can diffuse. Inside the sensor, CO2equilibrates with HCO3and CO32. For each

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[ln2][Hln]=RAε434HHnε6,20Hlnε620ln2RAε434ln2=Rln (A)

(b) From the mass balance (1) and the acid dissociation constant

, show that

[Hln]=F1nRln+1 (B)

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(c) Show that H+=Kln/Rln (D)

(d) From the carbonic acid dissociation equilibria, show that

[HCO3]=K1[CO(aq)]E[H+][CO32]=K1K2[CO(aq)]F[H+]2

(e) Write the charge balance for the solution in the sensor compartment. Substitute in expressions B, C, E, and F forHln,In2-,[HCO3], and[CO32]

(f) Suppose that the various constants have the following values:

ε4344HHn=8.00×103M1cm1    K1=3.0×107ε6620Hn=0    K2=3.3×1011ε434ln2=1.90×103M1cm1    Kln=2.0×107ε620ln2=1.70×104M1cm1    Kw=6.7×1015

From the measured absorbance ratio=2.84, findin the seawater.

(g) Approximately what is the ionic strength inside the sensor compartment? Were we justified in neglecting activity coefficients in working this problem?

19-C. The protein bovine serum albumin can bind several molecules of the dye methyl orange. To measure the binding constant K for one dye molecule, solutions were prepared with a fixed concentration \(\left( {{X_0}} \right)\) of dye and a larger, variable concentration of protein (P). The equilibrium is Reaction 19-18, with X 5 methyl

orange.

Experimental data are shown in cells A16-D20 in the spreadsheet on the next page. The authors report the increase in absorbance \((\Delta A)\) at 490 nm as P is added to X. X and PX absorb visible light, but P does not. Equilibrium expression 19-20 applies and (PX) is given by Equation 19-21. Before P is added, the absorbance is \({\varepsilon _X}{X_{0 - }}\). The increase in absorbance when P is added is

The spreadsheet uses Solver to vary K and \(\Delta E\) in cells B10:B11 to minimize the sum of squares of differences between observed and calculated \(\Delta A\) in solutions with different amounts of P. Cell E16 computes (PX) from Equation 19-21, which is Equation A on line 6 of the spreadsheet. Cells F16 and G16 find (X) and (P) from mass balances. Cell H16 computes \(\Delta {A_{calc}} = \Delta E(PX)\)which is Equation B on line 7.

To estimate a value of K in cell B10, suppose that 50% of X has reacted in row 20 of the spreadsheet. The total concentration of X is\({X_0} = 5.7\mu M\). If half is reacted, then \((X) = (PX) = 2.85\mu M\) and\((P) = {P_0} - (PX) = 40.4 - 2.85 = 37.55\mu M\). The binding

constant is \(K = (PX)/(P)(X)) = (2.85\mu M)/(37.55\mu M)(2.85\mu M)) = \)\(2.7 \times 1{0^4}\) which we enter as our guess for K in cell B10. We estimate \(\Delta \varepsilon \)in cell B11 by supposing that 50% of X has reacted in row 20. In Equation B on line 7, \(\Delta A = \Delta \varepsilon (PX)\).The measured

value of \((\Delta A)\)in row 20 is 0.0291 and we just estimated that \((PX) = \)\(2.85\mu M\). Therefore, our guess for \(\Delta \varepsilon \)in cell B11 is \(\Delta \varepsilon = \Delta A/\)\((PX) = (0.0291)/(2.85\mu M) = 1.0 \times 1{0^4}\)

Your assignment is to write formulas in columns E through J of the spreadsheet to reproduce what is shown and to find values in cells E17:J20. Then use Solver to find K and \(\Delta \varepsilon \)in cells B10:B11 to minimize \(\Sigma {\left( {{A_{oths\;}} - {A_{calc\;}}} \right)^2}\)in cell I21.

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