When are isosbestic points observed and why?

Short Answer

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When spectra of two compounds with a constant total concentration cross at any wavelength, the isosbestic points are observed.

Step by step solution

01

Define isosbestic points:

The identified wavelength or frequency will not change at total absorbance of a sample during the chemical reaction or physical change.

02

Determine the observation of isosbestic points:

When spectra of two compounds with a constant total concentration cross at any wavelength, the isosbestic points are observed. Hence, the mixtures that have same total concentration also will have same isosbestic points. The isosbestic points are observed which helps in finding, how one species is converted into other major species.

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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.

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

The figure shows spectra of1.00×10-4MMnO4-,1.00×10-4and an unknown mixture of both, all in1.000cmpath length cells. Absorbances are given in the table. Use the least squares procedure in Figure 19-3 to find the concentration of each species in the mixture.

Visible spectrum ofMnO4-,Cr2O72-, and an unknown mixture containing both ions.

Two ways to analyze a mixture. Figure 19-5 shows the spectrum of the indicator bromothymol blue adjusted to several pH values. The spectrum at pHis that of the pure blue form and the spectrum at pH 1.8is that of the pure yellow form. At other pHvalues, there is a mixture of the two forms. The total concentration isand the path length isin all spectra. For the purpose of calculation, assume that there are more than two significant digits in concentration and path length. Absorbance at the dots on three of the curves in Figure 19-5 is given in the table.

(a) Prepare a spreadsheet like Figure 19-3 to use absorption at all six wavelengths to find[In-]and[HIn]in the mixture. Comment on the sum[In-]+[HIn].

(b) From[In-]in the mixture, and frompKa=7.10for HIn,calculate theof the mixture. (This calculation is the source of pH labels in the figure.)

(c) Use Equations 19-6 at the peak wavelengths ofto findin the mixture. Compare your answers to those in (a). Which answers, (a) or (c), are probably more accurate? Why?

Figure 19-6 is a Scatchard plot for the addition of 0-20nM antigen X to a fixed concentration of antibodyP=(Po=10nM)Prepare a Scatchard plot from the data in the table and find K for the reactionP+XPX. The table gives measured concentrations of unbound X and the complex PX. It is recommended that the fraction of saturation should span the range ,-0.2-0.8. What is the range of the fraction of saturation for the data?

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