What is the advantage of a time-resolved emission measurement with Eu3 versus

This problem can be worked with Equations 18-6 on a calculator or with the spreadsheet in Figure 18-5. Transferrin is the iron-transport protein found in blood. It has a molecular mass of 81 000 and carries two Fe3 ions. Desferrioxamine B is a chelator used to treat patients with iron overload (Box 11-1). It has a molecular mass of about 650 and can bind one Fe3. Desferrioxamine can take iron from many sites within the body and is excreted (with its iron) through the kidneys. Molar absorptivities of these compounds (saturated with iron) at two wavelengths are given in the table. Both compounds are colorless (no visible absorption) in the absence of iron. (nm) Transferrin Desferrioxamine 428 3 540 2 730 470 4 170 2 290 (a) A solution of transferrin exhibits an absorbance of 0.463 at 470 nm in a 1.000-cm cell. Calculate the concentration of transferrin in milligrams per milliliter and the concentration of bound iron in micrograms per milliliter. (b) After adding desferrioxamine (which dilutes the sample), the absorbance at 470 nm was 0.424, and the absorbance at 428 nm was 0.401. Calculate the fraction of iron in transferrin and the fraction in desferrioxamine. Remember that transferrin binds two iron atoms and desferrioxamine binds only one. 1

The spreadsheet lists molar absorptivities of three dyes and the absorbance of a mixture of the dyes at visible wavelengths.

Compound P, which absorbs light at 305 nm, was titrated with X, which does not absorb at this wavelength. The product, PX, also absorbs at 305 nm. Absorbance was measured in a 1.000-cm cell, and the concentration of free X was determined independently, with results in the table. Prepare a Scatchard plot and find the equilibrium constant for the reaction .

Complex formation by 3-aminopyridine and picric acid in chloroform solution gives a yellow product with an absorbance maximum at 400 nm. Neither starting material absorbs significantly at this wavelength. Stock solutions containing 1.00  104 M of each compound were mixed as indicated, and the absorbances of the mixtures were recorded. Prepare a graph of absorbance versus mole fraction of 3-aminopyridine and find the stoichiometry of the complex. Picric acid (mL) 3-Aminopyridine (mL) Absorbance at 400 nm 2.70 0.30 0.106 2.40 0.60 0.214 2.10 0.90 0.311 1.80 1.20 0.402 1.50 1.50 0.442 1.20 1.80 0.404 0.90 2.10 0.318 0.60 2.40 0.222 0.30 2.70 0.110

This problem can be worked by calculator or with the spreadsheet in Figure 18-5. Consider compounds X and Y in the example labeled Analysis of a Mixture, Using Equations 18-6 in Section 18-1. Find [X] and [Y] in a solution whose absorbance is 0.233 at 272 nm and 0.200 at 327 nm in a 0.100-cm cell.

The figure shows spectra of 1.00  104 M MnO 4, 1.00  104 M Cr2O7 2, and an unknown mixture of both, all in 1.000-cm-pathlength cells. Absorbances are given in the table. Use the least-squares procedure in Figure 18-3 to find the concentration of each species in the mixture. Wavelength (nm) MnO 4 standard Cr2O7 2 standard Mixture 266 0.042 0.410 0.766 288 0.082 0.283 0.571 320 0.168 0.158 0.422 350 0.125 0.318 0.672 360 0.056 0.181 0.366

When are isosbestic points observed and why?

The metal ion indicator xylenol orange (Table 11-3) is yellow at pH 6 (max 439 nm). The spectral changes that occur as VO2 is added to the indicator at pH 6 are shown here. The mole ratio VO2/xylenol orange at each point is Trace Mole ratio Trace Mole ratio Trace Mole ratio 0 0 6 0.60 12 1.3 1 0.10 7 0.70 13 1.5 2 0.20 8 0.80 14 2.0 3 0.30 9 0.90 15 3.1 4 0.40 10 1.0 16 4.1 5 0.50 11 1.1 Suggest a sequence of chemical reactions to explain the spectral changes, especially the isosbestic points at 457 and 528 nm. A

Infrared spectra are customarily recorded on a transmittance scale so that weak and strong bands can be displayed on the same scale. The region near 2 000 cm1 in the infrared spectra of compounds A and B is shown in the figure. Note that absorption corresponds to a downward peak on this scale. The spectra were recorded from a 0.010 0 M solution of each, in cells with 0.005 00-cm pathlengths. A mixture of A and B in a 0.005 00-cm cell gave a transmittance of 34.0% at 2 022 cm1 and 38.3% at 1 993 cm1 . Find the concentrations of A and B

Spectroscopic data for the indicators thymol blue (TB), semithymol blue (STB), and methylthymol blue (MTB) are shown in the table. A solution of TB, STB, and MTB in a 1.000-cm cuvet had absorbances of 0.412 at 455 nm, 0.350 at 485 nm, and 0.632 at 545 nm. Modify the spreadsheet in Figure 18-5 to handle three simultaneous equations and find [TB], [STB], and [MTB] in the mixture.  (nm) TB STB MTB 455 4 800 11 100 18 900 485 7 350 11 200 11 800 545 36 400 13 900 4 450 SOURCE: S. Kiciak, H. Gontarz, and E. Krzyz. anowska, Monitoring the Synthesis of Semimethylthymol Blue and Methylthymol Blue, Talanta 1995, 42, 1245.

The spreadsheet gives the product b for four pure compounds and a mixture at infrared wavelengths. Modify Figure 18-5 to solve four equations and find the concentration of each compound. You can treat the coefficient matrix as if it were molar absorptivity because the pathlength was constant (but unknown) for all measurements.

The spreadsheet gives the product b for four pure compounds and a mixture at infrared wavelengths. Modify Figure 18-5 to solve four equations and find the concentration of each compound. You can treat the coefficient matrix as if it were molar absorptivity because the pathlength was constant (but unknown) for all measurements.

Chemical equilibrium and analysis of a mixture. (Warning! This is a long problem.) A remote optical sensor for CO2 in the ocean was designed to operate without the need for calibration.31 The sensor compartment is separated from seawater by a silicone membrane through which CO2, but not dissolved ions, can diffuse. Inside the sensor, CO2 equilibrates with HCO3  and CO3 2. For each measurement, the sensor is flushed with fresh solution containing 50.0 M bromothymol blue indicator (NaHIn) and 42.0 M NaOH. All indicator is in the form HIn or In2 near neutral pH, so we can write two mass balances: (1) [HIn] [In2] FIn 50.0 M and (2) [Na] FNa 50.0 M 42.0 M 92.0 M. HIn has an absorbance maximum at 434 nm and In2 has a maximum at 620 nm. The sensor measures the absorbance ratio RA A620/A434 reproducibly without need for calibration. From this ratio, we can find [CO2(aq)] in the seawater as outlined here: (a) From Beers law for the mixture, write equations for [HIn] and [In2] in terms of the absorbances at 620 and 434 nm (A620 and A434). Then show that (A) (b) From the mass balance (1) and the acid dissociation constant KIn, show that (B) (C) (c) Show that [H] KIn/RIn. (D) [In2

Figure 18-7 is a Scatchard plot for the addition of 020 nM antigen X to a fixed concentration of antibody P (P0 10 nM). Prepare a Scatchard plot from the data in the table and find K for the reaction . The table gives measured concentrations of unbound X and the complex PX. It is recommended that the fraction of saturation should span the range ~0.20.8. What is the range of the fraction of saturation for the data? [X] (nM) [PX] (nM) [X] (nM) [PX] (nM) 0.120 2.87 0.731 6.29 0.192 3.80 1.22 6.77 0.296 4.66 1.50 7.52 0.450 5.54 3.61 8.45

Scatchard plot for binding of estradiol to albumin. Data in the table come from a student experiment to measure the binding constant of the radioactively labeled hormone estradiol (X) to the protein, bovine serum albumin (P). Estradiol (7.5 nM) was equilibrated with various concentrations of albumin for 30 min at 37 C. A small fraction of unbound estradiol was removed by solid-phase microextraction (Section 23-4) and measured by liquid scintillation counting. Albumin is present in large excess, so its concentration in any given solution is essentially equal to its initial concentration in that solution. Call the initial concentration of estradiol [X]0 and the final concentration of unbound estradiol [X]. Then bound estradiol is [X]0  [X] and the equilibrium constant is which you can rearrange to A graph of [X]0/[X] versus [P] should be a straight line with a slope of K. The quotient [X]0/[X] is equal to the counts of radioactive

Iodine reacts with mesitylene to form a complex with an absorption maximum at 332 nm in CCl4 solution: Iodine Mesitylene Complex (a) Given that the product absorbs at 332 nm, but neither reactant has significant absorbance at this wavelength, use the equilibrium constant, K, and Beers law to show that where A is the absorbance at 332 nm, is the molar absorptivity of the complex at 332 nm, [mesitylene] is the concentration of free mesitylene, and [I2]tot is the total concentration of iodine in the solution ( [I2] [complex]). The cell pathlength is 1.000 cm. (b) Spectrophotometric data for this reaction are shown in the table. Because [mesitylene]tot >> [I2], we can say that [mesitylene] [mesitylene]tot. Prepare a graph of A/([mesitylene][I2]tot) versus A/[I2]tot and find the equilibrium constant and molar absorptivity of the complex. Absorbance [Mesitylene]tot (M) [I2]tot (M) at 332 nm 1.690 7.817  105 0.369 0.921 8 2.558  104 0.822 0.633 8 3.224  104 0.787 0.482 9 3.573  104 0.703 0.390 0 3.788  104 0.624 0.327 1 3.934  104 0.556 SOURCE: P. J. Ogren and J. R. Norton, Applying a Simple Linear Least-Squares Algorithm to Data with Uncertainties in Both Variables, J. Chem. Ed. 1992, 69, A130. 1

Now we use Solver to find K for the previous problem. The only absorbing species at 332 nm is the complex, so, from Beers law, [complex] A/ (because pathlength 1.000 cm). I2 is either free or bound in the complex, so [I2] [I2]tot [complex]. There is a huge excess of mesitylene, so [mesitylene] [mesitylene]tot. The spreadsheet shows some of the data. You will need to use all the data. Column A contains [mesitylene] and column B contains [I2]tot. K

Method of continuous variation. Make a graph of absorbance versus mole fraction of thiocyanate from the data in the table. mL Fe3 mL SCN Absorbance solutiona solutionb at 455 nm 30.00 0 0.001 27.00 3.00 0.122 24.00 6.00 0.226 21.00 9.00 0.293 18.00 12.00 0.331 15.00 15.00 0.346 12.00 18.00 0.327 9.00 21.00 0.286 6.00 24.00 0.214 3.00 27.00 0.109 0 30.00 0.002 a. Fe3 solution: 1.00 mM Fe(NO3)3 10.0 mM HNO3. b. SCN solution: 1.00 mM KSCN 15.0 mM HCl. SOURCE: Z. D. Hill and P. MacCarthy, Novel Approach to Jobs Method, J. Chem. Ed. 1986, 63, 162. (a) What is the stoichiometry of the predominant Fe(SCN)n 3n species? (b) Why is the peak not as sharp as those in Figure 18-8? (c) Why does one solution contain 10.0 mM acid and the other 15.0 mM acid? 18-1

The indicator xylenol orange (Table 11-3) forms a complex with Zr(IV) in HCl solution. Prepare a Job plot from the data   1 2 3 4 5 6 7 Guess for : 0.6338 0.9218 0.822 1.6900 0.369 7.380E-05 9.99282 1.644E-04 1.574E-04 [Mesitylene] [I2]tot A [Complex] = A / Keq A BC D E 1.50511 2.558E-04 7.817E-05 3.224E-04 0.787 Standard Dev = Stdev/Average = 4.77680 1.06552 Average = 4.48307 1.95128 5.000E+03 442 CHAPTER 18 Applications of Spectrophotometry in the table and suggest the stoichiometry of the complex (xylenol orange)xZrz. Mole fraction Relative Mole fraction Relative xylenol absorbance at xylenol absorbance at orange max550 nm orange max550 nm 0.100 0.110 0.600 0.236 0.200 0.235 0.700 0.145 0.300 0.352 0.800 0.088 0.400 0.412 0.900 0.045 0.500 0.348 SOURCE: L. Perring and B. Bourqui, An Alternative Method for Fluoride Determination by Ion Chromatography Using Post-Column Reaction, Am. Lab. News Ed., March 2002, p. 28.

Simulating a Jobs plot. Consider the reaction , for which K [AB2]/[A][B]2 . Suppose that the following mixtures of A and B at a fixed total concentration of 104 M are prepared: [A]total (M) [B]total (M) [A]total (M) [B]total (M) 1.00  105 9.00  105 5.00  105 5.00  105 2.00  105 8.00  105 6.00  105 4.00  105 2.50  105 7.50  105 7.00  105 3.00  105 3.00  105 7.00  105 8.00  105 2.00  105 3.33  105 6.67  105 9.00  105 1.00  105 4.00  105 6.00  105 (a) Prepare a spreadsheet to find the concentration of AB2 for each mixture, for equilibrium constants of K 106 , 107 , and 108 . One way to do this is to enter the values of [A]total and [B]total in columns A and B, respectively. Then put a trial (guessed) value of [AB2] in column C. From the mass balances [A]total [A] [AB2] and [B]total [B] 2[AB2], we can write K [AB2]/[A][B]2 [AB2]/{([A]total [AB2])([B]total 2[AB2])2 }. In column D, enter the reaction quotient [AB2]/[A][B]2 . For example, cell D2 has the formula =C2/((A2-C2)(B2-2*C2)^2). Then vary the value of [AB2] in cell C2 with Solver until the reaction quotient in cell D2 is equal to the desired equilibrium constant (such as 108 ). (b) Prepare a graph by the method of continuous variation in which you plot [AB2] versus mole fraction of A for each equilibrium constant. Explain the shapes of the curves. 18-17.

A study was conducted with derivatives of the DNA nucleotide bases adenine and thymine bound inside micelles (Box 25-1) in aqueous solution. Adenine derivative Thymine derivative Sodium dodecyl sulfate Sodium dodecyl sulfate forms micelles with the hydrocarbon tails pointed inward and ionic headgroups exposed to water. It was

Explain what is done in flow injection analysis and sequential injection. What is the principal difference between the two techniques? Which one is called lab-on-a-valve?

Explain how signal amplification is achieved in enzymelinked immunosorbent assays.

What is the advantage of a time-resolved emission measurement with Eu3 versus measurement of fluorescence from organic chromophores?

Here is an immunoassay to measure explosives such as trinitrotoluene (TNT) in organic solvent extracts of soil. The assay employs a flow cytometer, which counts small particles (such as living cells) flowing through a narrow tube past a detector. The cytometer in this experiment irradiates the particles with a green laser and measures fluorescence from each particle as it flows past the detector. 1. Antibodies that bind TNT are chemically attached to 5- mdiameter latex beads. + + + + + + + + + + + + + + + H N N (CH3)3 N N N N N H O O O + Problems 443 2. The beads are incubated with a fluorescent derivative of TNT to saturate the antibodies, and excess TNT derivative is removed. The beads are resuspended in aqueous detergent. 3. 5 L of the suspension are added to 100 L of sample or standard. TNT in the sample or standard displaces some derivatized TNT from bound antibodies. The higher the concentration of TNT, the more derivatized TNT is displaced. 4. An aliquot is injected into the flow cytometer, which measures fluorescence of individual beads as they pass the detector. The figure shows median fluorescence intensity  standard deviation. TNT can be quantified in the ppb to ppm range. Fluorescence of TNT-antibody-beads versus TNT concentration. [From G. P. Anderson, S. C. Moreira, P. T. Charles, I. L. Medintz, E. R. Goldman, M. Zeinali, and C. R. Taitt, TNT Detection Using Multiplexed Liquid Array Displacement Immunoassays, Anal. Chem. 2006, 78, 2279.] Draw pictures showing the state of the beads in steps 1, 2, and 3 and explain how this method works.

The graph shows the effect of pH on quenching of luminescence of tris(2,2-bipyridine)Ru(II) by 2,6-dimethylphenol. The ordinate, KSV, is the collection of constants, kq/(ke kd), in the Stern-Volmer equation 18-25. The greater KSV, the greater the quenching. Suggest an explanation for the shape of the graph and estimate pKa for 2,6-dimethylphenol.

Fluorescence quenching in micelles. Consider an aqueous solution with a high concentration of micelles (Box 25-1) 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 . 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: 32 (1) where n! is n factorial ( n[n  1][n  2] . . . [1]). The probability that there are no molecules of Q in a micelle is (2) because 0! 1. Let I0 be the fluorescence intensity of pyrene in the absence of Q and let IQ be the intensity in the presence of Q (both measured at Probability of 0 molecules of Q in micelle Q0 0! eQ e Q Probability of n molecules of Q in micelle Pn Qn n! e Q Q [Q]/[M] N Cl CH2(CH2)14CH3 Pyrene (fluorescent species) Cetylpyridinium chloride (quencher, Q) 1 500 1 000 500 0 1098 11 12 13 pH KSV 444 CHAPTER 18 Applications of Spectrophotometry the same concentration of micelles). The quotient IQ/I0 must be which is the probability that a micelle does not possess a quencher molecule. Substituting gives (3) Micelles are made of the surfactant molecule, sodium dodecyl sulfate (shown in Problem 18-17). 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 (4) where Nav is the average number of molecules of surfactant in each micelle. Combining Equations 3 and 4 gives an expression for fluorescence as a function of total quencher concentration, [Q]: (5) 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 3.8 M pyrene in a micellar solution with a total concentration of sodium dodecyl sulfate [S] 20.8 mM. Q ( M) I0/IQ Q ( M) I0/IQ Q ( M) I0/IQ 0 1 158 2.03 316 4.04 53 1.28 210 2.60 366 5.02 105 1.61 262 3.30 418 6.32 SOURCE: M. F. R. Prieto, M. C. R. Rodrguez, M. M. Gonzlez, A. M. R. Rodrguez, and J. C. M. Fernndez, Fluorescene Quenching in Microheterogeneous Media, J. Chem. Ed. 1995, 72, 662. (a) If micelles were not present, quenching would be expected to follow the Stern-Volmer equation 18-25. Show that the graph of I0/IQ versus [Q] is not linear. (b) The critical micelle concentration is 8.1 mM. Prepare a graph of ln(I0/IQ) 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, , when [Q] 0.200 mM. (d) Compute the fractions of micelles containing 0, 1, and 2 molecules of Q when [Q] 0.200 mM. Q ln I