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Chapter 19: Q1P (page 484)

This problem can be worked by calculator or with the spreadsheet in Figure 19-4. Consider compounds X and Y in the example labeled “Analysis of a Mixture, Using Equations 19-6” on page 464. 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.

Short Answer

Expert verified

The solution for [X] isx=8.034‐10-5M

The solution for [Y] isy=2.62‐10-4M

Step by step solution

01

Find the solution for [X] :

X=0.2333870.2006421640387399642X=0.233‐642-387‐0.2001640‐642-399‐387X=72.186898467X=8.034‐10-5M

02

Find the solution for [Y]:

The solution of [Y] can be calculated as follows

y=16400.2333990.2001640387399642y=1640·0.200-399·0.2331640·642-399·387y=235.033898467y=2.6.10-4M

03

Spreadsheet:

The formula use for the cell F5 and F6

= MMULT(MINVERSE(B5:C6); D5:D6)

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Most popular questions from this chapter

The spreadsheet gives the productεbfor four pure compounds and a mixture at infrared wavelengths. Modify Figure 19-4 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 path length was constant (but unknown) for all measurements.

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

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.

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(7.5nM)was equilibrated with various concentrations of albumin for 30minat37°C.A small fraction of unbound estradiol was removed by solid phase microextraction (Section24-4) 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]0and the final concentration of unbound estradiol [X]. Then bound estradiol is[X]0―[X]and the equilibrium constant is

X+P⇌PX   K=[PX][X][P]=[X]0-[X][X][P]

which you can rearrange to
localid="1663648487221" [X]0[X]=K[P]+1

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 estradiol extracted from a solution without albumin divided by the counts of estradiol extracted from a solution with estradiol. (b) What fraction of estradiol is bound to albumin at the first and last points?

This problem can be worked with Equations 19-6 on a calculator or with the spreadsheet in Figure 19-4. Transferrin is the iron-transport protein found in blood. It has a molecular mass of 81 000 and carries twoFe3+ions. Desferrioxamine B is a chelator used to treat patients with iron overload (see the opening of Chapter 12). It has a molecular mass of about 650 and can bind oneFe3+Fe31. 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.


(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.
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