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.

Short Answer

Expert verified

The value of\({\rm{\Delta }}\varepsilon \)and K which is in cell B10:B11 will minimize \({\rm{\Sigma }}{\left( {{A_{{\rm{oths\;}}}} - {A_{{\rm{calc\;}}}}} \right)^2}\)in cell I21. We get the spreadsheet as,

Step by step solution

01

Find K:

The value of\({\rm{\Delta }}\varepsilon \)and K which is in cell B10:B11 will minimize \({\rm{\Sigma }}{\left( {{A_{{\rm{oths\;}}}} - {A_{{\rm{calc\;}}}}} \right)^2}\)in cell I21. We get the spreadsheet as,

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Most popular questions from this chapter

Method of continuous variation. Make a graph of absorbance versus mole fraction of thiocyanate from the data in the table.

(c)Why does one solution containacid and the otheracid?

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”?

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 2000cm-1in 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 0.0100M solution of each, in cells with 0.00500 - cm path lengths. A mixture of A and B in a 0.00500 - cm cell gave a transmittance 34.0 % of at2022cm and 383% at 1093 cm. Find the concentrations of A and B.

Chemical equilibrium and analysis of a mixture. (Warning! This is a long problem.) A remote optical sensor for CO2in the ocean was designed to operate without the need for calibration.33


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

measurement, the sensor is flushed with fresh solution containingbromothymol blue indicator. All indicator is in the formnear neutral pH, so we can

write two mass balances:

[HIn]+[ln2]=FIn=50.0μMand[Na+]=F=50.0μM+42.0μM=92.0μM

has an absorbance maximum at 434 nm andhas a maximum at 620 nm. The sensor measures the absorbance ratio RA=A620/A434reproducibly without need for calibration. From this ratio, we can findin the seawater as outlined here:

(a).From Beer’s law for the mixture, write equations forin terms of the absorbance at 620 and 434 nmThen show that

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

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

, show that

[Hln]=F1nRln+1 (B)

[ln2]=KlnFln[H+](Rln+1) (C)

(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?

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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