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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

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?

Spectroscopic data for the indicators thymol blue (TB),semithymol blue (STB), and methylthymol blue (MTB) are shown in the table. A solution ofTB,STB,MTB in a1.000-cm cuvet had absorbances of 0.412at455nm,0.350 at485nm, and 0.632 at545nm. Modify the spreadsheet in Figure 19-4 to handle three simultaneous equations and find[TB],[STB]and[MTB]in the mixture.

Challenging your acid-base prowess. A solution was prepared by mixing 25.00mL of 0.800Maniline, 25.00mLsulfanilic acid, andand then diluting to 100.0mL. (stands for protonated indicator.)


The absorbance measured at550nmin 5.00 - cmwas 0.110.Find the concentrations ofHIn and In andpafor HIn

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?

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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