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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.

Short Answer

Expert verified

The concentration of each species in the mixture is.

MnO4=8.32·10-5MCr2O4=1.78·10-4M

Step by step solution

01

State Beer’s Law:

Beer's law states that through the sample and the concentration of the absorbing species, the absorbance is proportional to the path length.

A=εbC

A is the absorbance,εis the molar absorptivity,b is the length of light path, C is the concentration.

02

Using the beer’s Law find the concentration:

Make a spreadsheet,

Calculatevalues using Beer’s Law,

=Ab-standard

Thus, Column can be calculated by using the formula,

A=xbXguess+bYguess

Hence, in cell D10 and D11 we know the values.

Take the concentration is0.001 M for each compound.

Thus, we calculated column G and column H.

Calculate the sum in columnH8.

Use the solver to calculate the concentration of unknown and highlight the cellH8.

Select data tabSolver and enterH8 in Set Objective.

Select the min button and in by Changing Variables enterD10 and D11.

Thus, solving method should be GR G nonlinear.

Hence, set Constraint Precision to a small number such as1E-12.

Therefore, by clicking solve, the values appears in the cell D10 andD11.

Thus, the concentration of each species in the mixture is.

MnO4=8.32·10-5MCr2O4=1.78·10-4M.

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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.

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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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