The figure shows spectra of$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathbf{MMnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$and an unknown mixture of both, all in$\mathbf{1}\mathbf{.}\mathbf{000}\mathbf{‐}\mathbf{cm}\mathbf{‐}$path 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 of${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{,}{\mathbf{Cr}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{,}$ and an unknown mixture containing both ions.

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

$\mathrm{A}=\mathrm{\epsilon bC}$

A is the absorbance,$\mathrm{\epsilon }$is the molar absorptivity,b is the length of light path, C is the concentration.

Make a spreadsheet,

Calculatevalues using Beer’s Law,

$\partial =\frac{\mathrm{A}}{\mathrm{b}-\left[\mathrm{standard}\right]}$

Thus, Column can be calculated by using the formula,

$\mathrm{A}=\partial \mathrm{x}‐\mathrm{b}‐{\left[\mathrm{X}\right]}_{\mathrm{guess}}+\partial ‐\mathrm{b}‐{\left[\mathrm{Y}\right]}_{\mathrm{guess}}$

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 tab$\to$Solver 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.

$\left[{\mathrm{MnO}}_4\right]=8.32·{10}^{-5}\mathrm{M}\left[{\mathrm{Cr}}_2{\mathrm{O}}_4\right]=1.78·{10}^{-4}\mathrm{M}.$