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$\left[{\mathrm{In}}^{-}\right]\mathrm{and}\left[\mathrm{HIn}\right]$in the mixture. Comment on the sum$\mathbf{\left[}{\mathbf{In}}^{\mathbf{-}}\mathbf{\right]}\mathbf{+}\mathbf{\left[}\mathbf{HIn}\mathbf{\right]}$.

(b) From$\mathbf{\left[}{\mathbf{In}}^{\mathbf{-}}\mathbf{\right]}$in the mixture, and from$\mathit{p}{\mathit{K}}_{\mathbf{a}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{10}$for 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?

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

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

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

Hence, the spreadsheet is,

Calculatevalues using Beer’s Law,

$ϵ=\frac{A}{b\cdot [s\tan dard]}$

Thus, Column can be calculated by using the formula,

$\mathrm{A}={\mathrm{ϵ}}_{\mathrm{x}}\cdot \mathrm{b}\cdot [\mathrm{X}{]}_{\mathrm{guess}}+{\mathrm{ϵ}}_{\mathrm{y}}\cdot \mathrm{b}\cdot [\mathrm{Y}{]}_{\mathrm{Guess}}$

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

Take the concentration is 0.001 M for each compound.

Thus, we calculated column G and H column.

Calculate the sum in column H8.

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

Select data tab$\to$ Solver and enter H8 in Set Objective.

Select the min button and in by Changing Variables enter D10 and D11.

Thus, solving method should be GRG nonlinear.

Hence, set Constraint Precision to a small number such as 1E-12.

Click solve, the values appears in the cell D10 and D11.

Therefore, the values of$\left[{\mathrm{In}}^{-}\right]\mathrm{and}\left[\mathrm{HIn}\right]$ are $\left[{\mathrm{In}}^{-}\right]=3.28.{10}^{-6}\mathrm{M},\left[\mathrm{HIn}\right]=6.91.{10}^{-6}\mathrm{M}$.

Consider the Henderson-Hassel Balch equation,

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \left(\frac{\left[{\mathrm{I}}^{-}\right]}{\left[\mathrm{HIn}\right]}\right)$

$\mathrm{pH}=7.10+\log \left(\frac{7.10+\log 3.28\cdot {10}^{-6}}{6.91\cdot {10}^{-6}}\right)$

pH = 6.78

Calculate the concentration of${\mathrm{In}}^{-}$ by using the formula,

$[\ln ]=\frac{{\displaystyle \left|\begin{array}{lc}{\mathrm{A}}_{\mathrm{mixture}}& {\dot{\mathrm{o}}}_{\mathrm{HIn}}\cdot \mathrm{b}\\ {\mathrm{A}}_{\mathrm{mixture}}& {\dot{\mathrm{o}}}_{\mathrm{HIn}\text{'}}\cdot \mathrm{b}\end{array}\right|}}{\left|\begin{array}{l}{\dot{\mathrm{l}}}_{\mathrm{In}{}^{-}}\cdot \mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\dot{\mathrm{H}}}_{\mathrm{HIn}}\cdot \mathrm{b}\\ {\dot{\mathrm{o}}}_{\mathrm{In}{}^{-}}\cdot \mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\dot{\mathrm{o}}}_{{\mathrm{Hln}}^{\mathrm{\prime }}}\cdot \mathrm{b}\end{array}\right|}$

Use the values from the mention spreadsheet,

$$\begin{gathered} \left[{\ln }^{-}\right]=\frac{\left|\begin{array}{cc}0.265& 34400\\ 0.272& 500\end{array}\right|}{\left|\begin{array}{cc}7600& 34400\\ 81100& 500\end{array}\right|} \\ \left[{\ln }^{-}\right]=\frac{(0.265\cdot 500)-(34400\cdot 0.272)}{(7600\cdot 500)-(34400\cdot 81100)} \\ \left[{\ln }^{-}\right]=3.31\cdot {10}^{-6}\mathrm{M} \end{gathered}$$

Find the concentration of HIn,

$[\mathrm{Hln}]=\frac{\left|\begin{array}{cc}{\dot{\mathrm{o}}}_{{\mathrm{In}}^{-}}\cdot \mathrm{b}& {\mathrm{A}}_{\mathrm{mixture}}\\ {\widehat{\mathrm{o}}}_{\mathrm{l}}{\ln }^{-}\cdot \mathrm{b}& {\mathrm{A}}_{\mathrm{mixture}}\end{array}\right|}{\left|\begin{array}{cc}{\dot{\mathrm{o}}}_{{\mathrm{In}}^{-}}\cdot \mathrm{b}& {\dot{\mathrm{o}}}_{\mathrm{HIn}}\cdot \mathrm{b}\\ {\widehat{\mathrm{o}}}^{\mathrm{\prime }}{\ln }^{-}\cdot \mathrm{b}& {\dot{\mathrm{o}}}_{{\mathrm{HIn}}^{\mathrm{\prime }}}\cdot \mathrm{b}\end{array}\right|}$

Use the values from spreadsheet,

$\mathrm{Hln}]=\frac{\left|\begin{array}{cc}7600& 0.265\\ 81100& 0.272\end{array}\right|}{\left|\begin{array}{cc}7600& 34400\\ 81100& 500\end{array}\right|}$

$$\begin{gathered} [\mathrm{H}\ln ]=\frac{(7600\cdot 0.272)-(81100\cdot 0.265)}{(7600\cdot 500)-(34400\cdot 81100)} \\ \left[\mathrm{Hln}\right]=6.97\cdot {10}^{-6}\mathrm{M} \end{gathered}$$

Therefore, the answer (a) is probably more accurate.