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Chapter 18: Problem 6

semithymol blue (STB), and methylthymol blue (MTB) are shown in the table. A solution of TB, STB, and MTB in a \(1.000-\mathrm{cm}\) cuvet had absorbances of \(0.412\) at \(455 \mathrm{~nm}, 0.350\) at \(485 \mathrm{~nm}\), and \(0.632\) at \(545 \mathrm{~nm}\). Modify the spreadsheet in Figure 18-5 to handle three simultaneous equations and find [TB], [STB], and [MTB] in the mixture.

Short Answer

Expert verified
Set up equations using Beer-Lambert Law, input them into a spreadsheet, and solve using matrix algebra for dye concentrations.

Step by step solution

01

Understanding the Problem

You are given a mixture of dyes with three different absorbances at specific wavelengths. The task is to determine the concentration of each dye in the mixture using the measured absorbances and the linear relationship provided by the Beer-Lambert Law.
02

Beer-Lambert Law Introduction

The Beer-Lambert Law relates absorbance (A) to concentration (c) and path length (l), \[ A = \varepsilon \cdot c \cdot l \] where \( \varepsilon \) is the molar absorptivity. We need to relate this to each dye's contribution at each wavelength.
03

Setting Up Equations

Assume the following equations based on Beer-Lambert Law: \[ 0.412 = \varepsilon_{TB}^{455} [TB] + \varepsilon_{STB}^{455} [STB] + \varepsilon_{MTB}^{455} [MTB] \] \[ 0.350 = \varepsilon_{TB}^{485} [TB] + \varepsilon_{STB}^{485} [STB] + \varepsilon_{MTB}^{485} [MTB] \] \[ 0.632 = \varepsilon_{TB}^{545} [TB] + \varepsilon_{STB}^{545} [STB] + \varepsilon_{MTB}^{545} [MTB] \] Here, \( \varepsilon_{Dye}^{\lambda} \) is the molar absorptivity of each dye at respective wavelengths.
04

Spreadsheet Setup

Set up a spreadsheet where each row represents an equation and each column corresponds to a concentration or known absorbance value. Input \( \varepsilon_{Dye}^{\lambda} \) values in the cells to form a matrix alongside the concentration variables for TB, STB, and MTB.
05

Solving Using Matrix Algebra

Using the spreadsheet, set the equations in a matrix form as \ \[ A = \varepsilon \cdot [Dye] \] Use matrix algebra, preferably the inverse method, to solve for \[ [Dye] = \varepsilon^{-1} \cdot A \] Utilize spreadsheet functions for matrix multiplication and inversion to find concentrations \( [TB], [STB], [MTB] \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Absorptivity
Molar absorptivity, also known as the molar extinction coefficient, is a measure of how well a substance absorbs light at a given wavelength. It is denoted by the symbol \( \varepsilon \). This value is crucial in using the Beer-Lambert Law, which relates absorbance to concentration. Molar absorptivity has units of \( \mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{cm}^{-1} \), and it reflects how strongly a chemical species attenuates light. In practical terms, when you know the molar absorptivity, you can determine the concentration of a solution by measuring how much light it absorbs at a specific wavelength. The relationship also depends on path length, but in standard cuvettes, this is usually 1 cm, simplifying calculations. This property allows different substances to be distinguished and quantified in mixtures, provided their absorbance values and wavelengths are known.
Simultaneous Equations
Simultaneous equations are sets of equations containing multiple variables that are solved together. In the context of determining concentrations of different dyes in a mixture, simultaneous equations arise from applying the Beer-Lambert Law to several wavelengths.For example, consider a solution with three dyes whose combined absorbance can be expressed as a system of equations like:
  • \( 0.412 = \varepsilon_{TB}^{455} [TB] + \varepsilon_{STB}^{455} [STB] + \varepsilon_{MTB}^{455} [MTB] \)
  • \( 0.350 = \varepsilon_{TB}^{485} [TB] + \varepsilon_{STB}^{485} [STB] + \varepsilon_{MTB}^{485} [MTB] \)
  • \( 0.632 = \varepsilon_{TB}^{545} [TB] + \varepsilon_{STB}^{545} [STB] + \varepsilon_{MTB}^{545} [MTB] \)
By setting up these equations, we can solve for the concentrations \( [TB] \), \( [STB] \), \( [MTB] \) using methods such as substitution, elimination, or, more efficiently, matrix algebra.
Absorbance and Concentration Relationship
The relationship between absorbance and concentration is central to understanding how the Beer-Lambert Law can be applied in practice. According to this law, absorbance (A) is directly proportional to the concentration (c) of the absorbing species and the path length (l) of the cuvette:\[ A = \varepsilon \cdot c \cdot l \]This means that if you keep the path length constant, the absorbance measured at a certain wavelength will increase linearly with the concentration of the solution. Therefore, by measuring absorbance, you can back-calculate to find concentration if the molar absorptivity \( \varepsilon \) is known. This linear relationship simplifies analyzing mixtures with overlapping spectra, as different compounds will have distinct \( \varepsilon \) values at various wavelengths.
Matrix Algebra in Spreadsheets
Matrix algebra involves performing mathematical operations on arrays of numbers, specifically matrices, using functions and techniques such as matrix multiplication, addition, and inversion. In spreadsheets, this technique becomes very handy when we want to solve a system of simultaneous linear equations.Consider the Beer-Lambert equations presented earlier as a set of matrix equations: \[ A = \varepsilon \cdot [Dye] \]Here, \( A \) is a vector of absorbance values, \( \varepsilon \) is the matrix of molar absorptivity, and \( [Dye] \) is the vector containing dye concentrations we need to find. By rearranging to solve for \( [Dye] \), we get:\[ [Dye] = \varepsilon^{-1} \cdot A \]This approach uses the inverse of the \( \varepsilon \) matrix to compute concentrations. Most spreadsheet software, like Excel, provides built-in functions to handle matrices such as \( \text{MMULT} \) for multiplication and \( \text{MINVERSE} \) for inversion, allowing you to easily find the required concentrations and validate your results.

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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 (7.5 nM) was equilibrated with various concentrations of albumin for \(30 \mathrm{~min}\) at \(37^{\circ} \mathrm{C}\). A small fraction of unbound estradiol was removed by solid-phase microextraction (Section 23-4) 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 \([\mathrm{X}]_{0}\) and the final concentration of unbound estradiol \([X]\). Then bound estradiol is \([\mathrm{X}]_{0}-[\mathrm{X}]\) and the equilibrium constant is $$ \mathrm{X}+\mathrm{P} \rightleftharpoons \mathrm{PX} \quad K=\frac{[\mathrm{PX}]}{[\mathrm{X}][\mathrm{P}]}=\frac{[\mathrm{X}]_{0}-[\mathrm{X}]}{[\mathrm{X}][\mathrm{P}]} $$ which you can rearrange to $$ \frac{[\mathrm{X}]_{0}}{[\mathrm{X}]}=K[\mathrm{P}]+1 $$ A graph of \([\mathrm{X}]_{0} /[\mathrm{X}]\) versus \([\mathrm{P}]\) should be a straight line with a slope of \(K\). The quotient \([\mathrm{X}]_{0} /[\mathrm{X}]\) 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. (a) Prepare a graph of \(\left.[\mathrm{X}]_{0} / \mathrm{X}\right]\) versus \([\mathrm{P}]\) and find \(K\). From the standard deviation of the slope, find the \(95 \%\) confidence interval for \(K\). (b) What fraction of estradiol is bound to albumin at the first and last points? $$ \begin{array}{lc|ll} {[\mathrm{X}]_{0} /[\mathrm{X}]} & {[\mathrm{P}](\mu \mathrm{M})} & {[\mathrm{X}]_{0} /[\mathrm{X}]} & {[\mathrm{P}](\mu \mathrm{M})} \\ \hline 1.26 & 6.3 & 3.33 & 50.0 \\ 1.62 & 10.0 & 4.19 & 60.0 \\ 2.16 & 20.0 & 4.13 & 70.0 \\ 2.51 & 30.0 & 4.36 & 80.0 \\ 3.34 & 40.0 & & \\ \hline \end{array} $$

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