Chapter 9, Problem 9.89QP

Absorbance values for five standard solutions of a colored solute were determined at 410 nm with a 1.00-cm path length, giving the following table of data:

Solute concentration (M)	A
0.250	0.165
0.500	0.317
0.750	0.510
1.000	0.650
1.250	0.837

The absorbance of a solution of unknown concentration containing the same solute was 0.400. What is the concentration of the unknown solution (page 385)?

9.89 Referring to Problem 9.88, (a) determine the absorbance values you would expect for solutions with the following concentrations: 0.4 M, 0.6 M, 0.8 M, 1.1 M. (b) Using data in the table in Problem 9.88, calculate the average molar absorptivity of the compound and determine the units of molar absorptivity (page 385).

Interpretation Introduction

Interpretation:

The average molar absorptivity and its unit of the given compound should be determined.

Concept introduction:

Beer-Lambert law:

• The absorbance of electromagnetic radiation for a chemical compound is directly proportional to the concentration and path length of the compound.

$\text{A}\text{=}\text{εbc}$

Where,

$\begin{array}{l}\text{A=}\text{Absorbance}\\ \text{ε}\text{=}\text{Molar}\text{absorptivity}\\ \text{b=}\text{Path}\text{length}\text{of}\text{the}\text{solution}\text{(cm)}\\ \text{c}\text{=}\text{Molar}\text{comcentration}\text{of}\text{the}\text{solution}\end{array}$

Answer to Problem 9.89QP

• The concentration of the unknown solution is $\text{0}\text{.600 M}$.

Explanation of Solution

Record the give data.

Solute concentration (M)	Absorbance
0.250	0.165
0.500	0.317
0.750	0.510
1.000	0.650
1.250	0.837

The concentrations of unknown solutions are $\text{0}\text{.4M,0}\text{.6M,0}\text{.8M,1}\text{.1M}\text{.}$

• The given concentrations and absorbance’s of the standard and unknown solutions are recorded as shown above.

The absorbance’s of unknown solutions.

• According to the Beer-Lambert law the concentration is directly proportional to the absorbance of the solution so the plot of concentration and absorbance of standard solutions to give a slope.
• The plot of concentrations agenise absorbance of standard solutions is draws first.

Fig.1

• The concentrations of unknown solutions are $\text{0}\text{.4M,0}\text{.6M,0}\text{.8M,1}\text{.1M}\text{.}$ noted in the plot (Fig.1) and to measure the absorbance of unknown solution.
• In the plot (Fig.1) absorbance of unknown solutions are measured,

Solute concentration (M)	Absorbance
0.4	0.26
0.6	0.40
0.8	0.53
1.1	0.73

The average molar absorptivity and it unit of the given compound

$\begin{array}{l}\text{Rearranging}\text{equation}\text{1,}\\ \text{ε}\text{=}\frac{\text{A}}{\text{bc}}\\ \text{For}\text{0250M}\\ \text{=}\frac{\text{0}\text{.165}}{\text{(1cm)(0}\text{.250mol/L)}}\\ \text{=}\text{0}\text{.660}\raisebox{1ex}{$\text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\text{.cm}$}\right.\\ \text{For}\text{0}\text{.500M}\\ \text{=}\frac{\text{0}\text{.317}}{\text{(1cm)(0}\text{.500mol/L)}}\\ \text{=}\text{0}\text{.634}\raisebox{1ex}{$\text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\text{.cm}$}\right.\\ \text{For}\text{0}\text{.750M}\\ \text{=}\frac{\text{0}\text{.510}}{\text{(1cm)(0}\text{.750mol/L)}}\\ \text{=}\text{0}\text{.680}\raisebox{1ex}{$\text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\text{.cm}$}\right.\\ \text{For}1.000\text{M}\\ \text{=}\frac{\text{0}\text{.650}}{\text{(1cm)(1}\text{.000}\text{mol/L)}}\\ \text{=}\text{0}\text{.650}\raisebox{1ex}{$\text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\text{.cm}$}\right.\\ \text{For}1.250\text{M}\\ \text{=}\frac{\text{0}\text{.837}}{\text{(1cm)(1}\text{.250}\text{mol/L)}}\\ \text{=}\text{0}\text{.670}\raisebox{1ex}{$\text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\text{.cm}$}\right.\end{array}$

To take an average of calculated above molar absorptivity to give average molar absorptivity of the given compound.

$\begin{array}{l}\text{=}\frac{\text{0}\text{.660+0}\text{.634+0}\text{.680+0}\text{.650+0}\text{.670}}{\text{5}}\\ \text{=}\text{0}\text{.659}\raisebox{1ex}{$\text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\text{.}\text{cm}$}\right.\end{array}$

• The average molar absorptivity of the given compound is $\text{0}\text{.659}\raisebox{1ex}{$\text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\text{.}\text{cm}$}\right.$
• The unit of molar absorptivity of the given compound is $\raisebox{1ex}{$\text{L}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\text{.}\text{cm}$}\right.$

Conclusion

The average molar absorptivity and its unit of the given compound should be determined.