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Chapter 26: Problem 8

The logarithm of the molar absorptivity of phenol in aqueous solution is \(3.812\) at \(211 \mathrm{~nm}\). Calculate the range of phenol concentrations that can be used if the absorbance is to be greater than \(0.150\) and less than 1.500 with a \(1.25-\mathrm{cm}\) cell.

Short Answer

Expert verified
The concentration range is between \(1.85 \times 10^{-5} \text{ mol/L}\) and \(1.85 \times 10^{-4} \text{ mol/L}\).

Step by step solution

01

Understand the relationship

In this exercise, we are given the logarithm of molar absorptivity and we need to understand the relation between absorbance, concentration, and path length. The Beer-Lambert law is: \[ A = \varepsilon c l \] where \( A \) is absorbance, \( \varepsilon \) is molar absorptivity, \( c \) is concentration in mol/L, and \( l \) is the path length in cm.
02

Convert log of molar absorptivity

We know \( \log(\varepsilon) = 3.812 \). To find \( \varepsilon \), we use \( \varepsilon = 10^{3.812} \). This calculates \( \varepsilon \) as \[ \varepsilon = 6472.23 \text{ L/mol路cm} \].
03

Calculate the minimum concentration

To find the minimum concentration, use the minimum absorbance value (\( A = 0.150 \)). Plug this into the Beer-Lambert equation: \[ 0.150 = 6472.23 \cdot c \cdot 1.25 \]. Solve for \( c \) to get the minimum concentration: \[ c = \frac{0.150}{6472.23 \times 1.25} = 1.85 \times 10^{-5} \text{ mol/L} \].
04

Calculate the maximum concentration

To find the maximum concentration, use the maximum absorbance value (\( A = 1.500 \)). Plug this into the Beer-Lambert equation: \[ 1.500 = 6472.23 \cdot c \cdot 1.25 \]. Solve for \( c \) to get the maximum concentration: \[ c = \frac{1.500}{6472.23 \times 1.25} = 1.85 \times 10^{-4} \text{ mol/L} \].
05

Formulate the concentration range

Now that we have both concentrations, we can now state the range of concentrations of phenol that can give the desired absorbance, which is between \( 1.85 \times 10^{-5} \text{ mol/L} \) and \( 1.85 \times 10^{-4} \text{ mol/L} \).

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Key Concepts

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

Absorbance Calculation
The absorbance calculation is a key part of understanding how to measure the concentration of a solution using light.
Absorbance, indicated by the symbol \( A \), refers to the ability of a substance to absorb light at a specific wavelength.
In the Beer-Lambert Law, absorbance is calculated using the formula:
  • \( A = \varepsilon c l \)
where:
  • \( \varepsilon \) is the molar absorptivity (a measure of how strongly a chemical species absorbs light at a given wavelength),
  • \( c \) represents the concentration of the solution,
  • and \( l \) is the path length, or the distance the light travels through the solution, typically measured in centimeters.
The absorbance doesn't have units because it is a logarithmic measure of the ratio between transmitted light and incident light.
Using absorbance allows scientists to quantify how much a sample absorbs light and therefore deduce the concentration of a solute present in a solution based on how much light is absorbed.
Molar Absorptivity
Molar absorptivity is critical when interpreting results from spectroscopic analyses.
This constant \( \varepsilon \) helps determine how intensely a compound absorbs light at a specific wavelength.
In our problem, we started with the logarithm of the molar absorptivity, which was given as \( \log(\varepsilon) = 3.812 \).
To find \( \varepsilon \), you take the antilogarithm:
  • \( \varepsilon = 10^{3.812} \).
This calculation results in \( \varepsilon = 6472.23 \text{ L/mol路cm} \).
This value shows us how much light, per concentration unit, is absorbed by a 1 cm path length.
Understanding molar absorptivity allows chemists to predict how a compound at various concentrations affects the absorbance, which is crucial for quantitative analysis in spectrophotometry.
Concentration Range
Finding the concentration range is essential for determining the suitable concentrations for reliable absorbance measurements.
The Beer-Lambert Law allows us to set boundaries around the concentrations that provide absorbance within a detectable and effective range.
In this exercise, we are tasked with ensuring the absorbance remains between 0.150 and 1.500.
Using previously calculated molar absorptivity, \( \varepsilon = 6472.23 \text{ L/mol路cm} \), and a path length \( l = 1.25 \text{ cm} \), we solve for the concentration \( c \).
  • Minimum concentration: Using \( A = 0.150 \), \( c = \frac{0.150}{6472.23 \times 1.25} = 1.85 \times 10^{-5} \text{ mol/L}\).
  • Maximum concentration: Using \( A = 1.500 \), \( c = \frac{1.500}{6472.23 \times 1.25} = 1.85 \times 10^{-4} \text{ mol/L}\).
The concentration range is crucial to ensure accuracy and linearity in spectroscopic measurements, so staying within this range yields valid results.

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Most popular questions from this chapter

A standard solution was put through appropriate dilutions to give the concentrations of iron shown in the accompanying table. The iron(II)-1,10,phenanthroline complex was then formed in \(25.0-\mathrm{mL}\) aliquots of these solutions, following which each was diluted to \(50.0 \mathrm{~mL}\) (see color plate 15). The absorbances in the table (1.00-cm cells) were recorded at \(510 \mathrm{~nm}\). \begin{tabular}{cr} Fe(II) Concentration in & \\ Original Solution, Ppm & \(\boldsymbol{A}_{\text {se }}\) \\ \hline \(4.00\) & \(0.160\) \\ \(10.0\) & \(0.390\) \\ \(16.0\) & \(0.630\) \\ \(24.0\) & \(0.950\) \\ \(32.0\) & \(1.260\) \\ \(40.0\) & \(1.580\) \\ \hline \end{tabular} (a) Plot a calibration curve from these data. "(b) Use the method of least squares to find an equation relating absorbance and the concentration of iron(II). "(c) Calculate the standard deviation of the slope and intercept.

The molar absorptivity for aqueous solutions of phenol at \(211 \mathrm{~nm}\) is \(6.17 \times 10^{5} \mathrm{~L} \mathrm{~cm}^{-1} \mathrm{~mol}^{-1}\). Calculate the permissible range of phenol concentrations if the transmittance is to be less than \(85 \%\) and greater than \(7 \%\) when the measurements are made in \(1.00-\mathrm{cm}\) cells.

A photometer with a linear response to radiation gave a reading of \(690 \mathrm{mV}\) with a blank in the light path and \(169 \mathrm{mV}\) when the blank was replaced by an absorbing solution. Calculate *(a) the transmittance and absorbance of the absorbing solution. (b) the expected transmittance if the concentration of absorber is one half that of the original solution. "(c) the transmittance to be expected if the light path through the original solution is doubled.

Palladium(II) forms an intensely-colored complex at \(\mathrm{pH} 3.5\) with arsenazo III at \(660 \mathrm{~nm}{ }^{19} \mathrm{~A}\) meteorite was pulverized in a ball mill, and the resulting powder was digested with various strong mineral acids. The resulting solution was evaporated to dryness, dissolved in dilute hydrochloric acid, and separated from interferents by ion- exchange chromatography (see Section 33D). The resulting solution containing an unknown amount of \(\mathrm{Pd}(\mathrm{II})\) was then diluted to \(50.00 \mathrm{~mL}\) with \(\mathrm{pH} 3.5\) buffer. Ten-milliliter aliquots of this analyte solution were then transferred to six 50 -mL volumetric flasks. A standard solution was then prepared that was \(1.00 \times 10^{-5} \mathrm{M}\) in \(\mathrm{Pd}(\mathrm{II})\). Volumes of the standard solution shown in the table were then pipetted into the volumetric flasks along with \(10.00 \mathrm{~mL}\). of \(0.01 \mathrm{M}\) arsenazo III. Each solution was then diluted to \(50.00 \mathrm{~mL}\), and the absorbance of each solution was measured at \(660 \mathrm{~nm}\) in \(1.00-\mathrm{cm}\) cells. \begin{tabular}{cc} Volume Standard Solution, mL. & \(A_{\text {es0 }}\) \\ \hline \(0.00\) & \(0.209\) \\ \(5.00\) & \(0.329\) \\ \(10.00\) & \(0.455\) \\ \(15.00\) & \(0.581\) \\ \(20.00\) & \(0.707\) \\ \(25.00\) & \(0.833\) \\ \hline \end{tabular} (a) Enter the data into a spreadsheet and construct a standard additions plot. (b) Determine the slope and intercept of the line. (c) Determine the standard deviation of the slope and of the intercept. (d) Calculate the concentration of Pd(II) in the analyte solution. (e) Find the standard deviation of the measured concentration.

Molar absorptivity data for the cobalt and nickel complexes with 2,3-quinoxalinedithiol are \(\varepsilon_{\mathrm{Ce}}=\) 36,400 and \(\varepsilon_{\mathrm{Ni}}=5520\) at \(510 \mathrm{~nm}\) and \(\varepsilon_{\mathrm{Co}}=1240\) and \(\varepsilon_{\mathrm{Ni}}=17.500\) at \(656 \mathrm{~nm}\). A \(0.425-\mathrm{g}\) sample was dissolved and diluted to \(50.0 \mathrm{~mL}\). A \(25.0\) - \(\mathrm{mL}\) aliquot was treated to eliminate interferences; after addition of 2,3-quinoxalinedithiol, the volume was adjusted to \(50.0 \mathrm{~mL}\). This solution had an absorbance of \(0.446\) at \(510 \mathrm{~nm}\) and \(0.326\) at \(656 \mathrm{~nm}\) in a \(1.00-\mathrm{cm}\) cell. Calculate the concentration in parts per million of cobalt and nickel in the sample.
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