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Chapter 26: Problem 6

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.

Short Answer

Expert verified
The permissible phenol concentration range is \(1.14 \times 10^{-7}\) M to \(1.87 \times 10^{-6}\) M.

Step by step solution

01

Understand Beer-Lambert Law

The Beer-Lambert Law describes the absorption of light in a medium. It is expressed as \( A = \varepsilon c l \), where \( A \) is the absorbance, \( \varepsilon \) is the molar absorptivity, \( c \) is the concentration, and \( l \) is the path length. Transmittance (\( T \)) is related to absorbance by \( T = 10^{-A} \). From this, we can also express \( A \) as \( A = -\log_{10}(T) \).
02

Calculate Absorbance for Given Transmittance Range

Using the transmittance percentages provided, calculate the corresponding absorbance values using the formula \( A = -\log_{10}(T) \). For \( T = 0.85 \), \( A_{max} = -\log_{10}(0.85) \approx 0.0706 \). For \( T = 0.07 \), \( A_{min} = -\log_{10}(0.07) \approx 1.1549 \).
03

Apply Beer-Lambert Law to Find Concentration Range

Using the Beer-Lambert Law \( A = \varepsilon c l \), solve for concentration \( c \). The path length \( l \) is \( 1 \) cm, and \( \varepsilon = 6.17 \times 10^{5} \, \text{L cm}^{-1} \text{mol}^{-1} \). Thus, \( c = \frac{A}{\varepsilon l} \). Calculate \( c_{min} = \frac{1.1549}{6.17 \times 10^{5} \times 1} \approx 1.87 \times 10^{-6} \) M, and \( c_{max} = \frac{0.0706}{6.17 \times 10^{5} \times 1} \approx 1.14 \times 10^{-7} \) M.
04

Verify and Conclude Range of Concentrations

The permissible concentration of phenol, where transmittance is between 7% and 85%, is between \( 1.14 \times 10^{-7} \) M and \( 1.87 \times 10^{-6} \) M. Check that your calculations match expected physical constraints.

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Key Concepts

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

Molar Absorptivity
Molar absorptivity, often symbolized by \( \varepsilon \), is a key component in the Beer-Lambert Law, which describes how light is absorbed by a solution. It鈥檚 a measure of how well a chemical species absorbs light at a particular wavelength.

It鈥檚 typically expressed in units of \( \text{L} \, \text{cm}^{-1} \, \text{mol}^{-1} \). The higher the molar absorptivity, the better the substance is at absorbing light, which results in a higher absorbance for a given concentration and path length.

In the case of phenol, given its molar absorptivity at \( 211 \, \mathrm{nm} \) is \( 6.17 \times 10^{5} \, \text{L} \, \text{cm}^{-1} \, \text{mol}^{-1} \), this value is crucial for calculating how changes in concentration would influence the absorbance under these conditions.
Transmittance
Transmittance, represented as \( T \), is a measure of how much light passes through a sample compared to how much originally entered it. It is often expressed as a percentage (%). A transmittance of 85% means that 85% of the light passes through, and 15% is absorbed.

Transmittance is inversely related to absorbance, as shown by the equation \( A = -\log_{10}(T) \). This means that as the transmittance decreases, more light is absorbed, and the absorbance value increases.
  • Transmittance of 85% corresponds to minimal absorbance, meaning the solution is mostly transparent.
  • Transmittance of 7% indicates high absorbance, suggesting the solution is very opaque.
Understanding this relationship is vital for calculating the corresponding absorbance, which will then be used to determine the concentration range of phenol in the solution.
Concentration Range
The concentration range provides insight into the limits within which a chemical can be found to maintain a particular transmittance. We calculate it using the Beer-Lambert Law, which connects concentration \( c \), absorptivity \( \varepsilon \), path length \( l \), and absorbance \( A \) with the formula \( A = \varepsilon c l \).

For phenol, we aim to find the concentration range of phenol that allows transmittance between 7% and 85%. Calculated absorbances at those transmittance values lead us to determine the concentration limits:

  • Upper Limit: A transmittance of 85% implies a minimal concentration of \( 1.14 \times 10^{-7} \) \( \text{mol/L} \).
  • Lower Limit: A transmittance of 7% implies a maximum concentration of \( 1.87 \times 10^{-6} \) \( \text{mol/L} \).
These calculations ensure that with given molar absorptivity and path length, the concentration stays within practical limits to achieve the desired transmittance.
Absorbance Calculation
Absorbance, denoted by \( A \), is a dimensionless number representing the amount of light absorbed by a sample. It's essential in quantifying how opaque a solution is at a particular wavelength.

To calculate absorbance, use the formula \( A = -\log_{10}(T) \), where \( T \) is transmittance. For instance:
  • At a transmittance of 85% (or 0.85), the absorbance \( A_{max} \) is approximately 0.0706.
  • At a transmittance of 7% (or 0.07), the absorbance \( A_{min} \) is approximately 1.1549.
These calculated absorbance values serve as the foundation to apply the Beer-Lambert Law in deriving the concentration of a solute within the solution. Understanding how absorbance changes with concentration helps in designing experiments and calculations in chemical analysis.

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Most popular questions from this chapter

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.

Sketch a photometric titration curve for the titration of \(\mathrm{Sn}^{2+}\) with \(\mathrm{MnO}_{4}^{-}\). What color radiation should be used for this titration? Explain.

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