Spectrophotometry of chromate-dichromate equilibrium

  • Thread starter Thread starter 24forChromium
  • Start date Start date
  • Tags Tags
    Equilibrium
Click For Summary
SUMMARY

The discussion focuses on using spectrophotometry to analyze the chromate-dichromate equilibrium by measuring the absorbance of colored ions in solution. It emphasizes the importance of selecting a wavelength where absorbance is maximized for optimal sensitivity, following the Beer-Lambert law. The procedure involves determining total absorbance (A_total) as a sum of individual absorbances (A_1 and A_2) at chosen wavelengths, allowing for the calculation of concentrations (C_1 and C_2) of each ion in a mixture. The selection of wavelengths with distinct absorptivity ratios is crucial to minimize measurement errors.

PREREQUISITES
  • Understanding of Beer-Lambert law
  • Familiarity with spectrophotometry techniques
  • Knowledge of absorbance and transmittance concepts
  • Basic principles of chemical equilibrium
NEXT STEPS
  • Research the Beer-Lambert law and its applications in spectrophotometry
  • Learn about selecting optimal wavelengths for absorbance measurements
  • Explore methods for analyzing mixtures using spectrophotometric data
  • Investigate the impact of path length on absorbance in spectrophotometric analysis
USEFUL FOR

Chemists, laboratory technicians, and students studying analytical chemistry or spectrophotometry who are interested in quantitative analysis of chemical mixtures.

24forChromium
Messages
155
Reaction score
7
The goal is to determine the relative abundance of the ions using the spectrophotometer. As far as I understand, to determine the relative (or absolute) concentration of a coloured ion in solution, one must have the absorbance/transmittance of the solution at multiple wavelengths.

Once the absorbance/transmittance of a solution over various wavelengths of light is acquired, one should do something with the maximum point of the absorbance/transmittance over frequency graph, which I do not know.

Please explain the following steps, point out any mistakes in my procedure and suggest a wavelength range to examine the chromate-dichromate mixture in.
 
Chemistry news on Phys.org
24forChromium said:
As far as I understand, to determine the relative (or absolute) concentration of a coloured ion in solution, one must have the absorbance/transmittance of the solution at multiple wavelengths.

No, you select one wavelength for that. Typically we will select the one where absorbance is the highest, as it gives you the highest sensitivity (you can measure lower concentrations).

Google Beer-Lambert law.
 
Borek said:
No, you select one wavelength for that. Typically we will select the one where absorbance is the highest, as it gives you the highest sensitivity (you can measure lower concentrations).

Google Beer-Lambert law.
Thanks for the reply, so what happens is that, across a certain length of the liquid, the solution will absorb an amount of light that is related to its concentration. Is there a way to find out how much of each ion there is if there is a mixture of two liquids?

Here is what I think: first decide on a certain wavelength, find the total absorbance of the mixture (A_total)

A_total = A_1 + A_2 where A_1 and A_2 represent the absorbance of each chemical at this wavelength
A_total = a_1 * L * C_1 + a_2 * L * C_2 where a represent the absorptivity of each chemical, L is the length of the examined liquid, C represent the concentration of each solute.

Is that right?
 
OK so far. Now, if you select another wavelength you will have another measurement result. That will yield second equation in the same two unknowns (c1 and c2).

You just have to select such pair of wavelengths that a1/a2 ratios are as different as possible (that will help minimalize errors).
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
1
Views
4K
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 33 ·
2
Replies
33
Views
12K
Replies
5
Views
16K