UV-Visible Spectroscopy and Colorimetry: Principles and Applications
Expert reviewed •23 November 2024• 5 minute read
Introduction
UV-Visible spectroscopy and colorimetry are fundamental analytical techniques used to determine the concentration of colored compounds in solution. These methods are essential tools in modern chemical analysis, particularly in the analysis of inorganic compounds.
Principles of UV-Visible Spectrophotometry
UV-Visible spectrophotometry involves exposing samples to electromagnetic radiation in two specific ranges:
Ultraviolet (UV) radiation: 10-400 nm
Visible light: 400-700 nm
The technique is based on a simple principle: when radiation passes through a sample, molecules can absorb specific wavelengths of light, causing electrons to transition to higher energy levels. This absorption is selective and depends on the molecular structure of the compound.
Key Components of a UV-Vis Spectrophotometer
Light source (UV and visible)
Monochromator
Sample holder (cuvette)
Detector
Data processor
The Beer-Lambert Law
The fundamental relationship governing absorption spectroscopy is the Beer-Lambert Law:
A=εcl
Where:
A = Absorbance
ε = Molar extinction coefficient
c = Concentration
l = Path length
Colorimetry: Principles and Applications
Colorimetry is a specific application of spectrophotometry that focuses on visible light absorption. It relies on the relationship between a compound's color and its complementary absorbed wavelength.
Color Wheel and Absorption
A solution's color is complementary to the wavelength it absorbs. For example:
Red solutions absorb green light (490-560 nm)
Blue solutions absorb orange light
Yellow solutions absorb violet light
Practical Application: Iron Analysis
Here's a worked example of determining iron concentration in a tablet:
Given:
Sample mass: 0.200 g
Final volume: 200 mL
Measured absorbance: 0.6105
The reaction involved:
Fe(aq)3++SCN(aq)−−>[FeSCN](aq)2+
Calculation steps:
From calibration curve at A = 0.6105:
c=5.5×10−5 mol L−1
Calculate moles of Fe³⁺:
n=cV=5.5×10−5 mol L−1×0.2 L=1.1×10−5 mol
Calculate mass of iron:
m=55.85 g mol−1×1.1×10−5 mol=6.14×10−4 g