UV/VISIBLE SPECTROSCOPY

UV/VISSpectroscopy

What does a UV/Vis Spectrometer

Measure?

Transmittance (T) is defined as the amount of light passing through the sample solution (P) divided by

the amount of incident radiation (Po)

T = P/Po

In practice Po is estimated by a “blank” which accounts for reflection and scattering losses.

UV/VISSpectroscopy

UV/VISSpectroscopy

As the radiation is absorbed in the sample,the total intensity of radiation is reducedas it travels through the sample. This results in a non-linear relationship between transmittance and concentration.

Absorbance (A) is based on the amount of light absorbed by the solution and is defined as the log of the inverse of the transmittance.

A = log10(1/T)

UV/VISSpectroscopy

UV/VISSpectroscopy

Within limits, the relationship between absorbance and concentration is linear.

Beer’s law or the Beer-Lambert law

A = bc

is the molar absorptivity with units of (L/mol-cm)b is the path length of the sample ie. the inside cross section of the sample cuvette (cm)c is the concentration of the compound in solution in (mol/L)

UV/VISSpectroscopy

Increased interaction between the molecules can affect absorbance at high concentrations.

UV/VISSpectroscopy

Electrolytes can also alter the molar absorptivity of the analyte by electostatic interactions.

Molar absorptivity is also altered by the refractive index of the solution.

Hindicator H+ + Indicator-

When an analyte associates or dissociates in solution producing products with different absorption spectra,

deviations from Beer’s law can be observed if the equilibrium shifts at different concentrations.

UV/VISSpectroscopy

Other causes of deviation from Beer’s law:Radiation is not monochromatic

Stray radiation

Absorbance readings less than 0.10 and higher than 1.5 can contain significant error and should be

avoided.

UV/VISSpectroscopy

Quantification of Two Compounds

Spectra of Compound 1Spectra of Compound 2

Wavelength

Ab

sorb

ance

UV/VISSpectroscopy

Quantification of Two Compounds

The method to determine the concentrations of two compounds (a & b) in a mixture involves the simultaneously solution of the following two equations:

At wavelength 1 A1 = ea1ca + eb1cb

At wavelength 2 A2 = ea2ca + eb2cb

UV/VISSpectroscopy

UV/VISSpectroscopy

Quantification of Two Compounds

At wavelength 1 A1 = ea1ca + eb1cb

At wavelength 2 A2 = ea2ca + eb2cb

Where ea1, eb1, ea2, eb2 are the molar absorptivities for the two compounds at the two wavelengths which are determined from standard solutions - and ca and cb are the concentrations of the two unknown compounds.

A = bc

Quantification of Two Compounds

Sample Absorptivity

at λ465 Absorptivity at

λ540 Compound a 11636 26579 Compound b 17949 2667

Absorbance of the mixture at λ465 = 0.870 at λ540 = 0.362

EXAMPLE

UV/VISSpectroscopy

Quantification of two compounds

Set up equations:

(1) 0.870 = 11636ca + 17949cb

(2) 0.362 = 26579ca + 2667cb

(3) 0.870 = 11636ca + 17949cb

(4) 2.436 = 178877ca + 17949cb

multiply equation (2) by 6.73, giving equation (4):

UV/VISSpectroscopy

Quantification of two compounds

Set up equations:

(3) 0.870 = 11,636ca + 17,949cb

(4) 2.436 = 178,877ca + 17,949cb

Subtract (3) from (4), giving equation (5):

(5) 1.566 = 167,240ca

UV/VISSpectroscopy

Quantification of Two Compounds (5) 1.566 = 16724ca

Solve for ca:(6) ca = 9.36 x 10-6 moles/L = concentration of compound a

Substitute value from (6) into equation (1) and solve for cb:0.870 = 11636(9.36 x 10-6) + 17949cb

0.870 = 0.109 + 17949cb

0.760 = 17949cb

cb = 4.24 x 10-5 moles/L = concentration of compound b

UV/VISSpectroscopy