Question

3. Individual spectra of pure compounds A and B were taken using 1.00 x 10M solutions of each, using a cell with a 5.00 cm path length. The transmittance data for each solution at two key wavenumbers are as follows: Solution ) % Transmittance at Wavenumber (cm 2022 1993 31.0 79.7 97.4 20.0 Pure A Pure B An unknown mixture of A and B in a 5.00 cm cell gave transmittance values of 50.6% at 2022 cm? and 20.1% at 1993 cm . Find the concentrations of both A and B in the mixture.

Answer 1

Absorbance (A) is related to Transmittance (T) as:

A = -logT

According to lambert beer law, A = $\epsilon$ cl, where $\epsilon$ is the molar absorptivity coefficient, unit = M^-1cm^-1

                                                                   (Note that it measures how strongly a specie absorbs at a given wavelength),

                                                                   c = concentration of specie in M

                                                                   l = path length in cm.

In the above problem,

Step1 : calculate the absorbance of each specie using the transmittance data. (A = -logT)

	A (at 2022cm-1)	A (at 1993cm-1)
pure A	-log (0.31)   = 0.51	-log(.797)= 0.098
pure B	-log(0.974) = 0.011	-log (0.20) = 0.70

Step 2 : Apply lambert beer law, A = $\epsilon$ cl; to determine $\epsilon$ for each specie for both the given wavelengths.

Given that c = 1.00x 10^-5 M for both the species when pure, and l =5.00 cm,

for pure A and pure B; $\epsilon$ (in units of M^-1cm^-1) = A/cl   = A/(5.00x 10^-5 )

	$\epsilon$ x 105(at 2022cm-1) (in M-1cm-1)	$\epsilon$x 10-5 (at 1993cm-1) (in M-1cm-1)
pure A	0.102	0.02
pure B	0.0022	0.14

Step 3: In the mixture of A and B, note that Absorbance will be additive.

A (total) = -log (T) = A_A + A_B

for 2022 cm^-1, A = -log (0.506) = 0.30 = ( $\epsilon$ _Ac_A + $\epsilon$ _Bc_B) l = (0.102c_A + 0.011c_B) 5x 10⁵

^{$\Rightarrow$}        0.102c_A + 0.011c_B = 0.06 x 10^-5

for 1993 cm^-1, A = -log (0.201) = 0.70 = ( $\epsilon$ _Ac_A + $\epsilon$ _Bc_B) l = (0.02c_A + 0.14c_B) 5x 10⁵

^{$\Rightarrow$}         0.02c_A + 0.14 c_B = 0.14 x 10^-5

2 linear equations with 2 variables are obtained, solving which,

c_A = 0.49 x 10^-5 M
c_B =   0.93x 10^-5 M

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