Question

A solution contains 1.0mg of KMnO4/L. When measured in a 1.00cm cell at 525 nm, the transmittance was 0.300. When measured under similar conditions at 500 nm, the transmittance was 0.350. (a) Calculate the absorbance A at each wavelength, (b) Calculate the molar absorptivity at each wavelength. (c) What would T be if the cell length were in each case 2.00cm? (d) Calculate the absorptivity (if concentration is in mg/L) for the solution at each wave length.

Answer 1

The Beer-Lambert law (or Beer's law) is the linear relationship between absorbance and concentration of an absorbing species. The general Beer-Lambert law is usually written as:

A = a() * b * c (1)

where A is the measured absorbance, a() is a wavelength-dependent absorptivity coefficient, b is the path length, and c is the analyte concentration. When working in concentration units of molarity, theBeer-Lambert law is written as:

A = * b * c (2)

where is the wavelength-dependent molar absorptivity coefficient with units of M^-1 cm^-1. Data are frequently reported in percent transmission (I/I₀ * 100) or in absorbannce [A = log (I/I₀)].

Experimental measurements are usually made in terms of transmittance (T), which is defined as:

T = I / I_{o (3)}

where I is the light intensity after it passes through the sample and I_o is the initial light intensity. The relation between A and T is:

A = -log T = - log (I / I_o). (4)

Absorption of light by a sample

The Beer-Lambert law can be rearranged to obtain an expression for ϵ (the molar absorptivity):

ϵ=A/bc (5)

(A) Answer of (a) 525nm

On basis of equation (4)

A = -log T

= -log 0.300

   =0.5229

(b) On basis of equation (5)

ϵ=A/bc where,

A=0.5229 , I=1 ,C is conc. of KMnO4 in Molar i.e. 6.3263*10^-6

ϵ=0.5229/(1*6.3263*10^-6)=82632.53

ϵ=82632.53

(c) If l=2 then

A=ϵbc

=82632.53*2*6.3263*10^-6=1.0458

A=1.0458

Hence, A=-logT

i.e. 1.0458=-logT

   T=0.0448

And if conc. in mg/mL then ϵ=0.5229 as per equation (5)

(B) Answer of (a) 500nm

On basis of equation (4)

A = -log T

= -log 0.350

   =0.4559

(b) On basis of equation (5)

ϵ=A/bc where,

A=0.4559 , I=1 ,C is conc. of KMnO4 in Molar i.e. 6.3263*10^-6

ϵ=0.4559/(1*6.3263*10^-6)=72064.24

ϵ=72064.24

(c) If l=2 then

A=ϵbc

=72064.24*2*6.3263*10^-6=0.9118

A=0.9118

Hence, A=-logT

i.e. 0.9118=-logT

   T=0.0923

And if conc. in mg/mL then ϵ=0.4559 as per equation (5)