The transmittance for a sample of a solution, measured at #590nm# in a #1.5-cm# cuvette, was #76.2%#. What is the corresponding absorbance?

Question

Owen Ambrose · Accepted Answer

Just as loudness and sound intensity are related through a logarithmic relationship, so too does absorbance and transmittance.

#A = log(1/T)#

where #T# is the transmittance. Why must this be #1/T# since #0 < T < 1#?

We thus know what the absorbance is:

#color(blue)(A) = log(1/0.762) = color(blue)(0.118)#

At low concentrations, Beer's law can be used to see a linear relationship based on the concentration:

#A = epsilonbc#

where

The molar absorptivity would therefore be:

#color(blue)(epsilon) = A/(bc)#

#= 0.118/("1 cm" cdot "0.0802 mol/L")#

#= color(blue)("1.472 L/mol"cdot"cm")#

which is not too high.

Physically speaking, it makes sense because, despite the low concentration, the absorbance was relatively low.

If the absorptivity was already known to be #"1000 L/mol"cdot"cm"#, one could find the concentration to be:

#color(blue)(c) = A/(epsilonb)#

#= 0.118/("1000 L/mol"cdot"cm" cdot "1 cm")#

#= color(blue)(1.181 xx 10^(-4) "M")#

The absorbance was only this low because there was insufficient particle concentration to absorb the incoming excitation light source if the molar absorptivity was this high, which is quite large.

Peyton Adams · Answer

undefined To calculate the corresponding absorbance, you can use the formula:

Absorbance = -log10(Transmittance)

Given that the transmittance is 76.2%, plug this value into the formula:

Absorbance = -log10(0.762)

Calculating this yields the corresponding absorbance value.