A sample in a #1cm# cuvette gives an absorbance reading of #0.558#. If the absorptivity for this sample is #15000 L/(mol.cm)#, what is the molar concentration?

1 Answer
Mar 21, 2017

Answer:

The concentrations are (a) #3.72 × 10^"-5" color(white)(l)"mol·L"^"-1"# and (b) #9.33 × 10^"-6" color(white)(l)"mol·L"^"-1"#.

Explanation:

A. Molar concentration from absorbance

To solve this problem, we use Beer's Law:

#color(blue)(bar(ul(|color(white)(a/a) A = epsilonlc color(white)(a/a)|)))" "#

where

#A =# the absorbance
#epsilon =# the molar absorptivity constant
#l =# the path length of the cuvette
#c = # the molar concentration of the solution.

We can rearrange this equation to get

#c = A/(epsilonl)#

In this problem,

#A = 0.558#
#epsilon =color(white)(l) "15 000 L·mol"^"-1""cm"^"-1"#
#lcolor(white)(ll) = "1 cm"#

Then,

#c = 0.558/("15 000 L·mol"^"-1"color(red)(cancel(color(black)("cm"^"-1"))) × 1 color(red)(cancel(color(black)("cm")))) = 3.72 × 10^"-5" color(white)(l)"mol·L"^"-1"#

Molar concentration from transmittance

Recall that absorbance is defined as

#color(blue)(bar(ul(|color(white)(a/a)A = log(I_0/I)color(white)(a/a)|)))" "#

where

#I_0# and #I# are the intensities of the incident and transmitted light.

Transmittance #T# is defined as

#color(blue)(bar(ul(|color(white)(a/a)T = I/I_0color(white)(a/a)|)))" "#

#A = log(1/T)#

Also, percent transmittance #%T = 100T#

so

#T = (%T)/100#

#A = log(100/(%T))# or

#color(blue)(bar(ul(|color(white)(a/a)A = 2-log(%T)color(white)(a/a)|)))" "#

This last equation is worth remembering because it gives you an easy way to calculate absorbance from percent transmittance.

#A = 2 - log72.6 = "2 - 1.860" = 0.140#

Now, we can use Beer's Law:

#c = A/(epsilonl) = 0.140/("15 000 L·mol"^"-1"color(red)(cancel(color(black)("cm"^"-1"))) × 1 color(red)(cancel(color(black)("cm")))) = 9.33 × 10^"-6" color(white)(l)"mol·L"^"-1"#