# 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?

Mar 21, 2017

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:

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} A = \epsilon l c \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

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

$c = \frac{A}{\epsilon l}$

In this problem,

$A = 0.558$
$\epsilon = \textcolor{w h i t e}{l} \text{15 000 L·mol"^"-1""cm"^"-1}$
$l \textcolor{w h i t e}{l l} = \text{1 cm}$

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

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} A = \log \left({I}_{0} / I\right) \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

${I}_{0}$ and $I$ are the intensities of the incident and transmitted light.

Transmittance $T$ is defined as

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} T = \frac{I}{I} _ 0 \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

$A = \log \left(\frac{1}{T}\right)$

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.

$A = 2 - \log 72.6 = \text{2 - 1.860} = 0.140$

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"