Question

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?

Answer 1

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"`