Question

A substance in an aqueous solution at a concentration of 0.01M shows an optical transmittance of 28% with a path length of 2mm calculate the molar absorption coefficient of the solute? What would be the transmittance in a cell of 1cm thick?

Answer 1

(a) `epsilon = "280 L·mol"^"-1""cm"^"-1"`; (b) `T = 0.017`.

Explanation:

The absorption of radiation

Many compounds absorb ultraviolet (UV) or visible (VIS) light as it passes through a solution.

If a beam of radiation of power `P_0` passes through a solution, absorption takes place. The radiation leaving the sample has power `P`.

We can express the amount of radiation absorbed as transmittance `T`.

`color(blue)(bar(ul(|color(white)(a/a)T = P/P_0color(white)(a/a)|)))" "`

We can express the amount of radiation absorbed as absorbance `A`

`A = log(P_0/P) = log(1/T)`

`color(blue)(bar(ul(|color(white)(a/a)A = "-log"Tcolor(white)(a/a)|)))" "`

The Beer-Lambert Law

The Beer-Lambert Law states that the absorbance `A` of a solution is directly proportional to the molar concentration `"c"` of the solute and the path length `l` (usually expressed in centimetres).

`color(blue)(bar(ul(|color(white)(a/a)A = epsilonclcolor(white)(a/a)|)))" "`

`epsilon` is a proportionality constant called the molar absorption coefficient.

Now, we can write

`logT = -epsiloncl`

or

`epsilon = -logT/(cl)`

Part (a). Calculate the molar absorption coefficient

`epsilon = -logT/(cl) = -log0.28/("0.01 mol·L"^"-1" × "0.2 cm") = "280 L·mol"^"-1""cm"^"-1"`

Part (b). Calculate the transmittance

`logT = -epsiloncl = "-280" color(red)(cancel(color(black)("L·mol"^"-1""cm"^"-1")))× 0.01 color(red)(cancel(color(black)("mol·L"^"-1"))) × 1 color(red)(cancel(color(black)("cm"))) = "-2.8"`

`T = 10^"-2.8" = 0.0017`