# Beer-Lambert Law

Imagine a glass of pure water. You can easily see through it and light readily passes through it. Add a few drops of red food dye to the water and now it is more difficult to see through the water. Furthermore, less light passes through the red solution. As you continue to add more red food color, the less light can pass through the solution.

There is a relationship between the concentration of the solution and the amount of light that can pass through it. This relationship can be expressed using the Beer-Lambert Law and is given as

$A = \varepsilon bc$ where b is the path length of the sample/cuvette (cm), $$\varepsilon$$ is the molar absorptivity of the solution (in L mol–1 cm–1), and c is the concentration of the solution (mol L–1). A is the base-10 logarithm ratio of the incoming power of light and the outgoing power of light [given as log(P0/P)].

Let us assume an experimental setup where a light is shown upon a solution placed in our cuvette. The incoming power of light, P0 is constant. As the concentration, c, of the solution increases, A increases. For A to increase, P decreases since P0 is constant. Therefore, A is a measure (and proportional) to the concentration of the solution.

## Practice Problems

### Problem 1

A solution was found to absorb 75% of 275 nm light across a 1.5 cm cuvette.
The solute has a molar absorptivity of 8000 L mol–1 cm–1.
What is the concentration (in M) of the solution?

Solution

\begin{align*} A &= \varepsilon b c \\ \log \left ( \dfrac{P_0}{P} \right ) &= \varepsilon b c \\ c &= \dfrac{\log\left ( \dfrac{P_0}{P} \right )}{\varepsilon b} \\ &= \dfrac{\log\left ( \dfrac{1.0}{0.25} \right )}{\left ( 8000~\mathrm{L~mol^{-1}~cm^{-1}} \right ) \left ( 1.5~\mathrm{cm} \right )} \\ &= 5.02\times 10^{-5}~\mathrm{mol~L^{-1}} \end{align*}

### Problem 2

The solution from above was found to have a concentration of 5.50 × 10–5 M after some time had passed. What percentage of light does this new solution absorb?

Solution

Solve for A: \begin{align*} A &= \varepsilon b c \\ &= \left ( 8000~\mathrm{L~mol^{-1}~cm^{-1}} \right ) \left ( 1.5~\mathrm{cm} \right ) \left ( 1.50\times 10^{-5} \right ) = 0.66 \end{align*}

Solve for P: \begin{align*} \log \left ( \dfrac{P_0}{P} \right ) &= 0.66 \\[1ex] \dfrac{1}{P} &= 10^{0.66} = 4.57 \\[1ex] P &= \dfrac{1}{4.57} = 0.219 \end{align*}

Get % - Let 1 be the amount of incoming light. Let 0.219 be the amount of outgoing light. \begin{align*} (1-0.219) \times 100\% &= 78\% \end{align*}

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