# Question #19c27

Dec 18, 2016

There is not enough information to give an actual number. The best I can do is give a relative value.

${A}_{2} = {A}_{1} \left({V}_{1} / {V}_{2}\right)$

$= \left(\left(\text{6.00 mL")/("6.00 + 36.00 mL}\right)\right) {A}_{1}$

$= \frac{1}{7} {A}_{1}$

Beer's Law is:

$A = \epsilon b c$

where $A$ is absorbance, $\epsilon$ is molar absorptivity (or the extinction coefficient) in ${\text{L"cdot"mol"^(-1)"cm}}^{- 1}$, $b = \text{1 cm}$ is the path length of your cuvette, and $c$ is the concentration in $\text{mol/L}$.

The dilution that occurs is a state function (volume only depends on the initial and final state, and not on the path you take to change the volume), and so, the new concentration you have is based on:

${c}_{1} {V}_{1} = {c}_{2} {V}_{2}$

(which you may know as ${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$.)

Your final concentration in $\text{mol/L}$ is then (where the final volume includes the original $\text{6.00 mL}$):

(1) ${c}_{2} = {c}_{1} \left({V}_{1} / {V}_{2}\right)$

Similarly, we can write Beer's law as an initial and final state, where $\epsilon$ and $b$ do not change for the same compound in solution in the same cuvette:

(2a) ${A}_{1} = \epsilon b {c}_{1}$
(2b) ${A}_{2} = \epsilon b {c}_{2}$

Solve for ${c}_{1}$ in (2a) to get:

(3) ${c}_{1} = {A}_{1} / \left(\epsilon b\right)$

Next, plug ${c}_{2}$ from (1) into (2b) to get:

${A}_{2} = \epsilon b {c}_{1} \left(\frac{{V}_{1}}{{V}_{2}}\right)$

Plugging in ${c}_{1}$ from (3):

${A}_{2} = \cancel{\epsilon b} \left({A}_{1} / \cancel{\epsilon b}\right) \left(\frac{{V}_{1}}{{V}_{2}}\right)$

Therefore, the new absorbance is simply:

$\textcolor{b l u e}{{A}_{2} = {A}_{1} \left({V}_{1} / {V}_{2}\right)}$

That should make sense, since the less concentrated the solution in real life, the less colored it would look, and thus, the weaker the absorbance it would have.