# Question #19c27

##### 1 Answer

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

#A_2 = A_1 (V_1/V_2)#

#= (("6.00 mL")/("6.00 + 36.00 mL"))A_1#

#= 1/7 A_1#

**Beer's Law** is:

#A = epsilonbc# where

#A# isabsorbance,#epsilon# ismolar absorptivity(or the extinction coefficient) in#"L"cdot"mol"^(-1)"cm"^(-1)# ,#b = "1 cm"# is thepath lengthof your cuvette, and#c# is theconcentrationin#"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_1V_1 = c_2V_2# (which you may know as

#M_1V_1 = M_2V_2# .)

Your **final concentration** in *includes* the original

(1)#c_2 = c_1(V_1/V_2)#

Similarly, we can write *Beer's law* as an initial and final state, where

(2a)#A_1 = epsilonbc_1#

(2b)#A_2 = epsilonbc_2#

Solve for **(2a)** to get:

(3)#c_1 = A_1/(epsilonb)#

Next, plug **(1)** into **(2b)** to get:

#A_2 = epsilonbc_1((V_1)/(V_2))#

Plugging in **(3)**:

#A_2 = cancel(epsilonb)(A_1/cancel(epsilonb))((V_1)/(V_2))#

Therefore, the new absorbance is simply:

#color(blue)(A_2 = A_1 (V_1/V_2))#

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.