# Question 9ef4e

Use the Beer-Lambert Law to establish the unknown protein concentration using the absorbance values and the molar extinction coefficient for both the standard sample and the unknown sample:

A = εlc

#### Explanation:

Where:
$A$ = absorbance
ε = molar extinction coefficient
$l$ = length
$c$ = concentration

The absorbance value for the 0.25ml in 2.25ml 0.9% NaCl sample = 0.42

The absorbance for a 0.5mg/ml sample = 0.25

The length of the cuvette (the tube that the sample is placed in order to find the absorbance via spectrophotometry is usually 1cm) so we will use that for $l$, and for any given system, εl forms a constant.

Therefore we can then find $c$ of the unknown, using the standard that underwent spectrophotometry under the same conditions by setting up equations for both sets of data, and dividing the right side of each by $c$ to get:

A/c = εl

So we''ll have:

(A(Standard))/(c (Standard)) = εl

(A(Unknown))/(c (Unknown)) = εl

Since εl is going to be constant in both equations, the right side will be equal for both the unknown and the standard solution, so we can cancel it out, leaving us with:

$\frac{A \left(S \tan \mathrm{da} r d\right)}{c \left(S \tan \mathrm{da} r d\right)} = \frac{A \left(U n k n o w n\right)}{c \left(U n k n o w n\right)}$

We want to find $c \left(u n k n o w n\right)$, so rearrange the equation such that:

$c \left(u n k n o w n\right)$ = (A (unknown) * c(standard))/ (A (standard)#

Plugging in what we have:

$c \left(u n k n o w n\right) = \left(\frac{0.42 \cdot 0.5 \frac{m g}{m L}}{0.25}\right) = 0.84 \frac{m g}{m L}$

Since this concentration was from a dilution made from 0.25ml serum and 2.25ml of a 0.9% NaCl solution, which formed a total of 2.50ml we must find the original concentration using the dilution equation:

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

Rearranged, this gives:

${c}_{1} = \frac{{c}_{2} {V}_{2}}{{V}_{1}} = \left(0.84 \frac{m g}{m L} \cdot \frac{2.50 \left(m L\right)}{0.25 \left(m L\right)}\right) = 8.4 \frac{m g}{m L}$