Question

A 5.00 ml sample of blood was treated with trichloroacetic acid to precipitate proteine. After centrifugation, the resulting solution was brought to a PH of 3 and was extracted with two 5-ml, portions of methyl isobutyl ketone containing the organic lead complexing agent APCD. The extract was aspirated directly into an air-acetylene flame yielding an absorbance of 0.444 at 283.3nm. Five-milliliter aliquots of standard solutions containing 0.250 and 0.450ppm BP were treated in the same way and yielded absorbances of 0.396 and 0.599. Calculate the concentration(ppm) in the sample assuming that Beer's law is followed

Answer 1

we know that according to Beer lambert' s law

Absrobance = absrobptitvity X lenght X concentration

If we will plot a graph between absrobance and concentration then we will obtain a straight line

The slope of graph = (difference in absrobance at two points) / (difference in concentrations at those points)

Here

A1 = 0.396 (first point)

A2 = 0.599 (second point)

C1 = 0.250 ppm (first point)

C2 = 0.450 ppm (second point)

slope = 0.599- 0.396 / 0.45 - 0.25 = 1.015

For the graph, the line equation can be written as

y = mx + c

C = intercept

m = slope

so we can calculate the value of intercept as

0.396 = 0.25 X 1.015 + c

c = 0.142

So we have all the parameters to obtain the value of any unknown value

Here we need to calculate concentration ( the x-axis value)

We are given with the abarobance values 0.444 ( the y axis value)

Let us put the values

0.444 = x 1.015 + 0.142

x = 0.297 ppm = concentration