Question

How to graph:

I am tasked to graph a Beer-Lambert of BPB. It is explained that BPB that has it a molar extinction coefficient of basic form BPB is 31100M^-1cm^-1 at 590nm. Cuvette pathlength is 1cm. It transferred about 1ml of BPB solution into the cuvette for test and record. It was then made a serious of dilutions with 1 to 2, 1 to 4, 1 to 10, 1 to 20, and 1 to 40. They were tested and recorded the absorbance. It was explained that 1 to 4 dilutions means that 1 part of the original solution is diluted into a total of 4 parts e.g. 10ml sugar solution is diluted with 30ml water, making the final solution into 40 ml. The sugar concentration in the original solution should be 4 times as in the diluted one.

Example of data that was recorded from the dilutions.

using 590nm

Dilution	Reading 1	reading 2	Average
Original, no dilution	1.6	1.6	1.6
1 to 2	.8	.8	.8
1 to 4	.3	.3	.3
1 to 10	.1	.1	.1
1 to 20	.060	.060	.060
1 to 40	.05	.05	.05

How do you plot the average absorbance reading vs the dilution factor?

Where can you find out the liner range from the curve that was plotted?

Is the original solution within the linear range?

Calculate the concentration of the original BPB solution by Beer- lambert law

Answer 1

Use Beer’s Lambert’s law to determine the concentrations of BPB in the prepared solutions. The Beer-Lambert’s law is

A = ε*c*l

where A is the absorbance of a solution having concentration c and path length l. ε is molar absorptivity of the solution and is unique for the solution.It is given that ε = 31100 M^-1.cm^-1 and l = 1 cm.

Prepare a table of the average absorbance vs the dilution factor.

Dilution	Average Absorbance	Dilution Factor
Original, no dilution	1.6	1/1 = 1
1 to 2	0.8	2/1 = 2
1 to 4	0.3	4/1 = 4
1 to 10	0.1	10/1 = 10
1 to 20	0.06	20/1 = 20
1 to 40	0.05	40/1 = 40

Plot absorbance vs dilution factor as below.

The linear range of the plot occurs between dilution factor 10 to 40.

The original sample doesn’t fall in the linear range as evidenced from the curve.

Use Beer’ law. Put A = 1.6 and get

1.6 = (31100 M^-1.cm^-1)*c*(1 cm)

=====> 1.6 = (31100 M^-1)*c

=====> c = (1.6)/(31100 M^-1)

=====> c = 5.1447*10^-5 M.

The concentration of BPB in the original solution is 5.1447*10^-5 M (ans).