Question

The following data set describes the visible spectroscopy absorbance readings (at 565 nm) for a series of violet solutions. Use a computer (Excel, Google Sheets, etc.) to generate a calibration curve (10 points). Be sure your plot includes a title, axes labels, units, all of the data points, and a trend line. Recognize that when the concentration = 0, the solution is clear and colorless. Attach the plot as a separate page in this document to submit to the course website. After making the plot, use it to answer the questions below. Be sure to show your work where appropriate.

concentration/mM	absorbance/a.u
0.20	0.054
1.00	0.276
1.40	0.381
1.80	0.498
2.20	0.603

2.) What is the equation of the trend line through these data points? How did you determine this?

3.) What is the extinction co-efficient of this species? Assume a 1.00 cm cuvette.

4.) If an unknown solution has an absorbance reading of 0.337 a.u., what is the concentration of the solution?

5.) If a solution had a known concentration of 0.55 mM, what would the anticipated absorbance reading be?

Please include step by step for questions 4-5, thank you!

Answer 1

2. The equation of the line is y = mx+C, where m is the slope and C is the intercept, the linear line is fitted to this equation.

3. The extinction co-efficient of this species is calculated from the Beer Lambert's law A = Ecl, where E is extinction coefficient, c is the concentration of species, and l is the pathlength of cell

E = A/cl = 0.054/0.20 mM x 1 cm = 0.27 x 10³ M^-1 cm^-1 = 0.27 x 10³ M^-1 cm^-1 . The extinction coefficient is constant while increasing the concentration.

4. A = Ecl, c = A/El = 0.337/0.27 x 10³ M^-1 cm^-1 x 1 cm

                            c = 1.24 x 10^-3 M

The concentration of unknown solution of absorbance at 0.337 a.u is 1.24 x 10^-3 M

5. A = Ecl

       = 0.27 x 10³ M^-1 cm^-1 x 0.55 x 10^-3 M x 1 cm

      = 0.1485 (a.u)

The absorbance of unknown solution of concentration at 0.55 mM is 0.1485 (a.u)