Download presentation

Presentation is loading. Please wait.

Published byAnnice Long Modified about 1 year ago

1
Lab 8

2
Purpose Absorbance – Single Analyte Absorbance – Multiple Analyte Solving unknown concentrations Procedure Safety Concerns Waste Next Lab Reminder

3
The purpose of this lab is to demonstrate the additive property of absorbance. The molar absorptivity (ε) values for both Cu(II) and Ni(II) will be found at two analytical wavelengths by measuring the absorbance of both ions in solutions of known concentration. The two analytical wavelengths will then be utilized to find the concentrations of both Cu(II) and Ni(II) in an unknown mixture.

4
For a single analyte, we can use the Beer-Lambert Law: Abs = ε b c to determine any one of the variables, given that three of them are already known. Since molar absorptivity (ε ) is the same for a given compound (x), regardless of concentration, at a given wavelength, we can clarify Beer’s Law terms: A λ1 = ε λ1x b [x] Given that the path length (b) through our cuvets is 1.445 cm, we can define a new variable k λ1x = ε λ1x b to simplify calculations and modify Beer’s Law accordingly: A λ1 = k λ1x [x] After you make up your solutions today, you will simply use this modified Beer’s Law to determine the molar absorptivity (k λ1x ) of each solution at a given wavelength. The average k value for each set of solutions at each wavelength is calculated and used to solve for unknown concentrations.

5
For multiple analytes we exploit the additive property of absorbance: Abs T = Abs 1 + Abs 2 +... The same number of wavelengths are used as number of analytes that are analyzed. Therefore, in two analytes, we use: A λ1 = ε λ1x b [x] + ε λ1y b [y]; Define k λ1x = ε λ1x b and k λ1y = ε λ1y b; So A λ1 = k λ1x [x] + k λ1y [y] A λ2 = ε λ2x b [x] + ε λ2y b [y]; Define k λ2x = ε λ2x b and k λ2y = ε λ2y b; So A λ2 = k λ2x [x] + k λ2y [y]

6
Beer’s Law states: A λ1 = k λ1x [x] + k λ1y [y] At 395 nm, this equation becomes: A 395 = k 395Ni(II) [Ni 2+ ] + k 395Cu(II) [Cu 2+ ] = k 395 [Ni 2+ ] + 0 To solve for [Ni 2+ ] in our unknown solution: [Ni 2+ ] =

7
Beer’s Law states: A λ2 = k λ2x [x] + k λ2y [y] To solve for [y], [y] = To solve for [Cu 2+ ] in our unknown solution: [Cu(II)] =

8
Make up 4 solutions of Cu 2+ (aq) and 4 solutions of Ni 2+ (aq). Determine the k value of each solution at the indicated wavelengths. Find the average k values for Cu 2+ (aq) at 775 nm and for Ni 2+ (aq) at 775 nm and 395 nm. Solve for the unknown concentrations of your unknown solution using the derived equations.

9
Reagents: ◦ Cupric Sulfate ◦ Nickel Sulfate / Nickel Chloride Eye Contact: ◦ Irritation, redness, pain, conjunctivitis, ulceration, clouding of cornea Skin Contact: ◦ Irritation, redness, rash and itching. Sensitizer. Inhalation: ◦ Coughing, sore throat, shortness of breath, ulceration and perforation of the respiratory tract. Fumes from heating may cause symptoms similar to a cold. May cause metallic taste in mouth. Lung damage, allergy and asthma may occur. Ingestion: ◦ Burning of the mouth, esophagus, and stomach. Hemorrhagic gastritis, nausea, vomiting, abdominal pain, giddiness, myocardial weakness, metallic taste, and diarrhea. Systemic copper poisoning with capillary damage, headache, cold sweat, weak pulse, kidney and liver damage, CNS excitation and depression, jaundice, convulsions, blood effects, paralysis, coma and death.

10
Copper and Nickel are both toxic. Dispose of them in the appropriate container(s) in the fume hood.

11
Lab 9 is next.

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google