Lab 5: Spectrophotometric Analysis of Permanganate Solutions Pre-Laboratory Assi
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Question
Lab 5: Spectrophotometric Analysis of Permanganate Solutions Pre-Laboratory Assignment 1. Read an authoritative source for a discussion of spectrophotometry 2. The two components of the solution you will be making in this experiment must be used with great care. List these two substances and describe the problems associated with each. 3. 1.10-Phenanthroline forms a bright red complex with ironan. The complex has an analytical wavelength of 510 nm and has a high molar absorptivity. Hydroxylamine hydrochloride is added to keep the iron in a reduced +2 state, and the pH is controlled using an acetate buffer. A series of iron ID-phenanthroline solutions were made by adding various amounts of 4.300 x 10 M to 5 mL each of hydroxylamine hydrochloride, acetate buffer, and 1,10-phenanthroline. Distilled water was added to make the total volume of each solution 50.00 mL. The percent transmittance of each solution was read, using a blank containing everything but the iron (ID solution. Data for the determination are volume of stock volume of stock solution, mL solution, mL 21.78 7.00 80.35 14.09 64.71 2.00 10.00 11.32 33.65 5.00 2+ in each solution 1) Calculate the molar concentration of [Fe(phenanthroline) (2) Convert percent transmittances to equivalent absorbances, using Equation 4 custom page 81Explanation / Answer
1. Spectrophotometry experiment carried out here involves light being absorbed by the Fe-phenanthroline complex in the solution. the remaining light is transmitted which is measured as %T. The calculations are based on Beer's law.
2. The stock Fe2+ solution must be made with care so that the final concentration of Fe2+ remains the same as calculated theoretically.
The dilution of the solutions must be done with care not to exceed the volume or have lower volume for each solutions for consistency and lower error for the measurements.
3. For the given experiment.
absorbance = 2 - log(%T)
concentration of Fe(Phen)3^2+ = molarity of stock x volume of stock/total volume of solution
(1) we get for each solutions,
Stock solution = 1 ml, concentration of Fe(Phen)3^2+ = 4.3 x 10^-4 x 1/50 = 8.6 x 10^-6 M
Stock solution = 2 ml, concentration of Fe(Phen)3^2+ = 4.3 x 10^-4 x 2/50 = 1.72 x 10^-5 M
Stock solution = 5 ml, concentration of Fe(Phen)3^2+ = 4.3 x 10^-4 x 5/50 = 4.3 x 10^-5 M
Stock solution = 7 ml, concentration of Fe(Phen)3^2+ = 4.3 x 10^-4 x 7/50 = 6.02 x 10^-5 M
Stock solution = 9 ml, concentration of Fe(Phen)3^2+ = 4.3 x 10^-4 x 9/50 = 7.74 x 10^-5 M
Stock solution = 10 ml, concentration of Fe(Phen)3^2+ = 4.3 x 10^-4 x 10/50 = 8.6 x 10^-5 M
(2) For solutions,
Stock solution = 1 ml, absorbance = 2 - log(80.35) = 0.095
Stock solution = 2 ml, absorbance = 2 - log(64.71) = 0.189
Stock solution = 5 ml, absorbance = 2 - log(33.65) = 0.473
Stock solution = 7 ml, absorbance = 2 - log(21.78) = 0.662
Stock solution = 9 ml, absorbance = 2 - log(14.09) = 0.851
Stock solution = 10 ml, absorbance = 2 - log(11.32) = 0.946
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