For solution a 6a label 3 clean and dry sample vials

This preview shows page 3 - 7 out of 10 pages.

For solution A:6a. Label 3 clean and dry sample vials A1, A2, A3. Add solution A, KHP, and 0.20M HCl according to the following table, then measure the absorbance for the solution in A1 and A2 at λ1and λ2(the two peaks) and A3 through the spectrum.6b. Dilute A3 (bromophenol blue in its acid form) across three different test tubes, labeled A4, A5, A6 and measure their absorbance at λ1and λ2.For solution B: 7a. Label 3 clean and dry sample vials B1, B2, B3. Add solution , KHP, and 0.20M NaOH according to the following table, then measure the absorbance for the solution in B1 and B2 at λ1and λ2(the two peaks) and B3 through the spectrum.2
7b. Dilute B3 (bromophenol blue in its base form) across three different test tubes, labeled B4, B5, B6 and measure their absorbance at λ1and λ2.8. Remove the cuvette from the spectrophotometer, empty and rinse it twice with distilled water. Turn off the spectrophotometer. Discard all solutions into the liquid waste pails in the common fume hoods. Wash the volumetric flask and leave it on the bench. Rinse the sample vials and the large test tubes and place them in the common fume hoods. Wash any other glassware taken from the locker and return it to the locker.Collection of DataPlease refer to the attached data sheets for collected data as well as solution volumes and concentrations.Results and Calculations1.2.3
λ 1= 440nmpHAlog [(A – AIN-)/(AHIN– A)]pKINA14.190.2540.7764.97A23.480.1840.1363.62A32.060.290------B13.990.146-0.1293.86B24.410.122-0.3064.10B36.710.037------Table 1λ 2= 590nmpHAlog [(A – AIN-)/(AHIN– A)]pKINA14.190.497-0.1024.09A23.480.1890.6424.12A32.060.036------B13.990.3860.1304.12B24.410.548-0.2144.19B36.710.861------Table 2Example of calculation of log [(A – AIN-)/(AHIN– A)] (using A3 and B3 at wavelength 1):A = 0.219AIN-= 0.037AHIN= 0.288Thus log [(A – AIN-)/(AHIN– A)] = log [(0.219 – 0.037)/(0.288 – 0.219)] = 0.4213.In order to determine the molar absorptivity constant, the Beer – Lambert law was used. This law states A = ε×c×l was used, where ε is the molar absorptivity constant, c is the concentration, and l is the path length. As our path length was 1 centimeter for our experiment, this equation can be reduced to a linear relation between A and c, with constant ε, shown by A = ε×c. Based on the below tables and graphs, we have:Solutionε value at wavelength1Solutionε value at wavelength24
A324161.00g-1mol-1A33020.10g-1mol-1A418959.70g-1mol-1A42600.60g-1mol-1A511241.60g-1mol-1A50g-1mol-1A63438.60g-1mol-1A60g-1mol-1Table 3Solutionε value at wavelength1Solutionε value at wavelength2B33104.0g-1mol-1B372231.5g-1mol-1B41845.6g-1mol-1B455788.5g-1mol-1B51845.6g-1mol-1B532550.3g-1mol-1B61845.6g

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture