Chemical Analysis by Redox Titration PDF

Preview

Summary

Lab report...

Description

Chemical Analysis by Redox Titration

Purpose of Experiment: Analyze the percent hydrogen peroxide in a common “drugstore” solution by titrating it with potassium permanganate. Pre-Lab Questions: 1. Use the half-reaction method to write a balanced chemical equation for the overall reaction between H2O2 and MnO4 in acidic solution. a. MnO4-(aq) + 5e- + 8H+(aq) ⇒ Mn2+(aq) + 4H2O(l) b. H2O2(aq) ⇒ O2(g) + 2H+ (aq) + 2ec. 5H2O2 + 2MnO4- + 16H+ ⇒ 5O2 + 10H+ + 2Mn2+ + 8H2O d. 2MnO 4 -  + 5H2 O   + 6H+  ⇒ 2Mn2 +  + 5O2  + 8H 2 O  2  2. What is the mole ratio of hydrogen peroxide to permanganate ion in the balanced chemical equation you determined in Question #1? How many moles of hydrogen peroxide will be oxidized by 0.0045 moles of potassium permanganate in acidic solution? a. Mole ratio of hydrogen peroxide to permanganate ion is 5:2, respectively. b. 0.005625 moles of hydrogen peroxide will be oxidized by 0.0045 moles of potassium permanganate in an acidic solution. 3. Review the procedure. Is it necessary to know the exact mass of: a. Hydrogen peroxide solution added to the flask in steps 7 and 8? i. It is necessary to know the exact mass of hydrogen peroxide solution added to the flask in steps 7 and 8. b. Water added to the flask in steps 8 and 9? i. It is not necessary to know the amount of water added to the flask in steps 8 and 9. c. Why or why not?

i.

Water is only present to dilute the solution - it does not affect the number of moles in any way. Hydrogen peroxide, however, affects the number of moles in the reaction.

Post-Lab Questions 1. Determine the mass percent of H2O2 in the commercial sample for each trial. a. 0.03998 L * (.1mol/1L) ⇒ 0.003998 mol i. 0.003998 mol * (34g / 1mol) ⇒ 0.1359g ii. 100% * (0.1359g/10.6812)g ⇒ 1.264% b. 0.03610 L * (.1mol/1L) ⇒ 0.003610 mol i. 0.003610 mol * (34g / 1mol) ⇒ 0.12274g ii. 100% * (0.12274g/9.5497g) ⇒ 1.285% c. 0.03713 L * (.1mol/1L) ⇒ 0.003713 mol i. 0.003713 mol * (34g / 1mol) ⇒ 0.126242g ii. 100% * (0.126242g/9.1964g) ⇒ 1.373% 2. Calculate the average percent of H2 O2 in the commercial sample. Compare this value to the concentration of H2O2 listed as the active ingredient on the product label. a. (1.373 + 1.285 + 1.264) / 3 = 1.307% H2Ω∫errO  2 b. Concentration of H2O2 listed as active ingredient on product label is 3%....