Standardization of Kmno4 Essay

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Formal Lab Report Standardization of KMnO4 Abstract: Potassium permanganate, KMnO4, is a strong oxidizing agent. Permanganate, MnO4, is an intense dark purple color. Reduction of purple permanganate ion to the colorless Mn2+ ion, the solution will turn from dark purple to a faint pink color at the equivalence point. No additional indicator is needed for this titration. The reduction of permanganate requires strong acidic conditions. Introduction: In this experiment a purple colored solution of potassium permanganate, with an approximate concentration of 0.025M, will be added to a solution containing Fe2+ ions. The manganese is reduced from a 7+ oxidation state in the permanganate ion to form colorless Mn2+ ions. The equivalence point is indicated at the point when all of the Fe2+ ions in the solution are oxidized and the colorless mixture retains a purple tint. Procedure: Fill a buret and tip with KMnO4. Place approximately 1g of iron (ii) ammonium sulfate hexahydrate in a weighing boat and mass with an analytical balance to +/- 0.0001g. Transfer the FAS to an Erlenmeyer flask. Add 25mLof distilled water, 15mL of 3M H2SO4 to the flask and swirl to dissolve the FAS. Place about 50mL of water in a beaker and 1 drop of the permanganate solution for the color constant. Record the initial volume reading in the buret. Add the permanganate to the FAS solution until the equivalence point is reached. Record the final volume reading in the buret. Repeat the process 2 more times. Results: MnO4 + 5Fe + 8H = Mn +5Fe + 4H2O 5(Fe2+ = Fe3+ + 1e-) oxidized 5e + 8H + MnO4 =Mn + 4H2O 5Fe2+ = 5Fe3+ + 5e- MnO4 + 5Fe+ 8H = Mn + 5Fe + 4H2O 1.001g FAS x 1 mol/ 392.16 = .002553 moles FAS .002553 moles FAS x 1 mol Fe/ 1 mol FAS = .002553 moles Fe2+ .002553 moles Fe2+ x I mol MnO4/ 5 Fe2+ = .0005106 moles MnO4 .0005106 moles MnO4/ .0212 = .02408M 1.006 x 1/392.16 =

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