Kmno4
Results and Discussion
Potassium permanganate, KMnO4, is widely used as an oxidizing agent in volumetric analysis. In acid solution, MnO4- ion undergoes reduction to Mn2+ ion as shown in the equation:
8 H+(aq) + MnO4-(aq) + 5e- → Mn2+(aq) + 4 H2O(l).
Since the MnO4- ion is violet and the Mn2+ ion is nearly colorless, the end point in titrations using MnO4- as the titrant can be taken as the first permanent pink color that appears in the solution being titrated.1
The number of moles of potassium permanganate used in the titration is equal to the product of the molarity of the KMnO4 and the volume used. The number of moles of iron (II) present in the sample is obtained from the balanced equation for the reaction and the moles of permanganate used. The mass of iron in the unknown sample is then calculated from the number of moles and the atomic mass of iron and, finally, the corresponding percentage by weight of iron in the unknown is determined.
A blank titration was performed, just in case there are iron ions in the water used in the experiment, there is a need to be able to account for the effects of the material which is in the solvent. If you didn't run a blank titration, your results could come out high (the iron from the razor blade + the ions already in the water).
Standardization of KMnO4 Solution
M1 =
M1 =
= 0.0214 M
= 0.0214 M (0.2040 g N2C2O4)(1 mol KMnO4134 g N2C2O4) (2 MnO4-5 N2C2O4)(1000 mg1 g)
28.5
= 0.0197 M
= 0.0197 M
M2 =
M2 = (0.2030 g N2C2O4)(1 mol KMnO4134 g N2C2O4) (2 MnO4-5 N2C2O4)(1000 mg1 g)
30.7
= 0.0207 M
= 0.0207 M
M3 =
M3 = (0.2048 g N2C2O4)(1 mol KMnO4134 g N2C2O4) (2 MnO4-5 N2C2O4)(1000 mg1 g)
29.5
M = 0.0206 M N1 = M*h = 0.0214*5 = 0.11 N N2 = M*h = 0.0197*5 = 0.10 N N3 = M*h = 0.0207*5 = 0.10 N = 0.103 N The average normality shown by the different trials means that the standardization of KMnO4 performed is close to the 0.1N of the expected solution.
Analysis
Potassium permanganate, KMnO4, is widely used as an oxidizing agent in volumetric analysis. In acid solution, MnO4- ion undergoes reduction to Mn2+ ion as shown in the equation:
8 H+(aq) + MnO4-(aq) + 5e- → Mn2+(aq) + 4 H2O(l).
Since the MnO4- ion is violet and the Mn2+ ion is nearly colorless, the end point in titrations using MnO4- as the titrant can be taken as the first permanent pink color that appears in the solution being titrated.1
The number of moles of potassium permanganate used in the titration is equal to the product of the molarity of the KMnO4 and the volume used. The number of moles of iron (II) present in the sample is obtained from the balanced equation for the reaction and the moles of permanganate used. The mass of iron in the unknown sample is then calculated from the number of moles and the atomic mass of iron and, finally, the corresponding percentage by weight of iron in the unknown is determined.
A blank titration was performed, just in case there are iron ions in the water used in the experiment, there is a need to be able to account for the effects of the material which is in the solvent. If you didn't run a blank titration, your results could come out high (the iron from the razor blade + the ions already in the water).
Standardization of KMnO4 Solution
M1 =
M1 =
= 0.0214 M
= 0.0214 M (0.2040 g N2C2O4)(1 mol KMnO4134 g N2C2O4) (2 MnO4-5 N2C2O4)(1000 mg1 g)
28.5
= 0.0197 M
= 0.0197 M
M2 =
M2 = (0.2030 g N2C2O4)(1 mol KMnO4134 g N2C2O4) (2 MnO4-5 N2C2O4)(1000 mg1 g)
30.7
= 0.0207 M
= 0.0207 M
M3 =
M3 = (0.2048 g N2C2O4)(1 mol KMnO4134 g N2C2O4) (2 MnO4-5 N2C2O4)(1000 mg1 g)
29.5
M = 0.0206 M N1 = M*h = 0.0214*5 = 0.11 N N2 = M*h = 0.0197*5 = 0.10 N N3 = M*h = 0.0207*5 = 0.10 N = 0.103 N The average normality shown by the different trials means that the standardization of KMnO4 performed is close to the 0.1N of the expected solution.
Analysis