determine the eqoivalent weight of KMno4 in basic medium
31.608 g/mol
1/5 of molecular weight
Thio: 1 mole S2O32- = 2 equivalent So Eq. mass = 0.5 * (molar mass)
H2O2 was an equivalent weight of 34
The equivalent weight of boric acid is 20,61 g.
31.608 g/mol
1/5 of molecular weight
The oxidation state of 'Mn' in KMnO4 is +7 after reaction as oxidizing agent 'Mn; becomes +2 so change in oxidation number is '5' the formula mass divided by change in oxidation number is equal to equivalent mass or weight, 158/5 = 31.7
Formula weight of KMnO4 is 158.04. The equivalent weight depends on the reaction involved. In acidic media it is (formula weight) /5, in neutral although this is more difficult to control it is the formula weight. Knowing which type it is usually acidic as it is easier, you will need 158.04/5 X 0.2 = 6.32g. Sulfuric acid is usually added in the titration to ensure the correct oxidation reaction is carried out. Also note KMnO4 is not considered a primary standard so you should really filter the solution to remove any MnO2 caused by oxidation of organic impurites in the water you are using, and standardise the solution by titration against say oxalic acid. K
Thio: 1 mole S2O32- = 2 equivalent So Eq. mass = 0.5 * (molar mass)
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Equivalent weight ofc11h17n3os2 from molecular weight271.402
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