A study of the reactions given above which represent the oxidation of
ferrous compounds by potassium permanganate, shows that there are 2
molecules of KMnO_{4} and 10 molecules of FeSO_{4} on the
left-hand side, and 2 molecules of MnSO_{4} and 5 molecules of
Fe_{2}(SO_{4})_{5} on the right-hand side. Considering only these
compounds, and writing the formulas in such a way as to show the
oxides of the elements in each, the equation becomes:

K_{2}O.Mn_{2}O_{7} + 10(FeO.SO_{3}) --> K_{2}O.SO_{3} + 2(MnO.SO_{3})
+ 5(Fe_{2}O_{3}.3SO_{3}).

From this it appears that two molecules of KMnO_{4} (or 316.0 grams)
have given up five atoms (or 80 grams) of oxygen to oxidize the
ferrous compound. Since 8 grams of oxygen is the basis of normal
oxidizing solutions and 80 grams of oxygen are supplied by 316.0 grams
of KMnO_{4}, the normal solution of the permanganate should contain,
per liter, 316.0/10 grams, or 31.60 grams (Note 1).

The preparation of an approximately tenth-normal solution of the
reagent may be carried out as follows:

PROCEDURE.--Dissolve about 3.25 grams of potassium permanganate
crystals in approximately 1000 cc. of distilled water in a large
beaker, or casserole. Heat slowly and when the crystals have
dissolved, boil the solution for 10-15 minutes. Cover the solution
with a watch-glass; allow it to stand until cool, or preferably over
night. Filter the solution through a layer of asbestos. Transfer the
filtrate to a liter bottle and mix thoroughly (Note 2).