~Indirect Titration.~--The action of permanganate of potash upon a ferrous solution is one of oxidation, hence it is evident that if any other oxidising agent is present it will count as permanganate. In such a case the titration can be used (indirectly) to estimate the quantity of such oxidising agent, by determining how much less of the permanganate is used. For example, suppose that 1 gram of iron dissolved in sulphuric acid requires 100 c.c. of standard permanganate to fully oxidise it, but that the same amount of iron only requires 35.6 c.c. of the same standard permanganate if it has been previously heated with 0.5 gram of black oxide of manganese. Here it is evident that 0.5 gram of black oxide does the work of 64.4 c.c.[4] of the permanganate solution, and that these quantities are equivalent; moreover, if 64.4 c.c. correspond with 0.5 gram, then 100 c.c. correspond with 0.7764 which is the standard. On theoretical grounds, and by a method of calculation which will be explained further on (under the heading "Calculations from Formulæ"), it can be found that if the standard for iron is 1 gram, that for the black oxide will be 0.7764 gram. The principles of these indirect titrations become clearer when expressed in a condensed form. Thus, in the example selected, and using the formulæ Fe = Iron, KMnO_{4} = permanganate of potash, and MnO_{2} = oxide of manganese, we have:-- (1) 1 gram Fe = 100 c.c. KMnO_{4} (2) 1 gram Fe = 35.6 c.c. KMnO_{4} + 0.5 gram MnO_{2} .'. 100 c.c. KMnO_{4} = 35.6 c.c. KMnO_{4} + 0.5 gram MnO_{2} (100-35.6) c.c. KMnO_{4} = 0.5 gram MnO_{2} 64.4 c.c. KMnO_{4} = 0.5 gram MnO_{2} The iron does not enter into the calculation if the same quantity is present in the two experiments. An indirect titration thus requires three determinations, but if more than one assay is to be carried on, two of these need not be repeated. The standard is calculated in the usual way.