By the aid of the mean differences found we can cal- culate theoretically the change of volume which must correspond to the neutralisation of a certain base by a certain acid.
We know that for ammonium isobutyrate the expansion is — 19*27. Denoting by a and b the respective influences of the acid and of the base, we may write a +- b = — 19*27. If instead of isobutyric acid another acid be taken, then for butyric acid, a + 8*11 for acetic acid, and so on up to a + 18-39 for nitric acid.
If we take another base instead of ammonia, we shall have the following expansions :
To calculate now the change of volume which would take place by the neutralisation, for example, of mono- chloracetic acid by caustic soda, we find :
(a + 4-39) + (- a + 6-49) = 10-88 (instead of 10-63).
(for the acid) (for the base)
The factor a, the value of which is unknown, is, there- fore, eliminated ; and we clearly see that the change of volume which accompanies neutralisation is additively composed of two factors, one of which depends only on the nature of the acid, and the other only on the nature of the base. This recalls to us the law of thermo* neutrality, as well as the ratio of the electric conductivities of solutions.
Suppose now that we mix together three solutions (prepared as above directed), one containing an equivalent of hydrochloric acid, one an equivalent of caustic potash, and the third an equivalent of ammonia. The change of volume will depend on how much potash and how much ammonia seize the acid, and from the observed expansion the division of the acid between the two bases can be deduced.
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