The Principles of Chemistry, Volume IIMendeleyev, Dmitry Ivanovich
Science
The Principles of Chemistry, Volume II
Mendeleyev, Dmitry Ivanovich
Argon; Chemistry; Periodic law
[16] The researches of Thomsen showed that in very dilute aqueous
solutions the majority of monobasic acids--nitric, acetic,
hydrochloric, &c. (but hydrofluoric acid more and hydrocyanic
less)--HX evolve the following amounts of heat (in thousands of
calories) with caustic soda: NaHO + 2HX = 14; NaHO + HX = 14;
2NaHO + HX = 14; that is, if _n_ be a whole number _n_NaHO + HX =
14 and NaHO + _n_HX = 14. Hence reaction here only takes place
between one molecule of NaHO and one molecule of acid, and the
remaining quantity of acid or alkali does not enter into the
reaction. In the case of bibasic acids, H_{2}R´´ (sulphuric,
dithionic, oxalic, sulphuretted hydrogen, &c.), NaHO + 2H_{2}R´´ =
14; NaHO + H_{2}R´´ = 14; 2NaHO + H_{2}R´´ = 28; _n_NaHO +
H_{2}R´´ = 28; that is, with an excess of acid (NaHO + 2H´_{2}R´´)
14 thousand units of heat are developed, and with an excess of
alkali 28. When phosphoric acid is taken (but not all tribasic
acids--for instance, not citric) the general character of the
phenomenon is similar to the preceding, namely, NaHO +
2H_{3}PO_{4} = 14·7; NaHO + H_{3}PO_{4} = 14·8; 2NaHO +
H_{3}PO_{4} = 27·1; 3NaHO + H_{3}PO_{4} = 34·0; 6NaHO +
H_{3}PO_{4} = 35·3; or, in general terms, NaHO + _n_H_{3}PO_{4} =
14 (approximately) and _n_NaHO + H_{3}PO_{4} = 35 and not 42,
which shows a peculiarity of phosphoric acid. In the case of
energetic acids, when one equivalent (23 grams) of sodium (in the
form of hydroxide) replaces one equivalent (1 gram) of hydrogen
(with the formation of water and in dilute solutions), 14,000 heat
units are evolved; and this is true for phosphoric acid when in
H_{3}PO_{4}, Na or Na_{2} replaces H or H_{2}, but when Na_{3}
replaces H_{3} less heat is developed. This will be seen from the
following scheme based on the preceding figures: H_{3}PO_{4} +
NaHO = 14·8; NaH_{2}PO_{4} + NaHO = 12·3; Na_{2}HPO_{4} + NaHO =
5·9; with Na_{3}PO_{4} + NaHO, a very small amount of heat is
evolved, as may be judged from the fact that Na_{3}PO_{4} + 3NaHO
= 1·3, but still heat is evolved. It must be supposed that in
acting on phosphoric acid in the presence of a large quantity of
water, a certain portion of the sodium hydroxide remains as alkali
uncombined with the acid. Thus, on increasing the mass of the
alkali, heat is still evolved, and a fresh interchange between Na
and H takes place. Hence water shows a decomposing action on the
alkali phosphates. The same decomposing action of water is seen,
but to a less extent, with Na_{2}HPO_{4}, as may be judged both
from the reactions of this salt and from the amount of heat
developed by NaH_{2}PO_{4} with NaHO. Such an explanation is in
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