POTASSIUM TRIBROMIDE. 109
calculated from the data obtained should be practically equal.
Succinic acid is about six times as soluble in water as in ether.
II. Dissolve 10 g. of benzoic acid in a mixture of 100 g. of
water and 100 g. of benzene; shake a little longer after it is all
dissolved until the distribution equilibrium is reached, and titrate
10 c.c. from each layer. Add 50 c.c. of water and again shake
vigorously, repeating this several times in succession and titrating
both layers after each addition. Nearly constant values are
obtained in this experiment if the square of the solubility in water
is divided each time by the solubility in benzene; the molecular
weight of benzoic acid dissolved in benzene is nearly twice as large
as in water.B. Proof of the Existence of Potassium Tribromide in Aqueous
Solution.
Bromine dissolves more freely in a potassium bromide solution than in
water, this being due to the formation of the potassium salt of the complex
brom-hydrobromie acid, K [BrBr^]. The salt has never been obtained in a
solid form suitable for analysis; but its composition can be derived, with
the aid of the mass-action law and the law of distribution, by finding which
of the assumptions, x — 2, x = 4, etc., leads to conclusions agreeing with
the facts found experimentally. Calculating first on the basis of the com-
plex ion Br 3 ~ (i.e., x — 2), this ion would be partially broken down in solution
into simple bromine ions and free bromine:
Br 3 ~ <=i Br~ + Br 2.For the state of equilibrium, the mass-action law gives:
[Br~] [Br 2 ] fc
[Br,-]
The three concentrations which are in equilibrium according to the above
equation may be determined by preparing a concentrated solution of bro-
mine in carbon bisulphide and shaking one portion of it with water, and a
second portion with a potassium bromide solution of known molecular con-
centration,^1 A. In each experiment the molecular quantities of bromine,
D and B, which pass into the aqueous layer are determined; the excess of
bromine in the second case is equal to that combined in the complex: thus,
[Br 3 -] =B - D.
Furthermore, [BrJ = D; according to the distribution law, the amount
of free bromine existing in the potassium bromide solution, as well as that
(^1) The molecular concentration is the amount of substance dissolved in
one liter divided by its molecular weight.