The Elements of Qualitative Chemical Analysis, vol. 1, parts 1 and 2. / With Special Consideration of the Application of the Laws of Equilibrium and of the Modern Theories of Solution.

by Julius Stieglitz

New symbols, [Ba²⁺]₁, etc., are used for expressing the concentrations, as they are not the same as in the pure aqueous solutions. The much more soluble strontium sulphate makes the concentration of the sulphate-ion very much greater than it is in the saturated solution of pure barium sulphate and diminishes the concentration of the barium-ion proportionately (p. [145]). The concentration, [SO₄²⁻]₁, of the sulphate-ion, representing the actual (total) concentration of the ion in the solution saturated with both salts, appears in both of the new equations. Combining the two equations, we have, for the condition of equilibrium between the two precipitates and the supernatant liquid,

[Ba²⁺]₁ / [Sr²⁺]₁ = K_BaSO4 / K_SrSO4 = 1 / 2500.

That is, in a solution in equilibrium with both precipitates at 18°, the strontium-ion must be about 2500 times as concentrated as is the barium-ion. If we start with equivalent quantities of barium and strontium chlorides, say in 0.1 molar solutions, and gradually add sulphuric acid or ammonium sulphate, barium sulphate will be precipitated alone,[337] until the strontium-ion is in the excess [p165] indicated by the ratio given. After that, strontium sulphate will be precipitated, with traces of barium sulphate, the ratio expressed in the equilibrium equation being maintained in the supernatant liquid. On the other hand, if we start with a solution containing a very large excess of a strontium salt, more than is required by the equilibrium ratio, then strontium sulphate will be precipitated first, until the ratio given is reached.

Even should the more soluble salt be precipitated first from a solution containing, say, equal concentrations of barium and strontium ions, it could not remain in equilibrium with the supernatant liquid and would be converted into the less soluble one, before equilibrium was reached in the system. We can follow similar relations, experimentally, by using precipitates of different colors. Silver chromate Ag₂CrO₄ is an intensely red precipitate, that is rather difficultly soluble in water (exp.); a liter of water dissolves[338] 0.0252 gram or 8E−5 mole at 18°. The concentration of the silver-ion in the saturated solution is then 0.00016 mole.[339] Silver chloride AgCl, a white salt, is still less soluble in water, a liter of water at 18° dissolving 0.00134 gram or 1E−5 mole, and the concentration of the silver-ion in the saturated solution is, therefore, only 1E−5 mole, as compared with 1.6E−4 in the saturated silver chromate solution. If a mixture of potassium chromate and potassium chloride, containing approximately equal (0.01 molar) concentrations of the two salts, is prepared and silver nitrate solution added, drop by drop, to the mixture, the first permanent precipitate is the white silver chloride (exp.). However, as the silver nitrate solution strikes the surface of the liquid, a red precipitate of the chromate, mixed with chloride, is momentarily seen, where the silver nitrate temporarily produces a local excess of the precipitant. But the red precipitate disappears rapidly and gives way to the white precipitate of the less soluble chloride. The quantitative relations, which may be developed with the aid of the principle of the solubility-product (see below), are such that, if little chromate is used, it may serve as an [p166] indicator to determine quantitatively the moment when all the chloride, within the limits of allowed quantitative error, is precipitated, the first permanent tinge of pink (solid Ag₂CrO₄), mixed with the yellow color of the solution, being used as the indication that the precipitation of the chloride is complete. Potassium chromate is used as a favorite indicator in quantitative analysis, for this purpose.

The quantitative relations[340] for the precipitation may be developed as follows: For a supernatant liquid in which a precipitate of silver chromate just appears permanently, together with the chloride, i.e. for the condition of saturation with both silver salts at 18°, we have

K_AgCl = [Ag⁺] × [Cl⁻] = (1E−5)² = 1E−10

and

K_Ag2CrO4 = [Ag⁺]² × [CrO₄²⁻] =
(1.6E−4)² × (8E−5) = 2E−12,

and therefore:

[Cl⁻]² / [CrO₄²⁻] = (K_AgCl)² / K_Ag2CrO4 =
(1E−10)² / 2E−12 = 1 / 2E8.