«Fractional Precipitation.»[334]—If sulphuric acid is added to a 0.5 molar solution of the chloride either of barium, of strontium or of calcium, the corresponding sulphate is precipitated. We may ask which sulphate will be precipitated first, when sulphuric acid is added, drop by drop, to a solution containing the three chlorides. This problem, a case of the ‹fractional precipitation of salts containing a common ion›, may be treated from the point of view of the solubility-product principle as follows, the problem being limited, for the sake of simplicity, to only two of the sulphates, those of barium and strontium. For a saturated solution of barium sulphate,[335] at 18°, in contact with the solid salt, we have, according to the principle of the solubility-product, K_{BaSO_{4}} = [Ba^{2+}] × [SO_{4}^{2−}] = 1E−10, [p164] and for a saturated solution of strontium sulphate, we have similarly[336] K_{SrSO_{4}} = [Sr^{2+}] × [SO_{4}^{2−}] = 2.5E−7. Now, we may ask what the conditions are under which ‹both› precipitates can be present ‹together› in a condition of equilibrium with a supernatant saturated solution. In such a solution we have simultaneously K_{BaSO_{4}} = [Ba^{2+}]_{1} × [SO_{4}^{2−}]_{1}, K_{SrSO_{4}} = [Sr^{2+}]_{1} × [SO_{4}^{2−}]_{1}. New symbols, [Ba^{2+}]_{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, [SO_{4}^{2−}]_{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, [Ba^{2+}]_{1} / [Sr^{2+}]_{1} = K_{BaSO_{4}} / K_{SrSO_{4}} = 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.