Part III). Here some general principles, only, will be considered in

connection with the behavior of some of the most important of the precipitates.

There is a very wide range in the degree of the insolubility of such precipitates as are used in analysis. In the table at the end of Part IV, the exact solubilities of the most important [p163] precipitates of the alkaline earths are given, for 18°, in grams and moles per liter. The values are instructive in a number of respects.

Inspection of the table shows which are the least soluble (in molar terms), and therefore the best salts, for precipitating barium, strontium and calcium, in order to insure the use of the most sensitive tests for the ions of each of these metals. It also shows which salts must be treated with special precautions to escape error. It is further seen, that if the carbonates of these alkaline earths are precipitated by a moderate excess of ammonium carbonate, the addition of a sulphate (for instance ammonium sulphate), to the filtrate from the precipitated carbonates, will only precipitate barium sulphate, the only sulphate whose solubility (and solubility-product) is smaller than that of the corresponding carbonate. In the same way, calcium oxalate is the only oxalate of these three alkaline earths that will be precipitated by ammonium oxalate in the filtrate from the carbonates (see Part III in regard to precautions against difficultly soluble double oxalates of magnesium). Again, calcium sulphate is the only one of the sulphates, which is sufficiently soluble in water to give an immediate ‹heavy› precipitate, when the sulphates are shaken for a few moments with water and the filtered solution is treated with a few drops of ammonium oxalate solution.

«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.

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 Ag_{2}CrO_{4} 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 Ag_{2}CrO_{4}), 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_{Ag_{2}CrO_{4}} = [Ag^{+}]^2 × [CrO_{4}^{2−}] = (1.6E−4)^2 × (8E−5) = 2E−12,

and therefore:

[Cl^{−}]^2 / [CrO_{4}^{2−}] = (K_{AgCl})^2 / K_{Ag_{2}CrO_{4}} = (1E−10)^2 / 2E−12 = 1 / 2E8.

For a solution containing one or two drops (0.1 c.c.) of saturated potassium chromate solution per 100 c.c., a proportion frequently used in quantitative analysis, the concentration of the chromate-ion is approximately 2.5E−3 and the chloride-ion will consequently be precipitated until [Cl^{−}] = 3E−6. The chief source of error in the method, then, will not be due to the incompleteness of the prior precipitation of the chloride, but rather to the use of the small excess of silver nitrate required to precipitate sufficient chromate to be visible. This error may be avoided, and is avoided in very accurate work (‹e.g.› in water analysis), by determining, in a blank test, the amount of silver nitrate required to show the change of tint of a pure chromate solution of the concentration to be used in the titration and by titrating to this tint in the determination of the chloride: the volume of silver nitrate (‹e.g.› 0.2 c.c. of a 0.01 molar solution), required to produce the tint in the blank test, is subtracted from the total volume of silver nitrate used in the chloride determination.

We find thus, that the ‹order› of ‹precipitation› of ‹difficultly soluble salts, which contain a common ion› (fractional precipitation), is ‹subject to the equilibrium conditions derived from the application of the principle of the solubility-product to the salts in question›.[341]

It should be further noted that the condition of equilibrium between two precipitates, containing a common ion, and a supernatant liquid, depends on the concentrations ‹in the supernatant liquid›, in the ‹liquid phase›, and not on the quantities of the solids [p167] present. This conclusion was first reached by Guldberg and Waage, to whom we owe the law of mass action, and was fully confirmed by them. The modern treatment of the subject substitutes ion concentrations, i.e. the concentrations of the active components,[342] for the total concentrations used by these investigators.[343]

That the condition of equilibrium is dependent on the liquid phase can easily be demonstrated if mercurous chloride and mercurous hydroxide are selected as the two precipitates, in order that we may follow changes of concentration in the liquid phase by color changes. For the condition of equilibrium between the two precipitates and the supernatant liquid we may develop the relation

[OH^{−}] / [Cl^{−}] = K_{HgOH} / K_{HgCl} = K.

EXP. A few drops of phenolphthaleïn are added to 100 c.c. of a very dilute solution of potassium hydroxide (1 / 500 molar); the usual indication of the presence of the hydroxide-ion is shown and the intensity of the color will be a measure of its concentration. Three identical solutions are prepared and then a pinch of calomel is added to two of the solutions,—‹their pink color fades decidedly›.

Some of the calomel is decomposed into dark-colored mercurous oxide, which is precipitated, and sufficient ‹potassium chloride› is formed to bring the ratio [OH^{−}] / [Cl^{−}], in the solution, down to the value required by the constant K. Since no chloride-ion is present at the start, this result can be brought about only by the change indicated. Now, some chloride, say a little of a concentrated solution of potassium chloride, which reacts perfectly neutral, is added to one of the mixtures of the hydroxide and chloride. The concentration of the chloride-ion is increased in the solution, the ratio [OH^{−}] / [Cl^{−}] is made much ‹too small› and the condition of equilibrium is disturbed. Consequently, mercurous oxide reacts with potassium chloride, as expressed in the equation HgOH ↓ + KCl → HgCl ↓ + KOH, until the ratio [OH^{−}] / [Cl^{−}] [p168] has the value required by the constant K. The phenolphthaleïn is colored by the increased concentration of the hydroxide-ion and shows the direction of the change.[344]

«Precipitation by a Weak Base in the Presence of its Salts.»—Magnesium, in contrast to the other alkaline earth metals, forms very few difficultly soluble compounds. Magnesium-ammonium phosphate, Mg(NH_{4})PO_{4}, a double salt, is the most characteristic precipitate for identifying it. Magnesium hydroxide, as the table of solubilities shows, is also very difficultly soluble. It is readily precipitated by potassium or sodium hydroxide, but ammonium hydroxide, at best, precipitates it only incompletely[345] from solutions of its salts, and very commonly does not precipitate it at all—namely, when ammonium salts in sufficient quantity are present (‹exp.›). This peculiar behavior of ammonium salts, in interfering with the precipitation of magnesium hydroxide, puzzled chemists for many years, and a number of "theories" were offered in explanation of it. But they all proved untenable, and the first adequate explanation, the one in accord with all the facts, was only found after the development of the theory of ionization. Its quantitative application to the case in question gives a perfect insight into the relations, and brings important confirmation of the correctness of its fundamental assumptions.[346]

The fact that ammonium salts prevent the precipitation of magnesium hydroxide was formerly explained as being due to the formation of "double salts," such as MgCl_{2},NH_{4}Cl. It is true that such double salts exist, but, if their formation should prevent the precipitation of magnesium hydroxide, possibly by including the magnesium as part of a negative ion or radical,[347] MgCl_{3}, then the corresponding double salts of magnesium chloride with potassium and sodium chloride (‹e.g.› MgCl_{2},KCl), which are just as stable as the ammonium salts, should show the same behavior; but, as a matter of fact, the addition of potassium chloride does not interfere with the precipitation of the hydroxide by either [p169] potassium hydroxide or ammonium hydroxide (‹exp.›). So the explanation is untenable. Obviously, a ‹specific interference of ammonium salts with the precipitating power of ammonium hydroxide is involved›. But that is exactly what the law of chemical equilibrium, applied to the ionization of ammonium hydroxide, would demand: As a weak base, which is little ionized in pure aqueous solutions, it is ‹very much weaker as a base, produces a far smaller concentration of the hydroxide-ion, when readily ionizable salts of ammonium are added to the solution›,[348] and the precipitation of magnesium hydroxide, and of metal hydroxides in general, depends on the concentration of the hydroxide-ion, which is a factor in the solubility-products of bases.

The precipitation of magnesium hydroxide, in particular, depends on the relation of the product [Mg^{2+}] × [HO^{−}]^2 to the solubility-product constant of magnesium hydroxide. For the saturated aqueous solution, the product [Mg^{2+}] × [HO^{−}]^2 is equal to the solubility-product constant, and from the solubility of magnesium hydroxide (see the table) the value of the constant is found to be 15E−12 at 18°. The concentration of magnesium-ion is 0.000,154 in this solution. In a 0.1 molar solution of magnesium sulphate, which is ionized to the extent of 37.3%, the concentration of magnesium-ion is 0.0373, and it would require a concentration ‹greater› than 2E−5 of hydroxide-ion to precipitate any magnesium hydroxide.[349] Now, the concentration of the hydroxide-ion, in an ammoniacal solution, can easily be reduced far below this value by the addition of ammonium chloride, nitrate, sulphate or other readily ionizable ammonium salt to the solution, and then magnesium hydroxide will ‹not› be precipitated.

In solutions of ammonium hydroxide the concentration of hydroxide ion for 18° is found from the relation [NH_{4}^{+}] × [HO^{−}] : ([NH_{4}OH] + [NH_{3}]) = 18E−6 (p. 161). (1) For ammonium hydroxide, for instance in 0.2 molar solution, and in the absence of any ammonium salt, [HO^{−}] = [NH_{4}^{+}] = 0.0019.[350] This concentration of the hydroxide-ion should be more than sufficient to ‹precipitate› magnesium hydroxide in a 0.1 molar solution of the sulphate.[351]

EXP. 100 c.c. of a mixture which should contain magnesium sulphate in 0.1 molar solution would require 246 / (10 × 10) or 2.46 grams of MgSO_{4}, 7 H_{2}O. [p170] This weight of the salt is dissolved in 50 c.c. of water and 50 c.c. of 0.4 molar solution of ammonium hydroxide is added to it. Magnesium hydroxide is ‹precipitated›. (The precipitation is incomplete because the ammonium sulphate, formed in the reaction, reduces the ionization of ammonium hydroxide.)

(2) In the presence of 0.25 molar ammonium chloride, the chloride being dissociated to the extent of 80%,[352] the concentration of the hydroxide-ion in a 0.2 molar ammonium hydroxide solution is reduced to 18E−6.[353] According to the above calculation this is ‹too small› a concentration to precipitate magnesium hydroxide in a 0.1 molar solution of the sulphate.

EXP. 1.35 grams ammonium chloride (the weight corresponding to 100 c.c. of a mixture containing ammonium chloride in 0.25 molar solution) is dissolved in 50 c.c. of 0.4 molar ammonium hydroxide and 2.46 grams of magnesium sulphate, dissolved as before in 50 c.c. of water, is added to the mixture. ‹No precipitate› of magnesium hydroxide is formed.

In a similar fashion and for the same reason, ammonium salts interfere more or less with the precipitation of other hydroxides, for instance with the precipitation of manganous, nickelous, cobaltous, ferrous, zinc, cupric and cadmium hydroxides. But ammonium salts do not prevent the precipitation of aluminium, chromium and ferric hydroxides, which are much less soluble than the hydroxides just mentioned, and as much weaker bases, (Chap. X) are also much less readily ionized. They are precipitated by smaller concentrations of the hydroxide-ion than are the hydroxides of the first group, and their precipitation may be made quantitative.

EXP. Ferric hydroxide is readily precipitated when 2.7 grams of ferric chloride FeCl_{3}, 6 H_{2}O, the weight of the chloride required to give 100 c.c. of a 0.1 molar solution, is dissolved in as little water as possible. The solution is added to the mixture of magnesium sulphate, ammonium chloride and ammonium hydroxide obtained in the previous experiment, in which no magnesium hydroxide was precipitated.

FOOTNOTES:

[326] See Smith, ‹General Inorganic Chemistry›, p. 414, ‹General Chemistry for Colleges›, p. 277; Remsen, ‹Inorganic Chemistry (advanced course)›, p. 158.

[327] Mg^{2+} may also be precipitated as a carbonate with the other ions of the group (see Part III).

[328] In the case of ammonium nitrate a different form of decomposition, namely into nitrous oxide and water, predominates, and, in the case of ammonium nitrite, decomposition into nitrogen and water takes place so readily, that the decomposition of the salt takes that direction chiefly.

[329] Table, p. 106.

[330] T. S. Moore, ‹J. Chem. Soc.› (London), «91», 1379 (1907).

[331] From equation (I) we obtain directly, by the application of a simple mathematical transformation, that [NH_{3}] + [NH_{4}OH] / [NH_{4}OH] = ‹k›_{NH_{3}} + 1 = ‹k′›, and, therefore, [NH_{4}OH] = [NH_{3}] + [NH_{4}OH] / ‹k′›. Inserting this value for [NH_{4}OH] in equation (II) and transferring ‹k′›, we have [NH_{4}^{+}] × [HO^{−}] / ([NH_{3}] + [NH_{4}OH]) = ‹k›_{base} / ‹k′› = K.

[332] Water of crystallization, found in many of the precipitates used in qualitative analysis, will, as a rule, be indicated only in the formulas used in Part III, in the study of the reactions of ions.

[333] See Part III as to the method of vaporizing magnesium.

[334] ‹Cf.› Findlay, ‹Z. phys. Chem.›, «34», 409 (1900).

[335] The solubility of barium sulphate at 18°, according to the table, is 0.0023 / 233, or 1E−5 mole per liter, and the salt may be considered completely ionized at this dilution.

[336] The solubility (see the table) is 0.114 / 183.6, or 6.2E−4 mole per liter, 84% of which is ionized (according to conductivity determinations, Kohlrausch and Holborn, ‹loc. cit.›, p. 200). Hence [Sr^{2+}] = [SO_{4}^{2−}] = 0.00062 × 0.84 = 0.0005.

[337] To establish equilibrium, prolonged "digesting" is sometimes required. Double salts, solid solutions and mechanical enclosures are liable to interfere with the completeness of such separations by fractional precipitation. Resolution and reprecipitation will then usually effect a sufficiently accurate separation for most purposes. On account of the possibility of such complications, the conditions for a successful separation, within limits such as those described in the text, must in all cases be investigated.

[338] See the table at the end of Part IV.

[339] The salt may be considered completely ionized at this dilution. Each molecule of silver chromate forms two silver ions when it is ionized. See p. 141 in regard to the form the solubility-product takes in the case of a salt of this type.

[340] The complication, resulting from the hydrolysis of the chromate, is not included in this calculation.

[341] ‹Cf.› Findlay, ‹loc. cit.›

[342] Even Guldberg and Waage considered the ‹active› mass to be the fundamental factor and simply considered the total concentrations to be proportional to the "active masses," since they had no means of determining the proportion of "active" substances in the total concentrations.

[343] See Nernst, ‹Theoretical Chemistry›, p. 533, for a fuller discussion of the relations between the new and the old views on this subject.

[344] Dietrich and Wöhler, ‹Z. anorg. Chem.›, «34», 194 (1903).

[345] On account of the formation of an ammonium salt in the reaction.

[346] Loven, ‹Z. Anorg. Chem.›, «11», 404 (1896); Herz and Muks, ‹ibid.›, «38», 138 (1904).

[347] In that case the salts would be called "complex" salts, salts of a complex ion, MgCl_{3}^{−}. See Chapter XII. They are really "double salts"; ‹cf.› Smith, ‹General Inorganic Chemistry›, p. 536.

[348] See p. 114, and recall the laboratory experiments (Lab. Manual, p. 9, § 6), which may be given as lecture experiments at this point.

[349] Calling ‹x› the concentration of the hydroxide-ion, required to saturate a 0.1 molar magnesium sulphate solution with magnesium hydroxide, we have 0.0373 × ‹x›^2 = 15E−12 and ‹x› = 2E−5.

[350] Putting [HO^{−}] = [NH_{4}^{+}] = ‹y›, we have ‹y›^2 : (0.2 − ‹y›) = 18E−6. Then ‹y› = 0.0019.

[351] 0.0373 × 0.0019^2 = 0.13E−6, which is considerably larger than the solubility-product constant for magnesium hydroxide, 15E−12.

[352] Kohlrausch and Holborn, p. 159. Minor changes in the degrees of ionization of MgSO_{4} and NH_{4}Cl (and consequently of NH_{4}OH) occur, when the salts are present together. In a rigorous treatment, the ionization of each salt in the mixture would be calculated with the aid of Arrhenius's principle of isohydric solutions.

[353] ‹Vide› the analogous calculation on p. 113.

[p171]