CHAPTER VI

«CHEMICAL EQUILIBRIUM. THE LAW OF MASS ACTION»

The theory of ionization, as studied so far, gives us simple, rational explanations of many of our qualitative reactions—explanations which agree with phenomena taken from separate fields of investigation. But, if our study of the theory ceased at the present stage without further elaboration, we should fail to find in it a satisfactory explanation of a number of other important facts of analysis—notably, why certain reactions, the occurrence of which we might anticipate, ‹do not take place›. For instance, the addition of a soluble carbonate to a barium chloride solution precipitates almost all the barium as barium carbonate (‹exp.›); we have 2 Na^{+} + CO_{3}^{2−} + Ba^{2+} + 2 Cl^{−} → BaCO_{3} ↓ + 2 Na^{+} + 2 Cl^{−}. But the addition of carbonic acid to barium chloride solutions fails to produce the slightest precipitate (‹exp.›), although carbonic acid also gives rise to the carbonate-ion, CO_{3}^{2−}. In the same way silver nitrate readily precipitates silver phosphate from sodium phosphate solutions (‹exp.›), but not from a solution of phosphoric acid (‹exp.›). Hydrogen sulphide precipitates zinc sulphide from a zinc sulphate solution (‹exp.›), Zn^{2+} + SO_{4}^{2−} + 2 H^{+} + S^{2−} → ZnS ↓ + 2 H^{+} + SO_{4}^{2−}; but the addition of hydrochloric acid effectually prevents the precipitation (‹exp.›), although the hydrogen sulphide is still ionized, as is apparent from the precipitation of copper sulphide when copper sulphate is added to the mixture (‹exp.›). In the negative results, we have instances of a very large number of cases which require closer study, and a further development of the theory, if we wish to interpret them satisfactorily. The line of development to be followed is indicated perhaps most sharply by the following experiment.

EXP. Some sodium tetraborate (borax) is dissolved in a little water and silver nitrate is added to a small part of the solution. A pure ‹white› precipitate (silver borate) results. Another portion of the borate solution is diluted with a ‹large quantity of water›, and then silver nitrate is added; quite a different result is obtained—a ‹brown› precipitate (silver oxide) is formed. [p091]

The change in the quantity of water brought about the difference in result—the ‹quantitative› relations were altered thereby. In order to follow intelligently this and the other actions referred to, the study of reactions in solutions must be taken up from the ‹quantitative side›—the development heretofore has been essentially qualitative in character. On several occasions we have found that all electrolytes do not ionize equally well, and that the intensity of their action, demonstrated, for instance, for potassium and ammonium hydroxides, varies accordingly. We shall now have to study these relations in greater detail.

For our purpose, the study of two of the fundamental quantitative laws governing action in solution and of their application to analytical phenomena, will be sufficient: these are, ‹the law of chemical or homogeneous equilibrium›, in which the ‹law of mass action› is included, and the law of ‹physical› or ‹heterogeneous equilibrium›.

«The Law of Chemical Equilibrium.»—The law of chemical equilibrium may be expressed, for a simple case, by saying that if two substances ‹A› and ‹B› interact at a constant temperature to give two compounds ‹C› and ‹D› and, vice versa, ‹C› and ‹D› interact with each other to produce ‹A› and ‹B›, then equilibrium will be reached ‹when the ratio of the product of the concentrations of› ‹A› ‹and of› ‹B› ‹to the product of the concentrations of› ‹C› ‹and of› ‹D› ‹has a definite, constant value›, which is a value characteristic of the equilibrium between the compounds involved, at the given temperature. The action may be expressed in the chemical equation

‹A› + ‹B› ⇄ ‹C› + ‹D›,

in which ‹A›, ‹B›, ‹C› and ‹D› represent four different substances reacting in the molecular proportions indicated by their symbols, which as usual represent molecular weights. And the condition for equilibrium may be expressed in the mathematical equation

[‹A›] × [‹B›] / ([‹C›] × [‹D›]) = ‹k›.

[‹A›], [‹B›], [‹C›] and [‹D›] are used to represent the concentrations[166] of [p092] the four reacting substances and ‹k› is some definite number, called the equilibrium constant.

The law was discovered by Guldberg and Waage in 1867, and, with certain limiting conditions (see below) it has been fully established by extensive experimental work.[167] The significance of the law may be interpreted on the basis of the following considerations. If we start with the two substances ‹A› and ‹B› alone and have one mole of each in one liter (as gas or in solution) at a given temperature, then, all the conditions being given,—the temperature, the concentrations, and the nature of the substances,—the reaction ‹A› + ‹B› → ‹C› + ‹D›, leading to the formation of ‹C› and ‹D›, ‹will proceed with a perfectly definite velocity›. The molecules of ‹A› and of ‹B› move in all directions (kinetic theory of gases and solutions), and molecules of ‹A› will collide with molecules of ‹B› a definite number of times in unit time and will form a definite number[168] of molecules of ‹C› and ‹D› per minute. The velocity of chemical change of a given substance (‹chemical velocity›) is also measured in terms of moles, and is represented by the number of moles or the fraction of a mole changed per minute. If ‹v′›_{1} stands for the velocity of the action between ‹A› and ‹B›, under the given conditions, then

‹v′›_{1} = ‹k›_{1},

where ‹k›_{1} is some number. Now, if the concentration of one of the components, ‹e.g.› ‹A›, should be doubled, then the chances for collision and for action between molecules of ‹A› and ‹B› will be twice as great as before and the velocity of the action will be doubled. If only one-tenth of the concentration of ‹A› (one-tenth mole) is used, the velocity will only be one-tenth as great as originally, and, in general terms, if [‹A›] moles of ‹A› are used per liter, the ‹velocity of the change will be proportional to› [‹A›], ‹and equal to› ‹k›_{1} × [‹A›]. If the concentration of the other reacting component, ‹B›, is now doubled, the chances for action are again doubled, and, in general, the velocity of the action will be proportional also to the concentration [p093] [‹B›] of the second reacting substance. For the velocity, ‹v›_{1} of the action for any concentrations, [‹A›] and [‹B›], of ‹A› and ‹B› at any moment at a given temperature, we have

‹v›_{1} = ‹k›_{1} × [‹A›] × [‹B›].

Hence, if by the symbols [‹A›] and [‹B›] the concentrations at any given moment are represented, we may say that the velocity of the formation of ‹C› and ‹D› at that moment[169] ‹is proportional to the product of the concentrations of› ‹A› ‹and› ‹B›, ‹and to some constant›, which is characteristic of the interaction of ‹A› and ‹B›.

The validity of this conclusion has been fully verified by ‹experiment›.[170] The case is an instance of the ‹law of mass action, which states that in chemical changes the velocity of the action is proportional at any moment to the molecular concentrations›[171] ‹of the reacting components, and to a constant, which is characteristic of the chemical nature of the reacting components› (and of the temperature).

If we start with the reversed action

‹A› + ‹B› ← ‹C› + ‹D›,

the relation may be developed in the same way. Thus the two substances ‹C› and ‹D› will react upon each other, at the given temperature, with a velocity proportional to a constant, ‹k›_{2}, and, at any given moment, proportional also to their respective concentrations at that moment:

‹v›_{2} = ‹k›_{2} × [‹C›] × [‹D›].

Equilibrium will be reached when the substances ‹A› and ‹B› are formed at any moment from ‹C› and ‹D› just as rapidly as they are used up to produce ‹C› and ‹D›, and ‹vice versa›. Such is the case, [p094] when ‹the velocities of the two opposite reactions are equal to each other›. For the condition of equilibrium, then, ‹v›_{1} must be equal to ‹v›_{2} and therefore

‹k›_{1} × [‹A›] × [‹B›] = ‹k›_{2} × [‹C›] × [‹D›]

or

[‹A›] × [‹B›] / ([‹C›] × [‹D›]) = ‹k›_{2} / ‹k›_{1} = ‹k›_{equilibrium}.

In this way the meaning of the fundamental law of chemical equilibrium may be developed from the consideration of the velocities of the reversible actions, such as are involved in all conditions of equilibrium, and the ‹equilibrium constant represents the ratio of the velocity constants of the two opposite reactions›. This conclusion has been fully verified by experiment, the equilibrium constant being, as a matter of fact, found equal to the ratio of the velocity constants.[172]

The relations, so far considered, have been those of the simplest type of reversible reaction. We may now discuss the modifications required for other types of reaction by the law of equilibrium.

When two molecules of any reacting component take part in a reaction—for instance, in ‹A› + 2 ‹B› ⇄ ‹C› + ‹D›—‹the concentration of this component is raised to the second power in the mathematical expression of the law of equilibrium;› when three molecules of a component take part, its concentration is raised to the third power, etc.

For instance, hydrogen iodide is decomposed, reversibly, into hydrogen and iodine, according to 2 HI ⇄ H_{2} + I_{2}. A condition of equilibrium is reached, at a given temperature when

[H_{2}] × [I_{2}] / [HI]^2 = K.

At 440°, the results given in the following table were obtained by Bodenstein. The concentrations are expressed in moles per liter.[173] The constant is calcuated according to the equation just given. Analytical errors affect the value of the constant most in the first and last experiments, as a result of the very small concentrations of one component, I_{2} or H_{2}.

[p095]

[H_{2}] [I_{2}] [HI] K 0.0268 0.000190 0.0177 (0.016) 0.00986 0.00203 0.0328 0.019 0.00308 0.00783 0.0337 0.021 0.00175 0.0114 0.0315 0.020 0.000653 0.0204 0.0236 0.024 0.000265 0.0242 0.0202 (0.016)

The mechanical significance of the raised powers of the concentrations of components, two or more molecules of which take part in a reaction as indicated, will be discussed further on, in connection with a case of equilibrium between an electrolyte and its ions (Chapter VI, p. 102).

«Limitations to the Law of Chemical Equilibrium.»—Quite in agreement with the interpretation of the law of chemical equilibrium from the view-point of the kinetic theory, it is found that, in its applications, one must take into consideration the possibility, that molecular attractions (p. 27) or other important forces of attraction or repulsion (‹e.g.› electrical) of a more than ‹negligible› magnitude exist between the molecules of the components in a reversible reaction. If such forces are involved, suitable allowance must be made for them, so that the mathematical formulation of the law may express the facts of observation.[174] For these reasons, the law, in its simplest terms, which, alone, can be considered here, holds for the relations ‹obtaining in dilute systems›[175] (dilute solutions or gases that are not too strongly compressed), and in ‹systems involving only nonelectrolytes or only weak ionogens›, more generally than for the relations in concentrated solutions (or strongly compressed gases) or in systems, in which electrically charged particles (ions) are present in large proportions (see below, p. 108).

«The Factors of the Law of Chemical Equilibrium.»—Inspection of the mathematical expression of the law of chemical equilibrium (p. 91) shows that there are two significant kinds of factors in it: first, the equilibrium ‹constant›, whose value depends only on the nature of the substances involved and on the temperature.[176] In [p096] the second place, we have concentration factors, which, to a large extent,[177] may be varied at will.

The following experiments may be used to illustrate the significance of the two classes of factors: Phosphorus pentabromide is partially decomposed by heat into the tribromide and bromine (a case of gaseous dissociation):

PBr_{5} ⇄ PBr_{3} + Br_{2}.

Phosphorus trichlordibromide is decomposed more or less, in a similar fashion, into the components phosphorus trichloride and bromine, according to the equation

PCl_{3}Br_{2} ⇄ PCl_{3} + Br_{2}.

For the condition of equilibrium in the two cases we have

[PBr_{3}] × [Br_{2}]′ / [PBr_{5}] = ‹k›_{1} and [PCl_{3}] × [Br_{2}]″ / [PCl_{3}Br_{2}] = ‹k›_{2}.

EXP. Two tubes containing equivalent quantities of the two bromides are placed side by side in warm water.[178] The tube containing the trichlordibromide is found to be much more intensely colored by free bromine than that containing the pentabromide.

The intensity of the color of the bromine vapor shows that the concentration of bromine, [Br_{2}]″, in the PCl_{3}Br_{2} tube, is ‹greater than the corresponding concentration›, [Br_{2}]′, in the PBr_{5} tube. As a molecule of pentahalide PX_{5} dissociates into one molecule of PX_{3} and one molecule of X_{2}, [PCl_{3}] equals [Br_{2}]″ and is ‹greater› than [PBr_{3}], which is equal to [Br_{2}]′. Further, more of the pentabromide than of the trichlordibromide must be left undecomposed, ‹i.e.› [PCl_{3}Br_{2}] is ‹smaller› than PBr_{5}. Since the factors in the ‹numerator› of the second equation are ‹both larger›, and the factor in the ‹denominator smaller, than the corresponding factors› in the first equation, ‹k›_{2} ‹must be greater than› ‹k›_{1}. These ‹constants› are thus seen to be a ‹measure› of the ‹chemical stability› of these pentahalides. It is evident, too, that in reactions which depend on the presence of free bromine, such as the bromination of many organic compounds, the trichlordibromide should be more ‹effective› than the equivalent quantity of the pentabromide.

[p097]

In the second place, if we were to introduce into either tube, for instance into the tube containing the phosphorus trichlordibromide, an ‹excess of one of the dissociation products›, say an excess of phosphorus trichloride, then the condition of equilibrium would necessarily be disturbed:

‹y› [PCl_{3}] × [Br_{2}]′ / [PCl_{3}Br_{2}] > ‹k›_{2},

in which the bracketed symbols represent the concentrations of the first experiment. The velocities of the two opposite reactions would be no longer equal, the combination of trichloride with bromine would be accelerated by the increased concentration of the former. Here, equilibrium would only be reëstablished when the trichloride and bromine had combined to a sufficient extent to make

(‹y› [PCl_{3}] − ‹x›) × ([Br_{2}]′ − ‹x›) / ([PCl_{3}Br_{2}] + ‹x›) = ‹k›_{2},

in which x represents the number of moles of additional phosphorus trichlordibromide formed in unit volume by the combination of bromine with phosphorus trichloride. The ‹net result› is seen to be that an ‹increase in the concentration of the one dissociation product eo ipso reduces the concentration of the other dissociation product›.

EXP. A third tube charged with the same quantity of phosphorus trichlordibromide as the tube mentioned above, and with an added excess of phosphorus trichloride, is placed in the warm water next to the tube containing the trichlordibromide. Its color is much paler than that of the latter, owing to the suppression of free bromine.[179]

The concentration of the free bromine, ([Br_{2}]′ − ‹x›), under the new conditions of equilibrium, is smaller than the original concentration [Br_{2}]′—a result confirmed by experience. ‹It is in our power, therefore, arbitrarily to change the concentration of a reacting component›, in a case of equilibrium, ‹and thus to affect the reactivity of the system;› for instance, for brominating purposes, the new system would be less effective than the original one, and it might be of especial service where bromination is to be avoided.

In the cases studied, are found the two fundamentally important relations expressed by the law of equilibrium: ‹the equilibrium constant is a measure of the stability of a certain system› and, in a way, of its ‹reactivity› at a given temperature; and the [p098] ‹concentration factors are variables›, which we may change to a very considerable extent, so as, to a certain degree, to subject the system to our own purposes. We shall repeatedly have occasion to refer to these two fundamental relations and we shall use them again and again in our analytical work.

«Chemical Equilibrium of Electrolytes.»—Ionization of an electrolyte is a reversible chemical action and its relation to the law of chemical equilibrium will now be discussed. For acetic acid, ionization into hydrogen and acetate ions occurs thus: CH_{3}CO_{2}H ⇄ CH_{3}CO_{2}^{−} + H^{+}, and, in accordance with the law of equilibrium, at a given temperature, the following relation would hold:

[H^{+}] × [CH_{3}CO_{2}^{−}] / [CH_{3}CO_{2}H] = K_{ionization}.

If the total concentration of the acid is known, the concentrations of the ions and of the non-ionized acid may be calculated from the conductivity of the solution. For instance, if 60 grams of acetic acid (1 mole) is dissolved in sufficient water to make 10 liters, the equivalent conductivity of the solution (p. 50) is found to be 4.67 reciprocal ohms at 18°. The maximum conductivity of one mole of acetic acid, at infinite dilution, when all the acid would be ionized, would be 347. Therefore, in the acid under examination, 4.67 / 347, or 1.34 per cent, is ionized (p. 50). Since the total concentration of the acid is 0.1 mole ‹per liter› and 1.34 per cent is ionized, the concentration of the hydrogen-ion, [H^{+}], is 0.1 × 0.0134, and that of the acetate-ion, [CH_{3}CO_{2}^{−}], is the same. The concentration of the non-ionized acetic acid, [CH_{3}CO_{2}H], is 0.1 × 0.9866. If these values are inserted in the equation for the condition of equilibrium, we have

(0.1 × 0.0134)^2 / (0.1 × 0.9866) = K_{ionization} = 18.2E−6.

From this experimental result, the equilibrium constant, which is called the ‹ionization constant› of the acid, is found to have the value 18.2E−6. If the ratio [H^{+}] × [CH_{3}CO_{2}^{−}] / [CH_{3}CO_{2}H] really is a constant, the same value, within the limits of experimental errors, should be obtained from acetic acid in other concentrations. Now, if the above solution is diluted to ten times its volume, the concentration of the acid is made 0.01 mole ‹per liter›, the conductivity [p099] is found to have increased to 14.5 reciprocal ohms, and the percentage of ionized acid is then 14.5 / 347, or 4.17. Here, [H^{+}] and [CH_{3}CO_{2}^{−}] = 0.01 × 0.0417 and [CH_{3}CO_{2}H] = 0.01 × 0.9583. Inserting these values in our general equation and calculating the result, we obtain 18.1E−6 as the value of the constant. In the following table[180] are given the molar conductivities, Λ (column 2), of acetic acid of varying concentrations, ‹m› (column 1). The degrees of ionization, α, and the ionization constant, calculated according to the equilibrium equation, are given in columns 3 and 4.

IONIZATION OF ACETIC ACID. Λ_{∞} = 347.

m. Λ. 100 α. K. 0.1 4.67 1.34 18.2E−6 0.08 5.22 1.50 18.3E−6 0.03 8.50 2.45 18.5E−6 0.01 14.50 4.17 18.1E−6

It is evident, that a constant value is found for the ratio [H^{+}] × [CH_{3}CO_{2}^{−}] / [CH_{3}CO_{2}H] and that the ionization of acetic acid, in these dilute solutions, obeys the law of chemical equilibrium.[181] The equilibrium constant expresses in definite, quantitative terms the tendency of acetic acid to ionize in dilute solution. Examination of other acids shows that there is an enormous range in the values found for their respective ionization constants. The constants are the best ‹measure› of the ‹strength› of the ‹acids› as acids. Obviously, the more readily acids in equivalent solutions ionize, the greater will be the concentration of the hydrogen-ion to which the characteristic acid properties are due, and the more pronounced (stronger) will be the exhibition of these properties. From the ionization constants one may calculate, for instance, the proportion in which two competing acids will neutralize a base, when the latter is used in quantity insufficient to neutralize both acids. [p100]

Inspection of the equation for acetic acid, which is the typical equilibrium equation for all ‹monobasic› acids, shows that the greater the degrees of ionization of acids are in equivalent solutions, ‹i.e.› the greater the concentrations of the hydrogen-ion which their ionization produces in equivalent solutions, the larger will be the values of their ionization constants. The acids with the ‹larger› constants are, then, the ‹stronger› acids.

«The Ionization of Various Acids.»—The table given on page 104 shows the ionization constants for a number of acids of interest in analysis. Before proceeding to give the table, we must consider further two important points.

In the first place, for the strongest acids, such as hydrochloric, nitric, hydrobromic and similar acids, chemists have been unable to determine ionization constants on the basis of the law of chemical equilibrium. Strong acids, strong bases and most salts (see pp. 106–8, below), the three classes comprising ‹all› the ‹very readily ionizable electrolytes, do not give constants› when the values of the equilibrium ratio,[182] [Cation] × [Anion] / [Molecules], are calculated for different concentrations, and ‹they therefore do not ionize simply in accordance with the law of chemical equilibrium›. The reasons for this abnormal behavior will be discussed presently (p. 108), when other necessary facts are before us. In order to have, at least, a rough basis for comparison of these strong acids with the weak ones, which do obey the law of chemical equilibrium, the table will give for the strong acids ‹the value of the above ratio as calculated from their ionization in 0.1 molar solutions›.

«The Ionization of Polybasic Acids.»—In the second place, the meaning of the ‹constants for polybasic acids›, such as sulphuric, phosphoric, carbonic and similar acids, requires explanation. The relations for carbonic acid will be first developed, as representing a typical case. Carbonic acid, in ionizing, forms the carbonate-ion CO_{3}^{2−}, and the hydrogen-ion, as expressed in the [p101] equation[183] H_{2}CO_{3} ⇄ 2 H^{+} + CO_{3}^{2−}. According to the law of chemical equilibrium for the case where a product (here the hydrogen-ion) appears twice on one side of the reaction equation, we have, for the condition of equilibrium (p. 94)

[H^{+}]^2 × [CO_{3}^{2−}] / [H_{2}CO_{3}] = K. (1)

We may ask, however, whether both the hydrogen atoms of carbonic acid show the ‹same tendency to ionize›, or, since there is a vast difference in the ease of ionization of different acids, whether there is not also a difference in the ease of ionization of the different hydrogen atoms in a polybasic acid. As a matter of experiment, we find that a molecule of carbonic acid does ionize, first, and more readily, into one hydrogen ion and an ‹acid carbonate› ion HCO_{3}^{−}, according to H_{2}CO_{3} ⇄ H^{+} + HCO_{3}^{−}.

For this reversible reaction we have[184]

[H^{+}] × [HCO_{3}^{−}] / [H_{2}CO_{3}] = K_{1}. (2)

The value of this constant,[185] called the ‹primary ionization constant› of carbonic acid, is 0.3E−6.

The acid carbonate-ion HCO_{3}^{−}, in turn, is ionized to a certain extent, producing another hydrogen ion and the carbonate-ion, CO_{3}^{2−}. We have HCO_{3}^{−} ⇄ H^{+} + CO_{3}^{2−}, and

[H^{+}] × [CO_{3}^{2−}] / [HCO_{3}^{−}] = K_{2}. (3)

The value of this constant[186], called the ‹constant› of the [p102] ‹secondary ionization› of carbonic acid, is 0.07E−9, which has about one four-thousandth of the value of the constant for the primary ionization.

If we combine equations (2) and (3) we have

[H^{+}] × [HCO_{3}^{−}] × [H^{+}] × [CO_{3}^{2−}] / ([H_{2}CO_{3}] × [HCO_{3}^{−}]) = K_{1} × K_{2} (4)

or

[H^{+}]^2 × [CO_{3}^{2−}] / [H_{2}CO_{3}] = K.

This is equation (1), derived originally by the application of the law of mass action to the relation between the carbonate-ion, CO_{3}^{2−}, the hydrogen-ion, and carbonic acid, H_{2}CO_{3}.

This relation, and, in particular, the ‹significant squaring› of the concentration of the hydrogen-ion, an ion which appears ‹twice› in the equation for the formation of carbonic acid from carbonate and hydrogen ions, (2 H^{+} + CO_{3}^{2−} ⇄ H_{2}CO_{3}), may now be interpreted mechanically (p. 92) as follows: For the formation of carbonic acid from a carbonate ion and ‹two› hydrogen ions, a carbonate ion must collide and combine first with one hydrogen ion, and the velocity for the formation of this intermediate product, HCO_{3}^{−}, will be proportional to the (total) concentration of the hydrogen ions; the product, HCO_{3}^{−}, to form H_{2}CO_{3}, must collide and combine with a hydrogen ion once more, and this combination will proceed with a velocity ‹again proportional› to the (total) concentration of the hydrogen ions. So the velocity for the transformation of CO_{3}^{2−} into H_{2}CO_{3} will be proportional ‹twice over› to the (total) concentration of the hydrogen ions—as well as, in the usual fashion, to the concentration of the carbonate ions present at any moment.

‹It is a general principle that the primary ionization of polyvalent acids occurs more readily than the secondary›, and this in turn more readily than the tertiary (if a third ionizable hydrogen atom is present in the acid).

In the case of phosphoric acid, for instance, the primary ionization into the hydrogen-ion and the dihydrogen-phosphate-ion, H_{2}PO_{4}^{−}, takes place so readily that phosphoric acid reacts strongly acid[187] to methyl orange,[188] the [p103] concentration of hydrogen-ion being sufficiently great to affect this indicator (see EXP. below).

When phosphoric acid is neutralized by one equivalent of a base, say of sodium hydroxide, the salt formed, sodium dihydrogen-phosphate, NaH_{2}PO_{4}, yields sodium-ion and dihydrogen-phosphate-ion, H_{2}PO_{4}^{−}. The latter is ionized ‹somewhat› into H^{+} and the bivalent hydrogen-phosphate-ion, HPO_{4}^{2−}. The ionization of the ion H_{2}PO_{4}^{−} is now the chief source of supply of hydrogen-ion (the further ionization of HPO_{4}^{2−} is practically negligible here) and it is ionized so little that the solution of NaH_{2}PO_{4} no longer changes the color of methyl orange (see EXP. below). The solution is, however, acid to the indicator phenolphthaleïn, which is much more sensitive to the hydrogen-ion and will show the presence of much smaller concentrations of it than will methyl orange. The addition of a second equivalent of sodium hydroxide to the solution converts NaH_{2}PO_{4} into Na_{2}HPO_{4}. This salt gives sodium-ion and the hydrogen-phosphate-ion HPO_{4}^{2−}, which, in turn, is ionized only very slightly, producing phosphate-ion PO_{4}^{3−}, and again hydrogen-ion. The ionization of HPO_{4}^{2−} is so slight, however, and the concentration of the hydrogen-ion, therefore, so minute, that the solution does not react acid even to the sensitive indicator phenolphthaleïn.

EXP. Methyl orange (very little) is added to 10 c.c. of a 0.1 molar solution of phosphoric acid and 10 c.c. of 0.1 molar sodium hydroxide solution is added to the mixture; the color will be found to change from the acid to the neutral tint just as the last drop or two of the alkali are added. Phenolphthaleïn is then added to the mixture and 10 c.c. more of the 0.1 molar sodium hydroxide solution are required to change the color of the mixture to a pronounced pink (alkaline) tint.

Even sulphuric acid, although its two hydrogen atoms are ionized very easily, making sulphuric acid a strong acid, shows a difference in the ease of ionization of the two hydrogen atoms. Since ionization, in general, is favored by dilution, we find that in the case of such a strong acid the difference is most marked in more concentrated solutions, the smaller amount of water starting the ionization in the more favored direction and producing first, chiefly, hydrogen-sulphate ions, HSO_{4}^{−}. When the solution is diluted, the hydrogen-sulphate ions are to a very considerable extent dissociated into sulphate ions and hydrogen ions. The described change in ionization can be roughly followed with the aid of an insoluble sulphate like barium sulphate. Barium sulphate, while very insoluble in water, dissolves in rather strong sulphuric acid to form the acid sulphate, Ba(HSO_{4})_{2}, the SO_{4}^{2−} ion of the sulphate being more or less suppressed by uniting with hydrogen-ion. We have the action

BaSO_{4} ⇄ Ba^{2+} + SO_{4}^{2−} and

Ba^{2+} + SO_{4}^{2−} + H^{+} + HSO_{4}^{−} ⇄ Ba^{2+} + 2 HSO_{4}^{−}.

If the solution of the acid sulphate is poured into a large volume of water, barium sulphate is immediately reprecipitated, the hydrogen-sulphate-ion being dissociated, in the dilute solution, into hydrogen-ion and sulphate-ion, SO_{4}^{2−}, whose barium salt is so difficultly soluble:

Ba^{2+} + 2 HSO_{4}^{−} → Ba^{2+} + 2 H^{+} + 2 SO_{4}^{2−} → BaSO_{4} ↓ + 2 H^{+} + SO_{4}^{2−}.

[p104]

EXP. Finely divided barium sulphate is warmed for a moment with a few cubic centimeters of concentrated sulphuric acid in a test tube, the mixture is allowed to settle, and some of the clear acid is carefully decanted into a large beaker full of water.

It may be added, that while the primary ionization of sulphuric acid does not yield an equilibrium constant for the ratio [H^{+}] × [HSO_{4}] / [H_{2}SO_{4}], even such a strong acid as is sulphuric acid is found to give a fairly good constant[189] for [H^{+}] × [SO_{4}^{2−}] / [HSO_{4}^{−}]. The value of this constant[189] is 0.03.

«The Ionization Constants[A] of Acids»

Acid. Equilibrium Ratio. K. Hydrochloric [H^{+}]×[Cl^{−}]/[HCl] (1) Hydrobromic [H^{+}]×[Br^{−}]/[HBr] (1) Hydroiodic [H^{+}]×[I^{−}]/[HI] (1) Nitric [H^{+}]×[NO_{3}^{−}]/[HNO_{3}] (1) Chromic[B] [H^{+}]×[HCrO_{4}^{−}]/[H_{2}CrO_{4}] (1) [H^{+}]×[CrO_{4}^{2−}]/[HCrO_{4}^{−}] 0.6E−6 Sulphuric[C][D] [H^{+}]×[HSO_{4}^{−}]/[H_{2}SO_{4}] (1) [H^{+}]×[SO_{4}^{2−}]/[HSO_{4}^{−}] 0.3E−1 Oxalic[E] [H^{+}]×[C_{2}O_{4}^{−}]/[H_{2}C_{2}O_{4}] 3.8E−2 [H^{+}]×[C_{2}O_{4}^{2−}]/[HC_{2}O_{4}^{−}] 0.5E−4 Phosphoric[F] [H^{+}]×[H_{2}PO_{4}^{−}]/[H_{3}PO_{4}] 0.1E−1 [H^{+}]×[HPO_{4}^{2−}]/[H_{2}PO_{4}^{−}] 0.2E−6 [H^{+}]×[PO_{4}^{3−}]/[HPO_{4}^{2−}] 0.4E−12 Arsenic[D] [H^{+}]×[H_{2}AsO_{4}^{−}]/[H_{3}AsO_{4}] 0.5E−2 Nitrous[B] [H^{+}]×[NO_{2}^{−}]/[HNO_{2}] 0.5E−3 Acetic[G] [H^{+}]×[CH_{3}CO_{2}^{−}]/[CH_{3}CO_{2}H] 1.8E−5 Carbonic[H][I] [H^{+}]×[HCO_{3}^{−}]/([H_{2}CO_{3}] + [CO_{2}]) 0.3E−6 [H^{+}]×[CO_{3}^{2−}]/[HCO_{3}^{−}] 0.7E−10 H_{2}S[J][K] [H^{+}]×[SH^{−}]/[H_{2}S] 0.9E−7 [H^{+}]×[S^{2−}]/[SH^{−}] 0.1E−14 Boric[B] [H^{+}]×[H_{2}BO_{3}^{−}]/[H_{3}BO_{3}] 0.7E−9 Hydrocyanic [H^{+}]×[CN^{−}]/[HCN] 0.7E−9 Arsenious [H^{+}]×[H_{2}AsO_{3}^{−}]/[H_{3}AsO_{3}] 0.6E−9 Water[L][C] [H^{+}]×[HO^{−}]/[H_{2}O] at 25° 0.2E−15 at 100° 0.9E−14 [H^{+}]×[HO^{−}] at 25° 1.2E−14 at 100° 0.5E−11

TABLE NOTES:

A. As explained on p. 100, the bracketed values given for the strong acids ‹are not constants›, but express the values of the ratios [H^{+}] × [Anion] / [Acid] for 0.1 ‹molar solutions›.

B. See references, Noyes, ‹ibid.›, «32», 860 (1910).

C. Noyes and Eastman, ‹Carnegie Institution Publications›, «63», 274 (1907).

D. Luther, ‹Z. Elektroch›, «13», 296 (1907).

E. Chandler (McCoy), ‹J. Am. Chem. Soc.›, «30», 713 (1908).

F. Abbot and Bray, ‹ibid.›, «31», 760 (1909).

G. See above, p. 99.

H. Walker, ‹J. Chem. Soc.›, (London), «77», 5 (1900).

I. McCoy, ‹Am. Chem. J.›, «29», 455 (1903); Stieglitz, ‹Carnegie Institution Publications›, «107», 243 (1909).

J. Auerbach, ‹Z. phys. Chem.›, «49», 220 (1904).

K. Knox, in Abegg's laboratory, ‹Trans. Faraday Soc.›, «4», 43 (1908).

L. ‹Vide› p. 66.

[p105]

The difference in the tendencies of acids to ionize, as expressed in the table, may be recognized in equivalent solutions by any of the properties dependent on the ionization, such as the conductivity, the chemical activity, the osmotic pressure and allied effects, and so forth. If the conductivities of equal volumes of equivalent (‹e.g.› normal) solutions of hydrochloric, phosphoric and acetic acids are compared (‹exp.›)[190], it is readily seen that hydrochloric acid is the best conductor, phosphoric acid a much poorer one, and acetic acid an exceedingly poor one (the conductivity of normal acetic acid is about 1 / 200 that of normal hydrochloric acid, and the conductivity of normal phosphoric acid is about 1 / 14 that of normal hydrochloric acid). Since the conductivity is approximately proportional to the concentration of the hydrogen-ion[191] in each of the solutions, it is evident that the hydrochloric acid is ionized to a considerably greater extent than either of the other acids—than acetic acid, in particular. Similarly, if a drop (0.05 c.c.) of molar hydrochloric acid and a drop of molar acetic acid are added to equal volumes (50 c.c.) of a very dilute solution of methyl orange (‹exp.›), the color will be changed decidedly by the hydrochloric acid to a bright pink, but by the acetic acid only to an orange hue. Again, if a precipitate of barium chromate or calcium oxalate is treated with some strong acid, hydrochloric or nitric, for instance, it dissolves readily, while a considerable excess of acetic acid (‹exp.›) only dissolves traces of either precipitate.[192] In this way, the chemical behavior of these acids differs in degree, as a result of the different tendencies to ionize, which are expressed in the constants of the table. Advantage is taken, in analysis, of such differences. Acetic acid, for instance, is used when only a slight degree of acidity is desired—as in recognizing barium-ion by its chromate, or oxalic acid by means of its calcium salt. Hydrochloric or nitric acid is used when decided acidity is required—as in the separation of groups by hydrogen sulphide (Chap. XI).

«The Ionization of Bases.»—The same relations hold for bases as for acids: the weaker bases give ionization constants as [p106] do the weaker acids; the strong bases, again, as was mentioned above, do not give constants. The values for strong bases, stated in the following table, are bracketed, and refer to the ionization in 0.1 molar solutions of the bases. Polyvalent bases, like polybasic acids, ionize in stages, and the primary ionization is usually stronger than the secondary ionization, and so forth. For instance, a study of ferric chloride solution shows that the third hydroxide group of ferric hydroxide, Fe(OH)_{3}, must have the smallest ionization constant.[193]

«The Ionization Constants[A] of Bases»

Base. Ratio. K. Potassium hydroxide [K^{+}]×[HO^{−}]/[KOH] (1) Sodium hydroxide [Na^{+}]×[HO^{−}]/[NaOH] (1) Barium hydroxide [Ba^{2+}]×[HO^{−}]^2/[Ba(OH)_{2}] (0.03) Strontium hydroxide [Sr^{2+}]×[HO^{−}]^2/[Sr(OH)_{2}] (0.03) Calcium hydroxide[B] [Ca^{2+}]×[HO^{−}]^2/[Ca(OH)_{2}] (0.03) Ammonium hydroxide[C][D] [NH_{4}^{+}]×[HO^{−}]/ ([NH_{4}OH]+[NH_{3}]) 1.8E−5 Hydrazine[E] [N_{2}H_{5}^{+}]×[HO^{−}]/ ([N_{2}H_{5}OH]+[N_{2}H_{4}]) 0.3E−5 Aniline[F] [C_{6}H_{5}NH_{3}^{+}]×[HO^{−}]/ ([C_{6}H_{5}NH_{3}OH]+ [C_{6}H_{5}NH_{2}]) 0.5E−9 Water [H^{+}]×[HO^{−}]/[H_{2}O] at 25° 0.2E−15 at 100° 0.9E−14 [H^{+}]×[HO^{−}] at 25° 1.2E−14 at 100° 0.5E−11

TABLE NOTES:

A. As explained above, the bracketed values under K for strong bases are not constants, but express the values of the ratios for 0.1 molar solutions only. (For the alkaline earths 0.05 molar solutions are referred to.)

B. Estimated value.

C. ‹Vide› page 160.

D. Corrected value, from Bredig's data, ‹Z. phys. Chem.›, «13», 322 (1894). A. A. Noyes, ‹Carnegie Institution, Publication› No. 63, p. 178, finds the value 17.3E−6 for 18°.

E. Uncorrected value, Bredig, ‹ibid.›

F. Bredig, ‹ibid.›; Stieglitz and Derby, ‹Am. Chem. J.›, «31», 457 (1904).

While the constants of a large number of organic bases[194] have been determined few constants of inorganic bases are as yet known. The fact that the majority of ‹inorganic bases› are polyvalent and difficultly soluble has made their examination in this respect more difficult. From the study of the decomposition of salts by water (see