Prof. Ya. I. Mikhailenko and A. M. Chernukhin.

On the Question of Adsorption.

Adsorption of Organic Dyes by Clay.

(From the Laboratory of Organic Chemistry of the Tomsk Technological Institute).

Adsorption processes at the present time are acquiring ever greater and greater importance in various branches of industry.

Coal was formerly used almost exclusively as the adsorbent. But, in connection with the growth of production, the use of coal as an adsorbent became a major overhead expense. Therefore at the present time coal has begun to be replaced by various silicates, which occur in nature in enormous quantity in a ready-made form. Many of them, as has turned out, possess a great adsorbing capacity¹.

In view of the fact that the phenomena of adsorption are of considerable theoretical as well as practical interest, a great number of investigations have been and, especially in recent times, are being devoted to the study of adsorption phenomena. But it absolutely cannot be said that the essence of adsorption phenomena has been sufficiently elucidated.

The more closely any adsorption process is investigated, for example, the dyeing of fabrics, the more complex and complicated it proves to be, and the process which was previously designated by the name “adsorption” often turned out to consist of a series of separate processes: here there is chemism, and dissolution, and, finally, adsorption proper².

The original aim of our work was to try to elucidate the dependence between the capacity of a given substance to be adsorbed and the structure of its molecule; namely, to try to determine to what extent particular atomic groupings promote adsorption, or whether adsorption is not connected with the structure of the molecule.

In parallel it was intended to use the experiments carried out to clarify the apparently complex phenomenon of “adsorption” itself. In the course of the work, other tasks gradually emerged as well.

To clarify the dependence between the magnitude of adsorption and the structure of the molecule, it seemed to us that the most convenient object could be dyes, and the adsorbent—clay.

Clay was chosen, on the one hand, in view of its technical application as an adsorbent, and, on the other hand, as a substance of definite composition and more or less established chemical function³.

The dyes were taken from the best German factories: F. Kahlbaum, Merck, and Meister Lucius und Brüning, with the notation “for photographic purposes” or “for medical purposes.”

The experiments were carried out exclusively in aqueous solutions. The solutions were prepared on the assumption that 1 cu. cm of solution should contain 1 mg of dye.

Tomsk white clay was taken as the adsorbent. Before use it was twice elutriated in a Schöne apparatus. For the work a fraction with grain size $\leq 0.002$ mm was taken. Analysis showed the following composition of this fraction:

Average of two determinations:

Component	Content
Loss on ignition	11.78%
$\mathrm{SiO_2}$	52.40%
$\mathrm{Al_2O_3}$	32.92%
$\mathrm{Fe_2O_3}$	0.17%
$\mathrm{MnO}$	0.43%
$\mathrm{CaO}$	1.96%
$\mathrm{CO_2}$	0.16%
Sum	99.18%

In its mineralogical composition this clay is kaolinite with an admixture of parent rock and limestone. The latter circumstance is confirmed by the content of $\mathrm{CO_2}$. The following formula closely fits this clay:

\[ 3\mathrm{SiO_2}\cdot \mathrm{Al_2O_3}\cdot 2\mathrm{H_2O} \]

\[ 56.60\%\quad 32.07\%\quad 11.33\%. \]

The elutriated clay, in order to free it from $\mathrm{CaCO_3}$, was treated with a weak solution of $\mathrm{HNO_3}$, then carefully washed free of traces of adsorbed $\mathrm{HNO_3}$. The clay prepared in this way was dried to constant weight at $100^\circ\mathrm{C}$. All weighings were carried out on a chemical balance with an accuracy of 0.0002 g.

The first series of experiments was carried out in order to find the amount of dye, in mg, adsorbed by one gram of clay when the content in the aqueous phase was 0.5 to 1 mg per 100 cu. cm of liquid. All experiments were conducted exclusively at room temperature, with fluctuations of $\pm 5^\circ\mathrm{C}$ allowed.

This series of experiments was performed as follows:

A weighed portion of 2.0000 g of elutriated clay was poured into a dry flask of 125–150 cu. cm capacity. To this was added, from a burette, the dye solution in the required amount, determined approximately by a preliminary experiment. Then enough water was added to bring the total volume to 100 cu. cm. With this same water the adhering clay particles were washed from the watch glass on which the weighing had been carried out. The mixture was shaken for 15 min, after which it was left to settle for twenty-four hours. If the solution became completely decolorized, dye solution was again carefully added from the burette. The mixture was again shaken for 15 min and left until complete clarification, and so on until 0.5–1 mg of dye remained in the aqueous phase. At this point the experiment was stopped. The amount of dye remaining in the solution and not adsorbed was determined colorimetrically.

Subtracting the unabsorbed amount of dye from the total amount taken for the experiment, and dividing the resulting difference by the weighed portion of clay, we obtain the number of mg of dye adsorbed by one gram of clay under the condition that the dye content in the liquid phase is 0.5–1 mg in 100 cu. cm of volume. For explanation we give an extract from the journal:

A weighed portion of 2.0000 g of clay was taken, poured into a flask of 125 cu. cm capacity; 18 cu. cm of a solution of methylene blue containing 1 mg of dye in 1 cu. cm was added, and then 82 cu. cm of water. After shaking and settling, 0.05 mg of dye remained unabsorbed in 100 cu. cm. Added

more precisely, 4 cc of a methylene-blue solution. After adsorption, 0.6 mg remained in 100 cc. The total volume of liquid is:

\[ 18+82+4=104\ \text{cc}. \]

The total amount of dye taken for the experiment is:

\[ 18+4=22\ \text{mg}. \]

There remained unabsorbed:

\[ \frac{0.6\cdot 104}{100}=0.62\ \text{mg}. \]

Absorbed by one gram of clay:

\[ \frac{22.0-0.62}{2}=10.69\ \text{mg}. \]

Two other check experiments gave the numbers: 10.4 mg and 10.65 mg. Thus the average of the three experiments will be 10.6 mg.

The results of the experiments are brought together in Tables I, II, and III. In the tables each number is the average of three determinations, which differ from one another in most cases by 2–3 units in the first decimal place.

Table I.

Dyes of the triphenylmethane series.

No.	Name of dye	Structural formula	Adsorption in mg	Relation to dialysis
1	Crystal violet	$$\mathrm{C}\begin{cases}\mathrm{C_6H_4\!\cdot N(CH_3)_2}\\ \mathrm{C_6H_4\!\cdot N(CH_3)_2}\\ \mathrm{C_6H_4{=}N(CH_3)_2}\\ \mathrm{Cl}\end{cases}$$	12.23	Rapidly dialyze\(^1\).
2	Methyl violet	$$\mathrm{C}\begin{cases}\mathrm{C_6H_4\!\cdot N(CH_3)_2}\\ \mathrm{C_6H_4\!\cdot NHCH_3}\\ \mathrm{C_6H_4{=}N(CH_3)_2Cl}\end{cases}$$	12.98	Rapidly dialyze\(^1\).
3	Ethyl violet	$$\mathrm{C}\begin{cases}\mathrm{C_6H_4\!\cdot N(C_2H_5)_2}\\ \mathrm{C_6H_4\!\cdot N(C_2H_5)_2}\\ \mathrm{C_6H_4{=}N(C_2H_5)_2Cl}\end{cases}$$	12.67	Rapidly dialyze\(^1\).
4	Fuchsin	$$\mathrm{C}\begin{cases}\mathrm{C_6H_4\!\cdot NH_2}\\ \mathrm{C_6H_4\!\cdot NH_2}\\ \mathrm{C_6H_3{=}NH_2Cl}\\ \mathrm{CH_3}\end{cases}$$	9.79	Rapidly dialyze\(^1\).
5	Malachite green	$$\mathrm{C}\begin{cases}\mathrm{C_6H_4}\\ \mathrm{C_6H_4\!\cdot N(CH_3)_2}\\ \mathrm{C_6H_4{=}N(CH_3)_2Cl}\end{cases}$$	6.8	Rapidly dialyze\(^1\).
6	Victoria blue	$$\mathrm{C}\begin{cases}\mathrm{C_6H_4\!\cdot N(CH_3)_2}\\ \mathrm{C_6H_4\!\cdot N(CH_3)_2}\\ \mathrm{C_{10}H_6{=}NH{-}C_6H_5}\\ \mathrm{Cl}\end{cases}$$	12.02	Rapidly dialyze\(^1\).
7	Guinea green	$$\mathrm{OH{-}C}\begin{cases}\mathrm{C_6H_5}\\ \mathrm{C_6H_4{-}N{-}C_6H_4{-}SO_3Na}\\ \mathrm{C_6H_4{=}N{-}CH_2{-}C_6H_4{-}SO_3Na}\\ \mathrm{C_2H_5}\end{cases}$$	1.2	Rapidly dialyze\(^1\).
8	Patent blue	$$\mathrm{C_6H_2}\begin{cases}\mathrm{SO_3}\\ \mathrm{SO_3}\end{cases}\mathrm{Ca};\quad \mathrm{OH{-}C}\begin{cases}\mathrm{C_6H_4\!\cdot N(CH_3)_2}\\ \mathrm{C_6H_4\!\cdot N(CH_3)_2}\end{cases}$$	0.0	Rapidly dialyze\(^1\).

\(^1\) Freundlich. Kapillarchemie. 1909.

Of the 8 representatives of this series that were tested, patent blue is not adsorbed at all, and guinea green is adsorbed extremely weakly in comparison with other dyes. Both of these dyes, as is evident from their structural formulae, upon dissociation give a colored anion. All the remaining dyes, adsorbed to varying degrees, in aqueous solution, according to their structural formulae, dissociate according to the type of ammonium salts and thereby give colored cations, which are complex organic radicals, and anions of mineral or organic acids.

Table II.

Azo dyes.

No.	Name of dye	Structural formula	Adsorbed amount, in mgr.	Relation to dialysis
1	Chrysoidine	$C_6H_5 \cdot N{=}N \cdot C_6H_3(NH_2)_2$	6,72	Dialyzes rapidly.
2	Methyl orange	$(NaSO_3)\ C_6H_4 \cdot N{=}N \cdot C_6H_4N(CH_3)_2$	0,0
3	Fast yellow	$(NaSO_3)\ C_6H_4 \cdot N{=}N \cdot C_6H_4NH_2$ $+$ $(NaSO_3)\ C_6H_4 \cdot N{=}N \cdot C_6H_3\begin{matrix}\diagup NH_2\\ \diagdown SO_3Na\end{matrix}$	0,0
4	Bismarck brown	$C_6H_4\begin{matrix}\diagup N{=}N\cdot C_6H_3(NH_2)_2\\ \diagdown N{=}N\cdot C_6H_3(NH_2)_2\end{matrix}$	8,72	Dialyzes rapidly.
5	Crystal ponceau	$C_{10}H_7N{=}N{-}C_{10}H_4\begin{matrix}\diagup SO_3Na\ 6\\ {-}SO_3Na\ 8\\ \diagdown OH\end{matrix}$	0,0

All the dyes investigated are either sodium salts of azo-sulfonic acids (methyl orange, fast yellow, crystal ponceau) or salts of the ammonium type (chrysoidine, Bismarck brown). Only the latter are adsorbed.

Table III.

Dyes of the safranine, acridine, and thiazine series.

No.	Name of dye	Structural formula	Adsorbed amount, in mgr.	Relation to dialysis
1	Phenosafranine	$NH_2{-}C_6H_3\begin{matrix}N\\[-2pt]\diagup\ \ \diagdown\\[-2pt]N\end{matrix}C_6H_3\cdot NH_2$ $\ \ \ \ \ \ \ \ \ \ \ \vert{-}Cl$ $\ \ \ \ \ \ \ \ \ \ \ C_6H_5$	11,68	Dialyzes rapidly.
2	Safranine	$\begin{matrix}CH_3\\ NH_2\end{matrix}\!\!>\!C_6H_2\begin{matrix}N\\[-2pt]\diagup\ \ \diagdown\\[-2pt]N\end{matrix}C_6H_2\!<\!\begin{matrix}CH_3\\ NH_2\end{matrix}$ $\ \ \ \ \ \ \ \ \ \ \ \vert{-}Cl$ $\ \ \ \ \ \ \ \ \ \ \ C_6H_5$	11,01
3	Heliotrope according to tannin	$(CH_3)_2N{-}C_6H_3\begin{matrix}N\\[-2pt]\diagup\ \ \diagdown\\[-2pt]N\end{matrix}C_6H_2\!<\!\begin{matrix}CH_3\\ NH_2\end{matrix}$ $\ \ \ \ \ \ \ \ \ \ \ \vert$ $\ \ \ \ \ \ \ Cl\ \ C_6H_3\!<\!\begin{matrix}CH_3\\ CH_3\end{matrix}$	17,0

No.	Name of dye	Structural formula	Adsorbed amount, in mgr.	Relation to dialysis
4	Phosphine . . . . .	$(\mathrm{NO_3})\,\mathrm{H_3N{-}C_6H_3}\begin{matrix} & \mathrm{N}\\[-2pt] \diagup & \diagdown\\[-2pt] & \mathrm{C}\end{matrix}\mathrm{C_6H_4}$ $\left.\vphantom{\begin{matrix}a\\a\end{matrix}}\right|$ $\mathrm{C_6H_4{-}NH_2}$	13,78
5	Methylene blue.	$(\mathrm{CH_3})_2\mathrm{N\!\cdot C_6H_3}\begin{matrix} \diagup\!\!\!\!\diagup & \mathrm{N}\\[-2pt] & \diagdown\\[-2pt] & \mathrm{S}\end{matrix}\mathrm{C_6H_3\!\cdot N(CH_3)_2}$ $\left.\vphantom{\begin{matrix}a\\a\end{matrix}}\right|$ $\mathrm{Cl}$	10,6	Rapidly dialyzes.

As is evident from the structure of these dyes, they are quaternary-substituted ammonias (methylene blue is a quaternary-substituted sulfonium). They all dissociate according to the type of ammonium salts and give a colored cation. They are all well adsorbed.

Table IV.

Dyes of the phthalein series¹.

No.	Name of dye	Structural formula	Adsorbed amount, in mgr.	Relation to dialysis
1	Sodium salt of fluorescein (uranine) . . . . .	$\mathrm{C_6H_3}\begin{matrix} \diagup \mathrm{ONa}\\[-2pt] \diagdown \mathrm{O} \end{matrix}$ $\mathrm{C{=}C_6H_3{=}O}$ $\mathrm{C_6H_4\!\cdot CO.ONa}$	0,0	Rapidly dialyzes.
2	Tetrabromo-fluorescein (eosin).	Structure analogous to uranine.	0,0	Rapidly dialyzes.
3	Tetraiodofluorescein (erythrosin).	Structure analogous to uranine.	0,0	Rapidly dialyzes.
4	Methyl ether of tetrabromofluorescein . . .	Structure analogous to uranine.	0,0	Rapidly dialyzes.
5	Tetraiododichlorofluorescein.	Structure analogous to uranine.	0,0	Rapidly dialyzes.
6	Dibromofluorescein . . . .	Structure analogous to uranine.	0,0	Rapidly dialyzes.
7	Diiodofluorescein	Structure analogous to uranine.	0,0	Rapidly dialyzes.
8	Rhodamine B . . .	$\mathrm{C_6H_3{-}N(C_2H_5)_2}$ $\mathrm{C{=}C_6H_3{-}N(C_2H_5)_2}$ $\mathrm{C_6H_4\!\cdot COOH}$ with $\mathrm{O{-}Cl}$	0,0	Rapidly dialyzes.

All the dyes listed in Table IV, except rhodamine, by their structure are sodium salts of complex carboxylic acids or phenols. All of them are completely not adsorbed by clay.

Table V.

No.	Name of dye	Structural formula	Adsorbed quantity, in mg	Relation to dialysis
1	Diamine blue . . . . . .	\(\begin{aligned} &\mathrm{C_6H_5{-}N{=}N{-}C_{10}H_4(OH)(NH_2)(SO_3Na)}\\[-2pt] &\mid\\[-2pt] &\mathrm{C_6H_5{-}N{=}N{-}C_{10}H_4(OH)(NH_2)(SO_3Na)} \end{aligned}\)	0.0
2	Tartrazine . . . .	\(\begin{aligned} &\mathrm{COOH}\\[-2pt] &\mid\\[-2pt] &\mathrm{C{=}N{-}NH{-}C_6H_4SO_3Na}\\[-2pt] &\mid\\[-2pt] &\mathrm{C{=}N{-}NH{-}C_6H_4SO_3Na}\\[-2pt] &\mid\\[-2pt] &\mathrm{COOH} \end{aligned}\)	0.0
3	Naphthol yellow.	\(\mathrm{C_{10}H_4\begin{cases}OH\\NO_2\\NO_2\\SO_3Na\end{cases}}\)	0.0

These dyes are salts of sulfonated complex organic molecules containing “amino” groups, etc. All of them are completely not adsorbed by clay.

Thus, a series of organic dyes, varied in structure, of different degrees of complexity, with different physical and chemical properties and different degrees of dispersion, has been tested for the ability to be adsorbed by clay. Some of them are adsorbed by clay to a greater or lesser degree, while others are not adsorbed at all. From the material presented above, a connection is undoubtedly revealed between the ability of a dye to be adsorbed by clay and its acidic or basic character; namely, only basic dyes are adsorbed by clay. This circumstance has also been noted by other authors. It would seem to us most rational to formulate this connection as follows: those dyes are capable of being adsorbed by clay whose molecules, upon dissociation, give a complex colored cation and an anion of a mineral or organic acid. Such are dyes of the type of substituted ammonium salts. Those dyes are not adsorbed whose molecules, upon dissociation, give a colored anion and, as the cation, hydrogen or a metal.

The structure of the organic radicals, the presence in them of various atomic groupings, including “amino” groups, substituted or not substituted on hydrocarbon radicals, or other basic groups, provided only that they do not take part in the formation of a salt of the ammonium-salt type, is of absolutely no decisive significance. Thus, for example, patent blue and crystal violet belong to one and the same chromogen group of the triphenyl-methane series. Both have in their molecule “amino” groups substituted by hydrocarbon radicals. But in the molecule of crystal violet one of these “amino” groups participates in the formation of a salt of the substituted-ammonium type and therefore dissociates according to the scheme:

\[ \left[ \begin{array}{c} \mathrm{C_6H_4{=}N(CH_3)_2}\\ \mathrm{C{-}C_6H_4{-}N(CH_3)_2}\\ \mathrm{C_6H_4{-}N(CH_3)_2} \end{array} \right]\mathrm{Cl}' \]

i.e., the entire content of the dye molecule gives a cation. It is adsorbed by clay. In the molecule of patent blue, however, the “amino” group does not accept

participation in the formation of a salt. The salt here is formed at the expense of the sulfone groups, and the molecule dissociates according to the scheme:

\[ \left[ \begin{array}{c} \mathrm{C_6H_4} \begin{array}{c} \mathrm{SO_2{-}O}\\ \mathrm{SO_2{-}O}\\ \mathrm{OH} \end{array} \\[-2mm] \mathrm{(HO){-}C{-}C_6H_4{:}N{-}(CH_3)_2}\\ \mathrm{\phantom{(HO){-}C{-}}C_6H_4{:}N{-}(CH_3)_2} \end{array} \right]^{\prime\prime} \; \Big| \; \mathrm{Ca} \]

i.e., the content of the dye molecule gives an anion. Patent blue is not at all adsorbed by clay.

Another representative of the triphenylmethane series, Guinea green, which is constructed analogously to patent blue (see Table I), is adsorbed very weakly. Possibly this slight adsorption is due to impurities.

Rhodamine B is assigned two structural formulae¹:

\[ \begin{array}{ll} 1) & \begin{array}{c} \mathrm{C} \begin{array}{l} \diagup \ \mathrm{C_6H_3{-}N(C_2H_5)_2}\\ \mathrm{\ \ \ \ \ \ \ \ \ \ O}\\ \diagdown \ \mathrm{C_6H_3{=}N(C_2H_5)_2}\\ \mathrm{\ \ \ \ \ \ \ \ \ \ \ \ \ |}\\ \mathrm{\ \ \ \ \ \ \ \ \ \ \ \ \ Cl}\\ \diagdown \ \mathrm{C_6H_4.CO.OH} \end{array} \end{array} \end{array} \qquad \text{and} \qquad 2) \quad \begin{array}{c} \mathrm{C} \begin{array}{l} \diagup \ \mathrm{C_6H_3{-}N(C_2H_5)_2}\\ \mathrm{\ \ \ \ \ O{-}Cl}\\ \diagdown \ \mathrm{C_6H_3{-}N(C_2H_5)_2}\\ \diagdown \ \mathrm{C_6H_4.CO.OH} \end{array} \end{array} \end{array} \]

According to the first, rhodamine is a quaternary ammonium and OH, and, consequently, ought to be adsorbed by clay. According to the second, in aqueous solution it should rather dissociate as an acid than as a substituted-resorcin salt.

It is not adsorbed by clay and, consequently, the choice must be made in favor of the second.

Of the dyes of the second type tested by us, i.e., those which, on dissociating, give a colored anion—i.e., dissociate as organic acids or their salts—not one is adsorbed by clay, even to a slight degree.

It has also been stated in the literature that only colloidal dyes are adsorbed by clay². Apparently such an assumption is incorrect. Thus fuchsin, methylene blue, and others, which pass through a parchment membrane very rapidly, are at the same time well adsorbed. The same applies to many other dyes, such as chrysoidin, Bismarck brown, methylene blue, and many others.

Thus, the sole factor in deciding the question of whether a given dye will be adsorbed or not is its electrolytic dissociation of type I or type II. Only dyes giving a colored cation are adsorbed. Since this is so, the following assumption suggests itself: each suspended particle of clay, being negatively charged, attracts to itself the colored cations of the dye—as though becoming enveloped by them. The forces of adhesion are thus electrical forces. Such an interpretation of the phenomenon of adsorption is in agreement with the circumstance that adsorption increases with increasing dilution of the adsorbate.

But if this were so, then the solution would become enriched in anions and would lose electroneutrality:

\[ \underset{\text{dye}}{\mathrm{Clay} + n\,\mathrm{Kat}\,|\,\mathrm{Cl}^{\prime}} = \underset{\text{adsorption product}}{\mathrm{Clay}\!\cdot n\mathrm{Kat}.} + n\mathrm{Cl}^{\prime} \]

However, in all the literature on this question it has not been possible to find any indications that anyone has had occasion to establish a disturbance of electroneutrality during adsorption.

And we ourselves likewise were unable to detect this. Thus this supposition falls away.

The second supposition: a reaction of exchange decomposition takes place according to the scheme:

\[ \underset{\text{clay}}{\mathrm{An|H}} + \underset{\text{dye}}{\mathrm{Kat.|Cl}} = \mathrm{H\cdot Cl} + \underset{\text{adsorption product}}{\mathrm{Kat.An}}, \]

i.e. the adsorption product is the salt of silicic acid and the cation of the dye, which is insoluble in water. The reaction occurs only in the surface layer, but may probably also proceed into the interior if the experiment is continued for a long time. This point of view can be tested experimentally; for this, first of all, it is necessary to determine whether hydrochloric acid is formed during adsorption, and in what amount.

To resolve this question the following experiments were carried out. First of all, the clay used by us for adsorption was most thoroughly freed from the small amount of CaCO₃ contained in it. Otherwise the HCl formed in the process would be neutralized, and we would not be able to detect its presence.

It is possible that many authors for this reason were unable to establish an acid reaction after adsorption.

The clay was freed from CaCO₃ in the following manner: 100 gr. of clay, sifted through a sieve of 4900 openings per 1 sq. inch, was treated with 200 cc. of a 5% solution of HNO₃ for 1 day at 60°—70°C. After this the acid solution was decanted, and the clay was very thoroughly washed until the acid reaction in the wash waters had completely disappeared.

Thus, up to 20 liters of distilled water were used for washing 100 gr. of clay. The clay washed in this way was dried at 100°C and then used for experiments. With the clay prepared in this manner the following quantitative determinations were made.

Experiment with “Victoria” blue, \(\mathrm{C_{33}H_{32}N_3Cl}\).

A weighed portion of 0.2500 gr. of dye was dissolved in 1000 cc. of water. 45 gr. of clay were added, i.e. approximately three times the amount required for complete decolorization of the solution. After adsorption the liquid had an acid reaction and contained the chlorine ion (\(\mathrm{Cl'}\)). 600 cc. of the filtered transparent and colorless liquid were titrated with an \(\frac{N}{18.24}\) solution of \(\mathrm{NaOH}\).

For the titration 4.5 cc. of solution were required. Hence we calculate that for 1000 cc. of liquid there would have been required

\[ \frac{4.5}{600.0}\cdot 1000.0 = 7.5 \]

cc. of \(\mathrm{NaOH}\) solution.

The theory for the reaction:

\[ (\mathrm{C_{33}H_{32}N_3})|\mathrm{Cl} + \mathrm{An|H} = \mathrm{C_{33}H_{32}N_3An} + \mathrm{H'Cl'} \]

requires for the same weighed portion of dye 9.036 cc. of \(\mathrm{NaOH}\) solution of the above concentration. It is thus found that

\[ 100 \times \frac{7.5}{9.036} = 83.1\% \]

of the entire amount of dye taken for the experiment entered into the reaction.

The neutral solution obtained after titration was evaporated to a small volume and titrated with an \(\frac{N}{20}\) solution of \(\mathrm{AgNO_3}\). For the titration 5 cc. were required. Hence we calculate for 1000 cc. of solution:

\[ \frac{5.0}{600.0}\cdot 1000.0 = 8.3 \text{ cc.} \]

The theory for a weighed portion of 0.2500 g of Victoria blue, according to the equation given above, requires 9.89 cc. Thus, in the solution, chlorine ion was found:

\[ 100 \cdot \frac{8.3}{9.89} = 83.92\%. \]

These fairly concordant results show that not all the dye is absorbed by the adsorbent with the formation of HCl.

Experiment with methylene blue \(C_{16}H_{18}N_3SCl\)

A weighed portion of the dye, 0.3000 g, was taken and dissolved in 500 cc of water. Washed clay was added until the solution was completely decolorized. The colorless, transparent liquid was filtered off and titrated with an \(\frac{N}{18.24}\) solution of NaOH. 11 cc of NaOH solution were used.

Theory for the reaction:

\[ (C_{16}H_{18}N_3S)Cl' + An'H = C_{16}H_{18}N_3SAn + H \cdot Cl' \]

requires 17.08 cc of NaOH solution.

Thus, hydrogen ion was found:

\[ 100 \times \frac{11.0}{17.08} = 64.4\% \]

as against theory.

The liquid, neutralized with a solution of caustic soda, was titrated with an \(\frac{N}{20}\) solution of AgNO\(_3\). 12.0 cc of solution were used. The theory, according to the above equation, requires 18.75 cc of \(\frac{N}{20}\) AgNO\(_3\) solution. Thus, Cl′ ion was found:

\[ 100 \cdot \frac{12.0}{18.75} = 64\% \]

of the theoretical amount.

Experiment with phenosafranine \(C_{18}H_{15}N_4Cl\)

A weighed portion of the dye, 0.3000 g, was taken, dissolved in 500 cc of water, and decolorized with clay. For the titration of all the acid, 10 cc of an \(\frac{N}{18.24}\) NaOH solution were used. Theory for the reaction:

\[ (C_{18}H_{15}N_4)Cl' + An'H = (C_{18}H_{15}N_4)An + H \cdot Cl' \]

requires 17 cc. Thus hydrogen ions were found:

\[ \frac{10}{17} \cdot 100 = 58.7\% \]

as against theory. The neutralized solution was then titrated with an \(\frac{N}{20}\) AgNO\(_3\) solution. For the titration, 11.2 cc were used, as against 18.6 cc required by theory. Thus Cl′ ions were found:

\[ \frac{11.2}{18.6} \cdot 100 = 60.7\%. \]

Experiment with methyl violet \(C_{24}H_{28}N_3Cl\)

A weighed portion of the dye, 0.4000 g, was taken, dissolved in 250 cc of water, and decolorized with an excess of clay. For the neutralization of the acid formed, 7.8 cc of an \(\frac{N}{18.24}\) NaOH solution were used.

Theory for the reaction:

\[ (C_{24}H_{28}N_3)Cl' + H \cdot An' = (C_{24}H_{28}N_3)An + H \cdot Cl' \]

requires 18.64 cc.

Thus, hydrogen ions found:

\[ 100 \frac{7.8}{18.64} = 41.8\%. \]

The neutralized solution was titrated with \(\frac{N}{20}\) solution of \(\mathrm{AgNO_3}\), with 10.8 c.c. being used. Theory requires 20.3 c.c. Thus, chlorine ions found: 53.2%.

In the last two experiments there is no agreement between the amounts of \(\mathrm{H}^{\bullet}\) and \(\mathrm{Cl}'\) ions, and in both cases there is more of the latter. This is explained by the probable contamination of the dye with chloride salts of alkali metals, used, as is known, for salting out the dye.

Experiments with ethyl violet. \(\mathrm{C_{31}H_{42}N_3Cl}\).

I. A sample of dye of 0.4000 g was dissolved in 250 c.c. of water and decolorized with an excess of clay. For neutralization of the acid formed, 8.9 c.c. of \(\frac{N}{17.81}\ \mathrm{NaOH}\) were required.

Theory for the reaction:

\[ (\mathrm{C_{31}N_4N_3}) \cdot |\mathrm{Cl}'| + \mathrm{H}^{\bullet}|\mathrm{An}'| = \mathrm{C_{31}H_{42}N_3An} + \mathrm{H}^{\bullet}|\mathrm{Cl}'| \]

requires 14.1 c.c. Thus, hydrogen ions found:

\[ 100 \frac{8.9}{14.1} = 63.1\%. \]

The neutralized liquid was titrated with \(\frac{N}{25.54}\) solution of \(\mathrm{AgNO_3}\). For the titration, 13.5 c.c. were used, against 20.3 c.c. required by theory. This amounts to

\[ \frac{13.5}{20.3} 100 = 66.5\%. \]

II. A sample of 0.4350 g was taken. After dissolution in 250 c.c. of water, it was decolorized with clay. For neutralization of the acid formed, 9.8 c.c. of \(\frac{N}{17.31}\ \mathrm{NaOH}\) solution were required, against 15.4% theoretical, i.e., hydrogen ions found:

\[ 100 \cdot \frac{9.8}{15.4} = 63.7\%. \]

For titration of the neutralized liquid, 14.1 c.c. of \(\frac{N}{25.54}\ \mathrm{AgNO_3}\) solution were used, against 22.1 c.c. theoretical, i.e.,

\[ \frac{14.1}{22.1} \cdot 100 = 65.0\% \]

of chlorine ions. The reason for the small discrepancy in the figures for hydrogen and chlorine ions is here the same as in the preceding experiments, i.e., contamination of the dye with chlorides.

The experiments cited undoubtedly establish the presence of a reaction leading to the formation of HCl. The circumstance which stands out is that the amount of hydrochloric acid liberated in all cases except one is very close to half the amount required by the above reaction equations. This circumstance could be interpreted to mean that the exchange decomposition reaction proceeds not according to the equation, for example, for ethyl violet,

\[ (\mathrm{C_{31}H_{42}N_3}) \cdot |\mathrm{Cl}'| + \mathrm{H}^{\bullet}|\mathrm{An}'| = \mathrm{C_{32}H_{42}N_3Cl \cdot An} + \mathrm{H}^{\bullet}|\mathrm{Cl}'|, \]

but according to the equation:

\[ 2(\mathrm{C_{31}H_{42}N_3}) \cdot |\mathrm{Cl}'| + \mathrm{H}^{\bullet}|\mathrm{An}'| = \mathrm{C_{31}H_{42}N_3An \cdot C_{31}H_{42}N_3Cl} + \mathrm{H}^{\bullet}|\mathrm{Cl}'| \]

and the deviations from this equation obtained in the experiments should be ascribed to the insufficient purity of the dye with which the experiments were conducted.

But it is also possible, as will be shown below, that the following explanation is more probable:

At the first moment there proceeds a reaction of exchange decomposition, of course on the surface of the clay particles. As a result of this reaction a new surface is formed from the silicic-acid salt of the dye. This surface adsorbs neutral dye molecules according to the type of mechanical adsorption. In this process, at the first moment only a monomolecular layer is formed, i.e. each molecule of the silicic dye adsorbs only one molecule of dye. But if there is an excess of dye in the aqueous phase, then a second layer is formed on the surface of the particles, and so on. We should observe adsorption as a function of the content of dye in the aqueous phase. The following two experiments, carried out quantitatively, make one think that the reaction of exchange decomposition between a particle of acid silicate and the dye apparently proceeds monomolecularly.

I. 0.2000 g of methyl violet was taken in 200 c.c. of water. \(\frac{1}{10}\) N solution of \(\mathrm{Na_2SiO_3}\) was added in excess. All the dye passed into the precipitate. In the solution there was found: by titration, \(8.34\%\) chlorine of the weighed portion of methyl violet taken, as against \(8.75\%\) theoretical content.

II. 0.4000 g of methylene blue was taken in 200 c.c. of water. An excess of \(\frac{N}{10}\) solution of \(\mathrm{Na_2SiO_3}\) (40 c.c.) was added. All the dye passed into the precipitate; the filtrate showed a chlorine content of \(10.9\%\) of the weighed portion taken, as against \(11.11\%\) Cl of the theoretical content.

Thus, in both experiments all the chlorine of the dye, within the limits of experimental error, was found in the solution in the ionic state, i.e. the reaction would seem to have proceeded according to the equation:

\[ 2(C_{16}H_{18}N_3S)\cdot|Cl| + Na_2\cdot SiO_3 = (C_{16}H_{18}N_3S)_2SiO_3 + 2Na\cdot|Cl| \]

Methylene blue                                                     Silicic-acid salt                                                                      of methylene blue.

If the developed picture of adsorption is correct, then in the so-called phenomenon of “adsorption” by clay one must distinguish two essentially different processes: the first—a chemical reaction of exchange decomposition, the second—adsorption proper. The second process is adsorption, but no longer by the clay, rather by the surface layer of the silicic-acid dye.

This view is confirmed by the following experiments, carried out with clay that had adsorbed methylene blue, phenosafranine, and Victoria blue:

Clay that had adsorbed the dye, after washing with water, was dried at \(100^\circ\) C. Then, already in the dry state, it was extracted with 96% ethyl alcohol. Part of the dye passed into solution. Such treatment was continued until the alcohol began to be colored only to a very slight degree. This indicated that the clay had ceased to give up mechanically adsorbed dye, and only the silicic-acid compound of the dye remained, formed by the reaction of exchange decomposition. The alcoholic solution thus obtained was evaporated, and the residue was subjected to microscopic examination; under the microscope the residue fully resembled the appearance of the original material in the form of crystals, in color, etc. Then the residue obtained in this way was fused with \(\mathrm{NaON + KNO_3}\) in a silver cup, and the melt obtained was dissolved in water, after which the liquid was transferred into a glass beaker, where it was strongly acidified with nitric acid.

After addition of a solution of \(\mathrm{AgNO_3}\), a large precipitate of \(\mathrm{AgCl}\) appeared. Although quantitative determinations in this direction were not made by us, nevertheless from these qualitative tests it is clear that a hydrochloride salt of the dye is partly adsorbed, which can be removed with alcohol. Another part of the adsorbed dye is not removed by alcohol.

If the picture given above of the adsorption of basic dyes by clay is correct, then, in checking the adsorption law, one must take into account not the whole dye, but only the part adsorbed mechanically.

As is already known from the numerous works of Freundlich and his students, mechanical adsorption is well expressed by the following exponential equation:

\[ \frac{x}{m}=\beta \cdot c^{\frac{1}{n}}, \]

where

\(x\) is the amount of adsorbed substance, in thousandths of a gram-molecule.

\(m\) is the amount of adsorbent, in grams.

\(c\) is the concentration of the dye in the aqueous phase, established after adsorption and expressed in gram-molecules per liter.

\(\beta\) and \(\frac{1}{n}\) are constants.

In applying this equation to the adsorption of basic dyes by clay, the magnitude \(x\) must be taken equal to the difference between the total amount of dye absorbed and that amount which, according to the scheme given above, enters into the reaction of exchange decomposition.

As has already been indicated above, the amount of substance entering into the chemical reaction is easily determined from the amount of HCl formed. Taking this circumstance into account, we followed the adsorption law by means of the experiments given below:

Experiment with methylene blue.

By the method described above, measurements were made of the amounts of dye adsorbed by clay at various concentrations of the dye in the liquid phase. The following results were obtained:

Concentration of dye in the liquid phase	Amount of dye adsorbed by 1 g of clay
6.0 mg per liter	10.6 mg
24.2 “ “	12.58 “
114.0 “ “	13.60 “
250.0 “ “	15.00 “

In the preceding experiments it was found that methylene blue is absorbed by one gram of clay in the amount of 10.6 mg when the solution contains 0.6 mg of dye in 100 cu. cm. According to our experiments, 64% of the total absorbed amount of dye enters into the reaction of exchange decomposition. Consequently, in the present case this amount is equal to:

\[ 10.6 \cdot 0.64 = 6.76 \text{ mg}. \]

We assume that this amount will be constant and will not depend on the increasing concentration of dye in the liquid phase. Therefore, in order to determine the amount of dye absorbed mechanically, we must each time subtract from the total amount of dye adsorbed by one gram of clay 6.76 mg. Performing this subtraction here, we obtain the following values for the amounts of dye adsorbed mechanically:

Concentration of dye in the liquid phase, mg/liter	Amount of dye adsorbed by one gram of clay, in mg
6.0	3.84
24.2	5.82
114.0	6.84
250.0	8.24

Recalculating these data as applied to the adsorption equation, for the constants \(\beta\) and \(\frac{1}{n}\), we obtain the following values:

\[ \beta = 0{,}0484;\quad \frac{1}{n}=0{,}101. \]

A comparison of the adsorption magnitudes calculated with the aid of these constants and those found experimentally is given in the following table:

\(C\)	\(\dfrac{x}{m}\), found experimentally	\(\dfrac{x}{m}\), calculated
0,0000186	0,01378	0,01611
0,0000755	0,01879	0,01814
0,0003380	0,02190	0,02180
0,0012780	0,02629	0,02370

Experiment with phenosafranine.

Determination of the adsorption of phenosafranine by clay at various concentrations of the dye in the liquid phase:

Concentration of phenosafranine in the liquid phase, in mg per liter	Amount of phenosafranine adsorbed by one gram of clay
10,0	14,0
80,0	17,0
200,0	20,0

The amount of phenosafranine that entered into the reaction of exchange decomposition, as determined above, was \(66\%\) of 10,6 mg, i.e. equal to 7,01 mg.

Subtracting this amount from the total adsorbed amount, we obtain the following figures for the mechanically adsorbed phenosafranine:

Concentration of phenosafranine in the liquid phase, in mg per liter	Amount of phenosafranine adsorbed mechanically by one gram of clay
10,0	6,99
80,0	9,99
200,0	12,99

Recalculating these data as applied to the adsorption equation, for the constants we obtain the following values:

\[ \beta = 0{,}0172;\quad \frac{1}{n}=0{,}161. \]

A comparison of the adsorption magnitudes calculated with the aid of these constants and found experimentally is given in the following table:

\(C\)	\(\dfrac{x}{m}\), found experimentally	\(\dfrac{x}{m}\), calculated
0,0000310	0,02178	0,02312
0,0002480	0,03060	0,02964
0,0006250	0,04021	0,03827

Thus, both experiments indicate that the amounts of dye calculated by us as “properly adsorbed,” when the concentration in the liquid phase is varied within very wide limits (in one case 20-fold, and in the other 40-fold), apparently obey Freundlich’s law.

In conclusion we shall cite several observations on the displacement of an adsorbed substance from adsorption products.

If one takes the product of adsorption of safranine by clay, washes it well with water, then again stirs it with water and adds to the mixture a certain quan-

quantity of methylene blue, then immediately part of the adsorbed safranine will pass into solution and color it bright red. In exactly the same way methylene blue is adsorbed. Subsequent addition of methylene blue entails a further displacement of safranine.

But such displacement does not proceed quantitatively, and stops at a certain limit, after which, in the solution, besides the safranine displaced from the adsorption product, there still remains part of the methylene blue. Apparently an equilibrium is established between the adsorbed safranine and the methylene blue in solution. Quantitative determinations could not be made for the reason that the solution acquires intermediate colors between the two dyes; therefore the use of the colorimetric method becomes impossible.

In the same way one can observe the displacement of methylene blue by other dyes, such as, for example, fuchsin, phosphine, methyl violet, malachite green, and others among the dyes investigated by us. The greatest ability to displace dyes from their adsorption products with clay is possessed by methylene blue, then phosphine, safranine.

As the experiments show, displacement occurs only in part, of the adsorbed “mechanically.” Indeed, if from the adsorption product of clay with safranine one removes, as completely as possible, the “mechanically” adsorbed portion—first by washing with water and then, after preliminary drying, with 96° wine spirit—then, after removal of the spirit, methylene blue displaces nothing. It is adsorbed completely and the solution remains uncolored. With further addition of the solution it is colored a faint pink. This latter circumstance is probably explained by the incomplete removal of the mechanically adsorbed portion of safranine. The same was also done with respect to Victoria blue.

Thus, in observations of this kind, one must state that not all the adsorbed dye behaves uniformly toward different influences. One part of it is bound with the adsorbent differently than another.

The neutral salts of quinoline act in a special manner, for example quinoline hydrochloride or nitrate. Aqueous solutions of these salts displace all the dye from the products of its adsorption by clay, completely decolorizing the clay.

The scheme presented above for the complex phenomenon of the sorption of basic dyes by clay is thus confirmed from various sides. Nevertheless, for a more complete elucidation of the question, a series of experiments was undertaken with the aim of finding out whether the formation of hydrochloric acid in solution after adsorption by clay could be ascribed to hydrolysis of the hydrochloride salt of the dye. One of these experiments, apparently fully eliminating such an assumption, is given here.

Adsorption by clay in an acid solution.

A sample of 0.2000 g of methylene blue was dissolved in water containing a large excess of nitric acid. The solution was decolorized by clay. [[unclear: The filtrate]] was precipitated with silver nitrate. 0.0513 g of silver chloride was obtained. The chlorine contained here, 0.0204 g, amounts to 90.95% of all the chlorine contained in the 0.2000 g sample of methylene blue. A check experiment gave the same results. Thus, in an acid solution almost all the chlorine of methylene blue is liberated in the form of HCl. To ascribe this to hydrolysis is in any case impossible.

We do not consider the investigation of the questions touched upon here to be completed, and we propose to continue it in the near future.

1. Seyewetz et Sisley. Chimie des matières colorantes. 1898. Theory and Practice of Dyeing Substances, 1916. ↩↩↩

2. P. Rohland. Zeitschr. für Anorg. Chemie, vol. 77, pp. 116–118. ↩↩

3. Even such a material as coal is regarded by a whole series of authors (Glassner and Swida, Freundlich, and others) as contaminated with various little-investigated substances which strongly obscure the process of adsorption by coals. Glassner und Swida. Ch. Lib. Ann. 357, 95. ↩