[AlO+] × [HO−] / [Al(OH)3] = KBase.
The formation of traces of nonionized (basic) aluminium aluminate would satisfy the equilibrium requirements for AlO+ + AlO2− ⇄ AlO(AlO2), since the aluminate, like other aluminates, is presumably readily ionizable in aqueous solutions. Aluminium hydroxide, as a base and as an acid, would yield in the first moment greater concentrations of the hydroxide and hydrogen ions than would satisfy the equilibrium constant for water (p. [176]); the excess of these ions must combine to form water, until the product of their concentrations is equal to the ionization constant of water. The neutralization of these first quantities of hydrogen and hydroxide ions would destroy the momentary condition of equilibrium between aluminium hydroxide and its ions and would lead to its further ionization, both as a base and as an acid, and to the solution of some aluminium hydroxide (see the above equilibrium equations). However, since AlO+ and AlO2− remain practically uncombined and therefore accumulate in the solution, the concentrations of the hydroxide and hydrogen ions formed grow smaller and smaller; for an increasing excess of the ion AlO+ will allow only smaller and smaller values for [HO−], according to the equilibrium equation for KBase, and, similarly, an increasing excess of the ion AlO2− will permit [H+] to reach only smaller and smaller values, according to the equilibrium equation for KAcid. When the values for [HO−] and [H+] have in this way become small enough to make [HO−] × [H+] = KHOH, equilibrium is reached. It is evident that in such a solution, in the condition of equilibrium, [HO−] is not equal to [AlO+], as it would ordinarily be, according to the ionization equation Al(OH)3 ⇄ AlO+ + HO− + H2O, but is much smaller. Similarly, [H+] is much smaller than [AlO2−].
Just how much aluminium aluminate must be formed by a self-neutralization of the amphoteric hydroxide will depend on the values for KBase and KAcid and on the solubility of aluminium hydroxide (nonionized Al(OH)3). The two equilibrium equations may be combined:
| [AlO+] × [AlO2−] × [H+] × [HO−] | = KBase × KAcid. |
| [Al(OH)3]2 |
Since [H+] × [HO−] = KHOH, and since [AlO+] and [AlO2−] may be taken to represent each the concentration of the practically completely ionized aluminium aluminate AlO(AlO2), we have[379]
| [Alum. Aluminate]2 | = | KBase × KAcid | , |
| [Alum. Hydroxide]2 | KHOH |
or
| [Alum. Aluminate] | = √ ( | KBase × KAcid | ). |
| [Alum. Hydroxide] | KHOH |
It is clear, that the smaller the ionization constants KBase and KAcid are, and the smaller the solubility of nonionized aluminium hydroxide [Alum. Hydroxide] is, the smaller must be the concentration of the aluminate formed to satisfy the conditions for equilibrium.