An instance where a very weak insoluble acid will not combine, appreciably, with a very weak insoluble base, is found in the case of ‹aluminium hydroxide›. A development of the equilibrium equations for its ionization as a base and its ionization as an acid would show, that all the constants would be readily satisfied, when a very minute quantity of dissolved ionized aluminium aluminate is formed. [p187] «Self-Neutralization of Amphoteric Hydroxides.»—We may consider a saturated solution of aluminium hydroxide, in contact with the solid hydroxide. For the ‹acid ionization›,[377] Al(OH)_{3} ⇄ AlO_{2}^{−} + H^{+} + H_{2}O, we have [AlO_{2}^{−}] × [H^{+}] / [Al(OH)_{3}] = K_{Acid}. Similarly, for the ‹basic ionization›,[378] Al(OH)_{3} ⇄ (AlO)^{+} + HO^{−} + H_{2}O, we have [AlO^{+}] × [HO^{−}] / [Al(OH)_{3}] = K_{Base}. The formation of ‹traces of nonionized› (basic) aluminium aluminate would satisfy the equilibrium requirements for AlO^{+} + AlO_{2}^{−} ⇄ AlO(AlO_{2}), 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 AlO_{2}^{−} 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 K_{Base}, and, similarly, an increasing excess of the ion AlO_{2}^{−} will permit [H^{+}] to reach only smaller and smaller values, according to the equilibrium equation for K_{Acid}. When the values for [HO^{−}] and [H^{+}] have in this way become small enough to make [HO^{−}] × [H^{+}] = K_{HOH}, 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^{−} + H_{2}O, but is much ‹smaller›. Similarly, [H^{+}] is much smaller than [AlO_{2}^{−}].