Adding equations I and II we find
[Al3+] + [AlO33−] =
KBas.S.P. × [HO−]−3 + [HO−]3 × K′Ac.S.P. × KHOH−3
III
Aluminium hydroxide will be most completely precipitated when [Al3+] + [AlO33−] is a minimum, the values [Al3+] and [AlO33−] measuring the solubility of aluminium as aluminium-ion and as aluminate-ion. If we put [Al3+] + [AlO33−] = y and [HO−] = x, we can find the value x (the concentration of the hydroxide-ion) for which y is a minimum. We have y = KBas.S.P. × x−3 + x3 × K′Ac.S.P. × KHOH−3, and find, by means of the calculus,[393] that y is a minimum, when x = +(KHOH3 × KBas.S.P. / K′Ac.S.P.)1/6.
If aluminium hydroxide were as strong an acid as it is a base, i.e. if KBas.S.P. = K′Ac.S.P., we would have, simply, x = [HO−] = ((1.2E−14)3)1/6 = √(1.2E−14) (at 25°), which is the concentration of the hydroxide-ion in pure water at 25° (p. [176]). In other words, a perfectly neutral solution would then give us the conditions for as complete a precipitation as possible. But aluminium hydroxide is a stronger base than acid, KBas.S.P. > K′Ac.S.P., and consequently we find for x = [HO−] = (KHOH3 × KBas.S.P. / K′Ac.S.P.)1/6, a value somewhat greater than the concentration of the hydroxide-ion in pure water, i.e. we must use a slightly alkaline medium—which agrees with common practice. In other words, there is less danger of losing aluminium hydroxide in the form of aluminate, owing to the weaker acid character of the hydroxide, than there is of losing it in the form of aluminium-ion. The most favorable degree of alkalinity for the precipitation would depend on the relation of KBas.S.P.. and K′Ac.S.P..
The exact values for KBas.S.P. and K′Ac.S.P., the two solubility-product constants, and for the corresponding ionization constants, which would show the same ratio, are still not known. But, if, for the sake of an illustration, we take recourse to assumed values for these constants, we find that the solubility of aluminium, as aluminium-ion and as aluminate-ion, is, by calculation, as anticipated, a minimum for a solution, which contains the concentration of HO− calculated (for x) in the manner indicated above. And the further interesting conclusion is reached that this minimum loss of aluminium [p198] hydroxide would occur when [Al3+] = [AlO33−]—which would correspond to a saturated solution of aluminium aluminate, Al(AlO3).
When the ionization of aluminium hydroxide, as an acid, is considered to take place according to Al(OH)3 ⇄ AlO2− + H+ + H2O, which agrees best with its real behavior (p. [172]), we can find, similarly, that [Al3+] + [AlO2−] is a minimum, when aluminium hydroxide is precipitated in such a way, that an excess x of the hydroxide-ion is used, and x = [HO−] = (3 KHOH × KBas.S.P. / KAc.S.P.)0.25,—where KAc.S.P. represents the solubility-product constant for [AlO2−] × [H+]. That a minimum loss of aluminium hydroxide would be suffered when the favorable excess of the hydroxide-ion (x) is calculated on the basis of the equation as given, may readily be seen by again assuming definite values for KBas.S.P. and KAc.S.P.. It also appears that this minimum loss[394] of aluminium includes one-third as many Al3+ ions, as AlO2− ions—a relation corresponding, again, to a saturated solution of aluminium aluminate, Al(AlO2)3.
Chapter X Footnotes
[354] When all three of the hydrogen atoms in the hydroxide are ionized, an aluminate ion, AlO33− is formed: Al(OH)3 ⇄ AlO33− + 3 H+. But, as in the case of other weak polybasic acids, a single hydrogen atom is far more readily ionized than are the remaining two (p. [102]), and the ion Al(OH)2O−, which is formed by the primary ionization, readily loses water and forms the anhydride ion AlO2−. The most important aluminates are derivatives of this ion.
[355] See the table at the back of Smith's Inorganic Chemistry, or p. 149 of Remsen's Inorganic Chemistry.