Consider the dissolution of quartz (a very slow reaction) by the action of a strong base solution:

SiO₂(s) + 4OH^-(aq) ----> SiO₄^4-(aq) + H₂O(l)

1. What is the acid, and what is the base, in the reactants of this equation? SiO₂ is the acid; OH^- is the base.

2. Describe how the reaction can be understood as an acid-base reaction using at least three different acid-base definitions (Arrhenius, Brønsted-Lowry, Lewis, Lux-Flood, and Solvent System).

The Arrhenius definition of an acid is that it increases the concentration of H⁺ in aqueous solution. In this reaction, the concentration of OH^- is being decreased; if we remember that [H⁺][OH^-] = K_w, this means that the H⁺ concentration is being increased. In this way we can see SiO₂ as an Arrhenius acid. But H⁺ is never the dominant ion in the solutions represented by this equation, so Dr. Arrhenius may have had difficulty recognizing SiO₂ as an acid.

Because our solvent system in this example is water, the Solvent System definition in this case reduces to the Arrhenius definition; SiO₂ is an acid because it decreases the OH^- concentration and thereby increases the H⁺ (or H₃O⁺) concentration. Remember that the solvent system definition, it is the concentrations of the characteristic anions and cations that matter; in water, it is the H⁺ concentration and OH^- concentration that matter.

Neither the Brønsted-Lowry or Lewis definitions can be directly and easily applied to this reaction. SiO₂ has no protons to donate, and the SiO₄^4- product is not the direct adduct of Lewis acid and Lewis base expected for typical Lewis acid/Lewis base adduct formation. However, if a hypothetical intermediate of H₂SiO₄^2- is postulated, both the Lewis and Brønsted-Lowry roles become more clear:

SiO₂(s) + 4OH^-(aq) ----> H₂SiO₄^2- + 2OH^- ----> SiO₄^4-(aq) + H₂O(l)

The H₂SiO₄^4- is a simple Lewis acid/Lewis base adduct; the OH^- ions donate a lone pair of electrons to the electrophilic Si. This hypothetical adduct, then, has protons to donate in a classic Brønsted-Lowry sense to the remaining OH^-.

The Lux-Flood definition can be straightforwardly applied to the equation as written. SiO₂ gains two O^2- (oxide) ions to become SiO₄^4-, and therefore it is an unequivocal oxide acceptor. The OH^- is obviously, then, the oxide donor; the attachment of the leftover protons to the remaining hydroxides is not an intrinsic aspect of the Lux-Flood definition but is a fairly obvious step.

3. Compare how these different definitions aid or hinder our understanding of the reaction. In other words, different definitions may emphasize different aspects of the reaction. Some may be more complex or even give a confusing picture of what is happening. Decide which one is best for your understanding of the reaction.

The application of the Lux-Flood definition is very straightforward, in that it is clear that SiO₂ accepts oxides. No intermediates such as H₂SiO₄^2- need to be postulated to see how SiO₂ is acting as an acid. But oxide donation is not very universal, and not much mechanistic insight can be gained simply by labeling something as an "oxide acceptor."

The Lewis aspect of this reaction is a bit harder to see upon inspection of the original equation; it is best seen when imagining the initial adduct of OH^- and SiO₂. One advantage of viewing SiO₂ as a Lewis acid, however, is that it invites comparison to reactions with other Lewis bases, such as F^-.*

In this example, the Brønsted-Lowry definition doesn't immediately point to what is the acid and what is the base in this equation, because the acid in this case doesn't have any protons to donate. But if your "Arrhenius sense" of acids and bases is flagged by protonation of OH^- to make water, it is worth investigating just how proton transfer takes place. Postulating the H₂SiO₄^2- intermediate makes one think about the existence of equilibrium constants for the deprotonation of H₂SiO₄^2- to yield HSiO₄^3- and SiO₄^4-. Pauling's rules might even be brought to bear if one imagines the neutral acid H₄SiO₄. In fact, this reaction is likely to have different products depending on the pH of the final (aqueous) solution.

Arrhenius, Solvent System, and Usanovich definitions add little to our understanding of this reaction.

*HF will dissolve SiO₂ faster than OH^-. It is interesting to wonder why, as F^- is a weaker proton base than O^2-; the reasons are probably complicated and have to do with the fact that SiO₄^4- is not stable in aqueous solution and tends to polymerize.