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D15.2 Reaction Between Amphiprotic Species

Acid-base reactions can occur between two amphiprotic species. For example, mixing a solution containing hydrogen sulfate ions (HSO4-) and a solution containing hydrogen carbonate ions (HCO3-) results in an acid-base reaction. However, if both reactants can act as either an acid or a base, which reactant is the acid and which is the base? In the example mixture, there are two possibilities:

possibility I:          HSO4-(aq) + HCO3-(aq) ⇌ SO42-(aq) + H2CO3(aq)
possibility II:          HSO4-(aq) + HCO3-(aq) ⇌ H2SO4(aq) + CO32-(aq)

Qualitatively, a product-favored acid-base reaction involves a stronger acid reacting with a stronger base to form a weaker acid and a weaker base. Acid and base strengths are reflected in Ka and Kb values.

In possibility I, the acids are HSO4- (Ka = 1.1 × 10-2) and H2CO3 (Ka = 4.3 × 10-7) and the bases are HCO3- (Kb = 2.3 × 10-8) and SO42− (Kb = 9.1 × 10-13). While all the species are weak acids/bases, the stronger acid and the stronger base are on the reactant side of the equation. Therefore, this reaction is product-favored at equilibrium.

In possibility II, the acids are H2SO4 (Ka = 4.0 × 103) and HCO3- (Ka = 4.7 × 10-11) and the bases are HSO4- (Kb = 2.5 × 10-18) and CO32− (Kb = 2.1 × 10-4). The stronger acid and the stronger base are on the product side of the equation, so this reaction is reactant-favored at equilibrium. This reaction also indicates the formation of a strong acid, H2SO4 (a relatively higher energy reactive species), from a weak acid reacting with a weak base. This situation is not thermodynamically favorable.

Exercise: Acid-Base Reactions

We can also make use of the ionization constants to quantitatively determine which reaction occurs. In possibility I,

HSO4-(aq) + H2O(ℓ) ⇌ SO42-(aq) + H3O+(aq)          [latex]K_1 = K_{\text{a,HSO}_4^-} = 1.1\;\times\;10^{-2}[/latex]
HCO3-(aq) + H3O+(aq) ⇌ H2CO3(aq) + H2O(ℓ)          [latex]K_2 = \dfrac{1}{K_{\text{a,H}_2\text{CO}_3}} = \dfrac{1}{4.3\;\times\;10^{-7}}[/latex]

The sum of these two reactions gives the overall reaction for possibility I:

HSO4-(aq) + H2O(ℓ) + HCO3-(aq) + H3O+(aq) ⇌ SO42-(aq) + H3O+(aq) + H2CO3(aq) + H2O(ℓ)
HSO4-(aq) + HCO3-(aq) ⇌ SO42-(aq) + H2CO3(aq)

Therefore, the equilibrium constant for possibility I is:

[latex]K_{\text{overall, possibility I}} = K_1 \times K_2 = \dfrac{1.1\;\times\;10^{-2}}{4.3\;\times\;10^{-7}} = 2.6\;\times\;10^{4}[/latex]

Possibility I is product-favored at equilibrium because the equilibrium constant is much greater than 1.

In possibility II,

HSO4-(aq) + H3O+(aq) ⇌ H2SO4(aq) + H2O(ℓ)          [latex]K_1 = \dfrac{1}{K_{\text{a,H}_2\text{SO}_4}} = \dfrac{1}{4.0\;\times\;10^{3}}[/latex]
HCO3-(aq) + H2O(ℓ) ⇌ CO32-(aq) + H3O+(aq)          [latex]K_2 = K_{\text{a,HCO}_3^-} = 4.7\;\times\;10^{-11}[/latex]

The sum of these two reaction gives the overall reaction for possibility II:

HSO4-(aq) + H3O+(aq) + HCO3-(aq) + H2O(ℓ) ⇌ H2SO4(aq) + H2O(ℓ) + CO32-(aq) + H3O+(aq)
HSO4-(aq) + HCO3-(aq) ⇌ H2SO4(aq) + CO32-(aq)

Therefore, the equilibrium constant for possibility II is:

[latex]K_{\text{overall, possibility II}} = K_1 \times K_2 = \dfrac{4.7\;\times\;10^{-11}}{4.0\;\times\;10^{3}} = 1.2\;\times\;10^{-14}[/latex]

Possibility II is heavily reactant-favored at equilibrium (this reaction can be considered as not occurring). Therefore, of the two possibilities, the reaction that proceeds forward and produces products is possibility I, where HSO4- acts as an acid and HCO3- acts as a base.

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