Why do diacidic and triacidic alkalis dissociate in one step?

Question

My book states that dibasic and tribasic acids dissociate in water in two and three steps respectively but for diacidic and triacidic bases it states that they dissociate in one step. I do not know why.

For example:

$\ce{H2SO4}$ dissociates in water in two steps as follows: $$\ce{H2SO4(aq) <=> HSO4^{-} + H+}$$ $$\ce{HSO4^{-}(aq) <=> SO4^{2-} + H+}$$

but

$\ce{Cu(OH)2}$ dissociates in water in one step as follows:

$$\ce{Cu(OH)2(aq) <=> Cu^{2+} + 2OH-}$$

Why is this so. Why does not $\ce{Cu(OH)2}$ dissociate in two steps to form $\ce{CuOH+}$ in the first step and $\ce{Cu^{2+}}$ in the second step?

Answer 1

We do have separated dissociation constants for the heavier alkaline earth hydroxides whose solubility allows a high enough pH to access both equilibria.

Calcium hydroxide

Acidity ($pK_a$): 12.63 (first OH−), 11.57 (second OH−)[1][2]

(The $pK_b$ values are obtained by subtracting the above from $pK_w=14$.)

In contrast with hydroxy acids such as sulfuric or phosphoric acid, the bonding between the central atom and the hydroxyl groups is predominantly ionic, with little covalent overlap; so the induction effect we would expect to see with increasing charge buildup is diminished. This shows up as the two dissociation constants being only about one order of magnitude apart in calcium hydroxide, versus about five orders of magnitude for successive deprotonations of phosphoric acid. The closeness of the dissociation constants for hydroxide bases combined with the limited solubility in most cases makes separate dissociation constants usually hard to measure.

Cited References

1. "Sortierte Liste: pKb-Werte, nach Ordnungszahl sortiert. – Das Periodensystem online".
2. ChemBuddy dissociation constants pKa and pKb

Answer 2

Polyprotic acids

In most acids, even the strong acids, the bond to the acidic hydrogen atom is covalent (though frequently very polar). Thus, for acids, the more correct description of what happens in solution is ionization - a covalent bond is breaking heterolytically to form ions. The solvent is involved in solvating the proton, which is why we usually write $\ce{H3O+}$ for protons in water and not $\ce{H+}$.

Polyprotic acids ionize in separate steps because after the first ionization, the new species has a different strength covalent bond to break. Consider your example of sulfuric acid. Sulfuric acid is a strong acid that ionizes readily and (nearly) completely in water:

$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$

The resulting hydrogen sulfate anion is a different chemical species. Its bonds have different strengths. The second proton is not as easily ionized. In fact, the hydrogen sulfate anion is considered a weak acid. It does not completely ionize in water. An equilibrium is established with an equilibrium constant of $K_a = 1.02 \times 10^{-2}$. This means that at $\mathrm{pH = 2}$, the hydrogen sulfate anion is 50% ionized: half is hydrogen sulfate and half has ben converted to sulfate.

$$\ce{HSO4- + H2O <=> SO4^2- + H3O+}$$

What evidence do we have for this step-wise ionization? We have titrations! If we add small volumes of a solution of a stong base (say $\ce{NaOH}$) of known concentration to a solution of sulfuric acid and monitor the pH, we can see two distinct equivalence points (the inflection points on the graph where the pH change is very large over a small volume of base added) corresponding to the two ionization events. However, because the first ionization of sulfuric acid is complete before the base is added, a titration curve of sulfuric acid will only show one equivalence point when two equivalents of base are added (though there is a blip at 1 equivalent).

If we consider maleic acid $(\ce{H2C4H2O4})$, for which both ionizations are "weak" $(K_1=1.26\times 10^{-2},\ K_2=8.51\times 10^{-7})$, we get a different looking titration curve with inflections at 1 and 2 equivalents of base. If the ionization of both protons occurred simultaneously, we would not see a curve like this.

$$\ce{H2C4H2O4 +H2O <=> HC4H2O4- + H3O+}\\ \ce{HC4H2O4- + H2O <=> C4H2O4^2- + H3O+}$$

Bases

Bases that are the hydroxide salts of metals are ionic compounds. Though many of them have poor solubility in water, what is dissolved is completely ionized. Thus, for copper (II) hydroxide in water at equilibrium, there is only solid copper (II) hydroxide ad solubilized copper(II) and hydroxide ions:

$$\ce{Cu(OH)2 (S) <=> Cu^2+ (aq) + 2OH- (aq)}$$

A titration curve for copper (II) hydroxide with a strong acid like hydrochloric acid would show only one equivalence point at two equivalents of acid:

All ionic bases will behave similarly.

However, some bases are covalent and ionize by reacting with water:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

If there was to be a base of this type that had two ionizable groups, we would see two equivalence points.

Answer 3

They most definitely do. The problem with bases is that most of the common ones that are soluble are monobasic, polybasic ones are usually insoluble and thus, the intermediate states are hard to detect.

Various polyamines that are polybases most definitely react step-by-step. However inorganic polybases often form insoluble basic salts and usually are insoluble themselves and thus partially dissociated forms in solution exist only in miniscle amounts in fast equilibrium with precipitate. And no, basic salts such as various $\ce{Cu(OH)_xCl_y}$ can't surve as a proof, since stability of solids is ruled by very different set of rules.

Furthermore, when an inorganic polybase dissociates incompletely it usually forms various polynuclear particles with potentially very complex structures. This buries any attempts to create a simple and neat picture. This chemistry is extremely complex, studied only a little and completely unsuited for introductory courses. So most such courses keep such chemistry in mind but pretend it does not exist.