15.12: Polyprotic Acids

Acids are classified by the number of protons per molecule that they can give up in a reaction. Acids such as HCl, HNO₃, and HCN that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:

Even though it contains four hydrogen atoms, acetic acid, CH₃CO₂H, is also monoprotic because only the hydrogen atom from the carboxyl group (COOH) reacts with bases:

Similarly, monoprotic bases are bases that will accept a single proton.

Diprotic acids contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows:

This stepwise ionization process occurs for all polyprotic acids. Carbonic acid, H₂CO₃, is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts.

First ionization: H₂CO₃ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HCO₃⁻ (aq)

The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities.

Second ionization: HCO₃⁻ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + CO₃²⁻ (aq)

K_H2CO3 is larger than K_HCO3− by a factor of 10⁴, so H₂CO₃ is the dominant producer of hydronium ion in the solution. This means that little of the HCO₃⁻ formed by the ionization of H₂CO₃ ionizes to give hydronium ions (and carbonate ions), and the concentrations of H₃O⁺ and HCO₃⁻ are practically equal in a pure aqueous solution of H₂CO₃.

If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This approach is demonstrated in the following example exercise.

Ionization of a Diprotic Acid

“Carbonated water” contains a palatable amount of dissolved carbon dioxide. The solution is acidic because CO₂ reacts with water to form carbonic acid, H₂CO₃. What are [H₃O⁺], [HCO₃⁻], and [CO₃²⁻] in a saturated solution of CO₂ with an initial [H₂CO₃] = 0.033?

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As indicated by the ionization constants, H₂CO₃ is a much stronger acid than HCO₃⁻, so the stepwise ionization reactions may be treated separately.Using the provided information, an ICE table for this first ionization step is prepared:

Substituting the equilibrium concentrations into the equilibrium equation gives

Assuming x << 0.033 and solving the simplified equation yields

The ICE table defined x as equal to the bicarbonate ion molarity and the hydronium ion molarity:

Using the bicarbonate ion concentration computed above, the second ionization is subjected to a similar equilibrium calculation: HCO₃⁻ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + CO₃²⁻ (aq)

To summarize: at equilibrium [H₂CO₃] = 0.033 M; [H₃O⁺] = 1.2 × 10⁻⁴; [HCO₃⁻] = 1.2 × 10⁻⁴ M; and [CO₃²⁻] = 4.7 × 10⁻¹¹ M.

A triprotic acid is an acid that has three ionizable H atoms. Phosphoric acid is one example:

As for the diprotic acid examples, each successive ionization reaction is less extensive than the former, reflected in decreasing values for the stepwise acid ionization constants. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 10⁵ to 10⁶.

This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H₃PO₄ complicated. However, because the successive ionization constants differ by a factor of 10⁵ to 10⁶, large differences exist in the small changes in concentration accompanying the ionization reactions. This allows the use of math-simplifying assumptions and processes, as demonstrated in the examples above. Polyprotic bases are capable of accepting more than one hydrogen ion. The carbonate ion is an example of a diprotic base, because it can accept two protons, as shown below. Similar to the case for polyprotic acids, note the ionization constants decrease with ionization step. Likewise, equilibrium calculations involving polyprotic bases follow the same approaches as those for polyprotic acids.

This text is adapted from Openstax, Chemistry 2e, Section 14.5: Polyprotic Acids.

Monoprotic acids, like hydrofluoric acid, contain a single ionizable proton. In contrast, polyprotic acids contain two or more ionizable protons. For example, sulfurous acid has two ionizable protons, and phosphoric acid has three.

A polyprotic acid loses each of its protons sequentially, and each reaction has its own K_a.

It is easier to remove a proton from a neutral molecule than a negatively charged one because the negative charge increases the strength of the bond between the proton and the anion.

Therefore, for phosphoric acid, the K_a for the removal of the first proton, K_a1, is higher than the second one, K_a2, which is higher than K_a3.

The pH of a polyprotic acid can be estimated using only the first reaction if K_a1 is at least one thousand times larger than its subsequent K_a’s.

For example, the pH of a 0.050 molar ascorbic acid solution can be determined using its K_a1 and an ICE table.

When dissolved in water, ascorbic acid dissociates into hydronium and ascorbate ions. The K_a1 for this reaction is 8 × 10⁻⁵, and it is equal to the concentration of hydronium times the concentration of ascorbate monoanion, divided by the concentration of ascorbic acid.

An ICE table can be prepared for this reaction with the initial and equilibrium concentrations. Due to the small value of x, 0.050 minus x is approximately equal to 0.050.

Substituting these values into the expression for K_a1, the value for x is equal to 0.0020 molar, which is only 4% of the initial concentration of ascorbic acid. Thus, the approximation is valid.

The pH of the solution equals 2.70.

The concentration of ascorbate dianion formed in the second step of the reaction can also be calculated using K_a2 and an ICE table.

K_a2 is equal to 1.6 × 10⁻¹², and it can be expressed as the concentration of hydronium times the concentration of ascorbate dianion divided by the concentration of ascorbate monoanion.

For this reaction, the ICE table is filled in with the initial concentration of ascorbate monoanion and hydronium from the first reaction, 0.002 molar. Due to the small value of x, it can be omitted from the equilibrium concentrations of ascorbate monoanion and hydronium.

After substituting these values in the K_a2 expression and solving, x is 1.6 ×10⁻¹² molar.

Since x is less than 5% of 0.002 molar, the approximation is valid.

As the concentration of hydronium ions that were generated during the second step of ascorbic acid dissociation is negligible, the first ionizable proton determines the pH of the solution.