5.4 – Polyprotic Acids

We can classify acids 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:

HCl (aq) + H₂O (l) → H₃O⁺ (aq) + Cl^– (aq)

HNO₃ (aq) + H₂O (l) →H₃O⁺ (aq) + NO₃^– (aq)

HCN (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + CN^– (aq)

Note: the K_a of HCN is 6.2 x 10^-10. This means that the acid does not fully ionize in water and necessitates the equilibrium arrow.

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

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

Diprotic Acids

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:

First ionization: H₂SO₄ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HSO₄^– (aq)

For this reaction, K_a1 > 10², indicating ~100% complete ionization of the first proton.

Second ionization: HSO₄^– (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + SO₄^2- (aq)

For this reaction, K_a2 = 1.0×10⁻², indicating that the second ionization is only partial.

This stepwise ionization process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of 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)

Second ionization: HCO₃^– (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + CO₃^2- (aq)

Let’s confirm this very important concept mathematically. The ICE tables below display both ionization reactions, starting with a 10.0 M solution of H₂CO₃. Notice that with the first ionization reaction, we generate a small concentration of H₃O⁺ and HCO₃^– (even starting with 10.0 M of H₂CO₃!) – this is again due to the relatively small K_a1 value. Afterwards, the second ionization reaction which has an even smaller K_a2 value practically causes no change in the concentration of HCO₃^– and H₃O⁺ and generates a negligible amount of CO₃^2-.

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 can simplify our work considerably because we can determine the concentration of H₃O⁺ and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization.

Example 5.4.1 – Ionization of a Diprotic Acid

When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. 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 M?

H₂CO₃ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HCO₃^– (aq)   K_a1 = 4.5 x 10^-7

HCO₃^– (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + CO₃^2- (aq)    K_a2 = 4.7 x 10^-11

Solution

As indicated by the ionization constants, H₂CO₃ is a much stronger acid than HCO₃⁻, so H₂CO₃ is the dominant producer of hydronium ions in solution. Thus there are two parts in the solution of this problem: (1) Using the customary four steps, we determine the concentration of H₃O⁺ and HCO₃⁻ produced by ionization of H₂CO₃. (2) Then we determine the concentration of CO₃²⁻ in a solution with the concentration of H₃O⁺ and HCO₃⁻ determined in (1). To summarize:

Determine the concentrations of H₃O⁺ and HCO₃⁻.

H₂CO₃ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HCO₃^– (aq)          K_a1 = 4.5 x 10^-7

As for the ionization of any other weak acid:

An abbreviated table of changes and concentrations shows:

H₂CO₃ + H₂O⇌ H₃O⁺ + HCO₃^–

Substituting the equilibrium concentrations into the equilibrium gives us:

Solving the preceding equation making our standard assumptions gives:

x=1.2×10^-4

Thus:

[H₂CO₃]=0.033 M

[H₃O⁺]=[HCO₃^–]=1.2×10^-4 M

Determine the concentration of CO₃²⁻ in a solution at equilibrium with [H₃O⁺] and [HCO₃⁻] both equal to 1.2 × 10⁻⁴ M.

HCO₃^– (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + CO₃^2- (aq)

To summarize: In part 1 of this example, we found that the H₂CO₃ in a 0.033 M solution ionizes slightly and at equilibrium [H₂CO₃] = 0.033 M; [H₃O⁺] = 1.2 × 10⁻⁴ M; and [HCO₃⁻] = 1.2×10⁻⁴ M. In part 2, we determined that [CO₃²⁻] = 4.7×10⁻¹¹ M.

Check Your Learning 5.4.1 – Ionization of a Diprotic Acid

The concentration of H₂S in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate [H₃O⁺], [HS⁻], and [S²⁻] in the solution:

H₂S (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HS^– (aq)   K_a1 = 8.9 x 10^-8

HS^– (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + S^2- (aq)     K_a2 = 1.0 x 10^-19

Answer

[H₂S] = 0.1 M; [H₃O⁺] = [HS⁻] = 0.000094 M; [S²⁻] = 1 × 10⁻¹⁹ M ≈ 0 M

We note that the concentration of the sulfide ion is the same as K_a2. This is due to the fact that each subsequent ionization of a polyprotic acid occurs to a lesser degree.

Triprotic Acids

A triprotic acid is an acid that has three ionizable protons that undergo stepwise ionization: Phosphoric acid is a typical example:

First ionization: H₃PO₄ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + H₂PO₄^– (aq)   K_a1 = 7.5 x 10^-3

Second ionization: H₂PO₄^– (aq) + H₂O (l) ⇌  H₃O⁺ (aq) + HPO₄^2- (aq)   K_a2 = 6.2 x 10^-5

Third ionization: HPO₄^2– (aq) + H₂O (l) ⇌  H₃O⁺ (aq) + PO₄^3- (aq)   K_a3 = 4.2 x 10^-13

As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. 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 ionization 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, the calculations can be broken down into a series of parts similar to those for diprotic acids.

Polyprotic Bases

Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions:

First protonation:

H₂O (l) + CO₃^2- (aq) ⇌ HCO₃^– (aq) + OH^– (aq)    K_b1 = K_w/K_a2= 2.1 × 10^-4

Second protonation:

H₂O (l) + HCO₃^– (aq) ⇌ H₂CO₃ (aq) + OH^– (aq)   K_b2 = K_w/K_a1= 2.2 × 10^-8

Example 5.4.2 – Converting Between K_a and K_b – Polyprotic Acids-Bases

Given the following chemical equilibrium reaction, which of the following expressions correctly describes its equilibrium constant, K:

PO₄^3- + H₂O ⇌ OH^– + HPO₄^–

(a) 1/K_a2

(b) K_w/K_a3

(c) 1/K_a3

(d) K_w/K_a2

Solution

Notice how in this reaction, the main species PO₄^3- accepts a proton as it reacts with water – this means that it is the base in this reaction and hence, K is more specifically K_b. Recall that, for any conjugate acid-base pair, the relationship between K_a and K_b is: K_w = K_a × K_b. We can already rule out options (a) and (c) since they do not include K_w in the expression for K.

The reaction indicated starts with a phosphate ion (which is non-protonated) that is converted into its conjugate monoprotic acid HPO₄^–. PO₄^3- is in its most deprotonated form (no protons), so this reaction is K_b1 – the first reaction in which PO₄^3- picks up its very first proton. With that in mind, we can say that this reaction corresponds to K_a3. Since we’re working with HPO₄^– as the acid and PO₄^3- as its conjugate base, this equilibrium can be seen in one of two ways: either we have PO₄^3- as the base reacting with water to pick it up its very first proton and become HPO₄^3- (which is K_b1), or we have HPO₄^3- as the acid reacting with water to lose its last proton and become PO₄^3- (which is K_a3). We can see that both (b) and (d) both have the expression arranged correctly such that K_w/K_a = K_b, however, only (b) uses K_a3 to correctly convert to K_b1.

Therefore, the correct answer is (b).

Check Your Learning 5.4.2 – Converting Between K_a and K_b – Polyprotic Acids-

Bases

Given the information below, give the balanced chemical equations and calculate the equilibrium constants for the 3 successive protonation reactions of the hypothetical base, A^3– (aq).

H₃A (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + H₂A^– (aq)        K_a1 = 1.5 x 10^-4

H₂A^– (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HA^2– (aq)      K_a2 = 2.7 x 10^-9

HA^2– (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^3– (aq)        K_a3 = 8.8 x 10^-12

Answer:

A^3– (aq) + H₂O (l) ⇌ OH^– (aq) + HA^2– (aq)          K_b1 = K_W/K_a3 = 1.1 x 10^-3

HA^2– (aq) + H₂O (l) ⇌ OH^– (aq) + H₂A^– (aq).      K_b2 = K_W/K_a2 = 3.7 x 10^-6

H₂A^– (aq) + H₂O (l) ⇌ OH^– (aq) + H₃A (aq)        K_b3 = K_W/K_a1 = 6.7 x 10^-11

Questions

★ Questions

1. Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134 M solution of H₂CO₃, a diprotic acid: [H₃O⁺], [OH⁻], [H₂CO₃], [HCO₃⁻], [CO₃²⁻]? No calculations are needed to answer this question.

2. Calculate the concentration of each species present in a 0.050 M solution of H₂S.

3. Calculate the concentration of each species present in a 0.010 M solution of phthalic acid, C₆H₄(CO₂H)₂.

C₆H₄(CO₂H)₂(aq)+H₂O(l)↔H₃O(aq)⁺+C₆H₄(CO₂H)(CO₂)(aq)^–  K_a=1.1×10^-3

C₆H₄(CO₂H(CO₂)(aq)-+H₂O(l)↔H₃O(aq)⁺+C₆H₄(CO₂)₂(aq)^2-  K_a=3.9×10^-6

★★ Questions

4. Salicylic acid, HOC₆H₄CO₂H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.

(a) Both functional groups of salicylic acid ionize in water, with K_a = 1.0 × 10⁻³ for the—CO₂H group and 4.2 × 10⁻¹³ for the −OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g/L).

(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH₃CO₂C₆H₄CO₂H. The −CO₂H functional group is still present, but its acidity is reduced, Ka = 3.0 × 10⁻⁴. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).

(c) Under some conditions, aspirin reacts with water and forms a solution of salicylic acid and acetic acid:

CH₃CO₂C₆H₄CO₂H (aq) + H₂O (l) ⇌ HOC₆H₄CO₂H (aq) + CH₃CO₂H (aq)

(i) Which of the acids, salicylic acid or acetic acid, produces more hydronium ions in such a solution?

(ii) What are the concentrations of molecules and ions in a solution produced by the hydrolysis of 0.50 g of aspirin dissolved in enough water to give 75 mL of solution?

5. The ion HTe⁻ is an amphiprotic species; it can act as either an acid or a base.

(a) What is K_a for the acid reaction of HTe⁻ with H₂O?

(b) What is K_b for the reaction in which HTe⁻ functions as a base in water?

(c) Demonstrate whether or not the second ionization of H₂Te can be neglected in the calculation of [HTe⁻] in a 0.10 M solution of H₂Te.

Answers

1. [H₃O⁺] and [HCO₃⁻] are practically equal.

2. [HS] = 7.1 x 10⁻⁵ M, [H₃O⁺] = 7.1 x 10⁻⁵ M, [S⁻²] = 1.29 x 10⁻¹³ M, [H₂S] = 0.0499 M

[C₆H₄(CO₂)₂^2-(aq)]=3.9×10^-6 M, [OH-]=3.6×10^-12 M

4. (a) pH = 2.44, (b) pH = 2.70, (c) (i) salicylic acid (ii) The concentration of aspirin and acetic acid is 2.1 x 10⁻⁴ M

5. (a) K_a2=1.5×10^-11

(b) K_b=4.3×10^-12

(c) ([Te^2-][H₃O⁺(aq)])/[HTe^–]=((x)(0.0141+x))/(0.0141-x)≈((x)(0.0141))/(0.0141)=1.5×10^-11

Solving for x gives 1.5×10^-11 M. Therefore, compared with 0.014 M, this values is negligible (1.1×10^-7%)