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Equilibrium in Solutions of Weak Acids

We have seen that equilibrium exists when two opposing reactions occur at the same rate. In Chapter 10, we discussed the equilibrium between a liquid and its vapor, with the opposing reactions being evaporation and condensation:

Rateevaporation = Ratecondensation

In Chapter 11, we discussed the equilibrium between dissolved and undissolved solute in saturated solutions, in which the opposing reactions are precipitation and dissolution:

Ratedissolution = Rateprecipitation

The ionization of a weak acid is also an equilibrium situation. Here the equilibrium is between molecules and ions. We have implied the existence of this equilibrium with double arrows in the ionization equations for weak acids. In a solution of acetic acid, a weak acid, the two ongoing reactions are dissociation of molecules into ions and recombination of ions into molecules. This equilibrium is shown by the equation

HC2H3O2   H+ + C2H3O2-

The opposing reaction rates are

Ratedissociation = Raterecombination of ions

There is another criterion for equilibrium. For a system in equilibrium, a mathematical relationship exists between the concentration of the components of the equilibrium. This relationship is known as the equilibrium constant, Keq. If the ionization is that of a weak acid, the equilibrium constant is known as the acid dissociation constant and has the symbol Ka.

In general terms, for a reaction

aA + bB cC + dD

the equilibrium constant expression is

  Keq = [C]c[D]d
[A]a[B]b

where the square brackets mean molar concentration. Note that the numerator of the equilibrium constant expression is the product of the concentrations of the products, each raised to a power equal to its coefficient in the balanced equation for the equilibrium reaction. The denominator is the product of the concentrations of the reactants, each raised to a power equal to its coefficient in the balanced equation for the equilibrium reaction. A weak acid with the formula HA would show in solution the equilibrium

HA H+ + A-

The ionization constant expression for this equilibrium is

  Ka = [H+][A-]
[HA]

By convention, these ionization equilibria are always written with the ions as products. Thus, in an ionization constant expression, the concentrations of the ions are always in the numerator, that of the un-ionized molecules in the denominator.

Therefore, for acetic acid, with the ionization equilibrium

HC2H3O2  H+ + C2H3O2-

the equilibrium constant expression is

  Ka = [H+][C2H3O2-]
[HC2H3O2]

We have already pointed out that a weak acid is only slightly ionized and that its solution contains mostly molecules and many fewer ions. Therefore, the concentration value of the un-ionized molecules in the denominator of the equilibrium constant is much larger than those of the ions in the numerator. Consequently, values of the ionization constants for weak acids are always much less than 1. Table 12.4 lists several weak acids and shows the equations for their ionization and the expression and value of their acid dissociation constants. Note several points in this table:

1. In regard to the acid dissociation constants of polyprotic acids, we have already noted (Section 12.4B) that polyprotic acids ionize stepwise and that each anion is less ionized than its conjugate acid. Each of the ionizations has an equilibrium equation and an equilibrium constant. Phosphoric acid has three ionizable hydrogens. The acid molecule ionizes to yield a hydrogen ion and a dihydrogen phosphate ion:

H3PO4  H+ + H2PO4-       Ka = 7.5 X 10-3

The dihydrogen phosphate ion ionizes to yield another hydrogen ion and the monohydrogen phosphate ion:

H2PO4  H+ + HPO42-       Ka = 6.2 X 10-8

Finally, the monohydrogen phosphate ion ionizes to yield another hydrogen ion and the phosphate anion:

HPO42-  H+ + PO43-         Ka = 2.2 X 10-12

Notice that the acid dissociation constant becomes smaller with each ionization. For acids with more than one ionizable hydrogen, the first dissociation constant is always the largest; the value for each successive dissociation constant decreases. This progression goes along with the prediction in Section 12.4 that, in each successive ionization, the anion formed is a stronger base.

TABLE 12.4 Some common weak acids
Electrolyte Equilibrium equation Acid
dissociation
expression
Ka
acetic acid HC2H3O2 H+ + C2H3O2- [H+][C2H3O2-]
[HC2H3O2]
1.8 X 10-5
formic acid HCO2H H+ + HCO2- [H+][HCO2-]
[HCO2H]
1.8 X 10-4
nitrous acid HNO2 H+ + NO2- [H+][NO2-]
[HNO2]
4.6 X 10-4
hydrocyanic acid HCN H+ + CN- [H+][CN-]
[HCN]
4.9 X 10-10
carbonic acid CO2 + H2O H+ + HCO3- [H+][HCO3-]
[CO2]
4.3 X 10-7
  HCO3- H+ + CO32- [H+][CO32-]
[HCO3-]
5.6 X 10-11
phosphoric acid H3PO4 H+ + H2PO4- [H+][H2PO4-]
[H3PO4]
7.5 X 10-3
  H2PO4- H+ + HPO42- [H+][HPO42-]
[H2PO4-]
6.2 X 10-8
  HPO42- H+ + PO43- [H+][PO43-]
[HPO42-]
2.2 X 10-13
ammonium ion NH4+ H+ + NH3 [H+][NH3]
[NH4+]
5.5 X 10-10


2. Carbonic acid (H2CO3) is a solution of carbon dioxide in water. Molecules of carbonic acid are not stable. The mixture of carbon dioxide and water ionizes stepwise like phosphoric acid:

CO2 + H2O  H+ + HCO3-

HCO3-  CO32- + H+

3. Ammonium ion acts as a weak acid, ionizing to ammonia and a hydrogen ion:

NH4+  H+ + NH3


Example:

Ascorbic acid, vitamin C, is a weak electrolyte. Its molecular formula is C6H8O6. In aqueous solution, ascorbic acid ionizes to form H+ and ascorbate ion, C6H7O6-.

a. Write the equation for the equilibrium established in this ionization.

b. Write the acid dissociation constant expression for ascorbic acid.

Solution

a. The equation shows the loss of one hydrogen ion; the rest of the molecule is the ascorbate anion.

C6H8O6   H+ + C6H7O6-

b. An Acid dissociation constant expression has the concentrations of the ions in the numerator and that of the un-ionized acid molecules in the denominator. The acid dissociation constant expression of ascorbic acid is:

Ka =
[H+][C6H7O6-]
[C6H8O6]

 

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