Weak Acid Equilibrium

Acid Dissociation Constants

When an uncharged weak acid is added to water, a heterogeneous equilibrium forms in which aqueous acid molecules, HA(aq), react with liquid water to form aqueous hydronium ions and aqueous anions, A^-(aq). The latter are produced when the acid molecules lose H⁺ ions to water.

HA(aq) + H₂O(l)      H₃O⁺(aq) + A^-(aq)

In writing an equilibrium constant expression for this heterogeneous equilibrium, we leave out the concentration of the liquid water. The equilibrium constant for this expression is called the acid dissociation constant, K_a.

      = acid dissociation constant

When the equilibrium in question occurs in solution, the chemical formulas enclosed in brackets in the equilibrium constant expression represent the molarities of the substances (moles of solute per liter of solution).

Remember that H⁺ can be used to represent H₃O⁺, thus simplifying our depiction of the reaction between a weak acid and water and its acid dissociation constant expression:

HA(aq)      H⁺(aq) + A^-(aq)

      = acid dissociation constant

For example, acetic acid is a weak acid, because when it is added to water, it reacts with the water in a reversible fashion to form hydronium and acetate ions.

HC₂H₃O₂(aq) + H₂O(l)      H₃O⁺(aq) + C₂H₃O₂^-(aq)

or HC₂H₃O₂(aq)      H⁺(aq) + C₂H₃O₂^-(aq)

      = 1.8 × 10^-⁵

EXAMPLE 1 - Writing an Acid Dissociation Constant: Write the equation for the reaction between the weak acid nitrous acid and water, and write the expression for its acid dissociation constant.

Solution:

HNO₂(aq) + H₂O(l)      H₃O⁺(aq) + NO₂^-(aq)

or HNO₂(aq)      H⁺(aq) + NO₂^-(aq)

The table below lists acid dissociation constants for some common weak acids. These K_a values can be used to describe the relative strength of the acids. A stronger acid will generate more hydronium ions in solution. A larger K_a indicates a greater ratio of ions (including hydronium ions) to uncharged acid. Therefore, a larger K_a indicates a stronger acid. For example, the larger K_a for chlorous acid (1.2 × 10^-²) compared to acetic acid (1.8 × 10^-⁵) tells us that chlorous acid is stronger than acetic acid.

Acid Dissociation Constants, Ka, for Common Weak Acids

Weak Acid Equation K_a

acetic acid HC₂H₃O₂      H⁺ + C₂H₃O₂^- 1.8 × 10^-⁵

benzoic acid C₆H₅CO₂H
                  H⁺ + C₆H₅CO₂^- 6.4 × 10^-⁵

chlorous acid HClO₂      H⁺ + ClO₂^- 1.2 × 10^-²

formic acid HCHO₂      H⁺ + CHO₂^- 1.8 × 10^-⁴

hydrocyanic acid HCN      H⁺ + CN^- 6.2 × 10^-¹⁰

hydrofluoric acid HF      H⁺ + F^- 7.2 × 10^-⁴

hypobromous acid HOBr      H⁺ + OBr^- 2 × 10^-⁹

hypochlorous acid HOCl      H⁺ + OCl^- 3.5 × 10^-⁸

hypoiodous acid HOI      H⁺ + OI^- 2 × 10^-¹¹

lactic acid CH₃CH(OH)CO₂H
       H⁺ + CH₃CH(OH)CO₂^- 1.38 × 10^-⁴

nitrous acid HNO₂      H⁺ + NO₂^- 4.0 × 10^-⁴

phenol HOC₆H₅      H⁺ + OC₆H₅^- 1.6 × 10^-¹⁰

propionic acid CH₃CH₂CO₂H
            H⁺ + CH₃CH₂CO₂^- 1.3 × 10^-⁵

The following study sheet describes one procedure for calculating the pH of solutions of weak acids. If you take other chemistry courses, you will find that there are variations on this procedure for some weak acid solutions.

Study Sheet - Calculating pH for Weak Acid Solutions

Tip-off - You are given the concentration of a weak acid solution and asked to calculate its pH.

General Steps -

STEP 1 Write the equation for the ionization of the weak acid in water.

HA(aq)      H⁺(aq) + A^-(aq)

STEP 2 Write the K_a expression for the weak acid.

STEP 3 Describe each equilibrium concentration in terms of x.

x = [H⁺]_equilibrium = [A^-]_equilibrium

[HA]_equilibrium = [HA]_initial - x

STEP 4 Assume that the initial concentration of weak acid is approximately equal to the equilibrium concentration. (Weak acids are rarely ionized to a large degree. We can most often assume that the initial concentration added, [HA]_initial is much larger than x. Thus, the equilibrium concentration is approximately equal to the concentration added. You may learn how to deal with weak acid solutions for which this approximation is not appropriate in other chemistry courses.)

[HA]_equilibrium = [HA]_initial

STEP 5 Plug the concentrations described in terms of x into the Ka expression, and solve for x.

EXAMPLE 2 - pH Calculations for Weak Acid Solutions: Vinegar is a dilute water solution of acetic acid with small amounts of other components. Calculate the pH of bottled vinegar that is 0.667 M HC₂H₃O₂, assuming that none of the other components affect the acidity of the solution.

HC₂H₃O₂(aq)      H⁺(aq) + C₂H₃O₂^-(aq)

We get the value for the acid dissociation constant for this reaction from the table above.

x² = 1.2 × 10^-⁵ x = 3.5 × 10^-³

[H⁺] = 3.5 × 10^-³ M H⁺       pH = -log(3.5 × 10^-³) = 2.46