Introduction

Below are a weak acid and a weak base problem. These problems are straightforward in that there are no initial reactions that affect the concentration of the weak acid or base.

Sample Weak Acid Problem

What is the pH of a 0.100 M solution of acetic acid (K_a = 1.8x10^-5)?

1. First identify what species are present, and any reactions that could occur. (The title of this problem states that it is a weak acid problem, but not all problems will tell you the type of equilibrium that will be involved.) The only specie present is acetic acid. The [H⁺] of water (1.0x10^-7 M) can almost always be neglected. If the concentration of the weak acid is low, then you must solve a complex equilibria problem that includes the dissociation of water. The pre-equilibrium concentration of acetic acid is 0.100 M.

2. The balanced equilibrium reaction is: CH₃COOH_(aq) H⁺_(aq) + CH₃COO^-_(aq)

and the equilibrium constant expression is:

      [H+][CH3COO-]
Ka = ------------
      [CH3COOH]

3. Now calculate the reaction quotient, Q, to determine the direction in which the reaction will proceed to reach equilibrium. In acid-base problems, it is seldom necessary to calculate Q exactly. In this sample problem, the pre-equilibrium concentration of acetate ion, COO^-, is zero, so Q is zero and the reaction goes in the forward direction.

4. For each mol of CH₃COOH that dissociates, 1 mole each of H⁺ and CH₃COO^- forms. The pre-equilibrium concentrations, the concentration changes, and the equilibrium concentrations are summarized in the following table:

	CH3COOH	H+	COO-
[ ]o	0.100 M	~0	0
[ ]	-x M	+x M	+x M
[ ]eq	(0.100 - x) M	x M	x M

Where [ ]_o are the pre-equilibrium concentrations, [ ] are the changes in concentrations, and [ ]_eq are expressions for the equilibrium concentrations.

5. We can now calculate the equilibrium concentrations using the equilibrium constant expression:

      [H+][CH3COO-]
Ka = ------------
      [CH3COOH]

      (x)(x)
Ka = --------
    (0.100-x)

We could solve this expression exactly, but since we are dealing with a weak acid, x is much smaller than 0.100 and can be neglected in the denominator.

1.8x10^-5 = x² / 0.100

x² = 1.8x10^-6

x = 1.3x10^-3 M

Note that x << 0.100 M, and the approximation was valid.

This problem asked for the pH of the solution.

[H⁺] = x = 1.3x10^-3 M

pH = -log[H⁺] = -log(1.3x10^-3)

pH = 2.87

We can calculate Q to check that we are at equilibrium:
Q = (1.3x10^-3)(1.3x10^-3) / 0.100 = 1.810^-5. Q = K_a, so the system is at equilibrium.

Sample Weak Base Problem

What is the pH of a 0.100 M solution of ammonia (K_b = 1.8x10^-5)?

1. First identify what species are present, and any reactions that could occur. (The title of this problem states that it is a weak base problem.) The only specie present is ammonia. The [OH^-] of water (1.0x10^-7 M) can almost always be neglected. If the concentration of the weak base is low, then you must solve a complex equilibria problem that includes the dissociation of water. The pre-equilibrium concentration of ammonia is 0.100 M.

2. The balanced equilibrium reaction is: NH_3(aq) + H₂O NH₄⁺_(aq) + OH^-_(aq)

and the equilibrium constant expression is:

     [NH4+][OH-]
Kb = -----------
        [NH3]

3. Now calculate the reaction quotient, Q, to determine the direction in which the reaction will proceed to reach equilibrium. In acid-base problems, it is seldom necessary to calculate Q exactly. In this sample problem, the pre-equilibrium concentration of ammonium ion, NH₄⁺, is zero, so Q is zero and the reaction goes in the forward direction.

4. For each mol of NH₃ that reacts with water, 1 mole each of NH₄⁺ and OH^- forms. The pre-equilibrium concentrations, the concentration changes, and the equilibrium concentrations are summarized in the following table:

	NH3	NH4+	OH-
[ ]o	0.100 M	0	~0
[ ]	-x M	+x M	+x M
[ ]eq	(0.100 - x) M	x M	x M

Where [ ]_o are the pre-equilibrium concentrations, [ ] are the changes in concentrations, and [ ]_eq are expressions for the equilibrium concentrations.

5. We can now calculate the equilibrium concentrations using the equilibrium constant expression:

     [NH4+][OH-]
Kb = -----------
        [NH3]

      (x)(x)
Kb = ---------
     (0.100-x)

We could solve this expression exactly, but since we are dealing with a weak base, x is much smaller than 0.100 and can be neglected in the denominator.

1.8x10^-5 = x² / 0.100

x² = 1.8x10^-6

x = 1.3x10^-3 M

Note that x << 0.100 M, and the approximation was valid.

This problem asked for the pH of the solution.

[OH^-] = x = 1.3x10^-3 M

pOH = -log[OH^-] = -log(1.3x10^-3)

pOH = 2.87

pH = 14.00 - POH = 14.00 - 2.87

pH = 11.13

We can calculate Q to check that we are at equilibrium:
Q = (1.3x10^-3)(1.3x10^-3) / 0.100 = 1.810^-5. Q = K_a, so the system is at equilibrium.