IV. Weak Acids and Weak Bases
IV-1. Introduction and Background IV-2. Sample Problems IV-3. Problem List IV-4. Weak Acid Problems IV-5. Weak Base Problems

### 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 (Ka = 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- 0.100 M ~0 0 -x M +x M +x M (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)
```
Ka = x2 / (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 = Ka, so the system is at equilibrium.

### Sample Weak Base Problem

What is the pH of a 0.100 M solution of ammonia (Kb = 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: NH3(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- 0.100 M 0 ~0 -x M +x M +x M (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)
```
Kb = x2 / (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 = Ka, so the system is at equilibrium.