VI. Buffers and Polyprotic Acids
VI-1. Buffers VI-2. Polyprotic Acids VI-3. Problem List VI-4. Buffer Problems VI-5. Polyprotic Acid Problems

### Introduction

A buffer is a solution that can maintain a nearly constant pH when diluted, or when strong acids or bases are added. A buffer solution consists of a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid).

Example: Consider a solution containing both acetic acid, CH3COOH, and acetate ions, CH3COO-.

Any strong base that is added to the solution is neutralized by acetic acid:
CH3COOH (aq) + OH-(aq) CH3COO-(aq) + H2O (aq)

Any strong acid that is added to the solution is neutralized by acetate:
CH3COO-(aq) + H+(aq) CH3COOH (aq)

The amount of strong acid or base that a buffer can neutralize is called the buffer capacity. After the strong base or acid is neutralized, equilibrium will be reestablished.

Since the concentrations of the conjugate acid-base pair in a buffer are usually high, there is very little change in the acid-base pair concentrations as the system establishes equilibrium. Therefore, the [H+] concentration can be calculated from the equilibrium expression using the pre-equilibrium concentrations of the conjugate acid-base pair. For our problem-solving method, step 4 can be omitted.

In the example above how could we produce the solution containing CH3COOH and CH3COO-? In general a buffer can be prepared by:

1. Mixing a weak acid and a salt of its conjugate base in solution.
2. Mixing a weak acid and enough strong base to neutralize a portion of the weak acid.
3. Mixing a weak base and enough strong acid to neutralize a portion of the weak base.

### Sample Problem

What is the pH of a buffer solution containing 0.100 moles of both CH3COOH and CH3COO- in 0.100 L of water?

1. The only species present in solution that take part in the equilibrium are acetic acid and acetate ion. The pre-equilibrium concentrations of acetic acid and acetate ion are given as 1.00 M (0.100 moles/0.100 L). In buffers, the concentrations of the weak acid and conjugate base are high, so the [H+] of water (1.0x10-7 M) can always be neglected.

2. The balanced equilibrium reaction is: CH3COOH(aq) H+(aq) + CH3COO-(aq)

and the equilibrium constant expression is:

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

3. It is seldom necessary to calculate Q in buffer problems. See explanation in step 4.

4. As stated in the introduction above, this step is not necessary in buffer problems. The concentrations of CH3COOH and CH3COO- do not change appreciably as equilibrium is established.

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

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

(x)(1.00 M)
Ka = -----------
(1.00 M)
```
x = Ka

x = 1.8x10-5

This problem asked for the pH of the solution.

[H+] = x = 1.8x10-5 M

pH = -log[H+] = -log(1.8x10-5)

 pH = 4.74

Notice that when the concentrations of the weak acid and conjugate base are equal, the pH of the buffer solution equals pKa.

Return to the top of this page.

CHP Home Equilibrium Practice Problems

Copyright © 1997-2000 by Brian M. Tissue, all rights reserved.