CHP Home GenChem     Analytical     Instrumentation     Index

# Buffers (Acid-Base)

## Introduction

A buffer is a solution that can maintain a nearly constant pH if it is diluted, or if 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.

Since the buffer concentration is usually high, the [H+] concentration can be calculated from the equilibrium expression assuming that the concentrations of the conjugate acid-base pair do not change appreciably.

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

The equilibrium is:
CH3COOH(aq) CH3COO-(aq) + H+(aq)        Ka = 1.8x10-5 (at 25 oC)

 [ ]o [ ] [ ]eq CH3COOH CH3COO- H+ 1.00 M 1.00 M -- ~0 ~0 x 1.00 M 1.00 M x

What is Q for these initial conditions? In which direction will the reaction shift to reach equilibrium?

```       [H+][CH3COO-]
Ka  =  -------------
[CH3COOH]
```
1.8x10-5 = [H+] (1.00 M) / (1.00 M)
[H+] = 1.8x10-5
pH = -log(1.8x10-5) = 4.74

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

What happens if we add some strong acid to this buffer?

Example: What is the pH of the resulting solution when 10.0 mL of 1.00 M HCl is added to the above buffer?

 [ ]o [ ] [ ]eq CH3COOH CH3COO- H+ (1.00 M)(0.100 L)/0.110 L= 0.909 M (1.00 M)(0.100 L)/0.110 L= 0.909 M (1.00 M)(0.010 L)/0.110 L= 0.0909 M +0.0909 M -0.0909 M ~ -0.0909 M (0.909+0.0909) M= 1.00 M (0.909-0.0909) M= 0.818 M x

```       [H+][CH3COO-]
Ka  =  -------------
[CH3COOH]
```
1.8x10-5 = [H+] (0.818 M) / (1.00 M)
[H+] = 2.20x10-5
pH = -log(2.20x10-5) = 4.66

This problem can also be done using the number of moles of the acids and bases:

 moleso (moles) moleseq CH3COOH CH3COO- H+ 0.100 mol 0.100 mol (1.00 M)(0.010 L) = 0.010 mol +0.010 -0.010 ~ -0.010 0.100+0.010=0.110 0.100-0.010=0.090 x

```       [H+][CH3COO-]
Ka  =  -------------
[CH3COOH]
```
1.8x10-5 = [H+] (0.090) / (0.110)
[H+] = 2.20x10-5
pH = -log(2.20x10-5) = 4.66

In these problems how did we initially produce the 0.1 M CH3COOH and CH3COO-? In general how can we make a buffer?

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

 Top of Page