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


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:

Ka = -------------

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:

Ka = -------------

     (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.

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