|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:
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.