Buffers
A buffer solution is a solution of a weak acid and its salt (conjugate base) or a weak base and its salt (conjugate acid). Both components must be present. A buffer solution has the ability to resist change in pH upon the addition of small amounts of either acid or base. A buffer solution must have a relatively large concentration of acid to react with any OH- ions that are added to it, and it must contain similar concentration of base to react with any added H3O+ ions. The acid and base components of the buffer must not neutralize each other. These requirements are met by an acid-base conjugate pair.
To understand how a buffer works consider the acetic acid equilibrium
CH3COOH(aq) + H2O(l) <==> H3O+(aq) + CH3COO-(aq)
Suppose a strong acid is added to a solution contains CH3COO- ions and CH3COOH molecules in about equal concentrations. The newly added H3O+ ions transfer protons to the to a relatively large reservoir of CH3COO- ions to form CH3COOH molecules. Only a small fraction of these newly formed CH3COOH molecules ionize, so the pH is virtually unchanged. If a small amount of strong base is added instead, the OH- ions remove protons from the CH3COOH molecules. As a result, the concentration of CH3COO- increases, but the OH- concentration and thus the pH remains nearly unchanged. So the weak acid acts as a reservoir of hydrogen ions to react with OH- ions, whereas its conjugate base acts as a sink for H3O+.
The Henderson-Hasselbalch Equation
Calculating the pH of a buffer solution involves the equilibrium constant expression for the weak acid or weak base used in the buffer. We can rearrange the equilibrium constant expression into a form more convenient to use however. Using a weak acid for example and rearranging the expression for Ka
HA(aq) + H2O(l) <==> H3O+(aq) + A-(aq)
[H3O+][A-] Ka = ---------- [HA]
[HA] [H3O+] = Ka x ---- [A-]
take the negative logarithm of both sides:
[HA] -log[H3O+] = -log Ka -log ---- [A-]
Which gives,
[A-] pH = pKa + log ---- This is known as the Henderson-Hasselbalch equation [HA]
We are using equilibrium concentrations of acid and conjugate base in solution for the expression above. However only a tiny fraction of the acid donates protons to the conjugate base. Thus, a good approximation for the equilibrium concentrations is to just use the initial concentrations of the acid and salt.
Example 1
(a) What is the pH of a 0.30 M solution of ammonia, NH3, which is also 0.36 M in ammonium nitrate, NH4NO3? (b) What is the pH after the addition of 20.0 mL of 0.050 M HCl to 80.0 mL of the buffer solution?
(a) Use the Henderson-Hasselbalch equation (for a weak base).
[NH4+] pH = pKb + log ------ = 4.74 + log (0.36/0.30) = 4.82 [NH3]
(b) First we do some stoichiometric calculations:
mol NH4+ present initially 0.0800 L x 0.36 mol/L = 0.029 mol
mol NH3 present initially 0.0800 L x 0.30 mol/L = 0.024 mol
mol HCl added 0.0200 L x 0.40 mol/L = 0.0080 mol (this reacts with NH3 to become NH4+)
mol NH4+ present after HCl added 0.029 mol + 0.0080 mol = 0.030 mol
mol NH3 present after HCl added 0.024 mol - 0.008 mol = 0.023 mol
Since volume will cancel from the ratio we can use moles in the Henderson-Hassellbalch equation.
[NH4+] pH = pKb + log ------ = 4.74 + log (0.030/0.023) = 4.86 [NH3]
Example 2
What ratio of acetic acid to acetate must be used to prepare a buffer solution having a pH of 5.00?
[CH3COO-] [CH3COO-] pH = pKa + log --------- = 4.74 + log --------- = 5.00 [CH3COOH] [CH3COOH]
[CH3COO-] log --------- = 5.00 - 4.74 = 0.26 [CH3COOH]
Taking the antilog we obtain
[CH3COO-] --------- = 1.8 [CH3COOH]