Acid-Base Buffers


A buffer is a solution that is resistant to changes in pH. The desired buffer pH determines what compounds are used to make buffer. To make a buffer with an acidic pH, ( pH less than 7) the solution is made using a weak acid and a soluble salt of its anion. A buffer with a pH of 4.74 can be made using a equal volumes of 0.10 M acetic acid, HC2H3O2, and 0.10 M sodium acetate, NaC2H3O2. Typically the range for an acid buffer is centered around the value of the dissociation constant for the weak acid. The pH is adjusted by controlling the ratio of the mols of weak acid and the mols of soluble salt. The "recipe" for preparing buffers is illustrated below.

A rule of thumb is that an acid buffer can be made using a weak acid and its salt.

To make a buffer solution with a basic pH, (pH is more than 7) the solution is typically made using a weak base and a soluble salt of its cation.

Buffers have at least two limitations. The natural pH range that matches the properties of the solute acid or base and the capacity of the buffer. One limitation depends on the equlibrium constant for the weak acid or base. The second comes from the limited solubility of the acids, bases or their salts. The amount of dissolved acid, base and salt determines the mols available in the buffer that cn react with "added" acids or bases from the outside.



A buffer can stabilize pH and acidity because it contains a reservoir of acid and a reservoir of base. The acid can react with added base. The anion of the weak acid "behaves" as a base so it reacts with added acid. A buffer of acetic acid has molecules of HC2H3O2 that can react with any base added to the solution. The same buffer mixture contains acetate ions, C2H3O21- that can react with protons(hydronium ions) from any acid that is added to the buffer.

A famous ( in biochemistry circles) relationship fo predicting pH for acid buffers is the Henderson-Hasselbalch equation.

The equation shows that the pH range for the buffer is tied to the Ka for the acid. The pH can be adjusted by controlling the molarity of the acid, [HA], and the molarity for the anion of the acid, [A1-]. This really requires control of the ratio of mols of anion to mols of acid.

Henderson-Hasselbalch equation
pH = pKa + log [A1-]/[HA]


[A1-] is the molarity of the anion of the weak acid;

[HA] is the molarity of the weak acid;

pKa is the -log of the weak acid dissociation constant;

log means the base ten logarithm.

Example: How the pH for a buffer relates to pKa.

What is pH of a solution made using equal numbers of mols of acetic acid,HC2H3O2, and mols of sodium acetate, NaC2H3O2 ? The ratio of mols of each will be the same as the ratio of the molarities (concentrations) because they are dissolved in the same final volume of solution.

1. The Henderson-Hasselbalch equation relates the ratio of mols of anion and mols of weak acid to pH.

2. Recall pH = pKa + log [A1-]/[HA]

3. Collect the data

desired pH = ? ;

for acetic acid, HC2H3O, from references Ka = 1.8 x 10-5;

from references or by calculating pKa = -log Ka ; pKa= 4.7

pKa = -log(1.8 x 10-5) = 4.7

[C2H3O2 1-] = [HC2H3O2] so ratio [C2H3O2 1-]/[HC2H3O2] = 1




4. Substitute into the equation;

pH = pKa + log [A1-]/[HA]

pH = 4.7 + log [C2H3O2 1-]/[HC2H3O2]

pH = 4.7 + log (1) = 4.7 + 0 = 4.7



Recall the log of 1 is zero.

pH = 4.7



5. The ratio of the concentration of the anion and the weak acid can be adjusted to change the pH . Typically the pH can be set to one pH unit above or one pH unit below the pKa for the acid. This matches a limit for the ratio of 1 to 10


Dr. Walt Volland all rights reserved 1997-2005

Revised April 12, 2005