Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110	General Chemistry	Fall 2003
Lecture Notes::Lec3_29 August	© R. Paselk 2003
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Review of Aqueous Equilibria

Acid-Base Chemistry, cont.

Common Ions & Buffers

Buffer calculations: One of the most frequent calls for calculating acid equilibria is calculations involving buffers. What is a buffer?

• A buffer is a solution which resists changes in pH. Essentially it consists of an acid and its salt (an acid and its conjugate base) in solution together. Thus the solution has a proton donor and a proton acceptor, so pH is stabilized. That is, if we add strong base it will react completely with the proton donor and be removed, and vice versa.
• A buffer is simply an acid equilibrium system with significant amounts of both the acid and its conjugate base.

Another way to look at buffers is in terms of the common ion effect.

• Since a buffer consists of a mixture of an acid and its salt, the dissociation of the acid will be suppressed by the presence of the salt. In other words, by Le Châtelier's Principle the equilibrium mixture of the acid is shifted to the undissociated form by the presence of the dissociated form.
• As a result the equilibrium concentrations of the acid and its salt are very nearly the amounts of the chemicals added ("x" is very small).

Example: Calculate the pH of a "buffer" (a solution which resists changes in pH) made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. K_a = 1.8 x 10^-5

This is also why the Henderson-Hasselbalch equation is legitimate for buffer calculations (x is ignored).

Henderson-Hasselbalch equation: pH = pKa + log([A-] / [HA]) (Note that an analogous equation may be written for bases: pOH = pKb + log([B+] / [BH]))

Example: for the buffer above: pK_a = -logK_a = 4.74.

pH = pK_a + log[A^-] / [HA]

pH = 4.74 + log(0.250) / (0.125)

= 4.74 + 0.301

pH = 5.04

Notice that only the ratio matters, the actual concentrations of the acid and salt are not that important for determining pH!

Buffer Preparation

Consider the preparation of a buffer using phosphate ions. The reactions involved and K_a and pK_a values are shown below:

Rxn   Ka  pKa  H3PO4 ´ H+ + H2PO4-  7.5 x 10-3  2.2  H2PO4- ´ H+ + HPO42-  6.2 x 10-8  7.2 HPO42- ´ H+ + PO43-  4.8 x 10-13  12.7

How would you make up a 0.10 molar phosphate buffer with a pH of 6.5 assuming all three sodium "phosphate" salts are available?

• First need to chose the appropriate salts. Looking at the pK_a values we would choose H₂PO₄^- as the acid and HPO₄^2- as the salt in the Henderson-Hasselbalch equation since the pH is between the pK_a's for these reactions:
  • pH = pK_a + log[A^-] / [HA]
  • pH = pK_a + log[HPO₄^2-] / [H₂PO₄^-]
  • and inserting values as appropriate, 6.5 = 7.2 + log[HPO₄^2-] / [H₂PO₄^-]
  • solving

  log[HPO₄^2-] / [H₂PO₄^-] = 6.5 - 7.2

  [H₂PO₄^-] / [HPO₄^2-] = 10^0.7 = 5.01

  [H₂PO₄^-] = 5.01*[HPO₄^2-]

  and [H₂PO₄^-] + [HPO₄^2-] = 0.1

  substituting, 5.01*[HPO₄^2-] + [HPO₄^2-] = 0.1

  [HPO₄^2-] = 0.017 M

  [H₂PO₄^-] = 0.084 M

Aqueous Ion Solubility

The solubility of ionic compounds in aqueous solutions has immense impact on the world. In particular ionic solubility determines the fate and deposition of vast mineral deposits on Earth's surface, it is used by organisms to create and model mineral components such as shell and bone, it impacts mineral accessibility in soils, and it is utilized and manipulated by humans in all manner of processes.

© R A Paselk

Last modified 29 August 2003