Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109

General Chemistry

Spring 2009

Lecture Notes:: 27 April

© R. Paselk 2002


Acids and Bases, cont.

The pH Scale

The concentration of hydronium ion in water is extremely influential on all kinds of chemistry. The range of hydronium ion concentration in water is also vast, with extremes of about 10M to about 10-15M, and commonly ranging from 1M - 10-14M. Imagine plotting [H3O+] vs. volume of acid added to a base solution in a titration from 10-14M - 1M. If you had one cm on the graph paper = 10-14M, then you would need a piece of paper 109 km long (greater than the distance from the Sun to Jupiter) to plot this titration! Obviously a more convenient measure is needed. This is easily accomplished by looking instead at the logarithm of [H+] and defining a new term,

pH = -log[H+]

Because of the equilibrium dissociation of water to H+ + OH-, the concentration of hydrogen ion in water is related to the concentration of hydroxide ion:

H2O equilibrium arrow H+ + OH-, so

K = [H+][OH-] / [H2O]

But the concentration of water remains essentially the same in dilute solution, so by convention we define the dissociation constant or ion product for water:

Kw= [H+][OH-] = 1.0 x 10-14 @ 25 °C

Let's look at some general characteristics of pH in aqueous solution.


Note that the "p" has the more general meaning of "-log[]". Thus pOH is -log [OH-], pCa = -log [Ca2+], etc.

pH of weak acid solutions

Weak acid dissociations involve equilibria. The equilibrium constants have a specific symbol = Ka.

Example: What is the pH of a 0.10 M solution of acetic acid. Ka = 1.8 x 10-5

  HOAc  equilibrium arrow H+ + OAc-
Before reaction 0.10 M   0 0
@ Equilibrium
0.10 M- x
  x   x

Ka = [H+][OAc-] / [HOAc]

assume x << 0.1 since Ka =1.8 x 10-5, then [HOAc] = 0.10 M

Substituting, Ka = (x)(x) / 0.10 = 1.8 x 10-5,

x2 = 1.8 x 10-6

x = 1.34 x 10-3M; assumption OK.

pH = - log (1.34 x 10-3) = 2.87

Notice the significant figures. For a log function the number in front of the decimal is the exponent of ten, thus pH = 2.87 is a 2 significant figure number.

Acid Equilibria

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

With this in mind let's do some examples.

Example: Calculate the pH of a buffer 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. Ka = 1.8 x 10-5

  HOAc  equilibrium arrow H+ + OAc-
Before reaction 0.0125 moles/L   0 0.0250 moles/L
@ Equilibrium
(0.0125 - x) M
assume x is small,
= 0.0125
(0.0250 + x) M
assume x is small,
= 0.0250

Ka = [H+][OAc-] / [HOAc]

Substituting, Ka = [H+](0.0250) / (0.0125) = 1.8 x 10-5

Rearranging, [H+] = (1.8 x 10-5)(0.0125) / (0.0250) = 0.90 x 10-5

x is within experimental error (0.000009 < ±0.0001), so assumption OK

pH = 5.046

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Last modified 27 April 2009