Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110

General Chemistry

Summer 2006

Lecture Notes::Lec2 31 May

© R. Paselk 2006


Review of Aqueous Equilibria

Acid-Base Chemistry, 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. In order to accomodate this vast range of concentration conveniently we look instead at the logarithm of [H+] (recall that log 10x = x) and define a new term,

pH = -log[H+]

Remember also that the concentration of hydrogen ion in water is related to the concentration of hydroxide ion due to the equilibrium dissociation of water:

H2O H+ + OH-, so

They are calculated using 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.


Note that the significant figures are correct, 1 is the power of ten, only the figures to the right are significant.

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


Acid-Base Titration and Hydrolysis

(You can review titration and titrations curves on-line in my Chem 109 notes in my course archives:

When a weak acid is titrated with a strong base it is found that the equivalence point always occurs at a pH above neutrality (pH > 7), as seen in the figure for acetic acid titrated with hydroxide ion:

Similarly, when a weak base is titrated with a strong acid the equivalence point always occurs at a pH below neutrality (pH < 7).

Why? Essentially it is a consequence of the the nature of acids and bases as seen in the Brønsted-Lowry acid description: when a weak acid is titrated a strong base, the acid's conjugate base is formed. The conjugate base then competes with hydroxide ion for hydrogen ion, leading to a slightly higher concentration of hydroxide ion as the equilibrium is shifted:

A- + H2O OH- + HA

This process is sometimes referred to as hydrolysis, which is the term we will also use. (Notice that Zumdahl discusses this phenomena in section 14.8 Acid-Base Properties of Salts pp 683-688.)

Note that the anion of a strong acid, such as HCl, will not affect pH since the chloride ion has no tendency to react with water (as implied in the definition of "strong"). Similarly the cation of a strong base such as NaOH will not affect pH since Na has no tendency to react with water either.

So how is the pH affected by the presence of the salts of weak acids? Let's look at our favorite acid, acetic acid reacting with NaOH:

HC2H3O2 + OH- C2H3O2- + H2O

The resulting acetate ion can now react with water:

C2H3O2- + H2O HC2H3O2 + OH-

This reaction can be written as the sum of the association of acetic acid and the dissociation of water:

C2H3O2- + H+ HC2H3O2

H2O H+ + OH-

Notice that the the equilibrium constant for the association of acetic acid is the inverse of the dissociation constant (the reaction is backwards, inverting the equilibrium expression): K = 1/Ka

The overall equilibrium constant is then the product of the equilibrium constants for the two reactions, Kh:

Kh = (Kw)(1/Ka)

Kh is in fact Kb for the conjugate base. (Zumdahl use Kb in his discussion.) Kb, conj = Kw/Ka

A similar treatment is seen for weak bases.

Example: Calculate the pH of a 0.54 M solution of  NH4Cl. Kb = 1.8 x 10-5

   NH4+ + H2   NH3 + H3O+
at equilibrium 0.54 - x     x   x

Kh = (Kw)(1/Kb) = [H+] [NH3] / [NH4+]

Kh = (1.0 x 10-14) / (1.8 x 10-5) = 5.6 x 10-10

x2/(0.54 - x) = 5.6 x 10-10; assume x << 0.54

x = [H3O+] = 1.7 x 10-5 M

pH = 4.76

Common Ions & Buffers

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

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

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. Ka = 1.8 x 10-5

  HOAc   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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