Physical Chemistry Third Edition

7.3 Chemical Equilibrium in Solutions 325

K 1  7. 52 × 10 −^3 , K 2  6. 23 × 10 −^8 , K 3  2. 2 × 10 −^13

Three simultaneous equations must be solved. It is best to seek simplifying approximations, such

as neglecting the H+ions from the third ionization in discussing the first two ionizations. The

validity of such approximations should be checked at the end of the calculation.

Theconjugate baseof a weak acid is a weak base. The acetate ion is the conjugate

base of acetic acid. When sodium acetate is added to water, the acetate ions hydrolyze

(react with water):

A−+H 2 O
HA+OH− (7.3-13)

where we use the same abbreviations as before: HA for CH 3 CO 2 H and A−for CH 3 CO− 2.

The equilibrium relation for this reaction is

Kb

γ(HA)(meq(HA)/m◦)γ(OH−)(meq(OH−)/m◦)

γ(A−)(meq(A−)/m◦)γ(H 2 O)x(H 2 O)

We assume thatγ(H 2 O)x(H 2 O)1 and that all of the activity coefficients of the ions

are equal toγ. We multiply the right-hand side of this equation top and bottom by

γ(H+)(meq(H+)/m◦):

Kb

γ(HA)(meq(HA)/m◦)γ(OH−)(meq(OH−)/m◦)

γ(A−)(meq(A−)/m◦)γ(H+)(meq(H+)/m◦)

γ(H+)(meq(H+)/m◦)



Kw

Ka



1. 00 × 10 −^14

1. 752 × 10 −^5

 5. 71 × 10 −^10

whereKais the equilibrium constant for the ionization of acetic acid andKwis the

equilibrium constant for the ionization of water.

EXAMPLE7.14

Find the pH of a solution made from 0.100 mol of sodium acetate in 1.000 kg of water. Use

the Davies equation to estimate activity coefficients. Remember to include the sodium ions

when you calculate the ionic strength.

Solution

We have the value ofKbfrom the previous example. We assume that the activity of water is

nearly equal to unity, and we ignore any OH−ions contributed by water:

Kb 5. 71 × 10 −^10



γ(HA)(m(HA)/m◦)γ(OH−)(m(OH−)/m◦)

γ(A−)(m(A−)/m◦)

We letm(HA)/m◦m(H−)/m◦x, and we first assume that activity coefficients are equal

to unity:

5. 71 × 10 −^10 

x^2

0. 100 −x