Physical Chemistry Third Edition

7.3 Chemical Equilibrium in Solutions 321

EXAMPLE7.10

Find∆G◦andKat 298.15 K for the ionization of water.

Solution

The Gibbs energy change of this reaction is the negative of that of Example 7.8:

∆G◦ 79 .892 kJ mol−^1

Kwe−∆G

◦/RT

exp

(

−79892 J mol−^1

(8.3145 J K−^1 mol−^1 )(298.15 K)

)

 1. 008 × 10 −^14

The equilibrium constant for the ionization of water can be used to calculate the

equilibrium molalities of hydrogen and hydroxide ions:

Kw

aeq(H+)aeq(OH−)

aeq(H 2 O)



γ(H+)(meq(H+)/m◦)γ(OH−)(meq(OH−)/m◦)

γ(H 2 O)xeq(H 2 O)



γ±^2 (meq(H+)/m◦)meq(OH−)/m◦)

γ(H 2 O)xeq(H 2 O)

γ^2 ±(meq(H+)/m◦)(meq(OH−)/m◦) (7.3-9)

where we assume thatγ(H 2 O)xeq(H 2 O)1.

Exercise 7.7

Calculate the molalities of hydrogen and hydroxide ions in pure water at 298.15 K. Use the

Davies equation to estimate activity coefficients if necessary.

In any aqueous solution, the ionization of water will equilibrate along with any other

equilibria.

EXAMPLE7.11

Find the value ofmeq(H+) in the solution of Example 7.9.

Solution

We use the value ofγ±from Example 7.9.

meq

(

H+

)

m◦



Kw

γ^2 ±meq

(

OH−

)

/m◦



(

1. 008 × 10 −^14

)

( 0. 960 )^2 ( 0. 00140 )

 7. 80 × 10 −^12