7.3 Chemical Equilibrium in Solutions
321
E X A M P L E 7.10
Find ∆G ◦ and K at 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
K w e −∆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:
K w
a eq (H + )a eq (OH
− )
a eq (H 2 O)
γ(H + )(m eq (H + )/m ◦ )γ(OH
− )(m eq (OH
− )/m ◦ )
γ(H 2 O)x eq (H 2 O)
γ 2
± (m eq (H + )/m ◦ ) m eq (OH
− )/m ◦ )
γ(H 2 O) x eq (H 2 O)
γ
2
± (m eq (H
+ )/m
◦ )(m eq (OH
− )/m
◦ )
(7.3-9)
where we assume that γ(H 2 O)x eq (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.
E X A M P L E 7.11
Find the value of m eq (H + ) in the solution of Example 7.9.
Solution
We use the value of γ ± from Example 7.9.
m eq
H +
m ◦
K w
γ 2
± m eq
OH −
/m ◦
1.008 × 10 −14
(0.960) 2 (0.00140)
7.80 × 10 −12