324 7 Chemical Equilibrium
We must solve two simultaneous equations. At 298.15 K the equilibrium relation for the acetic
acid is:
Ka 1. 75 × 10 −^5
γ
(
H+
)
meq
(
H+
)
/m◦γ
(
A−
)
meq
(
A−
)
/m◦
γ(HA)(meq(HA)/m◦)
(7.3-11a)
The equilibrium relation for water is
Kw 1. 0 × 10 −^14
(
γ(H+)meq(H+)/m◦
)(
γ(OH−)meq(OH−)/m◦
)
(7.3-11b)
We assume thatγ(HA)≈1. Since all of the ions are univalent, we assume that all of their
activity coefficients are equal and denote them byγ. We let
z
meq
(
H+
)
m◦
, y
meq
(
OH−
)
m◦
We express the other molalities in terms ofzandy. For each OH−ion, there is one H+ion
released from water, and for each A−ion there is an H+ion released from the acid, so that
meq
(
A−
)
m◦
z−y,
meq(HA)
m◦
m
m◦
−(z−y)
wheremis the stoichiometric molality. From Eq. (7.3-11b),
y
Kw
γ^2 z
We use this expression to replaceyin the acid equilibrium expression and obtain a single
equation:
Ka
γ^2 z
[
z−Kw/
(
γ^2 z
)]
m/m◦−z+Kw/
(
γ^2 z
) (7.3-12a)
When we multiply this equation out, we obtain a cubic equation:
γ^2 z^3 +Kz^2 −(m/m◦)K+Kwz−KKw/γ^2 0 (7.3-12b)
For our first approximation, we assume thatγ1 and solve forz. A numerical approximation
to the root of this cubic equation givesmeq(H+) 1. 61 × 10 −^7 mol kg−^1 , only 61% higher
than in pure water. Use of the Davies equation for this ionic strength givesγ± 0. 999999
so that no second approximation is necessary. The value of the pH is 6.79. If the hydrogen
ions from water are ignored, the result is thatmeq(H+) 9. 9 × 10 −^8 mol kg−^1 , which is in
error by 38% and is smaller than the value ofm(H+) in pure water.
Another situation that requires the solution of simultaneous equations is the ioniza-
tion of a polyprotic acid, which means that one molecule of the acid ionizes successively
to give more than one hydrogen ion.
Exercise 7.10
Find the pH of a solution made from 1.000 kg of water and 0.100 mol of phosphoric acid, H 3 PO 4 ,
for which the three acid dissociation constants are