Physical Chemistry Third Edition

320 7 Chemical Equilibrium

Since the solution is dilute we assume thatγ(H 2 O)x(H 2 O)≈1 and thatγ(NH 3 )≈1. We

introduce the mean ionic activity coefficient defined in Eq. (6.4-6):

γ±

[

γ(NH+ 4 )γ(OH−)

] 1 / 2

We assume that the contribution of OH−from the ionization of water is negligible and let

x

meq(NH+ 4 )

m◦



meq(OH−)

m◦

so that

1. 80 × 10 −^5 

γ±^2 x^2

0. 100 −x

We use successive approximations. For our first approximation we assume thatγ±≈1:

x^2 (1. 80 × 10 −^5 )(0. 100 −x)

x^2 +(1. 80 × 10 −^5 )x− 1. 80 × 10 −^6  0

From the quadratic formula

x

− 1. 80 × 10 −^5 ±

√(

1. 80 × 10 −^5

) 2

+ 4

(

1. 80 × 10 −^6

)

2

 0. 00133

We disregard a negative root to the quadratic equation, which is physically impossible. It is

always the case in a chemistry problem that one of the solutions to a quadratic equation is

unusable. For our next approximation we estimate the mean ionic activity coefficient for this

molality using the Davies equation. Since NH+ 4 and OH−are the only ions present, the first

estimate of the ionic strength is 0.00133 mol kg−^1

log 10 (γ±)− 0. 510

( √

0. 00133

1 +

√

0. 00133

−(0.30)(0.00133)

)

− 0. 01774

γ± 10 −^0.^01774  0. 960

1. 80 × 10 −^5 

(0.960)^2 x^2

0. 100 −x

Use of the quadratic formula on this equation yields x 0 .00140, or meq(NH+ 4 )

meq(OH−) 0 .00140 mol kg−^1. A further approximation produces only a small change, so

we stop at this approximation. Even for the low ionic strength of 0.00140 mol kg−^1 , neglect

of the activity coefficient has produced an error of about 5%.

The Ionization of Water

Water ionizes according to the chemical equation

H 2 O
H++OH−