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Chapter 11
Reaction Equilibrium Let m(X ) x. Equation (11.13) gives m(HX) 0.200 mol/kg x and

in Nonideal Systems m(H O ) x, since the H O formed by the water ionization (11.10) is negligi-
3
3
ble compared with that from the acetic acid. Setting g 1 in (11.15), we have

x 2
1.75 10 mol>kg (11.16)
5
0.200 mol>kg x
We can solve (11.16) using the quadratic formula, but a faster method is an iter-
ative solution, as follows. Since K is much less than the acid’s stoichiometric mo-
a
lality (0.200 mol/kg), the degree of ionization will be slight and 0.200 mol/kg
x can be well approximated by 0.200 mol/kg. (In extremely dilute solutions the
degree of ionization is substantial, and this approximation cannot be made.)
2
Therefore x /(0.200 mol/kg) 1.75 10 5 mol/kg, and x 1.87 10 3 mol/kg.
With this value of x, the denominator in (11.16) becomes 0.200 mol/kg
2
0.002 mol/kg 0.198 mol/kg. Hence x /(0.198 mol/kg) 1.75 10 5 mol/kg,
and we get x 1.86 10 3 mol/kg.

Thus, with g taken as 1, we find m(H O ) m(X ) 1.86 10 3 mol/kg
3

and I 1 2 z m 1.86 10 3 mol/kg [Eq. (10.59)]. The Davies equation
2
m
j
j
j
(10.68) then gives g 0.953. Equation (11.15) becomes

2 2
10.9532 x
5
1.75 10 mol>kg
0.200 mol>kg x

Solving iteratively, as above, we get x 1.95 10 3 mol/kg m(H O ) I .
3
m
With this I , the Davies equation gives g 0.952. With this g , the equilibrium-

m

constant expression gives x 1.96 10 3 mol/kg m(H O ). This I gives
m
3
g 0.952 again, so the calculation is finished.

2

We have K g m(H O )m(OH ). The ionic strength is set by the acetic
3
w

acid ionization and is 1.96 10 3 mol/kg. The Davies equation gives the same

value for g of the pair H O and OH as for the pair H O and acetate, namely,
3

3
2
3

0.952. Hence, m(OH ) (1.00 10 14 )/[(0.952) (1.96 10 )] mol/kg
5.63 10 12 mol/kg.
An alternative solution method is to use a spreadsheet (Fig. 11.2). One sets
up the Solver to set the relative error Kerr to zero by varying x subject to the
constraint that x is between 0 and m. The initial guess for x is the value found by
hand calculation with g omitted.

A B C D E F
1 weak acid Ka = 1.75E-05 m = 0.2
2 x = 1.87E-03 I = 1.87E-03
3 g = 0.953117 Kcalc = 1.603E-05 Kerr = -0.0838
A B C D E F
1 weak acid Ka = 0.0000175 m = 0.2
2 x = 0.00187 I = =B2
3 g = =10^(-0.51*(I^0.5/(1+I^0.5)-0.3*I)) Kcalc = =g^2*x^2/(m-x) Kerr = =(D3-C1)/C1
Figure 11.2
Spreadsheet for weak-acid ionization. The lower part of this figure shows the formula view of the
spreadsheet (Sec. 5.6).