Page 354 - Physical Chemistry

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Section 11.3
Exercise Reaction Equilibrium
Set up the spreadsheet of Fig. 11.2 and solve for x. in Electrolyte Solutions
Exercise

Find m(H O ) in 1.00 mol/kg HC H O (aq) at 25°C. (Answer: 4.49 10 3
3 2 3 2
mol/kg.)

In Example 11.1, I is quite low so it doesn’t make much difference whether we
m
include g . This is not true in Example 11.2.

EXAMPLE 11.2 Buffer solution

Find m(H O ) in a 25°C aqueous solution with the stoichiometric molalities
3
[Eq. (10.48)] m(HC H O ) 0.100 mol/kg and m(NaC H O ) 0.200 mol/kg
2 3 2 2 3 2
(a buffer solution).
The salt NaC H O (which is a 1:1 electrolyte) exists virtually entirely in the
2 3 2
form of positive and negative ions in solution. Also, the ionization of HC H O
2 3 2
will contribute little to I in comparison with the contribution from the
m
NaC H O . Therefore, I 1 2 (0.200 0.200) mol/kg 0.200 mol/kg. The
3
m
2
2
Davies equation (10.68) then gives g 0.746. Substitution into (11.15) gives

2
g m1H O 2m1C H O 2

3
3
2
2
5
1.75 10 mol>kg
m1HC H O 2
2
2
3

2
10.7462 m1H O 210.200 mol>kg2
3

0.100 mol>kg

where m(C H O ) is well approximated by considering only the acetate ion from
2 3 2
the NaC H O and where we set m(HC H O ) equal to 0.100 mol/kg since
2 3 2 2 3 2
the degree of ionization of acetic acid is far less than in the previous example,
because of the added sodium acetate (common-ion effect). Solving, we find

m(H O ) 1.5 10 5 mol/kg. Note that, if g were omitted in this example,
3 7

we would get 8.7 10 6 mol/kg for m(H O ), which is in error by a whopping
5 3
44%. Except for solutions of quite low ionic strength, ionic equilibrium calcula-
tions that omit activity coefficients are likely to give only qualitatively correct
answers. Many of the ionic equilibrium calculations you did in first-year
chemistry are correct only in the exponent of 10, because of neglect of activity
coefficients (and neglect of ion-pair formation in salt solutions).
Exercise

Find m(H O ) in a 25°C aqueous solution with the stoichiometric molalities
3
m(HC H O ) 1.00 mol/kg and m(NaCl) 0.200 mol/kg. (Answer: 5.60
2 3 2
10 3 mol/kg.)
EXAMPLE 11.3 Very dilute weak-acid ionization

For HOI, K 2.3 10 11 mol/kg in water at 25°C. Find m(H O ) in a 1.0
a 3
10 4 mol/kg 25°C aqueous solution of HOI.
We have

HOI H O ∆ H O OI (11.17)
2
3

2
K g m1H O 2m1OI 2>m1HOI2 (11.18)

3
a