Physical Chemistry Third Edition

(C. Jardin) #1

7.6 The Temperature Dependence of Chemical Equilibrium. The Principle of Le Châtelier 335


b. Neglect the ionization of the acid and the hydrolysis of
the salt but use the Davies equation to estimate activity
coefficients.

7.42 Consider an acetic acid/acetate buffer solution made from
0.040 mol of acetic acid and 0.060 mol of sodium acetate
in 1.000 kg of water and maintained at 298.15 K.
a. Find the pH of the buffer, using the Henderson–
Hasselbalch equation.
b. Find the pH of the buffer using the Davies equation to
estimate activity coefficients.
c.Find the change in pH if 0.010 mol of solid sodium
hydroxide is added.
d.Find the change in pH if 0.010 mol of solid sodium
hydroxide is added to a solution of hydrochloric acid
and sodium chloride in 1.000 kg of water if the solution
has the same pH and ionic strength as the buffer
solution in part a.


7.43 How much solid NaOH must be added to 0.100 mol of
cacodylic acid in 1.000 kg of water at 298.15 K to make a
buffer solution with pH equal to 7.00? The acid
dissociation constant is equal to 6. 4 × 10 −^7.


a.Use the Henderson–Hasselbalch equation.
b.Use Eq. (7.5-3) and the Davies equation.
7.44 The two acid ionization constants ofortho-phthalic acid
areK 1  1. 3 × 10 −^3 andK 2  3. 9 × 10 −^6. Find the pH
of a solution made from 0.100 mol ofortho-phthalic acid,
0.100 mol of potassium hydrogen phthalate, 0.100 mol of
dipotassium phthalate, and 1.000 kg of water.
7.45 a.Find the value of the equilibrium constant at 298.15 K
for the reaction

NH 3 (aq)+H 2 O(l)NH+ 4 (aq)+OH−

b.0.1000 mol of NH 3 and 0.1500 mol of NH 4 Cl are
dissolved in 1.000 kg of H 2 O at a constant temperature
of 298.15 K. Write an equation analogous to the
Henderson–Hasselbalch equation that applies to this
case. Find the equilibrium molality of OH−, the pOH
(defined analogously to the pH), and the pH.
c.Use the Davies equation to estimate activity
coefficients and find the pOH. Neglect the ionization of
NH 3 and the hydrolysis of NH+ 4.

7.6 The Temperature Dependence of Chemical

Equilibrium. The Principle of Le Châtelier
We can calculate∆Gonly for a process that begins and ends at the same tempera-
ture. However, we can obtain the temperature dependence of∆Gfor such processes
considered at different temperatures. By differentiation of Eq. (7.1-13) we obtain

d∆G◦
dT



∑c

i 1

vi

(

∂μ◦i
∂T

)

P,n

−

∑c

i 1

viS

i−∆S

◦ (7.6-1)

We can also write

d(∆G◦/T)
dT

−

∆S◦

T


∆G◦

T^2

−

∆H◦

T^2

(7.6-2)

Exercise 7.15
Show that
d

(
∆G◦/T

)

d(1/T)
∆H◦ (7.6-3)
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