332
7 Chemical Equilibrium
and
m eq (A
− )
n s
1.000 kg
(7.5-2)
Assuming that γ HA equals unity, we obtain
K a
γ(H + )(m eq (H + )/m ◦ )γ(A
− )(m eq (A
− )/m ◦ )
m eq (HA)/m ◦
a(H + )γ(A
− )m eq (A
− )
m eq (HA)
a(H + )γ(A
− )n s
n a
(7.5-3)
If γ(A
− ) is also assumed to equal unity, this equation can be written in the form known
as the Henderson–Hasselbalch equation:
pH pK a + log 10
m s
m a
pK a + log 10
n s
n a
(7.5-4)
where pK a is defined by
pK a − log 10 (K a ) (definition of pK a )
(7.5-5)
and where m s and m a are the stoichiometric molalities of the salt and the acid. If
greater accuracy is desired than this equation provides, one can use Eq. (7.5-3) with
experimental or theoretical values of γ(A − ).
E X A M P L E 7.16
a. Using the Henderson–Hasselbalch equation, calculate the pH of a solution made from
0.400 mol of acetic acid and 0.600 mol of sodium acetate in 1.000 kg of water.
b. Repeat the calculation of part a using Eq. (7.5-3) and the Davies equation.
Solution
a. Using the value K a 1.75 × 10 −5 ,
pH 4.757 + log 10
0.600 mol
0.400 mol
4.932
b. Let m a be the molality of acid that would occur if the acid did not ionize and m s be
the molality of the anion that would occur if the salt did not hydrolyze. We assume that
the salt dissociates completely, that γ(H + ), γ(A − ), and γ(OH − ) are equal and can be
approximated by the Davies equation, and that γ(HA) 1. The ionization of the acid is
represented by
HA H + + A −
7 Chemical Equilibrium
and
m eq (A
− )
n s
1.000 kg
(7.5-2)
Assuming that γ HA equals unity, we obtain
K a
γ(H + )(m eq (H + )/m ◦ )γ(A
− )(m eq (A
− )/m ◦ )
m eq (HA)/m ◦
a(H + )γ(A
− )m eq (A
− )
m eq (HA)
a(H + )γ(A
− )n s
n a
(7.5-3)
If γ(A
− ) is also assumed to equal unity, this equation can be written in the form known
as the Henderson–Hasselbalch equation:
pH pK a + log 10
m s
m a
pK a + log 10
n s
n a
(7.5-4)
where pK a is defined by
pK a − log 10 (K a ) (definition of pK a )
(7.5-5)
and where m s and m a are the stoichiometric molalities of the salt and the acid. If
greater accuracy is desired than this equation provides, one can use Eq. (7.5-3) with
experimental or theoretical values of γ(A − ).
E X A M P L E 7.16
a. Using the Henderson–Hasselbalch equation, calculate the pH of a solution made from
0.400 mol of acetic acid and 0.600 mol of sodium acetate in 1.000 kg of water.
b. Repeat the calculation of part a using Eq. (7.5-3) and the Davies equation.
Solution
a. Using the value K a 1.75 × 10 −5 ,
pH 4.757 + log 10
0.600 mol
0.400 mol
4.932
b. Let m a be the molality of acid that would occur if the acid did not ionize and m s be
the molality of the anion that would occur if the salt did not hydrolyze. We assume that
the salt dissociates completely, that γ(H + ), γ(A − ), and γ(OH − ) are equal and can be
approximated by the Davies equation, and that γ(HA) 1. The ionization of the acid is
represented by
HA H + + A −