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The concepts of monoprotic acids, their equilibria, and their role in acid-base titrations. It includes discussions on strong and weak acids, buffers, and the calculation of pH using the Henderson-Hasselbalch equation. It also provides examples of acid-base reactions and their corresponding equilibrium constants.
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{
Table 6-
{
The concentration of H
will be 0.10 M.
{
The ionic strength of 0.10 M HCl is 0.10 M, at which theactivity coefficient of H
is 0.83 (Table 8-1).
{
The pH is –log
A
H+
{
pH = -log[H
]
γ
H+
= -log(0.10)(0.83)=1.
{
Because [H
][OH
{
AND pH + pOH = 14.
z
Weak acid dissociation:
z
Weak base dissociation:
Remember, a base is a proton acceptor. Just because you seeOH
-^
doesn’t imply base.
z
The conjugate base of a weak acid is a weak base.
The conjugate acid of a weak base is a weak acid.
b K
2
−
−
b
a K
−
−
a
w
b
a^
Implies
Consider a weak acid HA that has a given Ka. Find the pH of the solution:
z
What are the pertinent reactions?
z
What is the charge balance?
z
What is the mass balance?
Let’s call the formal concentration F.
z
Equilibria:
−
−
w a K
2
K
Formal concentration
is the total number of moles of a compound dissolved in a liter.
The formal concentration of a weak acid is the total amount of HA placed in the solution.
−
a
−
w
2
z
Unless the acid is very dilute or too weak.
The quadratic formula can always be used to solve weak acidproblems.
Unless the acid is very dilute or too weak.
z
However, the problem is even easier if you can neglect
x
from
the denominator.
z
This can
ONLY
be done if
x
is
MUCH
smaller than [HA].
How do I know if
x
is much smaller than [HA]?
z
Make the approximation and solve the problem.
z
If your answer supports your assumption then your answer is fine.
z
Suppose you are given that [HA] is 0.1 M and you find [A
-^ ] to be
1x
, then you are safe to say that
x
2
[B]
]
OH
][
BH [
K
OH
BH
O
H
B
b K
2
−
−
=
→
←
Find the pH of 0.10 M ammonia. {It’s not 13.} z
Pertinent reactions:
3
2
4
2
z
Woops, we have no K
b
tables in our text.
But, K
a
x K
b
w
, and Ka for NH
4
is listed in our table at the back of
the book.
b
w
a
z
To find the pH of 0.10 M NH
3
, we set up and solve the equation
If we let x = [NH
4
], the x also = [OH
-^ ] through stoichiometry.
Also, [NH
3
] = F – x where F = 0.10 M.
5
(^00).
14
w a
b
3
4
10
75 . 1
10
70 . 5
10
K K
K
]
[NH
]
][OH
[NH
10
−
−
−
× = × = = =
−
5
2
10
75 . 1
1 . 0
F
)
)(
(
−
×
=
−
=
x
x
x x
x
Find the pH now that we know [OH
= x
= 1.31 x
pOH = -log(1.31 x 10
pH = 14.00 – pOH = 14.00 – 2.88 = 11.
z
The solution is less basic than if the ammonia was totally dissociated.
