•
The greater the acid strength of HA, the less the base strength of A

6
Formal
Concentration
Formal Concentration (F):
•
total number of moles of a compound dissolved in a liter.
•
For a weak acid, it is the total amount of HA placed in the solution regardless of
how much has been converted to A

7
Formal concentration and weak acid
8
HA
⇄
A

+ H
+
HA
A

H
+
Initial Concentration
F
0
0
Final Concentration
Fx
x
x
Example:
Consider the composition of a solution containing 0.10 moles of weak acid in 1.0 L
of water.
We can say F = 0.10 M even if we do not know how much of the conjugate base
was actually formed.
Since it is a weak acid, it is only partially dissociated. At
equilibrium, the concentration of HA is less than the formal concentration.
From the mass balance:
F = 0.10 M = [HA] + [A

]
Let x = [A

]
F = [HA] + x
[HA] = F  x
pH depends on amount of x
Calculate pH of Weak Acid
9
•
We set up and solve the equation
where [H
+
] = [A

] =
x
, and
[HA] = F 
x
.
•
Solve for x by solving quadratic equation
HA
A

H
+
Initial Concentration
F
0
0
Final Concentration
Fx
x
x
HA
⇄
A

+ H
+
𝐾
a
=
H
+
A
−
[HA]
=
𝑥
𝑥
F − 𝑥
=
𝑥
2
F − 𝑥
Note:
For reasonable concentrations of weak acids (of reasonable strength), [H
+
] from
dissociation of acid >> contributions from dissociation of water
For every mole of HA
dissociated to A

, one
mole of H
+
is formed,
thus [H
+
] = [A

] =
x
,
Problem 1: Typical Weak Acid Problem
CH
3
COOH
CH
3
COO

H
3
O
+
Initial Concentration
F
0
0
Final Concentration
F  x
= 0.50
 x
x
x
CH
3
COOH + H
2
O
⇄
H
3
O
+
+ CH
3
COO

ax
2
+ bx + c = 0
x =
−?±
?
2
−4??
2?
10
Problem:
Calculate the pH of a solution of acetic acid (weak acid) if its formal
concentration is 0.50M given its K
a
is 1.75 x 10
5
.
K
a
=
[H
3
O
+
] [CH
3
COO
−
]
[CH
3
COOH]
= 1.75 x 10
5
K
a
=
(x)(x)
0.50
− x
= 1.75 x 10
5
x
2
+ (1.75 x 10
5
)(x)  0.875 x 10
5
= 0
x = 2.94 x 10
3
M = [H
+
]
pH = log[H
+
] = log(2.94 x 10
3
) = 2.53
Solution:
F = 0.50 M = [HA] + [A

]
Let x = [A

]
F = [HA] + x
[HA] = F
–
x = 0.50  x
Approximation:
Simplification of weakacid problem
11
Can always solve quadratic equation….
However, can first try to neglect x in the denominator (i.e. neglect x in F  x)
If x
≤ 1% of F (i.e. x
is 100 times ≤
F), then approximation is valid.
∴
As a simple rule:
•
If K < (10
4
)(F), then F  x
F (try to neglect x)
•
If K
≥ (
10
4
)(F), solve using the quadratic equation to reduce error on
calculation
Problem 2:
Weak Acid Dissociation  Simplification
Problem:
Calculate the pH of a 0.050 M trimethylammonium chloride solution.
12
Solution:
The pK
a
for trimethylammonium ion is 9.799 (see APPENDIX B).
Screen shot of ebook:
VitalSource Technologies LLC
13
(CH
3
)
3
NHCl
⟶
(CH
3
)
3
NH
+
+ Cl

Assume this chloride salt is completely dissociated. Cl

essentially has no acidic or
basic properties.
(CH
3
)
3
NH
+
⇄
(CH
3
)
3
N + H
+
pK
a
= logK
a
K
a
= 10
pKa
= 10
9.799
= 1.59x10
10
Since K
a
< (10
4
)(F): 1.59 x 10
10
< (10
4
)(0.050), try the assumption F  x
≅ 𝐹
(CH
3
)
3
NH
+
(CH
3
)
3
N
H
+
Initial Concentration
F
0
0
Final Concentration
F
–
x
≅
F = 0.050
x
x
Problem 2:
Weak Acid Dissociation
–
Simplification
(cont’d)
K
a
=
[H
+
] [(CH
3
)
3
N]
[(CH
3
)
3
NH
+
]
= 1.59 x 10
10
K
a
=
(x)(x)
0.050
= 1.59 x 10
10
x
2
 7.95 x 10
12
= 0
x =
7.95 𝑥 10
−12
= 2.8195 x 10
6
M = [H
+
]
pH = log[H
+
] = log(2.8195 x 10
6
) = 5.55
14
Determine if assumption is valid:
Is 100(x) ≤ F?
You've reached the end of your free preview.
Want to read all 108 pages?
 Fall '15
 CameronSkinner
 pH