March 181Weak Acids Weak Bases Titration
17.3 Weak Acids Weak Bases
Titration
Titration of Weak Acid with Strong Base
Titration of Base Acid with Strong Acid
Dr. Fred Omega Garces
Chemistry 201
Miramar College
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Titration of Weak Acids and Weak Bases: Equilibria and Calculations, Study notes of Stoichiometry
A detailed explanation of the titration process for weak acids and weak bases using strong acids and bases. It includes calculations for equilibrium, buffer regions, equivalence points, and excess titrant using the Henderson-Hasselbalch equation and Ka values. The document also includes examples of titration curves for various types of acids and bases.
What you will learn
- How does the presence of a buffer affect the titration of a weak acid or weak base?
- What is the difference between titrating a weak acid with a strong base and titrating a weak base with a strong acid?
- How does the Henderson-Hasselbalch equation apply to weak acid-strong base titrations?
- What is the role of Ka values in the titration of weak acids and weak bases?
- What happens at the equivalence point of a weak acid-strong base titration?
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1 March 18
17.3 Weak Acids Weak Bases
Titration
Titration of Weak Acid with Strong Base
Titration of Base Acid with Strong Acid
Dr. Fred Omega Garces
Chemistry 201
Miramar College
2 March 18
Weak Acid (or Weak Base) with Strong Base (or strong Acid)
Experimental technique and the concept is similar to that of the titration of a
strong acid with a strong base (or vice versa) with equilibrium concept applied.
pH calculation involves 4 different type of calculations.
i)
The analyte alone
(equilibrium calculation)
ii) Buffer region
(Henderson-Hasselbach eqn)
iii) Equivalence point
(Hydrolysis)
iv)
Excess titrant
(Stoichiometric calculation)
Equilb Buffer Hydrolysis Stoic Excess
Click for simulation
5 March 18
Type i: Weak Acid with Strong Base
0 % Type 1: Calculation EQUILBRIUM
0% KOH added (TYPE 1 Weak acid calc.) 0%, V
T
= 10.0 ml
Weak acid pKa - (Type 1 Calculation)
pH of solution is determined by the dissociation of the weak acid.
HOCl + H
O!
K
a
H
O
+ OCl
i 0.4M Ex 0 0
C -x -x +x +x
e 0.4M-x Ex +x +x
K
a
โ 8
x
2
.4M โx
, 0.4M โx โ 0.4M
x = .4 3 โข 10
โ 8
( )
H 30
[ ]
= x = 1.10โข 10
M
pH = 3.
6 March 18 Weak Acids Weak Bases Titration
Titration (4ii):
Weak Acid (or Weak Base) with Strong Base (or strong Acid)
50 % Type 2: Calculation BUFFER, Henderson- Hasselbalch Eqn
50% KOH added (TYPE 2 Buffer) 50%, V T = 15.0 ml
HOCl + KOH! H
2
O
- OCl
- K
s 4mmol 2mmol Ex 0 -
R - 2 - 2 +2 +2 -
f 2mmol 0 Ex+2 2mmol -
c .133M - Ex .133M
Notice that the concentration of the acid and its conjugate are equal.
In the mass action expression these two terms cancel. pH = pKa.
Buffer situation-(Type-2)
A. Long Approach
Note that the excess 2mmol of HOCl will dissociate in water to
HOCl + H 2
O "
K a
H 3
O
- OCl
i .133M Ex 0 0.133M
C - x - x +x +x
e .133-x Ex x .133+x
The calculation is a simple equilibrium analysis-
K
a
โ 8
x ( 0.133+ x)
( 0.133 - x)
, 0.133M ยฑ x โ 0.133M
x = 3 โ 10
โ 8
,Assumption checks!
H 30
[ ]
โ 8
M
pH = 7.
B Simple Approach
Buffer solution using Henderson Hasselbalch
equation and the sRfc table above
pH = pK
a
+ log
C
b
C
a
pH = 7.5 + log
.133M
.133M
pH = 7.
8 March 18
Titration (4iii):
Weak Acid (or Weak Base) with Strong Base (or strong Acid)
100% Type III Calculation HYDROLYSIS, conjugate back to original
100% KOH added (TYPE 3 Hydrolysis):
100%, V
T
= 20.0 ml
HOCl + KOH
H
O
+ OCl
+ K
s 4mmol 4mmol Ex 0 -
R - 4 - 4 +4 +4 -
f 0 0 Ex 4mmol -
c - - Ex 0.2M -
How is the pH or pOH calculated since there are no H 3
or OH
Actually the excess HOCl does react with water to form H 3
O
Equivalence point calculation, Hydrolysis (Type-3)
Note that all the HOCl acid is neutralize by the base. HOCl
cannot dissociate in water since there is no excess. But the
conjugate base OCl- can react with water in a hydrolysis reaction
according to-
OCl
- H
O
K b
OH
- HOCl
i 0.2M Ex 0 0
C - x - x +x +x
e 0.2-x Ex x x
K
b
K
w
K
a
x
( 0.2 - x)
0.2 - x โ 0.2M
x
x = 2.58โ 10
M
OH
M, pOH = 3. 59
pH = 10.
9 March 18
Titration (4iv):
Weak Acid (or Weak Base) with Strong Base (or strong Acid)
105% Type IV Calculation Stoichiometry
105% KOH added (TYPE 4 Strong Base)
HOCl + KOH
H
2
O + OCl- + K+
s 4mmol 4.2mmol Ex 0 -
R - 4 - 4 +4 +4 -
f 0 0.2mmol Ex 4mmol -
c - 9.76โข
M
Ex .195M -
Strong Base calculation (Type-4)
Since the excess is KOH, a strong base, then the pH (or pOH in
this case) is determine by the following dissociation of KOH:
KOH
OH
- + K
i 9.76โข10-3M 0 0
C - 9.76โข10-3M
- 3 M +9.76โข - 3 M
e - +9.76โข
- 3 M +9.76โข - 3 M
[OH
]
1
M
pOH = 2.
pH = 11.
Note that the 0.195 M of OCl- will contribute negligible
amounts of OH
- as it back reacts with water in a hydrolysis
type reaction.
A simple check shows that [OH
- ] 2 is negligible.
OCl
- H 2 O
!
K b
OH
- HOCl
i 0.195 M Ex
9.76e- 3 0
C - x - x +x +x
e 0.195 M-x Ex 9.76e- 3 +x x
K
b
=
K
w
K
a
=
1 โ 10
โ 14
3 โ 10
โ 8
=
( 9. 76 โ 10
- 3
- x) x
- 195 - x ( )
, 0. 195 - x โ 0. 195 M
( 9. 76 โ 10
- 3 x + x
2 )
= 3. 33 โ 10
โ 7 ( 0. 195 )
Solve Quadratic a= 1 , b= 9. 76 โ 10
- 3 , c= โ 6. 49 โ 10 - 8 x = [OH - ] 2
= 6. 64 โ 10
โ 6 M
Therefore you see that -
[OH
โ ]
Total
= [OH
- ] 1
- [OH
- ] 2
= 2. 54 โ 10
โ 4 M + 9. 76 โ 10
โ 3
[OH
โ ]
Total
= 1. 00 โ 10
โ 2 โ 9. 76 โ 10
โ 3 M
pOH = 2. 010 pH = 11. 990
11 March 18
Titration Curve Features: WA - SB
Titration of 40.00 mL 0.1000M
HCH
CH
OOO with 0.1000 M NaOH
Weak acid - Strong base
Titration curve for a weak acid by a strong
base: 40.00mL of 0.1000M CH 3
CH 2
OOOH by
0.1000M NaOH. When exactly one-half the
acid is neutralized, [CH 3
CH 2
COOH] =
[CH 3
CH 2
COO
]
and pH = pK a
= 8.
The equivalent point is above 7.00 because the
solution contains the weak base CH 3
CH 2
COO
.
Phenolphthalein is a suitable indicator for this
titration but Methyl red is not because its
color changes over a large volume range.
12 March 18
Titration Curve Features: Monoprotic Acid
Titration curve for a
series of acids (A - F)
being titrated with a
strong base
F
A
B
C
D
E
Acid F is the strongest acid, Acid
E is the next strongest acid
followed by acid D, acid C, Acid B
and acid A. Acid A is the
weakest among the weak acid.
The Kas of each acid is
determined by reading the pH
half way to the equivalent volume
for each acid.
14 March 18
Titration Curve: Polyprotic
Titration Curve of 0.100M H
SO
with 0.100 M NaOH
Curve for the titration of a weak polyprotic
acid. Titrating 40.00mL of 0.1000M H 2
SO 3
with 0.1000M NaOH leads to a curve with two
buffer regions and two equivalence points.
Because the K a
values are separated by
several orders of magnitude, in effect the
titration curve looks like two weak acid-strong
base curves attached. The pH of the first
equivalence point is below 7 because the
solution contains HSO 3
, which is a stronger
acid than it is a base.
K a
of HSO 3
= 6.5โข
;
K b
of HSO 3
= 7.1โข
15 March 18
Titration Curve: Poly-Basic
Titration Curve of 0.10 M Na
CO
with 0.10 M HCl.
1. CO
+ H
g HCO
+ H
O
2. HCO
+ H
g H
CO
+ H
O
17 March 18
Summary
There are two main type of titration problems.
The strategy to solve them are:
1 ) SA-SB: Strong acid being titrated with a strong base (or vice versa). This is a stoichiometry type of
problem. The pH at the equivalent point = 7.0.
2a) Weak acid being titrated with a strong base. In this type of problem, the there are four sub-problems
that must be solved.
a) At 0% titrant, the problem is an equilibrium type. The Ka of the weak acid will determine the extent
of ionization and therefore the pH.
b) At 1 - 99% titrant, the problem is a buffer type. Use the Henderson-Hasselbach Equation to solve for
the pH
c) At equivalence point, this is now a hydrolysis problem. Recall that the conjugate base of the weak acid
now reacts with water to produce OH-. The solution will be basic at the equivalent point.
d) Pass the equivalence point, this is now a stoichiometry problem. The excess titrant base will dictate
the pH of the solution.
2b) Weak base being titrated with a strong acid. This is the same type of problem as (2a) above except in this
case a weak base is being neutralized by the strong acid titrant.