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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?

Typology: Study notes

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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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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

โˆ’ 8

x

2

.4M โˆ’x
, 0.4M โˆ’x โ‰ˆ 0.4M
x = .4 3 โ€ข 10

โˆ’ 8

( )

H 30

[ ]

= x = 1.10โ€ข 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

โˆ’ 8

x ( 0.133+ x)

( 0.133 - x)

, 0.133M ยฑ x โ‰ˆ 0.133M
x = 3 โ‹… 10

โˆ’ 8

,Assumption checks!
H 30

[ ]

โˆ’ 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‰ˆ 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 - 9.76โ€ข

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 9.76โ€ข10-3M 0 0

C - 9.76โ€ข10-3M
  • 3 M +9.76โ€ข - 3 M

e - +9.76โ€ข

  • 3 M +9.76โ€ข - 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

โˆ’ 14

3 โ‹… 10

โˆ’ 8

=

( 9. 76 โ‹… 10

  • 3
    • x) x
  1. 195 - x ( )

, 0. 195 - x โ‰ˆ 0. 195 M

( 9. 76 โ‹… 10

  • 3 x + x

2 )

= 3. 33 โ‹… 10

โˆ’ 7 ( 0. 195 )

Solve Quadratic a= 1 , b= 9. 76 โ‹… 10

  • 3 , c= โˆ’ 6. 49 โ‹… 10 - 8 x = [OH - ] 2

= 6. 64 โ‹… 10

โˆ’ 6 M

Therefore you see that -

[OH

โˆ’ ]

Total

= [OH

  • ] 1
  • [OH
  • ] 2

= 2. 54 โ‹… 10

โˆ’ 4 M + 9. 76 โ‹… 10

โˆ’ 3

[OH

โˆ’ ]

Total

= 1. 00 โ‹… 10

โˆ’ 2 โ‰ˆ 9. 76 โ‹… 10

โˆ’ 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

= 6.5โ€ข

;

K b

of HSO 3

= 7.1โ€ข

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.