Homework
Fundamentals of
Analytical Chemistry
3,
99-15 odd, 18, 19, 21, 25, 27, 32, 34, 38,
43, 44, 45
Problems in green may or may not be part of
the assignment, depending upon the material
that is covered.
Chapter 14
Principles of Neutralization
Titrations
Indicators
Acid/Base
indicators
Acid form has a different color than the base
form of the substance
• 10x more acid (or greater) – acid color
• 10x more base (or greater) – base color
• Between these limits – color a combination
For ideal behavior, need transition through
colors quickly
• Weak acid/base substances
• From HendersonHenderson-Hasselbalch,
Hasselbalch, ∆pH = 2 for
transition
Indicators
Titration
• Minimize by using low concentration of indicator
any titration, we can consider three
phases
Before the equivalence point, the analyte is in
excess
At the equivalence point, the analyte and
titrant are present in stoichiometrically
equivalent amounts.
After the equivalence point, the titrant is in
excess
MUST BE highly colored for this to work
Ideally, less than 0.02 mL of titrant to cause change
• Can also correct with blank
Table
1414-1
List of indicators
pKa can be used to chose correct indicator
• Match pKa with pH of equivalence point
Titrations
Strong Acid/Strong Base Titrations
In
Error
Titrant reacts with analyte and indicator
To
determine pH, we have to look at the
composition of the solution after the
reaction has occurred.
I – starting conditions
• How much stuff do we have before any reaction
occurs
∆ – stoichiometric relationships
• HOW will the system react
• Must follow the stoichiometry
F – final conditions
• What’
What’s in solution after the reaction has gone to
completion
1
50.0 mL of 0.10 M HCl is titrated
with 0.10M NaOH
Strong Acid/Strong Base Titrations
Consider the following titration:
50.0 mL of
0.10 M HCl is titrated with 0.10M NaOH
Before any titrant is added:
• Solution is just the strong acid (HCl
(HCl))
• pH = -log 0.10
• pH = 1.00
50.0 mL of 0.10 M HCl is titrated
with 0.10M NaOH
After 10.00
mL of titrant is added:
HCl + NaOH H2O + Na+ +
I 5.00 1.00
~~
0
∆ -1.00 -1.00
~~
~~
F 4.00
~0
~~
~~
pH = -log (4.00/60) = 1.18
Cl0
~~
~~
50.0 mL of 0.10 M HCl is titrated
with 0.10M NaOH
After 45.00
mL of titrant is added:
HCl + NaOH H2O + Na+ +
I 5.00 4.50
~~
0
∆ -4.50 -4.50
~~
~~
F 0.50
~0
~~
~~
pH = -log (0.50/95) = 2.28
Cl0
~~
~~
After 5.00 mL of titrant is added:
HCl + NaOH H2O + Na+ + ClI 5.00 0.50
lots
0
0
∆ -0.50 -0.50
~~
~~
~~
F 4.50
~0
~~
~~
~~
pH = -log (4.50/55) = 1.09
[NOTE: ~~ means we don’
don’t care because these
substances do not affect the pH of the solution]
50.0 mL of 0.10 M HCl is titrated
with 0.10M NaOH
After 25.00
mL of titrant is added:
HCl + NaOH H2O + Na+ +
I 5.00 2.50
~~
0
∆ -2.50 -2.50
~~
~~
F 2.50
~0
~~
~~
pH = -log (2.50/75) = 1.48
Cl0
~~
~~
50.0 mL of 0.10 M HCl is titrated
with 0.10M NaOH
After 49.98
mL of titrant is added:
HCl + NaOH H2O + Na+ +
I 5.00 4.998
~~
0
∆ -4.998 -4.998
~~
~~
F 0.002
~0
~~
~~
pH = -log (0.002/99.98) = 4.70
Cl0
~~
~~
2
50.0 mL of 0.10 M HCl is titrated
with 0.10M NaOH
After 50.0
After 50.02
mL of titrant is added:
50.0 mL of 0.10 M HCl is titrated
with 0.10M NaOH
mL of titrant is added:
Strong Acid/Strong Base Titrations
Summary
HCl + NaOH H2O + Na+ +
I 5.00 7.00
~~
0
∆ -5.00 -5.00
~~
~~
F ~0
2.00 ~~
~~
pOH = -log (2.00/120) = 1.78
pH = 14.00 – 1.78 = 12.22
mL of titrant is added:
HCl + NaOH H2O + Na+ + ClI 5.00 5.002
~~
0
0
∆ -5.00 -5.00
~~
~~
~~
F ~0
0.002 ~~
~~
~~
pOH = -log (0.002/100.02) = 4.70
pH = 14.00 - 4.70 = 9.30
HCl + NaOH H2O + Na+ + ClI 5.00 5.00
lots
0
0
∆ -5.00 -5.00
~~
~~
~~
F ~0
~0
lots
~~
~~
What do we have affecting the pH?
After 70.0
50.0 mL of 0.10 M HCl is titrated
with 0.10M NaOH
Cl0
~~
~~
Prior to the equivalence point – excess
analyte
• pH determined by analyte (strong acid or strong
base)
At the equivalence point
• Nothing causing a ∆pH from neutral
• pH = 7.00
• ONLY FOR STRONG ACID/BASE TITRATION!!!
TITRATION!!!
Past the equivalence point – excess titrant
• pH determined by titrant
Titration Curve
Weak Acid/Strong Base Titration
pH
Strong acid/base
Consider the following titration:
50.0 mL of
a 0.10 M solution of a weak acid (HA, Ka =
1.0 x 10-5) titrated with 0.10 M NaOH.
NaOH.
14
12
10
8
6
4
2
Before any titrant
0
20
40
60
Volume
80
100
is added
Only a weak acid in solution
[ H + ] = K a C HA = (0.10)(1.0 x10−5 )
[ H + ] = 1.0 x10 −3 ; ∴ pH = 3.00
3
50.0 mL of a 0.10 M solution of a weak acid (HA,
Ka = 1.0 x 10-5) titrated with 0.10 M NaOH.
NaOH.
After 5.00
mL of titrant is added
Weak Acid/Strong Base Titration
Differences from strong acid/base
HA + NaOH A- + Na+ + H2O
I 5.00 0.50
0
~~ ~~
∆ -0.50 -0.50
+0.50 ~~ ~~
F 4.50 ~0
0.50 ~~ ~~
Buffer, pH = pKa + log b/a
pH = 5.00 + log 0.50/4.50 = 4.05
titration
Cannot ignore the products
• Will always generate the conjugate base of the
analyte
• This produces a buffer solution!
Reverse situation (weak base/strong acid)
• Will always generate the conjugate acid of the
analyte
• This produces a buffer solution!
Will always titrate
with either a strong acid
or a strong base
50.0 mL of a 0.10 M solution of a weak acid (HA,
Ka = 1.0 x 10-5) titrated with 0.10 M NaOH.
NaOH.
After 25.0
mL of titrant is added
HA + NaOH A- + Na+ + H2O
I 5.00 2.50
0
~~ ~~
∆ -2.50 -2.50
+2.50 ~~ ~~
F 2.50 ~0
2.50 ~~ ~~
Buffer, pH = pKa + log b/a
pH = 5.00 + log 2.50/2.50 = 5.00
50.0 mL of a 0.10 M solution of a weak acid (HA,
Ka = 1.0 x 10-5) titrated with 0.10 M NaOH.
NaOH.
After 50.0
mL of titrant is added
Weak Acid/Strong Base Titration
For this titration,
25.0 mL gets us ½ way to
the equivalence point
• Note that with a weak base/strong acid titration we
still have pH = pKa (and not pKb)
Weak Acid/Strong Base Titration
At
HA + NaOH A- + Na+ + H2O
I 5.00 5.00
0 ~~ ~~
∆ -5.00 -5.00
+5.00 ~~ ~~
F ~0
~0
5.00 ~~ ~~
Only a weak base in solution
At this point in any weak/strong titration, we
have the same amount of the conjugate acid
and the conjugate base in solution.
At this point pH will ALWAYS equal pKa!
the equivalence point
pH does not equal 7.00
• Because the conjugate base is in solution, the pH
is driven by this component and will be basic
• For a weak base/strong acid titration, you will have
the conjugate acid in solution at the equivalence
point, and the pH will be acidic
1.0 x10 −14 5.00
∗
1.0 x10 −5 100
−
−6
[OH ] = 7.1x10 ; pOH = 5.15; pH = 8.85
[OH − ] = K b C B =
4
50.0 mL of a 0.10 M solution of a weak acid (HA,
Ka = 1.0 x 10-5) titrated with 0.10 M NaOH.
NaOH.
After 70.0 mL of titrant is added
Weak Acid/Strong Base Titration
Past
HA + NaOH A- + Na+ + H2O
I 5.00 7.00
0
~~
~~
∆ -5.00 -5.00
+5.00 ~~
~~
F ~0
2.00
5.00 ~~
~~
Mixture of a weak base and a strong base
Ignore the weak base; [OH-] = 2.00/120
pOH = 1.78; pH = 12.22
the equivalence point
We will assume (and usually properly) that the
contribution to the hydroxide concentration
from the weak base is negligible relative to
the contribution from the strong base
LeChatlier’
LeChatlier’s Principle
Titration Curve
Strong Acid/
Strong Base
Titration Curve
0
20
40
60
Volume
80
100
80
100
Titration Curve
0
20
40
60
Volume
80
100
Weak Acid/
Strong Base
Titration Curve
Weak Acid - Strong Base
pH
pH
Weak Acid - Strong Base
14
12
10
8
6
4
2
pH
Strong acid/base
Titration Curve
14
12
10
8
6
4
2
14
12
10
8
6
4
2
0
20
40
60
Volume
5