CHEMISTRY 205 LECTURE
EXAM III Material
Chapter 10
PART 1 ACID - BASE CHEMISTRY
I. Calculation of pH, pOH, [H+], and [OH-]
Four Major Types of Systems:
A. Strong Acids and bases
a. 0.0001 M HCl
b. 25.0 ml 0.010 M HCl + 50.0 ml 0.010 M NaOH
B. Weak Acid or Weak Base
1. Weak Acid
a. Calculate the pH of a 0.10 F HAc solution Ka = 1.8 x 10 -5
HC2H3O2 + H2O  H3O+
+
C2H3O2Initial
Change
at Equilibrium
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b. Calculate the pH of a 0.10 F HClO2 solution
Initial
Change
at Equilibrium
(1) Method I - Quadratic
(2) Method II - Sucessive Approximation
C. Salts
1. Salts of Strong Acids and Bases
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2. Salts of Weak Acids and Bases
a Salt of a Weak Base
Calculate the pH of a 0.10M NH4Cl solution
Initial
Change
at Equilibrium
b Salt of a Weak Acid
Calculate the pH of 0.10 M KC2H3O2 solution
Initial
Change
at Equilibrium
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c.
Salt of a weak acid and weak base
D. Buffers
Weak Acid or Weak Base with its salt -Common Ion effect (more, latter)
Calculation of pH, pOH, [H+], and [OH-]
Four Major Types of Systems:
A. Strong Acids and bases
B. Weak Acid or Weak Base
C. Salts
a. Salt of a Strong Acid or Base
b. Conjugated Acid or Base - Salt of a Weak Acid or Base
D. Buffers - Weak Acid or Weak Base with its salt -Common Ion effect
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II. Fraction of Dissociation,
 =
x
[A  ]
= x+(F-x)

[ A ]  [HA]
x
= F
ex. Calculate the  for 0.10 M HClO2 (see page 2 "b") for [H+]
III Buffers
A. Background:
1. Resist Changes in pH
2. A mixture of conjugated species
Consider a mixture of HC2H3O2
+ NaC2H3O2
1. Buffering systems:
a. A Weak acid with its salt (a salt that contains the conjugated base of the weak acid)
Example: HClO and NaClO
b. A Weak base with its salt (a salt that contains the conjugated acid of the weak base)
Example: NH3 and NH4Cl
c. Acid Salts
Example: NaHCO3
d. A Salt of a weak acid & weak base
Example: NH4C2H3O2
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B. Buffer Calculations:
1. Calculate the [H3O+] for:a solution that is 0.500 M acetic acid and
sodium acetate. Ka = 1.8 x 10-5
HC2H3O2 + H2O 
H3O+
Initial
Change
at Equilibrium
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+
C2H3O2-
0.500 M
2. Calculate the pH when 25.0 ml of 0.150 M HCl is added to 40.0 ml of 0.200 M NH3 Kb = 1.8 x 10-5
Reaction:
Moles Before Reaction
Change
Moles After Reaction
Molarity After Reaction
TYPE 1 Calculation:
Equilibrium:
Initial
Change
at Equilibrium
TYPE 2 Calculation:
Henderson - Hasselbalch equation:
[H+][A-]
Ka= [HA]
[H+][A-]
log Ka = log [HA]
Derivation:
[A-]
log Ka = log [H+] + log [HA]
- log [H+]
 pH
Therefore:
[A-]
= -log Ka + log [HA]
pKa
[A-]
pH = pKa + log [HA]
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3 a. How many grams of NH4Cl must be added to 1 - Liter of a 0.10 M solution of NH3(aq) to adjust the pH
to 9.000. Kb = 1.8 x 10-5
NH3 + H2O 
NH4+ + OH-
Initial
Change
at Equilibrium
5.
A Strong acid is added to a salt to form a Buffer Solution
a. How many moles of a strong monoprotic acid must be added to 1-L of a 0.40 M
sodium formate, NaCO2H, solution to prepare a buffer of a pH = 4.35
Ka = 1.77 x 10-4
Reaction:
Equilibrium Eqn:
Initial
Change
at Equilibrium
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CHAPTER 11
PART II POLYPROTIC ACID-BASE EQUILIBRIA
I Calculation of the pH of diprotic and dibasic acids and bases
Diprotic acids are acids which can donate more than one protron whoses salts can accept more than one
Ex.
A. Background:
H2A  H+ + HAHA-  H+ + A2-
Ka1
Ka2
A2- + H2O  HA- + OHHA- + H2O  H2A + OH-
Kb1
Kb2
H2A  H+ + HAHA- + H2O  H2A + OH-
Ka1
Kb2
and
B. Acidic, Basic and Intermediate Forms
Acidic Form`
Basic form
Page 9
Intermediate Form
proton.
1.. Acidic Form
Calculation of pH of 0.10 M H2CO3
Equilibrium Eqn:
Initial
Change
at Equilibrium
2. Basic Form
Calculation of pH of 0.10 M Na2CO3
Equilibrium Eqn:
Initial
Change
at Equilibrium
Page 10
2. Intermediate Form
HA-  H+ + A2HA- + H2O  H2A + OH-
Ka2
Kb2
Page 11
Page 12
C. Amino Acids
Background
Calculations:
Glycine where R = H
1. Calculate the pH of 0.10 M Glycine as H2A+ (Acidic form)
Equilibrium Eqn:
Initial
Change
at Equilibrium
Page 13
2. Calculate the pH of 0.10 M Glycine as HA (neutral form)
Equilibrium Eqn:
Initial
Change
at Equilibrium
3. Calculate the pH of 0.10 M Glycine as A-(Basic Form)
Equilibrium Eqn:
Initial
Change
at Equilibrium
Page 14
Chapter 12
PART II NEUTRALIZATION TITRATIONS
I. Titration curve for a strong acid titrated with a strong base
50.0 mls of 0.10M HCl is titrated with 0.10 M NaOH
RXN: HCl + NaOH ---> NaCl + H2O
NOTE: Only for a strong acid-strong base titration the pH = 7 at the endpoint(equivalence, really)!
Construction of Titration Curve
Calculation of the volume of base needed to reach the equivalence point
0 ml NaOH - BEFORE ADDITION
pH calculation
25.00 ml NaOH = BEFORE THE EQUIVALENCE POINT
pH calculation
49.99 ml NaOH = BEFORE THE EQUIVALENCE POINT
pH calculation
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50.00 ml NaOH =
EQUIVALENCE POINT
pH calculation
75.00 ml NaOH = AFTER THE EQUIVALENCE POINT
pH calculation
Graph
Page 16
II. Titration curve for a strong base titrated with a strong acid
50.0 mls of 0.10M KOH is titrated with 0.10 M HCL
RXN: HCl + KOH ---> KCl + H2O
NOTE: Only for a strong acid-strong base titration the pH = 7 at the endpoint !
III. Indicators used in Titrations
A. Indicator selection
Titration of an unknown weak acid
1. End-point determination
Use an indicator that changes at the equivalence point
2. Calculation of error if a wrong indicator is used
B. Indicator Theory
1. Background
Page 17
2. Range calculation
Methyl Orange
a) In acid solution:
b) In basic solution:
c) Range for color change
d) Indicator Selection:
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IV. Titration curve for a Weak Acid titrated with a strong base
Four Regions
1.
2.
3.
4.
50.0 mls of 0.10M HAc is titrated with 0.10 M NaOH
RXN: HAc + NaOH ---> NaAc + H2O
NOTE: For a weak acid-strong base titration the pH
≠ 7 at the endpoint !
Construction of Titration Curve
Calculation of the volume of base needed to reach the equivalence point
Region 1 = Before addition of base
pH calculation
Region 2 = Buffer Region {BEFORE THE EQUIVALENCE POINT}
pH calculation
Method 1.
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Method 2
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Region 3 = THE EQUIVALENCE POINT
pH calculation
Region 4 = AFTER THE EQUIVALENCE POINT
pH calculation
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Graph
V. Titration curve for a Weak Base titrated with a strong acid
Four Regions
1.
2.
3.
4..
50.0 mls of 0.10M CH3NH2 (mehtyl amine) is titrated with 0.10 M HCl
RXN: HCl +
CH3NH2 --->
CH3NH3+ + Cl-
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VI Titration curves for polyfunctional acids and bases
Poly functional acids and bases will give multiple end points in a titration.
Construction of Titration Curve
25.0 mls of 0.1000M Maleic acid [C2H2(COOH)2 is titrated with 0.1000 M NaOH
Equilibria:
 HM- + H3O+
HM- + H2O  M2- + H3O+
H2 M + H 2 O
Ka1 = 1.20 x 10-2
Ka2 = 5.96 x 10-7
Calculation of the volume of base needed to reach the equivalence point
Region 1 = Before addition of base
pH calculation
Region 2 = First Buffer Region {BEFORE THE FIRST EQUIVALENCE POINT}
pH calculation
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Region 3 = THE FIRST EQUIVALENCE POINT
pH calculation
Region 4 = SECOND Buffer Region {BEFORE THE SECOND EQUIVALENCE POINT}
pH calculation
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Region 5 = THE SECOND EQUIVALENCE POINT
pH calculation
Region 6 = AFTER THE EQUIVALENCE POINT
pH calculation
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Graph
VI Titrations in Nonaqueous solvents
A. Strengths of acids and bases
B. The leveling effect
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PART III EDTA TITRATIONS
Chapter 13
I. CHELATING AGENTS
1.
2.
3.
4. Examples:
II. EDTA
A. Structure
Acid, protonated form
Base form (complexing form)
B. Complexes:
1.
2.
3.
4.
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5. Formation constants
a. Background - Fractional composition equations
b. Computation of 4 values for EDTA
4 =
K1K2K3K4
+
4
+
3
[H ] + K1[H ] + K1K2[H+]2 + K1K2K3[H+] + K1K2K3K4
The 4 equation shows that the tetrabasic form of EDTA,Y4-, is pH dependant!


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able 13-2 lists 4 at selected pH values
(1) Kf at various pH
(2) Calculate the molar Y4- concentration in a 0.33 M EDTA solution buffered at pH = 9.0
III. EDTA Titration curves
THREE REGIONS
1. Excess metal ion
2. Metal-EDTA dissociation
3. Excess EDTA
50.0 mls of 0.10M CaCl2 is titrated with 0.10 M EDTA
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RXN: Ca2+ + Y4- ---> CaY2Construction of Titration Curve
Calculation of the volume of base needed to reach the equivalence point
Region 1 = Before addition of EDTA
EXCESS Ca2+
DISSOCIATION OF CaY2- is negligable
Calculation of pCa2+
Region 2 = EQUIVALENCE POINT
FREE Ca2+is from the dissociation of CaY2Calculation of pCa2+
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Region 3 = AFTER THE EQUIVALENCE POINT
EXCESS EDTA - All Ca2+ is essentially complexed
FREE Ca2+is from the slight dissociation of CaY2Calculation of pCa2+
Graph
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