Download AP Ch 18 Acid Base Equilibria - Tenafly High School

Chapter 14
Acids and Bases
Chapter 14
Table of Contents
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
14.9
14.10
14.11
14.12
The Nature of Acids and Bases
Acid Strength
The pH Scale
Calculating the pH of Strong Acid Solutions
Calculating the pH of Weak Acid Solutions
Bases
Polyprotic Acids
Acid–Base Properties of Salts
The Effect of Structure on Acid–Base Properties
Acid–Base Properties of Oxides
The Lewis Acid–Base Model
Strategy for Solving Acid–Base Problems: A Summary
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Section 14.1
The Nature of Acids and Bases
Models of Acids and Bases
• Arrhenius: Acids produce H+ ions in solution,
bases produce OH- ions.
• Brønsted–Lowry: Acids are proton (H+) donors,
bases are proton acceptors.
HCl + H2O  Cl- + H3O+
acid base
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Section 14.1
The Nature of Acids and Bases
Brønsted–Lowry Reaction
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Section 14.1
The Nature of Acids and Bases
Acid in Water
HA(aq) +
acid
•
•
H2O(l)
base
H3O+(aq) + A-(aq)
conjugate
conjugate
acid
base
Conjugate base is everything that remains of the acid
molecule after a proton is lost.
Conjugate acid is formed when the proton is transferred
to the base.
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Section 14.11
The Lewis Acid–Base Model
Lewis Acids and Bases
• Lewis acid: electron pair acceptor
• Lewis base: electron pair donor
3+
Al3+ + 6 O
Al
H
Lewis acid
H
H
O
H
6
Lewis base
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Section 14.11
The Lewis Acid–Base Model
Three Models for Acids and Bases
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Section 14.1
The Nature of Acids and Bases
Acid Ionization Equilibrium
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Section 14.2
Atomic
Acid
Strength
Masses
• Strong acid:
 Ionization equilibrium lies far to the right.
 Yields a weak conjugate base.
 See AP strong acids list!
• Weak acid:
 Ionization equilibrium lies far to the left.
 Weaker the acid, stronger its conjugate
base.
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Section 14.2
Atomic
Acid
Strength
Masses
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Section 14.2
Atomic
Acid
Strength
Masses
Various Ways to Describe Acid Strength
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Section 14.2
Atomic
Acid
Strength
Masses
Water as an Acid and a Base
• Water is amphoteric:
 Behaves either as an acid or as a base.
• At 25°C:
Kw = [H+][OH–] = 1.0 × 10–14
• No matter what the solution contains, the
product of [H+] and [OH–] must always equal
1.0 × 10–14 at 25°C.
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Section 14.2
Atomic
Acid
Strength
Masses
Three Possible Situations
• [H+] = [OH–]; neutral solution
• [H+] > [OH–]; acidic solution
• [H+] < [OH–]; basic solution
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Section 14.2
Atomic
Acid
Strength
Masses
Self-Ionization of Water
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Section 14.2
Atomic
Acid
Strength
Masses
Concept Check
HA(aq) + H2O(l)
acid
base
H3O+(aq) + A-(aq)
conjugate conjugate
acid
base
What is the equilibrium constant expression
for an acid acting in water?


H3O   A 
K =
HA 
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Section 14.3
The pH
Mole
Scale
• pH = –log[H+]
• A compact way to represent solution acidity.
• pH decreases as [H+] increases.
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Section 14.3
The pH
Mole
Scale
pH Range
• pH = 7; neutral
• pH > 7; basic
 Higher the pH, more basic.
• pH < 7; acidic
 Lower the pH, more acidic.
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Section 14.3
The pH
Mole
Scale
The pH Scale and
pH Values of Some
Common
Substances
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Section 14.3
The pH
Mole
Scale
Exercise
Calculate the pH for each of the following
solutions.
a) 1.0 × 10–4 M H+
pH = 4.00
b) 0.040 M OH–
pH = 12.60
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Section 14.3
The pH
Mole
Scale
Exercise
The pH of a solution is 5.85. What is the [H+]
for this solution?
[H+] = 1.4 × 10–6 M
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Section 14.3
The pH
Mole
Scale
pH and pOH
• Recall:
Kw = [H+][OH–]
–log Kw = –log[H+] – log[OH–]
pKw = pH + pOH
14.00 = pH + pOH
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Section 14.3
The pH
Mole
Scale
Exercise
Calculate the pOH for each of the following
solutions.
a) 1.0 × 10–4 M H+
pOH = 10.00
b) 0.040 M OH–
pOH = 1.40
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Section 14.3
The pH
Mole
Scale
Exercise
The pH of a solution is 5.85. What is the
[OH–] for this solution?
[OH–] = 7.1 × 10–9 M
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Section 14.4
Calculating the pH of Strong Acid Solutions
Concept Check
Calculate the pH of a 1.5 x 10–11 M solution of
HCl.
pH = 7.00
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Equilibrium Constants
for Weak Acids
Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, H3O+, OAc-,
and the pH.
Step 1. Define equilibrium concs. in ICE
table.
[HOAc]
[H3O+]
[OAc-]
initial
1.00
0
0
change
-x
+x
+x
equilib
1.00-x
x
x
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.
Step 2. Write Ka expression
+ ][OAc- ]
2
[H
O
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x
This is a quadratic. Solve using quadratic
formula.
or you can make an approximation if x is very
small! (Rule of thumb: 10-5 or smaller is ok)
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.
Step 3. Solve Ka expression
+ ][OAc- ]
2
[H
O
x
3
Ka  1.8 x 10-5 =

[HOAc]
1.00 - x
First assume x is very small because
Ka is so small.
Ka  1.8 x 10-5 =
x2
1.00
Now we can more easily solve this
approximate expression.
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H3O+, OAc-, and the pH.
Step 3. Solve Ka approximate expression
Ka  1.8 x 10-5 =
x2
1.00
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37
Equilibria Involving A Weak Acid
Calculate the pH of a 0.0010 M solution of formic
acid, HCO2H.
HCO2H + H2O ↔ HCO2- + H3O+
Ka = 1.8 x 10-4
Approximate solution
[H3O+] = 4.2 x 10-4 M, pH = 3.37
Exact Solution
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.47
Equilibrium Constants
for Weak Bases
Weak base has Kb < 1
Leads to small [OH-] and a pH of 12 - 7
Relation
of Ka,
Kb,
[H3O+]
and pH
Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
Calculate the pH of a 0.50 M aqueous solution
of the weak acid HF.
(Ka = 7.2 x 10–4)
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Section 14.5
Calculating the pH of Weak Acid Solutions
Steps Toward Solving for pH
H3O+(aq) +
F–(aq)
0.50 M
~0
~0
–x
+x
+x
0.50–x
x
x
HF(aq) + H2O
Initial
Change
Equilibrium
Ka = 7.2 x 10–4
pH = 1.72
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Section 14.5
Calculating the pH of Weak Acid Solutions
Percent Dissociation (Ionization)
Percent dissociation =
amount dissociated (mol/L)
 100%
initial concentration (mol/L)
• For a given weak acid, the percent dissociation
increases as the acid becomes more dilute.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Exercise
A solution of 8.00 M formic acid (HCHO2) is
0.47% ionized in water.
Calculate the Ka value for formic acid.
Ka = 1.8 x 10–4
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Section 14.6
Bases
Arrhenius: bases produce OH– ions.
Brønsted–Lowry: bases are proton acceptors.
In a basic solution at 25°C, pH > 7.
Ionic compounds containing OH- are generally
considered strong bases.
 See strong bases list!
• pOH = –log[OH–]
• pH = 14.00 – pOH
•
•
•
•
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Section 14.6
Bases
Concept Check
Calculate the pH of a 1.0 x 10–3 M solution of
sodium hydroxide.
pH = 11.00
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Section 14.6
Bases
Concept Check
Calculate the pH of a 1.0 x 10–3 M solution of
calcium hydroxide.
pH = 11.30
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Section 14.6
Bases
• Equilibrium expression for weak bases uses Kb.
CN–(aq) + H2O(l)
Kb =
HCN(aq) + OH–(aq)


HCN
OH

  
CN 
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Section 14.6
Bases
• pH calculations for solutions of weak bases are
very similar to those for weak acids.
• Kw = [H+][OH–] = 1.0 × 10–14
• pOH = –log[OH–]
• pH = 14.00 – pOH
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Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O ↔ NH4+ + OHKb = 1.8 x 10-5
Step 1. Define equilibrium concs. in ICE table
[NH3]
[NH4+]
[OH-]
initial
0.010
0
0
change
-x
+x
+x
equilib
0.010 - x
x
x
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ + OHKb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
[NH4+ ][OH- ]
x2
-5
Kb  1.8 x 10 =
=
[NH3 ]
0.010 - x
Assume x is small, so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
and [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M
The approximation is valid !
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O  NH4+ + OHKb = 1.8 x 10-5
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14,
pH = 10.63
Section 14.6
Bases
Concept Check
Calculate the pH of a 2.0 M solution of
ammonia (NH3).
(Kb = 1.8 × 10–5)
pH = 11.78
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Ionization Constants of Conjugate Acid-Base Pairs
HA (aq)
A- (aq) + H2O (l)
H2O (l)
H+ (aq) + A- (aq)
OH- (aq) + HA (aq)
H+ (aq) + OH- (aq)
Ka
Kb
Kw
KaKb = Kw
Weak Acid and Its Conjugate Base
Kw
Ka =
Kb
Kw
Kb =
Ka
15.7
Section 14.7
Polyprotic Acids
• Acids that can furnish more than one proton.
• Always dissociates in a stepwise manner, one
proton at a time.
• The conjugate base of the first dissociation
equilibrium becomes the acid in the second
step.
• For a typical weak polyprotic acid:
Ka1 > Ka2 > Ka3
• For a typical polyprotic acid in water, only the
first dissociation step is important to pH.
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Section 14.7
Polyprotic Acids
Exercise
Calculate the pH of a 1.00 M solution of H3PO4.
Ka1 = 7.5 x 10-3
Ka2 = 6.2 x 10-8
Ka3 = 4.8 x 10-13
pH = 1.08
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Section 14.7
Polyprotic Acids
Sulfuric Acid
• AP always treats H2SO4 in this way:
– The first H+ is STRONG. Assume 100% ionization.
– The second H+ is WEAK. Use the Ka.
– Add the H+ Molarities together to find the pH.
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Types of Acid/Base Reactions:
Summary
Section 14.8
Acid–Base Properties of Salts
Salts
• Ionic compounds.
• When dissolved in water, break up into its ions
(which can behave as acids or bases).
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Section 14.8
Acid–Base Properties of Salts
Salts
• The salt of a strong acid and a strong base
gives a neutral solution.
 KCl, NaNO3
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Section 14.8
Acid–Base Properties of Salts
Salts
• A basic solution is formed if the anion of the salt
is the conjugate base of a weak acid.
 NaF, KC2H3O2
 Kw = Ka × Kb
 Use Kb when starting with base.
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Section 14.8
Acid–Base Properties of Salts
Salts
• An acidic solution is formed if the cation of the
salt is the conjugate acid of a weak base.
 NH4Cl
 Kw = Ka × Kb
 Use Ka when starting with acid.
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Section 14.8
Acid–Base Properties of Salts
Cation
neutral
neutral
Anion
neutral
conjugate
base of
weak acid
neutral
Acidic
or Basic
neutral
basic
conjugate
acidic
acid of
weak base
conjugate conjugate depends
acid of
base of on Ka & Kb
weak base weak acid
values
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Example
NaCl
NaF
NH4Cl
Al2(SO4)3
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Section 14.8
Acid–Base Properties of Salts
Qualitative Prediction of pH of Salt Solutions (from Weak Parents)
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Section 14.8
Acid–Base Properties of Salts
Exercise
HC2H3O2
HCN
Ka = 1.8 x 10-5
Ka = 6.2 x 10-10
Calculate the Kb values for: C2H3O2− and CN−
Kb (C2H3O2-) = 5.6 x 10-10
Kb (CN-) = 1.6 x 10-5
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Section 14.8
Acid–Base Properties of Salts
Concept Check
Arrange the following 1.0 M solutions from
lowest to highest pH.
HBr
NaCN
NaCl
NaOH
NH3
HF
NH4Cl
HCN
Justify your answer.
HBr, HF, HCN, NH4Cl, NaCl, NaCN, NH3, NaOH
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Section 14.8
Acid–Base Properties of Salts
Exercise
Calculate the pH of a 0.75 M aqueous solution
of NaCN.
Ka for HCN is 6.2 x 10–10.
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Section 14.8
Acid–Base Properties of Salts
Steps Toward Solving for pH
CN–(aq) + H2O
Initial
0.75 M
0
~0
–x
+x
+x
0.75–x
x
x
Change
Equilibrium
HCN(aq) + OH–(aq)
Kb = 1.6 x 10–5
pH = 11.54
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Section 14.9
The Effect of Structure on Acid–Base Properties
Models of Acids and Bases
• Two factors for acidity in binary compounds:
 Bond Polarity (high is good)
 Bond Strength (low is good)
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Section 14.9
The Effect of Structure on Acid–Base Properties
Bond Strengths and Acid Strengths for Hydrogen Halides
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Section 14.9
The Effect of Structure on Acid–Base Properties
Oxyacids
• Contains the group H–O–X.
• For a given series the acid strength increases
with an increase in the number of oxygen atoms
attached to the central atom.
• The greater the ability of X to draw electrons
toward itself, the greater the acidity of the
molecule.
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Section 14.9
The Effect of Structure on Acid–Base Properties
Several Series of Oxyacids and Their Ka Values
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Section 14.10
Acid–Base Properties of Oxides
Oxides
• Acidic Oxides (Acid Anhydrides):
 O-X bond is strong and covalent.
SO2, NO2, CO2
• When H-O-X grouping is dissolved in water, the
O-X bond will remain intact. It will be the polar
and relatively weak H-O bond that will tend to
break, releasing a proton.
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Section 14.10
Acid–Base Properties of Oxides
Oxides
• Basic Oxides (Basic Anhydrides):
 O-X bond is ionic.
K2O, CaO
• If X has a very low electronegativity, the O-X
bond will be ionic and subject to being broken in
polar water, producing a basic solution.
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Chapter 15
Acid–Base Equilibria
Chapter 15
Table of Contents
15.1
15.2
15.3
15.4
15.5
Solutions of Acids or Bases Containing a
Common Ion
Buffered Solutions
Buffering Capacity
Titrations and pH Curves
Acid–Base Indicators
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Section 15.1
Solutions of Acids or Bases Containing a Common Ion
Common Ion Effect
• Shift in equilibrium position that occurs because
of the addition of an ion already involved in the
equilibrium reaction.
• An application of Le Châtelier’s principle.
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Section 15.1
Solutions of Acids or Bases Containing a Common Ion
Example
HCN(aq) + H2O(l)
H3O+(aq) + CN-(aq)
• Addition of NaCN will shift the equilibrium to the
left because of the addition of CN-, which is
already involved in the equilibrium reaction.
• A solution of HCN and NaCN is less acidic than
a solution of HCN alone.
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Section 15.2
Atomic Masses
Buffered
Solutions
Key Points about Buffered Solutions
• Buffered Solution – resists a change in pH.
• They are weak acids or bases containing a salt
with common ion.
• After addition of strong acid or base, deal with
stoichiometry first, then the equilibrium.
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Which of the following are buffer systems? (a) KF/HF
(b) KCl/HCl, (c) Na2CO3/NaHCO3
(a) KF is a weak acid and F- is its conjugate base
buffer solution
(b) HCl is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution
16.3
Section 15.2
Atomic Masses
Buffered
Solutions
Adding an Acid to a Buffer
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Section 15.2
Atomic Masses
Buffered
Solutions
Buffers
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Section 15.2
Atomic Masses
Buffered
Solutions
Solving Problems with Buffered Solutions
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Section 15.2
Atomic Masses
Buffered
Solutions
Henderson–Hasselbalch Equation

 A 
pH = pK a + log
HA 
• For a particular buffering system (conjugate
acid–base pair), all solutions that have the
same ratio [A–] / [HA] will have the same pH.
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What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
Ka for HCOOH = 1.8 x 10 -4
[H+] [HCOO-]
Ka =
H+ (aq) + HCOO- (aq)
x = 1.038 X 10 -4
pH = 3.98
[HCOOH]
16.2
OR…… Use the Henderson-Hasselbach equation
Consider mixture of salt NaA and weak acid HA.
NaA (s)
Na+ (aq) + A- (aq)
HA (aq)
H+ (aq) + A- (aq)
[H+]
Ka [HA]
=
[A-]
-log [H+] = -log Ka - log
[HA]
[A-]
-]
[A
-log [H+] = -log Ka + log
[HA]
[H+][A-] pK = -log K
Ka =
a
a
[HA]
Henderson-Hasselbach
equation
[conjugate base]
pH = pKa + log
[acid]
[A-]
pH = pKa + log
[HA]
16.2
What is the pH of a solution containing 0.30 M HCOOH
and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
HCOOH (aq)
Initial (M)
Change (M)
Equilibrium (M)
Common ion effect
0.30 – x  0.30
0.52 + x  0.52
H+ (aq) + HCOO- (aq)
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
[HCOO-]
pH = pKa + log
[HCOOH]
[0.52]
= 4.01
pH = 3.77 + log
[0.30]
HCOOH pKa = 3.77
16.2
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer
system. What is the pH after the addition of 20.0 mL of
0.050 M NaOH to 80.0 mL of the buffer solution?
NH3 (aq) + H2O (l)
Kb =
[NH4+] [OH-]
[NH3]
Initial
Change
End
NH4+ (aq) + OH- (aq)
= 1.8 X 10-5
0.30
-x
0.30 - x
0.36
+x
0
+x
0.36 + x
x
(.36 + x)(x)
1.8 X 10-5 =
1.8 X
10-5 
(.30 – x)
0.36x
0.30
x = 1.5 X 10-5
pOH = 4.82
pH= 9.18
16.3
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer
system. What is the pH after the addition of 20.0 mL of
0.050 M NaOH to 80.0 mL of the buffer solution?
final volume = 80.0 mL + 20.0 mL = 100 mL
NH4+
0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M
OH-
0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M
NH3
0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M
start (M)
end (M)
Ka=
0.01
0.29
NH4+ (aq) + OH- (aq)
0.28
0.0
[H+] [NH3]
= 5.6 X 10-10
[NH4+]
[H+] 0.25
0.28
0.24
H2O (l) + NH3 (aq)
0.25
[H+] = 6.27 X 10 -10
= 5.6 X
10-10
pH = 9.20
16.3
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer
system. What is the pH after the addition of 20.0 mL of
0.050 M NaOH to 80.0 mL of the buffer solution?
NH4+ (aq)
[NH3]
pH = pKa + log
[NH4+]
H+ (aq) + NH3 (aq)
pKa = 9.25
[0.30]
pH = 9.25 + log
= 9.17
[0.36]
final volume = 80.0 mL + 20.0 mL = 100 mL
start (M)
end (M)
0.01
0.29
NH4+ (aq) + OH- (aq)
0.28
0.0
0.24
H2O (l) + NH3 (aq)
0.25
[0.25]
pH = 9.25 + log
= 9.20
[0.28]
16.3
Section 15.2
Atomic Masses
Buffered
Solutions
Buffered Solution Characteristics
• Buffers contain relatively large amounts of weak
acid and corresponding conjugate base.
• Added H+ reacts to completion with the
conjugate base.
• Added OH- reacts to completion with the weak
acid.
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83
Section 15.2
Atomic Masses
Buffered
Solutions
Buffered Solution Characteristics
• The pH in the buffered solution is determined by
the ratio of the concentrations of the weak acid
and weak base. As long as this ratio remains
virtually constant, the pH will remain virtually
constant. This will be the case as long as the
concentrations of the buffering materials (HA
and A– or B and BH+) are large compared with
amounts of H+ or OH– added.
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84
Section 15.3
The Mole Capacity
Buffering
• The amount of protons or hydroxide ions the
buffer can absorb without a significant change
in pH.
• Determined by the magnitudes of [HA] and [A–].
• A buffer with large capacity contains large
concentrations of the buffering components.
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85
Section 15.3
The Mole Capacity
Buffering
• Optimal buffering occurs when [HA] is equal to
[A–].
• It is for this condition that the ratio [A–] / [HA] is
most resistant to change when H+ or OH– is
added to the buffered solution.
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86
Section 15.4
Titrations and pH Curves
Titration Curve
• Plotting the pH of the solution being analyzed
as a function of the amount of titrant added.
• Equivalence (Stoichiometric) Point – point in the
titration when enough titrant has been added to
react exactly with the substance in solution
being titrated.
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Section 15.4
Titrations and pH Curves
Neutralization of a Strong Acid with a Strong Base
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Section 15.4
Titrations and pH Curves
The pH Curve for the Titration of 50.0 mL of 0.200 M HNO3 with
0.100 M NaOH
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Section 15.4
Titrations and pH Curves
The pH Curve for the Titration of 100.0 mL of 0.50 M NaOH
with 1.0 M HCI
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Section 15.4
Titrations and pH Curves
Weak Acid–Strong Base Titration
Step 1:
Step 2:
A stoichiometry problem (reaction is
assumed to run to completion) then
determine remaining species.
An equilibrium problem (determine
position of weak acid equilibrium and
calculate pH).
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Section 15.4
Titrations and pH Curves
Concept Check
Calculate the pH of a solution made by mixing
0.20 mol HC2H3O2 (Ka = 1.8 x 10–5) with 0.030
mol NaOH in 1.0 L of aqueous solution.
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Section 15.4
Titrations and pH Curves
Let’s Think About It…
HC2H3O2(aq) + OH– 
Before
Change
After
0.20 mol 0.030 mol
–0.030 mol –0.030 mol
0.17 mol
0
C2H3O2–(aq) + H2O
0
+0.030 mol
0.030 mol
K = 1.8 x 109
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93
Section 15.4
Titrations and pH Curves
Steps Toward Solving for pH
H3O+
+ C2H3O2-(aq)
0.170 M
~0
0.030 M
–x
+x
+x
0.170 – x
x
0.030 + x
HC2H3O2(aq) + H2O
Initial
Change
Equilibrium
Ka = 1.8 x 10–5
pH = 3.99
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94
Section 15.4
Titrations and pH Curves
Exercise
Calculate the pH of a 100.0 mL solution of
0.100 M acetic acid (HC2H3O2), which has a Ka
value of 1.8 x 10–5.
pH = 2.87
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Section 15.4
Titrations and pH Curves
Concept Check
Calculate the pH of a solution made by mixing
100.0 mL of a 0.100 M solution of acetic acid
(HC2H3O2), which has a Ka value of 1.8 x 10–5,
and 50.0 mL of a 0.10 M NaOH solution.
pH = 4.74
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Finding the Equivalence Point
(calculation method)
• Strong Acid vs. Strong Base
– 100 % ionized! pH = 7 No equilibrium!
• Weak Acid vs. Strong Base
– Acid is neutralized; Need Kb for conjugate base
equilibrium
• Strong Acid vs. Weak Base
– Base is neutralized; Need Ka for conjugate acid
equilibrium
• Weak Acid vs. Weak Base
– Depends on the strength of both; could be
conjugate acid, conjugate base, or pH 7
Exactly 100 mL of 0.10 M HNO2 are titrated with 100 mL of
a 0.10 M NaOH solution. What is the pH at the
equivalence point ?
start (moles)
0.01
0.01
HNO2 (aq) + OH- (aq)
NO2- (aq) + H2O (l)
end (moles)
0.0
0.0
0.01
Final volume = 200 mL
NO2- (aq) + H2O (l)
Initial (M)
Change (M)
0.01
= 0.05 M
0.200
OH- (aq) + HNO2 (aq)
[NO2-] =
0.05
0.00
0.00
-x
+x
+x
x
x
Equilibrium (M) 0.05 - x
[OH-][HNO2]
x2
-11
=
2.2
x
10
Kb =
=
[NO2-]
0.05-x
pOH = 5.98
0.05 – x  0.05 x  1.05 x 10-6 = [OH-]
pH = 14 – pOH = 8.02
Section 15.4
Titrations and pH Curves
The pH Curve for the Titration of 50.0 mL of 0.100 M HC2H3O2 with
0.100 M NaOH
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Section 15.4
Titrations and pH Curves
The pH Curve for the Titration of 100.0mL of 0.050 M NH3 with
0.10 M HCl
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Section 15.5
Acid–Base Indicators
• Marks the end point of a titration by changing
color.
• The equivalence point is not necessarily the
same as the end point (but they are ideally as
close as possible).
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Section 15.5
Acid–Base Indicators
The Acid and Base
Forms of the
Indicator
Phenolphthalein
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Section 15.5
Acid–Base Indicators
The Methyl Orange Indicator is Yellow in Basic Solution and Red
in Acidic Solution
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Section 15.5
Acid–Base Indicators
Useful pH Ranges for Several Common Indicators
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Chapter 16
Solubility and
Complex Ion
Equilibria
Chapter 16
Table of Contents
16.3
Equilibria Involving Complex Ions
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Section 16.3
The Mole Involving Complex Ions
Equilibria
Complex Ion Equilibria
• Charged species consisting of a metal ion
surrounded by ligands.
 Ligand: Lewis base
• Formation (stability) constant.
 Equilibrium constant for each step of the
formation of a complex ion by the addition of
an individual ligand to a metal ion or complex
ion in aqueous solution.
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Chapter 21
Transition Metals and
Coordination
Chemistry
Chapter 21
Table of Contents
21.1
21.2
21.3
The Transition Metals: A Survey
The First-Row Transition Metals
Coordination Compounds
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Section 21.1
The Transition Metals: A Survey
Forming Ionic Compounds
• More than one oxidation state is often found.
• Cations are often complex ions – species where
the transition metal ion is surrounded by a
certain number of ligands (Lewis bases).
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Section 21.1
The Transition Metals: A Survey
The Complex Ion Co(NH3)63+
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Section 21.1
The Transition Metals: A Survey
Ionic Compounds with Transition Metals
• Most compounds are colored because the
transition metal ion in the complex ion can
absorb visible light of specific wavelengths.
• Many compounds are paramagnetic.
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Section 21.1
The Transition Metals: A Survey
Electron Configurations
• First-row transition metal ions do not have 4s
electrons.
 Energy of the 3d orbitals is less than that of
the 4s orbital.
Ti: [Ar]4s23d2
Ti3+: [Ar]3d1
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Section 21.3
The Mole
Coordination
Compounds
Coordination Number
• Number of bonds formed between the metal ion
and the ligands in the complex ion.
 6 and 4 (most common)
 2 and 8 (least common)
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Section 21.3
The Mole
Coordination
Compounds
Ligands
• Neutral molecule or ion having a lone electron
pair that can be used to form a bond to a metal
ion.
 Monodentate ligand – one bond to a metal
ion
 Bidentate ligand (chelate) – two bonds to a
metal ion
 Polydentate ligand – more than two bonds to
a metal ion
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Complex Ion Formation
• These are usually formed from a
transition metal surrounded by ligands
(polar molecules or negative ions).
• As a "rule of thumb" you place twice the
number of ligands around an ion as the
charge on the ion... example: the dark
blue Cu(NH3)42+ (ammonia is used as a
test for Cu2+ ions), and Ag(NH3)2+.
• Memorize the common ligands.
Common Ligands
Ligands
Names used in the ion
H2O
NH3
aqua
ammine
OHClBrCNSCN-
hydroxy
chloro
bromo
cyano
thiocyanato (bonded through
sulphur)
isothiocyanato (bonded through
nitrogen)
Names
• Names: ligand first, then cation
Examples:
– tetraamminecopper(II) ion: Cu(NH3)42+
– diamminesilver(I) ion: Ag(NH3)2+.
– tetrahydroxyzinc(II) ion: Zn(OH)4 2-
• The charge is the sum of the parts
(2+) + 4(-1)= -2.
When Complexes Form
• The odd complex ion, FeSCN2+, shows up once in a
while
• Acid-base reactions may change NH3 into NH4+ (or vice
versa) which will alter its ability to act as a ligand.
• Visually, a precipitate may go back into solution as a
complex ion is formed. For example, Cu2+ + a little
NH4OH will form the light blue precipitate, Cu(OH)2. With
excess ammonia, the complex, Cu(NH3)42+, forms.
• Keywords such as "excess" and "concentrated" of any
solution may indicate complex ions in REACTIONS!
AgNO3 + HCl forms the white precipitate, AgCl. With
excess, concentrated HCl, the complex ion, AgCl2-,
forms and the solution clears.
Some Coordination Complexes
molecular
formula
Lewis
base/ligand
Lewis acid donor
atom
coordination
number
Ag(NH3)2+
NH3
Ag+
N
2
[Zn(CN)4]2- CN-
Zn2+
C
4
[Ni(CN)4]2-
CN-
Ni2+
C
4
[PtCl6] 2-
Cl-
Pt4+
Cl
6
Ni2+
N
6
[Ni(NH3)6]2+ NH3