George Mason University
General Chemistry 212
Chapter 18
Acid-Base Equilibria
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and
Change, 7th edition, 2011, McGraw-Hill
Martin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at George
Mason are intended for those students enrolled in a science
/engineering oriented curricula, with particular emphasis on
chemistry, biochemistry, and biology The material on these slides is
taken primarily from the course text but the instructor has modified,
condensed, or otherwise reorganized selected material.
Additional material from other sources may also be included.
Interpretation of course material to clarify concepts and solutions to
problems is the sole responsibility of this instructor.
1/19/2015
1
Chapter 18 – Acid-Base Equilibria



Acids & Bases in Water

Arrhenius Acid-Base Definition

Bronsted-Lowry Acid-Base Definition

Lewis Acid-Base Definition

Acid Dissociation Constant

Relative Strengths of Acids & Bases
Autoionization of Water and the pH Scale

Autoionization and Kw

The pH Scale
Proton Transfer and the Bronsted-Lowry Acid-Base
Definition

The Conjugate Acid-Base Pair

Met Direction of Acid-Base Reactions
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2
Chapter 18 – Acid-Base Equilibria



Solving Problems Involving Weak-Acid Equilibria

Finding Ka Given Concentrations

Finding Concentrations Given Ka

Extent of Acid Dissociation

Polyprotic Acids
Weak Bases and Their Relation to Weak Acids

Ammonia & The Amines

Anions of Weak Acids

The Relation Between Ka & Kb
Molecular Properties and Acid Strength

Nonmetal Hydrides

Oxoacids

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Acidity of Hydrated Metal Ions
3
Chapter 18 – Acid-Base Equilibria

Acid-Base Properties of Salt Solutions

Salts That Yield Neutral Solutions

Salts That Yield Acidic Solutions

Salts That Yield Basic Solutions

Salts of Weakly Acidic Cations and Weakly Basic Anions

Salts of Amphoteric Anions

Generalizing the Bronsted-Lowry Concept: The Leveling
Effect

Electron-Pair Donation and the Lewis Acid-Base Definition

Molecules as Lewis Acids

Metal Cations as Lewis Acids

Overview of Acid-Base Definitions
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4
Acid & Bases

Antoine Lavoisier was one of the first chemists to
try to explain what makes a substance acidic


In 1777, he proposed that oxygen was an
essential element in acids
The actual cause of acidity and basicity was
ultimately explained in terms of the effect these
compounds have in water by Svante Arrhenius in
1884
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5
Acids & Bases

Several concepts of acid-base theory:
The Arrhenius concept
The Bronsted-Lowry concept
The Lewis concept
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6
Acids & Bases

According to the Arrhenius concept of acids and
bases


An acid is a substance that, when
dissolved in water, increases the
concentration of Hydrogen ion, H+(aq)
The H+(aq) ion, called the Hydrogen ion is actually
chemically bonded to water, that is, H3O+, called
the Hydronium ion

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A base is a substance that when dissolved
in water increases the concentration of
the Hydroxide ion, OH-(aq)
7
Acids & Bases

More About the Hydronium Ion
The Hydronium Ion (H3O+) actually exists as a hydrogen
bonded H9O4+ cluster. The formation of Hydronium ions is
a complex process in aqueous solution
The Hydronium ion has trigonal
pyramidal geometry because all three HO-H dihedral bond angles are identical
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8
Acids & Bases

An Arrhenius acid is any substance which increases the
concentration of Hydrogen ions (H+) in solution. Acids
generally have a sour taste


An Arrhenius base is any substance which increases the
concentration of Hydroxide ions (OH-) in solution. Bases
generally have a bitter taste


KOH(s)  K+(aq) + OH-(aq)
Hydrogen and Hydroxide ions are important in aqueous
solutions because - Water reacts to form both ions


HBr(aq)  H+(aq) + Br-(aq)
H2O(l)  H+(aq) + OH-(aq)
HBr and KOH are examples of a strong acid and a
strong base; strong acids & bases dissociate completely
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9
Acids & Bases

Monoprotic acids are those acids that are able to donate
one proton per molecule during the process of dissociation
(sometimes called ionization) as shown below (symbolized
by HA):
HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq)
Common examples of monoprotic acids in mineral acids
include Hydrochloric Acid (HCl) and Nitric Acid (HNO3)

Polyprotic acids are able to donate more than one proton
per acid molecule.

Diprotic acids have two potential protons to donate
H2A(aq) + H2O(l) ⇌ H3O+(aq) + HA−(aq)

Triprotic acids have three potential protons to donate
H3A(aq) + H2O(l) ⇌ H3O+(aq) + H2A−(aq)
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10
Acids & Bases

Oxoacids

Oxoacids with one oxygen have the structure: HOY
HOCl

HOBr
HOI
Oxoacids with multiple oxygen have the structure:
(OH)mYOn
HOClO3 (HClO4)
HOClO2 (HOCLO3)
HOClO (HCLO2)
HOCL
(HClO)
The more oxygen atoms, the greater the acidity; thus
HOCLO3 is a stronger acid than HOCl
Acidity refers to the relative acid strength, i.e. the degree
to which the acid dissociates to form H+ (H3O+) & OH1/19/2015
11
Bronsted-Lowry Acids & Bases

Brønsted-Lowry Concept of Acids and Bases

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In the Brønsted-Lowry concept:

An Acid is a species that donates protons

A Base is a species that accepts protons

Acids and bases can be ions as well as molecular
substances

Acid-base reactions are not restricted to aqueous
solution

Some species can act as either acids or bases
(Amphoteric species) depending on what the
other reactant is
12
Bronsted-Lowry Acids & Bases

Amphoteric Species
 A species that can act as either an acid or base is:
Amphoteric
 Water is an important amphoteric species in the acidbase properties of aqueous solutions
 Water can react as an acid by donating a proton to a
base
N H 3 (a q ) + H 2 O (l)  N H 4 (a q ) + O H (a q )
+
-
H+

Water can also react as a base by accepting a proton
from an acid
H F (a q ) + H 2 O (l)  F  (a q ) + H 3 O + (a q )
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H+
13
Bronsted-Lowry Acids & Bases

Brønsted-Lowry Concept of Acids and Bases

Proton transfer as the essential feature of a

Brønsted-Lowry acid-base reaction
Lone pair
binds H+
Acid
donor
H+
Base
H+ acceptor
Lone pair
binds H+
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Acid
Base
H+ acceptor H+ donor
14
Acids & Bases

Bronsted-Lowery Concept – Conjugate Pairs

In Bronsted theory, acid and base reactants form:
Conjugate Pairs

The acid (HA) donates a proton to water leaving the
conjugate Base (A-)

The base (H2O) accepts a proton to form the conjugate
acid (H3O+)
HA(aq)
Acid
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+
H2O(l)
Base
⇌
A−(aq)
Conjugate
Base
+ H3O+(aq)
Conjugate
Acid
15
Bronsted-Lowry Acids & Bases

Bronsted-Lowery Acid-Base Conjugate Pairs
HNO2(aq) + H2O(l)  H3O+aq) + NO2-(aq)
acid
Bronsted
Acid





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base
Bronsted
Base
acid
base
H3O+ is conjugate NO2- is conjugate
acid of base H2O base of acid HNO2
Conjugate acid-base pairs are shown connected
Every acid has a conjugate base
Every base has a conjugate acid
The conjugate base has one fewer H and one more
“minus” charge than the acid
The conjugate acid has one more H and one fewer
minus charge than the base
16
Bronsted-Lowry Acids & Bases

Bronsted-Lowery Acid-Base Conjugate Pairs
NH3(aq) + H2O(l)  NH4+ aq) + OH-(aq)
base
acid
acid
base
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Bronsted
Base
Bronsted
Acid
Accepts
Proton
Donates
Proton
NH4+ is conjugate
acid of base NH3
OH- is conjugate
base of acid H2O
17
Bronsted-Lowry Acids & Bases

Bronsted-Lowery Acid-Base Conjugate Pairs
acid
H2S
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base
+
NH3
Bronsted
Acid
Bronsted
Base
Donates
Proton
Accepts
Proton
acid

NH4+
base
+
NH4+ is conjugate
acid of base NH3
HS-
HS- is conjugate
Base of acid H2S
18
Bronsted-Lowry Acids & Bases

Bronsted_Lowry Concept – The Leveling Effect

All Bronsted_Lowry acids yield H3O+ ions (Cation)

All Bronsted_Lowry bases yield OH- ions (Anion)

All strong acids are equally strong because all of them form
the strongest acid possible - H3O+

Similarly, all strong bases are equally strong because they
form the strongest base possible - OH-

Strong acids and bases dissociate completely yielding H3O+
and OH-

Any acid stronger than H3O+ simply donates a proton to H2O

Any base stronger than OH- simply accepts proton from H2O

Water exerts a leveling effect on any strong acid or base
by reacting with it to form water ionization products
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19
Practice Problem

Identify the conjugate acid-base pairs in the following:
H 2 PO 4 - (aq) + CO 3 -2 (aq)
Base
HCO 3 - (aq) + HPO4 2- (aq)
Conjugate Acid
Acid
Conjugate base
H 2 PO 4 - has one more H + than HPO 4 2- and
CO 3 2- has one fewer H + than HCO 3 H 2 PO 4 - & HCO 3 - are acids
HPO 4 2- & CO 3 2- are bases
H 2 PO 4 - / HPO 4 2- is a conjugate acid - base pair
HCO 3 - / CO 3 2- is a conjugate acid - base pair
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20
Lewis Acids & Bases

Lewis Acids & Bases

Some acid base reactions don’t fit the Bronsted-Lowry or
Arrhenius classifications

The Lewis acid-base concept expands the acid class

Such reactions involve a “sharing” of electron pairs
between atoms or ions

Lewis Acid – An electron deficient species (Electrophile)
that accepts an electron pair

Lewis Base – An electron rich species (Nucleophile) that
donates an electron pair
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21
Lewis Acids & Bases

Lewis Acids & Bases

The product of any Lewis acid-base reaction is called an
“Adduct”, a single species that contains a new covalent
bond (shared electron pair)
A+
+
Acid
 Electrophile 
:B
A-B
Base
 Nucleophile 
Adduct
F
B
F
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An acid is an
electron-pair
acceptor
F
F
H
+
H
N
HH
A base is an
electron-pair
donor
B
F
N
F
HH
Adduct
22
Lewis Acids & Bases

Species that do not contain Hydrogen in their formulas, such as
CO2 and Cu++ function as Lewis acids by accepting an electron
pair

The donated proton of a Bronsted-Lowry acid acts as a Lewis
acid by accepting an electron pair donated by the base:

The Lewis acids are Yellow and the Lewis bases are Blue in
following reaction
BF3

+ :NH3
Fe3+ + 6 H2O 
Fe(H2O)63+

F- + H3O+
HF
+ H2O
+
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BF3NH3
23
Lewis Acids & Bases

Solubility Effects

A Lewis acid-base reaction between nonpolar Diethyl
Ether and normally insoluble Aluminum Chloride
-
+

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+
The solubility of Aluminum Chloride in Dimethyl Ether
results from the Oxygen acting as a base by donating
an electron pair to the Aluminum acting as an acid
forming a water-soluble polar covalent bond
24
Practice Problem
The following shows ball-and-stick models of the reactants in
a Lewis acid-base reaction.
Write the complete equation for the reaction, including the
product
Identify each reactant as a Lewis acid or Lewis base
Donates
e- pair
Accepts
e- pair
AlCl3 + :NH3
Lewis acid
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Lewis base

AlCl3:NH3
Adduct
25
Acids & Bases (Summary)

Arrhenius acid concept
acid = proton (H+) producer;
base = OH- producer
HCl(aq)  H + (aq) + Cl - (aq)

KOH(s) 
K + (aq) + OH - (aq)
Formation of Water (H2O) from H+ & OH
Bronsted-Lowry concept

Stronger Acid (HA) transfers proton to a stronger base (H2O)
to form weaker acid (H3O+) and weaker base (A-)
HA(aq) + H 2O(l)  H 3O + (aq) + A - (aq)

Stronger base (NH3) accepts proton from stronger acid (H2O)
NH 3 (aq) + H 2O(l)

NH 4 + (aq) + OH - (aq)
Lewis Concept

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Donation and acceptance of an electron pair to form
a covalent bond in an adduct
26
Acids & Bases


Self Ionization of Water
Pure water undergoes auto-ionization to produce
Hydronium and Hydroxide ions
2 H2O(l)


H3O+(aq) + OH-(aq)
The extent of this process is described by an autoionization (ion-product) constant, Kw
K w = [H 3O+ ][OH- ] = 1.0  10-14


The concentrations of Hydronium and Hydroxide ions in
any aqueous solution must obey the auto-ionization
equilibrium
Classification of solutions using Kw:
Acidic:
[H3O+] > [OH-]
Basic:
[H3O+] < [OH-]
Neutral:
[H3O+] = [OH-]
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27
Acids & Bases

Because of the auto-ionization constant of water, the
amounts of Hydronium and Hydroxide ions are always
related
 For example, pure water at 25 oC,
K w = [H3O+ ][OH- ] = 1.0  10-14
[H 3O)+ ] = [OH- ] = 1.0  10-7

The Definition of pH
 The amount of Hydronium ion (H3O+) in solution is often
described by the pH of the solution
pH = - log[H 3O+ ]
[H3O+ ]

=
1  10-pH
The Definition of pOH
 Hydroxide (OH-) can be expressed as pOH
pOH = - log10 [OH- ]
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[OH - ] = 1  10-pOH
28
Acids & Bases

pH scale ranges from 0 to 14

The pH of pure water = -log(1.00 X 10-7) = 7.00
pH = 0.00 - 6.999+ = acidic
[H3O+] > [OH-]
pH = 7.00
[H3O+] = [OH-]
= neutral
pH = 7.00+ - 14.00 = basic

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[H3O+] < [OH-]
Acidic solutions have a lower pH (higher [H3O+] and
a higher pOH (lower [OH-] than basic solutions
29
Acids & Bases

Relations among pH, pOH, and pKw
K w = [H 3O+ ][OH- ] = 1.0  10-14
-log(K w ) = - log([H3O+ ][OH- ])
= - log(1  10-14 )
-log(K w ) = - log[H3O+ ] - log[OH- ] = - log(1  10-14 )
pK w =
1/19/2015
pH
+
pOH
=
14 (at 25o C)
30
Equilibrium & Acid-Base Conjugates

Equilibrium Constants (K) for acids & bases can also be
expressed as pK
 Reaction of an Acid (HA) with water as a base
HA(aq)
+
Acid
H2O(l)
H3O+ (aq)
Base
Conjugate Acid
[H 3O+ ][A- ]
Ka =
[HA]

Conjugate Base
pK a = - log10 [K a ]
Reaction of a Base (A-) with water as an acid
A- (aq)
Base
Kb =
1/19/2015
A-
+
+
H2O(l)
HA(aq)
Acid
Conjugate Acid
[HA][OH - ]
-
[A ]
+
OH- (aq)
Conjugate Base
pK b = - log10 [K b ]
31
Equilibrium & Acid-Base Conjugates

Relationship between:
Ka and HA
Kb and A Setup two dissociation reactions:
HA + H 2O
H 3O + + A A - + H 2O
HA
+ OH (autoionization)
2H 2O
H 3O + + OH  The sum of the two dissociation reactions is the
autoionization of water
 The overall equilibrium constant is the product of the
individual equilibrium constants
[H 3O + ][A - ] [HA][OH - ]
×
=
[H 3O + ][OH - ]
[HA]
[A ]
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Ka
×
Kb
=
Kw
32
Equilibrium & Acid-Base Conjugates

From this relationship, Ka of the acid in a conjugate pair
can be computed from Kb and vice versa

Reference tables typically have Ka & Kb values for
molecular species, but not for ions such as F- or CH3NH3+
Kw = Ka × Kb
-log(K w ) = - log([K a ][K b ])
-log(K w ) = - log[K a ] - log[K b ]
1/19/2015
pK w =
pK a
+
pK b = 14
(at 25o C)
pK w =
pH
+
pOH = 14
(at 25o C)
33
Equilibrium & Acid-Base Conjugates

Acid strength [H3O+] increases with increasing value of Ka
[H 3O+ ][A- ]
Ka =
[HA]

Base strength [OH-] increases with increasing value of kb
Kb =

[HA][OH - ]
[A - ]
A low pK corresponds to a high value of K
pK a = - log10 [K a ]
pK b = - log10 [K b ]

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A reaction that reaches equilibrium with mostly products
present (proceeds to far right) has a “low pK” (high K)
34
Practice Problem

Ex. Find Kb value for F-
HF + H 2O  H 3O + F
+
K a (HF) = 6.8  10-4
-
(K values of compounds from table)
K a (HF)  K b (F - ) = K w
-14
K
1.0

10
-11
w
K b (F - ) =
=
=
1.5

10
K a (HF)
6.8  10-4
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35
Acids & Bases

The Relationship Between Ka and pKa
Ka at 25oC
Hydrogen Sulfate ion (HSO4-) 1.0x10-2
Acid Strength
Acid Name (Formula)
1.99
Nitrous Acid (HNO2)
7.1x10-4
3.15
Acetic Acid (CH3COOH)
1.8x10-5
4.74
Hypobromous Acid (HBrO)
2.3x10-9
8.64
1.0x10-10
10.00
Phenol (C6H5OH)
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pKa
36
Acids & Bases
The pH Scale
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37
Acids & Bases

Strong Acids & Bases
 Strong acids and bases dissociate completely in aqueous
solutions
Strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4
Strong bases: all Group I & II Hydroxides
NaOH, KOH, Ca(OH)2, Mg(OH)2
 The term strong has nothing to do with the concentration
of the acid or base, but rather the extent of dissociation
 In the case of HCl, the dissociation lies very far to the
right because the conjugate base (Cl-) is extremely weak,
much weaker than water
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
 H2O is a much stronger base than Cl- and “out competes”
it for available protons
1/19/2015
38
Acids & Bases
Activity Series
Strengths of
acids and bases
Strength
determined by K,
not concentration
1/19/2015
39
Practice Problem

A strong monoprotic acid is dissolved in a beaker of water.
Which of the following pictures best represents the acid
solution
Ans: B
A solution of a
strong acid contains
equal amounts of
H3O+ and the acid
anion, such as a
“Cl-”
The monoprotic
acid is completely
dissociated
1/19/2015
A
B
C
40
Hydrohalic Acids – Weak vs Strong

The reason Hydrofluoric acid is weak relative to
the other Hydrohallic Acids (I, Br, Cl) is quite
complicated

The material being somewhat beyond the scope of
the discussions we have in the chem 211 & chem
212 courses

It starts out with the short length of the H - F
bond and the very high electronegativity of the
Fluoride ion

When you take into account the Enthalpy,
Entropy, and Free Energy properties (Chapter 19),
the numbers clearly show a distinct difference
between HF and the other Hydrohalic acids
1/19/2015
41
Hydrohalic Acids – Weak vs Strong

These properties can be summarized as follows:

Very strong Hydrogen Bonding between the
un-ionized Hydrogen Fluoride (HF) molecules
and water molecules

The energy required to break the H-F bond,
is much greater than for the bonds of the
other Hydrohalic Acid bonds

A large decrease in Entropy when the Hydrogen
Fluoride molecules react with water
1/19/2015
42
Hydrohalic Acids – Weak vs Strong

This is particularly noticeable with Fydrogen
Fluoride because the attractiveness of the very
small Fluoride ions produced imposes a lot of
order on the surrounding water molecules, and
also on nearby hydroxonium ions

The effect falls as the halide ions get bigger
1/19/2015
43
Hydrohalic Acids – Weak vs Strong

Note the values below. Hydrofluoric acid clearly
stands out as a weak acid

The Ka values clearly suggest a weak acid

The larger the value of the equilibrium constant
(Ka), the stronger the acid
ΔH
TΔS
(kJ mol-1) (kJ mol-1)
ΔG
Ka
(kJ mol-1)
(mol/L)
6.3 x 10-4
HF
-13
-29
+16
HCl
-59
-13
-46
2.00 x 10+6
HBr
-63
-4
-59
5.01 x 10+8
HI
-57
+4
-61
2.0 x 10+9
1/19/2015
44
Practice Problem
What are the concentrations of Hydronium Ions (H3O+) and
Hydroxide (OH-) ions in 1.2 M HBr?
HBr + H2O 
H3O+ + Br-
[HBr] = 1.2 mol / L
[H 3O+ ] = 1.2 mol / L
HBr is a strong acid
and is completely
dissociated
K w = [H3O+ ][OH- ] = 1  10-14
Autoionization of Water
Kw
1 x 10-14 mol / L
[OH ] =
=
+
1.2 mol / L
[H 3O ]
-
[OH- ] = 8.3 x 10-15 M
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45
Practice Problem
What are the concentrations of Hydronium Ions (H3O+) and
Hydroxide (OH-) ions in 0.085 M Ca(OH)2?
Ca(OH)2

Ca+2
+
2OH-
[Ca(OH)]2 = 0.085 mol / L
[OH ] = 2  0.085 = 0.17 mol / L
-
K w = [H3O+ ][OH- ] = 1  10-14
Ca(OH)2 is a strong
base and is completely
dissociated
Autoionization of Water
-14
K
1

10
w
[H 3O + ] =
=
[OH ] 0.17 mol / L
[H 3O+ ] = 5.8  10-14 mol / L
1/19/2015
46
Practice Problem
What is the pH of the following solution:?
[OH-] = 5.6 x 10-10 M
K w = [H 3O+ ][OH- ]
Kw
1.0  10-14
-5
[H 3O ] =
=
=
1.786

10
M
-10
[OH ]
5.6  10
+
pH = - log10 [H3O+ ] = - log10 (1.786  10-5 )
pH = -(log10 1.786 + log10 10-5 )
pH = - (0.252 - 5) = 5 - 0.252 = 4.75
1/19/2015
47
Practice Problem
What is the pOH of the following solution:?
[H3O+] = 9.3 x 10-4 M
K w = [H 3O+ ][OH- ]
-14
K
1.0

10
w
[OH - ] =
=
= 1.0753 M
+
-4
[H 3O ]
9.3  10
pOH = -log10 [OH- ] = -log10 (1.075268  10-11 )
pOH = -(log10 1.075268 + log10 10-11 )
pOH = - (0.031516 - 11.0) = 11.0 - 0.031516 = 11.0
1/19/2015
48
Practice Problem
What is the pH of a solution containing 0.00256 M HCl?
HCl + H 2O
H 3O+ + Cl -
Hydrochloric Acid (HCl) is a strong acid
It dissociates completely in aqueous solution
Thus : [HCl] = [H3O+ ] = [Cl - ] = 0.00256M
pH = - log[H 3O+ ]
pH = - log[0.00256]
pH = - (-2.59) = 2.59
1/19/2015
49
Practice Problem
What is the pH of a solution containing 0.00813 M Ca(OH)2?
Ca(OH)2
Ca+2 + 2OHCalcium Hydroxide (Ca(OH)2 ) is a strong base
It dissociates completely in aqueous solution
Thus : [Ca(OH)2 ] = [Ca+2 ] = 2[OH- ] = 0.00813 M
[OH- ] = 2× 0.00813 = 0.01626 M
K w = [H 3O+ ][OH- ]
-14
Kw
1.0

10
-13
[H 3O + ] =
=
=
6.15006

10
M
0.01626
[OH ]
pH = - log[H 3O+ ]
pH = - log[6.15006 x 10-13 ] = 12.2
1/19/2015
50
Practice Problem
What is [H3O+] in a solution with a pOH of 5.89?
pK w = pH + pOH = 14.00
pH = pK w - pOH = 14.00 - 5.89 = 8.11
pH = - log10 [H3O+ ]
log10[H 3O+ ] = -pH
[H3O+ ] = 10(-pH) = 10-8.11
[H3O+ ] = 7.76 x 10-9 M
1/19/2015
51
Weak Acid Equilibria Problems

Solving Problems involving Weak-Acid Equilibria


Types

Given equilibrium concentrations, find Ka

Given Ka and some concentrations, find other
concentrations
Approach

Determine “Knowns” and “Unknowns”

Write the Balanced Equation

Set up equilibrium expression –

Define “x” as the unknown change in [HA]

1/19/2015
[C ]c [ D ]d .....
Ka 
[ A]a [ B ]b ....
Typically, x = [A]diss (conc. of HA that dissociates)
x = [A]diss = [H3O+] = [A-]
52
Weak Acid Equilibria Problems

Construct a “Reaction” table

Make assumptions that simplify calculations

Usually, “x” is very small relative to initial
concentration of HA

Assume [HA]init ≈ [HA]eq & Apply 5% rule
If “x” < 5% [HA]init ; assumption is valid


Substitute values into Ka expression, solve for “x”
The Notation System

Brackets – [HA]; [HA]diss; [H3O+]
indicate “Molar” concentrations

1/19/2015
Bracket with “no” subscript refers to molar
concentration of species at equilibrium
53
Weak Acid Equilibria Assumptions

Assumptions
HA(aq)

+
H2O(l)
H3O+ (aq)
+
A-
The [H3O+] from the “autoionization” of water is
“Negligible”
It is much smaller that the [H3O+] from the dissociation
of HA
[H3O+] = [H3O+]from HA + [H3O+]from H2O ≈ [H3O+]from HA

A weak acid has a small Ka

Therefore, the change in acid concentration can be
neglected in the calculation of the equilibrium
concentration of the acid
1/19/2015
[HA] = [HA]init - [HA]dissoc ≈
[HA]init
54
Weak Acid Equilibria Assumptions

Assumption Criteria - 2 approaches
HA(aq) + H2O(l)
H3O+ (aq) + A H 3O +   A - 
K =
 HA 

For Kc relative to [HA]init (or Kp relative to PA(init))
[HA]init
if
> 400, the assumption [HA]  [Ha]init is justified
Kc
Neglecting x introduces an error < 5%
[HA]init
if
< 400, the assumption [HA]  [HA]init is not justified
Kc
Neglecting x introduces an error > 5%

For [HA]diss relative to [HA]init
if [HA]diss < 5% [HA]init , the assumption ([HA]  [HA]init ) is valid
1/19/2015
55
Practice Problem
What is the Ka of a 0.12 M solution of Phenylacetic acid (HPAc)
with a pH of 2.62
HPAc(aq) + H2O(l)  [H3O+](aq) + PAc-(aq)
[H3O+]from HPAc + [H3O+]from H2O ≈ [H3O+]from HPAc
pH = - log[H3O+]
[H3O+] = 10-pH = 10-2.62 = 2.4 x 10-3 M
+
H2O(l)

H3O+
+
PAc-
Concentration
HPAc(aq)
Initial Ci
0.12
-
0
0

-X
-
+X
+X
Equilibrium Cf
0.12 - X
-
X
X
Final Cf
0.12
2.4x10-3
2.4x10-3
[H3O+] = x = 2.4 x 10-3 M = [PAc]
HPAc is weak acid  HPAc = 0.12 M – x ≈ 0.12 M
[H 3O + ][PAc- ]
(2.4 x 10-3 )(2.4 x 10-3 )
Ka =

= 4.8×10-5
[HPAc]
0.12
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56
Practice Problem
What is the [H3O+] of a 0.10 M solution of Propanoic Acid
(HPr)?
Ka = 1.3 x 10-5
HPr(aq) + H2O
H 3O+ (aq) + Pr - (Aq)
[H 3O + ][Pr - ]
Ka =
= 1.3 x 10-5
[HPr]
HPr is a weak acid = [HPr]init = 0.10 M
Let x = [HPr]diss = [H3O+]from HPr = [Pr-]from HPr
+ H2O(l)

H3O+
+
Pr-
Concentration
HPr(aq)
Initial Ci
0.10
-
0
0

-X
-
+X
+X
Equilibrium Cf
0.10- X
-
X
X
Final Cf
0.10
X
X
Assumption: Ka is very small and x is small compared to
[HPr]init
 HPr = 0.10 M – x ≈ 0.10 M
Con’t
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57
Practice Problem (con’t)
Substituting into the Ka expression and solving for x
[H 3O + ][Pr - ]
(x)(x)
-5
Ka =
= 1.3  10 
[HPr]
0.10

x = 0.10  1.3  10-5

x = 1.1 x 10-3 M = [H 3O+ ] = [Pr - ] = [HPr]diss
Checking the Assumption: [HBr]diss < 5% [HBr]init
[HPr]diss 1.1  10-3 M
For [HPr]diss :
=
x 100 = 1.1%
[HPr]init
0.10M
1.1% < 5%  Assumption Valid
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58
Weak Acid Equilibria

Effect of Concentration on Extent of Acid Dissociation
Case 1: [HPr]init = 0.10 M
[H 3O + ][Pr - ]
(x)(x)
Ka =
= 1.3  10 -5 
[HPr]
0.10
x = 1.1 x 10-3 M = [H 3O+ ] = [Pr - ] = [HPr]diss
% HA diss
[HPr]diss
1.1 x 10-3 M
=
=
x 100 = 1.1%
-1
[HPr]init
1.0 x 10 M
Case 2: [HPr]init = 0.01 M
x = 3.6 x 10-4 M = [H 3O+ ] = [Pr - ] = [HPr]diss
% HAdiss
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3.6 x 10-4 M
=
x 100 = 3.6%
-2
1.0 x 10 M
59
Weak Acid Equilibria

Polyprotic Acids
 Acids with more than one ionizable proton are polyprotic
acids
 One proton at a time dissociates from the acid molecule
 Each dissociation step has a different Ka
H3PO4(aq) + H2O(l)  H2PO4- + H3O+(aq)
K a1
[H 2 PO 4- ][H 3O + ]
=
= 7.2 x 10-3
[H 3 PO 4 ]
H2PO4-(aq) + H2O(l)  HPO4-2 + H3O+(aq)
K a2
[HPO 4-2 ][H 3O + ]
-8
=
=
6.3
x
10
[H 2 PO 4- ]
HPO4-2(aq) + H2O(l)  PO4-3 + H3O+(aq)
K a3
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[PO 4-3 ][H 3O + ]
-13
=
=
4.2
x
10
[HPO 4- ]
K a1 >> K a2 >> K a3
60
Weak Acid Equilibria
Successive dissociation equilibrium constants (Ka)
for polyprotic acids have significantly lower values
Ka1 >> Ka2 >> Ka3
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61
Practice Problem

Calculate the following for 0.05 M Ascorbic Acid (H2Asc):
[H Asc], [HAsc-], [Asc2-], pH
2
Ka1 = 1.0 x 10-5
H2 Asc(aq) + H2O(l)
K a1
K a1 >> K a2
HAsc- (aq) + H3O+ (aq)
[HAsc- ][H 3O + ]
=
= 1.0  10-5
[H 2 Asc]
HAsc- (aq) + H2O(l)
K a2
Ka2 = 5 x 10-12
Asc2- (aq) + H3O+ (aq)
[Asc2- ][H 3O + ]
-12
=
=
5.0

10
[HAsc- ]
 [H3O+ ] from H2 Asc > > [H3O+ ] from HAsc-
Because K a1 is small, the amount of H2 Asc that dissociates can be neglected
relative to initial conc :
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[H2 Asc]init
 [H2 Asc]eq = 0.05 M
Con’t
62
Practice Problem (con’t)

Assumptions
Because K a2 << K a1 , [H 3O + ]from HAsc
[H 3O + ] from H Asc
-
2
[H 3O + ]from H Asc  [H 3O + ]
2
Because K a1 is small, [H 2 Asc]init - x = [H 2 Asc]  [H 2 Asc]init
[H 2 Asc] = 0.050M - x  0.050 M
K a1
[H 3O + ][HAsc- ]
=
= 1 x 10-5 
[H 2 Asc]
x
= [HAsc- ]  [H 3O+ ]  7.1  10-4 M
x2
0.050
pH = - log[H3O+ ] = - log(7.1  10-4 ) = 3.15
0.05
-
0
0
-
-
+X
+X
-
X
X
7.1x10-4
7.1x10-
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Equilibrium Cf
0.05 - X
Final Cf
0.05
+
HAsc-
Initial Ci
X
H2O(l)
H3O+
H2Asc(aq)

+

Concentration
4
Con’t
63
Practice Problem (con’t)

Calculate [Asc2-]
K a2
[Asc 2- ][H 3O + ]
-12
=
=
5.0
x
10
[HAsc- ]
(K a2 )[HAsc- ]
[Asc ] =
[H 3O + ]
2-
(5 x 10-12 )(7.1 x 10-4 )
-12
[Asc ] =
=
5

10
M
-4
7.1 x 10
2-
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64
Weak Bases & Relation to Weak Acids

Base – Any species that accepts a proton (Bronsted)

In water the base accepts a proton from water acting as
an acid (water is amphoteric) and leaves behind a OH- ion
B(aq) + H2O(l)
BH+ (aq) + OH- (aq)
[BH + ][OH - ]
Kc =
[B][H 2O]

The water term is treated as a constant in aqueous
reactions and is incorporated into the value of Kc
K c [H 2O] = K b
[BH + ][OH- ]
=
[B]
pK b = -logK b

Base strength increases with increasing Kb (pKb deceases)
1/19/2015
65
Weak Base Equilibria

Ammonia is the simplest Nitrogen-containing compound
that acts as a weak base in water
NH3 (aq) + H2O(l)
NH4+ (aq) + OH- (aq)

Ammonium Hydroxide (NH4OH) in aqueous solution
consists largely of unprotonated NH3 molecules, as its small
Kb indicates:
 NH 4 +  OH - 
 NH 4 +  OH - 
Kb =
=
 NH 3   H 2O
 NH 3 
K b = 1.76×10-5
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66
Weak Base Equilibria

In a 1.0 M NH3 solution:
[OH- ] = [NH4+ ] = 4.2 x 10-3 M
The NH3 is about 99.58% undissociated

If one or more of the Hydrogen atoms in ammonia is
replaced by an “organic” group (designated as R), an
“Amine” results:
Primary
Amine

Secondary
Amine
Tertiary
Amine
The Nitrogen atom in each compound has a “lone-pair” of
unbonded electrons, a Bronsted Base characteristic, i.e.,
the electron pair(s) can accept protons
1/19/2015
67
Weak Base Equilibria
1/19/2015
68
Practice Problem

What is the pH of a 1.5 M solution of Dimethylamine
(CH3)2NH? Kb = 5.9 x 10-4
(CH3 )2 NH2+ (aq) + OH- (aq)
(CH3 )2 NH(aq) + H2O(l)
Kb
[(CH 3 )2 NH 2 + ][OH - ]
=
[(CH 3 )2 NH]
Because Kb > > K w , the [OH- ] from the autoionization of water is negligible,thus,
[OH- ]from base = [(CH3 )2 NH2+ ] = [OH- ]
Because K b is small, assume the amount of Amine reacting is small, thus
[(CH3 )2 NH]init - [(CH3 )2 NH]reacting = [(CH3 )2 NH]  [(CH3 )2 NH]init
1/19/2015
Concentration
(CH3)2NH(aq
)
Initial Ci
1.5
-
0
0

-X
-
+X
+X
Equilibrium Cf
1.5 - X
-
X
X
+
H2O(l)

(CH3)2NH2(aq)
+
H3O+
69
Practice Problem (con’t)
Since Kb is small: [(CH3)2NH]init  [(CH3)2NH]: thus,
1.5 M - x 
1.5 M
[(CH 3 )2 NH 2 + ][OH - ]
x2
-4
Kb =
= 5.9 x 10 
[(CH 3 )2 NH]
1.5
x = [OH - ]  3.0 x 10-2
Checking the assumptions :
[B]init
Kb
3.0 x 10-2 M
x 100 = 2.0% (< 5%; assumption is justified)
1.5 M
1.5
3
=
=
2.5

10
> 400 (Assumption is justified
-4
5.9  10
Calculating pH:
Kw
1.0  10-14
-13
[H 3O ] =
=
=
3.3

10
M
-2
[OH ]
3.0  10
+
1/19/2015
pH = - [H 3 O + ] = - log(3.3 ×10-13 ) = 12.5
70
Anions of Weak Acids & Bases

The anions of weak acids are Bronsted-Lowry bases
A - (aq) + H 2O(l)
HA(aq) + OH - (aq)
Base (A- ) accepts a proton from water to form conjuate acid (HA)
F - (aq) + H 2O(l)
HF(aq) + OH - (aq)
-
[HF][OH ]
Kb =
[F - ]
-
F , anion of weak acid HF, acts as weak base
and accepts a proton to form HF
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71
Anions of Weak Acids & Bases

The acidity of HF vs F
HF is a weak acid, very little dissociates

Very little H3O+ and F- produced
HF(aq) + H 2O(l)

Water autoionization also contributes H3O+ and OH-,
but in much smaller amounts than H3O+ from HF
2H 2O(l)

H 3O + (aq) + F - (aq)
H 3O + (aq) + OH - (aq)
The Acidity of a solution is influenced mainly by the
H3O+ from HF and OH- from water
[H 3O + ]from HF >> [OH - ]from H O
(solution acidic)
2
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72
Anions of Weak Acids & Bases

Basicity of A-(aq)
 Consider 1 M solution of NaF
 Salt dissociates completely to yield stoichiometric
concentrations of F- , assume Na+ is spectator ion
 Some of the F reacts with water to form OH
F - (aq) + H 2O(l)
HF(aq) + OH - (aq)
[HF][OH - ]
Kb =
[F - ]

Water autoionization also contributes H3O+ and OH-, but
in much smaller amounts than OH- from reaction of Fand H2O
2H 2O(l)
-
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3/14/20
H 3O + (aq) + OH - (aq)
+
[OH ]from F- >> [H 3O ]from H O
2
(solution basic)
73
73
Anions of Weak Acids & Bases

Summary

The relative concentration of HA & A- determine the
acidity or basicity of the solution

In an HA solution, [HA] >> [A-] and
[H3O+]from HA >> [OH-]from H3O+
“Solution is Acidic”

In an A- solution, [A-] >> [HA] and
[OH-]from A-
>> [H3O+]from H3O+
“Solution is Basic”
1/19/2015
74
Practice Problem
What is the pH of 0.25 M Sodium Acetate (CH3COONa)?
Ka of Acetic Acid (HAc) is 1.8 x 10-5
Sodium Acetate, a Sodium salt, is water soluble
Initial concentration of Acetate, [Ac- ]init  [Ac- ] = 0.25 M
-
Ac (aq) + H 2O(l)
HAc(aq) + OH (aq)
Concentration
Ac-(aq)
Initial Ci
0.25
-
0
0

-X
-
+X
+X
Equilibrium Cf
0.25 - X
-
X
X
Kb
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[HAc][OH - ]
Kb =
[Ac- ]
-
+
H2O(l)

HAc(aq)
+
OH-
Kw
1.0 x 10-14
-10
=
=
=
5.6
x
10
Ka
1.8 x 10-5
75
Practice Problem

pH of 0.25 M Sodium Acetate (Con’t)
2
[HAc][OH - ]
X
-10
Kb =
=
5.6
x
10

0.25
[Ac ]
X = [OH - ]  1.2 x 10-5 M
Kw
1  10-14
-10
[H 3O ] =
=
=
8.3

10
M
-5
[OH ]
1.2  10
+
pH = - log[H3O+ ] = - log(8.3  10-10 ) = 9.08
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76
Acid Strength

Trends in Acid Strength of Nonmetal Hydrides

As the electronegativity of the
nonmetal (E) bonded to the
ionizable proton increases
(left to right), the acidity
increases

As the length of the E-H bond
increases (top to bottom), the
bond strength decreases, so
the acidity increases

The “strong” monoprotic
halide acids (HCl, HBr, HI) are
equally strong
Note: (HF) is a weak acid)
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77
Acid Strength in Oxoacids

Oxoacids

An Oxoacid is an acid that contains Oxygen

To be more specific, it is an acid that:
1/19/2015

contains oxygen

contains at least one other element (S, N, X)

has at least one Hydrogen atom (acid hydrogen)
bound to oxygen

forms an ion by the loss of one or more protons

acidity of an Oxoacid is not related to bond strength
as in nonmetal hydrides
78
Acid Strength in Oxoacids

Oxoacids with 1 Oxygen - HOE structure:
The more electronegative is E, the stronger the acid
Acid Strength:

HOCl > HOBr > HOI
Oxoacids with multiple Oxygens - (OH)mEOn structure
The more Oxygen atoms, the greater the acidity
HOClO3 (HClO4) > HOClO2 (HCLO3) >
HOClO (HCLO2) > HOCL
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(HClO)
79
Acid Strength in Oxoacids

For oxoacids with the same number of oxygen atoms
around E, acid strength increases with the electronegativity
of E
Ka(HOCl) = 2.9x10-8  Ka(HOBr) = 2.3x10-9  Ka(HOI) = 2.3x10-11

For oxoacids with different numbers of oxygen atoms
around a given E, acid strength increases with the number
of oxygen atoms.

Ka (HOCl
Hypochorous acid) = 2.9x10-8 (Weak)

Ka (HOCLO
Chlorous acid)

Ka (HOCLO2 Chloric acid)
≈ 1

Ka (HOCLO3 Perchloric acid)
= > 107
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= 1.12x10-2
(Strong)
80
Acidity of Hydrated Metal Ions

Aqueous solution of certain metal ions are acidic

As the MX species dissolves in water, the ions separate and
become bonded to a specific number of water molecules –
Hydration
M(NO3 )n (s) + xH2O(l)  M(H2O)xn+1 + nNO3- (aq)

If the metal ion, M+, is “small and highly charged (high
charge density), the hydrated metal ion transfers an H+ ion
from a bonded water molecule to water (Bronsted acid)
forming H3O+

The bound H2O that releases the proton, becomes a
bound OH- ion

The released proton forms Hydronium Ion with water
making solution Acidic
M(H2O)xn+1 (aq) + H2O(l)
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M(H2O)x-1 OH(n-1)+ (aq) + H3O+ (aq)
81
Acidity of Hydrated Metal Ions

The acidic behavior of the hydrated Al3+ ion.
Electron density
drawn toward Al3+
Nearby Base (H2O) attracts proton from one of the
Water molecules in the hydrated Al ion producing
a Hydronium ion (H3O+)
+
H 3 O+
OH-
Al(H2O)6
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3+
H 2O
Soln is
Acidic
Al(H2O)5OH2+
82
Practice Problem
Explain with equations and calculations whether an aqueous solution of
each salt is acidic, basic, or neutral.
1. Fe3+(HCOO1-)3 (Iron Formate)
Fe(HCOO)3(s) + n H2O(l) → Fe(H2O)n3+(aq) + 3 HCOO– (aq)
Fe(H2O)n3+(aq) + H2O(l) ⇆ Fe(H2O)n-1OH2+(aq) + H3O+(aq)
HCOO– (aq)
+ H2O(l) ⇆ OH– (aq) + HCOOH(aq)
Ka (Fe3+) = 6 x 10–3
Ka (HCOOH) = 1.8 X 10-4
(Appendix C)
Kb (HCOO–) = Kw / Ka (HCOOH) = (1.0 x 10–14) / (1.8 x 10–4)
Kb (HCOO–) = 5.6 x 10–11 (unrounded)
Ka (Fe3+) > Kb (HCOO–)
 Solution is Acidic
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(6 x 10–3 > 5.6 x 10–11)
83
Practice Problem
2. KHCO3 (Potassium Bicarbonate)
KHCO3(s)
+ H2O(l) → K+(aq)
+ HCO3– (aq)
HCO3– (aq) + H2O(l) ⇆ H3O+(aq) + CO32– (aq)
HCO3– (aq) + H2O(l) ⇆ OH– (aq) + H2CO3(aq)
Potassium (Group 1A) is fully hydrated in aqueous solution and does not
participate in an acid/base reaction
Only the Bicarbonate (anion of weak acid) is considered in acidity
determination
Ka (HCO3–) = 4.7 x 10-11 From Appendix C
Kb (HCO3–) = Kw / Ka (HCO3-) = (1.0 x 10–14) / (4.7 x 10–11)
Kb (HCO3–) = 2.1 x 10-4
Since Kb (HCO3–) > Ka (HCO3–)
(2.1 x 10-4 > 4.7 x 10-11)
Solution is basic
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Practice Problem
3. K2S (Potassium Sulfide)
K2S(s)
+ H2O(l) ⇆ 2 K+(aq) + S2– (aq)
S2– (aq) + H2O(l)
⇆ OH– (aq) + HS– (aq)
basic
Potassium (Group 1A) is fully hydrated in aqueous solution and does not
participate in an acid/base reaction
Only the Sulfide (S2-) (anion of weak acid) is considered in acidity
determination
Ka (S2-) = 9.0 x 10-8
(H2S from table)
Kb (S2-) = Kw / Ka (S2-) = (1.0 x 10–14) / (9.0 x 10–8)
Kb (S2-) = 1.1 x 10-7
Since Kb (S2-) > Ka (S2-) Solution is basic
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Acid-Base Properties – Salt Solutions

Neutral Salt Solutions
 Salts consisting of the Anion of a Strong Acid and the
Cation of a Strong Base yield a neutral solutions because
the ions do not react with water
HNO3 (l) + H 2O(l)  NO3- (aq) + H 3O+ (aq)
Nitrate (NO3-) is a weaker base than water (H2O) ;
reaction goes to completion as the NO3- becomes fully
hydrated; does not react with water
NaOH(s)

H O(l)
2
Na + (aq) + OH - (aq)
Sodium becomes full hydrated; does not react with
water
H2O
NaNO3 (s) 
 Na+ (aq) + NO3- (aq) + H 3O+ (aq) + OH- (aq)
Na+ & NO - do not react with water, leaving just the
3
“autoionization of water, i.e., a neutral solution
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86
Acid-Base Properties – Salt Soln

Salts that produce Acidic Solutions

A salt consisting of the anion of a strong acid and the
cation of a weak base yields an acidic solution

The cation acts as a weak acid

The anion does not react with water (previous slide)

In a solution of NH4Cl, the NH4+ ion that forms from
the weak base, NH3, is a weak acid

The Chloride ion, the anion from a strong acid does
not react with water
H2O
NH 4Cl(s) + H 2O 
 NH 4 + (aq) + Cl -
NH4+ (aq) + H2O
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NH3 (aq) + H3O+
(Acidic)
87
Acid-Base Properties – Salt Soln

Salts that produce Basic Solutions

A salt consisting of the anion of a weak acid and the
cation of a strong base yields a basic solution

The anion acts as a weak base

The cation does not react with water

The anion of the weak acid accepts a proton from water
to yield OH- ion, producing a “Basic” solution
H2O
CH 3COONa(s) + H 2O 
 Na + (aq) + CH 3COO- (aq)
(dissoluton & hydration)
CH 3COO - + H 2O(l)
CH 3COOH(aq) + OH - (Basic)
(reaction of weak base - Acetate Anion))
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Salts of Weakly Acidic Cations and
Weakly Basic Anions

Overall acidity of solution depends on relative acid
strength (Ka) or base strength (Kb) of the separated ions
Ex. NH4CN - Acidic or Basic?


Write equations for any reactions that occur between
the separated ions and water
NH 4 + (aq) + H 2O(l)
CN - (aq) + H 2O(l)
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NH 3 (aq) + H 3O +
HCN(aq) + OH -
89
Salts of Weakly Acidic Cations and
Weakly Basic Anions
Ex. NH4CN - Acidic or Basic?
(Con’t)
Compare Ka of NH4+ & Kb of CNRecall: Molecular compounds only in tables of Ka & Kb
Kw
1.0 x 10-14
-10
K a of NH 4 =
=
=
5.7
x
10
K b of NH 3
1.76 x 10-5
+
Kw
1.0 x 10-14
-5
K b of CN =
=
=
1.6
x
10
K a of HCN 6.2 x 10-5
-
Magnitude of Kb (Kb > Ka) (1.6 x 10-5 / 5.7 x 10-10 = 3 x 104)
Kb of CN- >> Ka of NH4+
Solution is Basic
Acceptance of proton from H2O by CN- proceeds much further than
the donation of a proton to H2O by NH4+
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Salts of Weakly Acidic Cations and
Weakly Basic Anions
Behavior of Salts in Water
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91
Equation Summary
H 3O + A HA + OH -
HA + H 2O
A - + H 2O
H 3O + + OH -
2H 2O
2H 2O(l)
H 3O + (aq) + OH - (aq)
K w = [H 3O+ ][OH- ] = 1.0 x 10-14
-log(K w ) = - log([H3O+ ][OH- ]) = 14
-log(K w ) = - log[H3O+ ] - log[OH- ]
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pK w = - log[K w ]
[K w ] = 1 x 10-pK
pH = - log[H 3O+ ]
[H 3O+ ] = 1 x 10-pH
pOH = -log10 [OH- ]
[OH - ] = 1 x 10-pOH
pK w = pH + pOH = 14
(at 25o C)
w
92
Equation Summary
HA + H 2O
A - + H 2O
2H 2O
H 3O + A HA + OH -
H 3O + + OH -
[H 3O+][A - ] [HA][OH - ]
+
×
=
[H
O
][OH
]
3
[HA]
[A ]
Ka
×
Kb
=
Kw
= 14
-log(K w ) = - log([K a ][K b ])
-log(K w ) = - log[K a ] - log[K b ]
[H 3O+ ][A- ]
Ka =
[HA]
pK a = -log10 [K a ]
[K a ] = 1 x 10
[BH + ][OH- ]
Kb =
[B]
pK b = -log[K b ]
[K b ] = 1  10
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-pK a
-pK b
93
Equation Summary
[HA]diss ]
Percent HA dissociated =
x 100
[HA]init
[A]init
if
> 400, the assumption [HA]  [Ha]init is justified.
Kc
Neglecting x introduces an error < 5%
if [HA]diss < 5% [HA]init , the assumption ([HA]  [HA]init ) is valid
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94