Acids and Bases
Chapters 14 and 15
Acid/Base Definitions

Arrhenius Model



Bronsted-Lowry Model



Acids produce hydrogen ions (H+ or H3O+) in
aqueous solutions
Bases produce hydroxide ions (OH-) in aqueous
solutions
Acids are proton (H+) donors
Bases are proton (H+) acceptors
Lewis Model


Acids are electron pair acceptors
Bases are electron pair donors
Example 1

Water acting as a base by accepting a proton,
HCl acitng as an acid by donating a proton.
HCl acting as an acid by producing H3O+ ions
in solution. H+ ions provided by the HCl act as
an acid by accepting a pair of electrons from
the oxygen atom in the water (a base) to form
a bond in H3O+.
HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
acid
base conjugate conjugate
acid
base
Example 2

Water acting as an acid by donating a
proton, NH3 acting as a base by
accepting a proton. NH3 acting as a
base by producing OH- ions in solution.
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
base
acid
conjugate conjugate
acid
base
Example 3

Electron rich ammonia acting as a base
by donating a pair of electrons to
electron deficient Boron, that in turn acts
as an acid by accepting the pair of
electrons.
NH3 + BF3  H3N-BF3
Conjugate Acid/Base Pairs
Related by a hydrogen ion on either side
of the equation
 Example 2 previously


NH3 and NH4+
Strength of Acids and Bases
Strong acids and bases are assumed
to completely ionize in solution.
 For a strong acid (HA) and a strong
base (B), the following reactions go to
completion (no reverse reaction
occurs)

HA(aq) + H2O(l)  H3O+ + A-(aq)
 B(aq) + H2O(l)  BH+(aq) + OH-(aq)

Strength of Acids and Bases

Weak acids and bases have very little
ionization
Equilibria are set up (reverse reactions are
possible)
 The equilibrium positions for each lie heavily
to the left hand side of the equation
 HA(aq) + H2O(l)
 H3O+ + A-(aq)
 B(aq) + H2O(l)
 BH+(aq) + OH-(aq)

Strength of Acids and Bases
Weak Acids
Organic acids, any acid not
listed as a strong acid
Strong Acids
HBr, HCl, HI, HNO3, H2SO4,
HClO4, HClO3
Weak Bases
Ammonia and organic bases
(ex. Methylamine)
Strong Acids
LiOH, NaOH, KOH, RbOH,
CsOH, Ba(OH)2, Sr(OH)2,
Ca(OH)2, Mg(OH)2
Define the following words:

Arrhenius Acid/Base
 Bronsted-Lowry Acid/Base
 Lewis Acid/Base
 Strong
 Concentrated
 Corrosive
 Weak
 Dilute
 Equilibrium
 Indicator
Self Ionization of Water
H2O + H2O

H3O+ + OH At 25, [H3O+] = [OH-] = 1 x 10-7
 Kw is a constant at 25 C:
 Kw = [H3O+][OH-]
 Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

Amphoterism
can accept H+ or donate H+
 This is called amphoterism
 Pure water does it by itself, although
not much
 Water will make both protons and
hydroxides without any outside work
 Water
Amphoterism
Reaction
 2H2O(l)
H3O+(aq) + OH-(aq)
 We use the typical chemist’s shortcut to
write this equation
 2H2O(l)
H+(aq) + OH-(aq)
 Kw = [H+][OH-]
 The value of Kw is known at room
temp

Amphoterism
If you increase either protons or
hydroxides (by adding acid or base) you
MUST decrease the other
mathematically in order to maintain the
constant value of Kw
 Example – calculate [H+] for a solution
with 1.0x10-5 M OH- at 25oC. Is it
neutral, acidic or basic?

The pH Scale
Used to indicate the strength of an acid
or base
 Ranges from 0 – 14
 Acids are less than 7
 Bases are greater than 7
 7 is considered neutral

The pH Scale
The pH Scale

Calculating pH, pOH
pH = -log10(H3O+)
 pOH = -log10(OH-)


Relationship between pH and pOH


pH + pOH = 14
Finding [H3O+], [OH-] from pH, pOH
[H3O+] = 10-pH
 [OH-] = 10-pOH

The pH Scale
[OH-] = 1 x 10-14
[H+]
pH
[H+] = 1 x 10-14
[OH-]
pOH = 14 - pH
pH = 14 - pOH
pOH = -log[OH-]
OH[OH-] = 10-pOH
pH = -log[H+]
[H+] = 10-pH
H+
pOH
The pH Scale

Examples
Calculate the pH of a solution of 0.030M
hydrochloric acid
 Calculate the pH of a 0.010M solution of
calcium hydroxide
 Calculate the H3O+ concentration in a
solution with a pH of 4.32
 Calculate the pH of a solution made by
dissolving 2.00g of potassium hydroxide in
distilled water to a total volume of 250.mL

Ka – The Equilibrium Constant for
Weak Acids

For a weak acid (HA) dissociation is incomplete so an
equilibrium must be set up
HA(aq) + H2O(l)
H3O+ + A-(aq)
Initial
X
0
0
Change
-x
+x
+x
Equilibrium X-x
0+x 0+x
 Can’t go directly to the pH equation because we don’t
know how much acid is dissociated
 Derive the following equilibrium expression to help
solve these problems
Ka = [H3O+][A-]
[HA]
Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
 Step #1: Write the dissociation
equation
 HC2H3O2  C2H3O2 - + H+

Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
 Step 2: ICE it!

HC2H3O2  C2H3O2 - + H+
I
0.50
0
0
C
-x
+x
+x
E 0.50 – x
x
x

Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
 Step 3: Set up the law of mass action
 HC2H3O2  C2H3O2 - + H+
1.8 x 10-5 = (x)(x) = x2
(0.50-x) 0.50

Weak Acid Equilibrium Problem
What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?
 Step 4: Solve for x, which is also [H+]
 [H+] = 3.0 x 10-3 M

Weak Acid Equilibrium Problem

What is the pH of a 0.50 M solution of
acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?

Step 5: Convert [H+] to pH
pH = -log(3.0 x 10-3) = 2.52
Kb – The Equilibrium Constant for
Weak Bases

For a weak base (B) dissociation is incomplete so an
equilibrium must be set up
B(aq) + H2O(l)
BH+ + OH-(aq)
Initial
X
0
0
Change
-x
+x
+x
Equilibrium X-x
0+x 0+x
 Can’t go directly to the pOH equation because we
don’t know how much base is dissociated
 Derive the following equilibrium expression to help
solve these problems
Kb = [BH+][OH-]
[B]
Weak Base Equilibrium Problem

What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?

Step #1: Write the equation for the
reaction
NH3 + H2O  NH4+ + OH-
Weak Base Equilibrium Problem

What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?

Step 2: ICE it!
NH3 + H2O  NH4+ + OHI 0.50
0
0
C
-x
+x
+x
E 0.50 – x
x
x
Weak Base Equilibrium Problem

What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?

Step 3: Set up the law of mass action
NH3 + H2O  NH4+ + OH1.8 x 10-5 =
(x)(x) = x2
(0.50-x)
(0.50)
Weak Base Equilibrium Problem

What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?

Step 4: Solve for x, which is also [OH-]
[OH-] = 3.0 x 10-3M
Weak Base Equilibrium Problem

What is the pH of a 0.50 M solution of
ammonia, NH3, Kb = 1.8 x 10-5 ?

Step 5: Convert [OH-] to pH
pOH = -log(3.0 x 10-3M) = 2.52
pH = 14.00 – pOH = 11.48
Percent Dissociation
%
dissociation =
amount dissociated(mol/L) x 100%
initial concentration(mol/L)
 Specifies
the amount of weak acid
that has dissociated to achieve
equilibrium
Example
 0.500
M uric acid (HC5H3N4O4) is
1.6% dissociated. Find Ka and pH.
 What
is the percent dissociation of
0.500 M HC2H3O2, Ka = 1.8x10-5?
Indicators
A substance that changes color
according to the pH of the solution
 Often are weak acids where the ionized
and unionized forms are different colors
HIn(aq)  H+(aq) + In-(aq)
color 1
color 2

Indicators

Common indicators for acids and bases
 Litmus paper: red for acids, blue for
bases
 Phenophthalein: colorless for acids,
pink for bases
 Methyl orange: red for acids, yellow for
bases
Selection of Indicators
Some Acid-Base Indicators
Strong Acid/Strong Base Titrations
Strong Acid/Strong Base Example

Calculate the pH after these volumes of
0.2500M HCl are added to 50.00 mL of
0.1500M NaOH.
a) 0.00 mL
b) 4.00 mL
c) 29.50 mL
d) 30.00 mL
e) 30.50 mL
f) 40.00 mL
Strong Acid/Strong Base Example
Notice the RAPID drop in pH.
 1.00 mL between points c and e moves
us from pH 11.20 to 2.81 (9 orders of
magnitude; a billion times difference!)

Weak Acid/Strong Base Titration
Strong Acid/Weak Base Titrations
Weak Acid/Strong Base Example

Find pH after these volumes of 0.400 M
NaOH are added to 50.00 mL of 0.200 M
HCOOH (Ka = 1.8x10-4)
a) 0.00 mL
e) 25.00 mL
b) 5.00 mL
f) 25.50 mL
c) 12.50 mL
g) 40.00 mL
d) 24.50 mL
Weak Acid/Strong Base Example
Notice weak acid/strong base has
endpoint (equivalence point) in the
BASIC region, not at pH of 7.00 like in
strong acid/strong base.
 The exact same logic applies (and the
same math is used) for strong
ACID/weak BASE

Monoprotic and Polyprotic Acids
 Monoprotic
acids contain only one
“ionizable” proton
 Polyprotic acids contain more than
one
 Usually only the first H+ comes off
easily as a strong ion
 Each comes H+ off in a stepwise
manner, each with its own Ka
Polyprotic acids

Example: Carbonic Acid

Stage 1:
H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq)
Ka1 = [H3O+][HCO3-]
[H2CO3]

Stage 2:
HCO3-(aq) + H2O(l)  H3O+(aq) + CO3-2(aq)
Ka2 = [H3O+][CO3-2]
[HCO3-]

Overall:
H2CO3(aq) + 2H2O(l)  2H3O+(aq) + CO3-2(aq)
Ka = [H3O+]2[CO3-2] = (Ka1 )(Ka2)
[H2CO3]
Example
Find [PO4-3], pH and [OH-] in 6.0 M
H3PO4
 Ka1 = 7.5x10-3
Ka2 = 6.2x10-8
Ka3 = 4.8x10-13
 Since Ka1 >> Ka2 >> Ka3 we can ignore
the H+ in reactions 2 and 3
 To find [PO4-3] we must work our way all
the way to the last equation

Buffered Solutions
A
solution that resists a change in
pH when either hydroxide ions or
protons are added.
 Buffered solutions contain either:
A weak acid and its salt
A weak base and its salt
Acid/Salt Buffering Pairs

The salt will contain the anion of the acid, and
the cation of a strong base (NaOH, KOH)
Weak Acid
Formula of the
acid
Example of a salt of the weak acid
Hydrofluoric
HF
KF – Potassium fluoride
Formic
HCOOH
KHCOO – Potassium formate
Benzoic
C6H5COOH
NaC6H5COO – Sodium benzoate
Acetic
CH3COOH
NaH3COO – Sodium acetate
Carbonic
H2CO3
NaHCO3 – Sodium bicarbonate
Propanoic
HC3H5O2
NaC3H5O2 – Sodium propanoate
Hydrocyanic
HCN
KCN – Potassium cyanide
Base/Salt Buffering Pairs

The salt will contain the cation of the base,
and the anion of a strong acid (HCl, HNO3)
Base
Formula of
the base
Example of a salt of the weak base
Ammonia
NH3
NH4Cl – ammonium chloride
Methylamine
CH3NH2
CH3NH2Cl - Methylammonium
chloride
Ethylamine
C2H5NH2
C2H5NH3NO3 - Ethylammonium
nitrate
Aniline
C6H5NH2
C6H5NH3Cl - Aniline hyrdrochloride
Pyridine
C5H5N
C5H5NHCl - Pyridine hydrochloride
Buffered Solutions - Example

A 0.100M solution of ethanoic acid (Ka =
1.8 x 10-5) is mixed with a solution of
0.100M potassium ethanoate. Calculate
the pH of the resulting solution.
Henderson-Hasselbalch Equation
pH = pKa + log [A-] = pKa + log [salt]
[HA]
[acid]
pOH = pKb + log [BH+] = pKb + log [salt]
[B]
[base]
Acid-Base Properties of Salts

These salts simply dissociate in water:
Type of Salt
Examples Comment
Cation is from
a strong base,
anion from a
strong acid
KCl,
KNO3,
NaCl,
NaNO3

KCl(s)  K+(aq) + Cl-(aq)
pH of
solution
Both ions neutral
are neutral
Acid-Base Properties of Salts
Type of Salt
Examples Comment
Cation is from NaC2H3O2
a strong base, KCN,
NaF
anion from a
weak acid

Cation is
neutral,
anion is
basic
pH of
solution
basic
The basic anion can accept a proton from
water:
C2H3O2- + H2O
HC2H3O2 + OH-
Acid-Base Properties of Salts
Type of Salt
Examples Comment
pH of
solution
acidic
NH4Cl,
Cation is the
Cation is
conjugate acid NH4NO3
acidic,
of a weak
anion is
base, anion is
neutral
from a strong
acid
 The acidic cation can act as a proton
donor:
NH4+
NH3 + H+
Acid-Base Properties of Salts
Type of Salt
Examples Comment
pH of
solution
Cation is the
NH4C2H3O2 Cation is
See below
conjugate acid of NH4CN
acidic, anion
a weak base,
is basic
anion is
conjugate base
of a weak acid



IF Ka for the acidic ion is greater than Kb for the
basic ion, the solution is acidic
IF Kb for the basic ion is greater than Ka for the
acidic ion, the solution is basic
IF Kb for the basic ion is equal to Ka for the acidic
ion, the solution is neutral
Acid-Base Properties of Salts
Type of Salt
Examples
Cation is a highly Al(NO3)2;
charged metal
FeCl3
ion; anion is from
a strong acid

Comment
Hydrated
cation acts
as an acid;
anion is
neutral
pH of
solution
acidic
Step #1:

AlCl3(s) + 6H2O  Al(H2O)63+(aq) + Cl-(aq)
Salt
 Step #2:

water Complex ion
anion
Al(H2O)63+(aq)  Al(OH)(H2O)52+(aq) + H+(aq)
Acid
Conjugate base
Proton