CHAPTER 14
ACID/ BASE DEFINITIONS
ARRHENIUS
BRONSTED-LOWRY
LEWIS
1. ARRHENIUS
 Most limited definition of acids and bases.
 Acids supply H+ in aqueous solution.
 Bases supply OH-1 in aqueous solution.
 Limited since many bases do not contain OH-1.

Ex. NH3
2. BRONSTED-LOWRY
 More general definition.
 Acids are proton (H+) donors.
o When an acid loses a proton, it becomes a conjugate
base.
 Bases are proton acceptors.
o Once the base accepts a proton, it becomes a
conjugate acid.
 General form of a BL acid base reaction:
HA
+
H2O  A-1 +
H3O+
EXAMPLES:
Formic acid dissociates (HCOOH)
Perchloric acid dissociates
Acetic acid dissociates
 Relationship between acids and their
conjugates:
 The stronger the acid, the weaker its
conjugate base.
3. LEWIS
 Most general definition.
 Acids are electron pair acceptors.
Ex. BF3
 Bases are electron pair donors.
Ex. NH3
ACID DISSOCIATION
EQUILIBRIUM
 Ka expressions can be written for an acid
dissociation reaction.
 Ka = [ products]power
[reactants]power
Write the Ka expressions for the sample BL
reactions:
 Higher Ka values, more dissociation, stronger acids.
 Strong acids do not have a Ka value s(extremely large)
since the equilibrium lies so far to the right due to
complete dissociation of a strong acid.
 The strong acids are:
Sulfuric
Nitric
Perchloric
Hydrochloric
Hydrobromic
Hydroiodic
 Other Ka values are given in the appendix.
 Take note of acetic acid’s Ka value and memorize it for the AP
exam.
WATER
 Water’s Ka value is 1.0 x 10-14
How?
 Consider two waters reacting…this is called
auto-ionization of water.
 What is the concentration of H+ and OH- in a
sample of water.
MEASURING ACID and BASE
STRENGTH
 pH scale
 ranges from 0-14
 pH = - log [H+]
 pOH = -log [OH-1]
 pH + pOH = 14
For water @ 25 °C
pH = 7
[H+] = 10-pH
pOH = 7
[OH-1] = 10-pH
Kw = [H+][OH-1]
and pKw = pH + pOH = 14
Solution
pH
A
6.88
pOH
[H+1]
B
C
D
[OH-1]
8.4 E -14
3.11
1.0 E -7
Acid/base/
neutral
ACID STRENGTH and
CHEMICAL STRUCTURE
 When a substance is dissolved in water, it may
behave as an ACID, a BASE, or produce a
NEUTRAL solution.
 How does the chemical structure of a substance
determine such behaviors?
FACTORS AFFECTING ACID
STRENGTH
 Strength measured by:
o Ka value
 Recall Ka meaning.
o pH
pH = -log [H+]
measures how much an acid dissociates
1. BOND POLARITY
 a proton is transferred only when the H is the
positive pole of the compound
H + X
o in hydrides, ex. NaH
o
the H is negatively charged, so no H+ could result.
o in non-polar molecule, ex. CH4
 the electrons are not being pulled from the H, so no
H+ could result.
2. BOND STRENGTH also
affects acid strength
o strong bond – less likely to dissociate, weaker
acid
o weak bond – stronger acid
3. CONJUGATE STRENGTH
o stable conjugates – come from a strong acid
TYPES of ACIDS and
STRENGTH TRENDS
 Acids that can donate more than one proton are
called POLYPROTIC acids
o 2 protons, ex. H2SO4 = diprotic
o 3 protons, ex. H3PO4 = triprotic
1. BINARY ACIDS
made of H and one other element. General form H-X.
 H-X bond strength determines strength of the
acid.
 Bond strength DECREASES as the size of X
increases.
 Changes more drastically down a group.
 Going across a period, look at ELECTRONEGATIVITY.
IV
V
VI
VII
2
CH4
NH3
H2O
HF
3
SiH4
PH3
H2S
HCl
2. OXYACIDS
 contain OH groups bound to a central atom.
 OH groups acting as bases:
 When OH is bound to a group with extremely low
ELECTRONEGATIVITY.
 Ex. metals
 Ca(OH)2
 KOH
OH groups acting as acids:
 Bound to a NON-METAL.
 Does not readily lose OH.
 As the ELECTRONEGATIVITY of Y increases, so
will the acidity.
 The O-H bond becomes more POLAR and loss of
H+ is favored.
 When an additional OXYGEN is bound to the
central Y, further increases the POLARITY of the
OH bond favoring loss of the H+.
Rules for comparing oxyacid
strength:
1. for oxyacids that have the same number of OH
groups and the same number of O atoms, acid
strength increases with increasing electronegativity
of the central atom
2. for oxyacids with the same central atom, acid
strength increases as the number of oxygens
attached to Y increases.
Example:
Place the following oxyacids in order of increasing
strength:
HClO
HClO2
HClO4
HClO3
HBrO
3. CARBOXYLLIC ACIDS
 contain a carboxyl group
 additional oxygens attached to the carboxyl
group – draws electron density from the OH
group, increasing its polarity.
 strength of the carboxylic acid increases when
the number of electronegative atoms increases.
Example:
 CH3OH – is this an acid?
 CH3COOH – what acid is this? How does its
strength compare to CH3OH?
MORE PRACTICE
 Arrange the compounds in each of the following
series in order of increasing acid strength:
 AsH3
 HI
 NaH
 H2O
 Arrange the compounds in each of the
following series in order of increasing acid
strength:
a. H2SeO3
b. H2SeO4
c. H2O
Explain why
a. HCl is stronger than H2S as an acid.
b. benzoic acid (C6H5COOH) is a stronger acid than
phenol (C6H5OH)
c. H2SO4 is stronger than HSO4-1
Calculating pH of strong acids
 Large Ka, so products are favored.
 Assume the [acid]0 = [ H+ ] due to complete dissociation.
 What else contributes [ H+ ] in solution?
 Auto- ionization of water
 Assumed to be negligible for strong acids (unless they are
dilute >10-6 M)
 So,
pH = - log [acid]o
pH example (strong)
 Calculate the pH and [OH-1] of a 5 x 10_3 M solution of
HClO4.
pH of weak acid solutions
 Ka is small
 Consider all sources of H+ in solution
 ICE table
 Ka expression
 Solve, approximate whenever possible (check <5%)
pH example (weak)
 Calculate the pH of a 0.500 M solution of formic acid
HCOOH (Ka = 1.77 x 10-4).
Percent dissociation
 Shows how much of the initial acid is turned into H+
in solution.
% dissociation = amount dissociated x 100
initial concentration
 Calculate by comparing H+ at equilibrium to [acid]0.
 Ex. What is the percent dissociation in the formic acid
problem?
pH of a weak acid mixture
 Consider all sources of H+ and determine the one that
is most significant. Assume the others to be
negligible.
 LeChatlier’s principle predicts that acids with lower Ka
values will not be able to dissociate due to an
abundance of H+ from the stronger acid’s dissociation,
so the weaker acid’s dissociation is suppressed.
pH example (mixture of acids)
 Calculate the pH of a mixture of 2.00 M formic acid
and 1.50 molar hypobromous acid (Ka = 2.06 x 10-9).
1. Write dissociation reactions for all species in
solution.
HCOOH
HOBR
H2O
2. Determine the source of H+ in solution
3. ICE
HCOOH
H+
COOH-
4. Ka expression
5. H+ at equilibrium, pH
Ka from percent dissociation
 A 0.500 M solution of uric acid is 1.6% dissociated.
Calculate the value of Ka for uric acid.
HA
H+
A-
BASES
 Basically, the problems are the same as acids.
 Focus on [OH-] at equilibrium.
 Consider all sources of [OH-] including auto-ionization of
water.
 pOH = - log [OH-]
pH example (strong base)
 Calculate the pH of a solution made by putting 4.63
grams of LiOH into water and diluting to a total
volume of 400 ml.
pH example (weak base)
 Calculate the pH of a 0.350 M solution of
methylamine, CH3NH2 (Kb = 4.38 x 10-4)
Polyprotic acids
 Supply more than one proton to the solution.
 Each step has its own Ka value.
 Ka1 >> Ka2 > Ka3 , so many times it is possible to
ignore the contribution of H+ from the second (and
third)dissociation.
 Sulfuric acid is the exception to this rule!!!
pH example (polyprotic acid)
 Calculate the pH, [PO4-3] and [OH-] of a 6.0 M
phosphoric acid solution.
1. Write all relevant dissociation reactions and find Ka
values for each.
2. ICE, assumptions, Ka1
3. Ka2 to find [HPO4-2]
4. Ka3 to find [PO4-3]
5. [OH-], pOH, pH
ACID/ BASE properties of SALTS
 Ionic compounds dissociate in water and the resulting
ions can make the solution acidic, basic, or neutral.
 Na+ and other alkali and alkaline earth metals do NOT
exhibit acid or base properties.
What about Na (s)?
 Conjugate of strong acids or bases do NOT exhibit acid
or base properties.
 Kw = Ka x Kb = 1x10 -14
 If Ka> Kb, acidic
 If Kb> Ka, basic
Salt examples
 Predict whether each of the following will create an acid,
base, or neutral aqueous solution.
 Na3PO4
 KI
 NH4F
pH example (of a salt)
 Calculate the pH of a 0.500 M NaNO2 solution (Ka =
4.0 x 10-4)
Reaction
Kb
ICE
ACID-BASE PROPERTIES of SALT
SOLUTIONS
 When most salts are dissolved in water they will form
acidic, basic, or neutral solutions.
 When both the cation and the anion have an effect,
the ion with the weakest conjugate acid or base will
have the greatest influence on pH.
ANIONS
Reaction of anion in water:
A-1
+
H2O

HA
+
OH-1
If a strong acid is predicted as a product, the
equilibrium will lie to the left. No considerable acidic/
basic effects will be noted.
 If a weak acid is predicted as a product, the
equilibrium will lie to the right. Due to the
production of hydroxide ion, the pH of the solution will
increase (become more basic).
CATIONS
Polyatomic cations are considered to be conjugates of
weak bases.
Ex.
NH4+
+
H2O 
hydroxide ions are formed and the pH is increased.
 Most
metal ions can also react with water to
decrease the pH of an aqueous solution.
 Some exceptions include the alkali metals and some
alkaline earth metals (usually produce bases in H20).
 Metal ions are considered Lewis acids.
LEWIS DEFINITION of ACIDS
and BASES
 Lewis base – electron pair donor
 Ex. NH3
 Lewis acid – electron pair acceptor
Metal ions
 Positively charged metal ions attract the oxygens from the
water molecules.
 This reaction is known as hydration.
usually coordinate with # waters = 2 x metal ion’s charge
Example: Fe+3 in water: (coordinates with 6 water molecules)
 Acid dissociation constants for hydrolysis reactions incerase
with increasing charge and decreasing radius on the metal ion.
 Ex. Which would form a more acidic solution Cu+2 or Fe+3?
COMMON ION EFFECT (add on)
 Adding a salt to an acid base equilibrium that





contributes to the initial conjugate concentration.
Ex. What is the pH of a solution that is 0.5M in acetic
acid and 2.5M in sodium acetate, NaCH3COO?
Consider both dissociation rxns.
What ions does the salt contribute to the acid’s
equlibrium? (common ion)
What does LeChatelier predict?
Would the acid pH increase or decrease due to the
common ion?
BUFFERS
 Solutions that contain weak conjugate acid-base
pairs.
 Resist changes in pH when small amounts of strong
acid or base are added.
 Examples of common buffer systems:
 Blood
 Hydrogen carbonate/ carbonic acid
 Contain an acid species to neutralize hydroxide and
a basic species to neutralize hydrogen ions.
WEAK ACID BUFFER
HA

H+ + A-
With equilibrium expression:
Ka = [H+] [A-] / [HA]
Adding OH-1 to the solution:
OH-1
+
HA

H2O
+
A-
HA = weak acid in the buffer – used to absorb excess base
Adding H+ to the solution:
H+
+
A-

HA
A- = conjugate base of the weak acid in the buffer – used to absorb excess acid
BUFFER CAPACITY and PH
 Buffer capacity – the amount of acid or base a buffer can
neutralize before pH changes to a considerable degree.
 PH depends on the Ka for the acid and the relative
concentrations of the acid and base that comprise the
buffer.
 Calcualting pH and relating to Ka
 Uses the Henderson - Hasselbalch equation.
pH = pKa + log ([A-] / [HA] )
Henderson Hasselbalch (*hoff)
 Because weak acids and bases only slightly ionize,
it is possible to use the initial amounts of acid and
base when using the Henderson – Hasselbalch
relationship.
pH = pKa + log ([A-] / [ HA])
 Example: Calculate the pH of a buffer composed of
0.12 M lactic acid (HC3H3O3) and 0.10 M sodium
lactate. Ka for lactic acid = 1.4 x 10-4.

 Example: What is the ratio of HCO3-1 to H2CO3 in
blood of pH 7.4?
Weak base/ conjugate acid buffer
 pOH = pKb + log ([BH+]/[B])
ADDITION of A STRONG ACID
or BASE to BUFFER SYSTEM
 Reactions between STRONG ACIDS and WEAK BASES or a
STRONG BASE and WEAK ACID occur essentially to completion.
So the buffer should completely consume the additional acid or
base.
 Calculating pH after the addition of an acid or a base:
1. consider the acid-base neutralization reaction and determine its
effect on [HA] and [A-]. (Use stoichiometry).
2. Use Ka and the new concentration of [HA] and [A-] to calculate [H+]
using an ICE table or the Henderson – Hasselbalch equation.
Neutralization of excess acid
H+
+
A-  HA
adding H+ or OH- to
HA, A- at equilibrium
OH- + HA  H2O +
Neutralization of excess base
Stoichiometry
determines new
concentrations of the
conjugates
A-
pKa + log of
the ratio of
the
[conjugate]
[original
acid or base]
 Example: A buffer is made by adding 0.300 mole
acetic acid to 0.3000 mole sodium acetate to make
a 1.00L solution. The pH of the buffer is 4.74.
 calculate the pH of the solution after 0.020 moles of
NaOH is added
 calculate the pH of the solution after .020 moles of
HCl is added.