Chapter 16
Acids and Bases
The Arrhenius Model
 An acid is any substance that produces hydrogen
ions, H+, in an aqueous solution.
 Example: when hydrogen chloride gas is dissolved
in water, the following ions are produced.
HCl(g)  H+(aq) + Cl-(aq)
 A base is an substance that produces hydroxide
ions, OH-, in an solution.
 Example: When solid sodium hydroxide is
dissolved in water the following ions are
produced.
NaOH(s)  Na+(aq) + OH-(aq)
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Another Theory: Brønsted-Lowry Model
 The Arrhenius model is limiting in its
classification of acids and bases, suggesting
there is only one kind of acid or base. A more
general definition was suggested by a Danish
chemist named Johannes Brønsted and an
English chemist named Thomas Lowry.
 The Brønsted-Lowry Model states:
 An acid is a proton (H+) donor
 A base is a proton (H+) acceptor.
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Brønsted-Lowry Model
 Let’s look at a general reaction for an acid (HA) in water.
HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
 In the reaction which reactant was the proton donor
(acid)?
 Which reactant was the proton acceptor (base)?
 If we look at the products, we see that now we have role
reversal.
 Which product is a proton donor?
 Which product is a proton acceptor?
 In this case, the proton donor in the products side is
known as the conjugate acid, and the proton acceptor is
known as the conjugate base.
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Practice
In the following reactions identify the acid, base, conjugate acid,
and conjugate base.
1.
2.
3.
4.
5.
HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
H2S(aq) + H2O(l)  H3O+(aq) + HS-(aq)
NH3(aq) + H2O(aq)  NH4+(aq) + OH-(aq)
HSO4- + C2O42-  SO42- + HC2O4Write the conjugate bases for the following acids.
a. HClO4
b. HNO3
c. HC2H3O2
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16.2 Acid Strength
 When we put an acid or base in water, the
compound breaks apart into it’s respective ions
is called dissociation.
 The degree which a compound dissociates in
water determines the strength of the acid or
base.
 Strong acids and bases are substances that are
completely ionized, or completely dissociated in
solution.
 Weak acids and bases are substances that only
ionize or dissociate partially in solution.
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Figure 16.1: Representation of the behavior of acids of
different strengths in aqueous solution.
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Acids vs Conjugate Bases
Acid
Strong Acids
Acid Strength
Increases
Weak Acids
Conjugate Base
HClO4
ClO4-
HI
I-
HBr
Br-
HCl
Cl-
H2SO4
HSO4-
HNO3
NO3-
H3O+
H2O
HSO4-
SO42-
HF
F-
HNO2
NO2-
HCOOH
HCOO-
CH3COOH
CH3COO-
NH4+
NH3
HCN
CN-
H2O
OH-
NH3
NH2-
Weak Bases
Base Strength
Increases
Strong Bases
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16.3 Water as an Acid and a Base
 A substance is said to be amphoteric if it can
behave either as an acid or as a base.
 Water is the most common amphoteric
substance. Let’s look at the ionization of 2 water
molecules.
H2O(l) + H2O(l) ↔ H3O+(aq) + OH-(aq)
 In this reaction one water molecule acts as an
acid, and one acts as a base.
 At 25 C the concentrations of these ions have
been calculated repeatedly to be
[H3O+]=[OH-]= 1.0 x 10-7 M
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Ion-Product Constant, Kw
• [H3O+][OH-] = 1.0 x 10-14 = Kw
• We call this constant, 1.0 x 10-14, Kw or the ion-product
constant for water.
• Kw = [H+][OH-] = 1.0 x 10-14
• In any aqueous solution at 25 C, no matter what it
contains, the product of [H+] and [OH-] must always
equal 1.0 x 10-14.
• This means if the [H+] goes up, the [OH-] must go down
so the product does not change.
• [H+] = the concentration of H ions
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
1.
2.
3.
3 Types of Situations
In an aqueous solution, there are 3 possible situations.
Each one always maintains a
Kw=1.0 x 10-14.
A neutral solution, where [H+]=[OH-]
An acidic solution, where [H+]>[OH-]
A basic solution, where [H+]<[OH-]
Example: Calculate the [H+] and determine the type of
solution if you have 1.0 x 10-5 M OHWhat information do you know from this equation?
Kw = [H+][OH-]
Rearrange to find [H+] = Kw = 1.0 x 10-14 = ____
[OH-] 1.0 x 10-5
Is this solution acidic, basic, or neutral?
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Practice Problems
Determine the [H+] and [OH-] and whether the solution is acidic,
basic or neutral.
1.
2.
3.
4.
10.0 M H+
1.0 x 10-7 M OH3.4 x 10-4 M H+
2.6 x 10-8 M H+
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16.4 The pH Scale
 Calculating the [H+] and [OH-] brings a lot of
very small numbers which is rather
inconvenient. In 1909 a French chemist, named
Soren Sorensen, came up with the pH scale.
 pH, literally means “power of hydrogen”





When calculating the pH we use the following:
pH = - log [H+]
Calculate the pH of a solution with [H+] = 1.0 x 10-7 M
Acidic solution pH < 7.00
Basic solution pH > 7.00
Neutral solution pH= 7.00
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The pH Scale
• On the pH scale, each increase of 1 unit equals a
power of ten change in the [H+].
• A solution with a pH of 3, has a [H+] = 10-3 M,
which is 10 greater than a solution with pH of 4,
or [H+] of 10-4 M and 100 times greater than a pH
of 5 or 10-5=[H+].
• As the [H+] increases, the pH decreases.
• The pH scale runs values from 0-14, zero being
very acidic, 14 being very basic, and 7 being
neutral.
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Figure 16.3: The pH scale.
Use the scale to determine if the
following are acidic, neutral or basic:
a. grapefruit pH=3.2
b. orange juice pH=3.5
c. urine pH=4.8-7.5 (depends on H2O)
d. saliva pH= 6.4-6.9
e. milk pH=6.5
f. gastric juice in stomach pH=2.0
g. blood pH=7.35-7.45
h. tears pH=7.4
i. milk of magnesia pH=10.6
j. ammonia pH=11.4
k. Draino pH=12.0
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Calculating pOH
• Sometimes we find it easier to calculate the pOH, since we
are given the [OH-].
pOH = -log [OH-]
• Calculate the pOH of [OH-] = 1.0 x 10-2 M
• The pOH is not commonly used to describe a solution, so
we must convert the pOH to the pH
• pH + pOH = 14
• What is the pH of our pOH solved at the top of this slide?
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Calculating pH and pOH Practice

1.
2.
3.
4.

Calculate the pH and pOH for the following:
[H+] = 1.0 x 10-4 M
[OH-] = 1.0 x 10-3 M
[H+] = 1.0 x 10-12 M
Determine whether each of the solutions above
is acidic, basic or neutral.
NOTE: When determining sig figs for logs, the
number of decimal places for a log must equal
the number of significant figures in the original
number.
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Calculating the [H+] from the pH
• Sometimes the pH is available, but we need to know the
[H+] of the solution. In this case we need to find the
inverse log of the –pH.
[H+] = inverse log (-pH)
• On your calculator you may enter the –pH, then push
the inverse key and then the log.
• Your calculator may also have a 10x key, typically above
the log key. With this key you would enter your –pH,
then push the 10x key.
• Calculate the [H+] if the pH is 5.00.
• The same sequence is used for [OH-] and pOH.
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Practice Problems
• The pH of a human blood sample was measured
to be 7.41. What is the [H+] in this blood?
• The pOH of a liquid drain cleaner was found to be
10.50. What is the [OH-] for this cleaner?
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• The pH curve for the
titration of 25.0 mL of
0.200 M HNO3 with 0.100
M NaOH. If you graphed
the results from the
computer simulation, it
would look like this
graph.
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Neutralization Reaction Net Ionic Equation
 Since the acid and base in the reaction are strong,
they will both dissociate completely in water. Write
the following as an ionic equation:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
H+(aq) +Cl-(aq) +Na+(aq) +OH-(aq) Na+ +Cl- + H2O(l)
We can then rewrite this as a NIE:
H+(aq) + OH-(aq)  H2O(l)
 The above is a neutralization reaction. It is called so
because if equal amounts of H+ and OH- are available
for reaction, a neutral solution will result.
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Calculating Volume in Neutralization Reactions
• What volume of 0.100 M HCl is needed to
neutralize 25.0 mL of a 0.350 M NaOH solution?
• MacidVacid(#H+) + Mbase Vbase (#OH- )
• (0.100M) ?V (1) = 0.35M(.025L)(1)
• V= (0.35M)(.025L)
0.100 M
V= 0.0875L or 87.5mL
OR YOU CAN CALCULATE USING STOICHIOMETRY!
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 Step 1: Write a balanced net ionic equation for the
reaction.
H+(aq) + OH-(aq)  H2O(l)
 Step 2: Calculate the moles of reactants.
25.0 mL NaOH x 1L x 0.350 mol NaOH
1
1000 mL
1 L NaOH
= 8.75 x 10-3 mol OH Step 3: Determine which reactant is limiting.
 This problem requires the addition of just enough H+ ions
to react exactly with the OH- ions present.
 The number of moles of OH- ions present determines the
number of moles of H+ ions.
 Therefore the OH- are limiting.
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 Step 4: Calculate the moles of H+ required.
 8.75 x 10-3 mol OH- x 1 mol H+ = 8.75 x 10-3 mol
H+
1
1 mol OH Step 5: Calculate the volume of 0.100 M HCl
required.
 We know that HCl:H+ is a 1:1 ratio, so the [H+] is
0.100 M.
 Volume x 0.100 mol H+ = 8.75 x 10-3 mol H+
1L
 Now we must solve for volume.
 Volume = 8.75 x 10-2 L or 87.5 mL
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Practice Problem
1. Calculate the volume of 0.10 M HNO3 needed
to neutralize 125 mL of 0.050 M KOH.
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