Acids andbases slides

AP Chemistry Rapid Learning Series - 21
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A id d BAcids and Bases
AP Ch i t R id L i S i
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© Rapid Learning Inc. All rights reserved.
AP Chemistry Rapid Learning Series
Wayne Huang, PhD
Kelly Deters, PhD
Russell Dahl, PhD
Elizabeth James, PhD
Debbie Bilyen, M.A.

Learning Objectives
Several definitions of “acid” and
“base”
By completing this tutorial you will learn about…
Strong versus weak acids and
bases
Conjugate acids and bases
How strong and weak relates to
equilibrium
How to find the pH of strong and
weak acids and bases
3/71
weak acids and bases
How salts can be acidic or basic
How buffers work
Titration curves
Concept Map
Chemistry
Studies
Previous content
New content
and Bases
Acids
and Bases Can be
Matter
pH EquilibriumEquilibrium
Constants
Conjugates
Form
Form
Scale to measure Have
Can be
4/71
Strong
or Weak
Strong
or Weak
Buffer
Common IonCommon Ion
Effect
Shown with
Affected by

Acid and Base
Definitions
5/71
Definition - Arrhenius
Arrhenius Acid – Produces the
hydronium ion in waterhydronium ion in water.
Arrhenius Base – Produces the
H3O+
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hydroxide ion in water.
OH-

O
HH Cl
Arrhenius Acids and Bases
Arrhenius Acid: HCl + H2O H3O+ + Cl-
O
HH Cl+1
-1
HH
O
HHHH
O
HH Acid
Water
Hydronium ion
Arrhenius Base: NH + H O NH + + OH-
7/71
Arrhenius Base: NH3 + H2O NH4
+ + OH-
HH
N
HH
HH
O
HHHH
HH
N
HH
HH
HH
O
HH
+1
-1
Definition - Brønsted-Lowry
Brønsted Lowry AcidBrønsted-Lowry Acid –
Donates a proton (H+).
Brønsted-Lowry Base –
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Accepts a proton (H+).

Definition - Lewis
Lewis Acid – Electron
tacceptor.
Lewis Base – Electron
donor.
9/71
Comparing the Definitions
How do the 3 definitions relate?
Acid Base Note
Produces
H3O+ in water
Donates a H+
Accepts
electrons
Produces
OH- in water
Accepts a H+
Donates
electrons
Requires water
Does not need to
be in water
Does not need to
use “H”
Arrhenius
Brønsted-
Lowry
Lewis
10/71
Many Arrhenius acids/bases are also Brønsted-Lowry and
Lewis acids/bases.

Properties of Acids and Bases
What are some common properties?
Acids Bases
Taste sour (e.g. citrus) Taste bitter (e.g. soap)
React with metals to
form H2 gas
Have pH levels below 7
Turn Litmus red
Feel slippery
Have pH levels above 7
Turn Litmus blue
11/71
StStrong versus
Weak
12/71

Definition - Strong and Weak
Strong – Most of the molecules perform
their “duty”.y
Strong Acid: Most of the molecules donate
their H+.
Weak – Very few of the molecules
perform their “duty”.
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perform their duty .
Weak Acid: Only a very few of the molecules
donate their H+.
Definition - Concentrated and Dilute
Concentrated – Many acid
b l l hor base molecules have
been added to the system.
Dilute – Only a few acid or
base molec les ha e been
14/71
base molecules have been
added to the system.

Possible Combinations
There are 4 combinations of strong, weak, concentrated
and dilute:
Concentrated Dilute
A lot added & almost
all do their “duty”.
A lot added, but very
Not many added, but
of what is there,
most will do their
“duty”.
Only a few added
and of those, only a
Strong
W k
15/71
A lot added, but very
few do their “duty”.
and of those, only a
very small % do their
“duty”.
Weak
Common Strong Acids and Bases
There are only a few common strong acids and bases to
remember - the rest will most likely be weak!
HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4
NaOH, KOH, Ca(OH)2, Ba(OH)2, Sr(OH)2
Strong Acids
Strong Bases
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Definition - Polyprotic Acids
Polyprotic acid – Acid
with more than one
proton to donate.
Examples …
Diprotic: H2CO3
17/71
Triprotic: H3PO4
Strength of Polyprotic Acids
Each proton is “weaker” than the one before.
A negatively
charged ion is
An acid donates
a proton
e.g: H2CO3
The results is a
negatively
charged ion
Becomes HCO3
-
less likely to
give up
another proton
and become -2
charged
Would be CO3
2-
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H2CO3 is a stronger acid than HCO3
-.
Sulfuric acid (H2SO4) is the only common strong polyprotic acid.
The 1st hydrogen is “strong” and the 2nd one is “weak”.

Conjugates of
Acids and Bases
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Definition - Conjugate Base
C j B Wh ’ l f fConjugate Base – What’s left after an
acid has donated its hydrogen.
HCl + H2O H3O+ + Cl-
Conjugate base
20/71
Acid
After the proton is donated, it can now act as a base…
it could accept a proton.

Definition - Conjugate Acid
Conjugate Acid What’s left after aConjugate Acid – What s left after a
base has accepted a proton.
Base Conjugate acid
21/71
After the proton is accepted, it can now act as an
acid… it could donate the extra proton.
Labeling Species
Example: For each of the following, label the acid (A), the base
(B), the conjugate acid (CA) and conjugate base (CB).
NH3 + H2O NH4
+ + OH-
3 2 4
H2SO4 + H2O H3O+ + HSO4
-
AB CA CB
A B CA CB
22/71
NH3 + H2SO4 NH4
+ + HSO4
-
AB CA CB

Definition - Amphoteric
Amphoteric – A molecule that can actp
as an acid or base.
Base
23/71
NH3 + H2O NH4
+ + OH-
Acid
Acid-Base
Equilibrium
24/71

Definition - Ka
Acid Dissociation Constant (Ka) –
Equilibrium constant for an acid
dissociation reaction.
General acid dissociation equation:
25/71
HA (aq) H+ (aq) + A- (aq)
][
]][[ 11
HA
AH
Ka
−+
=
Writing Ka Expressions
Example: Write the Ka expressions for the following:
HCl (aq) H+ (aq) + Cl- (aq)
]][[ 11
ClH −+
H2CO3 (aq) H+ (aq) + HCO3
- (aq)
][
]][[
HCl
ClH
Ka =
][
]][[
32
1
3
1
COH
HCOH
Ka
−+
=
26/71
HCO3
- (aq) H+ (aq) + CO3
2- (aq)
][
]][[
1
3
2
3
1
−
−+
=
HCO
COH
Ka

Definition - Kb
Base Dissociation
Constant (Kb) –( b)
Equilibrium constant
for a base
dissociation reaction.
General base dissociation equation:
BOH (aq) B+ (aq) + OH- (aq)
27/71
BOH (aq) B (aq) + OH (aq)
][
]][[ 11
BOH
OHB
Kb
−+
=
Writing Kb Expressions
Example: Write the Kb expressions for the following:
NaOH (aq) Na+ (aq) + OH- (aq)
]][[ 11
OHNa −+
Ca(OH)2 (aq) Ca2+ (aq) + 2 OH- (aq)
][
]][[
NaOH
OHNa
Kb =
])([
]][[
2
212
OHCa
OHCa
Kb
−+
=
28/71

Definition - Kw
Autoionization of Water – Water will
spontaneous dissociate into H+ and OH-.p
Water Dissociation Constant (Kw) –
Equilibrium constant for the autoionization of
water.
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Water Dissociation Equation: H2O (l) H+ (aq) + OH- (aq)
]][[ 11 −+
= OHHKw
H2O is not included because it’s a pure liquid!
Relating Ka, Kb and Kw
Ka and Kb of conjugates are related through Kw:
baw KKK ×= At 25°C, Kw = 1.0×10-14
Example: If the Ka for CH3COOH is 1.8×10-5, find the Kb for it’s
conjugate base CH3COO- at 25°C.
bK××=× −−
)108.1(100.1 514Ka = 1.8×10-5
Kw = 1.0×10-14
Kb = ?
bK=
×
−
−
)1081(
100.1
5
14
30/71
Kb = 5.6×10-10
b
× −
)108.1( 5

Relating Strength and Equilibrium
What do the Ka and Kb tell us about strength?
A large Ka means
the reaction
“lies to the
right”.
There is a large
ratio of products
: reactants.
For acid and base
reactions, the
products are
the dissociated
ions.
The more
dissociated ions,
the stronger the
acid or base.
31/71
The larger the Ka or Kb, the stronger the acid or base.
Strength and Conjugates
How does the strength of a species relate to the
strength of its conjugate?
If it donated the
A strong acid
donates it’s
proton easily.
After donating
the proton, it
then becomes
a (conjugate)
base.
If it donated the
proton easily, it
will not easily
grab and hold
onto another
one.
It will be a weak
base.
32/71
Strong Acid Weak Conjugate Base
Weak Acid
Strong Base
Weak Base
Strong Conjugate Base
Weak Conjugate Acid
Strong Conjugate Acid

pH
33/71
Definition - Logarithms
Logarithm – The number of
times a base must be
multiplied by itself to reach a
given number.
yx blog=
# of multiples
34/71
Base
# you’re trying to reach

Definition - pH
pH scale Logarithmic scalepH scale – Logarithmic scale
of the acidity of a solution.
The pH scale uses base “10”.
]log[ 1
3
+
−= OHpH
35/71
pH
OH −+
=10][ 1
3
pH has no units.
pH Scale
0 7 14
Some common items and their pH values:
2.0
Stomach
Contents
Acidic Neutral
6.5
~8.0
Tap water
Basic
36/71
7.4
Blood
Contents
3.0
Pop
4.5
Beer
5.5
Bread
Milk
Tap water

pH of Strong Acids
For strong acids, assume there is complete dissociation.
Example: Find the pH of a 0.25 M solution of HCl.
[H3O+] = 0.25 M
pH = ?
HCl is a strong acid:
If [HCl] = 0.25 M, then [H3O+] = 0.25 M
]log[ 1
3
+
−= OHpH
37/71
pH ?
pH = 0.60
)25.0log( MpH −=
pH of Strong Bases
When working base problems, you can find [OH-] by the
base concentration…but you must find [H3O+] to find
pH.
Example: Find the pH of a 0.15 M solution of NaOH .
NaOH is a strong base:
NaOH Na+ + OH-
If [NaOH] = 0.15 M, then [OH-] = 0.15 M
]][[ 11
3
−+
= OHOHKw
]10][[1001 114
O+
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[H3O+] = 6.7×10-14 M
pH = ?
pH = 13.18
]log[ 1
3
+
−= OHpH
)107.6log( 14
MpH −
×−=
]15.0][[100.1 1
3
14
MOH +−
=×

pH of Weak Acids - 1
When working with weak acids and bases, you must
use the equilibrium ICE chart techniques to find [H3O+].
Example: Find the pH of a 0.25 M solution of CH3COOH if
K = 1 8×10-5
CH3COOH H3O+ CH3COO-
Initial 0.25 M 0 M 0 M
Change
Equilibrium
-x +x +x
0.25 M - x 0 M + x 0 + x
Ka = 1.8×10 5 .
CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)
Approximations: 0.25 M 0 M + x 0 M + x
39/71
][
]][[
3
1
3
1
3
COOHCH
COOCHOH
Ka
−+
=
)25.0(
))((
108.1 5
M
xx
=× −
( )( ) xM =× −
25.0108.1 5
Ka is small enough to make approximations.
x = 0.0021 M
MMOH 0021.00][ 1
3 +=+
[H3O+] = 0.0021 M
pH of Weak Acids - 2
When working with weak acids and bases, you must
use the equilibrium ICE chart techniques to find [H3O+].
Example: Find the pH of a 0.25 M solution of CH3COOH if
K = 1 8×10-5
Initial 0.25 M 0 M 0 M
Change
Equilibrium
-x +x +x
0.25 M - x 0 M + x 0 + x
Ka = 1.8×10 5 .
40/71
x = 0.0021 M
[H3O+] = 0.0021 M
]log[ 1
3
+
−= OHpH
)021.0log( MpH −=
pH = 2.67

Acid-Base
Properties of Salts
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How Salts Have Acid/Base Properties
Understanding conjugates is a key to understanding
acid/base properties of salts.
A weak acid has
a strong
conjugate
base that can
form salts.
That strong
conjugate
base will
produce a
basic solution.
Salts made from
the conjugate
of a weak acid
will have a
basic pH.
42/71
The opposite is also true: Salts made from the conjugate of a
weak base will be acidic.

The Combinations to Form Salts
Cation from… Strong Acid Weak Acid
Anion from…
Strong Base
Weak Base
Neutral Salt
Acidic Salt
Basic Salt
Neutral Salt
Think of it as the “strong on wins” (e.g. strong acid & weak base = acidic)
43/71
Example: Determine if each salt will be acidic, basic or neutral:
NH4Cl
NaH2PO4
Weak base & strong acidCame from NH3 and HCl
Acidic salt
Came from NaOH and H3PO4 Strong base & weak acid
Basic salt
Buffers
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Definition - Buffer
Buffer – Solution containing a weak acid
or base and its conjugate that resistsj g
change in pH levels.
e.g. H3PO4 (weak acid) and PO4
3- (conjugate base)
Conjugates are added in the form of a salt
45/71
j g
(soluble ionic compound).
e.g. To add PO4
3- to a buffer, Na3PO4 salt is added.
How Does a Buffer Work?
Buffer: Weak Acid and
conjugate base
How does a buffer resist pH changes?
Conjugate base “absorbs”
acid and produces more weak
acid already in buffer.
Strong Acid added
Weak acid “absorbs” base and
produces more conjugate base
already in buffer.
Strong Base added
Eventually, so much strong acid or base could be added and all of
46/71
the buffer material would be used up.
Buffer Capacity – Amount of strong acid or base
that can be absorbed without large pH change.

Henderson-Hasselbach Equation
This equation is used to find the pH of a buffer system.
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+=
][
][
log
acid
base
pKapH
pKa = -log Ka
[acid] & [base] refer to molarity of a
weak acid or base and its conjugate.
Find the pH of a solution with 0.50 M HC2H3O2 (Ka =
1.8×10-5) and 0.30 M NaC2H3O2.
pKa = -log (1.8×10-5) = 4.74
[acid] = 0.50 M HC2H3O2
[base] = 0.30 M C2H3O2
-
⎟
⎠
⎞
⎜
⎝
⎛
+=
M
M
pH
50.0
30.0
log74.4
Example:
47/71
[base] 0.30 M C2H3O2
pH = ?
)22.0(74.4 −+=pH
pH = 4.52
Adding Acid to a Buffer - 1
When adding an acid or base to a buffer solution,
complete the stoichiometry first.
Then, with the remaining components, find the pH.
1.8×10-5) and 0.30 M NaC2H3O2 that has had HCl added to
equal 0.25 M.
Example:
HC2H3O2 C2H3O2
-
Buffer 0.50 M 0.30 M
Acid reacts + 0 25 M - 0 25 M
The acid will react with the conjugate base, producing more weak acid.
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Acid reacts + 0.25 M 0.25 M
New components 0.75 M 0.05 M
pKa = -log (1.8×10-5) = 4.74
[acid] = 0.75 M HC2H3O2
[base] = 0.05 M C2H3O2
-
pH = ?
⎟
⎠
⎞
⎜
⎝
⎛
+=
M
M
pH
75.0
05.0
log74.4 )17.1(74.4 −+=pH
pH = 3.57

Adding Acid to a Buffer - 2
When adding an acid or base to a buffer solution,
complete the stoichiometry first.
Then, with the remaining components, find the pH.
1.8×10-5) and 0.30 M NaC2H3O2 that has had HCl added
to equal 0.25 M.
Example:
Although the pH did drop…an earlier example showed adding
the same amount of HCl to plain water would drop the pH
49/71
pH = 3.57
p p p
from 7.0 to 0.60.
The plain water is affected to a much greater degree than the
buffer.
Adding Base to a Buffer
Likewise, adding a base to a buffer will raise the pH, but
much less than adding it to water.
Find the pH of a solution with 0 50 M HC H O (K =Example: Find the pH of a solution with 0.50 M HC2H3O2 (Ka =
1.8×10-5) and 0.30 M NaC2H3O2 that has had NaOH added
to equal 0.15 M.
Example:
HC2H3O2 C2H3O2
-
Buffer 0.50 M 0.30 M
Base reacts - 0.15 M + 0.15 M
N t 0 35 M 0 45 M
The base will react with the weak acid, producing more conjugate base.
50/71
New components 0.35 M 0.45 M
pKa = -log (1.8×10-5) = 4.74
[acid] = 0.35 M HC2H3O2
[base] = 0.45 M C2H3O2
-
pH = ?
⎟
⎠
⎞
⎜
⎝
⎛
+=
M
M
pH
35.0
45.0
log74.4 )11.0(74.4 +=pH
pH = 4.85

Titrations
51/71
Definition - Titrations
Titration Addition of a knownTitration – Addition of a known
volume of a known
concentration solution to a
known volume of unknown
concentration solution to
determine the concentration.
52/71
determine the concentration.

Definition - End Point
End Point (or Stoichiometric Point) –
When there is no reactant left over—they
h ll b t d d th l tihave all been reacted and the solution
contains only products.
I di t P li id th t h
The end point must be reached in order to use stoichiometry
to calculate the unknown solution concentration.
53/71
Indicators – Paper or liquid that change
color based on pH level.
If the pH of the products is known, the indicator can be
chosen to indicate the end point.
Definition - Titration Curve
Titration Curve –
Graph showing pHGraph showing pH
of unknown
solution verses
the volume of the
known solution
b i dd d
54/71
being added.

Titration Curve of Strong Acids
What does a titration of a strong acid look like?
Past the end point,
there is now
pH
End Point
pH changes dramatically as the
last molecule of acid reacts
there is now
excess base.
An indicator is
chosen that
changes color
at this pH
pKa
55/71
Volume of base added
pH changes
very slowly—
there is excess
acid remaining
½ way to end
point
Titration Curve of Weak Acids
How is the curve of a weak acid different from a strong
acid?
pH
W k
Higher starting pH (weak
acids are less “acidic”)
Less defined end point
inflection point
End point pH
above 7 (basic)
56/71
Volume of base addedStrong
Weak acids are less acidic )

Titration Curve of Polyprotic Acids
Diprotic acids have 2 protons to titrate.
pH
57/71
Volume of base added
There are 2 end points - usually one is more defined
than the other.
Common Ion
Effect
58/71

Definition - Common Ion Effect
Common Ion Effect – Effect an ion has on
the pH of an acidic or basic solution
containing the same ion.
59/71
Example: Adding NaCH3COO will affect the pH of
a solution of CH3COOH as they have a common
ion (CH3COO-).
Common Ion and Le Chatelier
Le Chatelier’s Principle says that adding a product will
push the reaction towards the reactants.
CH COOH (aq) + H O (l) H O+ (aq) + CH COO- (aq)CH3COOH (aq) + H2O (l) H3O (aq) + CH3COO (aq)
Adding the
common ion
(CH3COO-)
would push
the reaction
Fewer acid
molecules are
dissociated,
so there are
fewer H O+
pH will be higher
(less acidic)
with this
common ion
than without
60/71
back to the
reactants.
fewer H3O
ions.
than without.
Bases would work the opposite - adding the common ion of a
base would cause the solution to be “less basic” (have a lower
pH).

Solving Common Ion Problems
Common Ion problems are like weak acid problems -
except you have an initial amount of the conjugate.
Example: Find the pH of a 0.25 M solution of CH3COOH
(Ka = 1 8×10-5) if it is also 0 10 M NaCH COO
Initial 0.25 M 0 M 0.10 M
Change
Equilibrium
-x +x +x
0.25 M - x 0 M + x 0.10 + x
(Ka = 1.8×10 5) if it is also 0.10 M NaCH3COO.
Approximations: 0.25 M 0 M + x 0 10 M
61/71
][
]][[
3
1
3
1
3
COOHCH
COOCHOH
Ka
−+
=
)25.0(
)10.0)((
108.1 5
M
x
=× − ( )( ) x
M
M
=
× −
10.0
25.0108.1 5
Approximations: 0.25 M 0 M + x 0.10 M
Ka is small enough to make approximations
x = 4.5×10-5 M
MMOH 51
3 105.40][ −+
×+= [H3O+1] = 4.5×10-5 M
Solving Common Ion Problems
Common Ion problems are like weak acid problems -
except you have an initial amount of the conjugate.
Example: Find the pH of a 0.25 M solution of CH3COOH
(Ka = 1 8×10-5) if it is also 0 10 M NaCH COO
Initial 0.25 M 0 M 0.10 M
Change
Equilibrium
-x +x +x
0.25 M - x 0 M + x 0.10 + x
(Ka = 1.8×10 5) if it is also 0.10 M NaCH3COO.
Approximations: 0.25 M 0 M + x 0 10 M
62/71
Approximations: 0.25 M 0 M + x 0.10 M
x = 4.5×10-5 M
[H3O+1] = 4.5×10-5 M
]log[ 1
3
+
−= OHpH
)105.4log( 5
MpH −
×−=
pH = 4.35

Acids, Bases &
The AP Exam
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Acid/base in the Exam
Stoichiometry
Common acid/base problems:
Titrations
pH calculations
Ka and Kb
64/71

Multiple Choice Questions
Acid/base multiple choice questions tend to be more
qualitative.
Example: How can 100 mL of a 0.10 M basic solution with pH of
13.00 become a solution with a pH of 12.00?
A. Diluting solution with 8 more mL of distilled water
B. Diluting solution with 100 mL more of distilled water
C. Diluting solution with 900 mL more of distilled water
D. Adding 100 mL of 0.10 HA
E. Adding 100 mL of 0.10 base
65/71
pH is a log base 10 scale.
1 change in pH = 10 x change in acidity
You want the solution to be diluted by 1/10.
Making a 100 mL solution into a 1000 mL solution is a 1/10 dilution.
Answer: C
Free Response Questions
An acid-base free response question may contain sub-
questions such as:
a. Describe the features of the titration curve that indicate the acid is a
k idweak acid
b. Describe one way in which the acid dissociation constant can be
determined from the titration curve
c. A 25 mL sample of 0.10 M HA is titrated with 0.35 M NaOH
I. What is the volume of NaOH needed to reach the equivalence
point?
II. Is the pH at this equivalence point above or below 7?
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Answering Free Response Questions
a. Describe the features of the titration curve that indicate the acid
is a weak acid.
Weak acids begin at a higher pH and have a less defined
inflection point The stoichiometric point will be above pH
b. Describe one way in which the acid dissociation constant can be
determined from the titration curve.
inflection point. The stoichiometric point will be above pH
of 7
67/71
pKa = pH at volume added to get half-way to inflection
point. Then pKa = -log Ka
Answering Free Response Questions
c. A 25 mL sample of 0.10 M HA is titrated with 0.35 M NaOH
I. What is the volume of NaOH needed to reach the equivalence
point?
0.025 L HA 0.1 mole HA 1 mole NaOH 1 L NaOH
= _____ L NaOH
1 L HA 1 mole HA 0.35 mole NaOH
0.0071 L NaOH
68/71
II. Is the pH at this equivalence point above or below 7?
Weak acid + strong base = above 7

pH can be
determined from
pH can be
determined from
There are several
a s to define
There are several
a s to define
Acid/base
problems include
b ff tit ti
Acid/base
problems include
b ff tit ti
Learning Summary
Acid and baseAcid and base
determined from
the concentration
of hydronium
ions in a solution.
determined from
the concentration
of hydronium
ions in a solution.
ways to define
acids and bases.
ways to define
acids and bases.
buffers, titrations
and common ion
problems.
buffers, titrations
and common ion
problems.
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Acid and base
dissociation reactions
have dissociation
constants and can be
worked with ICE charts.
Acid and base
dissociation reactions
have dissociation
constants and can be
worked with ICE charts.
Some acids and bases
are strong, while
others are weak.
Some acids and bases
are strong, while
others are weak.
Congratulations
You have successfully completed
the core tutorial
Acids and Bases

Wh t’ N t
Chemistry :: Biology :: Physics :: Math
What’s Next …
Step 1: Concepts – Core Tutorial (Just Completed)
Step 2: Practice – Interactive Problem Drill
Step 3: Recap – Super Review Cheat Sheet
71/71
Go for it!
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Acids andbases slides

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Acids andbases slides