The document provides an overview of the chemistry of acids and bases, detailing their properties, nomenclature, and definitions according to various theories. It explains the pH scale, methods for calculating pH, and distinguishes between strong and weak acids and bases based on their ionization levels. The content also includes instructions for printing notes and practical examples for calculating concentrations in solutions.

1. 1
1
The Chemistry of
The Chemistry of
Acids and Bases
Acids and Bases Chemistry I – Chapter 19
Chemistry I – Chapter 19
Chemistry I HD – Chapter 16
Chemistry I HD – Chapter 16
ICP – Chapter 23
ICP – Chapter 23
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2. 2
2
Acid and Bases
Acid and Bases

3. 3
3
Acid and Bases
Acid and Bases

4. 4
4
Acid and Bases
Acid and Bases

5. 5
5
Acids
Have a sour taste. Vinegar is a solution of acetic acid. Citrus
Have a sour taste. Vinegar is a solution of acetic acid. Citrus
fruits contain citric acid.
fruits contain citric acid.
React with certain metals to produce hydrogen gas
React with certain metals to produce hydrogen gas.
.
React with carbonates and bicarbonates to produce carbon
React with carbonates and bicarbonates to produce carbon
dioxide gas
dioxide gas
Have a bitter taste.
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Feel slippery. Many soaps contain bases.
Bases

6. 6
6
Some Properties of Acids
 Produce H+
(as H3O+
) ions in water (the hydronium ion is a
hydrogen ion attached to a water molecule)
 Taste sour
 Corrode metals
 Electrolytes
 React with bases to form a salt and water
 pH is less than 7
 Turns blue litmus paper to red “Blue to Red A-CID”

7. 7
7
Anion
Ending Acid Name
-ide hydro-(stem)-ic acid
-ate (stem)-ic acid
-ite (stem)-ous acid
Acid Nomenclature Review
No Oxygen
No Oxygen

w/Oxygen
w/Oxygen
An easy way to remember which goes with which…
An easy way to remember which goes with which…
“
“In the cafeteria, you
In the cafeteria, you ATE
ATE something
something IC
ICky”
ky”

8. 8
8
• HBr
HBr (aq)
(aq)
• H
H2
2CO
CO3
3
• H
H2
2SO
SO3
3

 hydro
hydrobromic
bromic acid
acid

 carbon
carbonic
ic acid
acid

 sulfur
sulfurous
ous acid
acid
Acid Nomenclature Review

9. 9
9
Name ‘Em!
• HI
HI (aq)
(aq)
• HCl
HCl (aq)
(aq)
• H
H2
2SO
SO3
3
• HNO
HNO3
3
• HIO
HIO4
4

10. 10
10
Some Properties of Bases
 Produce OH
Produce OH-
-
ions in water
ions in water
 Taste bitter, chalky
Taste bitter, chalky
 Are electrolytes
Are electrolytes
 Feel soapy, slippery
Feel soapy, slippery
 React with acids to form salts and water
React with acids to form salts and water
 pH greater than 7
pH greater than 7
 Turns red litmus paper to blue “
Turns red litmus paper to blue “B
Basic
asic B
Blue”
lue”

11. 11
11
Some Common Bases
NaOH
NaOH sodium hydroxide
sodium hydroxide lye
lye
KOH
KOH potassium hydroxide
potassium hydroxide liquid soap
liquid soap
Ba(OH)
Ba(OH)2
2 barium hydroxide
barium hydroxide stabilizer for plastics
stabilizer for plastics
Mg(OH)
Mg(OH)2
2 magnesium hydroxide
magnesium hydroxide “MOM” Milk of magnesia
“MOM” Milk of magnesia
Al(OH)
Al(OH)3
3 aluminum hydroxide
aluminum hydroxide Maalox (antacid)
Maalox (antacid)

12. 12
12
Copyright McGraw-Hill 2009 12
All the other acids and bases are weak
electrolytes

13. 13
13
1. T/F. Acids pH 1-7. F 1-6.9
2. T/F. Bases taste bitter T
3. HF Hydrofluoric acid/ Hydroflouric
4. T/F. Strong bases, OH paired with elements
in Grp 3 – F Grp 1-2
5. NaOH – Sodium Hydroxide
6. H2SO4 – Sulfuric Acid
7. What is your basis of a strong acid? –
Complete dissociation, pH close to 1
8. HNO3, is it a strong acid? (Yes/No) - Yes
9. T/F Per if the no. oxygen increases T
10. Give one significance of acid or base?

14. 14
14
Copyright McGraw-Hill 2009 14
• Types of acids
–Monoprotic: one ionizable hydrogen
HCl + H2O  H3O+
+ Cl
–Diprotic: two ionizable hydrogens
H2SO4 + H2O  H3O+
+ HSO4

HSO4

+ H2O  H3O+
+ SO4
2

15. 15
15
Copyright McGraw-Hill 2009 15
–Triprotic: three ionizable hydrogens
H3PO4 + H2O  H3O+
+ H2PO4

H2PO4

+ H2O  H3O+
+ HPO4
2
HPO4
2
+ H2O  H3O+
+ PO4
3
–Polyprotic: generic term meaning
more than one ionizable hydrogen

16. 16
16
Copyright McGraw-Hill 2009 16
• Types of bases
–Monobasic: One OH
group
KOH  K+
+ OH
–Dibasic: Two OH
groups
Ba(OH)2  Ba2+
+ 2OH

17. 17
17
Acid/Base definitions
• Definition #1: Arrhenius (traditional)
Acids – produce H+
ions (or hydronium ions
H3O+
)
Bases – produce OH-
ions
(problem: some bases don’t have hydroxide
ions!)

18. 18
18
Arrhenius acid is a substance that produces H+
(H3O+
) in water
Arrhenius base is a substance that produces OH-
in water

19. 19
19
Acid/Base Definitions
• Definition #2: Brønsted – Lowry
Acids – proton donor
Bases – proton acceptor
A “proton” is really just a hydrogen
atom that has lost it’s electron!

20. 20
20
A Brønsted-Lowry acid is a proton donor
A Brønsted-Lowry base is a proton acceptor
acid
conjugate
base
base
conjugate
acid

21. 21
21

22. 22
22

23. 23
23
ACID-BASE THEORIES
ACID-BASE THEORIES
The Brønsted definition means NH
The Brønsted definition means NH3
3 is
is
a
a BASE
BASE in water — and water is
in water — and water is
itself an
itself an ACID
ACID
Base
Acid
Acid
Base
NH4
+
+ OH-
NH3 + H2O

24. 24
24
Conjugate Pairs
Conjugate Pairs

25. 25
25
How to determine a conjugate acid
and conjugate base

26. 26
26

27. 27
27

28. 28
28

29. 29
29

30. 30
30

31. 31
31

32. 32
32

33. 33
33

34. 34
34
Acids & Base Definitions
Acids & Base Definitions
Lewis acid - a
Lewis acid - a
substance that
substance that
accepts an electron
accepts an electron
pair
pair
Lewis base - a
Lewis base - a
substance that
substance that
donates an electron
donates an electron
pair
pair
Definition #3 – Lewis

35. 35
35
Formation of
Formation of hydronium ion
hydronium ion is also an
is also an
excellent example.
excellent example.
Lewis Acids & Bases
Lewis Acids & Bases
•Electron pair of the new O-H bond
Electron pair of the new O-H bond
originates on the Lewis base.
originates on the Lewis base.
H
H
H
BASE
••
•
•••
O—H
O—H
H+
ACID

36. 36
36
Lewis Acid/Base Reaction
Lewis Acid/Base Reaction

37. 37
37
Lewis Acid-Base Interactions
Lewis Acid-Base Interactions
in Biology
in Biology
• The heme group
The heme group
in hemoglobin can
in hemoglobin can
interact with O
interact with O2
2
and CO.
and CO.
• The Fe ion in
The Fe ion in
hemoglobin is a
hemoglobin is a
Lewis acid
Lewis acid
• O
O2
2 and CO can act
and CO can act
as Lewis bases
as Lewis bases
Heme group

38. 38
38
The
The pH scale
pH scale is a way of
is a way of
expressing the strength
expressing the strength
of acids and bases.
of acids and bases.
Instead of using very
Instead of using very
small numbers, we just
small numbers, we just
use the NEGATIVE
use the NEGATIVE
power of 10 on the
power of 10 on the
Molarity of the H
Molarity of the H+
+
(or
(or
OH
OH-
-
) ion.
) ion.
Under 7 = acid
Under 7 = acid
7 = neutral
7 = neutral
Over 7 = base
Over 7 = base

39. 39
39
pH of Common
pH of Common
Substances
Substances

40. 40
40
Calculating the pH
pH = - log [H+]
(Remember that the [ ] mean Molarity)
Example: If [H+
] = 1 X 10-10
pH = - log 1 X 10-10
pH = - (- 10)
pH = 10
Example: If [H+
] = 1.8 X 10-5
pH = - log 1.8 X 10-5
pH = - (- 4.74)
pH = 4.74

41. 41
41
Try These!
Try These!
Find the pH of
Find the pH of
these:
these:
1) A 0.15 M solution
1) A 0.15 M solution
of Hydrochloric
of Hydrochloric
acid
acid
2) A 3.00 X 10
2) A 3.00 X 10-7
-7
M
M
solution of Nitric
solution of Nitric
acid
acid

42. 42
42
pH calculations – Solving for H+
pH calculations – Solving for H+
If the pH of Coke is 3.12, [H
If the pH of Coke is 3.12, [H+
+
] = ???
] = ???
Because pH = - log [H
Because pH = - log [H+
+
] then
] then
- pH = log [H
- pH = log [H+
+
]
]
Take antilog (10
Take antilog (10x
x
) of both
) of both
sides and get
sides and get
10
10-pH
-pH
=
= [H
[H+
+
]
]
[H
[H+
+
] = 10
] = 10-3.12
-3.12
= 7.6 x 10
= 7.6 x 10-4
-4
M
M
*** to find antilog on your calculator, look for “Shift” or “2
*** to find antilog on your calculator, look for “Shift” or “2nd
nd
function” and then the log button
function” and then the log button

43. 43
43
pH calculations – Solving for H+
pH calculations – Solving for H+
• A solution has a pH of 8.5. What is the
A solution has a pH of 8.5. What is the
Molarity of hydrogen ions in the
Molarity of hydrogen ions in the
solution?
solution?
pH = - log [H
pH = - log [H+
+
]
]
8.5 = - log [H
8.5 = - log [H+
+
]
]
-8.5 = log [H
-8.5 = log [H+
+
]
]
Antilog -8.5 = antilog (log [H
Antilog -8.5 = antilog (log [H+
+
])
])
10
10-8.5
-8.5
= [H
= [H+
+
]
]
3.16 X 10
3.16 X 10-9
-9
= [H
= [H+
+
]
]

44. 44
44
More About Water
H
H2
2O can function as both an ACID and a BASE.
O can function as both an ACID and a BASE.
In pure water there can be
In pure water there can be AUTOIONIZATION
AUTOIONIZATION
Equilibrium constant for water = K
Equilibrium constant for water = Kw
w
K
Kw
w = [H
= [H3
3O
O+
+
] [OH
] [OH-
-
] =
] = 1.00 x 10
1.00 x 10-14
-14
at 25
at 25 o
o
C
C
HONORS ONLY!

45. 45
45
More About Water
K
Kw
w = [H
= [H3
3O
O+
+
] [OH
] [OH-
-
] = 1.00 x 10
] = 1.00 x 10-14
-14
at 25
at 25 o
o
C
C
In a
In a neutral
neutral solution [H
solution [H3
3O
O+
+
] = [OH
] = [OH-
-
]
]
so K
so Kw
w = [H
= [H3
3O
O+
+
]
]2
2
= [OH
= [OH-
-
]
]2
2
and so [H
and so [H3
3O
O+
+
] = [OH
] = [OH-
-
] = 1.00 x 10
] = 1.00 x 10-7
-7
M
M
OH-
H3O+
Autoionization
Autoionization
HONORS ONLY!

46. 46
46
pOH
• Since acids and bases are
Since acids and bases are
opposites, pH and pOH are
opposites, pH and pOH are
opposites!
opposites!
• pOH does not really exist, but it is
pOH does not really exist, but it is
useful for changing bases to pH.
useful for changing bases to pH.
• pOH looks at the perspective of a
pOH looks at the perspective of a
base
base
pOH = - log [OH
pOH = - log [OH-
-
]
]
Since pH and pOH are on opposite
Since pH and pOH are on opposite
ends,
ends,
pH + pOH = 14
pH + pOH = 14

47. 47
47
pH [H+
] [OH-
] pOH

48. 48
48
[H
[H3
3O
O+
+
], [OH
], [OH-
-
] and pH
] and pH
What is the pH of the
What is the pH of the 0.0010 M NaOH solution?
0.0010 M NaOH solution?
[OH-] = 0.0010 (or 1.0 X 10
[OH-] = 0.0010 (or 1.0 X 10-3
-3
M)
M)
pOH = - log 0.0010
pOH = - log 0.0010
pOH = 3
pOH = 3
pH = 14 – 3 = 11
pH = 14 – 3 = 11
OR K
OR Kw
w = [H
= [H3
3O
O+
+
] [OH
] [OH-
-
]
]
[H
[H3O
O+
+
] = 1.0 x 10
] = 1.0 x 10-11
-11
M
M
pH = - log (1.0 x 10
pH = - log (1.0 x 10-11
-11
) = 11.00
) = 11.00

49. 49
49
The pH of rainwater collected in a certain region of the
northeastern United States on a particular day was
4.82. What is the H+
ion concentration of the
rainwater?
The OH-
ion concentration of a blood sample is
2.5 x 10-7
M. What is the pH of the blood?

50. 50
50
[OH
[OH-
-
]
]
[H
[H+
+
]
] pOH
pOH
pH
pH 1
0
1
0
-
p
O
H
-
p
O
H
1
0
1
0
-
p
H
-
p
H
-
L
o
g
[
H
-
L
o
g
[
H
+
+
]
]
-
L
o
g
[
O
H
L
o
g
[
O
H
-
-
]
]
14
- pO
H
14
- pO
H
14
- pH
14
- pH
1.0
x
10
1.0
x
10
-14
-14
[O
H
[O
H
-
- ]
]
1.0
x
10
1.0
x
10
-14
-14
[H
[H
+
+ ]
]

51. 51
51
Calculating [H3O+
], pH, [OH-
], and pOH
Problem 1: A chemist dilutes concentrated
hydrochloric acid to make two solutions: (a) 3.0
M and (b) 0.0024 M. Calculate the [H3O+
], pH,
[OH-
], and pOH of the two solutions at 25°C.
Problem 2: What is the [H3O+
], [OH-
], and pOH
of a solution with pH = 3.67? Is this an acid,
base, or neutral?
Problem 3: Problem #2 with pH = 8.05?

52. 52
52
HNO3, HCl, H2SO4 and HClO4 are among the
only known strong acids.
Strong and Weak Acids/Bases
Strong and Weak Acids/Bases
The strength of an acid (or base) is
determined by the amount of
IONIZATION.
HONORS ONLY!

53. 53
53
Strong and Weak Acids/Bases
Strong and Weak Acids/Bases
• Generally divide acids and bases into STRONG or
Generally divide acids and bases into STRONG or
WEAK ones.
WEAK ones.
STRONG ACID:
STRONG ACID: HNO
HNO3
3 (aq) + H
(aq) + H2
2O (l) --->
O (l) --->
H
H3
3O
O+
+
(aq) + NO
(aq) + NO3
3
-
-
(aq)
(aq)
HNO
HNO3
3 is about 100% dissociated in water.
is about 100% dissociated in water.
HONORS ONLY!

54. 54
54
• Weak acids
Weak acids are much less than 100% ionized in
are much less than 100% ionized in
water.
water.
One of the best known is acetic acid = CH
One of the best known is acetic acid = CH3
3CO
CO2
2H
H
Strong and Weak Acids/Bases
Strong and Weak Acids/Bases
HONORS ONLY!

55. 55
55
• Strong Base:
Strong Base: 100% dissociated in
100% dissociated in
water.
water.
NaOH (aq) ---> Na
NaOH (aq) ---> Na+
+
(aq) + OH
(aq) + OH-
-
(aq)
(aq)
Strong and Weak Acids/Bases
Strong and Weak Acids/Bases
Other common strong
Other common strong
bases include KOH and
bases include KOH and
Ca(OH)
Ca(OH)2
2.
.
CaO (lime) + H
CaO (lime) + H2
2O -->
O -->
Ca(OH)
Ca(OH)2
2 (slaked lime)
(slaked lime)
CaO
CaO
HONORS ONLY!

56. 56
56
• Weak base:
Weak base: less than 100% ionized
less than 100% ionized
in water
in water
One of the best known weak bases is
One of the best known weak bases is
ammonia
ammonia
NH
NH3
3 (aq) + H
(aq) + H2
2O (l)
O (l) 
 NH
NH4
4
+
+
(aq) + OH
(aq) + OH-
-
(aq)
(aq)
Strong and Weak Acids/Bases
Strong and Weak Acids/Bases
HONORS ONLY!

57. 57
57
Weak Bases
Weak Bases
HONORS ONLY!

58. 58
58
Equilibria Involving
Equilibria Involving
Weak Acids and Bases
Weak Acids and Bases
Consider acetic acid, HC
Consider acetic acid, HC2
2H
H3
3O
O2
2 (HOAc)
(HOAc)
HC
HC2
2H
H3
3O
O2
2 + H
+ H2
2O
O 
 H
H3
3O
O+
+
+ C
+ C2
2H
H3
3O
O2
2
-
-
Acid
Acid Conj. base
Conj. base
Ka 
[H3O+
][OAc-
]
[HOAc]
 1.8 x 10-5
(K is designated K
(K is designated Ka
a for ACID)
for ACID)
K gives the ratio of ions (split up) to molecules
K gives the ratio of ions (split up) to molecules
(don’t split up)
(don’t split up)
HONORS ONLY!

59. 59
59
Ionization Constants for Acids/Bases
Ionization Constants for Acids/Bases
Acids
Acids Conjugate
Conjugate
Bases
Bases
Increase
strength
Increase
strength
HONORS ONLY!

60. 60
60
Equilibrium Constants
Equilibrium Constants
for Weak Acids
for Weak Acids
Weak acid has K
Weak acid has Ka
a < 1
< 1
Leads to small [H
Leads to small [H3
3O
O+
+
] and a pH of 2 - 7
] and a pH of 2 - 7
HONORS ONLY!

61. 61
61
Equilibrium Constants
Equilibrium Constants
for Weak Bases
for Weak Bases
Weak base has K
Weak base has Kb
b < 1
< 1
Leads to small [OH
Leads to small [OH-
-
] and a pH of 12 - 7
] and a pH of 12 - 7
HONORS ONLY!

62. 62
62
Relation
Relation
of K
of Ka
a, K
, Kb
b,
,
[H
[H3
3O
O+
+
]
]
and pH
and pH
HONORS ONLY!

63. 63
63
Equilibria Involving A Weak Acid
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the
You have 1.00 M HOAc. Calc. the
equilibrium concs. of HOAc, H
equilibrium concs. of HOAc, H3
3O
O+
+
, OAc
, OAc-
-
,
,
and the pH.
and the pH.
Step 1.
Step 1. Define equilibrium concs. in ICE
Define equilibrium concs. in ICE
table.
table.
[HOAc]
[HOAc] [H
[H3
3O
O+
+
]
] [OAc
[OAc-
-
]
]
initial
initial
change
change
equilib
equilib
1.00
1.00 0
0 0
0
-x
-x +x
+x +x
+x
1.00-x
1.00-x x
x x
x
HONORS ONLY!

64. 64
64
Equilibria Involving A Weak Acid
Equilibria Involving A Weak Acid
Step 2.
Step 2. Write K
Write Ka
a expression
expression
You have 1.00 M HOAc. Calc. the equilibrium concs.
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H
of HOAc, H3
3O
O+
+
, OAc
, OAc-
-
, and the pH.
, and the pH.
Ka 1.8 x 10-5
=
[H3O+
][OAc-
]
[HOAc]

x2
1.00 - x
This is a quadratic. Solve using quadratic
This is a quadratic. Solve using quadratic
formula.
formula.
or you can make an approximation if x is very
or you can make an approximation if x is very
small! (Rule of thumb: 10
small! (Rule of thumb: 10-5
-5
or smaller is ok)
or smaller is ok)
HONORS ONLY!

65. 65
65
Equilibria Involving A Weak Acid
Equilibria Involving A Weak Acid
Step 3.
Step 3. Solve K
Solve Ka
a expression
expression
You have 1.00 M HOAc. Calc. the equilibrium concs.
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H
of HOAc, H3
3O
O+
+
, OAc
, OAc-
-
, and the pH.
, and the pH.
Ka 1.8 x 10-5 =
[H3O+][OAc-]
[HOAc]

x2
1.00 - x
First assume x is very small because
First assume x is very small because
K
Ka
a is so small.
is so small.
Ka 1.8 x 10-5 =
x2
1.00
Now we can more easily solve this
Now we can more easily solve this
approximate expression.
approximate expression.
HONORS ONLY!

66. 66
66
Approximating
Approximating
If K is really small, the equilibrium
If K is really small, the equilibrium
concentrations will be nearly the same as
concentrations will be nearly the same as
the initial concentrations.
the initial concentrations.
Example: 0.20 – x is just about 0.20 if
Example: 0.20 – x is just about 0.20 if
x is really small.
x is really small.
If the K is 10
If the K is 10-5
-5
or smaller (10
or smaller (10-6
-6
, 10
, 10-7
-7
, etc.), you
, etc.), you
should approximate. Otherwise, you have
should approximate. Otherwise, you have
to use the quadratic.
to use the quadratic.

67. 67
67
Equilibria Involving A Weak Acid
Equilibria Involving A Weak Acid
Step 3.
Step 3. Solve K
Solve Ka
a approximate
approximate expression
expression
You have 1.00 M HOAc. Calc. the equilibrium concs.
You have 1.00 M HOAc. Calc. the equilibrium concs.
of HOAc, H
of HOAc, H3
3O
O+
+
, OAc
, OAc-
-
, and the pH.
, and the pH.
Ka 1.8 x 10-5
=
x2
1.00
x =
x = [
[H
H3
3O
O+
+
] = [
] = [OAc
OAc-
-
] = 4.2 x 10
] = 4.2 x 10-3
-3
M
M
pH = - log [
pH = - log [H
H3
3O
O+
+
] = -log (4.2 x 10
] = -log (4.2 x 10-3
-3
) =
) = 2.37
2.37
HONORS ONLY!

68. 68
68
Equilibria Involving A Weak Acid
Equilibria Involving A Weak Acid
Calculate the pH of a 0.0010 M solution of
Calculate the pH of a 0.0010 M solution of
formic acid, HCO
formic acid, HCO2
2H.
H.
HCO
HCO2
2H + H
H + H2
2O
O 
 HCO
HCO2
2
-
-
+ H
+ H3
3O
O+
+
K
Ka
a = 1.8 x 10
= 1.8 x 10-4
-4
Approximate solution
Approximate solution
[H
[H3
3O
O+
+
] = 4.2 x 10
] = 4.2 x 10-4
-4
M,
M, pH = 3.37
pH = 3.37
Exact Solution
Exact Solution
[H
[H3
3O
O+
+
] = [HCO
] = [HCO2
2
-
-
] = 3.4 x 10
] = 3.4 x 10-4
-4
M
M
[HCO
[HCO2
2H] = 0.0010 - 3.4 x 10
H] = 0.0010 - 3.4 x 10-4
-4
= 0.0007 M
= 0.0007 M
pH = 3.47
pH = 3.47
HONORS ONLY!

69. 69
69
Equilibria Involving A Weak Base
Equilibria Involving A Weak Base
You have 0.010 M NH
You have 0.010 M NH3
3. Calc. the pH.
. Calc. the pH.
NH
NH3
3 + H
+ H2
2O
O 
 NH
NH4
4
+
+
+ OH
+ OH-
-
K
Kb
b = 1.8 x 10
= 1.8 x 10-5
-5
Step 1.
Step 1. Define equilibrium concs. in ICE table
Define equilibrium concs. in ICE table
[NH
[NH3
3]
] [NH
[NH4
4
+
+
]
] [OH
[OH-
-
]
]
initial
initial
change
change
equilib
equilib
0.010
0.010 0
0 0
0
-x
-x +x
+x +x
+x
0.010 - x
0.010 - x x
x x
x
HONORS ONLY!

70. 70
70
Equilibria Involving A Weak Base
Equilibria Involving A Weak Base
You have 0.010 M NH
You have 0.010 M NH3
3. Calc. the pH.
. Calc. the pH.
NH
NH3
3 + H
+ H2
2O
O 
 NH
NH4
4
+
+
+ OH
+ OH-
-
K
Kb
b = 1.8 x 10
= 1.8 x 10-5
-5
Step 1.
Step 1. Define equilibrium concs. in ICE table
Define equilibrium concs. in ICE table
[NH
[NH3
3]
] [NH
[NH4
4
+
+
]
] [OH
[OH-
-
]
]
initial
initial
change
change
equilib
equilib
0.010
0.010 0
0 0
0
-x
-x +x
+x +x
+x
0.010 - x
0.010 - x x
x x
x
HONORS ONLY!

71. 71
71
Equilibria Involving A Weak Base
Equilibria Involving A Weak Base
You have 0.010 M NH
You have 0.010 M NH3
3. Calc. the pH.
. Calc. the pH.
NH
NH3
3 + H
+ H2
2O
O 
 NH
NH4
4
+
+
+ OH
+ OH-
-
K
Kb
b = 1.8 x 10
= 1.8 x 10-5
-5
Step 2.
Step 2. Solve the equilibrium expression
Solve the equilibrium expression
Kb  1.8 x 10-5
=
[NH4
+][OH-]
[NH3]
=
x2
0.010 - x
Assume x is small, so
Assume x is small, so
x = [OH
x = [OH-
-
] = [NH
] = [NH4
4
+
+
] = 4.2 x 10
] = 4.2 x 10-4
-4
M
M
and [NH
and [NH3
3] = 0.010 - 4.2 x 10
] = 0.010 - 4.2 x 10-4
-4
≈ 0.010 M
≈ 0.010 M
The approximation is valid
The approximation is valid !
!
HONORS ONLY!

72. 72
72
Equilibria Involving A Weak Base
Equilibria Involving A Weak Base
You have 0.010 M NH
You have 0.010 M NH3
3. Calc. the pH.
. Calc. the pH.
NH
NH3
3 + H
+ H2
2O
O 
 NH
NH4
4
+
+
+ OH
+ OH-
-
K
Kb
b = 1.8 x 10
= 1.8 x 10-5
-5
Step 3.
Step 3. Calculate pH
Calculate pH
[OH
[OH-
-
] = 4.2 x 10
] = 4.2 x 10-4
-4
M
M
so pOH = - log [OH
so pOH = - log [OH-
-
] = 3.37
] = 3.37
Because pH + pOH = 14,
Because pH + pOH = 14,
pH = 10.63
pH = 10.63
HONORS ONLY!

73. 73
73
Types of Acid/Base Reactions:
Types of Acid/Base Reactions:
Summary
Summary
HONORS ONLY!

74. 74
74
pH testing
• There are several ways to test pH
There are several ways to test pH
–Blue litmus paper (red = acid)
Blue litmus paper (red = acid)
–Red litmus paper (blue = basic)
Red litmus paper (blue = basic)
–pH paper (multi-colored)
pH paper (multi-colored)
–pH meter (7 is neutral, <7 acid, >7
pH meter (7 is neutral, <7 acid, >7
base)
base)
–Universal indicator (multi-colored)
Universal indicator (multi-colored)
–Indicators like phenolphthalein
Indicators like phenolphthalein
–Natural indicators like red cabbage,
Natural indicators like red cabbage,
radishes
radishes

75. 75
75
Paper testing
• Paper tests like litmus paper and pH
Paper tests like litmus paper and pH
paper
paper
– Put a stirring rod into the solution
Put a stirring rod into the solution
and stir.
and stir.
– Take the stirring rod out, and place
Take the stirring rod out, and place
a drop of the solution from the end
a drop of the solution from the end
of the stirring rod onto a piece of
of the stirring rod onto a piece of
the paper
the paper
– Read and record the color change.
Read and record the color change.
Note what the color indicates.
Note what the color indicates.
– You should only use a small portion
You should only use a small portion
of the paper. You can use one
of the paper. You can use one
piece of paper for several tests.
piece of paper for several tests.

76. 76
76
pH paper

77. 77
77
pH meter
• Tests the voltage of the
Tests the voltage of the
electrolyte
electrolyte
• Converts the voltage to
Converts the voltage to
pH
pH
• Very cheap, accurate
Very cheap, accurate
• Must be calibrated with
Must be calibrated with
a buffer solution
a buffer solution

78. 78
78
pH indicators
• Indicators are dyes that can be
added that will change color in
the presence of an acid or base.
• Some indicators only work in a
specific range of pH
• Once the drops are added, the
sample is ruined
• Some dyes are natural, like radish
skin or red cabbage

79. 79
79
ACID-BASE REACTIONS
ACID-BASE REACTIONS
Titrations
Titrations
H
H2
2C
C2
2O
O4
4(aq) + 2 NaOH(aq) --->
(aq) + 2 NaOH(aq) --->
acid
acid base
base
Na
Na2
2C
C2
2O
O4
4(aq) + 2 H
(aq) + 2 H2
2O(liq)
O(liq)
Carry out this reaction using a
Carry out this reaction using a TITRATION
TITRATION.
.
Oxalic acid,
Oxalic acid,
H
H2
2C
C2
2O
O4
4

80. 80
80
Setup for titrating an acid with a base
Setup for titrating an acid with a base

81. 81
81
Titration
Titration
1. Add solution from the buret.
1. Add solution from the buret.
2. Reagent (base) reacts with
2. Reagent (base) reacts with
compound (acid) in solution
compound (acid) in solution
in the flask.
in the flask.
3.
3. Indicator shows when exact
Indicator shows when exact
stoichiometric reaction has
stoichiometric reaction has
occurred. (Acid = Base)
occurred. (Acid = Base)
This is called
This is called
NEUTRALIZATION.
NEUTRALIZATION.

82. 82
82
35.62 mL of NaOH is
35.62 mL of NaOH is
neutralized with 25.2 mL of
neutralized with 25.2 mL of
0.0998 M HCl by titration to
0.0998 M HCl by titration to
an equivalence point. What
an equivalence point. What
is the concentration of the
is the concentration of the
NaOH?
NaOH?
LAB PROBLEM #1: Standardize a
LAB PROBLEM #1: Standardize a
solution of NaOH — i.e., accurately
solution of NaOH — i.e., accurately
determine its concentration.
determine its concentration.

83. 83
83
PROBLEM: You have 50.0 mL of 3.0 M
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH.
NaOH and you want 0.50 M NaOH.
What do you do?
What do you do?
Add water to the 3.0 M solution to lower
Add water to the 3.0 M solution to lower
its concentration to 0.50 M
its concentration to 0.50 M
Dilute the solution!
Dilute the solution!

84. 84
84
PROBLEM: You have 50.0 mL of 3.0 M
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
NaOH and you want 0.50 M NaOH. What do
you do?
you do?
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute
But how much water
But how much water
do we add?
do we add?

85. 85
85
PROBLEM: You have 50.0 mL of 3.0 M
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
NaOH and you want 0.50 M NaOH. What do
you do
you do?
?
How much water is added?
How much water is added?
The important point is that --->
The important point is that --->
moles of NaOH in ORIGINAL solution =
moles of NaOH in ORIGINAL solution =
moles of NaOH in FINAL solution
moles of NaOH in FINAL solution

86. 86
86
PROBLEM: You have 50.0 mL of 3.0 M NaOH and
PROBLEM: You have 50.0 mL of 3.0 M NaOH and
you want 0.50 M NaOH. What do you do?
you want 0.50 M NaOH. What do you do?
Amount of NaOH in original solution =
Amount of NaOH in original solution =
M • V
M • V =
=
(3.0 mol/L)(0.050 L) = 0.5 M NaOH X V
(3.0 mol/L)(0.050 L) = 0.5 M NaOH X V
Amount of NaOH in final solution must also =
Amount of NaOH in final solution must also =
0.15 mol NaOH
0.15 mol NaOH
Volume of final solution =
Volume of final solution =
(0.15 mol NaOH) / (0.50 M) = 0.30 L
(0.15 mol NaOH) / (0.50 M) = 0.30 L
or
or 300 mL
300 mL

87. 87
87
PROBLEM: You have 50.0 mL of 3.0 M
PROBLEM: You have 50.0 mL of 3.0 M
NaOH and you want 0.50 M NaOH. What do
NaOH and you want 0.50 M NaOH. What do
you do?
you do?
Conclusion:
Conclusion:
add 250 mL
add 250 mL
of water
of water to
to
50.0 mL of 3.0
50.0 mL of 3.0
M NaOH to
M NaOH to
make 300 mL
make 300 mL
of 0.50 M
of 0.50 M
NaOH.
NaOH.
3.0 M NaOH 0.50 M NaOH
H2O
Concentrated Dilute

88. 88
88
A shortcut
A shortcut
M
M1
1 • V
• V1
1 = M
= M2
2 • V
• V2
2
Preparing Solutions by
Preparing Solutions by
Dilution
Dilution

89. 89
89
You try this dilution problem
• You have a stock bottle of hydrochloric acid,
which is 12.1 M. You need 400. mL of 0.10 M
HCl. How much of the acid and how much
water will you need?