The Chemistry of
Acids and Bases
Chapter 17
Strong and Weak Acids/Bases
• Acids and bases into STRONG or WEAK ones.
STRONG ACID
Or
HNO3(aq) ---> H+(aq) + NO3- (aq)
HNO3 is about 100% dissociated in water.
2
Strong and Weak Acids/Bases
HNO3, HCl, H2SO4 and HClO4 are
among the strong acids.
H—Cl
Cl
H2O
H3O
+
-
3
Strong and Weak Acids/Bases
• Weak acids are much less than 100%
ionized in water.
• One of the best known is acetic acid =
HC2H3O2 = CH3CO2H = HOAc =
CH3COOH
HC2H3O2 + H2O
<--> C2H3O2-
or
HC2H3O2 <--> C2H3O2- + H+
+
H3O+
4
Strong and Weak Acids/Bases
5
Strong Bases
100% dissociated in water
NaOH(aq) ---> Na+(aq) + OH-(aq)
Other common strong bases
include KOH and Ca(OH)2.
CaO (lime) + H2O -->
Ca(OH)2 (slaked lime)
CaO
Strong and Weak Acids/Bases
Weak Bases
Less than 100% ionized in water
One of the best known weak bases is
ammonia.
NH3 (aq) + H2O (l)
<-->
NH4+ (aq) + OH- (aq)
6
ACID-BASE THEORIES
• The most general theory for common
aqueous acids and bases is the
BRØNSTED - LOWRY theory
• ACIDS DONATE H+ IONS
• BASES ACCEPT H+ IONS
7
8
ACID-BASE THEORIES
NH 3
Base
+
H 2O
Acid
NH 4 + + OH Acid
Base
The Brønsted definition means NH3
is a BASE in water — and water is
itself an ACID
ACID-BASE THEORIES
NH3 is a BASE in water — and water is
itself an ACID.
NH3 / NH4+ is a conjugate pair —
related by the gain or loss of H+
Every acid has a conjugate base and vice-versa.
9
ACID-BASE THEORIES
10
A strong acid is 100% dissociated.
Therefore, a STRONG ACID — a good H+ donor —
must have a WEAK CONJUGATE BASE —
a poor H+ acceptor.
HNO3(aq) + H2O(l) <-->
STRONG A
base
H3O+ (aq) + NO3- (aq)
acid
weak B
Notice that every A-B reaction has two
acids and two bases!
ACID-BASE THEORIES
STRONG
ACID
WEAK
BASE
We know that HNO3 is a strong acid.
1. It is a stronger acid than H3O+
2. H2O is a stronger base than NO3-
11
ACID-BASE THEORIES
12
Acetic acid is only 0.42% ionized when [HC2H3O2] = 1.0
M. It is a WEAK ACID
Because [H3O+] is small, this must mean
1.
H3O+ is a stronger acid than HC2H3O2
2.
C2H3O2- is a stronger base than H2O
ACIDS AND BASES
13
• Monoprotic acids can donate only one proton,
while polyprotic acids can donate two or more
protons.
• Monoprotic bases can accept only one proton,
while polyprotic bases can accept two or more
protons.
• Amphiprotic substances can behave as either
acids or bases.
Acid-Base Reactions
• Now we can describe
reactions of acids with
bases and the
direction of such
reaction.
• Consider the acid HF
reacting with the base
NH3.
• HF + NH3 <--> NH4+ + F-
14
Predicting the Direction of
Acid-Base
Reactions
Based on experiment, we can put acids and
bases on a chart.
See Table 17.3 or Appendix H tables 16
ACIDS
STRONG
CONJUGATE BASES
weak
weak
STRONG
This chart can be used to predict the
direction of reactions between any A-B pair.
Reactions always go from the stronger A-B
pair to the weaker A-B pair.
15
ACID-BASE THEORIES
16
Predicting the direction of an acid-base reaction.
ACID 1 + BASE 2
STRONG
ACID 2 + BASE 1
weak
Reactions always go from the
stronger A-B pair
weaker A-B pair.
MORE ABOUT WATER
H2O can function as both an ACID and a BASE.
In pure water there can be AUTOIONIZATION
17
MORE ABOUT WATER
Autoionization
2H2O
H 3O +
OH-
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
In a neutral solution [H3O+] = [OH-]
and so [H3O+] = [OH-] = 1.00 x 10-7 M
18
Calculating
[H+]
&
[OH-]
You add 0.0010 mol of NaOH to 1.0 L of
pure water. Calculate [H3O+] and [OH-].
Solution
HOH
<-->
H+ + OH-
Le Chatelier predicts equilibrium shifts
left
to the ____________.
[H+] < 10-7 at equilibrium.
Set up a concentration table.
19
Calculating
[H+]
&
[OH-]
You add 0.0010 mol of NaOH to 1.0 L of
pure water. Calculate [H3O+] and [OH-].
Solution
H2O(l)
<--> H+ +
OHinitial
0
0.0010
change
+x
+x
equilib
x
0.0010 + x
Kw = [H+][OH-] = (x) (0.0010 + x)
Because x << 0.0010 M, assume [OH-] = 0.0010 M
[H3O+] = Kw / 0.0010 = 1.0 x 10-11 M
This solution is BASIC because [H3O+] < [OH-]
20
21
[H+], [OH-] and pH
A common way to express acidity and
basicity is with pH.
pH = log (1/ [H+]) = - log [H+]
In a neutral solution,
[H+] = [OH-] = 1.00 x 10-7 at 25 oC
pH = -log (1.00 x 10-7) = - (-7.000)
= 7.000
[H+],
[OH-]
and pH
What is the pH of the 0.0010 M NaOH
solution?
[H+] = 1.0 x 10-11 M
pH = - log (1.0 x 10-11) = 11.00
General conclusion —
Basic solution
Neutral
Acidic solution
pH > 7
pH = 7
pH < 7
22
[H+], [OH-] and pH
If the pH of Coke is 3.12, it is ____________.
Acidic
Because pH = - log [H+] then
log [H+] = - pH
Take antilog and get
+
[H ]
=
-pH
10
[H+] = 10-3.12 =
7.6 x 10-4 M
23
24
Other pX Scales
In general
and so
pX = - log X
pOH = - log [OH-]
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
Take the - log of both sides
- log (10-14) = - log [H3O+] + (- log [OH-])
pKw = 14.00 = pH + pOH
Equilibria Involving
Weak Acids and Bases
Aspirin is a good
example of a weak acid,
Ka = 3.2 x 10-4
25
Equilibria Involving
Weak Acids and Bases
Acid
Conjugate Base
acetic, HC2H3O2
C2H3O2-, acetate
ammonium, NH4+
NH3, ammonia
bicarbonate, HCO3-
CO32-, carbonate
A weak acid (or base) is one that
ionizes to a VERY small extent (< 5%).
26
Equilibria Involving
Weak Acids and Bases
• For a weak acid, HA, and its
conjugate base, A-,
Ka . Kb = Kw
• Thus, one can calculate Ka from Kb
and Kb from Ka.
• Notice as Ka increases, Kb
decreases and visa versa.
• The stronger the acid the weaker
the conjugate base.
27
Equilibria Involving
Weak Acids and Bases
Consider acetic acid
HC2H3O2 + H2O
<-->
H3O+
Acid
+
C2H3O2-
Conj. base
[H 3 O + ][C2H3O2- ]
Ka =
=
[HC2H3O2]
1.8 x 10 -5
(K is designated Ka for an ACID)
Because [H3O+] and [C2H3O2-] are SMALL,
Ka << 1.
28
Equilibria Involving
Weak Acids and Bases
Values of Ka for acid and Kb for bases
are found in TABLE 17.4 — page 799
Notice the relation of TABLE 17.4 to
the table of relative acid/base
strengths (Table 17.3).
29
Increasing Base Strength
Increasing Acid Strength
30
Partial table 17.3
Equilibria Involving A Weak Acid
You have 1.00 M HOAc.
Calculate the equilibrium concentrations of
HOAc, H3O+, OAc-, and the pH.
Step 1. Define equilibrium concentrations.
[HOAc]
initial
[H3O+]
[OAc-]
1.00
0
0
change
-x
+x
+x
equilib
1.00-x
x
x
Note that we neglect [H3O+] from H2O.
31
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calculate the
equilibrium concentrations of HOAc, H3O+,
OAc-, and the pH.
Step 2. Write Ka expression
[H3O+ ][OAc- ]
-5
Ka = 1.8 x 10
=
=
[HOAc]
x2
1.00 - x
This is a quadratic. Solve using quadratic
formula or method of approximations
(see Appendix A.4).
32
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calculate the
equilibrium concentrations of HOAc, H3O+,
OAc-, and the pH.
Step 3. Solve Ka expression
+ ][OAc- ]
2
[H
O
x
3
Ka = 1.8 x 10-5 =
=
[HOAc]
1.00 - x
First assume x is very small because Ka is
so small.
2
Ka = 1.8 x 10-5 =
x
1.00
And so x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
33
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calculate the
equilibrium concentrations of HOAc, H3O+,
OAc-, and the pH.
Step 3. Solve Ka approximate expression
2
x
Ka = 1.8 x 10-5 =
1.00
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = - log [H3O+] = -log (4.2 x 10-3) = 2.37
34
Equilibria Involving A Weak Acid
35
Consider the approximate expression
2
x
Ka = 1.8 x 10-5 =
1.00
+
x = [H 3 O ] = [K a • 1.00]
For many weak acids
[H3O+] = [conj. base] = [Ka • Co]1/2
where C0 = initial concentration of acid
Useful Rule of Thumb:
If 1000 • Ka < Co, then [H3O+] = [Ka • Co]1/2
1/2
36
Approximation Rules
10 times K for .1M, so 10K < [X]
100 times K for .10M, so 100K < [X]
1000 times K for .100M , so 1000K < [X]
Look at the number of decimal places to
determine factor to multiply K by
Equilibria Involving A Weak Acid
Calculate the pH of a 0.0010 M solution of
formic acid, HCO2H.
HCO2H + H2O
HCO2- + H3O+
Ka = 1.8 x 10-4
Approximate solution
[H3O+] = [Ka • Co]1/2 = 4.2 x 10-4 M, pH = 3.37
Exact Solution
[H3O+] = [HCO2-] = 3.4 x 10-4 M
[HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M
pH = 3.47 NOT Valid
37
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calculate the pH.
NH4+ +
NH3 + H2O
OH-
Kb = 1.8 x 10-5
Step 1. Define equilibrium concentrations.
initial
change
equilib
[NH3]
[NH4+]
[OH-]
0.010
0
0
-x
+x
+x
x
x
0.010 - x
38
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calculate the pH.
NH3 + H2O
NH4+ +
OH-
Kb = 1.8 x 10-5
Step 2. Solve the equilibrium expression
2
+ ][OH- ]
[NH
x
4
Kb = 1.8 x 10-5 =
=
[NH3 ]
0.010 - x
Assume x is small (100•Kb < Co), so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
and [NH3] = 0.010 - 4.2 x 10-4 = 0.010 M
The approximation is valid !
39
40
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calculate the pH.
NH3 + H2O
NH4+ +
Kb = 1.8 x 10-5
Step 3. Calculate pH
[OH-] = 4.2 x 10-4 M
so pOH = - log [OH-] = 3.37
Because pH + pOH = 14,
pH = 10.63
OH-
Equilibria Involving
Weak Acids and Bases
1. The pH of a 0.10 M nicotinic acid solution
is 2.92. Calculate Ka and the % ionization.
1.4x10-5 1.2%
2. A solution of propionic acid,
CH3CH2COOH, is 0.20 M. Ka = 1.3x10-5.
Calculate the pH and % ionization.
2.79
0.80%
3. A solution of hydrazine, N2H4, is 0.025 M.
Kb = 8.5x10-5. Calculate the pH and %
ionization.
11.15 5.6%
(4.) 
41
42
4. Calculate the pH and
concentrations of the arsenate
containing species in a 0.15 M
solution of H3AsO4. K1= 2.5x10-4,
K2 = 5.6x10-8, K3 = 3.0x10-13
2.21, 0.14, 6.1x10-3, 5.8x10-8, 2.8x10-18
Acid-Base Properties of Salts
MX + H2O ----> acidic or basic solution?
Consider NH4Cl
NH4Cl(aq) ----> NH4+(aq) + Cl-(aq)
(a)
Reaction of Cl- with H2O
Cl- +
base
H2O <---->
acid
X
HCl
acid
+
OHbase
Cl- ion is a VERY weak base because its conjugate
acid is strong.
Therefore, Cl- ----> neutral solution
43
Acid-Base Properties of Salts
NH4Cl(aq) ----> NH4+(aq) + Cl-(aq)
(b)
Reaction of NH4+ with H2O
NH4+ + H2O <----> NH3
acid
base
base
+
H3O+
acid
NH4+ ion is a moderate acid because its
conjugate base is weak.
Therefore, NH4+ ----> acidic solution
See TABLE 17.5 for a summary of
acid-base properties of ions.
44
45
Determine if the following solutions are
acidic, basic, or neutral. KBr,CrCl3, NaNO2,
KHCO3, Na2CO3, NH4C2H3O2
Acid-Base Properties of Salts
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O ---> neutral
CO32base
Step 1.
initial
change
equilib
+
H2O
HCO3-
acid
acid
+
base
Kb = 2.1 x 10-4
Set up concentration table
[CO32-]
[HCO3-]
[OH-]
0.10
-x
0.10 - x
0
+x
x
OH-
0
+x
x
46
Acid-Base Properties of Salts
47
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O ---> neutral
CO32- +
H2O
HCO3+
OHbase
acid
acid
base
Kb = 2.1 x 10-4
Step 2.
Solve the equilibrium expression
2
- ][OH- ]
[HCO
x
3
K b = 2.1 x 10-4 =
=
2
0.10 - x
[CO3 ]
Assume 0.10 - x 0.10, because 100•Kb < Co
x = [HCO3-] = [OH-] = 0.0046 M
Acid-Base Properties of Salts
48
Calculate the pH of a 0.10 M solution of Na2CO3.
Na+ + H2O ---> neutral
CO32- +
H2O
HCO3+
OHbase
acid
acid
base
Kb = 2.1 x 10-4
Step 3.
Calculate the pH
[OH-] = 0.0046 M
pOH = - log [OH-] = 2.34
pH + pOH = 14,
so pH = 11.66, and the solution is ________.
Basic
Review Problems
49
• Calculate the pH and concentrations of
the arsenate containing species in a
0.15 M solution of H3AsO4. K1= 2.5x10-4,
K2 = 5.6x10-8, K3 = 3.0x10-13
2.21, 0.14, 6.1x10-3, 5.8x10-8, 2.8x10-18
• Calculate the pH of a 0.050 M K2CO3 solution.
11.50
Lewis Acids & Bases
• Lewis acid =
electron pair
acceptor (BF3)
• Lewis base =
electron pair donor
(NH3)
50
51
Lewis Acids & Bases
A Lewis acid and base can interact
by sharing an electron pair.
Lewis Acids & Bases
A Lewis acid and base can interact
by sharing an electron pair.
Formation of hydronium ion is an
excellent example.
H
+
ACID
••
•• O—H
H
BASE
•• +
H O—H
H
52
Lewis Acids & Bases
Other good examples involve metal ions.
Colorless Mx(NO3)y form colorful
hydrated [Mx(H2O)y]z+ compounds
53
Lewis Acids & Bases
54
Other good examples involve metal ions.
Co
2+
ACID
••
•• O—H
H
BASE
Co
2+
•• ••
O—H
H
Such bonds as the H2O ---> Co bond are often
called COORDINATE COVALENT BONDS
because both electrons are supplied by one of
the atoms of the bond.
Lewis Acids & Bases
The combination of metal
ions (Lewis acids) with
Lewis bases such as H2O
and NH3 ------>
COMPLEX IONS
All metal ions form
complex ions with water
— and are of the type
[M(H2O)x]n+ where x = 4
and 6.
2+
[Cu(NH3)4]
55
Lewis Acids & Bases
Add NH3 to light blue [Cu(H2O)4]2+ ------>
light blue Cu(OH)2 and then deep blue
[Cu(NH3)4]2+
56
57
Lewis Acids & Bases
[Ni(H2O)6]2+ + 6 NH3 ---> [Ni(NH3)6]2+
+ DMG
Lewis Acids & Bases
The Fe2+ in heme can interact with O2 or
CO in a Lewis acid-base reaction.
58
Lewis Acids & Bases
Many complex ions containing water undergo
HYDROLYSIS to give acidic solutions.
[Cu(H2O)4]2+ + H2O ---> [Cu(H2O)3(OH)]+ + H3O+
59
Lewis Acids & Bases
Many complex ions containing water undergo
HYDROLYSIS to give acidic solutions.
This explains why water solutions of Fe3+,
Al3+, Cu2+, Pb2+, etc. are acidic.
60
Lewis Acids & Bases
61
This explains AMPHOTERIC nature of some
metal hydroxides.
Al(OH)3(s) + 3 H+ --> Al3+ + 3 H2O
Here Al(OH)3 is a Brønsted base.
Al(OH)3(s) + OH- --> Al(OH)4-
Here Al(OH)3 is a Lewis acid.
••
O—H ••
••
Al
3+
Amphoterism of Al(OH)3
Al(OH)3 on right
Add NaOH
Add HCl
See Kotz / Treichel 5th, page 722
62
63
1. The electronegativity of
the O atoms causes the
H attached to O to be
highly positive.
2. The O—H bond is
highly polar.
3. The H atom of O—H is
readily attracted to polar
H2O.
Neutral Lewis Acid
64
Carbon dioxide is a neutral (molecular) Lewis acid.
-0.75
+1.5
-0.75
Lewis Acids & Bases
Many complex ions are very
stable.
Cu2+ + 4 NH3 <-->
[Cu(NH3)4]2+
K for the reaction is called
Kformation
or a “formation constant”.
Here K = 6.8 x 1012.
Rxn. is strongly product-favored.
65
Lewis Acids & Bases
Formation of complex ions explains why
you can dissolve a precipitate by forming a
complex ion.
AgCl(s) + 2 NH3
Ag(NH3)2+ + Cl-
AgCl(s)
66
Lewis Acids & Bases
67
Formation of complex ions explains why
you can dissolve a precipitate by forming a
complex ion.
Ag+ + Cl-
Ksp = 1.8 x 10-10
Ag+ + 2 NH3 <--> Ag(NH3)2+
Kform = 1.6 x 107
AgCl(s)
<-->
------------------------------------AgCl(s) + 2 NH3
<-->
Ag(NH3)2+ + Cl-
Knet = Ksp • Kform = 2.9 x 10-3
Why?
• Why are some
compounds acids?
• Why are some
compounds bases?
• Why do acids and bases
vary in strength?
• Can we predict
variations in acidity or
basicity?
69
Acetic acid
Ka = 1.8 x 10-5
Trichloroacetic acid
Ka = 0.3
Trichloroacetic acid is much stronger acid
owing to the high electronegativity of Cl,
which withdraws electrons from the rest of
the molecule. This makes the O—H bond
highly polar. The H of O—H is very positive.
Oxyacid Strength
• For oxyacids, XOn(OH)m, acid
strength increases as n increases
and is independent of m.
• Rank in order of increasing
acidity: HBrO3, H3BO3, HIO4
BrO2(OH),
B(OH)3
B(OH)3, IO3(OH)
BrO2(OH)
IO3(OH)
70
Basicity of Oxoanions
NO3-
CO32These ions are BASES.
PO43-
They become more and more basic as the
negative charge increases.
As the charge goes up, they interact more
strongly with polar water molecules.
71
72
Practice Problems
1. Write the equation for the reaction of
hypochlorous acid and ammonia. Label the
acids and bases. Indicate the conjugate pairs.
2. A 0.15 M weak acid solution is determined to
be 2.0% dissociated. Calculate the Ka.
3. Calculate the pH of the following solutions:
a) 0.50 M nitric acid
b) 0.25 M potassium hydroxide
c) 0.15 M phosphoric acid
d) 0.22 M sodium sulfite
e) 0.010 potassium cyanide
73
Practice Problems
4. Calculate the carbonate ion concentration
in a 0.10 M carbonic acid solution.
5. Determine whether a solution of Cs2SO3
is acidic, neutral or basic.
6. Write the Lewis acid/base reaction
between PH3 and BF3 using dot structures.
Indicate the acid and the base.
74
Practice Problems Answers
HClO + NH3 <--> NH4+ + ClOacid base
acid
base
2. 6.0 x 10-5
3. a) .30
b) 13.40
c) 1.52
d) 10.28
e) 10.70
4. 4.8 x 10-11
5. basic
1.
75
Practice Problems Answers
F
+
F
.. B .. F
-->
F
F
.. B .. F
The end.
H
.. P .. H
.. P .. H
H
H
H
6.
76
ammonia
NH3 + H2O <--> NH4+ + OH-
or
NH4OH
<-->
NH4+ + OH-
Return
Sample Questions
77
Give the conjugate base of HCN.
CNGive the conjugate acid of NO2-.
HNO2
Give the conjugate acid of NH3.
NH4+
The ion HCO3- has both a conjugate acid and
conjugate base. Give the formula of each.
H2CO3
CO32-
Sample Questions
Label the acid-base pairs:
CN- + HOH <--> HCN + OHbase acid
acid base
HCO31- + HOH <--> OH- + H2CO3
base
acid
base
acid
HCO31- + HOH <--> CO32- + H3O+
acid
base
base
acid
78
Acid-Base Reactions
Consider the acid HF reacting with
the base NH3.
HF + NH3 <--> NH4+ + Facid base
acid
Acids: HF > NH4+
Bases: NH3 > FReaction favors products
base
79
Increasing Base Strength
Increasing Acid Strength
80
Acid-Base Reactions
Does the reaction between nitrous acid
and ammonia favor reactants or
products?
HNO2 + NH3 <-->
NH4+
acid
acid
base
+
Acids: HNO2 > NH4+
Bases: NH3 > NO2-
Reaction favors products
NO2-
base
81
Acid-Base Reactions
Does the reaction between hypochlorous
acid and fluoride ion favor reactants or
products?
HClO
acid
+
F-
<-->
base
HF
+
acid
Acids: HF > HClO
Bases: ClO- > F-
Reaction favors reactants
ClO-
base
82
Acid-Base Reactions
Does the reaction between ammonium
chloride and sodium sulfate favor
reactants or products?
NH4Cl + Na2SO4 <--> ???
NH4+
acid
+
SO42-
<-->
base
NH3
base
Acids: HSO41- > NH4+
Bases: NH3 > SO42-
Reaction favors reactants
+
HSO41-
acid
83
Sample Questions
84
1. Determine the pH, pOH, and [OH-]
[H+] = 2.5 x 10-6
pH = - log [H+] = 5.60
pOH = 14 - 5.60 = 8.40
pOH = - log [OH-]
or 1.00 x 10-14 = [H+][OH-]
8.40 = - log [OH-] 1.00 x 10-14 = (2.5 x 10-6)[OH-]
[OH-] = 10-8.40
[OH-] = 4.0 x 10-9 M
Sample Questions
2. Determine the pOH, [H+], and [OH-]
pH = 6.52
pOH = 7.48
[H+] = 3.0 x 10-7 M
[OH-] = 3.3 x 10-8 M
How can we determine the pH of a solution?
Acid/Base indicators is one method.
85
Name of Indicator
Methyl violet
Thymol blue
Orange IV
Methyl orange
Bromophenol blue
Congo red
Bromocresol green
Methyl red
Bromocresol purple
Bromothymol blue
Phenol red
Thymol blue
Phenolphthalein
Thymolphthalein
Alizarin yellow
Indigo carmine
pH Interval
0.2 - 3.0
1.2 - 2.8
1.3 - 3.0
3.1 - 4.4
3.0 - 4.6
3.0 - 5.0
3.8 - 5.4
4.4 - 6.2
5.2 - 6.8
6.0 - 7.6
6.8 - 8.2
8.0 - 9.6
8.3 - 10.0
9.3 - 10.5
10.0 - 12.0
11.4 - 13.0
86
Color Change
yellow to blue-violet
red to yellow
red to yellow
red to orange to yellow
yellow to blue-violet
blue to red
yellow to blue
red to yellow
yellow to purple
yellow to blue
yellow to red
yellow to blue
colorless to red
colorless to blue
yellow to red
blue to yellow
Sample Questions
3. Determine the pH, pOH, [H+], and [OH-]
0.035 M HCl.
Strong acid --> [H+] = 0.035
pH = 1.46
pOH = 12.54
[OH-] = 2.9 x 10-13 M
87
Sample Questions
4. Determine the pH, pOH, [H+], and [OH-]
0.15 M NaOH.
Strong base -->
[OH-] = 0.15 M
pOH = 0.82
pH = 13.18
[H+] = 6.7 x 10-14 M
88
Equilibria Involving A Weak Acid
You have 0.10 M HC6H4NO2.
Its pH is measured to be 2.92
Calculate the equilibrium constant.
Step 1. Write the equation.
HA <-->
H+
+
A-
89
Equilibria Involving A Weak Acid
Step 2. Make a chart.
HA
<-->
H+
+
A-
0.10
0
0
- 0.0012
0.0012
0.0012
0.10
0.0012
0.0012
90
Equilibria Involving A Weak Acid
Step 3. Write Ka expression, without and
with numbers.
Ka =
[H+][A-]
[HA]
(0.0012)(0.0012)
Ka =
(0.10 )
91
Equilibria Involving A Weak Acid
Step 4. Solve.
(0.0012)(0.0012)
Ka =
(0.10 )
Ka = 1.4 x 10-5
Step 5. Answer question(s).
Ka = 1.4 x 10-5
92
93
Sample Questions
1. You have 0.050 M HCN. Its pH is
measured to be 5.4. Calculate the
equilibrium constant.
HCN
<-->
H+
+
CN0.050
0
0
- 4 x10-6
+ 4 x10-6
+ 4 x10-6
0.050
4 x10-6
4 x10-6
Ka =
[H+][CN-]
[HCN]
=
(4 x 10-6)2
(0.050)
= 3 x 10-10
Sample Questions
94
2. You have 0.1000 M HA which is 1.30%
dissociated.
Calculate the equilibrium constant.
HA
<-->
H+
+
A0.1000
0
0
- 0.00130
+ 0.00130 + 0.00130
0.0987
Ka =
0.00130
[H+][A-]
[HA]
=
0.00130
(0.00130)2
(0.0987)
= 1.71 x 10-5
Equilibria Involving A Weak Acid
You have 1.00 M HC2H3O2.
Calculate the equilibrium concentrations
of HC2H3O2, H+, C2H3O2 -, and the pH.
Step 1. Write the equation.
HC2H3O2
<-->
H+
+
C2H3O2-
95
Equilibria Involving A Weak Acid
Step 2. Make a chart.
HC2H3O2
<-->
H+
+
C2H3O2-
1.00
0
0
-x
+x
+x
1.00-x
x
x
96
Equilibria Involving A Weak Acid
Step 3. Write Ka expression, without and
with numbers.
Ka =
[H+][C2H3O2-]
[HC2H3O2]
(x)(x)
1.8 x 10-5 =
(1.00 - x)
97
Equilibria Involving A Weak Acid
Step 4. Solve.
(x)(x)
1.8 x 10-5 =
(1.00 - x)
1.8 x 10-5(1.00 - x) = x2
x2 + 1.8 x 10-5x - 1.8 x 10-5 = 0
x = 4.2 x 10-3 M
98
Equilibria Involving A Weak Acid
Step 5. Answer question(s).
x = [H+] = [C2H3O2-] = 0.0042 M
[HC2H3O2] = 1.00 - 0.0042 = 1.00 M
pH = 2.38
99
Equilibria Involving A Weak Acid
Note: Assume x is very small because Ka
is so small.
(x)(x)
1.8 x 10-5 =
(1.00 - x)
(x)(x)
1.8 x 10-5 =
(1.00)
1.8 x 10-5(1.00) = x2
x = 4.2 x 10-3 M
General Rule: Simplfy if [HA] / Ka > 1000
100
101
Sample Questions
1. Calculate the pH of a 0.100 M HF solution.
HF
<-->
H+
+
F0.100
0
0
-x
+x
+x
0.100 - x
Ka =
x
[H+][F-]
[HF]
=
X = 0.0081 M = {H+]
x
x2
(0.100 - x)
= 7.2 x 10 - 4
pH = 2.09
102
Sample Questions
2. Calculate the pH of a 0.20 M HClO solution.
HClO
<-->
H+
+
ClO0.20
0
0
-x
+x
+x
0.20
Ka =
x
[H+][ClO-]
[HClO]
=
X = 8.4 x 10 - 5 M = {H+]
x
x2
(0.20)
= 3.5 x 10 - 8
pH = 4.08
Sample Questions
103
3. Calculate the pH of a 0.20 M HCOOH solution.
HA
<-->
H+
+
A0.20
0
0
-x
+x
+x
0.20 - x
Ka =
x
[H+][A-]
[HA]
=
X = 0.0059 M = {H+]
x
x2
(0.20 - x)
= 1.8 x 10-4
pH = 2.23
104
Sample Questions
4. Calculate the pH of a 0.15 M H2CO3 solution.
H2CO3
<-->
H+
+
HCO30.15
0
0
-x
0.15
Ka =
[H+][HCO3-]
[H2CO3]
+x
+x
x
x
=
X = 2.5 x 10-4 M
x2
0.15
= 4.2 x 10-7
Sample Questions
105
Does the second equilibrium have a
significant effect on the pH? NO!
HCO31-
<-->
2.5 x 10-4
-y
2.5 x 10-4
Ka =
[H+][CO32-]
[HCO31-]
y = 4.8 x 10-11 M
H+
+
CO32-
2.5 x 10-4
0
+y
+y
2.5 x 10-4
y
(2.5 x 10-4)y
-11
=
=
4.8
x
10
2.5 x 10-4
pH = 3.60
Equilibria Involving A Weak Base
You have 0.10 M BOH. Its pH is measured
to be 10.11
Calculate the equilibrium constant.
Step 1. Write the equation.
BOH <-->
B+
+
OH-
[OH]- = 10-pOH = 10-3.89 = 1.3 x 10-4 M
106
Equilibria Involving A Weak Base
Step 2. Make a chart.
BOH
<-->
B+
+
OH-
0.10
0
0
- 0.00013
0.00013
0.00013
00.10
0.00013
0.00013
107
Equilibria Involving A Weak Base
Step 3. Write Kb expression, without and
with numbers.
Kb =
[B+][OH-]
[BOH]
(0.00013)(0.00013)
Kb =
(0.10)
108
Equilibria Involving A Weak Base
Step 4. Solve.
(0.00013)(0.00013)
Kb =
(0.10)
Kb = 1.7 x 10-7
Step 5. Answer question(s).
Kb = 1.7 x 10-7
109
110
Sample Question
You have 0.0200 M B.
Its is 3.0% dissociated.
Calculate the equilibrium constant.
B + HOH <--> HB+
+
OH0.0200
0
0
- 6.0 x 10-4
+ 6.0 x 10-4 + 6.0 x 10-4
6.0 x 10-4
0.0194
Kb =
[HB+ ][OH-]
[B]
6.0 x 10-4
(6.0 x 10-4)2
=
= 1.9x10-5
(0.0194)
Equilibria Involving A Weak Base
You have 0.100 M NH3
Calculate all equilibrium concentrations
and the pH.
Step 1. Write the equation.
NH3 + HOH
<-->
NH4+
+
OH-
111
112
Equilibria Involving A Weak Base
Step 2. Make a chart.
NH3 + HOH <-->
0.100
NH4+
+
OH-
0
0
-x
+x
+x
0.100
x
x
Equilibria Involving A Weak Base
Step 3. Write Kb expression, without and
with numbers.
Kb =
[NH4+][OH-]
[NH3]
(x)(x)
1.8 x 10-5 =
(0.100)
113
Equilibria Involving A Weak Base
Step 4. Solve.
(x)(x)
1.8 x 10-5 =
(0.100)
x = 1.3 x 10-3 M
114
Equilibria Involving A Weak Base
Step 5. Answer question(s).
[NH4+] = [OH-] = 0.0013 M
[NH3] = 0.099 M
pOH = 2.89
pH = 11.11
115
116
Sample Questions
Calculate the pH of a 0.015 M B solution.
Kb = 2.5 x 10-7
B + HOH <--> HB+
0.015
0
-x
+x
0.015
Kb =
+
x
[HB+ ][OH-]
[B]
=
OH0
+x
x
x2
(0.015)
X = 6.5 x 10-5 M = [OH-]
= 2.5x10-7
pH = 9.79
Sample Problems: Weak Acids and Bases
1. The pH of a 0.10 M nicotinic acid solution
is 2.92. Calculate Ka and the % ionization.
HA
<-->
H+
+
A0.10
0
0
- 0.0012
+ 0.0012
+ 0.0012
0.10
Ka =
0.0012
[H+][A-]
[HA]
=
0.0012
(0.0012)2
(0.10)
= 1.4 x 10-5
% ionization = 0.0012/0.10 x 100 = 1.2%
117
Sample Problems: Weak Acids and Bases
118
2. A solution of propionic acid,
CH3CH2COOH, is 0.20 M. Ka = 1.3x10-5.
Calculate the pH and % ionization.
HA
0.20
-x
<-->
0.20
H+
0
+x
+
x
Ka =
[H+][A-]
[HA]
=
A0
+x
x
x2
(0.20)
= 1.3 x 10-5
x = 1.6 x 10-3 = [H+], pH = 2.80
% ionization = 0.0016/0.20 x 100 = 0.80%
Sample Problems: Weak Acids and Bases
119
3. A solution of hydrazine, N2H4, is 0.025 M.
Kb = 8.5x10-5. Calculate the pH and % ionization.
N2H4 + HOH
0.025
-x
<-->
0.025
[N2H5+][OH-]
Kb =
=
[N2H4]
N2H5+
0
+x
x
+
OH0
+x
x
x2
(0.025)
= 8.5 x 10-5
x = 1.5 x 10-3 = [OH-], pOH = 2.82 pH = 11.18
% ionization = 0.0015/0.025 x 100 = 6.0%
4. Calculate the pH and concentrations of the
arsenate containing species in a 0.15 M
solution of H3AsO4.
K1=
2.5x10-4, K2 = 5.6x10-8, K3 = 3.0x10-13
H3AsO4
0.15
-x
<-->
H+
0
+x
+
120
H2AsO40
+x
0.15 - x
x
x
[H+][H2AsO4-]
x2
Ka =
=
= 2.5 x 10-4
[H3AsO4]
(0.15 - x)
x = 6.0 x 10-3 M = [H+],
pH = 2.22
[H3AsO4 ] = 0.14 M [H2AsO4-] = 6.0 x 10-3 M
4. Calculate the pH and concentrations of
the arsenate containing species in a
0.15 M solution of H3AsO4.
K1= 2.5x10-4, K2 = 5.6x10-8, K3 = 3.0x10-13
H2AsO4<--> H+ +
6.0 x 10-3
6.0 x 10-3
-y
+y
121
HAsO420
+y
6.0 x 10-3
6.0 x 10-3
y
[H+][HAsO42-] 6.0 x 10-3 y
-8
Ka =
=
=
5.6
x
10
[H2AsO4-]
6.0 x 10-3
y = 5.6 x 10-8 M = [HAsO42-]
4. Calculate the pH and concentrations
of the arsenate containing species in a
0.15 M solution of H3AsO4.
K1= 2.5x10-4, K2 = 5.6x10-8, K3 = 3.0x10-13
HAsO42<--> H+ +
5.6 x 10-8
6.0 x 10-3
-z
+z
122
AsO430
+z
5.6 x 10-8
6.0 x 10-3
z
[H+][AsO43-]
6.0 x 10-3 z
-13
Ka =
=
=
3.0
x
10
[HAsO42-]
5.6 x 10-8
z = 2.8 x 10-18 M = [AsO43-]
Acid-Base Properties of Salts
123
Determine if the following solutions are
acidic, basic, or neutral.
KBr
K+ + HOH <--> KOH + H+
neutral
Br + HOH <--> HBr + OH-
X
X
CrCl3
Cr3+ + HOH <-->
Cl- + HOH <-->
X
CrOH2+ + H+
HCl + OH-
acidic
Acid-Base Properties of Salts
124
Determine if the following solutions are
acidic, basic, or neutral.
NaNO2
Na+ + HOH <--> NaOH + H+
basic
NO2- + HOH <--> HNO2 + OHKHCO3
K+ + HOH <--> KOH + H+
HCO3- + HOH <--> H2CO3 + OH- Kb = 2.4x10-8
HCO3- + HOH <--> CO32- + H3O+ Ka = 4.8x10-11
X
X
basic
Acid-Base Properties of Salts
125
Determine if the following solutions are
acidic, basic, or neutral.
Na2CO3
Na+ + HOH <--> NaOH + H+
basic
CO32- + HOH <--> HCO3- + OH-
X
NH4C2H3O2
NH4+ + HOH <--> NH3 + H3O+ Ka = 5.6x10-10
C2H3O2- + HOH<-->HC2H3O2 +OH- Kb =5.6x10-10
neutral
126
Acid-Base Properties of Salts
You have 0.10 M Na2CO3
Calculate the pH.
Step 1. Write the equation.
CO32- + HOH
<-->
HCO3-
+
OH-
127
Acid-Base Properties of Salts
Step 2. Make a chart.
CO32- + HOH <--> HCO3-
+
OH-
0.10
0
0
-x
+x
+x
0.10 - x
x
x
Acid-Base Properties of Salts
Step 3. Write Kb expression, without and
with numbers.
Kb =
[HCO3-][OH-]
[CO32-]
(x)(x)
2.1 x 10-4 =
(0.10 - x)
128
Acid-Base Properties of Salts
Step 4. Solve.
(x)(x)
2.1 x 10-4 =
(0.10 - x)
x = 4.5 x 10-3 M
129
Acid-Base Properties of Salts
Step 5. Answer question(s).
[OH-] = 0.0045 M
pOH = 2.35
pH = 11.65
130
131
Sample Questions
1. Calculate the pH of a 0.10 M K3PO4
solution.
PO43- + HOH <--> HPO42- +
OH0.10
0
0
-x
+x
+x
0.10 - x
Kb =
x
[HPO42- ][OH-]
3-
[PO4 ]
X = 0.041
=
x
x2
(0.10 - x)
pH = 12.61
= 0.028
132
Sample Questions
2. Calculate the pH of a 0.10 M Na2C2O4
solution.
C2O42- + HOH <--> HC2O4- +
OH0.10
0
0
-x
+x
+x
0.10
Kb =
x
[HC2O4- ][OH-]
[C2O42- ]
=
X = 4.0 x 10-6
x
x2
(0.10)
pH = 8.60
=1.6x10-10
133
Sample Questions
3. Calculate the pH of a 0.045 M KCN
solution.
CN- + HOH <--> HCN
+
OH0.045
0
0
-x
+x
+x
x
x
0.045
Kb =
[HCN][OH-]
[CN- ]
x = 1.1 x 10-3 = [OH-]
=
x2
(0.045)
= 2.5x10-5
pOH = 2.96, pH = 11.04