Post on 04-Jun-2018
1
AP Chemistry: Acids, Bases, and Salts
Unit Objectives 1. Understand the acid-base theories of Arrhenius, Brønsted-Lowry, and Lewis.
2. Identify strong acids and bases and calculate their pH’s.
3. Calculate the pH of a weak acid or base.
4. Calculate the concentration of a strong or weak acid or base from its pH.
5. Calculate the pH and ion concentrations in a polyprotic acid.
6. Predict the pH of a salt from its formula and then calculate the pH of the salt.
7. Identify the components of a buffer and perform calculations involving the preparation of a
buffer and the addition of a strong acid or strong base to a buffer.
8. Perform calculations involving strong acid-strong base titrations as well as weak acid-strong base
and weak base-strong acid calculations.
9. Be familiar with titration curves and selection of an acid-base indicator.
Key Terms acid-base indicatior
acid dissociation constant
Acidic oxides
ampohoteric substance
Arrhenius Concept
autoionization
basic oxides
Brønsted-Lowry model
buffered solution
buffering capacity
conjugate base
conjugate acid
conjugate acid-base pair
diprotic acid
equivalence point
Henderson-Hasselbalch equation
hydronium ion
ion-product (dissociation) constant
Lewis acid
Lewis base
major species
monoprotic acids
oxyacids
organic acids
percent dissociation
pH scale
polyprotic acid
salt
strong acids
strong bases
titration curve
triprotic acid
weak acids
weak bases
Equations and Constants
K Ka3
+ -
a
+ -H O ][A
HA] which is often simplified as
H ][A
HA]
[ ]
[
[ ]
[
Kb [ ]
[
HB ][OH
B]
+ -
K
K K
w
a b
[OH ][H ] @ 25 C
=
10 10 14.
pH = - log [H pOH = - log [OH
14 = pH + pOH
+ -], ]
pH = p + log [A
HA]
-
Ka]
[
pOH = p + log [HB
B]
+
Kb]
[ p = - log p = - log K K K Ka a b b,
2
AP Chemistry: Acid-Base Chemistry
Text References
Pages 658-660
Note: H+ does
not really exist
in aqueous
solutions. It is
hydrated in
water to form
H3O+
(hydronium
ion). However
H+ is often used
to make things
easier.
I. Properties of Aqueous Solutions of Acids and Bases
A. Acids
1. Sour taste.
Ex. Vinegar
2. Change colors of indicators.
Ex. Acids turn blue litmus red.
3. Non-oxidizing acids react with active metals to liberate
hydrogen gas.
Ex. Zn + 2HCl → ZnCl2 + H2
4. React with metal oxides, metal hydroxides, yielding a salt and
water.
Ex. CaO + HCl → CaCl2 + H2O
5. React with carbonates, yielding CO2.
Ex. 2HCl + Na2CO3 → 2NaCl + H2O +CO2
6. Electrolytes, because they ionize in water.
B. Bases
1. Bitter taste
Ex. Alka selter and other antacids
2. Slippery (slimy) to the touch.
Ex. Soap
3. Change colors of indicators.
Ex. Bases turn red litmus blue.
4. React with acids to produce a salt and water.
Ex. HCl + NaOH → NaCl + H2O
5. Electrolytes, because they ionize in water.
II. Acid-Base Theories
A. Arrhenius – this is the most simplistic definition of acids and bases
1. Acid – produces H+ in solution. Ex. HCl, HBr, H3PO4
a. monoprotic acid – contains one ionizable hydrogen.
Ex. HCl
b. diprotic acid – contains two ionizable hydrogens.
Ex. H2SO4, H2CO3
c. triprotric acid – contains three ionizable hydrogens.
Ex. H3PO4
Note: Only one hydrogen ionizes at a time
2. Base – produces OH- in solution. Ex. NaOH
Ammonia is not defined as a base according to the Arrhenius
definition of bases.
B. Brønsted-Lowry
1. Acid – donates a proton in water.
2. Base – accepts a proton in water.
Ammonia is defined as a base under this theory.
3. Conjugate Acid (CA) – acid formed when a base accepts a
proton.
4. Conjugate base (CB) – remaining part of an acid after it has
released a proton.
5. Identifying Conjugate Acid-Base Pairs
NH3 + H2O ⇄ NH4+ + OH
-
base acid CA CB
3
a. Label the acid, base, CA, and CB
HNO3 + H2O ⇄ H3O+ + NO3
-
acid base CA CB
C2H3O2-
+
H2O ⇄ HC2H3O2 + OH-
base acid CA CB
b. amphoteric – substances that can act as both an acid
and a base.
Write two equations to show that HS- is amphoteric.
HS- + H2O ⇄ H2S
+ OH
-
HS-
+ H2O ⇄ H3O+
+ S2-
6. Writing Disssociation Reactions and Equilibrium Expressions
a. The general reaction that occurs when an acid is
dissolved in water can best be represented as
HA(aq) + H2O(l) ⇄ H3O+(aq) + A
-(aq)
Or HA(aq) ⇄ H+(aq) + A
-(aq)
The equilibrium expression for the reaction is
Ka
3+ - + -H O A
HA]
H A
HA]
[ ][ ]
[
[ ][ ]
[ b. The general reaction that occurs when a base is
dissolved in water can be represented as
B(aq) + H2O(l) ⇄ HB+(aq) + OH
-(aq)
The equilibrium expression for the reaction is
Kb
[ ]
[
HB ][OH
B]
+ -
c. Write the dissociation reaction and the corresponding
Ka equilibrium expression for each of the following
acids in water.
a. HCN
HCN(aq) + H2O(l) ⇄ H3O+(aq) + CN
-(aq)
Or HCN(aq) ⇄ H+(aq) + CN-(aq)
Ka
3+ - + -H O CN
HCN]
H CN
HCN]
[ ][ ]
[
[ ][ ]
[
b. HOC6H5
HOC6H5(aq) + H2O(l) ⇄ H3O+(aq) + OC6H5
-(aq)
Or HOC6H5(aq) ⇄ H+(aq) + OC6H5-(aq)
Ka3
+6 5
-
6 5
+6 5
-
6 5
H O OC H
HOC H ]
H OC H
HOC H ]
[ ][ ]
[
[ ][ ]
[
c. C6H5NH3+
C6H5NH3+
(aq) + H2O(l) ⇄ H3O+(aq) + C6H5NH2(aq)
Or C6H5NH3+
(aq) ⇄ H+(aq) + C6H5NH2(aq)
Ka3
+6 5 2
6 5 3+
+6 5 2
6 5 3+
H O C H NH
C H NH ]
H C H NH
C H NH ]
[ ][ ]
[
[ ][ ]
[
4
Pages 704-707
Pages 661-665
701-704
Strong Acids
HCl, HBr, HI,
HNO3, H2SO4,
HClO4
Strong Soluble
Bases
Hydroxides or
oxides of IA and
IIA metals
C. Lewis Model
1. Acid – Electron-pair acceptor
A Lewis acid has an empty atomic orbital that it can use to
accept (share) an electron pair from a molecule that has a lone
pair of electrons (Lewis Base). The bond formed is called a
coordinate covalent bond.
2. Base – Electron-pair donor
3. Examples
a. Reaction between boron trifluoride and ammonia.
b. Hydration of a metal ion, such as Al3+
.
4. Tell whether each of the following is a Lewis acid or base.
a. PH3 Base
b. BCl3 Acid
c. H2S Base
5. Zinc hydroxide is an amphoteric substance. Write equations
that describe Zn(OH)2 acting as a Brønsted-Lowry base
toward H+ and as a Lewis acid toward OH
-.
Brønsted-Lowry Base:
Zn(OH)2(s) + 2H+(aq) → Zn
2+(aq) + 2H2O(l)
Lewis Acid:
Zn(OH)2(s) + 2OH-(aq) Zn(OH)4
2-(aq)
III. Strengths of Acids and Bases (Do not confuse concentration with strength)
The strength of an acid or base is determined by the extent to which it ionizes
or dissociates in aqueous solutions.
A. Strong Acids and Bases
1. Strong acids and bases dissociate (ionize) completely in water.
2. They have very large equilibrium constants (K values). The
dissociation (ionization) equilibrium position lies all the way
to the right.
3. The stronger the acid the weaker its conjugate base. The
stronger the base the weaker its conjugate acid.
4. The strength of binary acids (HA) depends upon two factors.
a. The larger the molecule A is the stronger the acid will
be.
b. The more polar the bond between H and A is, the
weaker the acid will be.
Ex. HF << HCl < HBr < HI
5
Pages 664-668
5. The strength of oxyacids (HOX) depends upon two factors.
a. The more electronegative element X is, the stronger
the acid HOX will be. Ex. HClO3 > HBrO3.
b. The more oxygen present in the polyatomic ion, the
stronger the acid is within that group.
Ex. HClO4 > HClO3 > HClO2 > HClO.
6. For polyprotic acids (acids where more than one proton can be
removed), each successive proton becomes more difficult to
remove. Therefore, for polyprotic acids, the anions formed by
the dissociation of the acidic hydrogen are always less acidic
than their parent molecule.
B. Weak Acids and bases
1. Weak acids and bases only partially ionize in water.
2. They have small equilibrium constants (K values). The
dissociation (ionization) equilibrium position lies far to the
left.
3. Weak acids have relatively strong conjugate bases. Weak
bases have relatively strong conjugate acids.
4. The percent dissociation of a weak acid or base can be
calculated using the following formula.
Percent Dissociation =
amount dissociated (mol / L)
initial concentration (mol / L)100
a. The smaller the percent dissociation, the weaker the
acid or base.
b. Dilution of a weak acid increases its percent
dissociation.
IV. The pH Concept
A. Self-Ionization of Water
1. Since the water molecule is amphoteric, it may dissociate with
itself to a slight extent. In the self-ionization of water, two
water molecules collide producing a hydronium ion and a
hydroxide ion.
H2O(l) + H2O(l) ⇄ H3O+(aq) + OH
-(aq)
H2O(l) ⇄ H+(aq) + OH
-(aq)
2. The equilibrium expression used here is referred to as
ionization constant for water, Kw
Kw = [H3O+][OH
-] = 1.0 × 10
-14 (at 25°C)
Kw = Ka × Kb
3. All aqueous solutions have H3O+ and OH
- present.
a. Neutral: [H3O+] = [OH
-] = 1.0 × 10
-7
b. Acidic: [H3O+] > [OH
-]
c. Basic (alkaline): [H3O+] < [OH
-]
4. Calculate the [OH-] if the [H3O
+] is 1.8 × 10
-8. Is the solution
acidic, basic or neutral?
Kw 3+ -H O OH [ ][ ] .10 10 14
[ ][ ]
.
..OH
H O
-
3+
KMw 10 10
18 1056 10
14
8
7
H3O+] < [OH
-] basic
6
Pages 666-669
Reminder: [H
+]
and [H3O+] are
used
interchangeably.
Pages 669-671
Pages 671-675
B. pH Scale
1. The pH scale was developed by Sorenson in 1909. The pH
scale is an easy way to express very small [H3O+]. The pH
scale ranges from 0 to 14.
2. pH<7: acidic solution
pH = 7 neutral solution
pH >7: basic solution
3. pH = -log [H3O+] [H3O
+] = antilog (-pH)
pOH = -log [OH-] [OH
-] = antilog (-pOH)
pH + pOH = 14
4. When expressing the pH or pOH you should use as many
decimal places as there are significant figures in the problem.
5. The pH of a sample of human blood was measured to be 7.41
at 25°C. Calculate pOH, [H+] and [OH
-] for the sample.
pOH = 14 – 7.41 = 6.59
[H+] = 3.9×10
-8 M
[OH-] = 2.6×10
-7 M
C. Calculating the pH of Strong Acid (or Base) Solutions
1. Calculating the pH of a solution of a strong monoprotic acid is
straightforward because [H+] equals the original concentration
of the acid.
2. What is the pH of a 0.040 M solution of HClO4?
pH = -log(0.040) = 1.40
3. An aqueous solution of HNO3 has a pH of 2.34. What is the
concentration of the acid?
[H+] = antilog(-2.34) = 10
-2.34 = 0.0046 M
4. What is the pH of a 0.50 M solution of NaOH?
pOH = -log(0.50) = 0.30
pH = 14 – 0.30 = 13.70
D. Calculating the pH of Weak Acid (or Base) Solutions
1. Calculating the pH of a weak acid (or base) involves setting up
an equilibrium expression. Always start by writing the
equation, setting up the acid (or base) equilibrium expression
(Ka or Kb), defining initial concentrations, changes and final
concentrations in terms of x, substituting values and variable
into the Ka (or Kb) expression and solving for x. We will use
RICE tables.
R = reaction
I = initial concentrations
C = change taking place
E = equilibrium concentrations
7
Often the –x in a
Ka expression
can be treated as
negligible if the
Ka value is
small. When
you assume that
it is negligible,
you must check
the validity of
this assumption.
To be valid, x
must be less
than 5% of the
original
number.
2. Calculate the pH of a 1.00×10-4
M solution of acetic acid. The
Ka of acetic acid is 1.8×10-5
.
R HC2H3O2 ⇄ H
+ C2H3O
2-
I 1.00×10-4
0 0
C -x +x + x
E 1.00×10-4
– x x x
Ka [ ][
[ ].
H C H O ]
HC H O
+2 3 2
-
2 3 2
18 10 5
18 10100 10
5
4.
( )( )
( . )
x x
x
Let’s assume that –x is negligible.
We will assume that 1.00×10-4
–x ≈ 1.00×10-4
18 10
100 104 2 105
4
5.( )( )
( . ).
x xx
We must now check the validity of our assumption.
4 2 10
10 10100 42%
5
4
.
.
Unfortunately, this is greater than 5% so the assumption is
invalid and we must use the quadratic equation to solve for x.
18 10
100 10
5
4.
( )( )
( . )
x x
x 1.8×10
-5(1.00×10
-4 – x) = x
2
1.8×10-9
– 1.8×10-5
x = x2
x2 + 1.8×10
-5x – 1.8×10
-9 =0
x = 3.5×10-5
and -5.2×10-5
(discard)
pH = - log(3.5×10-5
) = 4.56
8
Pages 675-677
Since the Ka for
HNO2 is the
largest it will be
the primary
source of H+.
3. The hypochlorite ion (OCl-) is a strong oxidizing agent often found in
household bleaches and disinfectants. It is also the active ingredient that
forms when swimming pool water is treated with chlorine. In addition to
its oxidizing abilities, the hypochlorite ion has a relatively high affinity for
protons (it is a much stronger base than Cl-, for example) and forms the
weakly acidic hypochlorous acid (HClO, Ka = 3.5×10-8
). Calculate the pH
of a 0.100 M aqueous solution of hypochlorous acid.
R HClO ⇄ H+ ClO
-
I 0.100 0 0
C -x +x + x
E 0.100 – x x x
Ka [ ][
[ ].
H ClO ]
HClO
+ -
35 10 8
35 10
0100
8.( )( )
( . )
x x
x Assume 0.100 – x ≈ 0.100
35 100100
59 108 5.( )( )
( . ). x x
x
Check the validity of the assumption.
59 10
0100100 0 059%
5.
..
Since it is less than 5% the assumption is valid.
pH = - log(5.9×10-5
) = 4.23
E. Determination of the pH of a Mixture of Weak Acids
1. Only the acid with the largest Ka value will contribute an appreciable [H+].
Determine the pH based on this acid and ignore any others.
2. Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2×10-10
)
and 5.00 M HNO2 (Ka = 4.0×10-4
). Also calculate the concentration of
cyanide ion (CN-) in this solution at equilibrium.
R HNO2 ⇄ H+ NO2
-
I 5.00 0 0
C -x +x + x
E 5.00 – x x x
Ka [ ][
[ ].
H NO ]
HNO
+2
-
2
4 0 10 4
4 0 10
500
4.( )( )
( . )
x x
x Assume 5.00 – x ≈ 5.00
4 0 10
5000 454.
( )( )
( . ). x x
x
Check the validity of the assumption.
0 45
500100 0 90%
.
..
Since it is less than 5% the assumption is valid. pH = - log(0.45) = 1.35
Now we need to calculate [CN-].
Ka [ ][
[ ].
H CN ]
HCN
+ -
6 2 10 2
( . )[
( . ).
0 45
1006 2 10 2CN ]-
[ ]
( . )( . )
..CN-
6 2 10 100
0 4514 10
108M
9
Pages 677-681
Pages 688-694
F. Calculating Percent Dissociation (also known as percent ionization)
1. It is often useful to specify the amount of weak acid that
dissociated in achieving equilibrium in an aqueous solution.
The percent dissociation is defined as follows:
Percent Dissociation =
amount dissociated (mol / L)
initial concentration (mol / L)100
2. Calculate the percent dissociation of acetic acid in a 1.00 M
HC2H3O2 solution. The Ka of acetic acid is 1.8×10-5
.
R HC2H3O2 ⇄ H
+ C2H3O2
-
I 1.00 M 0 0
C -x +x + x
E 1.00 – x x x
Ka [ ][
[ ].
H C H O ]
HC H O
+2 3 2
-
2 3 2
18 10 5
18 10
100
5.( )( )
( . )
x x
x Let’s assume that 1.00 –x ≈ 1.00
18 10
1004 2 105 3.
( )( )
( . ). x x
x
We must now check the validity of our assumption.
4 2 10
100100 0 42%
3.
..
This is less than 5% so the assumption is valid.
Percent Dissociation =
[H
HC H O ]
+
2 3 2
]
[
.
..
1004 2 10
100100 0 42%
3
G. Calculating the pH of a Polyprotic Acid
1. A polyprotic acid always dissociates in a stepwise manner,
one proton at a time.
2. Each step has a characteristic Ka value.
3. Typically for a weak polyprotic acid, Ka1 > Ka2 > Ka3
4. Write out the stepwise Ka reactions for the diprotic acid
H2SO3.
H2SO3 + H2O ⇄ H3O+ + HSO3
-
HSO3- + H2O ⇄ H3O
+ + SO3
2-
5. For a typical polyprotic acid in water, only the first
dissociation step is important in determining the pH.
10
Sulfuric Acid is
unique. It is a
strong acid in the
first dissociation
step (Ka1 is very
large). It is a
weak acid in the
second step.
For relatively
concentrated
solutions of H2SO4
(1.0 M or higher),
the large
concentration of
H+ from the first
dissociation step
represses the
second step, which
can be neglected
as a contributor of
H+ ions. For
dilute solutions of
sulfuric acid, the
second step does
make a significant
contribution, and
the quadratic
equation must be
used to obtain the
total H+
concentration.
6. Calculate the pH of a 5.0 M H3PO4 solution and the
equilibrium concentrations of the species H3PO4, H2PO4-,
HPO42-
, and PO43-
. Ka1 = 7.5×10-3
, Ka2 = 6.2×10-9
, 4.8×10-13
.
The dominant equilibrium is the dissociation of H3PO4.
R H3PO4 ⇄ H
+ H2PO4
-
I 5.0 0 0
C -x +x + x
E 5.0 – x x x
Substituting the equilibrium concentrations into the expression
for Ka1 and making the usual approximation gives
K
x x
x
xxa
+2 4
3 41
H H PO
H PO
7 5 105 0 50
01932
.[ ][ ]
[ ]
( )( )
( . ) ..
Check the validity of the assumption.
019
50100 38%
.
..
The assumption is valid.
Therefore [H+] = [H2PO4
-] = 0.19
pH = -log(0.19) = 0.72
The concentration of HPO42-
can be obtained by using the
expression for Ka2.
K Ma
+4
2 4
442
H HPO
H PO
HPO HPO
6 2 10
019
0196 2 108
2 22 8.
[ ][ ]
[ ]
( . )[ ]
( . )[ ] .
The concentration of PO43- can be obtained by using
the expression for Ka3.
Ka
+4
4
4
3
H PO
HPO
PO
4 8 10
019
6 2 10
133
2
3
8.
[ ][ ]
[ ]
( . )[ ]
( . )
[ ]
( . )( . )
..PO4
313 8
194 8 10 6 2 10
01916 10
M
7. Calculate the pH of a 1.0 M H2SO4 solution.
The initial concentration of H+ is at least 1.0 M. We must
determine if the HSO4- ion dissociates enough to produce a
significant contribution to the concentration of H+.
R HSO4
- ⇄ H
+ SO4
2-
I 1.00 M 1.0 0
C -x +x + x
E 1.00 – x x x
Substituting the equilibrium concentrations into the expression
for Ka2 and making the usual approximation gives
Kx x x
x Ma
+4
42
H
HSO
(1.0)( ) = 1.2
12 1010
10 10102
22.
[ ][SO ]
[ ]
( . )( )
( . ) ( . )
Check the validity of the assumption.
0 012
10100 12%
.
..
The assumption is valid.
[H+] = 1.0 + x = 1.0 + 0.012 = 1.0 M (to the correct number of
significant figures)
pH = -log(1.0 M) = 0.00
11
Pages 694-700
V. Acid Base Properties of Salts
A. A salt is made by neutralizing an acid with a base. When a salt
dissolves in water, it releases ions having an equal number of positive
and negative charges. Thus a solution of a salt should be neither
acidic nor basic.
B. Some salts do form neutral solutions, but others react with water
(hydrolye) to form acidic or basic solutions.
C. Types of Salts
1. Neutral Salts – Salts that are formed from the cation of a
strong base and the anion of a strong acid form neutral
solutions when dissolved in water. A salt such as NaNO3
gives a neutral solution.
2. Basic Salts – Salts that are formed from the cation of a strong
base and the anion of a weak acid form basic solutions when
dissolved in water. The anion hydrolyzes the water molecule
to produce hydroxide ions and thus a basic solution. K2S
should be basic since S2-
is the conjugate base of the very
weak acid HS-, while K+ does not hydrolyze appreciably.
S2-
+ H2O ⇄ OH- + HS
-
strong base weak acid
3. Acidic Salts – Salts that are formed from the cation of a weak
base and the anion of a strong acid form acidic solutions when
dissolved in water. The cation hydrolyzes the water molecule
to produce hydronium and thus an acidic solution. NH4Cl
should be weakly acidic, sinc NH4+ hydrolyzes to give an
acidic solution, while Cl- does not hydrolyze.
NH4+ + H2O ⇄ H3O
+ + NH3
Strong acid weak base
4. The salt produced from the anion of a weak acid and the
cation of a weak base may form an acidic, basic, or neutral
solution. The Ka and Kb must be compared to predict the
nature of the salt.
Ka > Kb pH < 7 (acidic)
Kb > Kb pH > 7 (basic)
Ka = Kb pH = 7 (neutral)
D. Use the information at the right to predict whether an aqueous
solution of each of the following salts will be acidic, basic or neutral.
1. NaCl
neutral: anion of a strong acid
and cation of a strong base
2. NH4SO4
acidic: anion of a strong acid and cation of a weak base
3. NH4C2H3O2
neutral: Ka = Kb
4. NH4CN
basic: Kb > Ka
KK
Kb
w
a
(for CN - ).
..
10 10
6 2 1016 10
14
10
5
NH4+ Ka = 5.6×10
-10
C2H3O2- Kb = 5.6×10
-10
HCN Ka = 6.2×10-10
12
NH4
+ is a
conjugate acid
of a weak base
Cl- is a
conjugate base
of a strong acid
NH4Cl is the salt
of a strong acid
and weak base. It
will be acidic.
F
- is a conjugate
base of a weak
acid
Na+ is the
conjugate acid
of a strong base NaF is the salt of
a weak acid and a
strong base. It
will be basic.
E. Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3
is 1.8×10-5
.
The major species in solution are NH4+, Cl
-, and H2O. Both the NH4
+
and the H2O can produce H+.
NH4+ ⇄ NH3 + H
+
Ka [ ][ ]
[ ]
NH H
NH
3+
4+
We must determine the value for the Ka of NH4
+ from the relationship
Ka × Kb = Kw
K
K
Ka
w
b
(for NH (for NH
4+
3
))
.
..
10 10
18 1056 10
14
5
10
This means that NH4
+ is a stronger acid than H2O (Ka = 1.0×10
-14) and
will dominate in the production of H+
R NH4+ ⇄ H
+ NH3
I 0.10 0 0
C -x +x + x
E 0.10 – x x x
Substituting the equilibrium concentrations into the expression for Ka
and making the usual approximation gives
K
x x
x
xa
56 10
010 010
102
.[ ][ ]
[ ]
( )( )
. .
NH H
NH
3+
4+
x ≈7.5×10
-6
Check the validity of the assumption.
7 5 10
010100 0 0075%
6.
..
The assumption is valid
[H+] = 7.5×10-6
M and pH = 5.13
F. Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is
7.2×10-4
.
The major species in solution are Na+, F
- (from the weak acid HF) and
H2O. Both the F- and the H2O can produce OH
-.
F- + H2O ⇄ HF + OH
-
We must determine the value for the Kb of HF from the relationship
Ka × Kb = Kw
K
K
Kb
w
a
(for HF (for HF
))
.
..
10 10
7 2 1014 10
14
4
11
This means that F
- is a stronger base than H2O (Kb = 1.0×10
-14) and
will dominate in the production of OH-
R F- ⇄ HF OH
-
I 0.30 0 0
C -x +x + x
E 0.30 – x x x
Substituting the equilibrium concentrations into the expression for Kb
and making the usual approximation gives
K
x x
x
xb
14 10
0 30 0 30
112
.[ ][ ]
[ ]
( )( )
. .
HF OH
F
-
-
x ≈ 2.0×10
-6 The approximation is valid by the 5% rule.
[OH-] = 2.0×10
-6 M pOH = 5.69; pH = 14 - 5.69 = 8.31
13
Pages 722-736
VI. Buffers
A. A buffer is a solution of a weak acid or base and its salt which resists
changes in pH when either OH- or H
+ ions are added.
B. Buffer capacity – The amount of acid or base that can be absorbed by
a buffer system without a significant change in pH. In order to have a
large buffer capacity, a solution should have large concentrations of
both buffer components.
C. Calculating the pH of a Buffered Solution
1. A buffered solution contains 0.50 M acetic acid (HC2H3O2,
Ka = 1.8×10-5
) and 0.50 M sodium acetate (NaC2H3O2).
Calculate the pH of this solution.
The major species in solution are
HC2H3O2 Weak acid
Na+
Neither acid nor base
C2H3O2- Conjugate Base of HC2H3O2
H2O very weak acid or base
The acetic acid dissociation equilibrium, which involves both
HC2H3O2 and C2H3O2-, will control the pH of the solution.
R HC2H3O2 ⇄ H+ C2H3O2
-
I 0.50 0 0.50
C -x +x + x
E 0.50 – x x 0.50 + x
Substituting the equilibrium concentrations into the expression
for Ka and making the usual approximation gives
K
x x
x
xa
18 10
050
050
050
050
5.[ ] ]
[ ]
( )( . )
.
( . )
.
H [C H O
HC H O
+2 3 2
-
2 3 2 x ≈ 1.8×10
-5
Check the validity of the approximation.
18 10
050100 0 0036%
5.
..
The assumption is valid.
[H+] = 1.8×10
-5 pH = 4.74
2. Another way to calculate the pH of a buffer system is with the
Henderson-Hasselbach equation.
pH = p + log
[A
HA]
-
Ka]
[ This equation must be used with caution. This equation is
only valid for solutions that contain weak monoprotic acids
and their salts or weak bases and their salts. The buffered
solution cannot be too dilute and the Ka/Kb cannot be too large.
Let’s calculate the pH of the buffer system in question 1 using
the Henderson-Hasselbach equation.
pKa = -log(1.8×10-5
) = 4.74
pH = p + log
[A
HA]
-
Ka]
[. log
.
.. 4 74
050
0504 74
14
Page 736
Ideally, the end
point and the
equivalence
point in a
titration should
coincide.
D. Calculating pH Changes in Buffered Solutions.
1. Calculate the change in pH that occurs when 0.010 mol solid
NaOH is added to 1.0 L of the buffered solution described in
the example above. Compare this pH change with that which
occurs when 0.010 mol solid NaOH is added to 1.0 L of water.
The major species in solution are HC2H3O2, Na+, C2H3O2
-,
OH-, and H2O.
HC2H3O2 + OH
- → C2H3O2
- + H2O
Before 0.50 M 0.010 M 0.50 M
After 0.50-0.010
=0.49 M
0.010-0.010
= 0
0.50 + 0.010
=0.51 M
pH = p + log
[A
HA]
-
Ka]
[. log
.
.. 4 74
051
0 494 76
ΔpH = 4.76 -4.74 = + 0.02
[ ]
.
..OH
mol
L
0 010
100 010 M
pOH = -log(0.010) = 2 pH = 14-2 = 12
ΔpH = 12.00 – 7.00 = + 5.00
VII. Titrations
A. Titration – the process by which the volume of a standard solution
required to react with a specific amount of a substance is determined.
B. Titration Curve – plot of the pH versus the amount (usually volume)
of acid or base added.
C. End Point – the point at which the color of an indicator changes in a
titration. It is determined by the Ka value for the indicator.
D. Equivalence Point – the point at which chemically equivalent amounts
of acid and base have reacted.
E. Steps in carrying out a titration to determine the concentration of an
unknown acid solution.
1. A measured amount of an acid of unknown concentration is
added to an Erlenmeyer flask.
2. An appropriate indicator (such as phenolphthalein) is added to
the solution.
3. Measured amounts of a base of known concentration are
mixed into the acid. The solution of known concentration is
called the standard solution. The addition of the base is
carried out using a buret. This process is continued until the
indicator indicates that the end point has been reached.
4. The point at which the two solutions used in a titration are
present in chemically equivalent amounts is the equivalence
point.
F. Choosing an Indicator - Choose an indicator that will change color as
close as possible to the end point.
15
Pages 737-740
VIII. Titration Problems
A. Titration of a Strong Acid with a Strong Base
1. A titration is performed by adding 0.600 M KOH to 40.0 mL of
0.800 M HCl. The net ionic equation for this reaction is:
H+ + OH
- → H2O
a. Calculate the pH before the addition of any KOH.
Remember before you add any base all you have is the
strong acid in the flask. Strong acids completely ionize in
water.
HCl + H2O → Cl- + H3O
+
pH = -log [H3O+]
pH = -log[0.800] = 0.097
b. Calculate the pH after the addition of 5.0 mL of the base.
H+ + OH
- → H2O
(.040 L)(0.8 M) = 0.032 moles (0.005 L)(0.6 M) = 0.003 moles
-0.003 moles -0.003 moles
0.029 moles 0 moles
The total volume of solution is 45 mL (40.0 mL + 5.0 mL).
pH = - log[H+]
pH log0.029 moles
0.045 L LNM
OQP 0191.
c. Calculate the pH after the addition of 20.0 mL of the
base.
H+ + OH
- → H2O
(.040 L)(0.8 M) = 0.032 moles (0.020L)(0.6 M) = 0.012 moles
-0.012 moles -0.012 moles
0.020 moles 0 moles
The total volume of solution is 60 mL (40.0 mL + 20.0 mL).
pH = - log[H+]
pH log0.020 moles
0.060 L LNM
OQP 0 477.
16
d. Calculate the pH after the addition of 40.0 mL of the
base.
H+ + OH
- → H2O
(.040 L)(0.8 M) = 0.032 moles (0.040L)(0.6 M) = 0.024 moles
-0.024 moles -0.024 moles
0.008 moles 0 moles
The total volume of solution is 80 mL (40.0 mL + 40.0 mL).
pH = - log[H+]
pH log0.008 moles
0.080 L LNM
OQP 100.
e. Calculate the pH after the addition of 52.0 mL of the
base.
H+ + OH
- → H2O
(.040 L)(0.8 M) = 0.032 moles (0.052L)(0.6 M) = 0.0312 moles
-0.0312 moles -0.0312 moles
0.0008 moles 0 moles
The total volume of solution is 92 mL (40.0 mL + 52.0 mL).
pH = - log[H+]
pH log0.0008 moles
0.092 L LNM
OQP 2 06.
f. Calculate the volume of base needed to reach the
equivalence point.
The equivalence point is where the moles of acid equals
the moles of base.
We know we have 0.032 moles of acid. The volume of
base needed is equal to:
Molarity
mol
L( )M
Lmol
= 0.032 mol
= 0.533 L = 53.3 mL
M M0 6.
g. What is the pH at the equivalence point?
H+ + OH
- → H2O
pH = 7 (Strong acid and a strong base)
17
h. Calculate the pH after adding 5.00 mL of NaOH past the
equivalence point.
H+ + OH
- → H2O
(.040 L)(0.8 M) = 0.032 moles (0.0583 L)(0.6 M) = 0.03498
moles
-0.032 moles -0.032 moles
0 moles 0.00298 moles
The total volume of solution is 98.3 mL (40.0 mL + 58.3 mL).
pOH = - log[OH-]
pOH log0.00298 moles
0.0983 L LNM
OQP 152.
pH = 14.00 – 1.52 = 12.48
2. A sketch of the titration curve from the problem just
completed.
3. What are some characteristics of the curve?
a. Extreme pH at the beginning
b. No Buffering
c. Sharp rise in pH at equivalence point
d. The equivalence point is at 7
18
Pages 740-749
Why is it okay
to use the moles
of the acids and
base instead of
the
concentrations?
The volume was
the same for
both.
B. Titration of a Weak Acid with a Strong Base
1. A titration is performed by adding 0.200 M NaOH to 24.0 mL
of 0.350 M HOCl. The net ionic equation for this reactions
is:
HOCl + OH- OCl
- + H2O
a. Calculate the pH before the addition of any NaOH. The
Ka of HOCl is 3 x 10-8
.
0.350 – x ≈ 0.350 (This assumption is made because the Ka
value for HOCl is so small – less than 10-4
.)
HOCl + H2O OCl- + H3O
+
Ka [H O ][OCl ]
[HOCl]
3
3 100 350
82
= = 1.02 10 -4xx M
. pH = -log[H3O
+] = -log[1.02×10
-4] = 3.99
b. Calculate the pH after the addition of 5.0 mL of the
base.
The Henderson-Hasselbach equation can be used because a weak
acid and a salt are left over.
pH p log [base]
[acid]
FHGIKJKa
pH 0.0010
0.0074 6.65
FHG
IKJ
log ( ) log3 10 8
R HOCl + H2O → OCl- + H3O
+
I 0.350 M
0
0
C - x
+ x
+ x
E 0.350-x
x
x
HOCl + OH- → OCl
- + H2O
(0.350 M)(0.024L)=
0.0084 moles
(0.200M)(0.0050 L)
= 0.0010 moles
0
- 0.0010 - 0.0010
+ 0.0010
0.0074 0
0.0010
19
c. Calculate the pH after the addition of 15.0 mL of the
base.
HOCl + OH- → OCl
- + H2O
(0.350 M)(0.024L)=
0.0084 moles
(0.200M)(0.0150 L)
= 0.0030 moles
0
-0.0030 -0.0030
+0.0030
0.0054 0
0.0030
The Henderson-Hasselbach equation can be used because a weak
acid and a salt are left over.
pH = p + log [base]
[acid]Ka
FHGIKJ
pH = - log (3 10 + log 0.0030
0.0054 = 7.27-8
FHG
IKJ)
d. Calculate the pH after the addition of 25.0 mL of the
base.
HOCl + OH- → OCl
- + H2O
(0.350 M)(0.024L)=
0.0084 moles
(0.200M)(0.0250 L)
= 0.0050 moles
0
-0.0050 -0.0050
+0.0050
0.0034 0
0.0050
The Henderson-Hasselbach equation can be used because a weak
acid and a salt are left over.
pH = p + log [base]
[acid]Ka
FHGIKJ
pH = - log (3 10 + log 0.0050
0.0034 = 7.69-8
FHG
IKJ)
e. Calculate the pH after the addition of 35.0 mL of the
base.
HOCl + OH- → OCl
- + H2O
(0.350 M)(0.024L)=
0.0084 moles
(0.200M)(0.0350 L)
= 0.0070 moles
0
-0.0070 -0.0070
+0.0070
0.0014 0
0.0070
The Henderson-Hasselbach equation can be used because a weak
acid and a salt are left over.
pH = - log (3 10 log 0.0070
0.0014 = 8.22-8
FHG
IKJ)
20
f. Calculate the volume of base needed to reach the
equivalence point.
The equivalence point is where the moles of acid equals
the moles of base.
We know we have 0.0084 moles of acid. The volume
of base needed is equal to:
Molarity
mol
L( )M
Lmol
= 0.0084 mol
= 0.0420 L = 42.00 mL
M M0 2. g. What is the pH at the equivalence point?
HOCl + OH- → OCl
- + H2O
(0.350 M)(0.024L)=
0.0084 moles
(0.200M)(0.0420 L)
= 0.0084 moles
0
-0.0084 -0.0084
+0.0084
0 0
0.0084
Molarity
mol
L =
0.0084 mol
L + 0.042 L ( )
..M M
0 0240127
R OCl- + H2O → HOCl + OH
-
I 0.127 M
0
0
C - x
+ x
+ x
E 0.127-x
x
x
Kw = Ka × Kb = 1 × 10-14
KK
Kb
w
a
=
1 10
3 10
-14
-8333 10 7.
0.127 – x ≈ 0.127 (since the Ka and Kb are so small)
Kb [ ][
[ ]
OH HOCl]
OCl
-
-
333 100127
7.[ ][ ]
. x x
x = [OH
-] = 2.056 × 10
-4 M
pOH = -log[2.056 × 10-4
] = 3.687
pH = 14 – 3.687 = 10.31
21
At the midpoint
(half-
equivalence
point) the pH =
pKa. Why?
This is because
pH = p + log [A ]
[HA]
-
Ka
and since [A-] =
[HA], the
log[A
HA]
-]
[ 0
.
h. Calculate the pH after adding 5.00 mL of NaOH past
the equivalence point.
HOCl + OH- → OCl
- + H2O
(0.350 M)(0.024L)=
0.0084 moles
(0.200M)(0.0470 L)
= 0.0094 moles
-0.0084 -0.0084
0 0.0010
The total volume of solution is 71.0 mL (24.0 mL + 47.0 mL).
pOH = - log[OH-]
pOH log0.0010 moles
0.0710 L LNM
OQP 185.
pH = 14.00 – 1.85 = 12.15
2. A sketch of the titration curve from the problem just
completed.
3. What are some characteristics of the curve?
a. Not as an extreme of start in pH
b. Buffering occurs
c. Not as sharp of a rise in pH at equivalence point
d. The equivalence point is not at 7