Chapter 7 Acids and Bases
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
The Nature of Acids and Bases
Acid Strength
The pH Scale
Calculating the pH of Strong Acid Solutions Midterm Exam 2
Calculating the pH of Weak Acid Solutions
Bases
Polyprotic Acids
Acid-Base Properties of Salts
Acid Solutions in Which Water Contributes to the
H+ Concentration
7.10 Strong Acid Solutions in Which Water Contributes
to the H+ Concentration
7.11 Strategy for solving Acid-Base Problems: A Summary
1
Arrhenius (or Classical) Acid-Base Definition
An acid is a substance that contains hydrogen and dissociates
in water to yield a hydronium ion: H3O+ (H+)
A base is a substance that contains the hydroxyl group and
dissociates in water to yield a hydroxide ion: OH Neutralization is the reaction of an H+ (H3O+) ion from the
acid and the OH - ion from the base to form water, H2O.
An indicator is a substance that changes color with acidity.
The neutralization reaction is exothermic and releases approximately
56 kJ per mole of acid and base that react.
H+(aq) + OH-(aq)
H2O(l)
+ 56 kJ/mol
2
1
Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species that donates an H+ ion.
An acid must contain H in its formula; HNO3 and H2PO4- are two
examples; all Arrhenius acids are Brønsted-Lowry acids.
A base is a proton acceptor, any species that accepts an H+ ion.
A base must contain a lone pair of electrons to bind the H+ ion; a
few examples are NH3, CO32-, F -, as well as OH-. Brønsted-Lowry
bases are not Arrhenius bases, but all Arrhenius bases produce the
Brønsted-Lowry base OH-.
Therefore, in the Brønsted-Lowry perspective, one species donates a
proton and another species accepts it: an acid-base reaction is a
proton transfer process.
Acids donate a proton to water
Bases accept a proton from water
3
Acids donate a proton to water
Bases accept a proton from water
Water molecules can behave as a base or as an acid !
4
2
Identify the Brønsted acids and bases in the
following equation (A = Brønsted acid, B =
Brønsted base):
HPO42– + HSO4–
H2PO4– + SO42–
1.
2.
3.
4.
B, A, B, A
B, A, A, B
A, B, A, B
A, B, B, A
5
Molecular Model of Conjugate Acids & Bases
The reaction of an acid HA with water to form H3O+ and
a conjugate base
HA
Acid
A
Base
Conjugate
acid
Conjugate
base
• Conjugate acid is formed when a proton is transferred to the base.
• Conjugate base is everything that remains of the acid molecule
after a proton is lost.
• Conjugated acid-base pair consists of two substances related to
each other by donating and accepting a single proton.
6
3
The Conjugate Pairs in Some Acid-Base Reactions
Conjugate Pair
Acid
+
Base
Base
+
Acid
Conjugate Pair
Reaction 1
HF
+
H2O
F–
+
H3O+
Reaction 2
HCOOH
+
CN–
HCOO–
+
HCN
NH3
+
HCO3–
Reaction 3
NH4+
+
CO32–
Reaction 4
H2PO4–
+
OH–
HPO42–
+
H2O
Reaction 5
H2SO4
+
N2H5+
HSO4–
+
N2H62+
Reaction 6
HPO42–
+
SO32–
PO43–
+
HSO3–
7
Identifying Conjugate Acid-Base Pairs
Problem: The following chemical reactions are important for industrial
processes. Identify the conjugate acid-base pairs.
(a) HSO4-(aq) + CN-(aq)
SO42-(aq) + HCN(aq)
(b) ClO-(aq) + H2O(l)
HClO(aq) + OH-(aq)
2(c) S (aq) + H2O(aq)
HS-(aq) + OH-(aq)
Plan: To find the conjugate acid-base pair, we find the species that
donates H+ and the one that accepts it. The acid (or base) on the left
becomes the conjugate base (or acid) on the right.
Solution:
(a) The proton is transferred from the sulfate to the cyanide so:
HSO4-(aq)/SO42-(aq) and CN-(aq)/HCN(aq )
(b) The water gives up one proton to the hypochlorite anion so:
ClO-(aq)/HClO(aq) and H2O(l) / OH-(aq )
(c) One of water’s protons is transferred to the sulfide ion so:
S2-(aq)/HS-(aq) and H2O(l)/OH-(aq)
8
4
The conjugate acid and conjugate base of
bicarbonate ion (HCO3–) are, respectively,
1.
2.
3.
4.
5.
H3O+ and OH–
H3O+ and CO32–
H2CO3 and OH–
H2CO3 and CO32–
CO32– and OH–
9
The Acid-Dissociation Constant (Ka)
Strong acids dissociate completely in water to form ions:
H3O+(aq) + A-(aq)
HA(aq) + H2O(l)
In a dilute solution of a strong acid, almost no HA molecules exist:
[H3O+] = [HA]init or [HA]eq = 0
K = Ka =
[H3O+][A-]
[HA]
at equilibrium, and Ka >> 1
Nitric acid is an example: HNO3 (aq) + H2O(l)
H3O+(aq) + NO3-(aq)
Weak acids dissociate very slightly into ions in water:
HA(aq) + H2O(l)
⇌
H3O+(aq) + A-(aq)
In a dilute solution of a weak acid, the great majority of HA
molecules are not dissociated: [H3O+] << [HA]init or [HA]eq = [HA]init
K = Ka =
[H3O+][A-]
[HA]
at equilibrium, and Ka << 1
10
5
The Meaning of Ka, the Acid Dissociation Constant
For the ionization of an acid, HA:
HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
Therefore:
Ka =
[H3O+] [A-]
[HA]
Stronger acid
The stronger the acid, the higher the [H3O+]
at equilibrium, and the larger the Ka:
higher [H3O+]
larger Ka
• For a weak acid with a relative high Ka (~10-2 ), a 1 M solution has
~10% of the HA molecules dissociated.
• For a weak acid with a moderate Ka (~10-5 ), a 1 M solution has
~ 0.3% of the HA molecules dissociated.
• For a weak acid with a relatively low Ka (~10-10 ), a 1 M solution has ~
0.001% of the HA molecules dissociated.
11
The Extent of
Dissociation for
Strong and
Weak Acids
12
6
Review
Selected Acids and Bases
Acids
Strong: H+(aq) + A-(aq)
hydrochloric, HCl
hydrobromic, HBr
hydroiodoic, HI
nitric acid, HNO3
sulfuric acid, H2SO4
perchloric acid, HClO4
Weak
hydrofluoric, HF
phosphoric acid, H3PO4
acetic acid, CH3COOH
(or HC2H3O2)
Bases
Strong: M+(aq) + OH-(aq)
lithium hydroxide, LiOH
sodium hydroxide, NaOH
potassium hydroxide, KOH
calcium hydroxide, Ca(OH)2
strontium hydroxide, Sr(OH)2
barium hydroxide, Ba(OH)2
*(M is Group I or II metal)
Weak
ammonia, NH3
accepts proton from water to make
NH4+(aq) and OH-(aq)
13
Classifying the Relative Strengths of Acids and Bases–I
Strong acids. There are two types of strong acids:
1. The hydrohalic acids: HCl, HBr, and HI
2. Oxyacids in which the number of O atoms exceeds the number of
ionizable H atoms by two or more, such as HNO3, H2SO4, HClO4
Weak acids. There are many more weak acids than strong ones.
Four types, with examples, are:
1. The hydrohalic acid, HF
2. Those acids in which H is NOT bound to O or to a halogen, such
as HCN and H2S
3. Oxyacids in which the number of O atoms equals or exceeds by
one the number of ionizable H atoms, such as HClO, HNO2, and H3PO4
4. Organic acids (general formula RCOOH), such as CH3COOH and
C6H5COOH
14
7
Classifying the Relative Strengths of Acids and Bases - II
Strong bases. Soluble compounds containing O2- or OH- ions are strong
bases. The cations are usually those of the most active metals:
1) M2O or MOH, where M= Group 1A(1) metals (Li, Na, K, Rb, Cs)
2) MO or M(OH)2, where M = Group 2A(2) metals (Ca, Sr, Ba)
(MgO and Mg(OH)2 are only slightly soluble, but the soluble
portion dissociates completely.)
:
:
:
Weak bases. Many compounds with an electron-rich nitrogen are weak
bases (none are Arrhenius bases). The common structural feature
is an N atom that has a lone electron pair in its Lewis structure.
1) Ammonia (:NH3)
2) Amines (general formula RNH2, R2NH, R3N), such as
CH3CH2NH2, (CH3)2NH, (C3H7)3N, and C5H5N
:
:
:
:
15
16
8
Relative strengths of
acid/conjugate base
partners
Examples:
HCl and ClCH3COOH and CH3COOHCN and CN-
17
18
9
Among Proton Donors
Strong acid
Conj. acid of weak base
Water
Conj. acid of strong base
Higher Ka, stronger acid
Strength
Trends in Acid/Base Strength
Among Proton Acceptors
Strong base
Conj. base of weak acid
Water
Conj. base of strong acid
Higher Kb, stronger base
How do we know for sure?
Compare the Ka and Kb values.
19
20
10
OXYACIDS
•
•
•
•
•
•
most acids are oxyacids
acidic proton is attached to oxygen!
organic acids contain carboxyl (C-O) group
organic acids are weak acids
many inorganic acids are strong acids
acids can be mono- or polyprotic
– monoprotic acid dissociates one proton
– polyprotic acid dissociates several protons
21
22
11
23
Acid and Base Character and the pH Scale
To handle the very large variations in the concentrations of the
hydrogen ion in aqueous solutions, a scale called the pH scale
is used for which:
pH = - log [H+]
What is the pH of a solution that is 10-2 M in hydronium ion ?
pH = -log[H+] = (-1)log 10-2 = 2
pH of a neutral solution = 7.00
pH of an acidic solution < 7.00
pH of a basic solution > 7.00
24
12
Figure 7.3: The pH
scale and pH values of
some common
substances
25
The definition of pH
pH = -log[H3O+] = -log[H+]
Significant Figures for pH
The number of decimal places in the log is
equal to the number of significant figures in
the original number.
26
13
Calculating pH from its definition
pH = - log[H+]
What is the pH of a solution that is 1 x 10-12 M in hydronium ion?
pH = -log[H3O+] = -log (1 x 10-12) = -(-12.0) = 12.0
What is the pH of a solution that is 7.3 x 10-9 M in H+?
pH = -log (7.3 x 10-9) = -(-8.14) = 8.14
What is the pH of a 1 x 10-1 M H3O+ (aq) solution?
pH = -log[H3O+] = -log (0.1) = -(-1.0) = 1.0
27
pH of a Strong Acid
Calculate pH of 1.0 M HClO4
• major species : H+, ClO4- and H2O
• perchloric acid is a strong acid: FULLY dissociated
Therefore, 1.0 M HClO4 → 1.0 M H+
pH = - log [H+] = - log [1.0] = 0.00
Calculate pH of 0.01 M HClO4
pH = - log [H+] = - log [0.01] = 2.0
Calculate pH of 0.02 M HClO4
pH = - log [H+] = - log [0.02] = 1.7
28
14
Water itself is a weak acid and weak base:
Two water molecules react to form H3O+ and OH(but only slightly)
H2O
H2O
H 3 O+
OH-
29
Autoionization of Water
H2O(l) + H2O(l) ⇌
K=
[H3O+][OH-]
[H2O]2
The ion-product for water, Kw:
H3O+(aq) + OH-(aq)
= [H+][OH-] = Kw
Set = 1.0 since nearly pure liquid
water even when salts, acids, or
bases present.
K = Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25 °C)
For pure water the concentration of hydroxyl and hydronium ions
must be equal:
[H3O+] = [OH-] =
√1.0 x 10-14 = 1.0 x 10 -7 M (at 25 °C)
30
15
What is the pH of (a) pure water at 25 °C
([H+] = 1.0 x 10-7 M) and (b) 0.010 M HCl?
1.
2.
3.
4.
5.
6.
a) -7.00, b) -1.00
a) 7.00, b) 1.00
a) -7.00, b) -2.00
a) 7.00, b) 2.00
a) -7.00, b) -3.00
a) 7.00, b) 3.00
31
pH Calculations
What is the pH of (a) pure water at 25oC and (b) 0.010 M HCl?
(a) For pure water, [H+] = 1.0 x 10-7 M, so
pH = -log (1.0 x10-7) = 7.00
(b) The pH of 0.010 M HCl (a STRONG ACID) is
pH = -log (1 x 10-2) = 2.00
32
16
The Meaning of Kb, the Base Dissociation Constant
For the generalized reaction between a base, B(aq), and water:
⇌
B(aq) + H2O(l)
BH+(aq) + OH-(aq)
Equilibrium constant:
K = Kb = [BH+][OH-]
[B]
The stronger the base, the higher the [OH-] at
equilibrium and the larger the Kb.
33
The Relation Between Ka and Kb for a
Conjugate Acid-Base Pair
Acid
HA + H2O
H3O+ + A-
Base
A - + H2 O
HA + OH-
2 H2 O
[H3O+] [A-]
[HA]
Ka
x
x
H3O+ + OH-
[HA] [OH-]
[A-]
Kb
= [H3O+] [OH-]
=
Kw
For HNO2: Ka = 4.5 x 10-4 and for NO2- : Kb = 2.2 x 10-11
Ka x Kb = (4.5 x 10-4)(2.2 x 10-11) = 9.9 x 10-15
or ~ 10 x 10-15 = 1 x 10 -14 = Kw
34
17
Two New Definitions
Remember: Kw = [H+][OH-] so that [OH-] = Kw / [H+]
1)
pOH = -log [OH-]
= - log (Kw/[H+]) = - log Kw – (- log [H+])
= pKw – pH
= - log (1.0 x 10-14) – pH
pOH = 14.00 – pH
2) It is convenient to express equilibrium constants for acid
dissociation in the form
pKa = - log Ka
35
pH of a Strong Base
Calculate pH of 1.0 M NaOH
• major species : Na+, OH- and H2O
• sodium hydroxide is a strong base: FULLY dissociated
Therefore, 1.0 M NaOH → 1.0 M [OH-]
pH = - log [H+]; pOH= - log [OH-]; pKw= pH + pOH
= 14
pH = 14 – pOH
pH = 14 – log 1.0 = 14.0
Calculate pH of 0.01 M NaOH
pH= 14 – log 0.01 = 14 – 2 = 12.0
Calculate pH of 0.02 M NaOH
pH= 14 – log 0.02 = 14 - 1.70 = 12.3
36
18
Calculating [H3O+], pH, [OH-], and pOH
Problem: 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 25oC.
Plan:
Complete dissociation:
[H+] = initial HCl
concentration
HCl is Strong Acid
Use [H+] to calculate pH,
OH- and pOH
37
Calculating [H3O+], pH, [OH-], and pOH
Problem: 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 25oC.
Solution:
(a) [H+] = 3.0 M
pH = -log[H+] = -log(3.0) = -0.48
-14
[OH-] = Kw+ = 1 x 10
[H ]
3.0
= 3.3 x 10-15 M
pOH = - log(3.3 x 10-15) = 14.48
(b) [H+] = 0.0024 M
[OH-] = Kw+
[H ]
pH = -log[H+] = -log(0.0024) = 2.62
= 1 x 10
0.0024
-14
= 4.2 x 10-12 M
pOH = -log(4.2 x 10-12) = 11.38
38
19
What is the pH of a 2.0 x 10-3 M solution
of NaOH?
1.
2.
3.
4.
5.
6.
7.
5.0 x 10-12
2.0 x 10-3
2.70
3.00
11.00
11.30
12.00
39
Review Problem - pH of a Strong Base
Calculate the pH of a 2.0 x 10-3 M solution of NaOH.
Since NaOH is a strong base, it will dissociate 100% in water.
NaOH(aq)
Na+(aq) + OH-(aq)
Since [NaOH] = 2.0 x 10-3 M , [OH-] = 2.0 x 10-3 M
The concentration of [H+] can be calculated from Kw at 25oC:
[H+] =
-14
Kw
= 1.0 x 10
[OH-]
2.0 x 10-3
= 5.0 x 10-12 M
pH = - log [H+] = - log(5.0 x 10-12) = 11.30
OR
pH = pKw – pOH
pH = 14.00 – (-log [OH-]) = 14.00 – 2.70 = 11.30
40
20
pH= -log[H+]
pOH= - log [OH-]
pKw= 14 = pH + pOH
Kw= [H+][OH-] 41
pH Box
14-pOH
pH
pOH
14-pH
10-pH
-log([H3O+])
[H3O+]
Kw/[OH-]
10-pOH
-log([OH-])
[OH-]
Kw/[H3O+]
42
21
The Acid-Dissociation Constant (Ka)
Remember
Acids dissociate into ions in water:
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
HA(aq)
Ka =
[H3O+][A-]
[HA]
H+(aq) + A-(aq)
=
[H+][A-]
[HA]
In a dilute solution of a WEAK acid, most of HA remains
undissociated and therefore
[HA]init = [HA]eq
Ka = [H+][A-]/[HA] = [H+]2 / [HA]init
43
Calculate the pH of a 1.50 M HNO2 Solution
Problem: Calculate the pH of a 1.00 M solution of nitrous acid HNO2.
Solution:
HNO2 (aq)
H+(aq) + NO2-(aq)
Ka = 4.0 x 10-4
Initial concentrations: [H+] = 0, [NO2-] = 0, [HNO2] = 1.00 M
Final concentrations: [H+] = x , [NO2-] = x , [HNO2] = 1.00 M - x
[H+] [NO2-]
(x) (x)
= 4.0 x 10-4 =
[HNO2]
1.00 - x
Assume 1.00 – x ≈ 1.00 to simplify the problem.
Ka =
x2
1.00
≈ 4.0 x 10-4
or x2 ≈ 4.0 x 10-4
x = 2.0 x 10-2 = 0.02 M = [H+] = [NO2-]
pH = - log[H+] = - log(2.0 x 10-2) = 1.70
44
22
Hypochlorous acid is a weak acid formed in
laundry bleach. What is the [H+] of a 0.125 M
HClO solution?
Ka = 3.5 x 10-8, Kb = 2.9 x 10-7
1.
2.
3.
4.
4.35 x 10-9 M
6.61 x 10-5 M
3.63 x 10-8 M
1.90 x 10-4 M
45
Determining Concentrations from Ka and Initial [HA]
Problem: Hypochlorous acid is a weak acid formed in laundry bleach.
What is the [H+] and pH of a 0.125 M HClO solution? Ka = 3.5 x 10-8
Plan: We need to find [H+]. First we write the balanced equation and
the expression for Ka and solve for the hydronium ion concentration.
Solution:
HClO(aq) + H2O(l)
H3O+(aq) + ClO-(aq)
Ka =
[H+] [ClO-]
= 3.5 x 10-8
[HClO]
Concentration (M)
Initial
Change
Equilibrium
Ka =
HClO
H2O
0.125
-x
0.125 - x
----------
(x)(x)
= 3.5 x 10-8
0.125-x
H+
0
+x
x
ClO0
+x
x
Assume (0.125 – x) ≈ 0.125
x2 = 0.4375 x 10-8
x = 0.66 x 10-4
pH = -log(H+) = -log(0.66 x 10-4) = 4.18
46
23
Finding Ka for a Weak Acid from the pH – I
Problem: The weak acid hypochlorous acid is formed in bleach
solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the
value of the Ka of this weak acid?
Plan: We are given [HClO]initial and the pH, which will allow us to find
[H3O+] and then the hypochlorite anion concentration, so we can
write the reaction and expression for Ka and solve directly.
Solution:
Calculating [H3O+] : [H3O+] = 10-pH = 10-4.19 = 6.46 x 10-5 M
Concentration (M) HClO(aq)
Initial
Change
Equilibrium
0.12
-x
0.12 -x
+
H2O(l)
H3O+(aq)
----------
1 x 10 -7
+x
x + 1 x 10 -7
+
ClO -(aq)
0
+x
x
Assumptions: [H3O+] = [H3O+]HClO
47
since HClO is a weak acid, we assume (0.12 M – x) ≈ 0.12 M
Finding the Ka for a Weak Acid from the pH – II
H3O+(aq) + ClO -(aq)
HClO(aq) + H2O(l)
x = [H3O+] = [ClO-] = 6.46 x 10-5 M
Ka =
[H3O+] [ClO-]
[HClO]
=
(6.46 x 10-5 M) (6.46 x 10-5 M)
0.12 M
Ka = 3.48 x 10-8
In text books it is found to be: 3.5 x 10-8
Checking:
1. For [H3O+]from water :
2. For [HClO]dissoc :
= 3.48 x 10-8
1 x 10-7 M
x 100 = 0.155%
6.46 x 10-5 M
assumptions are OK
6.46 x 10-5 M
0.12 M
x 100 = 0.0538 %
48
24
Overview: Solving Weak Acid Equilibrium Problems
• List the major species in the solution.
• Choose the species that can produce H+ and write balanced equations for the
reactions producing H+.
• Comparing the values of the equilibrium constants for the reactions you have
written, decide which reaction will dominate in the production of H+.
• Write the equilibrium expression for the dominant reaction.
• List the initial concentrations of the species participating in the dominate
reaction.
• Define the change needed to achieve equilibrium; that is, define x.
• Write the equilibrium concentrations in terms of x.
• Substitute the equilibrium concentrations into the equilibrium expression.
• Solve for x the “easy” way… assume that [HA]0 – x ≈ [HA]0
• Verify the approximation is valid (5% rule).
• Calculate [H+] and pH.
49
Mixtures of Several Acids - I
Calculate the pH of a solution that contains 1.00 M HF
(Ka = 7.2 x 10-4) and 5.00 M HOCl (Ka = 3.5 x 10-8). Also calculate the
concentrations of the F- and OCl- ions at equilibrium.
Three components produce H+:
HF(aq)
⇌
H+(aq) + F-(aq)
Ka = 7.2 x 10-4
HOCl(aq)
⇌
H+(aq) + OCl-(aq)
Ka = 3.5 x 10-8
H2O(aq)
⇌
H+(aq) + OH-(aq)
Ka = 1.0 x 10-14
Even though HF is a weak acid, it has by far the greatest Ka, so it will be the
dominant producer of H+. Solve for its I.C.E. table first, then use those
concentrations in the second most important equilibrium. Move down the
list, in each case setting
Ka =
[H+] [A-]
[HA]
= specific value
50
25
Mixtures of Several Acids - II
Initial Concentration
(mol/L)
Equilibrium Concentration
(mol/L)
[HF]0 = 1.00
[F-] = 0
[H+] = ~ 0
Ka =
[HF] = 1.00 – x
[F-] = x
[H+] = x
x mol HF
dissociates
[H+] [F-]
[HF]
= 7.2 x 10-4 =
(x) (x)
1.00-x
≈
x2
1.00
x = 2.7 x 10-2
Therefore,
[F- ] = [H+] = x = 2.7 x 10-2 and pH = 1.56
51
Mixtures of Several Acids - III
The concentration of H+ comes from the first part of this problem:
Initial Concentration
(mol/L)
[HOCl]0 = 5.00
[OCl-] = 0
[H+] = 2.7 x 10-2
Ka =
Equilibrium Concentration
(mol/L)
x mol HOCl
dissociates
[HOCl] = 5.00 – x
[OCl-] = x
[H+] = 2.7 x 10-2 + x
[H+] [OCl-]
(2.7 x 10-2 + x) (x)
= 3.5 x 10-8 =
[HOCl]
5.00-x
≈
(2.7 x 10-2) x
5.00
x = 6.5 x 10-6 M = [OCl-]
Therefore,
pH = 1.56
[F- ] = 2.7 x 10-2 M
[OCl-] = 6.5 x 10-6 M
52
26
PERCENT DISSOCIATION
% dissociation = amount dissociated (mol/L) x 100
initial concentration (mol/L)
x
% dissociation = [HA] x 100
o
•for a given acid, % dissociation increases as the acid is diluted
•for a weak acid, [H+] decreases with increasing dilution as [HA] does,
but % dissociation increases as dilution increases
53
Figure 7.5: Effect of dilution on the percent
dissociation and [H+]
Ka =
[H+] [A-]
[HA]
54
27
Calculate the percent dissociation of a 1.00 M
hydrocyanic acid solution, Ka = 6.20 x 10-10
1.
2.
3.
4.
2.49 x 10-5 %
2.49 x 10-3 %
6.20 x 10-10 %
6.20 x 10-8 %
55
Problem: Calculate the percent dissociation of a 1.00 M
hydrocyanic acid solution, Ka = 6.20 x 10-10
HCN(aq) + H2O(l)
HCN
Initial 1.00 M
Change
-x
Final
1.00 –x
H3O+
0
+x
x
⇌
H3O+(aq) + CN- (aq)
CN0
+x
x
Ka =
Ka =
[H3O+][CN-]
[HCN]
(x)(x)
= 6.20 x 10-10
(1.00-x)
Assume 1.00-x ≈ 1.00
Ka =
x2
1.00
= 6.2 x 10-10
x = 2.49 x 10-5
% dissociation =
2.49 x 10-5
x 100 = 0.00249%
1.00
56
28
Weak Bases
• Many compounds with an electron-rich nitrogen are
weak bases.
•The common structural feature is an N atom that has
a lone electron pair in its Lewis structure.
EXAMPLES:
1) Ammonia (:NH3)
2) Amines (general formula RNH2, R2NH, R3N)
..
N
..
N
H3C H CH3
..
N
H3C H H
Methylamine
..
N
H3C CH CH3 H
3
H C2H5
Dimethylamine
Trimethylamine
Ethylamine
57
General reaction (R is a hydrocarbon group or a H):
NR3(aq) + H2 O(l) ⇌ HNR3+(aq) + OH-(aq)
+
Kb = [HNR3 ] [OH ]
[NR3]
58
29
Determining pH from Kb and Initial [B]–I
Problem: Ammonia is commonly used cleaning agent and is a weak base,
with a Kb of 1.8 x 10-5. What is the pH of a 1.5 M NH3 solution?
Plan: Ammonia reacts with water to form [OH-]. Calculate [H3O+] and the
pH. The balanced equation and Kb expression are:
NH4+(aq) + OH-(aq)
NH3 (aq) + H2O(l)
Kb =
Concentration (M)
NH3
Initial
Change
Equilibrium
1.5
-x
1.5 - x
[NH4+] [OH-]
[NH3]
H2O
NH4+
----------
0
+x
x
OH0
+x
x
making the assumption: since Kb is small, (1.5 M – x) ≈ 1.5 M
59
Determining pH from Kb and Initial [B]–II
Substituting into the Kb expression and solving for x:
Kb =
[NH4+] [OH-]
[NH3]
=
(x)(x)
1.5
= 1.8 x 10-5
x2 = 2.7 x 10-5
x = 5.20 x 10-3 = [OH-] = [NH4+]
60
30
What is the pH of the 1.5 M
NH3 solution? [OH-]=5.20 x 10-3 M
1.
2.
3.
4.
2.28
2.284
11.72
11.716
61
Determining pH from Kb and Initial [B]–III
Calculating pH:
[H3O+] =
Kw
[OH-]
=
1.0 x 10-14
5.20 x
10-3
= 1.92 x 10-12
pH = -log[H3O+] = - log (1.92 x 10-12)
pH = 11.716
62
31
Polyprotic Acids
63
The Stepwise Dissociation of Phosphoric Acid
Calculating ions concentrations in a certain molarity of phosphoric acid
H3PO4 (aq) + H2O(l)
⇌ H2PO4-(aq) + H3O+(aq)
Ka1 = 7.5x10-3
H2PO4-(aq) + H2O(l)
⇌ HPO42-(aq) + H3O+(aq)
Ka2 = 6.2x10-8
HPO42-(aq) + H2O(l)
⇌ PO43-(aq) + H3O+(aq)
Ka3 = 4.8x10-13
Net: H3PO4 (aq) + 3 H2O(l)
⇌ PO43-(aq) + 3 H3O+(aq)
Order from biggest Ka to smallest:
Just like in the example with multiple weak acids, each subsequent
Ka is so much smaller then the one above it that its contribution can
be neglected with respect to changing any concentrations of
previous products!!
64
32
Dissociation of Phosphoric Acid - Problem I
Calculate the pH of a 5.0 M H3PO4 solution and determine the
equilibrium concentrations of the species:
H3PO4 , H2PO4-, HPO4-2, and PO4-3
Solution:
H3PO4 (aq)
⇌
Ka1 = 7.5 x 10-3 =
Initial Concentration (mol/L)
[H+] [H2PO4-]
[H3PO4]
Equilibrium Concentration (mol/L)
[H3PO4]0 = 5.0
[H2PO4-]0 = 0
[H+]0 = 0
Ka1 = 7.5 x 10-3 =
H+(aq) + H2PO4-(aq)
[H3PO4] = 5.0 - x
[H2PO4-] = x
[H+] = x
(x)(x)
5.0-x
≈
x2
5.0
x = 1.9 x 10-1 M
65
Dissociation of Phosphoric Acid - Problem II
Since 1.9 x 10-1 is less than 5% of 5.0, the approximation is acceptable:
x = [H+] = [H2PO4-] = 0.19 M
pH = 0.72, [H3PO4] = 5.0 – x = 4.8 M
The concentration of HPO42- can be obtained from Ka2:
Ka2 = 6.2 x 10-8 =
where:
[H+] [HPO42-]
[H2PO4-]
2= (0.19 M) [HPO4 ]
(0.19 M)
[H+] = [H2PO4-] = 0.19 M, so [HPO42-] = Ka2 = 6.2 x 10-8 M
To calculate [PO43-], we use the expression for Ka3 , and the values
obtained from the other calculations:
+
334 ]
Ka3 = 4.8 x 10-13 = [H ] [PO2-4 ] = (0.19 M) [PO
[HPO4 ]
(6.2 x 10-8 M)
[PO43-] =
(4.8 x 10-13)(6.2 x 10-8 M) = 1.6 x 10-19 M
(0.19 M)
66
33
Calculate the pH of 3.00 x 10-3 M Sulfuric Acid
Polyprotic acid, more than one proton to lose!!! Multiple Ka’s
H2SO4 (aq)
⇌ HSO4-(aq) + H+(aq)
HSO4-(aq) ⇌ SO4-(aq) + H+(aq)
Ka1 = LARGE
Ka2 = 1.2 x 10-2
Because first dissociation has a very large Ka it can be treated as
a strong acid (~100% dissociated).
67
Calculate the pH of 3.00 x 10-3 M Sulfuric Acid
HSO4-(aq) ⇌ SO4-(aq) + H+(aq)
Initial Concentration (mol/L)
[HSO4-]0 = 0.00300
[SO42-]0 = 0
[H+]0= 0.00300
Equilibrium Concentration (mol/L)
[HSO4-] = 0.00300 – x
[SO42-] = x
[H+] = 0.00300 + x
x mol/L HSO4dissociates
to reach
equilibrium
From dissociation of H2SO4
Ka = 1.2x10-2
[H+][SO42-]
=
[HSO4-]
(0.00300 + x)(x)
(0.00300 – x)
No obvious approximation looks valid, so we must solve with the
quadratic formula.
Ka2 = 1.2 x 10-2 =
0 = x2 + 0.015x – 3.6 x 10-5
x = 2.10 x 10-3
x=
[H+] = 0.00300 + x = 0.00510
-b +-
b2 – 4ac
2a
pH = 2.292
a=1
b = 0.015
c = -3.6 x 10-5
Note: Change in [H+] does not violate huge Ka for 1st dissociation: [H2SO4] = 0.
68
34
SALTS and their pH
sodium acetate (CH3COONa) solution is basic
CH3COO- + H2O ⇌ CH3COOH + OH-
ammonium chloride (NH4Cl) solution is acidic
NH4+ + H2O ⇌
NH3 + H3O+
sodium chloride solution is neutral
NaCl + H2O ⇌ Na+ + Cl- + H2O
69
Effects of Salts on pH and Acidity
Salts that consist of cations of strong bases and the anions of strong
acids have no effect on the [H+] when dissolved in water.
Examples: NaCl, KNO3, Na2SO4, NaClO4, KBr, etc.
A salt whose cation alone has neutral properties (such as Na+ or K+)
and whose anion is the conjugate base of a weak acid, produces a basic
solution in water, since anion captures some H+.
Examples: NaF, KCN, NaC2H3O2, Na3PO4, Na2CO3, K2S, Na2C2O4, etc.
A salt whose anion alone has neutral properties and whose cation is the
conjugate acid of a weak base OR a small and highly charged metal ion
will produce an acidic solution in water.
Examples: NH4Cl, AlCl3, Fe(NO3)3, etc.
70
35
See
Table 7.6
71
If the solid sodium cyanide (NaCN) is
dissolved in pure water, will the resulting
solution be acidic, basic, or neutral?
1. Acidic
2. Basic
3. Neutral
72
36
Example 7.15: Additional complex salts
Predict whether an aqueous solution of each of the following salts will be
acidic, basic, or neutral.
a) NH4C2H3O2
b) NH4CN
c) Al2(SO4)3
Solution: get needed K from Ka = Kw/Kb or Kb = Kw/Ka
a) ammonium ion (NH4+): Ka from Kb
Kw
Ka (for NH4+) =
= 5.6 x 10-10
Kb (for NH3)
acetate ion (C2H3O2-): Kb from Ka is 5.6 x 10-10
Since Ka = Kb the solution will be neutral and the pH close to 7.
b) ammonium ion (NH4+): Ka is 5.6 x 10-10
cyanide ion (CN-): Kb from Ka is 1.6 x 10-5
Since Kb >> Ka the solution will be basic (pH > 7).
c) hydrated aluminum ion: Ka for Al(H2O)63+ = 1.4 x 10-5
sulfate ion: Kb for sulfate (from Ka for HSO4-) = 8.3 x 10-13
Since Ka >> Kb the solution will be acidic (pH < 7).
73
Determining the pH of a Solution of A- – I
Problem: Sodium cyanide in water will produce a basic solution. What
is the pH of a 0.25 M solution of NaCN? Ka of HCN = 4.0 x 10-10
Plan: We have to find the pH of a solution of the cyanide ion, CN -,
which acts as a base in water:
CN-(aq) + H2O(l)
Kb =
HCN(aq) + OH-(aq)
[HCN] [OH-]
[CN-]
Solution: Setting up the reaction table:
Concentration (M)
Initial
Change
Equilibrium
CN0.25
-x
0.25 - x
H2O
----------
HCN
0
+x
x
OH0
+x
x
74
37
Determining the pH of a Solution of A- – II
Solving for Kb :
Kw
Kb =
Ka
=
1.0 x 10-14
4.0 x
10-10
= 2.5 x 10-5
Substituting into the expression for Kb and solving for x.
Make the assumption: since Kb is small, (0.25 M – x) ≈ 0.25 M
Kb =
[HCN] [OH-]
[CN-]
x2 = 6.25 x 10-6
solving for pH:
= 2.5 x 10-5 =
(x)(x)
0.25
x = 2.50 x 10-3 = [OH-]
pH = 14.00 – (-log [OH-])
pH = 14.00 – (-log 2.50 x 10-3)
pH = 14.00 – 2.60 = 11.40
75
Small and highly charged cations produce acidic solutions in water.
Examples: Al3+, Fe3+, etc..
76
38
pH of a small highly charged cation - I
Calculate the pH of a 0.010 M AlCl3 solution. The Ka value for the
Al(H2O)63+ ion is 1.4 x 10-5.
Solution:
Since the Al(H2O)63+ ion is a stronger acid than water, the
dominate equilibrium will be:
Al(H2O)63+(aq)
Al(OH)(H2O)52+(aq) + H+(aq)
[Al(OH)(H2O)52+][H+]
[Al(H2O)63+]
1.4 x 10-5 = Ka =
Initial Concentration (mol/L)
[Al(H2O)63+]0 = 0.010
[Al(OH)(H2O)52+] = 0
[H+]0 = ~ 0
Equilibrium Concentration (mol/L)
x mol/L
[Al(H2O)63+] = 0.010 – x
[Al(OH)(H2O)52+] = x
[H+] = x
77
pH of a small highly charged cation - II
Thus:
Ka =
[Al(OH)(H2O)52+] [H+]
[Al(H2O)63+]
Ka =
(x) (x)
0.010 - x
=
= 1.4 x 10-5
x2
= 1.4x10-5
0.010
x = 3.7 x 10-4
Since the approximation is valid by the 5% rule:
[H+] = x = 3.7 x 10-4 M
and pH = 3.43
78
39
Predicting the Relative Acidity of Salt Solutions
Problem: Determine whether an aqueous solution of iron(III) nitrite,
Fe(NO2)3, is acidic, basic, or neutral.
Plan: The formula consists of the small, highly charged, and therefore
weakly acidic, Fe3+ cation and the weakly basic NO2- (anion of the weak
acid HNO2). To determine the relative acidity of the solution, we write
equations that show the reactions of the ions with water, and then find Ka
and Kb of the ions to see which ion reacts to form H+ or OH- to a greater
extent.
Solution: Writing the reactions with water:
Fe(H2O)63+(aq) + H2O(l)
Fe(H2O)5OH2+(aq) + H3O+(aq)
NO2-(aq) + H2O(l)
HNO2(aq) + OH -(aq)
Obtaining Ka and Kb of the ions: for Fe3+(aq) Ka = 6 x 10-3
for NO2-(aq), Kb must be determined:
Kw
1.0 x 10-14
Kb of NO2- =
=
= 2.5 x 10-11
Ka of HNO2 4.1 x 10-4
79
Since Ka of Fe3+ > Kb of NO2-, the solution is acidic.
If the solid solid ammonium fluoride (NH4F) is
dissolved in pure water, will the resulting
solution be acidic, basic, or neutral?
NH3, Kb = 1.8 × 10–5
HF, Ka = 7.2 × 10–4
1. Acidic
2. Basic
3. Neutral
80
40
Summary: Solving Acid-Base Equilibria Problems
• List the major species in solution.
•Look for reactions that can be assumed to go to completion, such as a strong
acids/bases dissociating or H+ reacting with OH-.
•For a reaction that can be assumed to go to completion:
a) Determine the concentrations of the products.
b) Write down the major species in solution after the reaction.
•Look at each major component of the solution and decide whether it is an acid or
a base.
•Pick the equilibrium that will control the pH. Use known values of the
dissociation constants for the various species to determine the dominant
equilibrium.
a) Write the equation for the reaction and the equilibrium expression.
b) Set up the I.C.E. table and define x.
c) Compute the equilibrium concentrations in terms of x.
d) Substitute the [ ]eq into the equilibrium expression and solve for x.
e) Check the validity of the approximation.
f) Calculate the pH and other concentrations as required.
81
Summary: General Strategies for Solving Acid-Base Problems
Think Chemistry. Focus on the solution components and their reactions. It
will almost always be possible to choose one reaction that is the most
important.
Be systematic. Write down all the things you know, including things like the
equilibrium expression, list what you can calculate, and list what’s needed.
Be flexible. Although all acid-base problems are similar in many ways,
important differences do occur. Treat each problem as a separate entity. Do
not try to force a given problem to match any you have solved before. Look
for both the similarities and the differences.
Be patient. The complete solution to a complicated problem cannot be seen
immediately in all its detail. Pick the problem apart into its workable steps.
Be confident. Look within the problem for the solution, and let the problem
guide you. Assume that you can think it out. Do not rely on memorizing
solutions to problems. In fact, memorizing solutions is usually detrimental
because you tend to try to force a new problem to be the same as one you
have seen before.
82
Understand and think - don’
don’t just memorize.
memorize
41