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Ch. 14: Acid, Bases, and Salts
Sections 14.1 – 14.5
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Sinkholes such as this are the result of reactions between acidic groundwaters and
basic rock formations, like limestone. (credit: modification of work by Emil Kehnel)
Chemistry: OpenStax
1
Why study acids and bases?
2
Measuring acidity, basicity
!  Acids and bases are everywhere.
Figure 14.4
pH meters
!  What examples in your home can you think of for acids? For
bases?
!  How do we identify acids and bases in lab?
!  Acids: most food and drink, low pH, litmus paper turns/stays red.
!  Bases: most cleaning products, high pH, litmus paper turns/stays
blue.
Figure 14.5: Indicators
and pH paper
3
Acid and Base Definitions
4
14.1 BrØnsted-Lowry Acids and Bases
When a Brønsted acid donates a proton, what remains of the acid is
known as a conjugate base.
!  CHM 151 taught Arrhenius theory
!  Acids produce hydrogen ions in water
!  HCl (aq) ! H+ (aq) + Cl- (aq)
!  Bases produce hydroxide ions in water
!  NaOH (aq) ! Na+ (aq) + OH- (aq)
!  152 uses the Brønsted-Lowry theory
!  Acid: Substance that can donate H+ (HCl works for both
definitions)
!  Base: Substance that can accept H+ (NaOH doesn’t work with
this definition)
!  Identify the acid and base below:
!  HCl(aq) + NH3(g) " NH4+(aq) + Cl-(aq)
The two species HCl and Cl– are known as a conjugate acid-base
pair or simply a conjugate pair.
5
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6
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BrØnsted-Lowry Acids and Bases
BrØnsted-Lowry Acids and Bases
When a Brønsted base accepts a proton, the newly formed
protonated species is known as a conjugate acid.
What is the:
Conjugate base of:
HNO3?
Conjugate acid of:
O2-?
-
NO3
HSO4SO4-2
H2SO4?
HSO4-?
CO3-2?
HCO3-?
OHHCO3H2CO3
Label each of the species in the following equations as acid, base,
conjugate acid, or conjugate base:
(a) HF(aq) + NH3(aq) ⇌ F-(aq) + NH4+(aq)
Write an equation for the dissociation of HCN in water. Identify
the acid, the base, the conjugate acid, and the conjugate base.
(b) CH3COO-(aq) + H2O(l) ⇌ CH3COOH(aq) + OH-(aq)
7
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Dissociation of Water
8
Acid-Base Properties of Water
A species that can behave either as a Brønsted acid or a Brønsted
base is called amphoteric (or amphiprotic).
K w = [H 3O+ ][OH - ] = 1.0x10 -14 (at 25oC)
Depending on which ion concentration is in excess, the solution
will be considered acidic or basic.
Table 14.2: Summary of Relations for Acidic, Basic and
Neutral Solutions
Relative Ion
Classification
pH at 25 °C
Concentrations
O+
OH–
The autoionization of water produces H3 and
ions in
equilibrium with water.
The equilibrium expression for the autoionization of water is:
−
pH < 7
+
−
pH = 7
+
−
pH > 7
[H3O ] > [OH ]
[H3O ] = [OH ]
basic
[H3O ] < [OH ]
O+]
If a solution has [H3
= 3.4 x
M, what is [OH-] at 25oC?
Is the solution acidic or basic?
[OH-] = 2.9 x 10-11 M; acidic
K w = [H 3O+ ][OH - ] = 1.0x10 -14 (at 25oC)
9
10-4
Examples 14.1 – 14.3
14.2 pH and pOH
10
pH and Sig Figs
The pH of a solution is the negative base-10 logarithm of the
hydronium ion concentration (in mol/L).
(French “puissance d’hydrogen” (“power of hydrogen”) or from the
German Potenz power + H (symbol for hydrogen)).
pH = -log [H 3O+ ]
+
acidic
neutral
!  Calculate pH if [H3O+] = 1.2 x 10-11 M.
!  pH = -log (1.2 x 10-11) = 10.9208
!  Sig figs: the exponent (-11) is exact, 1.2 determines how
many decimal places can be in pH.
[H 3O+ ] = 10 -pH
!  The number of sig figs in concentration determines the
number of decimal places in pH. And vice versa! The
number of decimal places in pH determines the number
of sig figs in concentration.
In pure water at 25oC, [H3O+] = 1.0 x 10–7 M
pH = –log(1.0 x 10–7) = 7.00
(pH is dimensionless) ☺
Calculate the pH of a solution with [H3O+] = 1.2 x 10-11 M.
pH = 10.93
!  pH = 10.92
Examples 14.4, 14.5
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Figure 14.2
pH – Calculation vs Experimental
Table of HCl Solutions with corresponding pH values:
Data taken from Christopher G. McCarty and Ed Vitz, Journal of Chemical Education, 83(5), 752
(2006) and G.N. Lewis, M. Randall, K. Pitzer, D.F. Brewer, Thermodynamics (McGraw-Hill: New
York, 1961; pp. 233-34).
[HCl], M
13
The pOH Scale
Relative
Deviation
0.00050
0.0100
3.30
2.00
3.31
2.04
0.3%
1.9%
0.100
0.40
1.00
0.39
1.10
0.52
9%
25%
7.6
-0.88
-1.85
52%
CC BY-NC-SA 3.0: http://chemwiki.ucdavis.edu/Wikitexts/Solano_Community_College/
Chem_160/Chapter_11%3A_Acids_and_Bases/11.3%3A_The_pH_Scale
14
Equations Summary
A pOH scale analogous to the pH scale and is the negative base-10
logarithm of the hydroxide ion concentration.
[OH − ] = 10 -pOH
pOH = -log [OH − ]
Calculated pH Experimental
( -log [H3O+])
pH
From the definition of pH and pOH: pH + pOH = 14.00
If a solution has a measured pH of 7.82, calculate [H3O+],
[OH-], and pOH.
[H3O+] = 1.5x10-8 M, [OH-] = 6.7 x 10-7 M, pOH = 6.18
Example 14.6
!  Kw = [H3O+][OH-] = 1.0 x 10-14
!  pH = -log [H+] or
pH = -log [H3O+]
!  [H3O+] = 10-pH
!  pOH = -log [OH-]
!  [OH-] = 10-pOH
!  pH + pOH = 14
!  Kw = Ka · Kb = 1.0 x 10-14
!  pKa = -log Ka pKb = -log Kb
!  pKa + pKb = 14
15
Equations Summary
14.3 Acid and Base Strength
!  [H3O+] = 7.2 x 10-4 M, calculate [OH-].
1.4 x 10-11 M
! 
[H3O+]
! 
[H3O+]
= 1.4 x
10-6
M, calculate pH.
5.85
= 1.5 x
10-6
M, calculate pH.
5.82
!  pH = 4.24, calculate [H3
O+].
A strong acid or base is one that dissociates almost 100%.
(There are only 7 of these.) Strong tells us nothing about
concentration, pH, or how reactive they are.
For example, HF is a weak acid, but can dissolve a human
body (and porcelain bath tub). Source: Breaking Bad.
5.8 x 10-5 M
!  [OH-] = 8.6 x 10-2 M, calculate pOH.
1.07
!  pH = 10.97, calculate pOH.
3.03
In this chapter we will focus on the chemical reactions of acids
and bases. First we have to differentiate between “strong” and
“weak”.
A weak acid or base typically dissociates less than ~5%. Again,
they can be reactive, it just means they don’t break apart in water
to a large extent.
17
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Strength of Acids: Strong vs Weak
HA
H+
HA
A-
A-
H+ H+
H+
AA-
HA
HA
[HA] = 1.0 M
HA
AH+
HA
Strong Acids
Strong acids do not set up equilibrium. They dissociate nearly 100%
(we assume 100% to make calculations easier) and reactions (almost)
go to completion. Strong acids have weak conjugate bases.
7 strong acids: HCl (aq), HBr (aq), HI (aq), HNO3 (aq), HClO4
(aq), HClO3 (aq), and H2SO4 (aq)
HA
H+
HA
A-
HA
Memorize these!!!!!
[H+] = [A-] = 1.0 M
We can write equations for reactions in water and draw the acids in
solution: Hydrochloric acid HCl(aq) + H2O(l)
H3O+(aq) + Cl–(aq)
HA
HA
Draw HNO3 and H2SO4 in water.
10 H3O+
HA
10
HA
HA
[HA] = 1.0 M
[H+] = [A‑] = 0.050 M
Cl-
19
20
Strong Bases
Weak Acids
Strong bases do not set up equilibrium. They dissociate nearly
100% (we assume 100% to make calculations easier) and reactions
(almost) go to completion. Strong bases don’t have conjugate acids.
8 strong bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2,
Sr(OH)2, and Ba(OH)2
Weak acids DO NOT dissociate completely. They typically only
dissociate about 5% and set up equilibrium. Weak acids have strong
conjugate bases.
What do you notice about these? Memorize these also!!!!!
We can also draw weak acids in solution. Since they don’t dissociate
very much, what will be the majors species in solution?
We can write equations for reactions in water and draw the bases in
Li+(aq) + OH–(aq)
solution: Lithium hydroxide: LiOH(aq)
HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)
or HF (aq) ⇌ H+(aq) + F-(aq)
Draw RbOH and Ca(OH)2 in water.
10 Li+
Draw CH3COOH in water.
H3O+
9 HF
10 OHF-
21
22
Relative Acid-Base Strengths
Weak Bases
Figure 14.8
Weak bases DO NOT dissociate completely. They typically only
dissociate about 5% and set up equilibrium. Weak bases have strong
conjugate acids.
We can draw weak bases in solution. Since they don’t dissociate
very much, what will be the major species in solution?
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
NH4+
9 NH3
OH23
24
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Conjugate Acid-Base Pairs
Conjugate Acid-Base Pairs
A strong acid ionizes completely in water:
A weak acid ionizes to a limited degree in water:
HF(aq) + H2O ⇌ H3O+(aq) + F–(aq)
HCl(aq) + H2O (l) " H3O+(aq) + Cl–(aq)
No affinity for
the H+ ion
Strong affinity
for the H+ ion
The chloride ion is a weak conjugate base.
The fluoride ion is a (relatively) strong conjugate base.
Cl–(aq) + H2O(l) X
" HCl(aq) + OH–(aq)
F–(aq) + H2O(l) ⇌
HF(aq) + OH–(aq)
25
Conjugate Acid-Base Strength
26
Relationship Between Ka and Kb
A simple relationship between the ionization constant of a weak acid
(Ka) and the ionization constant of a weak base (Kb) can be derived:
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq)
⇌
CH3COOH(aq)
2 H2O(l)
⇌
H3O+(aq)
+
+
OH–(aq)
OH–(aq)
[H3O ][CH 3COO ] [CH3COOH][OH ]
x
= [H3O + ][OH − ]
[CH3COOH]
[CH3COO − ]
Figure 14.7: This diagram shows the relative strengths of
conjugate acid-base pairs, as indicated by their ionization
constants in aqueous solution.
27
Relationship Between Ka and Kb
−
−
Ka × Kb = Kw
28
Calculating pH’s of Strong Acids/Bases
Calculate the following:
!  Calculate the pH of 0.035 M HNO3.
!  pH = 1.46
Kb = 5.6 x 10-10
(b) Ka of the methylammonium ion (CH3NH3+),
Ka = 2.2 x 10-11
(Kb (CH3NH2) = 4.6 x 10-4)
(c) Kb of the fluoride (F-),
(Ka (HF) = 7.2 x 10-4)
Kb = 1.4 x 10-11
(d) Ka of the ammonium ion (NH4+)
(Kb (NH3) = 1.8 x 10-5)
Ka = 5.6 x 10-10
Example 14.8
CH3COO–(aq)
CH3COO– (aq) + H2O(l)
+
(a)  Kb of the acetate ion (CH3COO-),
(Ka (CH3COOH) = 1.8 x 10-5)
+
!  Calculate the pH of 0.15 M NaOH.
!  pH = 13.18
!  Calculate the pH of a 0.013 M Ba(OH)2 solution.
!  pOH = 1.585; pH = 12.41
29
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Weak Acid Equilibria
Calculating Equilibrium Concentrations
Weak acids DO NOT dissociate almost completely. They typically
only dissociate about 5% and set up equilibrium. Therefore we can
use an equilibrium constant, Ka, to describe to what extent a weak
acid dissociates. The ionization of a weak monoprotic acid HA in
water is represented by:
Calculate the pH of a 0.50 M CH3COOH solution at 25oC.
Ka = 1.8x10-5.
+
–
+
HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) K a =
-
[H 3O ][A ]
[HA]
Ka is called the acid ionization constant. The larger the value of Ka,
the stronger the acid. Which is stronger, HF or CH3COOH?
Solution (at 25 oC)
Ka
0.10 M HF
7.2 x 10–4
0.10 M CH3COOH
1.8 x 10–5
31
Weak Acids; Acid-Ionization Constants (Ka)
Table 16.5.1
Acid
hydroiodic acid
Formula
HI
Ka
2x109
pKa
-9.3
sulfuric acid
nitric acid
H2SO4
HNO3
1x102
2.3x101
-2.0
-1.37
hydronium ion
hydrogen sulfate
O+
H3
HSO4-
1.0
1.0x10-2
0.00
1.99
hydrofluoric acid
nitrous acid
HF
HNO2
7.2x10-4
5.6x10-4
3.14
3.25
HCO2H
C6H5CO2H
CH3CO2H
1.78x10-4
3.75
4.20
4.75
formic acid
benzoic acid
acetic acid
6.3x10-5
1.8x10-5
Change
Equilibrium
0
–x
+x
+x
Equilibrium concentration (M)
0.50 – x
x
x
Solving for x = [H3O+] = 0.003 M
Check x: [H3O+]eq / [HA]i x 100 = 0.6% ( ☺ )
pH = - log [H3O+] = 2.52
Worked Example 14.9, 14.12, 14.14
32
Weak Bases; Kb Constants
B(aq) + H2O(l)
Kb =
⇌ HB+(aq)
+
OH–(aq)
[HB+ ][OH - ]
[B]
Kb is called the base ionization constant.
The larger the value of Kb, the stronger the base.
What is the pH of a 0.040 M NH3 solution at 25oC. Kb = 1.8x10-5
33
Example 14.10, 14.13
34
Weak Bases; Kb Constants
25oC. Kb = 1.8x10-5
!  B(aq) + H2O(l) ⇌ BH+(aq) + OH-(aq)
Initial
0
Change in concentration (M)
Weak bases ionize very little (~5%, like weak acids) and calculations
are similar to those of weak acids.
Weak Bases; Kb Constants
!  What is the pH of a 0.040 M NH3 solution at
HA(aq) + H2O(l) ⇌ H3O (aq) + A (aq)
0.50
Initial concentration (M)
Table 16.5.2
Base
Formula
Kb
pKb
hydroxide ion
phosphate ion
OHPO43-
1.0
2.1x10-2
0.00
1.68
NH3(aq) +
H2O(l) ⇌
NH4+(aq) +
0.040 M
-
0
0
-x
-
+x
+x
dimethylamine
methylamine
(CH3)3NH
CH3NH2
5.4x10-4
4.6x10-4
3.27
3.34
0.040 – x
-
x
x
trimethlyamine
(CH3)3N
6.3x10-5
4.20
ammonia
pyridine
NH3
C5H5N
1.8x10-5
1.7x10-9
4.75
8.77
aniline
C6H5NH2
7.4x10-10
9.13
OH-(aq)
!  Kb = x2 / (0.040 – x) = 1.8 x 10-5; x = 8.4853x10-4 M = [OH-]
!  pOH = 3.0713; pH = 10.93
35
36
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[H3O+], pH, Ka, and pKa trends
Equilibrium Constants: Acids and Bases
Strong acids: complete dissociation; all products; large Ka values
HCl(aq) + H2O(l) " H3O+(aq) + Cl–(aq)
Ka = 1.3 x 106
pH = -log [H3O+]
As [H3O+] increases, what happens to pH of an acid?
Weak acids: very little dissociation; mostly reactants; small Ka values
HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)
Ka = 7.2 x 10-4
Strong bases: complete dissociation; all products; large-ish Kb values
CsOH(aq) " Cs+(aq) + OH-(aq)
Kb = 1.7 x 10-2
Weak bases: very little dissociation; mostly reactants; small Kb values
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
Kb = 1.8 x 10-5
As acid strength increases, what happens to Ka values of acids?
pKa = - log Ka
As Ka increases, what happens to pKa?
37
38
Weak Acids; Percent Ionization
Effect of Dilution on % diss., pH
A quantitative measure of the degree of ionization or dissociation is
percent ionization (or dissociation).
percent ionization =
[H3O +
[H 3O+ ]eq
x 100%
[HA]i
This is the same calculation as
checking the assumption of a
small x!
O+]
In the previous problem, [H3
= 0.018974 M; [HF]i = 0.50 M.
Calculate percent ionization.
Example 14.7
Concentration of
CH3COOH (M)
]
0.10
0.010
0.0010
0.00010 0.000010
1.3x10-3
4.1x10-4
1.3x10-4
4.1x10-5
1.3
4.1
13
41
100
pH
2.89
3.39
3.89
4.39
4.89
HCl
Concentration
of acid
eq
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pH
39
Weak Acids; Ka Constants
Degree of
Ionization (%)
CH3COOH
0.10
0.010
0.0010
0.10
0.010
0.0010
1
2
3
2.89
3.39
3.89
100
100
100
1.3
4.1
13
40
Weak Acids; Ka Constants
Calculate the Ka of a 0.10 M HCN solution at 25oC if the
pH measured to be 5.15.
We can use ICE tables to solve three variations of weak acid
(or base) problems.
HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN–(aq)
1)  Given [HA]i and Ka, we can find [H3O+] and then pH.
Examples 14.9, 14.12, 14.14
2) Given [HA]i and pH, we can find [H3O+] and then Ka.
Example 14.11
3) Given [HA]i and percent dissociation, we can find Ka.
•  % diss. = ([H3O+] / [HA]i) x 100
•  (same as checking assumption of small x)
Example 14.7
Ka values can be found in Appendix H – in the back of the
book. Kb values are in Appendix I.
1.3x10-5
Degree of
Ionization (%)
0.10
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.10 – x
x
x
Initial concentration (M)
Solving for x: [H3O+] = 10-pH = 7.07946 x 10-6 M
Ka = x2 / (0.10 – x) = 5.0 x 10-10 (Ka is unitless!)
41
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Weak Acids; Ka Constants
Practice Problems!
Calculate the Ka of a 0.050 M CH3COOH solution at 25oC
if the percent dissociation is 1.9 %.
HA(aq) + H2O(l) ⇌ H3
O+(aq)
+
A–(aq)
0.050
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.050 – x
x
x
Initial concentration (M)
!  The acid-dissociation constant for benzoic acid, HC7H5O2,
is 6.3 x 10-5. Calculate pH and % dissociation if initial
concentration of the acid is 0.050 M.
!  Ans: [H3O+] = 1.8 x 10-3 M; pH = 2.74; % dissoc. = 3.5 %
!  A 0.200 M solution of a weak acid is 9.4% dissociated.
Using this information, calculate [H3O+]eq, [A-]eq, [HA]eq,
and Ka.
!  Ans: [H3O+] = [A-] = 0.019 M, [HA] = 0.181 M, Ka = 2.0 x 10-3
!  The pH of a formic acid solution is measured to be 2.58.
% diss. = [H3O+] / [HA]i) x 100 = 1.9 %;
[H3O+] = x = 9.5 x 10-4 M
Ka = x2 / (0.050 – x) = 1.8 x 10-5 (Ka is unitless!)
Calculate the Ka of this acid if initial acid is 0.0411 M.
!  Ans: [H3O+] = 2.6303x10-3 M = x; Ka = x2 / (0.0411 – x) = 1.8x10-4
43
Factors That Affect Acid Strength
Molecular Structure/Acid Strength
The strength of an acid is measured by its tendency to ionize.
HX
→
Binary (hydrohalic) acid strength:
HF << HCl < HBr < HI
As we move down the group, the biggest factor in bond strength is
anion size.
H+ + X–
Two factors influence ionization:
Only HF is a weak acid. Why?
Even though HCl, HBr, and HI are all strong acids, they are not
equally strong acids.
1) The strength of the H—X bond; weaker bonds result in
stronger acids. Why???
Relative Acid
Strength
2) The polarity of the H—X bond; polar bonds result in
stronger acids. Why???
δ+
δ–
H—X
45
Molecular Structure/Acid Strength
HF
<
HCl
<
HBr
<
HI
H-X Bond Energy
(kJ/mol)
570
432
366
298
Ka
6.3x10-4
1.3x106
7.9x108
2.0x109
pKa
3.20
-6.1
-8.9
-9.3
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46
Molecular Structure/Acid Strength
As we move left to right across a row, we can examine
electronegativities to compare acid strength (for binary acids).
C is least electronegative, C-H is least polar (non-polar), least acidic.
F is most electronegative, F-H is most polar, most acidic (easiest to
ionize).
Oxoacids: An oxoacid contains hydrogen, oxygen, and a central
nonmetal atom (acids made from polyatomic ions).
Figure 14.14
Figure 14.13
47
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Molecular Structure/Acid Strength
Molecular Structure/Acid Strength
To compare oxoacid strength, we can divide the oxoacids into two
groups:
2) Oxoacids having the same central atom, but different numbers of
oxygen atoms. Different oxidation #’s. (# ox. #, # acid strength)
1)  Oxoacids having different central atoms that are from the same
group of the periodic table and have the same oxidation number. The
more electronegative central atom results in a stronger acid.
As the e- density
on the oxygen atoms
decreases, so does
their affinity for a
proton, making the
anion less basic.
HOX
Electronegativity of X
Ka
pKa
HOCl
3.0
4.0 x 10-8
7.40
HOBr
2.8
2.8 x 10-9
8.55
HOI
2.5
3.2 x 10-11
10.5
Cl is more electronegative; the O—H bond is more polar.
49
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Molecular Structure/Acid Strength
14.5 Diprotic and Polyprotic Acids
Determine which acid in each pair or group is stronger/strongest.
Explain why that acid is stronger/strongest.
1)  HCl
Diprotic and polyprotic acids undergo successive ionizations, losing
one proton at a time, and each has a Ka value associated with it.
H2CO3(aq) + H2O(l) ! HCO3-(aq) + H3O+(aq)
HF
2) HClO3
HClO
3)  HBrO
HClO
4) H2Se
HBr
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⎡H+ ⎤ ⎡HCO3− ⎤⎦
K a1 = ⎣ ⎦ ⎣
[H2CO3 ]
= 4.2 x 10-7
HCO3-(aq) + H2O(l) ! CO32-(aq) + H3O+(aq)
⎡H+ ⎤ ⎡CO32− ⎤
K a2 = ⎣ ⎦ ⎣ − ⎦ = 4.8 x 10-11
⎣⎡HCO3 ⎦⎤
Ka1 > Ka2 > Ka3 > Ka4 …
For a given acid, the first ionization constant is much larger than the
second which is larger than the third and so on.
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Diprotic and Polyprotic Acids
Diprotic and Polyprotic Acids
Appendix H: Ionization Constants of Weak Acids
Acid
Formula
Ka1
Formula
Ka2
Formula
Ka3
arsenic
acid
H3AsO4
5.5x10-3
H2AsO4-
1.7x10-7
HAsO42-
5.1x10-12
carbonic
acid
H2CO3
4.3x10-7
HCO3-
5.6x10-11
citric acid
C6H8O7
7.4x10-4
C6H7O7-
1.7x10-5
oxalic acid
C2H2O4
6.0x10-2
C2HO4-
6.1x10-5
phosphoric
acid
H3PO4
7.5x10-3
H2PO4-
6.2x10-8
sulfuric
acid
H2SO4
Strong
HSO4-
1.0x10-2
C6H6O72-
4.0x10-7
HPO42-
4.2x10-13
!  Find pH of a 0.15 M carbonic acid (H2CO3) solution.
!  How many ICE Tables?
H2CO3(aq)
+ H2O(l) !
HCO3-(aq) +
I
0.15 M
-
0
0
C
-x
-
+x
+x
E
0.15 - x
-
E
0.15 M
H3O+(aq)
x
X
2.51 x 10-4 M
2.51 x 10-4 M
!  Ka1 = [HCO3-][H3O+] / [H2CO3] = 4.3 x 10-7
!  Assume small x.
!  Use [HCO32-]eq and [H3O+]eq in 2nd ICE table
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Diprotic and Polyprotic Acids
14.4 Hydrolysis of Salt Solutions
!  HCO3-(aq) + H2O(l) ! CO32-(aq) + H3O+(aq)
!  Ka2 = [CO32-][H3O+] / [HCO3-] = 5.6 x 10-11
HCO3
-(aq)
+ H2O(l) !
CO3
2-(aq)
+
H3
2.51 x 10-4 M
-
0
C
-x
-
+x
+x
E
2.51 x 10-4 - x
-
x
2.51 x 10-4 + x
E
2.51 x 10-4 M
I
Salts are produced from acid-base reactions! But not all
salts have a neutral pH.
O+(aq)
2.51 x 10-4 M
NaOH (aq) + HCl (aq) " ?
NaOH (aq) + HF (aq) " ?
5.6 x 10-11 M 2.51 x 10-4 M
/(2.51x10-4 - x) = 5.6 x 10-11
!  Ka2 =
+ x)
!  Approximate small x; x = 5.6 x 10-11 M
!  Total [H3O+] = 2.51x10-4 + 5.6x10-11 = 2.51x10-4 M
!  pH = 3.60
Example 14.19
(x)(2.51x10-4
NH3 (aq) + HCl (aq) " ?
NH3 (aq) + CH3COOH (aq) ⇌ ?
What is a salt? How do we recognize one?
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Acid-Base Properties of Salt Solutions
Acid-Base Properties of Salt Solutions
•  Examples: NaCl, NaF, NH4Cl, NH4CH3COO
Salts are ionic compounds that contain a cation and an anion.
We can label ions as acidic, basic, or neutral based on the
behavior of their ions in water.
•  We can label salts based on the acid-base combination they are made
from OR by labeling each ion based on its behavior in water.
•  What will anions from weak acids do in water (e.g., F-)?
•  Can they be acidic?
•  What labels are possible for anions?
Cations from strong bases are neutral and don’t hydrolyze (Li+,
Na+, K+, Rb+, Cs+, Ca2+, Sr2+, Ba2+). They are spectator ions!
Anions from strong acids are neutral and don’t hydrolyze (Cl-,
Br-, I-, NO3-, ClO4-, ClO3-, HSO4-). They are spectator ions!
• What will cations from weak bases do in water (e.g., NH4+)?
•  Can they be basic?
•  What labels are possible for cations?
If the cation or anion comes from a weak base or a weak
acid, then these ions will hydrolyze (react with water) and
cause a shift in pH.
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Acid-Base Properties of Salt Solutions
!  Most salts cause pH shifts when dissolved in water. These acidic
or basic salts contain the conjugate acid (or base) of a weak base
(or acid). Salt hydrolysis occurs for acidic or basic salts. Nonneutral ions react with water to produce either hydroxide ions or
hydronium ions.
NaF: Na+ is neutral; F- is basic
Dissociation: NaF(s) " Na+(aq) + F-(aq)
Hydrolysis: F- + H2O ! HF + OH-; pH > 7
• 
Acid-Base Properties of Salt Solutions
The pH of salt solutions can be qualitatively predicted by
determining which ions will hydrolyze.
Examples
A cation that will not affect the pH of a solution is
# A Group 1A or heavy Group 2A cation (except Be2+)
# The conjugate base of a strong acid
Cl-
NH4Cl:
is acidic;
is neutral
Dissociation: NH4Cl(s) " NH4+(aq) + Cl-(aq)
Hydrolysis: NH4+ + H2O ! NH3 + H3O+ pH < 7
Cl– , NO3– , ClO4–
An anion that will make a solution basic is
# The conjugate base of a weak acid
NH4+
Li+ , Na+ , Ba2+
An anion that will not affect the pH of a solution is
CN– , NO2– , CH3COO–
A cation that will make a solution acidic is
# The conjugate acid of a weak base
NH4+ , CH3NH3+ , C2H5NH3+
# A small, highly charged metal ion (other than Group 1A
or 2A)
Al3+ , Cr3+ , Fe3+ , Bi3+
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Acid-Base Properties of Salt Solutions
Acid-Base Properties of Salt Solutions
Small, highly charged metal ions can react with water to produce an
acidic solution.
The contours show the electron density on the O atoms and the H atoms in both a free
water molecule (left) and water molecules coordinated to Na+, Mg2+, and Al3+ ions.
These contour maps demonstrate that the smallest, most highly charged metal ion
(Al3+) causes the greatest decrease in electron density of the O–H bonds of the water
molecule. Due to this effect, the acidity of hydrated metal ions increases as the charge
on the metal ion increases and its radius decreases.
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Group Work
!  Classify the following salts as acidic, basic, or neutral.
!  For acidic or basic salts, write the hydrolysis equation.
KBr
*Mg(NO3)2
NaClO4
NH4I
*AlCl3
SrF2
LiCN
KNO2
NaCH3COO
RbI
NH4NO3
NH4CH3COO
*: do not write hydrolysis equations for these salts
Example 14.17
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Calculating the pH of Salt Solutions
!  Calculate the pH of a 0.10 M NaCl solution.
!  7.00
!  Calculate the pH of a 0.10 M NaF solution. Ka (HF) =
7.2 x 10-4
!  8.07
!  Calculate the pH of a 0.10 M NH4Cl solution. Kb (NH3)
= 1.8 x 10-5
!  5.13
Examples 14.15, 14.16, 14.18
Calculating the pH of Salt Solutions
!  Calculate the pH of a 0.10 M NH4Cl solution.
Kb (NH3) = 1.8 x 10-5
!  Ka = Kw / Kb = 5.5556 x 10-10
!  Ka = x2 / (0.10 – x)
!  x = √(5.5556 x 10-10 x 0.10) = 7.454x10-6 M = [H3O+]
!  pH = - log (7.454x10-6)
!  5.13
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