Acids & Bases
CHAPTER 16
(& part of CHAPTER 17)
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
CHAPTER 16: Acids & Bases
Learning Objectives:
 Define Brønsted-Lowry Acid/Base
 Define Lewis Acid/Base
 Evaluate the strength of acids/bases
 Strong vs weak acids/bases
 Periodic trends
 Conjugate acids/bases
 Identify likely compounds that will form acids and
bases from the periodic table
 Acidic metal ions
 Acid/Base equilibrium:
 pH, pOH
 Ka, Kb, pKa, pKb
 Kw of water
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 16: Acids & Bases
Lecture Road Map:
① Brønsted-Lowry Acids/Bases
② Trends in acid strength
③ Lewis Acids & Bases
④ Acidity of hydrated metal ions
⑤ Acid/Base equilibrium
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Molecular Nature of Matter, 6E
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CHAPTER 16 Acids & Bases
Acid/Base
Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Weak Acids & Bases
• Incompletely ionized
• Molecules and ions exist in equilibrium
• HA = any weak acid; B = any weak base
HA(aq) + H2O
A–(aq) + H3O+(aq)
B (aq) + H2O
B H+(aq) + OH–(aq)
CH3COOH(aq) + H2O
HSO3–(aq) + H2O
NH4+(aq) + H2O
CH3COO–(aq) + H3O+(aq)
SO32–(aq) + H3O+(aq)
NH3 (aq) + H3O+ (aq)
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Weak Acids & Bases
• Often simplify as
• HA (aq)
A –(aq) + H+(aq)
-
+
[A ][H ]
Ka =
[HA]
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Weak Acids & Bases
Acid + Water
Or generally
HA(aq) + H2O
K c¢ =
Conjugate Base + Hydronium Ion
A–(aq) + H3O+(aq)
[ A - ][H3O+ ]
[HA ][H2O]
• But [H2O] = constant (55.6 M ) so rewrite as
K c¢ ´ [H2O] =
[ A ][H3O ]
-
[HA]
+
= Ka
• Where Ka = acid ionization constant
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Weak Acids & Bases
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Weak Acids & Bases
CH3COO–(aq) + H2O
NH4+(aq) + OH–(aq)
NH3(aq) + H2O
• Or generally
B (aq) + H2O
B H+(aq) + OH–(aq)
[BH ][OH ]
K c¢ =
[B ][H2O]
+
-
[BH ][OH ]
Kb =
[B ]
+
CH3COOH(aq) + OH–(aq)
But [H2O] = constant
so rewrite as
-
Where Kb = base ionization
constant
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Weak Acids & Bases
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
pH
• Lots of weak acids and bases
– How can we quantify their relative strengths?
• Need reference
– Choose H2O
• Water under right voltage
– Slight conductivity
– Where does conductivity come from?
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
pH
• Trace ionization  self-ionization of water
• H2O + H2O
H3O+(aq) + OH–(aq)
acid
base
acid
base
• Equilibrium law is:
Kc =
+
-
[H3O ][OH ]
[H2O]2
æ 1 mol ö
• But [H2O]pure = 1000 g ç
÷÷ = 55.6 M
ç
è 18.0 g ø
1.00 L
[H2O] = constant even for dilute solutions
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
pH
H2O + H2O
H3O+(aq) + OH–(aq)
• Since [H2O] = constant, equilibrium law is
K w = [H3O ][OH ]
+
-
• K w = is called the ion product of water
• Often omit second H2O molecule and write
• H2O
H+(aq) + OH–(aq)
K w = [H ][OH ]
+
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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13
Acid/Base
Equilibrium
Kw
H2O
H+(aq) + OH–(aq)
• For pure H2O at 25 °C
– [H+] = [OH–] = 1.0 × 10–7 M
– Kw = (1.0 × 10–7)(1.0 × 10–7) = 1.0 × 10–14
– See Table 17.1 for K w at various temperatures
• H2O auto-ionization occurs in all solutions
– When other ions present
• [H+] is usually NOT equal to [OH–]
• But Kw = [H+][OH–] = 1.0 × 10–14
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Definition of Acidic & Basic
• In aqueous solution,
– Product of [H3O+] and [OH–] equals K w
– [H3O+] and [OH–] may not actually equal each other
– Solutions are classified on the relative
concentrations of [H3O+] and [OH–]
Solution Classification
Neutral
[H3O+] = [OH–]
Acidic
[H3O+] > [OH–]
Basic
[H3O+] < [OH–]
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Weak Acids & Bases: Example
Ex. 1 In a sample of blood at 25 °C,
[H+] = 4.6  10–8 M. Find [OH–] and determine if the
solution is acidic, basic or neutral.
K w = [H ][OH ] = 1 ´10
+
-
[OH ] =
Kw
+
[H ]
-
=
1.0 ´ 10-14
4.6 ´ 10
-8
-14
= 2.2 ´ 10
-7
•So 2.2 × 10–7 M > 4.6 × 10–8 M
•[OH–] > [H3O+] so the solution is basic
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
pH
• Arrhenius (of kinetics fame)
– Sought an easy way to write the very small
numbers associated with [H+] and [OH–]
– Developed the “p” notation where p stands for
the –log mathematical operation
pX = -log X
– Result is a simple number
• pH is defined as:
+
pH = -log[H ]
– Define pOH as:
Jesperson, Brady,
Hyslop.
Chemistry: The
Molecular
Nature of Matter,
6E
-
pOH = -log[OH ]
– Define pKw as: pKw = -log K w = 14.00
• Take anti-log to obtain [H+], [OH–] or Kw
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Acid/Base
Equilibrium
General Properties of Logarithms
log(a ´ b ) = log a + log b
log a b = b ´ log a
æa ö
log çç ÷÷ = log a - log b
èb ø
Using Logarithms
• Start with K w = [H+ ][OH- ]
• Taking –log of both sides of eqn. gives
-log([H+ ][OH- ]) = -log K w = -log(1.0 ´10-14 )
-log[H ] - log[OH ] = -log K w = -(-14.00)
+
-
• So at 25 °C:
pH + pOH = pK w = 14.00
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Definition of Acidic and Basic
• As pH increases, [H+] decreases; pOH
decreases, and [OH–] increases
• As pH decreases, [H+] increases; pOH
increases, and [OH–] decreases
Neutral
pH = 7.00
Acidic
pH < 7.00
Basic
pH > 7.00
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Measuring pH
1.
pH meter
– Most accurate
– Calibrate with solutions of known pH before use
– Electrode sensitive to [H+]
– Accurate to  0.01 pH unit
2. Acid-base indicator
– Dyes, change color depending on [H+] in solution
– Used in pH paper and titrations
– Give pH to  1 pH unit
3. Litmus paper
– Red pH  4.7 acidic
– Blue pH  4.7 basic
– Strictly acidic vs. basic
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Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
Acid/Base
Equilibrium
Jesperson, Brady, Hyslop. Chemistry: The
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Acid/Base
Equilibrium
Example pH Calculations
Calculate pH and pOH of blood in Ex. 1.
We found [H+] = 4.6 × 10–8 M
[OH–] = 2.2 × 10–7 M
pH = –log(4.6 × 10–8) = 7.34
pOH = –log(2.2 × 10–7) = 6.66
14.00 = pKw
Or
pOH = 14.00 – pH = 14.00 – 7.34 = 6.66
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Example pH Calculations
What is the pH of NaOH solution at 25 °C in
which the OH– concentration is 0.0026 M?
[OH–] = 0.0026 M
pOH = –log(0.0026) = 2.59
pH = 14.00 – pOH
= 14.00 – 2.59
= 11.41
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Strong Acids
• Assume 100% dissociated in solution
– Good ~ if dilute
• Makes calculating [H+] and [OH] easier
• 1 mole H+ for every 1 mole HX
– So [H+] = CHX for strong acids
• Thus, if 0.040 M HClO4
• [H+] = 0.040 M
• And pH = –log (0.040) = 1.40
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
Strong
Acids
HCl
HBr
HI
HNO3
H2SO4
HClO3
HClO4
HX (general
term for a
strong acid)24
Acid/Base
Equilibrium
Strong Bases
Strong Bases
NaOH
KOH
LiOH
Ca(OH)2
Ba(OH)2
• 1 mole OH– for every 1 mole M OH
• [OH–] = CMOH for strong bases
• 2 mole OH– for 1 mole M(OH)2
• [OH–] = 2
for strong bases
C M (OH)
2
Sr(OH)2
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Effect of Auto ionization of Water with
Strong Bases
• The auto-ionization of H2O will always add to [H+] and
[OH–] of an acid or base. Does this have an effect on the
last answer?
– The previous problem had 0.00022 M [OH–] from the
Ca(OH)2 but the [H+] must have come from water. If it
came from water an equal amount of [OH–] comes
from water and the total [OH–] is
– [OH–]total = [OH–]from Ca(OH)2 + [OH–]from H2O
– [OH–]total = 0.00022 M + 4.6 × 10–11 M
= 0.00022 M (when properly rounded)
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Example pH Calculations
Kw
-14
1.0
´
10
-13
[OH ] = + =
= 5.0 ´ 10
0.0200
[H ]
– So [H+] from H2O must also be 5.0  10–13 M
• [H+]total = 0.020 M + (5.0  10–13 M)
= 0.020 M (when properly rounded)
• So we see that [H+]from H2O will be negligible except
in very dilute solutions of acids and bases
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
-
Looking at Weak Acids Again
+
[A ][H ]
Ka =
[HA]
pK a = -log K a
K a = 10
-pK a
pK b   logK b
What is the pKa of HF if Ka = 3.5 × 10–4?
HF(aq) + H2O
F–(aq) + H3O+(aq)
or
HF(aq)
F–(aq) + H+(aq)
-
+
[F ][H ]
Ka =
[HF]
= 3.5 × 10–4
pKa = –log Ka
= –log(3.5 × 10–4) = 3.46
K b  10 pK b
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Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Conjugate Acid-Base Pairs
1. Consider ionization reaction of generic acid and water
HA(aq) + H2O
A–(aq) + H3O+(aq)
[A ][H ]
Ka =
[HA]
-
+
2. Consider reaction of a salt containing anion of this acid (its
conjugate base) with water
A–(aq) + H2O
HA(aq) + OH–(aq)
Kb =
[HA][OH- ]
[A - ]
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Acid/Base
Equilibrium
Conjugate Acid-Base Pairs
HA(aq) + H2O
A–(aq) + H2O
2H2O

A–(aq) + H3O+(aq)
HA(aq) + OH–(aq)
H3O+(aq) + OH–(aq)


[A ][H ] [HA][OH ]


K a Kb 

 [H ][OH ]  K w

[HA]
[A ]
For any conjugate acid base pair:
K a  K b  K w  1.0  1014
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
(at 25 °C)
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Acid/Base
Equilibrium
More About Logarithms
log(a ´ b ) = log a + log b
K a ´ K b = K w = 1.0 ´10
-14
• Then taking –log of both sides of equation gives:
 log(K a  K b )   logK w   log(1.0  1014 )
 logK a  logK b   logK w  (14.00)
So
pK a + pK b = pK w = 14.00
(at 25 °C)
• Earlier we learned the inverse relationship of conjugate acidbase strengths, now we have numbers to illustrate this.
• The stronger the conjugate acid, the weaker the conjugate
base.
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Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E