Chapter 16:
Acid-Base Equilibria and
Solubility Equilibria
CHEM 101
Chemistry for Scientists and Engineers
-Spring 2024-2025Sesil Çınar, Çimen Özgüç Önal
1
Brønsted Acids and Bases
A Brønsted acid is a proton donor
A Brønsted base is a proton acceptor
NH3 (aq) + H2 O(l) ⇌ NH4+ (aq) + OH– (aq)
base
acid ⇄ acid
base
conjugate conjugate
base
acid
acid
base
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2 15-2
Example 15.1
Identify the conjugate acid-base pairs in the reaction
between ammonia and hydrofluoric acid in aqueous
solution
NH3 (aq) + HF(aq) ⇌ NH4+ (aq) + F – (aq)
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3 15-3
Acid-Base Properties of Water
H2 O(l) ⇄ H + (aq) + OH – (aq)
autoionization of water
conjugate
base
acid
H2 O + H2 O ⇄ H3 O+ + OH–
conjugate
acid
base
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4 15-4
The Ion Product of Water
H2 O l ⇄ H + aq + OH – aq
H + OH –
Kc =
H2 O = constant
H2 O
K c [H2O] = Kw = H+  OH– 
The ion-product constant (Kw) is the product of the
molar concentrations of H+ and OH− ions at a particular
temperature.
Solution Is
At 25°C
Kw = H+  OH–  = 1.0 × 10 –14
H+  = OH– 
H+  > OH– 
H+  < OH– 
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neutral
acidic
basic
5 15-5
Example 15.2
The concentration of OH− ions in a certain
household ammonia cleaning solution is 0.0025 M.
Calculate the concentration of H+ ions.
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6 15-6
pH – A Measure of Acidity
pH = – log H+ 
Solution Is
At 25°C
neutral
H+  = OH– 
H+  = 1.0 × 1.0 –7 pH = 7
acidic
H+  > OH– 
H+  > 1.0 × 1.0 –7 pH < 7
basic
H+  < OH– 
H+  < 1.0 × 1.0 –7 pH > 7
pH ↑
H+  ↓
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7 15-7
Other Important Relationships
Table 15.1
The pHs of Some
Common Fluids.
Sample
pOH = – log OH– 
pH Value
Gastric juice in the stomach
1.0−2.0
Lemon juice
2.4
Vinegar
3.0
Grapefruit juice
3.2
Orange juice
3.5
Urine
4.8−7.5
Water exposed to air*
5.5
Saliva
6.4−6.9
Milk
6.5
Pure water
7.0
Blood
7.35−7.45
Tears
7.4
Milk of magnesia
10.6
Household ammonia
11.5
H+  OH–  = Kw = 1.0 × 10 –14
– log H+  – log OH–  = 14.00
pH + pOH = 14.00
pH Meter
*Water exposed to air for a long
period of time absorbs atmospheric
CO2 to form carbonic acid, H2CO3.
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8 15-8
Example 15.3
The concentration of H+ ions in a bottle of table wine
was 3.2×10−4 M right after the cork was removed.
Only half of the wine was consumed. The other half,
after it had been standing open to the air for a month,
was found to have a hydrogen ion concentration
equal to 1.0×10−3M. Calculate the pH of the wine on
these two occasions.
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9 15-9
Example 15.3
Solution
pH = – log[H+ ].
+
–4
When the bottle was first opened, [H ] = 3.2 ×10 M,
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1015-10
Example 15.5
–
–4
In a NaOH solution OH  is 2.9 × 10 M.
Calculate the pH of the solution.
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1115-11
Example 15.5
Solution
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1215-12
Electrolytes
Strong Electrolyte −100% dissociation
H2O
NaCl(s ) 
→ Na+ (aq ) + Cl– (aq )
Weak Electrolyte − not completely dissociated
CH3 COOH ⇄ CH3 COO– (aq) + H + (aq)
Strong Acids are strong electrolytes
HCl (aq )+ H2O (l ) → H3O+ (aq )+Cl– (aq )
HNO3 (aq )+ H2O(l ) → H3O+ (aq )+NO3– (aq )
HClO4 (aq )+H2O(l ) → H3O+ (aq )+ClO4– (aq )
H2SO4 (aq )+ H2O(l ) → H3O+ (aq )+HSO4– (aq )
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1315-13
Electrolytes
Weak Acids are weak electrolytes
HF(aq) + H2 O(l) ⇄ H3 O+ (aq) + F – (aq)
HNO2 (aq) + H2 O(l) ⇄ H3 O+ (aq) + NO–2 (aq)
2–
+
HSO−
4 (aq) + H2 O(l) ⇄ H3 O (aq) + SO4 (aq)
H2 O(l) + H2 O(l) ⇄ H3 O+ (aq) + OH – (aq)
Strong Bases are strong electrolytes
H2O
NaOH(s ) 
→ Na+ (aq )+OH– (aq )
H2O
KOH(s ) 
→ K + (aq )+OH– (aq )
H2O
Ba(OH)2 (s ) 
→ Ba2+ (aq )+ 2OH– (aq )
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1415-14
Electrolytes
Weak Bases are weak electrolytes
F – (aq) + H2 O(l) ⇄ OH – (aq) + HF(aq)
NO–2 (aq) + H2 O(l) ⇄ OH – (aq) + HNO2 (aq)
Conjugate acid-base pairs:
The conjugate base of a strong acid has no measurable
strength.
H3O+ is the strongest acid that can exist in aqueous
solution.
The OH− ion is the strongest base that can exist in
aqueous solution.
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1515-15
Strong vs. Weak Acids
Strong Acid (HCl)
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Weak Acid (HF)
1615-16
Example 15.7
Predict the direction of the following reaction in
aqueous solution:
HNO2 aq + CN – aq ⇌ HCN aq + NO–2 aq
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1715-17
Example 15.7
Solution
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1815-18
Weak Acids (HA) and Acid Ionization
Constants
HA aq + H2 O l ⇄ H3 O+ aq + A– aq
HA aq ⇄ H + aq + A– aq
H + A–
Ka =
HA
Ka is the acid ionization constant
Ka ↑
weak acid strength ↑
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1915-19
Ionization Constants of Some Weak Acids
Table 15.3 Ionization Constants of Some Weak Acids
and Their Conjugate Bases at 25°C
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2015-20
Weak Bases and Base Ionization
Constants
NH3 aq + H2 O l ⇄ NH4+ aq + OH – aq
NH4+ OH –
Kb =
NH3
Kb is the base ionization constant
Kb ↑
weak base strength ↑
Solve weak base problems like weak acids except
solve for OH–  instead of H+  ⋅
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2115-21
Ionization Constants of Conjugate AcidBase Pairs
HA ( aq ) ⇄ H ( aq ) + A
–
( aq )
Ka
–
–
A ( aq ) + H2O ( l ) ⇄ OH ( aq ) + HA ( aq ) K b
Kw
H2O ( l ) ⇄ H+ ( aq ) + OH– ( aq )
+
K aK b = Kw
Weak Acid and Its Conjugate Base
Ka =
Kw
Kb
Kb =
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Kw
Ka
2215-22
Diprotic and Triprotic Acids
May yield more than one hydrogen ion per molecule.
Ionize in a stepwise manner; that is, they lose one proton at a time.
An ionization constant expression can be written for each ionization stage.
Consequently, two or more equilibrium constant expressions must often be
used to calculate the concentrations of species in the acid solution.
H2 CO3 aq ⇌ H + aq + HCO–3 aq
HCO–3
aq ⇌ H
+
aq
+ CO2–
3
aq
H + HCO–3
K a1 =
H2 CO3
H+ CO2–
3
K a2 =
HCO–3
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2315-23
Common Ion Effect
The common ion effect is the shift in equilibrium caused by the
addition of a compound having an ion in common with the
dissolved substance.
The presence of a common ion suppresses the ionization of
a weak acid or a weak base.
Consider mixture of CH3 COONa (strong electrolyte) and
CH3 COOH (weak acid).
CH3 COONa 𝑠𝑠 → Na+ 𝑎𝑎𝑎𝑎 + CH3 COO− (𝑎𝑎𝑎𝑎)
CH3 COOH 𝑎𝑎𝑎𝑎 ⇄ H + 𝑎𝑎𝑎𝑎 + CH3 COO− (𝑎𝑎𝑎𝑎)
common
ion
24
Example 16.1
(a) Calculate the pH of a 0.20 𝑀𝑀 CH3 COOH solution.
(b) What is the pH of a solution containing both 0.20 𝑀𝑀
CH3 COOH and 0.30 𝑀𝑀 CH3 COONa? The 𝐾𝐾a of CH3 COOH
is 1.8 x 10−5 .
25
Example 16.1
Solution
(a)
26
Example 16.1
27
Buffers
A buffer solution is a solution of:
1. A weak acid or a weak base and
2. The salt of the weak acid or weak base
Both must be present!
A buffer solution has the ability to resist
changes in pH upon the addition of small
amounts of either acid or base.
Consider an equal molar mixture of CH3 COOH and CH3 COONa
Add strong acid
H + 𝑎𝑎𝑎𝑎 + CH3 COO− (𝑎𝑎𝑎𝑎) → CH3 COOH(𝑎𝑎𝑎𝑎)
Add strong base
OH− 𝑎𝑎𝑎𝑎 + CH3 COOH 𝑎𝑎𝑎𝑎 → CH3 COO− 𝑎𝑎𝑎𝑎 + H2 O(𝑙𝑙)
28
Example 16.2
Which of the following solutions can be classified as buffer
systems?
(a) KH2 PO4 /H3 PO4
(b) NaClO4 ⁄HClO4
(c) C5 H5 N⁄C5 H5 NHCl (C5 H5 N is pyridine)
Strategy:
What constitutes a buffer system? Which of the preceding solutions contains a
weak acid and its salt (containing the weak conjugate base)? Which of the
preceding solutions contains a weak base and its salt (containing the weak
29
conjugate acid)?
Example 16.2
Solution
(a) KH2 PO4 /H3 PO4
(b) NaClO4 ⁄HClO4
(c) C5 H5 N⁄C5 H5 NHCl
30
Example 16.3
(a) Calculate the pH of a buffer system containing 1.0 𝑀𝑀
CH3 COOH and 1.0 𝑀𝑀 CH3 COONa.
(b) What is the pH of the buffer system after the addition of 0.10
mole of gaseous HCl to 1.0 L of the solution? Assume that
the volume of the solution does not change when the HCl is
added.
31
Example 16.3
Solution (a)
CH3 COOH(𝑎𝑎𝑎𝑎) ⇄ H + 𝑎𝑎𝑎𝑎 + CH3 COO− (𝑎𝑎𝑎𝑎)
Initial (𝑀𝑀):
Change (𝑀𝑀):
Equilibrium(𝑀𝑀):
H + CH3 COO−
𝐾𝐾a =
CH3 COOH
32
Example 16.3
33
Example 16.3
(b)
Initial (mol):
Change (mol):
Final(mol):
HCl(𝑎𝑎𝑎𝑎)
→
H+ 𝑎𝑎𝑎𝑎
Cl− (𝑎𝑎𝑎𝑎)
34
Example 16.3
The neutralization reaction is summarized next:
Initial (mol):
Change (mol):
Final(mol):
CH3 COO− (𝑎𝑎𝑎𝑎)
+ H + (𝑎𝑎𝑎𝑎) → CH3 COOH(𝑎𝑎𝑎𝑎)
35
Example 16.3
Initial (𝑀𝑀):
Change (𝑀𝑀):
Equilibrium(𝑀𝑀):
CH3 COOH(𝑎𝑎𝑎𝑎) ⇌ H + 𝑎𝑎𝑎𝑎 + CH3 COO− 𝑎𝑎𝑎𝑎
H + CH3 COO−
𝐾𝐾a =
CH3 COOH
36
Example 16.3
37
Solubility Equilibria
𝐾𝐾𝑠𝑠𝑠𝑠 = Ag + [Cl− ]
AgCl 𝑠𝑠 ⇄ Ag + 𝑎𝑎𝑎𝑎 + Cl− (𝑎𝑎𝑎𝑎)
𝑲𝑲𝒔𝒔𝒔𝒔 is the solubility product constant
MgF2 𝑠𝑠 ⇄ Mg 2+ 𝑎𝑎𝑎𝑎 + 2F − (𝑎𝑎𝑎𝑎)
Ag 2 CO3 𝑠𝑠 ⇄ 2Ag + 𝑎𝑎𝑎𝑎 + CO2−
3 (𝑎𝑎𝑎𝑎)
Ca3 PO4 2 𝑠𝑠 ⇄ 3Ca2+ 𝑎𝑎𝑎𝑎 + 2PO3−
4 (𝑎𝑎𝑎𝑎)
𝐾𝐾𝑠𝑠𝑠𝑠 = Mg 2+ [F − ]2
𝐾𝐾𝑠𝑠𝑠𝑠 = [Ag + ]2 [CO2−
3 ]
2
𝐾𝐾𝑠𝑠𝑠𝑠 = [Ca2+ ]3 PO3−
4
Dissolution of an ionic solid in aqueous solution:
𝑄𝑄 < 𝐾𝐾𝑠𝑠𝑠𝑠
𝑄𝑄 = 𝐾𝐾𝑠𝑠𝑠𝑠
𝑄𝑄 > 𝐾𝐾𝑠𝑠𝑠𝑠
Unsaturated solution
No precipitate
Saturated solution
Supersaturated solution
Precipitate will form
38
Solubility Products
39
Solubility
Molar solubility (mol/L) is the number of moles of solute
dissolved in 1 L of a saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in
1 L of a saturated solution.
40
Example 16.8
The solubility of calcium sulfate (CaSO4 ) is found to be
0.67 g⁄L. Calculate the value of 𝐾𝐾sp for calcium sulfate.
Strategy
solubility of
CaSO4 in g/L
⟶ molar solubility of ⟶
CaSO4
[Ca2+ ] and
[SO2−
4 ]
⟶
𝐾𝐾sp of
CaSo4
41
Example 16.8
Solution
Let s be the molar solubility (in mol/L) of CaSO4 .
Initial (M):
Change (M):
Equilibrium (M):
CaSO4 𝑠𝑠 ⇌ Ca2+ 𝑎𝑎𝑎𝑎 + SO2−
4 (𝑎𝑎𝑎𝑎)
-s
0
+s
s
0
+s
s
The solubility product for CaSO4 is
2
𝐾𝐾sp = Ca2+ SO2−
=
𝑠𝑠
4
42
Example 16.8
First, we calculate the number of moles of CaSO4 dissolved in
1 L of solution:
From the solubility equilibrium we see that for every mole of
CaSO4 that dissolves, 1 mole of Ca2+ and 1 mole of SO2−
4 are
produced. Thus, at equilibrium,
43
Example 16.8
Now we can calculate 𝐾𝐾sp :
𝐾𝐾sp = Ca2+ [SO2−
4 ]
44
Example 16.9
Calculate the solubility of copper(II) hydroxide, Cu(OH)2 , in
g/L. Ksp: 2.2 × 10−20
𝐾𝐾sp of ⟶
Cu(OH)2
solubility of
[Cu2+ ] and
molar solubility
⟶ Cu(OH)2 in g⁄L
[OH− ] ⟶ of Cu(OH)2
45
Example 16.9
Consider the dissociation of Cu(OH)2 in water:
Initial (M):
Change (M):
Equilibrium (M):
Cu OH 2 (𝑠𝑠) ⇌ Cu2+ 𝑎𝑎𝑎𝑎 + 2OH − (𝑎𝑎𝑎𝑎)
-s
0
+s
s
0
+2s
2s
𝐾𝐾sp = Cu2+ [OH− ]2
=
46
Example 16.9
47
Relationship of 𝐾𝐾sp and Molar Solubility
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Table 16.3 Relationship Between 𝑲𝑲𝒔𝒔𝒔𝒔 and Molar Solubility (s)
Compound
AgCl
BaSO4
𝑲𝑲𝒔𝒔𝒔𝒔 Expression
Ag + [Cl− ]
Ba2+ [SO2−
4 ]
Cation
Anion
S
S
S
S
Ag 2 CO3
[Ag + ]2 [CO2−
3 ]
2s
s
PbF2
[Pb2+ ][F − ]2
S
2s
Al(OH)3
[Al3+ ][OH − ]3
S
3s
Ca3 (PO4 )2
2
[Ca2+ ]3 [PO3−
4 ]
3s
2s
Relation Between 𝑲𝑲𝒔𝒔𝒔𝒔 and s
1
𝐾𝐾sp = 𝑠𝑠 2 ; 𝑠𝑠 = 𝐾𝐾sp 2
1
𝐾𝐾sp = 𝑠𝑠 2 ; 𝑠𝑠 = 𝐾𝐾sp 2
1
𝐾𝐾sp 3
𝐾𝐾sp = 4𝑠𝑠 3 ; 𝑠𝑠 =
4
2
𝐾𝐾sp 3
𝐾𝐾sp = 4𝑠𝑠 3 ; 𝑠𝑠 =
4
1
𝐾𝐾sp 4
𝐾𝐾sp = 27𝑠𝑠 4 ; 𝑠𝑠 =
27
1
𝐾𝐾sp 5
𝐾𝐾sp = 108𝑠𝑠 5 ; 𝑠𝑠 =
108
48
Example 16.10
Exactly 200 mL of 0.0040 𝑀𝑀 BaCl2 are mixed with exactly
600 mL of 0.0080 𝑀𝑀 K 2 SO4 . Will a precipitate form?
49
Example 16.10
Strategy
50
Example 16.10
Solution
The number of moles of Ba2+ present in the original 200 mL of
solution is
The total volume after combining the two solutions is
800 mL. The concentration of Ba2+ in the 800 mL volume is
51
Example 16.10
The number of moles of SO2−
4 in the original 600 mL solution is
The concentration of SO2−
4 in the 800 mL of the combined
solution is
52
Example 16.10
Now we must compare Q and 𝐾𝐾sp .
BaSO4 𝑠𝑠 ⇌ Ba2+ 𝑎𝑎𝑎𝑎 + SO2−
4 (𝑎𝑎𝑎𝑎)
As for Q,
𝐾𝐾sp = 1.1 × 10−10
Q = Ba2+ 0 [SO2−
4 ]0
53
Extraction and Separation
Processes
• Type of heterogeneous equilibrium: partitioning
of a solute between two immiscible solvents.
• Equilibrium constant = partition coefficient.
• Used in many separation processes in chemical
industry.
5454
Extraction Separatory funnel
• Partitioning of solute
(iodine) between two
immiscible solvents (H2O
and CCl4).
• Extraction of dissolved
iodine from one solvent
into another.
• Can be used in the
purification of water
containing iodine.
I2 (aq)
K=
[I 2 ]CCl 4
[I 2 ]aq
I2 (CCl4)
= 85 at 25°C
55
Chromatography
• Due to Partition Equilibria
• Tube packed with a porous material
(silica gel) onto which water has
been adsorbed. Water is the
stationary phase. With the addition of
solvent (mobile phase), different
solutes travel at different rates.
• Partition coefficient K of the solute:
[solute]stationary phase
𝐾𝐾 =
[solute]mobile phase
56
Problem
Citric acid (C6H8O7) dissolves in water to the extent of
720 g L-1 at 15°C and in diethyl ether to the extent of
22 g L-1 at the same temperature.
(a) Calculate the equilibrium constants at 15°C for:
C6H8O7 (s)
C6H8O7 (aq)
C6H8O7 (s)
C6H8O7 (ether)
(b) Calculate the partition coefficient K for the transfer
of citric acid from water into ether:
C6H8O7 (aq)
C6H8O7 (ether)
57
Solution
(a) Calculate the equilibrium constants at 15°C for:
C6H8O7 (s)
C6H8O7 (s)
C6H8O7 (aq) Ksp1 = [C6H8O7 ]aq
C6H8O7 (ether) Ksp2 = [C6H8O7 ]ether
Ksp1 = [C6H8O7 ]aq =
Ksp2 = [C6H8O7 ]ether =
(b) C6H8O7 (aq)
C6H8O7 (ether) K = ?
58
Review Questions- Chapter 7
1. For H2O (l)
H2O (g), when equilibrium is reached
a. H2O (g) formation rate is more than H2O (l) formation rate
b. H2O (g) formation rate is less than H2O (l) formation rate
c. H2O (g) formation rate stops but H2O (l) continues to form
d. H2O (g) formation rate is equal to H2O (l) formation rate
e. Both H2O (g) formation and H2O (l) formation stops
2. What is the unit of the equilibrium constant K for 2A (aq)
a. 1/atm
b. atm
c. atm/(mol/L)2
d. Pa/(mol/L)2
e. unitless
B (g)?
For the reaction, 2A (g)
3. K for this reaction is
a. 4 atm b. 2 atm c. 1
B (g) at equilibrium, PB = 4 atm and PA = 2 atm
d. 2
e. 1/2
4. If we add B at constant temperature, such that PB = 5 atm, the reaction will
a. Shift to the left
b. Shift to the right
c. Not shift
d. Shift first to the left, then to the right
e. Shift first to the right, then to the left
5. If we add B at constant temperature, such that PB = 5 atm, the equilibrium constant K
a. K < 1
b. K > 1
c. K = 1
d. Cannot be determined without a calculation
e. K = 5/4
6. Ammonium carbamate decomposes into ammonia and carbon dioxide in
accord with the equilibrium,
NH4OCONH2(s)
2 NH3(g) + CO2(g)
A sample of ammonium carbamate is placed in an evacuated glass vessel at
25°C, which is then sealed. When the System reaches equilibrium, the
total pressure of gases is found to be 0.115 atm. Thus the equilibrium
constant for this reaction at 25°C is
a. 2.25 x 10-4
b. 1.52 x 10-3
d. impossible to determine from the given information
c. 6.08 x 10-3
d. none of these