To calculate the new pH, use the HendersonHasselbalch equation:









[CH3CO2
]
pH  pKa  log
[CH3CO2H]









1141
To calculate the new pH, use the HendersonHasselbalch equation:









[CH3CO2
]
pH  pKa  log
[CH3CO2H]







 4.7  log 0.90 M
1.10 M]
















= 4.6
1142
To calculate the new pH, use the HendersonHasselbalch equation:









[CH3CO2
]
pH  pKa  log
[CH3CO2H]







 4.7  log 0.90 M
1.10 M]
















= 4.6
Note in this example that the [H+] changed from
[H+] = 10-pH = 10-4.7 = 2.0 x 10-5 M
to
[H+] = 10-pH = 10-4.6 = 2.5 x 10-5 M
1143
To appreciate the effectiveness of the buffer in this
example, consider what happens to the pH when
0.10 mols of gaseous HCl is added to 1.0 liter of
H2O.
1144
To appreciate the effectiveness of the buffer in this
example, consider what happens to the pH when
0.10 mols of gaseous HCl is added to 1.0 liter of
H2O.
The initial [H+] = 1.0 x 10-7 M
(from self-dissociation of water)
1145
To appreciate the effectiveness of the buffer in this
example, consider what happens to the pH when
0.10 mols of gaseous HCl is added to 1.0 liter of
H2O.
The initial [H+] = 1.0 x 10-7 M
(from self-dissociation of water)
After the HCl is added, [H+] = 1.0 x 10-1 M , so there
is a million fold increase in [H+]!
1146
To appreciate the effectiveness of the buffer in this
example, consider what happens to the pH when
0.10 mols of gaseous HCl is added to 1.0 liter of
H2O.
The initial [H+] = 1.0 x 10-7 M
(from self-dissociation of water)
After the HCl is added, [H+] = 1.0 x 10-1 M , so there
is a million fold increase in [H+]!
Whereas for the buffer, [H+] changes from
2.0 x 10-5 M to 2.5 x 10-5 M.
1147
Distribution Curve
1148
Distribution Curve
Distribution curve (for a buffer): Gives the fraction
of the acid component and the base component
present in solution as a function of the pH.
1149
Distribution Curve
Distribution curve (for a buffer): Gives the fraction
of the acid component and the base component
present in solution as a function of the pH.
Example: Consider the acetic acid/acetate buffer.
1150
Distribution Curve
Distribution curve (for a buffer): Gives the fraction
of the acid component and the base component
present in solution as a function of the pH.
Example: Consider the acetic acid/acetate buffer.
At low pH: H+ + CH3CO2CH3CO2H
1151
Distribution Curve
Distribution curve (for a buffer): Gives the fraction
of the acid component and the base component
present in solution as a function of the pH.
Example: Consider the acetic acid/acetate buffer.
At low pH: H+ + CH3CO2CH3CO2H
At high pH: OH- + CH3CO2H
CH3CO2-
1152
Buffer Range: The pH range in which a buffer is
effective. The range is sometimes defined as
follows:
1153
Buffer Range: The pH range in which a buffer is
effective. The range is sometimes defined as
follows:
pH  pKa  1
1154
Buffer Range: The pH range in which a buffer is
effective. The range is sometimes defined as
follows:
pH  pKa  1
For the acetic acid/acetate ion buffer (pKa =4.7) the
buffer range is pH = 3.7 – 5.7 .
1155
Buffer Range: The pH range in which a buffer is
effective. The range is sometimes defined as
follows:
pH  pKa  1
For the acetic acid/acetate ion buffer (pKa =4.7) the
buffer range is pH = 3.7 – 5.7 . Note that the buffer
functions best at pH = 4.7, i.e. when
[CH3CO2H] = [CH3CO2-]
1156
1157
1158
Exercise: Comment on the following as possible
buffer systems.
(a) NaCl(aq)/HCl(aq)
(b) NH3(aq)/NH4Cl(aq)
(c) H2PO4-(aq)/ HPO42-(aq)
(d) NaHCO3(aq)
1159
(a) NaCl(aq)/HCl(aq)
To be a buffer, it needs to be able to react with
both added H+ and added OH-. If H+ is added,
the following reaction does not occur to any
significant extent: H+(aq) + Cl-(aq)
HCl(aq)
(because HCl is a strong acid).
That means that (a) cannot be a buffer system.
1160
(b) NH3(aq)/NH4Cl(aq)
Addition of H+: H+(aq) + NH3(aq)
NH4+(aq)
Addition of OH-:
OH-(aq) + NH4+(aq)
NH3(aq) + H2O
Hence (b) is a buffer system.
1161
(b) NH3(aq)/NH4Cl(aq)
Addition of H+: H+(aq) + NH3(aq)
NH4+(aq)
Addition of OH-:
OH-(aq) + NH4+(aq)
NH3(aq) + H2O
Hence (b) is a buffer system.
(c) H2PO4-(aq)/ HPO42-(aq)
Addition of H+: H+(aq) + HPO42-(aq)
H2PO4-(aq)
Addition of OH-:
OH-(aq) + H2PO4-(aq)
HPO42-(aq) + H2O
Hence (c) is a buffer system.
1162
(d) NaHCO3(aq)
1163
(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note
we are dealing with reactions in net ionic form.)
1164
(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note
we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3-(aq)
H2CO3(aq)
1165
(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note
we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3-(aq)
H2CO3(aq)
Addition of OH-:
OH-(aq) + HCO3-(aq)
CO32-(aq) + H2O
1166
(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note
we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3-(aq)
H2CO3(aq)
Addition of OH-:
OH-(aq) + HCO3-(aq)
CO32-(aq) + H2O
Hence (d) is a buffer system.
1167
(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note
we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3-(aq)
H2CO3(aq)
Addition of OH-:
OH-(aq) + HCO3-(aq)
CO32-(aq) + H2O
Hence (d) is a buffer system.
Note that in this example, the HCO3- ion functions
as both the acid and the base.
1168
(d) NaHCO3(aq)
(The Na+ cation is not involved in the chemistry. Note
we are dealing with reactions in net ionic form.)
Addition of H+: H+(aq) + HCO3-(aq)
H2CO3(aq)
Addition of OH-:
OH-(aq) + HCO3-(aq)
CO32-(aq) + H2O
Hence (d) is a buffer system.
Note that in this example, the HCO3- ion functions
as both the acid and the base.
Various anions of multi-proton acids can function
as buffer systems with a single species present.
1169
IONIC EQUILIBRIUM
Solubility Products
1170
IONIC EQUILIBRIUM
Solubility Products
A saturated solution of an insoluble salt is a
heterogeneous equilibrium.
1171
IONIC EQUILIBRIUM
Solubility Products
A saturated solution of an insoluble salt is a
heterogeneous equilibrium.
Example: In a saturated solution of AgCl solution,
the following equilibrium is present:
1172
IONIC EQUILIBRIUM
Solubility Products
A saturated solution of an insoluble salt is a
heterogeneous equilibrium.
Example: In a saturated solution of AgCl solution,
the following equilibrium is present:
+
AgCl(s) 
Ag
+
Cl
(aq)
(aq)

1173
IONIC EQUILIBRIUM
Solubility Products
A saturated solution of an insoluble salt is a
heterogeneous equilibrium.
Example: In a saturated solution of AgCl solution,
the following equilibrium is present:
+
AgCl(s) 
Ag
+
Cl
(aq)
(aq)

The equilibrium constant is:

-]
[Ag
][Cl
K
[AgCl ]
1174
Now
K[AgCl ]  [Ag][Cl-]
1175
Now
K[AgCl ]  [Ag][Cl-]
but [AgCl] is a constant, recall
1176
Now
K[AgCl ]  [Ag][Cl-]
but [AgCl] is a constant, recall
[AgCl] 
density AgCl
molar mass AgCl
1177
Now
K[AgCl ]  [Ag][Cl-]
but [AgCl] is a constant, recall
[AgCl] 
density AgCl
molar mass AgCl
We set Ksp  K[AgCl ] where the subscript sp stands
for solubility product.
1178
Now
K[AgCl ]  [Ag][Cl-]
but [AgCl] is a constant, recall
[AgCl] 
density AgCl
molar mass AgCl
We set Ksp  K[AgCl ] where the subscript sp stands
for solubility product. Hence,
Ksp  [Ag ][Cl-]
1179
2+
Examples: MgF2(s) 
Mg
+
2
F
(aq)
(aq)

1180
2+
Examples: MgF2(s) 
Mg
+
2
F
(aq)
(aq)

Ksp  [Mg2][F-]2
1181
2+
Examples: MgF2(s) 
Mg
+
2
F
(aq)
(aq)

Ksp  [Mg2][F-]2
Ca3(PO4)2(s)

 3 Ca2+(aq) + 2 PO43-(aq)
1182
2+
Examples: MgF2(s) 
Mg
+
2
F
(aq)
(aq)

Ksp  [Mg2][F-]2
Ca3(PO4)2(s)

 3 Ca2+(aq) + 2 PO43-(aq)
Ksp  [Ca2]3[PO34-]2
1183
2+
Examples: MgF2(s) 
Mg
+
2
F
(aq)
(aq)

Ksp  [Mg2][F-]2
Ca3(PO4)2(s)

 3 Ca2+(aq) + 2 PO43-(aq)
Ksp  [Ca2]3[PO34-]2
Note that the pure solids do not occur in the
expression for Ksp.
1184
Very small values for Ksp indicate very insoluble salts.
For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
1185
Very small values for Ksp indicate very insoluble salts.
For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-(aq) at 25 oC, we
have one of the following situations:
1186
Very small values for Ksp indicate very insoluble salts.
For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-(aq) at 25 oC, we
have one of the following situations:
[Ag][Cl-]  1.6x1010 unsaturated solution
1187
Very small values for Ksp indicate very insoluble salts.
For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-(aq) at 25 oC, we
have one of the following situations:
[Ag][Cl-]  1.6x1010 unsaturated solution
[Ag][Cl-]  1.6x1010 saturated solution
1188
Very small values for Ksp indicate very insoluble salts.
For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-(aq) at 25 oC, we
have one of the following situations:
[Ag][Cl-]  1.6x1010 unsaturated solution
[Ag][Cl-]  1.6x1010 saturated solution
[Ag][Cl-]  1.6x1010 supersaturated solution
1189
Very small values for Ksp indicate very insoluble salts.
For example, Ksp = 1.6 x 10-10 for AgCl at 25 oC.
In a solution containing Ag+(aq) and Cl-(aq) at 25 oC, we
have one of the following situations:
[Ag][Cl-]  1.6x1010 unsaturated solution
[Ag][Cl-]  1.6x1010 saturated solution
[Ag][Cl-]  1.6x1010 supersaturated solution
Some AgCl precipitate will form until the product of
the ionic concentrations is equal to 1.6 x 10-10.
1190
Sample problem 1: In a saturated solution of
Ag2CO3, the concentrations of the ions are [Ag+] =
2.54 x 10-4 M and [CO32-] = 1.27 x 10-4 M. Calculate
the Ksp and the solubility of Ag2CO3 in g/liter.
1191
Sample problem 1: In a saturated solution of
Ag2CO3, the concentrations of the ions are [Ag+] =
2.54 x 10-4 M and [CO32-] = 1.27 x 10-4 M. Calculate
the Ksp and the solubility of Ag2CO3 in g/liter.
+
2Ag2CO3(s) 
2
Ag
+
CO
(aq)
3 (aq)

1192
Sample problem 1: In a saturated solution of
Ag2CO3, the concentrations of the ions are [Ag+] =
2.54 x 10-4 M and [CO32-] = 1.27 x 10-4 M. Calculate
the Ksp and the solubility of Ag2CO3 in g/liter.
+
2Ag2CO3(s) 
2
Ag
+
CO
(aq)
3 (aq)

Ksp  [Ag ]2[CO23-]
= (2.54 x 10-4)2 (1.27 x 10-4)
= 8.19 x 10-12
1193
Sample problem 1: In a saturated solution of
Ag2CO3, the concentrations of the ions are [Ag+] =
2.54 x 10-4 M and [CO32-] = 1.27 x 10-4 M. Calculate
the Ksp and the solubility of Ag2CO3 in g/liter.
+
2Ag2CO3(s) 
2
Ag
+
CO
(aq)
3 (aq)

Ksp  [Ag ]2[CO23-]
= (2.54 x 10-4)2 (1.27 x 10-4)
= 8.19 x 10-12
The concentration of CO32-(aq) is equal to the number
of moles of Ag2CO3(s) that have dissolved.
1194
molar mass of Ag2CO3 = 275.8 g/mol
The solubility of Ag2CO3
= 1.27 x 10-4 mol l-1 275.8 g mol-1
= 0.0350 g/l
1195
Sample problem 2: Calculate the solubility of PbF2 in
g/l given Ksp = 4.1 x 10-8.
1196
Sample problem 2: Calculate the solubility of PbF2 in
g/l given Ksp = 4.1 x 10-8.

PbF2(s)  Pb2+(aq) + 2 F-(aq)
1197
Sample problem 2: Calculate the solubility of PbF2 in
g/l given Ksp = 4.1 x 10-8.

PbF2(s)  Pb2+(aq) + 2 F-(aq)
The abbreviated ICE table for this problem looks like
Pb2+
F-
1198
Sample problem 2: Calculate the solubility of PbF2 in
g/l given Ksp = 4.1 x 10-8.

PbF2(s)  Pb2+(aq) + 2 F-(aq)
The abbreviated ICE table for this problem looks like
Pb2+
Fy
2y
1199
Sample problem 2: Calculate the solubility of PbF2 in
g/l given Ksp = 4.1 x 10-8.

PbF2(s)  Pb2+(aq) + 2 F-(aq)
The abbreviated ICE table for this problem looks like
Pb2+
Fy
2y
Ksp  4.1x108  [Pb2][F-]2
1200