Download chemistry 103 - chem.uwec.edu

y   0.20 Kb   1.1x10 10
1081
y   0.20 Kb   1.1x10 10
y = 1.0 x 10-5 M
[OH-] = 1.0 x 10-5 M
1082
y   0.20 Kb   1.1x10 10
Hence
y = 1.0 x 10-5 M
[OH-] = 1.0 x 10-5 M
pOH = 5.0
1083
y   0.20 Kb   1.1x10 10
Hence
y = 1.0 x 10-5 M
[OH-] = 1.0 x 10-5 M
pOH = 5.0
Since pOH + pH = 14.0, therefore
pH = 9.0
1084
Acid-base titrations: The impact of
hydrolysis
1085
Acid-base titrations: The impact of
hydrolysis
Salt hydrolysis has an important effect on the pH
profile of acid-base titrations.
1086
Acid-base titrations: The impact of
hydrolysis
Salt hydrolysis has an important effect on the pH
profile of acid-base titrations. The equivalence point
may be above or below neutral conditions (i.e. pH =
7).
1087
Acid-base titrations: The impact of
hydrolysis
Salt hydrolysis has an important effect on the pH
profile of acid-base titrations. The equivalence point
may be above or below neutral conditions (i.e. pH =
7).
For the titration of a strong acid and a strong base,
the equivalence point should be at pH = 7.
1088
Acid-base titrations: The impact of
hydrolysis
Salt hydrolysis has an important effect on the pH
profile of acid-base titrations. The equivalence point
may be above or below neutral conditions (i.e. pH =
7).
For the titration of a strong acid and a strong base,
the equivalence point should be at pH = 7.
Example: HCl(aq) + NaOH(aq)
NaCl(aq) + H2O
1089
1090
1091
1092
1093
Indicators
1094
Indicators
Indicators are often very weak organic acids. We
will represent an indicator as HIn.
1095
Indicators
Indicators are often very weak organic acids. We
will represent an indicator as HIn.
During a titration such as (where we assume NaOH
is being added)
1096
Indicators
Indicators are often very weak organic acids. We
will represent an indicator as HIn.
During a titration such as (where we assume NaOH
is being added)
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O
1097
Indicators
Indicators are often very weak organic acids. We
will represent an indicator as HIn.
During a titration such as (where we assume NaOH
is being added)
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O
The first drop of excess NaOH then reacts with the
indicator that is present:
1098
Indicators
Indicators are often very weak organic acids. We
will represent an indicator as HIn.
During a titration such as (where we assume NaOH
is being added)
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O
The first drop of excess NaOH then reacts with the
indicator that is present:
HIn(aq) + OH-(aq)
H2O + In-(aq)
1099
Indicators
Indicators are often very weak organic acids. We
will represent an indicator as HIn.
During a titration such as (where we assume NaOH
is being added)
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O
The first drop of excess NaOH then reacts with the
indicator that is present:
HIn(aq) + OH-(aq)
H2O + In-(aq)
Now HIn and In- have different colors, so we can
detect that the acid-base reaction is complete.
1100
+
For the equilibrium: HIn(aq) 
H
+
In
(aq)
(aq)

1101
+
For the equilibrium: HIn(aq) 
H
+
In
(aq)
(aq)


[H
][In
]
KIn 
[HIn ]
1102
+
For the equilibrium: HIn(aq) 
H
+
In
(aq)
(aq)


[H
][In
]
KIn 
[HIn ]
Midway in the transition of the indicator color
change: [HIn] = [In-], and hence KIn = [H+] (midway
point).
1103
+
For the equilibrium: HIn(aq) 
H
+
In
(aq)
(aq)


[H
][In
]
KIn 
[HIn ]
Midway in the transition of the indicator color
change: [HIn] = [In-], and hence KIn = [H+] (midway
point). Take the log of both sides of this
relationship, leads to
pKIn = pH (midway point).
1104
1105
IONIC EQUILIBRIUM
1106
IONIC EQUILIBRIUM
Buffers
1107
Buffers
Buffer: A solution whose pH remains approximately
constant despite the addition of small amounts of
either acid or base.
1108
Buffers
Buffer: A solution whose pH remains approximately
constant despite the addition of small amounts of
either acid or base.
A buffer is a combination of species in solution that
maintains an approximately constant pH by virtue
of a pair of chemical reactions.
1109
Buffers
Buffer: A solution whose pH remains approximately
constant despite the addition of small amounts of
either acid or base.
A buffer is a combination of species in solution that
maintains an approximately constant pH by virtue
of a pair of chemical reactions. One reaction
describes a reaction of a buffer component with
added acid, the other reaction describes the
reaction of a buffer component with added base.
1110
Example: acetic acid/sodium acetate buffer
1111
Example: acetic acid/sodium acetate buffer
A solution containing these two substances has the
ability to neutralize both added acid and added
base.
1112
Example: acetic acid/sodium acetate buffer
A solution containing these two substances has the
ability to neutralize both added acid and added
base.
If base is added to the buffer, it will react with the
acid component:
CH3CO2H(aq) + OH-(aq)
CH3CO2-(aq) + H2O
1113
Example: acetic acid/sodium acetate buffer
A solution containing these two substances has the
ability to neutralize both added acid and added
base.
If base is added to the buffer, it will react with the
acid component:
CH3CO2H(aq) + OH-(aq)
CH3CO2-(aq) + H2O
If acid is added to the buffer, it will react with the
base component:
CH3CO2-(aq) + H+(aq)
CH3CO2H(aq)
1114
Example: acetic acid/sodium acetate buffer
A solution containing these two substances has the
ability to neutralize both added acid and added
base.
If base is added to the buffer, it will react with the
acid component:
CH3CO2H(aq) + OH-(aq)
CH3CO2-(aq) + H2O
If acid is added to the buffer, it will react with the
base component:
CH3CO2-(aq) + H+(aq)
CH3CO2H(aq)
Note that the Na+ is not directly involved in the
buffer chemistry.
1115
Quantitative treatment of Buffers:
The Henderson-Hasselbalch Equation
1116
Quantitative treatment of Buffers:
The Henderson-Hasselbalch Equation
Consider the equilibrium:
CH3CO2H(aq)
CH3CO2-(aq) + H+(aq)
1117
Quantitative treatment of Buffers:
The Henderson-Hasselbalch Equation
Consider the equilibrium:
CH3CO2H(aq)
CH3CO2-(aq) + H+(aq)
[H ][CH3CO2- ]
Ka 
[CH3CO2H]
1118
Quantitative treatment of Buffers:
The Henderson-Hasselbalch Equation
Consider the equilibrium:
CH3CO2H(aq)
CH3CO2-(aq) + H+(aq)
[H ][CH3CO2- ]
Ka 
[CH3CO2H]
Now take the log of both sides of the preceding
equation, to obtain
1119
Quantitative treatment of Buffers:
The Henderson-Hasselbalch Equation
Consider the equilibrium:
CH3CO2H(aq)
CH3CO2-(aq) + H+(aq)
[H ][CH3CO2- ]
Ka 
[CH3CO2H]
Now take the log of both sides of the preceding
equation, to obtain
-]
[CH
CO
3
2
logKa  log [H ]
[CH3CO2H]














1120
That is,
[CH
CO
]

3
2
logKa  log[H ]  log
[CH3CO2H]
















1121
That is,
[CH
CO
]

3
2
logKa  log[H ]  log
[CH3CO2H]








[CH
CO
]

3
2
log[H ]   logKa  log
[CH3CO2H]
























1122
That is,
[CH
CO
]

3
2
logKa  log[H ]  log
[CH3CO2H]








[CH
CO
]

3
2
log[H ]   logKa  log
[CH3CO2H]
























[CH3CO2- ]
pH  pKa  log
[CH3CO2H]
















1123
That is,
[CH
CO
]

3
2
logKa  log[H ]  log
[CH3CO2H]








[CH
CO
]

3
2
log[H ]   logKa  log
[CH3CO2H]
























[CH3CO2- ]
pH  pKa  log
[CH3CO2H]
















This is the Henderson-Hasselbalch equation for the
acetic acid system.
1124
We could repeat the previous approach for the
weak acid HA to obtain:
1125
We could repeat the previous approach for the
weak acid HA to obtain:
[A
]
pH  pKa  log
[HA]














1126
We could repeat the previous approach for the
weak acid HA to obtain:
[A
]
pH  pKa  log
[HA]














In a more general form it would be:







pH  pKa  log [conjugate base]
[conjugate acid]







1127
We could repeat the previous approach for the
weak acid HA to obtain:
[A
]
pH  pKa  log
[HA]














In a more general form it would be:







pH  pKa  log [conjugate base]
[conjugate acid]







Either of the preceding two equations are called the
Henderson-Hasselbalch equation.
1128
Example: Calculate the pH of a buffer system
containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2.
What is the pH of the buffer after the addition of
0.10 moles of gaseous HCl to 1.00 liter of the buffer
solution? The Ka for acetic acid is 1.8 x 10-5.
1129
Example: Calculate the pH of a buffer system
containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2.
What is the pH of the buffer after the addition of
0.10 moles of gaseous HCl to 1.00 liter of the buffer
solution? The Ka for acetic acid is 1.8 x 10-5.
Because acetic acid is a weak acid, we can ignore
the small amount of dissociation and assume at
equilibrium that
[CH3CO2H] = 1.0 M
1130
Example: Calculate the pH of a buffer system
containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2.
What is the pH of the buffer after the addition of
0.10 moles of gaseous HCl to 1.00 liter of the buffer
solution? The Ka for acetic acid is 1.8 x 10-5.
Because acetic acid is a weak acid, we can ignore
the small amount of dissociation and assume at
equilibrium that
[CH3CO2H] = 1.0 M
It is also important to keep in mind that there is a
lot of acetate ion present, and this will suppress the
dissociation of the acetic acid (Le Châtelier’s
Principle).
1131
A similar situation applies to the acetate ion, that is,
we can ignore the hydrolysis of this ion. Also, the
acetic acid present will suppress the hydrolysis of
the acetate ion (Le Châtelier’s Principle), so that
[CH3CO2-] = 1.0 M
1132
A similar situation applies to the acetate ion, that is,
we can ignore the hydrolysis of this ion. Also, the
acetic acid present will suppress the hydrolysis of
the acetate ion (Le Châtelier’s Principle), so that
[CH3CO2-] = 1.0 M
Now Ka = 1.8 x 10-5 so that pKa = 4.7
1133
A similar situation applies to the acetate ion, that is,
we can ignore the hydrolysis of this ion. Also, the
acetic acid present will suppress the hydrolysis of
the acetate ion (Le Châtelier’s Principle), so that
[CH3CO2-] = 1.0 M
Now Ka = 1.8 x 10-5 so that pKa = 4.7
Hence, from the Henderson-Hasselbalch equation:









[CH3CO2
]
pH  pKa  log
[CH3CO2H]







 4.7  log 1.0 M
1.0 M
= 4.7
















1134
Upon addition of 0.10 moles of HCl to 1.0 liter of
the buffer solution (we make the assumption that
the total volume does not change), the following
neutralization reaction occurs:
1135
Upon addition of 0.10 moles of HCl to 1.0 liter of
the buffer solution (we make the assumption that
the total volume does not change), the following
neutralization reaction occurs:
CH3CO2- + H+
CH3CO2H
1136
Upon addition of 0.10 moles of HCl to 1.0 liter of
the buffer solution (we make the assumption that
the total volume does not change), the following
neutralization reaction occurs:
CH3CO2- + H+
CH3CO2H
0.10 mols
0.10 mols
0.10 mols
(from the HCl)
1137
Upon addition of 0.10 moles of HCl to 1.0 liter of
the buffer solution (we make the assumption that
the total volume does not change), the following
neutralization reaction occurs:
CH3CO2- + H+
CH3CO2H
0.10 mols
0.10 mols
0.10 mols
(from the HCl)
At equilibrium, the concentrations of the buffer
components are (keep in mind the total volume is
1.0 liter):
1138
Upon addition of 0.10 moles of HCl to 1.0 liter of
the buffer solution (we make the assumption that
the total volume does not change), the following
neutralization reaction occurs:
CH3CO2- + H+
CH3CO2H
0.10 mols
0.10 mols
0.10 mols
(from the HCl)
At equilibrium, the concentrations of the buffer
components are (keep in mind the total volume is
1.0 liter):
[CH3CO2H] = 1.0 + 0.10 = 1.10 M
[CH3CO2-] = 1.0 - 0.10 = 0.90 M
1139
To calculate the new pH, use the HendersonHasselbalch equation:
1140