Chemistry: McMurry and Fay, 6th Edition
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Transcript Chemistry: McMurry and Fay, 6th Edition
C H E M I S T R Y
Chapter 15
Applications of Aqueous Equilibria
Neutralization Reaction
General Formula
Acid + Base Water + Salt
Neutralization Reactions
Strong Acid-Strong Base
HCl(aq)
+
NaOH(aq)
H2O(l) +
NaCl(aq)
Assume complete dissociation:
H3O+(aq) + OH–(aq)
2H2O(l)
(net ionic equation)
After neutralization: pH = 7
Strong acid-Strong base neutralization
When the number moles of acid and base are mixed
together
[H3O+] = [-OH] = 1.0 x 10-7M
Reaction proceeds far to the right
Neutralization Reactions
Weak Acid - Strong Base
CH3CO2H(aq) +
NaOH(aq)
H2O(l) +
NaCH3CO2(aq)
Assume complete dissociation:
CH3CO2H(aq) + OH–(aq)
H2O(l) + CH3CO2–(aq)
(net ionic equation)
After neutralization: pH > 7
Weak acid-strong base neutralization
Neutralization of any weak acid by a strong base goes 100%
to completion
-OH has a great infinity for protons
Neutralization Reactions
Weak Acid CH3CO2H(aq) +
Strong Base
NaOH(aq)
H2O(l) +
NaCH3CO2(aq)
Neutralization Reactions
Strong Acid - Weak Base
HCl(aq)
+ NH3(aq)
NH4Cl(aq)
Assume complete dissociation:
H3O+(aq) + NH3(aq)
H2O(l) +
(net ionic equation)
After neutralization: pH < 7
NH4+(aq)
Strong acid-weak base neutralization
Neutralization of any weak base by a strong acid goes 100%
to completion
H3O+ has a great infinity for protons
Neutralization Reactions
Strong Acid - Weak Base
HCl(aq) + NH3(aq)
NH4Cl(aq)
Neutralization Reactions
Weak Acid-Weak Base
CH3CO2H(aq) + NH3(aq)
NH4CH3CO2(aq)
Assume complete dissociation:
CH3CO2H(aq) + NH3(aq)
NH4+(aq) + CH3CO2–(aq)
(net ionic equation)
After neutralization: pH = ?
Weak acid-weak base neutralization
Less tendency to proceed to completion than neutralization
involving strong acids and strong bases
The Common-Ion Effect
Common-Ion Effect: The shift in the position of an equilibrium
on addition of a substance that provides an ion in common
with one of the ions already involved in the equilibrium.
• Example of Le Chatelier’s principle
• The addition of acetate ion ,CH3CO21-, to a solution of
acetic acid suppresses the dissociation of the acid. The
equilibrium shifts to the left.
CH3CO2H(aq) + H2O(l)
H3O1+(aq) + CH3CO21-(aq)
The Common-Ion Effect
The Common-Ion Effect
Le Châtelier’s Principle
CH3CO2H(aq) + H2O(l)
H3O1+(aq) + CH3CO21-(aq)
E.g Adding HCl and NaOH to a solution of acetic acid would
shift the equilibrium to which direction?
Example
The pH of a 0.10M of acetic is 2.87 at 25oC. Determine the
pH at the same temperature of a solution by adding 0.050
mole of sodium acetate to 1.0 L of 0.10 M acetic acid.
(Assume that the addition of sodium acetate does not
change the volume of the solution) Ka = 1.8 x 10-5
Example
Dtermine the pH at 25oC of a solution prepared by
dissolving 0.35 mol of ammonium chloride in 1.0 L of 0.25
M aqueous ammonia solution. No volume changed. Kb =
1.8 x 10-5
Buffer Solutions
Buffer Solution: A solution which contains a weak acid and its conjugate
base and resists drastic changes in pH
Weak acid
For
example:
+
Conjugate base
© 2012 Pearson Education, Inc.
CH3CO2H + CH3CO2–
HF + F–
NH4+ + NH3
H2PO4– + HPO42–
Chapter
15/18
Buffer Solutions
CH3CO2H(aq) + H2O(l)
Weak acid
Addition of OH1– to a buffer:
CH3CO2H(aq) +
OH–(aq)
H3O+(aq) + CH3CO2–(aq)
Conjugate base
(NaCH3CO2)
100%
H2O(l) + CH3CO2–(aq)
Addition of H3O1+ to a buffer:
CH3CO2
–(aq)
+ H3
O+(aq)
100%
H2O(l) + CH3CO2H(aq)
Buffer Solutions
Add a small amount of base (-OH) to a buffer solution
Acid component of solution neutralizes the added base
Add a small amount of acid (H3O+) to a buffer solution
Base component of solution neutralizes the added acid
The addition of –OH or H3O+ to a buffer solution will
change the pH of the solution, but not as drastically as the
addition of –OH or H3O+ to a non-buffered solution
Buffer Solutions
Buffer capacity
A measure of the amount of acid or base that a buffer
solution can absorb without a significant change in pH
Depends on how much weak acid and conjugated base is
present
For equal volume of solution, the more concentration the
solution, the greater the buffer capacity
For solution with the same concentration, increasing the
volume increases the buffer capacity
Example
pH of human blood (pH = 7.4) controlled by conjugated
acid-base pairs (H2CO3/HCO3-). Write a neutralization for
each condition
With addition of H3O+
With addition of -OH
Example
Calculate the pH of the buffer that results from mixing
60.0mL of 0.250M HCHO2 and 15.0 mL of 0.500M
NaCHO2
Ka = 1.8 x 10-4
Example
Calculate the pH of 0.100L of a buffer solution that is
0.25M in HF and 0.50 M in NaF,
Ka = 6.3 x 10-4
What is the change in pH on addition of 0.002 mol HCl
What is the change in pH on addition of 0.010 moles KOH
Calculate the pH after addition of 0.080 moles HBr
Example
calculate the pH of a 50.0 ml buffer solution that is 0.50 M
in NH3 and 0.20 M NH4Cl. For ammonia, pKb = 4.75
Calculate the pH after addition of 150.0 mg HCl
The Henderson-Hasselbalch
Equation
CH3CO2H(aq) + H2O(l)
Weak acid
Conjugate base
Acid(aq) + H2O(l)
Ka =
H3O1+(aq) + CH3CO21-(aq)
[H3O1+][Base]
[Acid]
H3O1+(aq) + Base(aq)
[Acid]
[H3O1+] = Ka
[Base]
The Henderson-Hasselbalch
Equation
[Acid]
[H3O1+] = Ka
[Base]
[Acid]
-log([H3O1+]) = -log(K[Base]
a) - log
[Base]
pH = pKa + log
[Acid]
Example
Calculate the pH of a buffer solution that is 0.50 M in
benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate
(NaC7H5O2). Ka = 6.5 x 10-5
How would you prepare a NaHCO3-Na2CO3 buffer solution
that has pH = 10.40 Ka2 = 5.6 x 10-11