Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, 2007 (John Wiley)
ISBN: 9 78047081 0866
CHEM1002 [Part 2]
A/Prof Adam Bridgeman (Series 1)
Dr Feike Dijkstra (Series 2)
Weeks 8 – 13
Office Hours:
Room:
e-mail:
e-mail:
Monday 2-3, Friday 1-2
543a
adam.bridgeman@sydney.edu.au
feike.dijkstra@sydney.edu.au
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Acids & Bases 3
Lecture 2:
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Strong/Weak Acids and Bases
Calculations
Polyprotic Acids
Lecture 3:
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Salts of Acids and Bases
Reproduced from ‘The
Extraordinary Chemistry
Buffer systems
of Ordinary Things, C.H.
Snyder, Wiley, 2002
Blackman Chapter 11, Sections 11.3-11.6 (Page 245)
Slide 3/18
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Salts of Weak Acids and Bases
• Is a solution of NaCN acidic or basic?
 NaCN is the salt of NaOH (strong base) and HCN (weak
acid). The base “wins”: pH > 7
 Overall reaction is
H2O(aq) + CN–(aq)
OH–(aq) + HCN(aq)
• Does a solution of NH4Cl have pH > 7 or < 7?
 Salt of NH4OH (weak base) and HCl (strong acid)
 acid “wins”: pH < 7 and reaction is
H2O(aq) + NH4+(aq)
NH3(aq) + H3O+(aq)
Slide 4/18
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The Common Ion Effect
• If you add the salt of an acid to a solution of the same
acid then the equilibrium will shift towards neutral.
CH3COOH(aq) + H2O(l)
CH3COO-(aq) + H3O+(aq)
• Addition of CH3COO-:
 By Le Chatelier’s principle the equilibrium will shift to the left
to remove CH3COO- and therefore decrease [H3O+].
• Addition of CH3COOH:
 By Le Chatelier’s principle the equilibrium will shift to the
right to remove CH3COOH and therefore increase [H3O+].
Slide 5/18
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Buffer System
Buffer after addition
of H3O+
CH3COO- CH3COOH
Buffer with equal
concentrations of
conjugate base and acid
H 3 O+
Buffer after addition
of OH-
OHCH3COO- CH3COOH
H2O + CH3COOH  H3O+ + CH3COO-
CH3COO- CH3COOH
CH3COOH + OH- H2O + CH3COO-
• A solution containing both a weak acid and its salt
withstands pH changes when acid or base (limited
amounts) are added.
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Buffer systems and pH Change
• Consider change in pH of pure water (pH = 7) if we add an equal
amount of 10–3 M HCl:
 [H+ ] = 1 x 10–3 M
 pH goes from 7 to 3!
Huge change!
What about buffers?
Slide 7/18
x
Buffer systems and pH Change
• Consider a buffer solution with 0.1 M each of sodium
acetate (NaAc) & acetic acid (HAc):
 What is the pH when 10–3 M HCl is added?
H+(aq) + Ac–(aq)
HAc(aq)
Initial:
Neutralization (I):
Change (C):
Equilibrium (E):
0.1 +10-3
0
+x
-x
0.1+10-3-x
Ka  10
cf slide 6
10-3
0.1
-4.7

x
Ka = 10-4.7
0.1
0.1 - 10-3
+x
0.1- 10-3+x

x 0.1 - 10-3  x
0.1  10-3 - x

Slide 8/18
x
Buffer systems and pH Change
Ka  10
•
-4.7


x 0.1 - 10-3  x
0.1  10
-3

-x


x 0.1 - 10-3

0.1  10-3
x = 1.02  Ka = 0.000020 << 0.001
 pH = – log x = 4.69
the pH hardly changes from 4.7!
Solution is buffered against pH change
Slide 9/18
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Henderson - Hasselbalch Equation
• For a buffer solution, which contains similar concentrations of
a conjugate acid/base pair of a weak acid:
[H  ][A - ] [H  ][initial base]
Ka 

[HA]
[initial acid]
• The dissociation of HA or protonation of A- does not lead to a
significant change in the concentrations of these species.
• Taking logs and rearranging gives:
[initial base]
pH  pK a  log
[initial acid]
Slide 10/18
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Buffer Preparation and Capacity
• Buffer Preparation
 If the pH of a required buffer is pKa of available acid then
use equimolar amounts of acid and conjugate base
 If the required pH differs from the pKa then use the
Henderson-Hasselbalch equation.
• Buffer Capacity
 Buffer capacity is related to the amount of strong acid or base
that can be added without causing significant pH change.
 Depends on amount of acid & conjugate base in solution:
 highest when [HA] and [A–] are large.
 highest when [HA]  [A–]
Slide 11/18
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Buffer Preparation and Capacity
Most effective buffers have acid/base ratio
less than 10 and more than 0.1  pH range is ±1
Slide 12/18
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Buffers in Natural Systems
• Biological systems, e.g. blood, contain buffers:
pH control essential because biochemical reactions are
very sensitive to pH.
• Human blood is slightly basic, pH  7.39 – 7.45.
• In a healthy person, blood pH is never more than 0.2 pH
units from its average value.
• pH < 7.2, “acidosis”; pH > 7.6, “alkalosis”.
• Death occurs if pH < 6.8 or > 7.8.
Slide 13/18
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Buffer System in Blood
• “Extracellular” buffer (outside cell)
H+(aq) + HCO3–(aq)
H2CO3(aq)
H2CO3(aq)
H2O(l) + CO2 (g)
 Removal of CO2 shifts equilibria to right, reducing
[H+], i.e., raising the pH
 The pH can be reduced by:
H2CO3(aq) + OH–(aq)
HCO3–(aq) + H2O(l)
Slide 14/18
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Example
• In the H3PO4 / NaH2PO4 / Na2HPO4 / Na3PO4 system, how
could you make up a buffer with a pH of 7.40?
DATA: Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, Ka3 = 4.2 x 10-13
• To make up a buffer, we need pH near pKa
pKa1 = 2.14, pKa2 = 7.20, pKa3 = 12.38
 must use mixture of H2PO4- and HPO4• Could go through whole procedure...
or simply use Henderson-Hasselbalch equation ...
Slide 15/18
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Example (Continued)
pH  pK a2
HPO24 - 

 log 
H2PO4- 


original
amounts
• Require buffer with a pH of 7.40
HPO24 - 

7.40  7.20  log 
H2PO4- 


HPO24 - 
  100.20  1.58
 
H2PO4- 


 the required ratio of Na2HPO4 to NaH2PO4 = 1.58:1
Slide 16/18
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1
Practice Examples
What is the pH of a 0.045 M solution of KOBr?
The pKa of HOBr is 8.63.
(a) 4.74
2.
b) 4.99
c) 8.25
d) 9.01
e) 10.64
A buffered solution is 0.0500 M CH3COOH and 0.0400 M
NaCH3CO2. If 0.0100 mol of gaseous HCl is added to 1.00 L of
the buffered solution, wahat is the final pH of the solution?
For acetic acid, pKa = 4.76
(a) 4.76
(b) 4.46
(c) 4.66
(d) 4.86
(e) 4.54
Slide 17/18
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Summary: Acids & Bases 3
Learning Outcomes - you should now be able to:
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Complete the worksheet
Understand acid and base equilibria
Identify conjugate acid/base pairs
Perform calculations with strong acids/bases
Use the buffer concept and construct buffers
Apply the Henderson-Hasselbalch equation
Answer Review Problems 11.36-11.37 and 11.96106 in Blackman
Next lecture:
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Titrations
Slide 18/18