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
e
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 and Bases
Lecture 1:
•
Common Acids and Bases
•
Definitions
•
Equilibria
•
Conjugate acid-base pairs
•
Autoionisation of water
•
pH
Lecture 2:
•
Strong Acids and Bases
•
Weak Acids and Bases
•
Polyprotic Acids
Reproduced from ‘The
Extraordinary Chemistry
of Ordinary Things, C.H.
Snyder, Wiley, 2002
(Page 245)
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Strong Acids and Bases
• Completely ionise in water:
e.g. HCl(aq)
H+(aq) + Cl–(aq)
• Equilibrium lies completely to the right, Ka  .
Strong acids
H2SO4, HCl, HBr, HI, HNO3, HClO4
Strong bases
All hydroxides of Groups 1 & 2 (except Be):
NaOH, Ca(OH)2, …
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x
Examples
• What is the pH of a 0.1 M HCl solution?
HCl(aq
H+(aq) + Cl-(aq)
[H+] = 0.1 M
 Thus [H+] = 0.1 M and pH = – log10 [H+] = 1.0
• What is the pH of a 0.002 M NaOH solution?
 Completely ionised, so [OH– ] = 0.002 M,
 pOH = – log10 [OH– ] = – log10 (0.002) = 2.7
 pH = 14 – 2.7 = 11.3
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Weak Acids
• Most acids or bases are weak
 they do not completely ionise in water
H+(aq) + A–(aq)
[H ][A  ]
Acid dissociation constant: K a 
[HA]
pKa = – log10Ka
HA(aq)
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Relationship between Ka and pKa
• The larger the value of Ka, the stronger the acid and the
lower the value of pKa.
Acid Name (Formula)
-
o
K a at 25 C
-2
pK a
Bisulfate ion (HSO4 )
1.02 x 10
Nitrous acid (HNO2)
7.1 x 10
3.15
Acetic Acid (CH3COOH)
1.8 x 10-5
4.74
Hypobromous acid (HBrO)
Phenol (C6H5OH)
2.3 x 10
-10
1.0 x 10
-4
-9
1.991
8.64
10.00
 Ka = 1.02 x 10-2 then pKa = -log10(1.02 x 10-2) = 1.991
 pKa = 1.991 then Ka = 10-1.991 = 1.02 x 10-2
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Example
• Find the pH of 0.1 M acetic acid (CH3COOH (HAc))
DATA:
pKa = 4.7,
HAc(aq)
Ka = 10-4.7
H+(aq) + Ac–(aq)
Initial (I):
0.1
0
0
Change (C):
-x
+x
+x
x
x
Equilibrium (E): 0.1 - x
Ka  104.7
Ka = 10-4.7
x2

0.10  x
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Example (Continued)
• Since the equilibrium constant is very small we assume
x << 0.1, i.e. ( 0.1 – x )  0.1
10-4.7  x2 / 0.1
x2  0.1  10-4.7 = 10-5.7
x   10-5.7 = 10-2.85
• As pH = – log10 [H+ ]
pH = – log10 x = – log10 10-2.85
= 2.9
• Check: x = 10-2.85 = 1.4  10-3 << 0.1 or
or (0.1 – x ) = 0.0986 M ~ 0.1 M
5 % rule
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Weak Bases
• Ionisation of a weak base:
NH3(aq) + H2O(l)
NH4+(aq) + OH–(aq)
• Equilibrium constant is called base ionisation constant, Kb :
Kb 
[NH4  ][OH ]
[NH3 ]
• Tactic: calculate pOH and then pH, given Kb.
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Relationship Between pKa and pKb
• For conjugate systems (Brønsted-Lowry acid-base pairs)
 Acid (HA): HA(aq)
H+(aq)
+
A–(aq)
 Conjugate base (A– ):
A–(aq) + H2O(l)
HA(aq) + OH–(aq)
[H ][A  ]
Ka 
[HA]
[HA][OH  ]
Kb 
[A  ]
[H ][A  ] [HA][OH  ]
Ka  Kb 

[HA]
[A  ]
 [H ] [OH  ]  Kw  10 14
pKa + pKb = 14
 We only need values of pKa , since pKb = 14 – pKa
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Example
• Find the pH of 1.0 x 10–2 M NaHCO2
(pKa of formic acid (HCO2H) is 4.1)
 pKb = 14 – pKa = 14 – 4.1 = 9.9
HCO2–(aq) + H2O(l)
OH–(aq) + HCO2H(aq)
Initial (I):
0.01
Change (C):
-x
+x
+x
Equilibrium (E):
0.01 - x
x
x
Kb 
[OH ][HCO2H]

[HCO2 ]
large
so 10
9.9
0

0
x2
10
2
 x

x2
102
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x
Example (Continued)
Kb 
x 
[OH ][HCO2H]

[HCO2 ]
so 10 9.9 
102  109.9 
x2
10
2
 x

x2
102
1011.9  105.95
• x = [OH–] so pOH = -log10(10-5.95) = 5.95
 pH = 14 – 5.95 = 8.05
= 8.1 (one significant figure)
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Polyprotic Acids
H3PO4(aq)
H+(aq) + H2PO4–(aq)
pKa1 = 2.2
H2PO4– (aq)
H+(aq) + HPO42–(aq)
pKa2 = 7.2
HPO42-(aq)
H+(aq) + PO43–(aq)
pKa3 = 12.4
• removing more protons is harder:
increasing pKa = decreasing Ka : Ka1 > Ka2 > Ka3
 reason: harder to remove +ve charge against
increasing -ve charge
• large difference in pKa values
 only need to consider one equilibrium at a time
(simplifies maths!)
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Practice Examples
1.
Which one of the following is NOT a conjugate acid-base pair?
(a)
HCO3– and CO32–
(b)
H3O+ and H2O
(c)
OH– and O2–
(d)
SO3 and HSO3–
(e)
NH2OH2+ and NH2OH
2.
What is the pH of a 0.20 M solution of boric acid? The pKa of boric acid is 9.24.
(a) 0.70 (b) 2.73 (c) 4.97 (d) 5.12 (e) 5.87
3.
Rank the following solutions (all 1.0 M) in DECREASING order of pH.
NaCl, NaCN, KOH, HCl, CH3COOH
(a)
(b)
(c)
(d)
(e)
HCl > CH3COOH > NaCN > NaCl > KOH
KOH > NaCl > NaCN > CH3COOH > HCl
KOH > NaCN > NaCl > HCl > CH3COOH
KOH > NaCl > NaCN > HCl > CH3COOH
KOH > NaCN > NaCl > CH3COOH > HCl
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Summary: Acids & Bases 2
Learning Outcomes - you should now be able to:
•
•
•
•
•
Complete the worksheet
Complete acid/base calculations
Use pH, pKw, pKa and pKb
Define strong and weak acids & bases
Answer Review Problems 11.12-11.35 in Blackman
Next lecture:
•
Buffer systems
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