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
Slide 2/17
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Acids & Bases
Lecture 3:
•
•
•
Salts of Acids and Bases
Buffer systems
Blackman Chapter 11,
Sections 11.3-11.6
Lecture 4:
•
•
Titrations
Blackman Chapter 11, Section 11.7
Reproduced from ‘The
Extraordinary Chemistry
of Ordinary Things, C.H.
Snyder, Wiley, 2002
(Page 245)
Slide 3/17
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Strong Acid/Strong Base Titration
• Strong acids completed dissociates in solution:
HA + H2O  A-(aq) + H3O+(aq)
pH = -log10([strong acid]initial)
• If strong base is added, it reacts with the H3O+ so [H3O+(aq)] is reduced:
H3O+(aq) + OH-(aq)  H2O(l)
pH = -log10([strong acid]remaining)
• Equivalence point: when the amount of added base = initial amount of
acid:
[H+(aq)] = 10-7.0 M
pH = 7.00
• After the equivalence point, any added base increases [OH-(aq)]:
 pH = 14.00 - pOH = 14.00 - (-log10([excess base]))
Slide 4/17
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Strong Base/Strong Acid Titration
• Initial pH is
 pH = 14 - pOH
= 14 -(-log10([strong base]initial)
• At the equivalence point,
 [H+(aq)] = 10-7.0 M
 pH = 7.00 (at 25 °C)
• After the equivalence point,
 pH = -log10([excess acid])
Slide 5/17
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Strong Base/Strong Acid Titration
Add HCl(g) in 0.10 mol amounts to 1.0 L of 0.5 M NaOH(aq)
16
14
12
10
pH
8
6
4
2
0
0
0.1
0.2
0.3
0.4
0.5
amount of added acid
0.6
0.7
0.8
Slide 6/17
x
Weak Acid/Strong Base Titration
Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.5 M acetic acid
Initial pH: initially the solution contains just acetic acid and its pH must be
calculated using the procedure outlined in slide 7 of lecture 2. pKa = 4.76.
initial (I)
CH3COOH
H 2O
H 3 O+
CH3COO–
0.5
large
0
0
change (C)
negligible
equilibrium (E)
large
Ka =
Slide 7/17
x
Weak Acid/Strong Base Titration
Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acid
Acidic region: in the acidic region, OH- has reacted with some of the acetic
acid to make acetate. The solution then contains both acetic acid and its
conjugate base and the Henderson-Hasselbalch equation can be used to
calculate the pH.
CH3COOH(aq) + OH-(aq)  CH3COO-(aq) + H2O(aq)
 [ base ] 
pH = pKa  log10 

[
acid
]


cf lecture 3
Slide 8/17
x
Weak Acid/Strong Base Titration
Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acid
Equivalence point: all of the acetic acid has reacted with OH- to form
CH3COO-. At this point, the solution contains a weak base so is basic. The
pOH must be calculated.
initial (I)
CH3COO-
H 2O
CH3COOH
OH-
0.5
large
0
0
change (C)
negligible
equilibrium (E)
large
pKb  14.00  pKa  9.24
Kb =
cf lecture 2
Slide 9/17
x
Weak Acid/Strong Base Titration
Add NaOH(s) in 0.10 mol amounts to 1.0 L of 0.50 M acetic acid
Alkaline region: after the equivalence point, the major source of OH- is that
from the excess NaOH. Very little is contributed from the acetate ion.
After the equivalence point, the calculations are then exactly the same as
for the strong acid/strong base titration.
When 0.60 mol of NaOH(s) is added, 0.50 mol react with the CH3COOH
leaving a 0.10 M solution of OH-:
Slide 10/17
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Weak Acid/Strong Base Titration
• Initial pH is higher than for strong acid / strong base
 A weak acid is present initially
• In acidic region, weak acid and conjugate base present
 Buffering region with slow change in pH
 At the 1/2 equivalence point, pH = pKa
• At equivalence point, conjugate base present
 pH > 7
• In alkaline region, excess strong base present
 Curve is the same as for strong acid/strong base titration
Slide 11/17
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Weak Acid/Strong Base Titrations
Slide 12/17
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Weak Base/Strong Acid Titration
Slide 13/17
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Titrations
• Equivalence Point:
 When number of moles of added base = original number
of moles of acid
 Strong acid/strong base pH = 7
 Weak acid/strong base pH > 7
 Strong acid/weak base pH < 7
• End Point:
 When a colour change in the indicator is observed
 Choose an indicator that changes colour close to the
equivalence point
Slide 14/17
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Indicators
weak acid
base
– each form has a different colour
• The pH at which acid  base depends on the pKa
of the indicator
pH 3.2
pH 4.4
methyl orange
pH 4.8 pH 5.4
pH 6.0
pH 7.6
methyl purple
bromothymol blue
pH 8.2
pH 10.0
phenolphthalein
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Practice Examples
1.Which one of the following combinations does the
titration curve to the right represent?
A.Addition of a strong base to a weak acid
B.Addition of a weak base to a strong acid
C.Addition of a weak acid to a strong base
D.Addition of a strong acid to a strong base
E.Addition of a strong acid to a weak base
12
10
8
pH
2. What is the value of the pKa that can be obtained from
this titration curve?
A.11.3
B.10.0
C.9.3
D.5.3
E.1.8
6
4
2
10
20
30
40
50
Amount of solution added (mL)
Slide 16/17
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Summary: Acids & Bases 4
Learning Outcomes - you should now be able to:
•
•
•
•
•
Complete the worksheet
Understand strong acid/strong base, strong
base/strong acid, weak acid/strong base and weak
base/strong acid titrations
Be able to extract pKa from the half equivalence
point
Be able to work at the pH at the equivalence point
Answer Review Problems 11.38-11.44 in Blackman
Next lecture:
•
Periodic Trends
Slide 17/17