Chapter 18
Acid-base Equilibrium II :
Buffers and Indicators
18.1 Buffers
18.2 Calculations Involving Composition
and pH of Buffer Solutions
18.3 Acid-base Indicators
18.4 Acid-base Titrations
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New Way Chemistry for Hong Kong A-Level Book 2
18.1 Buffers (SB p.151)
Buffers
A buffer solution is a solution that tends to resist change in
pH when a small amount of acid or base is added to it.
2 types of buffers:
1. Acidic buffer
--- mixing a weak acid and its salt of a strong base.
(e.g. CH3COOH & CH3COONa)
2. Basic buffer
--- mixing a weak base and its salt of a strong acid.
(e.g. NH3 & NH4Cl)
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18.1 Buffers (SB p.151)
Action of Acidic Buffers
CH3COOH(aq) \-----==\ CH3COO-(aq) + H+(aq)
CH3COO-Na+(aq) --------> CH3COO-(aq) + Na+(aq)
Note that the buffer contains a large amount of the weak
acid (CH3COOH) and its conjugate base (CH3COO-).
If a small amount of acid is added to this system,
H+(aq) + CH3COO-(aq) \==------\ CH3COOH(aq)
If a small amount of alkali is added to this system,
OH-(aq) + CH3COOH(aq) \==------\ CH3COO-(aq) + H2O(l)
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New Way Chemistry for Hong Kong A-Level Book 2
18.1 Buffers (SB p.152)
If a small amount of acid is added to this system,
H+(aq) + CH3COO-(aq) \==------\ CH3COOH(aq)
If a small amount of alkali is added to this system,
OH-(aq) + CH3COOH(aq) \==------\ CH3COO-(aq) + H2O(l)
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New Way Chemistry for Hong Kong A-Level Book 2
18.1 Buffers (SB p.153)
The pH of an Acidic Buffer Solution
CH3COOH(aq) \-----==\ CH3COO-(aq) + H+(aq)
CH3COO-Na+(aq) --------> CH3COO-(aq) + Na+(aq)
pH = pKa + log
pH = pKa +
5
[salt]
log [acid]
New Way Chemistry for Hong Kong A-Level Book 2
initial conc’s
18.1 Buffers (SB p.154)
Action of Basic Buffers
NH3(aq) + H2O \-----==\ NH4+(aq) + OH-(aq)
NH4+Cl-(aq) --------> NH4+(aq) + Cl-(aq)
Note that the buffer contains a large amount of the weak
base (NH3) and its conjugate acid (NH4+).
If a small amount of acid is added to this system,
H+(aq) + NH3(aq) \==------\ NH4+(aq)
If a small amount of alkali is added to this system,
OH-(aq) + NH4+(aq) \==------\ NH3(aq) + H2O(l)
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New Way Chemistry for Hong Kong A-Level Book 2
18.1 Buffers (SB p.154)
If a small amount of acid is added to this system,
H+(aq) + NH3(aq) \==------\ NH4+(aq)
If a small amount of alkali is added to this system,
OH-(aq) + NH4+(aq) \==------\ NH3(aq) + H2O(l)
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New Way Chemistry for Hong Kong A-Level Book 2
18.1 Buffers (SB p.155)
The pH of an Basic Buffer Solution
NH3(aq) + H2O \-----==\ NH4+(aq) + OH-(aq)
NH4+Cl-(aq) --------> NH4+(aq) + Cl-(aq)
pOH = pKb + log
pOH = pKb +
[salt]
log [base]
pH = 14 – pKb - log
8
[salt]
[base]
New Way Chemistry for Hong Kong A-Level Book 2
initial conc’s
18.2 Calculations Involving Composition and pH of Buffer Solutions (SB p.156)
Calculations Involving Composition and pH
of Buffer Solution
Example 18-1
A buffer is made by adding 4.1g of sodium ethanoate to 1dm3
of a 0.01 M solution of ethanoic acid. Calculate the pH of the
buffer.
(Given: Ka of CH3COOH at 298 K = 1.74 x 10-5 mol dm-3;
molar mass of CH3COONa = 82 g mol-1; assume there is no
volume change on mixing)
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New Way Chemistry for Hong Kong A-Level Book 2
18.2 Calculations Involving Composition and pH of Buffer Solutions (SB p.156)
Solution
Number of moles of CH3COONa = 4.1 g/ 82 g mol-1 = 0.05 mol
[CH3COO-(aq)] = 0.05 mol/1 dm3 = 0.05M
[CH3COOH(aq)] = 0.01M
pH = pKa
[CH 3 COO  (aq)]
+ log [CH COOH( aq)]
3
pH = -log(1.74x10-5) + log 0.05/0.01
= 4.76 + 0.70
Answer
= 5.46
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New Way Chemistry for Hong Kong A-Level Book 2
18.2 Calculations Involving Composition and pH of Buffer Solutions (SB p.159)
Example 18-3
How many grams of ammonium chloride would you add to 100 cm3
of 0.1 M NH3(aq) in order to prepare a basic buffer of pH 9.0?
(Given: Kb of NH3 at 298K = 1.74 x 10-5 mol dm-3; molar mass of
NH4Cl = 53.5 g mol-1)
Solution
Let x M be the concentration of ammonium chloride in
the buffer solution.
pH = 14 – pKb – log[salt]/[base]
9 = 14 – [-log(1.74 x 10-5)] – log(x / 0.1)
9 = 9.24 – log(x/0.1)
x = 0.174
∴ [NH4Cl(aq)] = 0.174 M (continued)
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Answer
18.2 Calculations Involving Composition and pH of Buffer Solutions (SB p.159)
Number of moles of NH4Cl used
= 0.174 M x 100/1000 dm3
= 0.0174 mol
Mass of NH4Cl used
= 0.0174 mol x 53.5 g mol-1
= 0.931 g
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18.3 Acid-base Indicators (SB p.160)
Acid-base Indicators
HIn(aq) + H2O(l)
H3O+(aq) + In-(aq)
weak acid
conjugate base
(colour 1)
(colour 2)
KIn =
pH = pKIn + log
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New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.161)
HIn(aq) + H2O(l)
H3O+(aq) + In-(aq)
weak acid
conjugate base
(colour 1)
(colour 2)
pH = pKIn + log
14
When

When
 10 , colour 2 is observed
, colour 1 is observed
New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.161)
Thus the indicator changes from colour 1 to colour 2
over a range of:
from

to
 10
which corresponds to pH =(pKIn-1) to pH = =(pKIn+1)
Generally speaking, the colour change takes place
over a range of 2 pH units (pH range of the indicator).
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18.3 Acid-base Indicators (SB p.162)
Phenolphthalein
HPh(aq) + H2O(l)
colourless
Ph-(aq) + H3O+(aq)
pink
KIn = 7 x 10-10 mol dm-3
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New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.162)
Phenolphthalein
≦ 1/10
When
pH ≦ pKIn + log 1/10
≦ pKIn - 1
≦ 8.15
When [Ph-(aq)] = [HPh(aq)],
pH = pKIn
= -log (7 x 10-10)
= 9.15

When [Ph (aq)] ≧ 10
[HPh(aq)]
pH ≧ pKIn + log 10
≧ pKIn + 1
≧ 10.15
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18.3 Acid-base Indicators (SB p.162)
Methyl Orange
HMe+(aq) + H2O(l)
red
Me(aq) + H3O+(aq)
yellow
KIn = 2 x 10-4 mol dm-3
pH = pKIn +
18
[Me(aq)]
log [HMe (aq)]
New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.162)
When
[Me(aq)]
[HMe (aq)]
1/10
pH ≦ pKIn + log 1/10
≦ pKIn - 1
≦2.7
When [Me(aq)] = [HMe+(aq)],
pH = pKIn
= -log (2 x 10-4)
= 3.7
When
[Me(aq)]
[HMe (aq)]
≧ 10
pH ≧ pKIn + log 10
≧ pKIn + 1
≧ 4.7
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18.3 Acid-base Indicators (SB p.163)
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18.4 Acid-base Titrations (SB p.164)
Acid-base Titrations
base
acid +
indicator
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Remarks
1. Titration is the determination of the
equivalence point, the point at which
equivalent quantities of the acid and
base have reacted.
2. There are often sharp changes in pH
near the equivalence points of the
titrations.
3. The point at which an indicator
change colour during titration is
called the end point.
4. A good indicator is one whose end pt.
matches with the equivalence pt.
New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.165)
pH Titration Curves
Strong Acid-Strong Base Titration
Sharp change in pH at the
equivalence point: 3-11
Both methyl orange &
phenolphthalein can indicate
the equivalence point
accurately.
(Their pH ranges lie within
the sharp change in pH.)
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New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.165)
Strong Acid-Weak Base Titration
Sharp change in pH at the
equivalence point: 2-6
Methyl orange can indicate
the equivalence point
accurately.
(Its pH range lies within the
sharp change in pH.)
Phenolphthalein cannot
indicate the equivalence
point accurately.
(Its pH range does NOT lie
within the sharp change in
pH.)
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New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.165)
Weak Acid-Strong Base Titration
Sharp change in pH at the
equivalence point: 8-12
Phenolphthalein can indicate
the equivalence point
accurately.
(Its pH range lies within the
sharp change in pH.)
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Methyl orange cannot
indicate the equivalence
point accurately.
(Its pH range does NOT lie
within the sharp change in
pH.)
New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.165)
Weak Acid-Weak Base Titration
NO Sharp change in pH at
the equivalence point.
Both methyl orange &
phenolphthalein CANNOT
indicate the equivalence
point accurately.
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New Way Chemistry for Hong Kong A-Level Book 2
18.3 Acid-base Indicators (SB p.167)
Double Indicator Method
Na2CO3 + HCl  ……..
Na2CO3 + HCl 
NaHCO3 + NaCl
NaHCO3 + HCl 
NaCl + CO2 + H2O
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The END
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New Way Chemistry for Hong Kong A-Level Book 2