Applications of
Aqueous Equilibria
Buffered Solutions
• A solution that resists a change in
pH when either hydroxide ions or
protons are added.
• Buffered solutions contain
either:
– A weak acid and its salt
– A weak base and its salt
Acid/Salt Buffering Pairs
The salt will contain the anion of the acid,
and the cation of a strong base (NaOH, KOH)
Weak Acid
Formula
of the acid
Hydrofluoric
HF
Formic
HCOOH
Benzoic
C6H5COOH
Acetic
Carbonic
Propanoic
Hydrocyanic
CH3COOH
H2CO3
HC3H5O2
HCN
Example of a salt of the
weak acid
KF – Potassium fluoride
KHCOO – Potassium formate
NaC6H5COO – Sodium benzoate
NaH3COO – Sodium acetate
NaHCO3 - Sodium bicarbonate
NaC3H5O2 - Sodium propanoate
KCN - potassium cyanide
Base/Salt Buffering Pairs
The salt will contain the cation of the base,
and the anion of a strong acid (HCl, HNO3)
Formula of
the base
Example of a salt of the weak
acid
NH3
NH4Cl - ammonium chloride
Methylamine
CH3NH2
CH3NH2Cl – methylammonium chloride
Ethylamine
C2H5NH2
C2H5NH3NO3 - ethylammonium nitrate
Aniline
C6H5NH2
C6H5NH3Cl – aniline hydrochloride
Base
Ammonia
Pyridine
C5H5N
C5H5NHCl – pyridine hydrochloride
Titration of an Unbuffered Solution
13
12
11
10
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH
9
pH
8
7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Titration of a Buffered Solution
13
12
11
A solution that is
0.10 M CH3COOH and
0.10 M NaCH3COO is
titrated with
0.10 M NaOH
10
9
pH
8
7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Comparing Results
Buffered
13
13
12
12
11
11
10
10
9
9
8
8
7
pH
pH
Unbuffered
6
6
5
5
4
4
3
3
2
1
0.00
7
2
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
1
0.00
5.00
10.00
15.00
20.00
25.00
30.00
35.00
milliliters NaOH (0.10 M)
In what ways are the graphs different?
In what ways are the graphs similar?
40.00
45.00
Comparing Results
pH
Gra ph
Buffered
Unbuffered
mL 0.10 M NaOH
Buffer capacity
• The best buffers have a ratio
[A-]/[HA] = 1
• This is most resistant to change
• True when [A-] = [HA]
• Make pH = pKa (since log1=0)
General equation
• Ka = [H+] [A-]
[HA]
• so [H+] = Ka [HA]
[A-]
• The [H+] depends on the ratio [HA]/[A-]
• taking the negative log of both sides
• pH = -log(Ka [HA]/[A-])
• pH = -log(Ka)-log([HA]/[A-])
• pH = pKa + log([A-]/[HA])
This is called the
Henderson-Hasselbalch Equation
pH
 [A  ] 
 pKa  log 
  pKa

[
HA
]


 [base ] 

 log 

[
acid
]


 [ BH  ] 
 [acid ] 
pOH  pKb  log 
  pKb  log 

 [base] 
 [ B] 
Using the
Henderson-Hasselbalch Equation
• pH = pKa + log([A-]/[HA])
• pH = pKa + log(base/acid)
• Calculate the pH of the following
mixtures
• 0.75 M lactic acid (HC3H5O3) and 0.25
M sodium lactate (Ka = 1.4 x 10-4)
• 0.25 M NH3 and 0.40 M NH4Cl
• (Kb = 1.8 x 10-5)
Prove they’re buffers
• What would the pH be if .020 mol of
HCl is added to 1.0 L of both of the
following solutions.
– 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
– 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)
• What would the pH be if 0.050 mol of
solid NaOH is added to each of the
proceeding.
Buffer capacity
• The pH of a buffered solution is
determined by the ratio [A-]/[HA].
• As long as this doesn’t change much the
pH won’t change much.
• The more concentrated these two are
the more H+ and OH- the solution will be
able to absorb.
• Larger concentrations bigger buffer
capacity.
Buffer Capacity
• Calculate the change in pH that occurs
when 0.010 mol of HCl(g) is added to
1.0L of each of the following:
• 5.00 M HAc and 5.00 M NaAc
• 0.050 M HAc and 0.050 M NaAc
• Ka= 1.8x10-5
Weak Acid/Strong Base Titration
13
12
11
10
9
Endpoint is above
pH 7
pH
8
7
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Strong Acid/Strong Base Titration
13
12
11
10
9
pH
8
7
A solution that is
0.10 M HCl is
titrated with
0.10 M NaOH
Endpoint is at
pH 7
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters NaOH (0.10 M)
30.00
35.00
40.00
45.00
Strong Base/Strong Acid Titration
13
12
A solution that is
0.10 M NaOH is
titrated with
0.10 M HCl
11
10
9
pH
8
7
Endpoint is at
pH 7
It is important to
recognize that
titration curves are
not always
increasing from left
to right.
6
5
4
3
2
1
0.00
5.00
10.00
15.00
20.00
25.00
milliliters HCl (0.10 M)
30.00
35.00
40.00
45.00
Weak Base/Strong Acid Titration
Summary
• Strong acid and base just stoichiometry.
• Determine Ka, use for 0 mL base
• Weak acid before equivalence point
–Stoichiometry first
–Then Henderson-Hasselbach
• Weak acid at equivalence point Kb
• Weak base after equivalence - leftover
strong base.
Summary
• Determine Ka, use for 0 mL acid.
• Weak base before equivalence point.
–Stoichiometry first
–Then Henderson-Hasselbach
• Weak base at equivalence point Ka.
• Weak base after equivalence - leftover
strong acid.
Selection of Indicators