RELATIVE CONCENTRATIONS OF BUFFER
COMPONENTS
CH3COOH(aq) + H2O(l)
[CH3COO−][H3O+]
Ka =
[CH3COOH]

CH3COO-(aq) + H3O+(aq)
[H3
O+]
[CH3COOH]
= Ka x
[CH3COO−]
Since Ka is constant, the [H3O+] of the solution depends on
the ratio of buffer component concentrations.
[HA]
+] increases.
If the ratio
increases,
[H
O
3
[A−]
[HA]
+] decreases.
If the ratio
decreases,
[H
O
3
[A−]
1
BUFFER CAPACITY AND RANGE

The buffer capacity is a measure of the
“strength” of the buffer, its ability to maintain
the pH following addition of strong acid or base.

The buffer range is the pH range over which the
buffer works. This is equal to pKa±1.
2
LEARNING CHECK: USING MOLECULAR
SCENES TO EXAMINE BUFFERS

The molecular scenes below represent samples of four
HA/A− buffers. (HA is blue and green, A− is green,
and other ions and water are not shown.)
(a) Which buffer has the highest pH?
(b) Which buffer has the greatest capacity?
(c) Should we add a small amount of concentrated
strong acid or strong base to convert sample 1 to
sample 2 (assuming no volume changes)?
3
PREPARING A BUFFER
1.
Choose the conjugate acid-base pair.

2.
Working pH = pKa ± 1
Calculate the ratio of buffer component
concentrations.
[base]
pH = pKa + log
3.
4.
[acid]
Determine the buffer concentration, and
calculate the required volume of stock solutions
and/or masses of components.
Mix the solution and correct the pH.
4
PRACTICE EXERCISE

You wish to prepare a 1.0 L buffer with a pH of
3.90. A list of possible acids and their conjugate
bases is shown below. Which combination should
be selected? Suppose the total concentration of
the buffer ([HA] + [A–]) is 0.5 M. What is the
concentration of each component?
Acid
Conjugate
Base
Ka
pKa
HSO4–
SO42–
1.2 x 10–2
1.92
CH3COOH
CH3COO–
1.8 x 10–5
4.74
HCO3–
CO32–
4.8 x 10–11
10.32
HF
F–
6.6 x 10–4
3.18
[CH3COOH]=0.437 M
[CH3COO–]=0.063 M
5
PRACTICE EXERCISE

An environmental chemist needs a carbonate
buffer of pH 10.00 to study the effects of the acid
rain on limestone-rich soils. How many grams of
Na2CO3 must she add to 1.5 L of freshly prepared
0.20 M NaHCO3 to make the buffer? Ka of HCO3−
is 4.7x10−11.
15 g Na2CO3
6
CHAPTER OUTLINE
1.
2.
3.
4.
Equilibria of Acid-Base Buffers
Acid-Base Titration Curves
Equilibria of Slightly Soluble Ionic Compounds
Equilibria Involving Complex Ions
Skipped topics: Titration of Weak Acid-Strong Base
and Titration of Weak Base-Strong Acid
Ionic Equilibria and Acid-Rain Problem
Equilibria Involving Complex Ions
8
TITRATION

A solution of accurately known
concentration is added gradually
added to another solution of
unknown concentration until the
chemical reaction between the two
solutions is complete.
Equivalence point – the
point at which the reaction is
complete.
Indicator – substance that
changes color at (or near) the
equivalence point.
9
ACID-BASE INDICATORS
Phenolphthalein
Bromothymol blue
The color of an indicator changes over a specific, narrow pH range, a
range of about 2 pH units.
ACID-BASE INDICATORS
11
ACID-BASE INDICATORS
HIn (aq)
Color of acid
[HIn]
(HIn)
 10
[In ]
predominates
H+ (aq) + In- (aq)
[HIn]  10
[In-]
Color of conjugate
base (In-)
predominates
12
STRONG ACID-STRONG BASE TITRATION
NaOH (aq) + HCl (aq)
OH- (aq) + H+ (aq)
H2O (l) + NaCl (aq)
H2O (l)
COMMON STRONG ACIDS AND STRONG
BASES
1 mol H+
per mol
acid
1 mol
OH– per
mol base
2 mol H+
per mol
acid
2 mol
OH– per
mol base
14
PRACTICE EXERCISE

40.00 mL of 0.1000 M HCl is titrated with
0.1000 M NaOH. Calculate the following:
1.00

Initial pH
1.48

+20.00 mL of NaOH solution
7.00

+40.00 mL of NaOH solution
12.05

+50.00 mL of NaOH solution
15
STRONG ACID-STRONG BASE TITRATION
Initial pH
[H3O+] = [HA]init
pH = -log[H3O+]
pH before equivalence point
initial mol H3O+ = Vacid x Macid
mol OH- added = Vbase x Mbase
mol H3O+remaining = (mol H3O+init) – (mol OH−added)
[H3
O+]
=
mol H3O+remaining
Vacid + Vbase
pH = -log[H3O+]
16
STRONG ACID-STRONG BASE TITRATION
pH at the equivalence point
pH = 7.00 for a strong acid-strong base titration.
pH beyond the equivalence point
initial mol H3O+ = Vacid x Macid
mol OH− added = Vbase x Mbase
mol OH−excess = (mol OH−added) – (mol H3O+init)
[OH−]
=
mol OH−excess
Vacid + Vbase
pOH = -log[OH−] and pH = 14.00 - pOH
17
WEAK ACID-STRONG BASE TITRATION
CH3COOH (aq) + OH- (aq)
At equivalence
CH3COO- (aq) + H2O (l)
point (pH > 7):
CH3COO- (aq) + H2O (l)
OH- (aq) + CH3COOH (aq)
STRONG ACID-WEAK BASE TITRATION
H+ (aq) + NH3 (aq)
At equivalence
NH4+ (aq) + H2O (l)
point (pH < 7):
NH4Cl (aq)
NH3 (aq) + H+ (aq)