Drill: A 0.100 M
solution of HZ
ionizes 20.0 %.
Calculate: KaHZ
Buffer
Solutions
Buffer Solution
•A solution that
resists changes in
pH
Buffer Solution
•Made from the
combination of a
weak acid & its
salt
Buffer Solution
•Made from the
combination of a
weak base & its
salt
Buffer Examples
•Mix acetic acid &
sodium acetate
•Mix ammonia &
ammonium chloride
Buffer Solution
•A buffer solution
works best when the
acid to salt ratio is
1:1
Buffer Solution
•A buffer solution
works best when the
base to salt ratio is
1:1
Buffer Solution
•The buffering capacity
of a solution works best
when the pH is near the
pKa
pKa or pKb
•pKa = - log Ka
•pKb = - log Kb
Buffer
Equilibria
To solve buffer
equilibrium
problems, use
the same 5 steps
5 Steps of
Equilibrium Problems
1) Set up & balance
reaction
5 Steps of
Equilibrium Problems
2) Assign Equilibrium
amounts in terms of x
(ICE)
5 Steps of
Equilibrium Problems
3) Write the equilibrium
expression (K = ?)
5 Steps of
Equilibrium Problems
4) Substitute Equilibrium
amounts into the K
5 Steps of
Equilibrium Problems
5) Solve for x
Buffer Problems
•Calculate the pH of a
solution containing
0.10 M HAc in 0.10 M
-5
NaAc: Ka = 1.8 x 10
Buffer Problems
•Calculate the pH of
0.10 M NH3 in
0.20 M NH4NO3:
-5
•Kb = 1.8 x 10
Buffer Problems
Calculate the pH of a
solution containing
0.10 M HBz in 0.20 M
-5
NaBz: Ka = 6.4 x 10
Drill:
Calculate the pH of a
solution containing
0.30 M HZ in 0.10 M
-5
NaZ: Ka = 3.0 x 10
Buffer Problem
Calculate the pH of a
solution containing
0.50 M R-NH2 in 0.10 M
-5
R-NH3I: Kb = 4.0 x 10
Derivations
from an
equilibrium
constant
HA
+
H
+
A
[ ][ ]
Ka =
[HA]
+
H
A
HA
+
H
+A
[ ][ ]
Ka =
[HA]
+
H
A
Cross multiply to
+
isolate [H ]
HA
+
H
+A
[
K
][
HA
]
a
+
[H ]=
[A ]
HA
+
H
+A
[
HA
]
+
[H ] = (Ka)
[A ]
HA
+
H
+A
[HA]
+
[H ] = (Ka)
[A ]
Take –log of each side
pH =
[HA]
pKa - log
[A ]
HendersonHasselbach Eq
[A ]
pH = pKa + log [HA]
HendersonHasselbach Eq
+
[B ]
pOH = pKb+ log [B]
Buffer Problems
•Calculate the salt to acid
ratio to make a buffer
solution with pH = 5.0
-5
•Ka for HBZ = 2.0 x 10
Derivations
from an
equilibrium
constant
HA
+
H
+A
[ ][ ]
Ka =
[HA]
+
H
A
[ ][ ]
Ka =
[HA]
+
H
A
Divide both sides by
+
[H ]
Ka
[ ]
=
+
[H ]
[HA]
A
You Get the Salt to
Acid Ratio
Drill:
•Calculate the salt to acid
ratio to make a buffer
solution with pH = 5.0
-5
•Ka for HBZ = 2.0 x 10
Buffer Problems
Calculate the salt to base
ratio to make a buffer
solution with pH = 9.48
-5
•Kb for MOH = 2.0 x 10
Equivalence Point
Point at which the #
of moles of the two
titrants are equal
Titration
Curves
14
12
10
8
6
4
2
0
0.00
10.00
20.00
30.00
40.00
50.00
14
12
10
8
[HA]=[OH-]
6
4
[HA]=[A-]
2
0
0.00
10.00
20.00
30.00
40.00
50.00
Drill:
Calculate the pH of a
buffer solution
containing 0.50 M HX
in 0.25 M KX.
-5
Ka = 2.5 x 10
14
12
[OH-] = [A-2]
10
8 [HA-] = [A-2]
6
[H2A] = [OH-]
4
2
[H2A] = [HA-]
0
0
20
40
60
80
100
-
Calculate the HCO3 to
H2CO3 ratio in blood
with pH = 7.40
-7
•Ka1 for H2CO3 = 4.4 x 10
150 ml of 0.10 M NaOH
is added to 100.0 ml of
0.10 M H2CO3.
Calculate pH.
-7
•Ka1 for H2CO3 = 4.4 x 10
-11
•Ka2 for H2CO3 = 4.8 x 10