Transcript Chapter

Chapter 16
Aqueous Ionic
Equilibrium
Buffers
 buffers are solutions that resist changes in pH when
an acid or base is added
 they act by neutralizing the added acid or base
 but just like everything else, there is a limit to what
they can do, eventually the pH changes
 many buffers are made by mixing a solution of a
weak acid with a solution of soluble salt containing its
conjugate base anion
2
Demo of a pH Buffer
Aqueous Solution
pH
Δ pH
Bottled Water (30 mL)
5.6
0.7
Bottled Water + 1 drop of 0.1M HNO3
4.9
Deionized Water (30 mL)
6.0
Deionized Water + 1 drop of 0.1M HNO3
3.8
pH Buffer
7.8
pH Buffer + 1 drop of 0.1M HNO3
7.8
2.2
0.0
What should the pH be when we add 1 drop of 0.1M HNO3 to 30mL
of deionized water ?
Assume 1 drop = 0.05mL
When this 1 drop is dissolved in 30 mL of H2O
moles H + = Molarity ´ Volume
moles
1L
= 0.1
´ 0.05mL ´
L
1000 mL
moles H + 5 ´ 10 -6 moles
[H ] =
=
= 1.66 ´ 10 -4 M
-3
V
30.05 ´ 10 L
+
pH = - log[H ] = - log(1.66 ´ 10 -4 M ) = 3.8
= 5 ´ 10 -6 moles H + in 1 drop
+
Why are pH buffers important?
Life on Earth is water-based
 Human 48-75% water
 Plants as high as 95% water
 H+ and OH- are chemically and structurally reactive in
cells
The functioning of the cell is very pH dependent
 Blood plasma pH = 7.4
 pH < 6.9 fatal (acidosis)
 pH > 7.9 fatal (alkalosis)
 Fish die if the pH of the water goes below 5 or above
9
Making an Acid Buffer
5
How Acid Buffers Work
HA(aq) + H2O(l)
A−(aq) + H3O+(aq)
 buffers work by applying Le Châtelier’s Principle to weak acid
equilibrium
 buffer solutions contain significant amounts of the weak acid
molecules, HA – these molecules react with added base to neutralize
it
HA(aq) + OH - (aq) ® A- (aq) + H 2O(l)
 the buffer solutions also contain significant amounts of the conjugate
base anion, A− - these ions combine with added acid to make more
HA and keep the H3O+ constant
6
How Buffers Work
H2O
new
HA
A−−
HA
HA
+
H3O+
Added
H3O+
7
How Buffers Work
H2O
new
A−
A−−
HA
HA
+
H3O+
Added
HO−
8
Common Ion Effect
HA(aq) + H2O(l)
A−(aq) + H3O+(aq)
 adding a salt containing the anion, NaA, that is the
conjugate base of the acid (the common ion) shifts
the position of equilibrium to the left
 this causes the pH to be higher than the pH of the
acid solution
 lowering the H3O+ ion concentration
9
Common Ion Effect
10
What is the pH of a buffer that is 0.100 M HC2H3O2 and
0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
HC2H3O2 + H2O
Write the reaction for
the acid with water
Construct an ICE table
for the reaction
initial
C2H3O2- + H3O+
[HA]
[A-]
[H3O+]
0.100
0.100
≈0
change
equilibrium
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
11
What is the pH of a buffer that is 0.100 M HC2H3O2 and
0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
HC2H3O2 + H2O
represent the change
in the concentrations in
terms of x
sum the columns to
find the equilibrium
concentrations in terms
of x
substitute into the
equilibrium constant
expression
initial
change
equilibrium
C2H3O2- + H3O+
[HA]
[A-]
[H3O+]
0.100
0.100
0
-x
+x
+x
0.100 -x 0.100 + x
x
[C2H 3O -2 ][H 3O + ] (0.100 + x )( x )
Ka =
=
[HC2H 3O 2 ]
(0.100 - x )
12
What is the pH of a buffer that is 0.100 M HC2H3O2 and
0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
determine the value of Ka
since Ka is very small,
approximate the
[HA]eq = [HA]init and [A−]eq
= [A−]init solve for x
initial
change
equilibrium
[HA]
[A-]
[H3O+]
0.100
0.100
≈0
-x
+x
+x
0.100
0.100
0.100
-x 0.100
+x
x
[C2H 3O -2 ][H 3O + ] (0.100 + x )( x )
Ka =
=
[HC2H 3O 2 ]
(0.100 - x )
[C2H 3O -2 ][H 3O + ] (0.100)( x )
Ka =
=
[HC2H 3O 2 ]
(0.100)
1.8 ´10-5 = x
13
What is the pH of a buffer that is 0.100 M HC2H3O2 and
0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
check if the
approximation is
valid by seeing if x
< 5% of
[HC2H3O2]init
initial
change
equilibrium
[HA]
[A-]
[H3O+]
0.100
0.100
≈0
-x
+x
+x
0.100
0.100
x
x = 1.8 x 10-5
-5
1.8 ´10
´100% = 0.018% < 5%
-1
1.00 ´10
the approximation is valid
14
What is the pH of a buffer that is 0.100 M HC2H3O2 and
0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
substitute x into the
equilibrium
concentration
definitions and solve
initial
[HA]
[A-]
[H3O+]
0.100
0.100
≈0
-x
+x
+x
change
equilibrium
0.100-x 0.100
0.100
0.100
+ x 1.8E-5
x
x = 1.8 x 10-5
-5
HC
H
O
=
0.100
x
=
0.100
1.8
´10
[ 2 3 2]
(
) = 0.100 M
[C2H 3O 2 ] = 0.100 + x = 0.100 + (1.8 ´10-5 ) = 0.100 M
-
[H 3O + ] = x =1.8 ´10-5 M
Tro, Chemistry: A
15
What is the pH of a buffer that is 0.100 M HC2H3O2 and
0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
substitute [H3O+] into
the formula for pH and
solve
initial
change
equilibrium
[HA]
[A-]
[H3O+]
0.100
0.100
≈0
-x
+x
+x
0.100
0.100
1.8E-5
pH = -log(H 3O + )
= -log(1.8 ´10-5 ) = 4.74
16
What is the pH of a buffer that is 0.100 M HC2H3O2 and
0.100 M NaC2H3O2 given Ka for HC2H3O2 = 1.8 x 10-5
check by substituting the
equilibrium concentrations
back into the equilibrium
constant expression and
comparing the calculated
Ka to the given Ka
initial
change
equilibrium
[HA]
[A-]
[H3O+]
0.100
0.100
≈0
-x
+x
+x
0.100
0.100
1.8E-5
[C2H 3O -2 ][H 3O + ]
Ka =
[HC 2H 3O 2 ]
the values match
(0.100)(1.8 ´10-5 )
=
= 1.8 ´10-5
(0.100)
17
What is the pH of a buffer that is 0.14 M
HF (pKa = 3.15) and 0.071 M KF?
Tro, Chemistry: A
18
Practice - What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
HF + H2O
Write the reaction for
the acid with water
Construct an ICE table
for the reaction
initial
F- + H3O+
[HA]
[A-]
[H3O+]
0.14
0.071
≈0
change
equilibrium
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
19
What is the pH of a buffer that is 0.14 M
HF (pKa = 3.15) and 0.071 M KF?
HF + H2O
represent the change
in the concentrations in
terms of x
initial
change
sum the columns to
find the equilibrium
concentrations in terms
of x
substitute into the
equilibrium constant
expression
equilibrium
F- + H3O+
[HA]
[A-]
[H3O+]
0.14
0.071
0
-x
+x
+x
0.14 -x
0.071 + x
x
[F - ][H 3O + ] (0.071+ x )( x )
Ka =
=
[HF]
(0.14 - x )
20
What is the pH of a buffer that is 0.14 M
HF (pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10-4
determine the value of Ka
since Ka is very small,
approximate the
[HA]eq = [HA]init and [A−]eq
= [A−]init solve for x
initial
change
equilibrium
K a = 10- pK a = 10-3.15
K a = 7.0 ´10-4
[HA]
[A-]
[H3O+]
0.14
0.071
≈0
-x
+x
+x
0.012
0.14
x
0.100
0.071
+x
x
[F - ][H 3O + ] (0.071+ x )( x )
Ka =
=
[HF]
(0.14 - x )
K a = 7.0 ´10-4 =
(0.071)( x )
(0.14 )
1.4 ´10-3 = x
21
What is the pH of a buffer that is 0.14 M
HF (pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x
10-4
check if the
approximation is valid by
seeing if x < 5% of
[HC2H3O2]init
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.14
0.071
≈0
-x
+x
+x
0.14
0.071
x
x = 1.4 x 10-3
-3
1.4 ´10
´100% = 1% < 5%
-1
1.4 ´10
the approximation is valid
22
Practice - What is the pH of a buffer that is
0.14 M HF (pKa = 3.15) and 0.071 M KF?
substitute x into the
equilibrium
concentration
definitions and solve
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.14
0.071
≈0
-x
+x
+x
0.14x 0.071
0.072
+ x 1.4E-3
x
0.14
x = 1.4 x 10-3
-3
HF
=
0.14
x
=
0.14
1.4
´10
[ ]
(
) = 0.14 M
[C2H 3O 2 ] = 0.071+ x = 0.071+ (1.4 ´10-3 ) = 0.072 M
-
[H 3O + ] = x =1.4 ´10-3 M
Tro, Chemistry: A
23
What is the pH of a buffer that is 0.14 M
HF (pKa = 3.15) and 0.071 M KF?
substitute [H3O+]
into the formula for
pH and solve
initial
change
equilibrium
[HA]
[A-]
[H3O+]
0.14
0.071
≈0
-x
+x
+x
0.14
0.072
1.4E-3
pH = -log(H 3O + )
= -log(1.4 ´10-3 ) = 2.85
24
What is the pH of a buffer that is 0.14 M
HF (pKa = 3.15) and 0.071 M KF?
Ka for HF = 7.0 x 10-4
check by substituting the
equilibrium concentrations
back into the equilibrium
constant expression and
comparing the calculated
Ka to the given Ka
the values are
close enough
initial
change
equilibrium
[HA]
[A-]
[H3O+]
0.14
0.071
≈0
-x
+x
+x
0.14
0.072
1.4E-3
[F - ][H 3O + ]
Ka =
[HF]
(0.072)(1.4 ´10-3 )
=
= 7.2 ´10-4
(0.14 )
25
Henderson-Hasselbalch Equation
 calculating the pH of a buffer solution can be simplified by
using an equation derived from the Ka expression called
the Henderson-Hasselbalch Equation
 the equation calculates the pH of a buffer from the Ka and
initial concentrations of the weak acid and salt of the
conjugate base
 as long as the “x is small” approximation is valid
[A- ]initial
pH = pK a + log10
[HA]initial
26
Deriving the Henderson-Hasselbalch
Equation
- ö
æ [A
æ
æ]æ æ [HA]
æ
öö ööö
æ [HA]
öö ææ[HA]
[HA]
+
+
pH O
= pH
pK
+pK
log
ç+KKaçaç-a++log
-log[H
- log[H
]]===a=-log
-log
log
-÷--çlog
log
ç
ç
÷÷ çç- ÷÷- - ÷÷÷
33 OpH
aç K
[HA]
è
ø
][A
øø] ]øøø
è èè [A
ø èè[A
èè ]è ø[A
pH = - log[H3O+ ] pK = [HA]
- log K a [A- ]
a
-log
-
[A ]
= log
[HA]
27
What is the pH of a buffer that is 0.050 M
HC7H5O2 and 0.150 M NaC7H5O2 where
Ka=6.5x10-5 for HC7H5O2?
28
What is the pH of a buffer that is 0.050 M
HC7H5O2 and 0.150 M NaC7H5O2 where
Ka=6.5x10-5 for HC7H5O2?
Assume the [HA] and [A-]
equilibrium
concentrations are the
same as the initial
Substitute into the
Henderson-Hasselbalch
Equation
Check the “x is small”
approximation
HC7H5O2 + H2O
C7H5O2- + H3O+
pK a = -log(K a )
= -log(6.5 ´10-5 ) = 4.187
- ö
ö
æ æ[A
0.150
(
)
]
pH = pK
4.187
+ log
÷ ÷
a + logç ç
0.050
è
è [HA]
ø ø
pH = 4.66
+ -5
-pH
[H
O
]
=
10
2.2 3´10
´100%
= 0.044%-5< 5%
+
-4.66
[H0.050
= 2.2 ´10
3O ] = 10
29
What is the pH of a buffer that is 0.050 M
HC7H5O2 and 0.150 M NaC7H5O2 where
Ka=6.5x10-5 for HC7H5O2?
HC7H5O2 + H2O
-
+
[A ][H 3O ]
Ka =
[HA]
0.150 - x ) x
(
Ka =
( 0.050 - x )
0.150 ) x
(
Ka ∼
= 3x
( 0.050 )
Ka
= x = 2.17 ´10 -5
3
initial
change
equilibrium
C7H5O2- + H3O+
[HA]
[A-]
[H3O+]
0.050
0.150
≈0
-x
+x
+x
0.050-x
0.150-x
x
pH = - log10 (2.17 ´ 10 -5 ) = 4.66
According to Henderson-Hasselbalch
pH=4.66
Do I Use the Full Equilibrium Analysis or the
Henderson-Hasselbalch Equation?

It is not clear that the Henderson-Hasselbalch Equation is an
approximation, but the way we use it, (by placing the initial
concentration of [HA]o and [A-]o into the equation assuming they
are equilibrium values), makes it an approximation

For this reason the Henderson-Hasselbalch equation is generally
good enough when the “x is small” approximation is applicable

generally, the “x is small” approximation will work when both of the
following are true:
a) the initial concentrations of acid and salt are not very dilute
b) the Ka is fairly small

for most problems, this means that the initial acid and salt
concentrations should be over 1000x larger than the value of31Ka