Transcript Chapter 15 Acids and Bases

Relationship between pH and pOH
 the sum of the pH and pOH of a solution = 14.00
 at 25°C
 can use pOH to find pH of a solution
-14
[H 3O ][OH ] = K w = 1.0 ´ 10
+
-14
- log [H O ][OH ] = - log 1.0 ´ 10
+
))
(
)
( (
( - log ([H O ])) + ( - log ([OH ])) = 14.00
3
+
3
pH + pOH = 14.00
1
pK
 a way of expressing the strength of an acid or base is pK
pKa = -log(Ka), Ka = 10-pKa
pKb = -log(Kb), Kb = 10-pKb
 the stronger the acid, the smaller the pKa
 larger Ka = smaller pKa
 because it is the –log
2
Finding the pH of a Strong Acid
 there are two sources of H3O+ in an aqueous solution of a
strong acid – the acid and the water
 for the strong acid, the contribution of the water to the total
[H3O+] is negligible
 for a monoprotic strong acid [H3O+] = [HA]
 0.10 M HCl has [H3O+] = 0.10 M and pH = 1.00
3
Finding the pH of a Weak Acid
 there are also two sources of H3O+ in and aqueous
solution of a weak acid – the acid and the water
 however, finding the [H3O+] is complicated by the fact that
the acid only undergoes partial ionization
 calculating the [H3O+] requires solving an ICE problem for
the reaction that defines the acidity of the acid (the top
equation)
 For a weak acid Ka << 1 so we may be able to apply
approximations to avoid solving a quadratic
4
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Write the reaction for
the acid with water
Construct an ICE table
for the reaction
HNO2 + H2O
initial
initial
NO2- + H3O+
[HNO22]
[NO22--]
[H33O++]
0.200
0
≈0
change
change
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
equilibrium
equilibrium
since no products initially, Qc = 0, and the reaction is proceeding forward
5
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
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
[HNO2]
[NO2-]
[H3O+]
0.200
0
0
+x
+x
x
x
-x
0.200 -x
x )( x )
[NO-2 ][H 3O+ ]
(
Ka =
=
[ HNO2 ] ( 0.200 - x )
6
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
determine the value of
Ka from Table 15.5
initial
change
since Ka is very small,
approximate the
[HNO2]eq = [HNO2]init and
solve for x
equilibrium
[HNO2]
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
0.200-x
0.200
x
x
éë NO--22 ùû éë H 33O++ ùû ( x()x()x()x )
K aa =
=
( 0.200 )- x )
[ HNO22 ]
4.6 ´10
-4
=
x2
2.00 ´10 -1
x=
( 4.6 ´10 )( 2.00 ´10 )
-4
x = 9.6 ´10 -3
-1
7
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
check if the
approximation is valid
by seeing if x < 5% of
[HNO2]init
initial
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
0.200
x
x
change
equilibrium
x = 9.6 x 10-3
[HNO2]
9.6 ´10
-3
2.00 ´10
-1
´100% = 4.8% < 5%
the approximation is valid
8
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
substitute x into the
equilibrium
concentration
definitions and solve
initial
change
equilibrium
[HNO2]
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
x
x
0.200-x
0.190 0.0096
0.0096
x = 9.6 x 10-3
-3
HNO
=
0.200
x
=
0.200
9.6
´10
[ 2]
(
) = 0.190 M
éë NO-2 ùû = éë H 3O+ ùû = x = 9.6 ´10 -3 M
9
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
substitute [H3O+] into
the formula for pH and
solve
initial
change
equilibrium
[HNO2]
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
0.190
0.0096 0.0096
pH = -log ( H 3O
+
= - log ( 9.6 ´10
)
-3
) = 2.02
10
Find the pH of 0.200 M HNO2(aq) solution @ 25°C
Ka for HNO2 = 4.6 x 10-4
check by substituting the
equilibrium concentrations
back into the equilibrium
constant expression and
comparing the calculated
Ka to the given Ka
initial
change
equilibrium
though not exact,
the answer is
reasonably close
[HNO2]
[NO2-]
[H3O+]
0.200
0
≈0
-x
+x
+x
0.190
0.0096 0.0096
éë NO-2 ùû éë H 3O + ùû
Ka =
=
[ HNO2 ]
9.6 ´ 10 )
(
=
-3 2
( 0.190 )
= 4.9 ´ 10 -4
11
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
Tro, Chemistry: A Molecular
Approach
12
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
Write the reaction for
the acid with water
Construct an ICE table
for the reaction
HC6H4NO2 + H2O
initial
C6H4NO2- + H3O+
[HA]
[A-]
[H3O+]
0.012
0
≈0
change
equilibrium
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
13
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
HC6H4NO2 + 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
C6H4NO2- + H3O+
[HA]
[A-]
[H3O+]
0.012
0
0
+x
x
+x
-x
0.012 -x
x
x )( x )
[C6 H 4 NO-2 ][H 3O+ ]
(
Ka =
=
[ HC6H 4 NO2 ] (1.2 ´10 -2 - x )
14
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
C6H4NO2- + H3O+
HC6H4NO2 + H2O
determine the value of
Ka
since Ka is very small,
approximate the [HA]eq =
[HA]init and solve for x
initial
change
equilibrium
éë A- ùû éë H 3O+ ùû ( x()x()x()x )
Ka =
=
( 0.012 )- x )
[ HA]
x
1.4 ´10 =
0.012
-5
2
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012-x
0.012
x
x
x=
(1.4 ´10 )( 0.012 )
-5
x = 4.1´10 -4
15
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
Ka for HC6H4NO2 = 1.4 x 10-5
check if the
approximation is
valid by seeing if x
< 5% of
[HC6H4NO2]init
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012
x
x
x = 4.1 x 10-4
4.1´10 -4
1.2 ´10
´
100
%
=
3
.
4
%
<
5
%
-2
the approximation is valid
16
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
substitute x into the
equilibrium
concentration
definitions and solve
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012-x
x
x
x = 4.1 x 10-4
-4
HC
H
NO
=
0.012
x
=
0.012
4.1´10
[ 6 4 2]
(
) = 0.012 M
+
éëC6 H 4 NO ùû = éë H 3O ùû = x = 4.1´10
2
-4
M
17
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
substitute [H3O+] into
the formula for pH and
solve
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012 0.00041 0.00041
pH = -log ( H 3O
+
= - log ( 4.1´10
)
-4
) = 3.39
18
What is the pH of a 0.012 M solution of nicotinic
acid, HC6H4NO2 given Ka = 1.4 x 10-5 @ 25°C?
check by substituting the
equilibrium concentrations
back into the equilibrium
constant expression and
comparing the calculated
Ka to the given Ka
the values match
initial
change
equilibrium
[HA]
[A2-]
[H3O+]
0.012
0
≈0
-x
+x
+x
0.012 0.00041 0.00041
2
+
[C6 H 4 NO ][H 3O ]
Ka =
[ HC6 H 4 NO2 ]
4.1 ´ 10 )
(
=
(1.2 ´ 10 )
-4 2
-2
= 1.4 ´ 10 -5
19
Find the pH of 0.100 M HClO2(aq) solution @ 25°C
given Ka for HClO2 = 1.1 x 10-2
Write the reaction for
the acid with water
Construct an ICE table
for the reaction
Enter the initial
concentrations –
assuming the [H3O+]
from water is ≈ 0
HClO2 + H2O
initial
ClO2- + H3O+
[HClO2]
[ClO2-]
[H3O+]
0.100
0
≈0
change
equilibrium
20
Find the pH of 0.100 M HClO2(aq) solution @ 25°C
given Ka for HClO2 = 1.1 x 10-2
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
[HClO2]
[ClO2-]
[H3O+]
0.100
0
≈0
-x
+x
+x
0.100-x
x
x
x )( x )
[ClO-2 ][H 3O+ ]
(
Ka =
=
[ HClO2 ] (1.00 ´10-1 - x )
21
Find the pH of 0.100 M HClO2(aq) solution @ 25°C
given Ka for HClO2 = 1.1 x 10-2
Wirte an expression for Ka
and use the ICE table to
obtain an equation for x
since Ka is very small,
approximate the [HClO2]eq
= [HClO2]init and solve for x
initial
[ClO2-]
[H3O+]
0.100
0
≈0
-x
+x
+x
0.100-x
x
x
change
equilibrium
éëClO-2 ùû éë H 3O+ ùû
x )( x )
(
Ka =
=
-1
1.00
´10
[ HClO2 ]
(
)
x=
x
1.1´10 =
1.00 ´10 -1
-2
[HClO2]
2
Tro, Chemistry: A Molecular
Approach
(1.1´10 )(1.00 ´10 )
-2
-1
x = 3.3 ´10 -2
22
Find the pH of 0.100 M HClO2(aq) solution @ 25°C
given Ka for HClO2 = 1.1 x 10-2
check if the
approximation is valid
by seeing if x < 5% of
[HNO2]init
initial
change
equilibrium
[HClO2]
[ClO2-]
[H3O+]
0.100
0
≈0
-x
+x
+x
0.100-x
x
x
x = 3.3 x 10-2
3.3 ´10 -2
1.00 ´10
´
100
%
=
33
%
>
5
%
-1
the approximation is invalid
23
Find the pH of 0.100 M HClO2(aq) solution @ 25°C
given Ka for HClO2 = 1.1 x 10-2
if the approximation is
invalid, solve for x
using the quadratic
formula
éëClO-2 ùû éë H 3O+ ùû
x )( x )
(
Ka =
=
[ HClO2 ]
(1.00 ´10-1 - x )
2
x
1.1´10-2 =
-1
1.00
´10
- x)
(
0 = x 2 + 0.011x - 0.0011
x=
-0.011±
(0.011)
2
- 4(1)(-0.0011)
2(1)
x = 0.028 or - 0.039
24
Find the pH of 0.100 M HClO2(aq) solution @ 25°C
given Ka for HClO2 = 1.1 x 10-2
substitute x into the
equilibrium
concentration
definitions and solve
initial
change
equilibrium
[HClO2]
[ClO2-]
[H3O+]
0.100
0
≈0
-x
+x
+x
0.100-x
0.072
0.028
x
0.028
x
x = 0.028
[ HClO ] = 0.100 - x = 0.100 - ( 0.028 ) = 0.072 M
2
éëClO-2 ùû = éë H 3O+ ùû = x = 0.028 M
25
Find the pH of 0.100 M HClO2(aq) solution @ 25°C
given Ka for HClO2 = 1.1 x 10-2
substitute [H3O+] into
the formula for pH and
solve
initial
change
equilibrium
[HClO2]
[ClO2-]
[H3O+]
0.100
0
≈0
-x
+x
+x
0.072
0.028
0.028
pH = -log(H 3O
+
)
= -log(0.028) = 1.55
26
Find the pH of 0.100 M HClO2(aq) solution @ 25°C
given Ka for HClO2 = 1.1 x 10-2
check by substituting the
equilibrium concentrations
back into the equilibrium
constant expression and
comparing the calculated
Ka to the given Ka
initial
change
equilibrium
[HClO2]
[ClO2-]
[H3O+]
0.100
0
≈0
-x
+x
+x
0.072
0.028
0.028
Ka =
the answer matches
+
ClO
H
O
[ 2 ][ 3 ]
=
[HClO 2 ]
2
0.028)
(
=
= 1.1 ´10-2
(0.072)
27
What is the Ka of a weak acid if a 0.100 M solution
has a pH of 4.25?
+
[H3O ] = 10
Use the pH to find the
equilibrium [H3O+]
Write the reaction for the
acid with water
Construct an ICE table
for the reaction
Enter the initial
concentrations and
[H3O+]equil
-pH
= 10
-4.25
-5
= 5.6 ´10 M
A- + H3O+
HA + H2O
initial
[HA]
[A-]
[H3O+]
0.100
0
≈0
change
equilibrium
5.6E-05
28
What is the Ka of a weak acid if a 0.100 M solution
has a pH of 4.25?
fill in the rest of the
table using the [H3O+]
as a guide
initial
if the difference is
insignificant, [HA]equil =
[HA]initial
substitute into the Ka
expression and
compute Ka
A- + H3O+
HA + H2O
change
[HA]
[A-]
[H3O+]
0.100
0
0
−5.6E-05 +5.6E-05 +5.6E-05
0.100 
0.100
5.6E-05
equilibrium
-
+
[A ][H 3O ]
Ka =
=
[HA]
K a = 3.1´10-8
5.6E-05 5.6E-05
-5
-5
5.6
´10
5.6
´10
(
)(
)
(0.100)
29
Percent Ionization
 another way to measure the strength of an acid is to
determine the percentage of acid molecules that ionize when
dissolved in water – this is called the percent ionization
 the higher the percent ionization, the stronger the acid
molarity of ionized acid
Percent Ionization =
´100%
initial molarity of acid
•
since [ionized acid]equil = [H3O+]equil
Percent Ionization =
[H 3O + ]equil
[HA]init
´100%
30
What is the percent ionization of a 2.5 M HNO2
solution if Ka for HNO2 = 4.6 x 10-4 ?
Write the reaction for the
acid with water
Construct an ICE table
for the reaction
Enter the Initial
Concentrations
HNO2 + H2O
NO2- + H3O+
[HNO2]
[NO2-]
[H3O+]
initial
2.5
0
≈0
change
-x
+x
x
+x
x
equilibrium
2.5 - x
Define the Change in
Concentration in terms
of x
Sum the columns to
define the Equilibrium
Concentrations
31
What is the percent ionization of a 2.5 M HNO2
solution if Ka for HNO2 = 4.6 x 10-4 ?
determine the value of Ka
from Table 15.5
initial
since Ka is very small,
approximate the [HNO2]eq
= [HNO2]init and solve for x
change
equilibrium
éë NO-2 ùû éë H 3O+ ùû ( x ) ( x )
Ka =
=
( 2.5 )
[ HNO2 ]
[HNO2]
[NO2-]
[H3O+]
2.5
0
≈0
-x
+x
+x
2.5-x ≈2.5
x
x
x=
-4
4.6
´10
(
)(2.5)
x = 3.4 ´10-2
x2
4.6 ´10 =
2.5
-4
32
What is the percent ionization of a 2.5 M HNO2
solution if Ka for HNO2 = 4.6 x 10-4 ?
NO2- + H3O+
HNO2 + H2O
substitute x into the
Equilibrium
Concentration
definitions and solve
initial
change
equilibrium
[HNO2]
[NO2-]
[H3O+]
2.5
0
≈0
-x
+x
+x
2.52.5
-x
0.034
x
0.034
x
x = 3.4 x 10-2
[HNO2 ]= 2.5 - x = 2.5 - (0.034)= 2.5 M
[NO ] = [H O ] = x = 0.034 M
2
+
3
33
What is the percent ionization of a 2.5 M HNO2
solution if Ka for HNO2 = 4.6 x 10-4 ?
HNO2 + H2O
Apply the Definition
and Compute the
Percent Ionization
initial
change
equilibrium
since the percent
ionization is < 5%, the
“x is small”
approximation is valid
NO2- + H3O+
[HNO2]
[NO2-]
[H3O+]
2.5
0
≈0
-x
+x
+x
2.5
0.034
0.034
Percent Ionization =
[H 3O + ]equil
[HNO 2 ]init
´100%
3.4 ´10 -2
=
´100% = 1.4%
2.5
34
Relationship Between [H3O+]equilibrium and [HA]initial
 What happens to [H3O+] at equilibrium if we dilute [HA]?
 decreasing/increasing the initial concentration of acid
results in increased/decreased percent ionization
 this means that the increase in H3O+ concentration is
slower than the increase in acid concentration
35
Finding the pH of Mixtures of Acids
 generally, you can ignore the contribution of the weaker
acid to the [H3O+]eq
 for a mixture of a strong acid with a weak acid, the
complete ionization of the strong acid provides more than
enough [H3O+] to shift the weak acid equilibrium to the left
 so far that the weak acid’s added [H3O+] is negligible
 for mixtures of weak acids, generally only need to consider
the stronger for the same reasons
 as long as one is significantly stronger than the other,
and their concentrations are similar
36
Find the pH of a mixture of 0.150 M HF(aq) solution
and 0.100 M HClO2(aq)
HF + H2O
Write the reactions for the
acids with water and
determine their Kas
F- + H3O+
Ka = 3.5 x 10-4
HClO + H2O
ClO- + H3O+ Ka = 2.9 x 10-8
H2O + H2O
OH- + H3O+ Kw = 1.0 x 10-14
If the Kas are sufficiently
different, use the strongest
acid to construct an ICE
table for the reaction
initial
Enter the initial
concentrations – assuming
the [H3O+] from water is ≈ 0
equilibrium
[HF]
[F-]
[H3O+]
0.150
0
≈0
change
37
Find the pH of a mixture of 0.150 M HF(aq) solution
and 0.100 M HClO2(aq)
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
[HF]
[F-]
[H3O+]
0.150
0
0
+x
x
+x
x
-x
equilibrium
0.150 -x
x )( x )
[F - ][H 3O+ ]
(
Ka =
=
-1
1.50
´10
- x)
[ HF ]
(
38
Find the pH of a mixture of 0.150 M HF(aq) solution
and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
determine the value of
Ka for HF
since Ka is very small,
approximate the [HF]eq =
[HF]init and solve for x
initial
change
equilibrium
éë F-- ùû éë H33O++ ùû ( x()x()x()x )
Kaa =
=
(0.150 )- x )
[ HF ]
3.5 ´10 -4 =
x2
[HF]
[F-]
[H3O+]
0.150
0
≈0
-x
+x
+x
0.150-x
0.150
x
x
x=
-4
-1
3.5
´10
1.50
´10
(
)(
)
x = 7.2 ´10-3
1.50 ´10 -1
39
Find the pH of a mixture of 0.150 M HF(aq) solution
and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
check if the
approximation is valid
by seeing if x < 5% of
[HF]init
initial
[HF]
[F-]
[H3O+]
0.150
0
≈0
-x
+x
+x
0.150
x
x
change
equilibrium
x = 7.2 x 10-3
7.2 ´10 -3
1.50 ´10
-1
´100% = 4.8% < 5%
the approximation is valid
40
Find the pH of a mixture of 0.150 M HF(aq) solution
and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
substitute x into the
equilibrium
concentration
definitions and solve
initial
[HF]
[F-]
[H3O+]
0.150
0.150
0
≈0
-x
+x
+x
change
equilibrium
x
x
0.150-x
0.143 0.0072
0.0072
x = 7.2 x 10-3
-3
HF
=
0.150
x
=
0.150
7.2
´10
[ ]
(
) = 0.143 M
[F ] = [H O ] = x = 7.2 ´10
+
-
-3
3
41
M
Find the pH of a mixture of 0.150 M HF(aq) solution
and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
substitute [H3O+]
into the formula for
pH and solve
initial
change
equilibrium
[HF]
[F-]
[H3O+]
0.150
0
≈0
-x
+x
+x
0.143
0.0072
0.0072
pH = -log(H 3O + )
= -log( 7.2 ´10-3 ) = 2.14
42
Find the pH of a mixture of 0.150 M HF(aq) solution
and 0.100 M HClO2(aq)
Ka for HF = 3.5 x 10-4
check by substituting the
equilibrium concentrations
back into the equilibrium
constant expression and
comparing the calculated
Ka to the given Ka
initial
change
equilibrium
[HF]
[F-]
[H3O+]
0.150
0
≈0
-x
+x
+x
0.143
0.0072
0.0072
Ka =
+
F
H
O
[ ][ 3 ]
[HF]
7.2 ´10 )
(
=
=
-3 2
though not exact, the answer is reasonably
close
(0.143)
= 3.6 ´10-4
43
Strong Arrhenius Bases
 the stronger the base, the more willing it is to
accept H+
NaOH  Na+ + OH-
 use water as the standard acid
 for strong bases, practically all molecules are
dissociated into OH– or accept H’s
 multi-OH strong bases completely
dissociated
 [HO–] = [strong base] x (# OH)
44
Calculate the pH at 25°C of a 0.0015 M Sr(OH)2 solution
and determine if the solution is acidic, basic, or neutral
Given:
Find:
Concept Plan:
Relationships:
[Sr(OH)2] = 1.5 x 10-3 M
pH
[OH-]
[Sr(OH)2]
[OH-]=2[Sr(OH)2]
[H3O+]
K w = [H 3O + ][OH- ]
pH
pH = - log[H 3O + ]
Solution:
K w = [H 3O ][OH - ]
+
[OH-]
= 2(0.0015)
= 0.0030 M
Check:
-14
1.0 ´10
[H 3O ] =
3.0 ´10-3
+
[H 3O + ] = 3.3 ´10-12 M
pH = - log( 3.3 ´10-12 )
pH = 11.48
pH is unitless. The fact that the pH > 7 means the
solution is basic
Practice: Calculate the pH of a 0.0010 M Ba(OH)2
solution and determine if it is acidic, basic, or neutral
46
Practice: Calculate the pH of a 0.0010 M Ba(OH)2
solution and determine if it is acidic, basic, or neutral
Ba(OH)2 = Ba2+ + 2 OH- therefore
[OH-] = 2 x 0.0010M = 0.0020M = 2.0 x 10-3 M
Kw = [H3O+][OH-]
-14
1.00
x
10
[H3O+] =
= 5.0 x 10-12M
-3
2.0 x 10
pH = -log [H3O+] = -log (5.0 x 10-12)
pH = 11.30
pH > 7 therefore basic
47