Chapter 14
Acids and Bases
Section 14.1
The Nature of Acids and Bases
Models of Acids and Bases
●
Es ions in solution, bases
Arrhenius: Acids produce H+
produce OH- ions.
-
↳
●
B donors,
Brønsted–Lowry: Acids are proton (H+)
bases are proton acceptors.
M
HCl + H2O
Cl-w + H3O+
e
acid base
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Section 14.1
The Nature of Acids and Bases
Acid in Water
HA(aq)
(aq)
+
0
●
●
H2O(l)
t
H3O+(aq)
+
↓
A-
000
-
Conjugate base is everything that remains of the acid
molecule after a proton is lost.
Conjugate acid is formed when the proton is transferred to
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the
base.
Section 14.2
Acid Strength
⑧
Strong acid: Large k
Ionization equilibrium lies far to the right.
Yields a weak conjugate base.
Weak acid: Small k
Ionization equilibrium lies far to the left.
Weaker the acid, stronger its conjugate base.
8
⑧
⑧
O
O
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Section 14.2
Acid Strength
HA
-
H+ A
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+
Sis
I
ionized
HB
-
very
is
0
littleHB
ionized
Section 14.2
Acid Strength
Various Ways to Describe Acid Strength
-
Equilibrium Constant;Ka=
acid
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Section 14.2
Acid Strength
Water as an Acid and a Base
H20(1)
kw
H
-
(aq)
+
04
+
[H+][0n-]
=
Water is amphoteric:
Behaves either as an acid or as a base.
At 25°C:
I
⑰
Kw = [H+][OH–]
= 1.0 × 10–14
No matter what the solution contains, the product of
- and [OH–] must always equal 1.0 × 10–14
- at
[H+]
25°C.
Kw will
⑧
⑧
⑧
&
>
If the temperature
*
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-
changes,
14
change.
0
-
(aa)
Section 14.2
Acid Strength
Three Possible Situations
·
[H+]
⑧
d
= [OH–];
neutral solution
↓ > [OH–];
[H+]
acidic solution
- > [H+];
⑰ basic solution
[OH–]
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Section 14.2
Acid Strength
CONCEPT CHECK!
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
acid
base
conjugate conjugate
acid
base
What is the equilibrium constant expression for
an acid acting in water?
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Section 14.2
Acid Strength
CONCEPT CHECK!
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
acid
base
conjugate conjugate
acid
base
What is the equilibrium constant expression for
an acid acting in water?
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Section 14.2
Acid Strength
CONCEPT CHECK!
If the equilibrium lies to the right, the value for Ka
is __________.
large (or >1)
If the equilibrium lies to the left, the value for Ka is
___________.
small (or <1)
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Section 14.2
Acid Strength
CONCEPT CHECK!
HA(aq) + H2O(l)
H3O+(aq) + A–(aq)
If water is a better base than A–, do products
or reactants dominate at equilibrium?
Does this mean HA is a strong or weak acid?
Is the value for Ka greater or less than 1?
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Section 14.2
Acid Strength
CONCEPT CHECK!
Consider a 1.0 M solution of HCl.
Order the following from strongest to weakest base and
explain:
H2O(l)
A–(aq) (from weak acid HA)
Cl–(aq)
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Section 14.2
Acid Strength
Let’s Think About It…
How good is Cl–(aq) as a base?
Is A–(aq) a good base?
The bases from strongest to weakest are:
A–, H2O, Cl–
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)
Section 14.2
Acid Strength
CONCEPT CHECK!
Consider a solution of NaA where A– is the anion from
weak acid HA:
A–(aq) + H2O(l)
base
acid
acid
HA(aq) + OH–(aq)
conjugate
conjugate
base
Which way will equilibrium lie?
left
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Section 14.2
Acid Strength
CONCEPT CHECK!
Consider a solution of NaA where A– is the anion from
weak acid HA:
A–(aq) + H2O(l)
HA(aq) + OH–(aq)
base
acid
conjugate
conjugate
acid
base
b) Is the value for Kb greater than or less than 1?
less than 1
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Section 14.2
Acid Strength
CONCEPT CHECK!
Consider a solution of NaA where A– is the anion from
weak acid HA:
A–(aq) + H2O(l)
base
acid
acid
HA(aq) + OH–(aq)
conjugate
conjugate
base
c) Does this mean A– is a strong or weak base?
weak base
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Section 14.2
Acid Strength
CONCEPT CHECK!
Acetic acid (HC2H3O2) and HCN are both weak acids.
Acetic acid is a stronger acid than HCN.
Arrange these bases from weakest to strongest and
explain your answer:
H2O
A
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I
Cl–
=
CN–
C2H3O2–
B g
A
0
Section 14.2
Acid Strength
Let’s Think About It…
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
acid
base
conjugate conjugate
acid
base
At 25°C, Kw = 1.0 × 10–14
The bases from weakest to strongest are:
Cl–, H2O, C2H3O2–, CN–
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Section 14.2
Acid Strength
CONCEPT CHECK!
Discuss whether the value of K for the reaction:
HCN(aq) + F–(aq)
CN–(aq) + HF(aq)
is
>1
<1
=1
(Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.)
Explain your answer.
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Section 14.2
Acid Strength
CONCEPT CHECK!
Calculate the value for K for the reaction:
HCN(aq) + F–(aq)
CN–(aq) + HF(aq)
(Ka for HCN is 6.2×10–10; Ka for HF is 7.2×10–4.)
K = 8.6 × 10–7
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Section 14.3
The pH Scale
High
*
⑧
O
⑧
⑧
⑧
concentration
ofHT lower pH
=
pH = –log[H+]
I
pH changes by 1 for every power of 10 change in [H+].
A compact way to represent solution acidity.
# increases.
pH decreases as [H+]
Significant figures:
The number of decimal places in the log is equal to
the number of significant figures in the original
number.
O
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Section 14.3
The pH Scale
pH Range
pH = 7; neutral
pH > 7; basic
Higher the pH, more basic.
pH < 7; acidic
Lower the pH, more acidic.
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Section 14.3
The pH Scale
The pH Scale and pH
Values of Some Common
Substances
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Section 14.3
The pH Scale
EXERCISE!
Calculate the pH for each of the following solutions.
I
significant
figures
pH
f
t M H+
1.0 × 10–4
7
a)
pH = 4.00
W
pH
=
=
[H +]
log(10x10-4)
-log
-
pH 4.00
=
I
decimal
places
&
b) 0.040 M OH–
Kr]]
significantpH = 12.60
figures
1.0x
i
=
I
I decimal
kw
(n +][0.040]
=
places
The
*
figures
significant
the
places
in
in the concentration should
be
the
same
number of
decimal
pH.
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Section 14.3
The pH Scale
EXERCISE!
The pH of a solution is 5.85. What is the [H+] for this
solution?
[H+] = 1.4 × 10–6 M
pH =-log[Ht]
[H+]
10
-
=
[H+] 10
-
PH
5.85
=
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Section 14.3
The pH Scale
pH and pOH
Recall:
Kw = [H+][OH–]
–log Kw = –log[H+] – log[OH–]
pKw = pH + pOH
14.00 = pH + pOH
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Section 14.3
The pH Scale
[H+] [0n]
EXERCISE!
1.0
=
10-1
x
Calculate the pOH for each of the following solutions.
pH
a)
=
-log(1.0
10
x
-
4)
1.0 × 10–4 M H+
pOH = 10.00
b) 0.040 M OH–
pOH = 1.40
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Section 14.3
The pH Scale
EXERCISE!
The pH of a solution is 5.85. What is the [OH–] for
this solution?
[OH–] = 7.1 × 10–9 M
[H+] 10
-
5.05
=
[H+] [On-T
1.0x10-in
=
](anfoss
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=
e
0
Section 14.4
Calculating the pH of Strong Acid Solutions
Thinking About Acid–Base Problems
⑧
⑧
What are the major species in solution?
What is the dominant reaction that will take place?
Is it an equilibrium reaction or a reaction that will go
essentially to completion?
React all major species until you are left with an
equilibrium reaction.
Solve for the pH if needed.
⑧
⑧
&
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Section 14.4
Calculating the pH of Strong Acid Solutions
CONCEPT CHECK!
Consider an aqueous solution of 2.0 × 10–3
s M HCl.
H20(e)
Hz0(e)=9*o-z
+
What are the major species in solution?
H+, Cl–, H2O
HCI will disassociate into
*
HC1
-
H
H+ and CI-
+
Cl-
+
What is the pH?
pH = 2.70
[H+] 20
=
bH
pH
=
-log[H+]
=-log [2.0 10-3]
pH
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10-3
x
x
2.70
=
0
Section 14.4
Calculating the pH of Strong Acid Solutions
CONCEPT CHECK!
Calculate the pH of a 1.5 × 10–11 M solution of HCl.
2.0
10-3
x
M
-
(130)
HC1 1.0x10 M
-
=
2.0x10-32t+eeyone
=
pH = 7.00
H20
HC1
+
-
Hz0
+
Cl-
1.5x10-"1.5 x 1 0 - M
H20 H20
+
Hz +OM1.0x10-M
pH =-log
pH
(1.0
1.0 x
10
x
10-
-
M
1)
7.00
=
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Section 14.4
Calculating the pH of Strong Acid Solutions
CONCEPT CHECK!
Calculate the pH of a 1.5 × 10–2 M solution of
HNO3.
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Section 14.4
Calculating the pH of Strong Acid Solutions
Let’s Think About It…
When HNO3 is added to water, a reaction takes place
immediately:
HNO3 + H2O H3O+ + NO3–
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Section 14.4
Calculating the pH of Strong Acid Solutions
Let’s Think About It…
Why is this reaction not likely?
NO3–(aq) + H2O(l)
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HNO3(aq) + OH–(aq)
0
Section 14.4
Calculating the pH of Strong Acid Solutions
Let’s Think About It…
What reaction controls the pH?
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
In aqueous solutions, this reaction is always taking place.
But is water the major contributor of H+ (H3O+)?
pH = 1.82
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.
.
.
.
Section 14.5
Calculating the pH of Weak Acid Solutions
List the major species in the solution.
Choose the species that can produce H+, and write
balanced equations for the reactions producing H+.
Using the values of the equilibrium constants for the
reactions you have written, decide which equilibrium will
dominate in producing H+.
Write the equilibrium expression for the dominant
equilibrium.
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.
.
.
.
Section 14.5
Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
List the initial concentrations of the species
participating in the dominant equilibrium.
Define the change needed to achieve equilibrium; that
is, define x.
Write the equilibrium concentrations in terms of x.
Substitute the equilibrium concentrations into the
equilibrium expression.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
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Section 14.5
Calculating the pH of Weak Acid Solutions
CONCEPT CHECK!
Consider a 0.80 M aqueous solution of the weak acid
HCN (Ka = 6.2 × 10–10).
What are the major species in solution?
HCN, H2O
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Section 14.5
Calculating the pH of Weak Acid Solutions
Let’s Think About It…
Why aren’t H+ or CN– major species?
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Section 14.5
Calculating the pH of Weak Acid Solutions
Consider This
KaL Kw;so
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we
use
Ka
6.2
=
x
10-10
0
Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
Calculate the pH of a 0.50 M aqueous solution of the weak
acid HF.
(Ka = 7.2 × 10–4)
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Section 14.5
Calculating the pH of Weak Acid Solutions
Let’s Think About It…
What are the major species in solution?
HF, H2O
Why aren’t H+ and F– major species?
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Section 14.5
Calculating the pH of Weak Acid Solutions
Let’s Think About It…
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Section 14.5
Calculating the pH of Weak Acid Solutions
Steps Toward Solving for pH
HF(aq) + H2O
Initial
Change
Equilibrium
ka
H3O+(aq) +
0.50 M
~0
~0
–x
+x
+x
0.50–x
x
x
3(5-
-
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F–(aq)
0
Section 14.5
Calculating the pH of Weak Acid Solutions
Percent Dissociation (Ionization)
For a given weak acid, the percent dissociation
increases/decreases (?) as the acid becomes more dilute.
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Section 14.5
Calculating the pH of Weak Acid Solutions
Percent Dissociation (Ionization)
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.5
Calculating the pH of Weak Acid Solutions
CONCEPT CHECK!
The percent ionization of a 4.00 M formic acid solution
should be:
higher than
lower than
the same as
the percent ionization of an 8.00 M formic acid solution.
Explain.
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Section 14.5
Calculating the pH of Weak Acid Solutions
CONCEPT CHECK!
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
Calculate the percent ionization of a 4.00 M formic
acid solution in water.
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
Calculate the percent ionization of a 4.00 M formic
acid solution in water.
% Ionization = 0.67%
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
Calculate the percent ionization of a 4.00 M formic
acid solution in water.
% Ionization = 0.67%
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Section 14.5
Calculating the pH of Weak Acid Solutions
EXERCISE!
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Section 14.6
Bases
Arrhenius: bases produce OH– ions.
Brønsted–Lowry: bases are proton acceptors.
In a basic solution at 25°C, pH > 7.
Ionic compounds containing OH- are generally
considered strong bases.
LiOH, NaOH, KOH, Ca(OH)2
pOH = –log[OH–]
pH = 14.00 – pOH
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Section 14.6
Bases
CONCEPT CHECK!
Calculate the pH of a 1.0 × 10–3 M solution of sodium
hydroxide.
pH = 11.00
NaOH
NaOH
1.0 × 10–3 M
Na+
Na+ + OH+
1.0 × 10–3 M
OH1.0 × 10–3 M M
pOH = - log [OH-]
pH = 14.00 - pOH
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Section 14.6
Bases
CONCEPT CHECK!
Calculate the pH of a 1.0 × 10–3 M solution of calcium
hydroxide.
pH = 11.30
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Section 14.6
Bases
Equilibrium expression for weak bases uses Kb.
CN–(aq) + H2O(l)
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HCN(aq) + OH–(aq)
0
Section 14.6
Bases
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Section 14.6
Bases
CONCEPT CHECK!
Calculate the pH of a 2.0 M solution of ammonia (NH3).
(Kb = 1.8 × 10–5)
pH = 11.78
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Section 14.7
Polyprotic Acids
⑧
⑧
⑳
·
Acids that can furnish more than one proton.
Always dissociates in a stepwise manner, one proton at a
time.
The conjugate base of the first dissociation equilibrium
becomes the acid in the second step.
For a typical weak polyprotic acid:
ply
dissociation.
Ka1a
> Ka2 > Ka3
al
For a typical polyprotic acid in water, only the first
dissociation step is important to pH.
At
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* count the
from the first
0
Section 14.7
Polyprotic Acids
EXERCISE!
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Section 14.7
Polyprotic Acids
EXERCISE!
Calculateusing Conjugate
are
Kal
acid
I
X
acid
-
X
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-
conjugate base
-conjugate
base
0
Section 14.7
Polyprotic Acids
CONCEPT CHECK!
Calculate the equilibrium concentration of PO43D in a
As in
1.00 M solution of H3PO4.
Ka1 = 7.5 × 10-3
Ka2 = 6.2 × 10-8
Ka3 = 4.8 × 10-13
Be
g
-
3
EM
[PO43-]
At = 3.6 × 10-19
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Section 14.7
Polyprotic Acids
CONCEPT CHECK!
From the previous exercise, the students should see that [H+] = [H2PO4-] = 0.0829 M (from the first
dissociation reaction).
Using the Ka2 expression, students can solve for [HPO42-] = 6.2 x 10-8 M.
H2PO4-(aq) + H2O
I
C
0.0829
-x
E 0.0829 - x
H3O+(aq) + HPO42-(aq)
0.0829
+x
0
+x
0.0829 + x
x
Many students will use “0” as the initial value of [H3O+]. [WRONG]
Note that [HPO42-] is equal to Ka2, which should make sense since [H+] = [H2PO4-] (“x” is negligible).
Using the Ka3 expression, students can solve for the phosphate ion concentration.
HPO42-(aq) + H2O
I
6.2 x 10-8
H3O+(aq) + PO43-(aq)
0.0829
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0
Section 14.8
Acid-Base Properties of Salts
saltis product
*
Salts
⑧
⑧
a
ofa n
acid-based
Ionic compounds.
When dissolved in water, break up into its ions (which
can behave as acids or bases).
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Section 14.8
Acid-Base Properties of Salts
Salts
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Section 14.8
Acid-Base Properties of Salts
Salts
↳Fi
weak
acid
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tet
+
gete
Nat
basic
solution
0
Section 14.8
Acid-Base Properties of Salts
Salts
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NH3
M
weak
base
+
H20
NHnt+OHd
conjugate
acid
0
Section 14.8
Acid-Base Properties of Salts
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Section 14.8
Acid-Base Properties of Salts
Qualitative Prediction of pH of Salt Solutions (from Weak
Parents)
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Section 14.8
Acid-Base Properties of Salts
EXERCISE!
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Section 14.8
Acid-Base Properties of Salts
EXERCISE!
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Section 14.8
Acid-Base Properties of Salts
CONCEPT CHECK!
basic
acidic
from
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acidic
to
0
basic
strong
Rank
*
Ka
acid
strong
neutral
using
values
salt
weak
acid
acidic
basic
-salt
salt
weak
base
base
Section 14.8
Acid-Base Properties of Salts
CONCEPT CHECK!
NaF
-
Nat
&
will have the
Consider a 0.30 M solution of NaF.
The Ka↳
for HF is 7.2 × 10-4.
What are the major species?
Na+, F-, H2O
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+
same
F
M
as
Nart and
F
Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
Why isn’t NaF considered a major species?
What are the possibilities for the dominant reactions?
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.
.
.
.
Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
The possibilities for the dominant reactions are:
F–(aq) + H2O(l)
H2O(l) + H2O(l)
Na+(aq) + H2O(l)
Na+(aq) + F–(aq)
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HF(aq) + OH–(aq)
H3O+(aq) + OH–(aq)
NaOH + H+(aq)
NaF
0
Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
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Section 14.8
Acid-Base Properties of Salts
EXERCISE!
Calculate the pH of a 0.75 M aqueous solution of NaCN.
KaA for HCN is 6.2 × 10–10.
W
-
-
a
10
-
pH
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will
be
7
Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
What are the major species in solution?
Na+, CN–, H2O
Why isn’t NaCN considered a major species?
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.
.
.
.
Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
What are all possibilities for the dominant reaction?
The possibilities for the dominant reaction are:
CN–(aq) + H2O(l)
HCN(aq) + OH–(aq)
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
Na+(aq) + H2O(l)
NaOH + H+(aq)
Na+(aq) + CN–(aq)
NaCN
Which of these reactions really occur?
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Section 14.8
Acid-Base Properties of Salts
Let’s Think About It…
How do we decide which reaction controls the pH?
CN–(aq) + H2O(l)
HCN(aq) + OH–(aq)
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
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Section 14.8
Acid-Base Properties of Salts
Steps Toward Solving for pH
CN–(aq) + H2O
Initial
Change
Equilibrium
HCN(aq) + OH–(aq)
0.75 M
0
~0
–x
+x
+x
0.75–x
x
x
Kb = 1.6 × 10–5
pH = 11.54
Section 14.9
The Effect of Structure on Acid-Base Properties
Models of Acids and Bases
⑧
Two factors for acidity in binary compounds:
polar,
stronger
Bond Polarity (high is good) -)
Bond Strength (low is good) lower strength, stronger
The
⑧
⑳
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the
more
the
acid.
=>
the
bond
0
acid
Section 14.9
The Effect of Structure on Acid-Base Properties
Bond Strengths and Acid Strengths for Hydrogen Halides
High
*
bond
strength,
weaker acid.
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Section 14.9
The Effect of Structure on Acid-Base Properties
Oxyacids
⑧
⑧
Contains the group H–O–X.
For a given series the acid strength increases with an
increase in the number of oxygen atoms attached to the
central atom.
The greater the ability of X to draw electrons toward
itself, the greater the acidity of the molecule.
↑The
higher electronegativity,
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the
stronger
the acid
0
Section 14.9
The Effect of Structure on Acid-Base Properties
The
*
one
with
more
otygenwilda
varie
Several Series of
Oxyacids and Their
Ka Values
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Section 14.9
The Effect of Structure on Acid-Base Properties
Comparison of Electronegativity of X and Ka Value
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Section 14.10
Acid-Base Properties of Oxides
Oxides
⑧
Acidic Oxides (Acid Anhydrides):
O—X bond is strong and covalent.
&
* NO2,
SO2,
A CO2
A
⑳
When H—O—X grouping is dissolved in water, the O—
X bond will remain intact. It will be the polar and
relatively weak H—O bond that will tend to break,
releasing a proton.
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Section 14.10
Acid-Base Properties of Oxides
Oxides
Basic Oxides (Basic Anhydrides):
O—X bond is ionic.
K2O,
A CaO
If X has a very low electronegativity, the O—X bond
will be ionic and subject to being broken in polar water,
producing a basic solution.
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Section 14.11
The Lewis Acid-Base Model
Lewis Acids and Bases
Lewis acid: electron pair acceptor
Lewis base: electron pair donor
Lewis acid
Strpairs
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Lewis base
rates
elections
pairs
0
Section 14.11
The Lewis Acid-Base Model
Three Models for Acids and Bases
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Section 14.12
Strategy for Solving Acid-Base Problems: A
Summary
When analyzing an acid-base equilibrium problem:
Ask this question: What are the major species in the
solution and what is their chemical behavior?
What major species are present?
Does a reaction occur that can be assumed to go to
completion?
What equilibrium dominates the solution?
Let the problem guide you. Be patient.
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