Transcript Slide 1

1
2
Arrhenius Theory
Acids release hydrogen ions
+
(H )
+
HCl → H + Cl
Bases release hydroxide
ions (OH )
+
NaOH → Na + OH
6
ACID-BASE THEORIES
• The most general theory for
common aqueous acids and bases
is the BRØNSTED - LOWRY
theory
• ACIDS DONATE H+ IONS
• BASES ACCEPT H+ IONS
Why is Water Neutral?
[H3O+ ] = [OH-]
When one water gains,
another loses a H+
Relative ion
concentrations
 pH
is a relative measure of the
hydrogen ion concentration
pH is a rating;
ranges from 0 – 14
0 = most, 7 equal, 14 = least
Any pX Scales
In general
pX = -log X
pOH = - log
pH = - log
[OH ]
+
[H ]
Determining pOH
pH + pOH = 14
If know one can determine
the other.
If pH = 13, what is the pOH?
13 + pOH = 14
pOH = 14 – 13 = 1
Why at a pH = 7 ?

Determined by concentration [ X ] of each ion
[H+ ]= [OH-] = 10-7M
[H+ ]= 10-pHM
[OH-] = 10-pOHM
Water
AUTOIONIZATION
H2O can function as both an ACID
and a BASE.
Equilibrium constant for water
= Kw
Kw = [H3O+] [OH-] =
1.00 x 10-14 at 25 oC
18
Water
19
OH-
H3O+
Autoionization
Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC
In a neutral solution [H3O+] = [OH-]
so [H3O+] = [OH-] = 1.00 x 10-7 M
Acid-Base Reactions
Chapter 17
20
ACIDS
Nonmetal oxides can be acids
CO2(aq) + H2O(liq) ---> H2CO3(aq)
SO3(aq) + H2O(liq) ---> H2SO4(aq)
and can come from
burning coal and oil.
21
BASES
22
Metal oxides are
bases
CaO(s) + H2O(liq) --> Ca(OH)2(aq)
CaO in water. Indicator shows
solution is basic.
Hydrolysis
23
24
Acid-Base Reactions
• sometimes called NEUTRALIZATIONS
because the solution is neither acidic nor
basic at the end.
• The other product of the A-B reaction is a
SALT, MX.
HX + MOH ---> MX + H2O
Mn+ comes from base & Xn- comes from acid
This is one way to make compounds!
25
Acid-Base Reactions
The “driving force” is the
formation of water.
NaOH(aq) + HCl(aq) ---> NaCl(aq) + H2O(liq)
Net ionic equation
OH-(aq) + H+(aq) ---> H2O(liq)
This applies to ALL reactions of
STRONG acids and bases.
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Weak Acids and Bases
Acid
acetic, CH3CO2H
ammonium, NH4+
bicarbonate, HCO3-
Conjugate Base
CH3CO2-, acetate
NH3, ammonia
CO32-, carbonate
A weak acid (or base) is one that
ionizes to a VERY small extent
(< 5%).
Equilibrium Constants
for Weak Acids
30
Weak acid has Ka < 1
Leads to small [H3O+] and a pH of 2 - 7
Equilibrium Constants
for Weak Bases
31
Weak base has Kb < 1
Leads to small [OH-] and a pH of 12 - 7
Kw = Ka* Kb
32
Acids
Conjugate
Bases
Weak Acids and Bases
34
acetic acid, CH3CO2H (HOAc)
HOAc + H2O D H3O+ + OAcAcid
Conj. base
+
-
[H3O ][OAc ]
-5
Ka 
 1.8 x 10
[HOAc]
(K is designated Ka for ACID)
[H3O+] and [OAc-] are SMALL, Ka << 1.
Equilibria Involving A Weak Acid
You have 1.00 M HOAc. Calc. the pH.
And the equilibrium concs. of EACH
Step 1. ICE table.
[HOAc]
I
C
E
[H3O+]
[OAc-]
35
Equilibria Involving A Weak Acid
[HOAc]
I
1.00
C
-x
E
1.00-x
[H3O+]
[OAc-]
0
0
+x
+x
x
x
Note that we neglect [H3O+] from H2O.
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Equilibria Involving A Weak Acid
Assume x is very small because Ka is so small.
+
-
[H3O ][OAc ]
-5
Ka  1.8 x 10
=

[HOAc]
-5
Ka  1.8 x 10 =
2
x
1.00 - x
2
x
1.00
Now we can more easily solve this
approximate expression.
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38
Equilibria Involving A Weak Acid
Step 3. Solve Ka approximate expression
-5
Ka  1.8 x 10 =
2
x
1.00
x = [H3O+] = [OAc-] = [Ka • 1.00]1/2
x = [H3O+] = [OAc-] = 4.2 x 10-3 M
pH = -log (4.2 x 10-3) = 2.37
Chapter 17 – Acids and Bases
What is the [H+] of a 1.0 M acetic acid if
Ka = 1.8 x 10-5 for the solution:
A.
B.
C.
D.
1.0 M
0.3 M
0.0042 M
1.8 x 10-5 M
© 2011 Pearson Education, Inc.
Chapter 17 – Acids and Bases
A 0.100 M solution of HX has a pH of is 4.5
Determine the Ka for the acid, HX.
A.
B.
C.
D.
1.0 x 10-8
3.0 x 10-7
2.5 x 10-8
4.5 x 10-9
© 2011 Pearson Education, Inc.
Weak Bases
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calc. the pH.
NH3 + H2O D NH4+ +
OH-
Kb = 1.8 x 10-5
Step 1; ICE table
[NH3]
I
C
E
[NH4+]
[OH-]
42
Weak Base
43
Step 1. ICE table
[NH3]
[NH4+]
[OH-]
0
0
I
0.010
C
-x
+x
+x
E
0.010 - x
x
x
Weak Base
44
Step 2. Solve the equilibrium expression
2
+ ][OH- ]
[NH
x
-5
4
Kb  1.8 x 10 =
=
[NH3 ]
0.010 - x
Assume x is small (100•Kb < Co), so
x = [OH-] = [NH4+] = 4.2 x 10-4 M
[NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M
The approximation is valid
!
Neutralization Reactions
NaOH + HCl → NaCl + H20
REACT
a base with an acid.
forms
salt and water
19.4 Neutralization Reactions >
Acid-Base Reactions
A reaction between an acid and a base will
go to completion when the solutions
contain equal moles of hydrogen ions and
hydroxide ions.
• The balanced equation provides the
correct ratio of acid to base.
46
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19.4 Neutralization Reactions >
Titration
Neutralization occurs when
the number of moles of
hydrogen ions is EQUAL to the
number of moles of hydroxide
ions.
47
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19.4 Neutralization Reactions >
Acid-Base Reactions
For hydrochloric acid and sodium
hydroxide, the mole ratio is 1:1.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
1 mol
48
1 mol
1 mol
1 mol
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19.4 Neutralization Reactions >
Acid-Base Reactions
For sulfuric acid and sodium hydroxide, the
ratio is 1:2.
•
Two moles of the base are required to neutralize one mole of the acid.
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
1 mol
49
2 mol
1 mol
2 mol
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19.4 Neutralization Reactions >
Acid-Base Reactions
Similarly, hydrochloric acid and calcium
hydroxide react in a 2:1 ratio.
2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)
2 mol
50
1 mol
1 mol
2 mol
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19.4 Neutralization Reactions >
Sample Problem 19.7
Finding the Moles Needed for Neutralization
How many moles of sulfuric acid are
required to neutralize 0.50 mol of
sodium hydroxide? The equation for
the reaction is
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O.
51
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19.4 Neutralization Reactions >
1
Analyze
Sample Problem 19.7
List the knowns and the unknown.
To determine the number of moles of
acid, you need to know the number of
moles of base and the mole ratio of acid
to base.
KNOWNS
UNKNOWN
mol NaOH = 0.50 mol
mol H2SO4 = ? mol
1 mol H2SO4/2 mol NaOH
(from balanced equation)
52
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19.4 Neutralization Reactions >
2
Sample Problem 19.7
Calculate Solve for the unknown.
Use the mole ratio of acid to base to
determine the number of moles of acid.
0.50 mol NaOH ×
53
1 mol H2SO4
2 mol NaOH
= 0.25 mol H2SO4
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19.4 Neutralization Reactions >
3
Sample Problem 19.7
Evaluate Does the result make sense?
Because the mole ratio of H2SO4 to
NaOH is 1:2, the number of moles of
H2SO4 should be half the number of the
moles of NaOH.
54
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19.4 Neutralization Reactions >
Titration
You can use a neutralization reaction
to determine the concentration of an acid
or base.
• The process of adding a measured
amount of a solution of known
concentration to a solution of
unknown concentration is called a
titration.
55
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19.4 Neutralization Reactions >
56
A flask with a known
volume of acids (and an
indicator) is placed
beneath a buret that is
filled with a base of
known concentration.
Titration
The base is slowly
added from the
buret to the acid.
A change in the color
of the solution is the
signal that
neutralization has
occurred.
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19.4 Neutralization Reactions >
57
19.4 Neutralization Reactions >
Titration
The indicator that is chosen for a titration
must change color at or near the pH of the
equivalence point.
• The point at which the
indicator changes color
is the end point of the
titration.
58
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19.2 Hydrogen IonsIndicators
and Acidity >
Acid-Base
An indicator (HIn) is a weak acid or base that
dissociates in a known pH range.
• Indicators work because their acid form and
base form have different colors in solution.
• The acid form of the indicator (HIn) is dominant at low pH
and high [H+].
• The base form (In−) is dominant at high pH and high [OH−].
59
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19.4 Neutralization Reactions >
The pH change of a solution during the
titration of a strong acid (HCl) with a strong
base (NaOH).
• The equivalence point for
this reaction occurs at a
pH of 7.
• As the titration nears the
equivalence point, the pH
rises dramatically
because hydrogen ions
are being used up.
60
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19.4 Neutralization Reactions >
Sample Problem 19.8
Determining Concentration by Titration
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l).
A 25-mL solution of H2SO4 is
neutralized by 18 mL of 1.0M NaOH.
What is the concentration of the
H2SO4 solution?
61
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19.4 Neutralization Reactions >
1
Analyze
Sample Problem 19.8
List the knowns and the unknown.
The conversion steps are as follows:
L NaOH → mol NaOH → mol H2SO4 → M H2SO4.
KNOWNS
UNKNOWN
[NaOH] = 1.0M
[H2SO4] = ?M
VNaOH = 18 mL = 0.018 L
VH
SO
2
62
4
= 18 mL = 0.018 L
Convert volume to
liters because molarity
is in moles per liter.
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19.4 Neutralization Reactions >
2
Sample Problem 19.8
Calculate Solve for the unknown.
Use the molarity to convert the volume
of base to moles of base.
0.018 L NaOH ×
63
1.0 mol NaOH
1 L NaOH
= 0.018 mol NaOH
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19.4 Neutralization Reactions >
2
Sample Problem 19.8
Calculate Solve for the unknown.
Use the mole ratio to find the moles
of acid.
0.018 mol NaOH ×
64
1.0 mol H2SO4
= 0.0090 mol H2SO4
2 mol NaOH
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19.4 Neutralization Reactions >
2
Sample Problem 19.8
Calculate Solve for the unknown.
Calculate the molarity by dividing moles
of acid by liters of solution.
0.0090 mol
mol of solute
molarity =
=
L of solution
0.025 L
= 0.36M H2SO4
65
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19.5 Salts in Solution >
Interpret Graphs
• One curve is for the
addition of sodium
hydroxide, a strong
base, to ethanoic acid,
a weak acid.
• An aqueous solution of
sodium ethanoate
exists at the
equivalence point.
CH3COOH(aq) + NaOH(aq) →
CH3COONa(aq) + H2O(l)
66
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Acetic acid titrated with NaOH
Weak acid titrated
with a strong base
Figure 18.5
67
Acid-Base Titration
Section 18.3
You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the
final solution?
HBz + NaOH ---> Na+ + Bz- + H2O
Kb = 1.6 x 10-10
C6H5CO2H = HBz
Benzoate ion = Bz-
68
69
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
pH at
half-way
point?
Benzoic acid
+ NaOH
pH at
equivalence
point?
pH of solution of
benzoic acid, a weak
acid
Acid-Base Reactions
QUESTION: You titrate 100. mL of a 0.025 M
solution of benzoic acid with 0.100 M NaOH to the
equivalence point. What is the pH of the final
solution?
Strategy — find the conc. of the conjugate
base Bz- in the solution AFTER the
titration, then calculate pH.
This is a two-step problem
1.
2.
stoichiometry of acid-base
reaction
equilibrium calculation
70
STOICHIOMETRY PORTION
M * V = mol
1. Calc. moles of NaOH req’d
(0.100 L)(0.025 M) = 0.0025 mol HBz
This requires 0.0025 mol NaOH
2.Calc. volume of NaOH req’d
0.0025 mol (1 L / 0.100 mol) = 0.025 L
25 mL of NaOH req’d
71
STOICHIOMETRY PORTION
3. Moles Bz- produced = moles HBz = 0.0025 mol
4. Calc. conc. of BzThere are 0.0025 mol of Bz- in a TOTAL
SOLUTION VOLUME of
125 mL
[Bz-]
= 0.0025 mol / 0.125 L
= 0.020 M
72
Equivalence Point
73
Most important species in solution is benzoate ion,
Bz-. It will react to form the weak conjugate base,
benzoic acid, HBz.
Bz- + H2O D HBz + OH-
Kb = 1.6 x 10-10
I
[Bz-]
0.020
[HBz]
0
[OH-]
0
C
-x
+x
+x
E
0.020 - x
x
x
74
Acid-Base Reactions
-10
Kb  1.6 x 10
x=
[OH ]
2
x
=
0.020 - x
= 1.8 x
-6
10
pOH = 5.75 -----> pH = 8.25
75
QUESTION: You titrate 100. mL of a
0.025 M solution of benzoic acid with
0.100 M NaOH to the equivalence point.
What is the pH at half-way point?
pH at
half-way
point?
Equivalence
point
pH = 8.25
Acid-Base Reactions
76
You titrate 100. mL of a 0.025 M solution of
benzoic acid with 0.100 M NaOH.
What is the pH at the half-way point?
Both HBz and Bzare present.
This is a BUFFER!

[H 3 O ] =
[HBz]
-
[Bz ]
• Ka
At the half-way point, [HBz] = [Bz-]
Therefore, [H3O+] = Ka = 6.3 x 10-5
pH = 4.20 = pKa of the acid
80
Weak base (NH3)
titrated with a
strong acid (HCl)
Figure 18.7
Acid-Base Reactions
• Strong acid + strong base
HCl + NaOH ----> SALT WATER
• Strong acid + weak base
HCl + NH3 ---> ACID
• Weak acid + strong base
HOAc + NaOH ---> BASIC
• Weak acid + weak base
HOAc + NH3 ---> Ka / Kb
81