The Nature of Acids and Bases

Transcription

Chapter 15
Acid-Base Equilibria
Section 15.1
Solutions of Acids or Bases Containing a
Common Ion
Alkaline water
pH 8-10
Section 13.1
The Equilibrium Condition
Section 14.1
The Nature
of Acidsto
andWatch
Bases
Videos
http://www.bozemanscience.com/ap-chemistry/
Videos to Watch
Section 14.1
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Which indicator is best suited for the weak acid below?
Section 14.1
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Watch this video for an explanation:
https://www.youtube.com/watch?v=tvCyrBHJfBk
Section 15.1
Common Ion
Common Ion Effect
 Shift in equilibrium position that occurs because of
the addition of an ion already involved in the
equilibrium reaction.
 An application of Le Châtelier’s principle.
Copyright © Cengage Learning. All rights reserved
43
Section 15.1
Common Ion
Example
HCN(aq) + H2O(l)
H3O+(aq) + CN-(aq)
 Addition of NaCN will shift the equilibrium to the
left because of the addition of CN-, which is already
involved in the equilibrium reaction.
 A solution of HCN and NaCN is less acidic than a
solution of HCN alone.
Section 15.2
Buffered Solutions
Key Points about Buffered Solutions
 Buffered Solution – resists a change in pH.
 They are weak acids or bases containing a common ion.
 After addition of strong acid or base, deal with
stoichiometry first, then the equilibrium.
45
Section 15.2
Buffered Solutions
Adding an Acid to a Buffer
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46
Section 15.2
Buffered Solutions
Buffers
47
Section 15.2
Buffered Solutions
Solving Problems with Buffered Solutions
48
Section 15.2
Buffered Solutions
Buffering: How Does It Work?
49
Section 15.2
Buffered Solutions
Buffering: How Does It Work?
50
Section 15.2
Buffered Solutions
Henderson–Hasselbalch Equation
 A  
pH = pK a + log
HA 
 For a particular buffering system (conjugate acid–base
pair), all solutions that have the same ratio [A–] / [HA]
will have the same pH.
51
Section 15.2
Buffered Solutions
EXERCISE!
What is the pH of a buffer solution that is 0.45 M acetic
acid (HC2H3O2) and 0.85 M sodium acetate (NaC2H3O2)?
The Ka for acetic acid is 1.8 × 10–5.
pH = 5.02
52
Section 15.2
Buffered Solutions
53
Section 15.2
Buffered Solutions
Buffered Solution Characteristics
 Buffers contain relatively large concentrations of a weak
acid and corresponding conjugate base.
 Added H+ reacts to completion with the weak base.
 Added OH- reacts to completion with the weak acid.
54
Section 15.2
Buffered Solutions
Buffered Solution Characteristics
 The pH in the buffered solution is determined by the
ratio of the concentrations of the weak acid and weak
base. As long as this ratio remains virtually constant, the
pH will remain virtually constant. This will be the case as
long as the concentrations of the buffering materials
(HA and A– or B and BH+) are large compared with
amounts of H+ or OH– added.
55
Section 15.3
Buffering Capacity
 The amount of protons or hydroxide ions the buffer can
absorb without a significant change in pH.
 Determined by the magnitudes of [HA] and [A–].
 A buffer with large capacity contains large
concentrations of the buffering components.
56
Section 15.3
Buffering Capacity
 Optimal buffering occurs when [HA] is equal to [A–].
 It is for this condition that the ratio [A–] / [HA] is most
resistant to change when H+ or OH– is added to the
buffered solution.
57
Section 15.3
Buffering Capacity
Choosing a Buffer
 pKa of the weak acid to be used in the buffer should be
as close as possible to the desired pH.
58
Section 15.4
Titrations and pH Curves
Titration Curve
 Plotting the pH of the solution being analyzed as a
function of the amount of titrant added.
 Equivalence (Stoichiometric) Point – point in the
titration when enough titrant has been added to react
exactly with the substance in solution being titrated.
59
Section 15.4
Neutralization of a Strong Acid with a Strong Base
60
Section 15.4
The pH Curve for the Titration of 50.0 mL of 0.200 M HNO3
with 0.100 M NaOH
61
Section 15.4
The pH Curve for the Titration of 100.0 mL of 0.50 M NaOH
with 1.0 M HCI
62
Section 15.4
Weak Acid–Strong Base Titration
Step 1:
Step 2:
A stoichiometry problem (reaction is
assumed to run to completion) then
determine concentration of acid
remaining and conjugate base formed.
An equilibrium problem (determine
position of weak acid equilibrium and
calculate pH).
63
Section 15.4
CONCEPT CHECK!
Consider a solution made by mixing 0.10 mol of HCN
(Ka = 6.2 × 10–10) with 0.040 mol NaOH in 1.0 L of
aqueous solution.
What are the major species immediately upon mixing
(that is, before a reaction)?
HCN, Na+, OH–, H2O
64
Section 15.4
Let’s Think About It…
 Why isn’t NaOH a major species?
 Why aren’t H+ and CN– major species?
 List all possibilities for the dominant reaction.
65
Section 15.4
The possibilities for the dominant reaction are:
1.
2.
3.
4.
5.
H2O(l) + H2O(l)
H3O+(aq) + OH–(aq)
HCN(aq) + H2O(l)
H3O+(aq) + CN–(aq)
HCN(aq) + OH–(aq)
CN–(aq) + H2O(l)
Na+(aq) + OH–(aq)
NaOH
Na+(aq) + H2O(l)
NaOH + H+(aq)
66
Section 15.4
 How do we decide which reaction controls the pH?
H2O(l) + H2O(l)
HCN(aq) + H2O(l)
H3O+(aq) + CN–(aq)
HCN(aq) + OH–(aq)
CN–(aq) + H2O(l)
Section 15.4
HCN(aq) + OH–(aq)
CN–(aq) + H2O(l)
 What are the major species after this reaction occurs?
HCN, CN–, H2O, Na+
68
Section 15.4
 Now you can treat this situation as before.
 List the possibilities for the dominant reaction.
 Determine which controls the pH.
69
Section 15.4
CONCEPT CHECK!
Calculate the pH of a solution made by mixing 0.20
mol HC2H3O2 (Ka = 1.8 × 10–5) with 0.030 mol NaOH in
1.0 L of aqueous solution.
70
Section 15.4
 What are the major species in solution?
Na+, OH–, HC2H3O2, H2O
 Why isn’t NaOH a major species?
 Why aren’t H+ and C2H3O2– major species?
71
Section 15.4
 What are the possibilities for the dominant reaction?
1.
2.
3.
4.
5.
H2O(l) + H2O(l)
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2–(aq)
HC2H3O2(aq) + OH–(aq)
C2H3O2–(aq) + H2O(l)
Na+(aq) + OH–(aq)
NaOH(aq)
Na+(aq) + H2O(l)
NaOH + H+(aq)
 Which of these reactions really occur?
72
Section 15.4
 Which reaction controls the pH?
H2O(l) + H2O(l)
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2–(aq)
HC2H3O2(aq) + OH–(aq)
C2H3O2–(aq) + H2O(l)
 How do you know?
73
Section 15.4
HC2H3O2(aq) + OH–
Before
Change
After
0.20 mol 0.030 mol
–0.030 mol –0.030 mol
0.17 mol
0
C2H3O2–(aq) + H2O
0
+0.030 mol
0.030 mol
K = 1.8 × 109
74
Section 15.4
Steps Toward Solving for pH
H3O+
+ C2H3O2-(aq)
0.170 M
~0
0.030 M
–x
+x
+x
0.170 – x
x
0.030 + x
HC2H3O2(aq) + H2O
Initial
Change
Equilibrium
Ka = 1.8 × 10–5
pH = 3.99
75
Section 15.4
EXERCISE!
Calculate the pH of a 100.0 mL solution of 0.100 M
acetic acid (HC2H3O2), which has a Ka value of 1.8 × 10–
5.
pH = 2.87
76
Section 15.4
CONCEPT CHECK!
Calculate the pH of a solution made by mixing 100.0
mL of a 0.100 M solution of acetic acid (HC2H3O2),
which has a Ka value of 1.8 × 10–5, and 50.0 mL of a
0.10 M NaOH solution.
pH = 4.74
77
Section 15.4
CONCEPT CHECK!
Calculate the pH of a solution at the equivalence point
when 100.0 mL of a 0.100 M solution of acetic acid
(HC2H3O2), which has a Ka value of 1.8 × 10–5, is titrated
with a 0.10 M NaOH solution.
pH = 8.72
78
Section 15.4
The pH Curve for the Titration of 50.0 mL of 0.100 M
HC2H3O2 with 0.100 M NaOH
79
Section 15.4
The pH Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids
with Various Ka Values with 0.10 M NaOH
80
Section 15.4
The pH Curve for the Titration of 100.0 mL of 0.050 M NH3
with 0.10 M HCl
81
Section 15.5
Acid-Base Indicators
 Marks the end point of a titration by changing color.
 The equivalence point is not necessarily the same as the
end point (but they are ideally as close as possible).
82
Section 15.5
The Acid and Base
Forms of the Indicator
Phenolphthalein
83
Section 15.5
The Methyl Orange Indicator is Yellow in Basic Solution and
Red in Acidic Solution
84
Section 15.5
Useful pH Ranges for Several Common Indicators
85
Section 15.5
Complex Ion
Equilibria
Section 15.5
Acid-Base
Indicators
Videos
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Section 15.5
Section 16.1
Solubility Equilibria and the Solubility Product
Solubility Equilibria
 Solubility product (Ksp) – equilibrium constant; has
only one value for a given solid at a given
temperature.
 Solubility – an equilibrium position.
Bi2S3(s)
2Bi3+(aq) + 3S2–(aq)
3+ 2
2 3
K sp = Bi  S 
89
Section 16.1
Section 16.1
Section 16.1
Section 16.1
Section 16.1
Section 16.1
Section 16.1
CONCEPT CHECK!
In comparing several salts at a given temperature,
does a higher Ksp value always mean a higher
solubility?
Explain. If yes, explain and verify. If no, provide a
counter-example.
No
96
Section 16.1
EXERCISE!
Calculate the solubility of silver chloride in water.
Ksp = 1.6 × 10–10
1.3×10-5 M
Calculate the solubility of silver phosphate in water.
Ksp = 1.8 × 10–18
1.6×10-5 M
97
Section 16.1
CONCEPT CHECK!
How does the solubility of silver chloride in water
compare to that of silver chloride in an acidic
solution (made by adding nitric acid to the
solution)?
Explain.
The solubilities are the same.
98
Section 16.1
CONCEPT CHECK!
How does the solubility of silver phosphate in water
compare to that of silver phosphate in an acidic
solution)?
Explain.
The silver phosphate is more soluble in an acidic
solution.
99
Section 16.1
CONCEPT CHECK!
How does the Ksp of silver phosphate in water
compare to that of silver phosphate in an acidic
solution)?
Explain.
The Ksp values are the same.
100
Section 16.1
EXERCISE!
Calculate the solubility of AgCl in:
Ksp = 1.6 × 10–10
a) 100.0 mL of 4.00 x 10-3 M calcium chloride.
2.0×10-8 M
b) 100.0 mL of 4.00 x 10-3 M calcium nitrate.
1.3×10-5 M
101
Section 16.2
Precipitation and Qualitative Analysis
Precipitation (Mixing Two Solutions of Ions)
 Q > Ksp; precipitation occurs and will continue until
the concentrations are reduced to the point that
they satisfy Ksp.
 Q < Ksp; no precipitation occurs.
102
Section 16.2
Selective Precipitation (Mixtures of Metal Ions)
 Use a reagent whose anion forms a precipitate with only
one or a few of the metal ions in the mixture.
 Example:
 Solution contains Ba2+ and Ag+ ions.
 Adding NaCl will form a precipitate with Ag+ (AgCl),
while still leaving Ba2+ in solution.
103
Section 16.2
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
 At a low pH, [S2–] is relatively low and only the very
insoluble HgS and CuS precipitate.
 When OH– is added to lower [H+], the value of [S2–]
increases, and MnS and NiS precipitate.
104
Section 16.2
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
105
Section 16.2
Separating the Common
Cations by Selective
Precipitation
106
Section 16.3
Equilibria Involving Complex Ions
Complex Ion Equilibria
 Charged species consisting of a metal ion surrounded by
ligands.
 Ligand: Lewis base
 Formation (stability) constant.
 Equilibrium constant for each step of the formation
of a complex ion by the addition of an individual
ligand to a metal ion or complex ion in aqueous
solution.
107
Section 16.3
Complex Ion Equilibria
Be2+(aq) + F–(aq)
BeF+(aq)
K1 = 7.9 × 104
BeF+(aq) + F–(aq)
BeF2(aq)
K2 = 5.8 × 103
BeF2(aq) + F–(aq)
BeF3– (aq)
K3 = 6.1 × 102
BeF3– (aq) + F–(aq)
BeF42– (aq)
K4 = 2.7 × 101
108
Section 16.3
Complex Ions and Solubility
 Two strategies for dissolving a water–insoluble ionic
solid.
 If the anion of the solid is a good base, the solubility
is greatly increased by acidifying the solution.
 In cases where the anion is not sufficiently basic, the
ionic solid often can be dissolved in a solution
containing a ligand that forms stable complex ions
with its cation.
109
Section 16.3
CONCEPT CHECK!
Calculate the solubility of silver chloride in 10.0 M ammonia
given the following information:
Ksp (AgCl) = 1.6 × 10–10
Ag+ + NH3
AgNH3+ + NH3
AgNH3+
Ag(NH3)2+
K = 2.1 × 103
K = 8.2 × 103
0.48 M
Calculate the concentration of NH3 in the final equilibrium
mixture.
9.0 M
110

The Nature of Acids and Bases

Transcription

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