Chapter 14 Acids and Bases - Corona-Norco / District … 14.1 The Nature of Acids and Bases Videos...
-
Upload
nguyendung -
Category
Documents
-
view
248 -
download
1
Transcript of Chapter 14 Acids and Bases - Corona-Norco / District … 14.1 The Nature of Acids and Bases Videos...
Chapter 14
Acids and Bases
Section 13.1 The Equilibrium Condition
Section 14.1 The Nature of Acids and Bases
http://www.bozemanscience.com/ap-chemistry/
Videos to Watch
Section 14.1 The Nature of Acids and Bases
Videos to Watch
https://www.youtube.com/watch?v=OA-
f_8F5iNU&app=desktop
https://www.youtube.com/watch?v=R2sqOiVwQdI&app
=desktop
Section 13.1 The Equilibrium Condition
https://phet.colorado.edu/sims/html/acid-base-solutions/latest/acid-base-solutions_en.html
Section 13.1 The Equilibrium Condition
Titration Simulation
http://www.mhhe.com/physsci/chemistry/animation
s/chang_7e_esp/crm3s5_5.swf
Section 13.1 The Equilibrium Condition
Section 13.1 The Equilibrium Condition
Section 14.1 The Nature of Acids and Bases
Copyright © Cengage Learning. All rights reserved 9
Models of Acids and Bases
Arrhenius: Acids produce H+ ions in solution, bases produce OH- ions.
Brønsted–Lowry: Acids are proton (H+) donors, bases are proton acceptors.
HCl + H2O Cl- + H3O+
acid base
Section 14.1 The Nature of Acids and Bases
Brønsted–Lowry Reaction
Copyright © Cengage Learning. All rights reserved 10
To play movie you must be in Slide Show Mode
PC Users: Please wait for content to load, then click to play
Mac Users: CLICK HERE
Section 14.1 The Nature of Acids and Bases
Copyright © Cengage Learning. All rights reserved 11
Acid in Water
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
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 the base.
Section 14.1 The Nature of Acids and Bases
Acid Ionization Equilibrium
Copyright © Cengage Learning. All rights reserved 12
To play movie you must be in Slide Show Mode
PC Users: Please wait for content to load, then click to play
Mac Users: CLICK HERE
Section 14.2 Acid Strength
Strong acid:
Ionization equilibrium lies far to the right.
Yields a weak conjugate base.
Weak acid:
Ionization equilibrium lies far to the left.
Weaker the acid, stronger its conjugate base.
Copyright © Cengage Learning. All rights reserved 13
Section 14.2 Acid Strength
Copyright © Cengage Learning. All rights reserved 14
Section 14.2 Acid Strength
Various Ways to Describe Acid Strength
Copyright © Cengage Learning. All rights reserved 15
Section 14.2 Acid Strength
Water as an Acid and a Base
Water is amphoteric:
Behaves either as an acid or as a base.
At 25°C:
Kw = [H+][OH–] = 1.0 × 10–14
No matter what the solution contains, the product of [H+] and [OH–] must always equal 1.0 × 10–14 at 25°C.
Copyright © Cengage Learning. All rights reserved 16
Section 14.2 Acid Strength
Three Possible Situations
[H+] = [OH–]; neutral solution
[H+] > [OH–]; acidic solution
[OH–] > [H+]; basic solution
Copyright © Cengage Learning. All rights reserved 17
Section 14.2 Acid Strength
Self-Ionization of Water
Copyright © Cengage Learning. All rights reserved 18
To play movie you must be in Slide Show Mode
PC Users: Please wait for content to load, then click to play
Mac Users: CLICK HERE
Section 14.2 Acid Strength
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?
Copyright © Cengage Learning. All rights reserved 19
3H O A
= HA
K
CONCEPT CHECK!
Section 14.2 Acid Strength
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)
Copyright © Cengage Learning. All rights reserved 20
CONCEPT CHECK!
Section 14.2 Acid Strength
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?
Copyright © Cengage Learning. All rights reserved 21
CONCEPT CHECK!
Section 14.2 Acid Strength
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)
Copyright © Cengage Learning. All rights reserved 22
CONCEPT CHECK!
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–
Copyright © Cengage Learning. All rights reserved 23
Section 14.2 Acid Strength
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
a) Which way will equilibrium lie?
left Copyright © Cengage Learning. All rights reserved 24
CONCEPT CHECK!
Section 14.2 Acid Strength
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
Copyright © Cengage Learning. All rights reserved 25
CONCEPT CHECK!
Section 14.2 Acid Strength
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
c) Does this mean A– is a strong or weak base?
weak base
Copyright © Cengage Learning. All rights reserved 26
CONCEPT CHECK!
Section 14.2 Acid Strength
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 Cl– CN– C2H3O2–
Copyright © Cengage Learning. All rights reserved 27
CONCEPT CHECK!
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–
Copyright © Cengage Learning. All rights reserved 28
Section 14.2 Acid Strength
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.
Copyright © Cengage Learning. All rights reserved 29
CONCEPT CHECK!
Section 14.2 Acid Strength
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
Copyright © Cengage Learning. All rights reserved 30
CONCEPT CHECK!
Section 14.3 The pH Scale
pH = –log[H+]
pH changes by 1 for every power of 10 change in [H+].
A compact way to represent solution acidity.
pH decreases as [H+] increases.
Significant figures:
The number of decimal places in the log is equal to the number of significant figures in the original number.
Copyright © Cengage Learning. All rights reserved 31
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.
Copyright © Cengage Learning. All rights reserved 32
Section 14.3 The pH Scale
The pH Scale and pH Values of Some Common Substances
Copyright © Cengage Learning. All rights reserved 33
Section 14.3 The pH Scale
Calculate the pH for each of the following solutions.
a) 1.0 × 10–4 M H+
pH = 4.00
b) 0.040 M OH–
pH = 12.60
Copyright © Cengage Learning. All rights reserved 34
EXERCISE!
Section 14.3 The pH Scale
The pH of a solution is 5.85. What is the [H+] for this solution?
[H+] = 1.4 × 10–6 M
Copyright © Cengage Learning. All rights reserved 35
EXERCISE!
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
Copyright © Cengage Learning. All rights reserved 36
Section 14.3 The pH Scale
Calculate the pOH for each of the following solutions.
a) 1.0 × 10–4 M H+
pOH = 10.00
b) 0.040 M OH–
pOH = 1.40
Copyright © Cengage Learning. All rights reserved 37
EXERCISE!
Section 14.3 The pH Scale
The pH of a solution is 5.85. What is the [OH–] for this solution?
[OH–] = 7.1 × 10–9 M
Copyright © Cengage Learning. All rights reserved 38
EXERCISE!
Section 14.4 Calculating the pH of Strong Acid Solutions
Thinking About Acid–Base Problems
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.
Copyright © Cengage Learning. All rights reserved 39
Section 14.4 Calculating the pH of Strong Acid Solutions
Consider an aqueous solution of 2.0 × 10–3 M HCl.
What are the major species in solution?
H+, Cl–, H2O
What is the pH?
pH = 2.70
Copyright © Cengage Learning. All rights reserved 40
CONCEPT CHECK!
Section 14.4 Calculating the pH of Strong Acid Solutions
Calculate the pH of a 1.5 × 10–11 M solution of HCl.
pH = 7.00
Copyright © Cengage Learning. All rights reserved 41
CONCEPT CHECK!
Section 14.4 Calculating the pH of Strong Acid Solutions
Calculate the pH of a 1.5 × 10–2 M solution of HNO3.
Copyright © Cengage Learning. All rights reserved 42
CONCEPT CHECK!
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–
Copyright © Cengage Learning. All rights reserved 43
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) HNO3(aq) + OH–(aq)
Copyright © Cengage Learning. All rights reserved 44
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
Copyright © Cengage Learning. All rights reserved 45
Section 14.5 Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
1. List the major species in the solution.
2. Choose the species that can produce H+, and write balanced equations for the reactions producing H+.
3. Using the values of the equilibrium constants for the reactions you have written, decide which equilibrium will dominate in producing H+.
4. Write the equilibrium expression for the dominant equilibrium.
Copyright © Cengage Learning. All rights reserved 46
Section 14.5 Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
5. List the initial concentrations of the species participating in the dominant equilibrium.
6. Define the change needed to achieve equilibrium; that is, define x.
7. Write the equilibrium concentrations in terms of x.
8. Substitute the equilibrium concentrations into the equilibrium expression.
Copyright © Cengage Learning. All rights reserved 47
Section 14.5 Calculating the pH of Weak Acid Solutions
Solving Weak Acid Equilibrium Problems
9. Solve for x the “easy” way, that is, by assuming that [HA]0 – x about equals [HA]0.
10.Use the 5% rule to verify whether the approximation is valid.
11.Calculate [H+] and pH.
Copyright © Cengage Learning. All rights reserved 48
Section 14.5 Calculating the pH of Weak Acid Solutions
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
Copyright © Cengage Learning. All rights reserved 49
CONCEPT CHECK!
Section 14.5 Calculating the pH of Weak Acid Solutions
Let’s Think About It…
Why aren’t H+ or CN– major species?
Copyright © Cengage Learning. All rights reserved 50
Section 14.5 Calculating the pH of Weak Acid Solutions
Consider This
HCN(aq) + H2O(l) H3O+(aq) + CN–(aq)
Ka = 6.2 × 10-10
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
Kw = 1.0 × 10-14
Which reaction controls the pH? Explain.
Copyright © Cengage Learning. All rights reserved 51
Section 14.5 Calculating the pH of Weak Acid Solutions
Calculate the pH of a 0.50 M aqueous solution of the weak acid HF.
(Ka = 7.2 × 10–4)
Copyright © Cengage Learning. All rights reserved 52
EXERCISE!
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?
Copyright © Cengage Learning. All rights reserved 53
Section 14.5 Calculating the pH of Weak Acid Solutions
Let’s Think About It…
What are the possibilities for the dominant reaction?
HF(aq) + H2O(l) H3O+(aq) + F–(aq)
Ka=7.2 × 10-4
H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Kw=1.0 × 10-14
Which reaction controls the pH? Why?
Copyright © Cengage Learning. All rights reserved 54
Section 14.5 Calculating the pH of Weak Acid Solutions
HF(aq) + H2O H3O+(aq) + F–(aq)
Initial 0.50 M ~ 0 ~ 0
Change –x +x +x
Equilibrium 0.50–x x x
Steps Toward Solving for pH
Ka = 7.2 × 10–4
pH = 1.72
Copyright © Cengage Learning. All rights reserved 55
Section 14.5 Calculating the pH of Weak Acid Solutions
Percent Dissociation (Ionization)
For a given weak acid, the percent dissociation increases as the acid becomes more dilute.
Copyright © Cengage Learning. All rights reserved 56
amount dissociated (mol/L)Percent dissociation = 100%
initial concentration (mol/L)
Section 14.5 Calculating the pH of Weak Acid Solutions
A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water.
Calculate the Ka value for formic acid.
Ka = 1.8 × 10–4
Copyright © Cengage Learning. All rights reserved 57
EXERCISE!
Section 14.5 Calculating the pH of Weak Acid Solutions
A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water. Calculate the Ka value for formic acid.
Ka = 1.8 × 10–4
Copyright © Cengage Learning. All rights reserved 58
EXERCISE!
If 8.00 M of the acid is 0.47% ionized, then 0.038 M
dissociates.
HCHO2(aq) + H2O H3O+(aq) + CHO2
-(aq)
I 8.00 0 0
C -0.038 +0.038 +0.038
E 7.96 0.038 0.038
Section 14.5 Calculating the pH of Weak Acid Solutions
Calculate the pH of an 8.00 M solution of formic acid. Use the data from the previous slide to help you solve this problem.
pH = 1.42
Copyright © Cengage Learning. All rights reserved 59
EXERCISE!
pH = –log[H+] = –log[0.038 M] = 1.42
Major species: HCHO2, H2O
Dominant reaction: HCHO2(aq) + H2O H3O+(aq) + CHO2-(aq)
Section 14.5 Calculating the pH of Weak Acid Solutions
The value of Ka for a 4.00 M formic acid solution should be:
higher than lower than the same as
the value of Ka of an 8.00 M formic acid solution.
Explain.
Copyright © Cengage Learning. All rights reserved 60
EXERCISE!
Section 14.5 Calculating the pH of Weak Acid Solutions
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.
Copyright © Cengage Learning. All rights reserved 61
CONCEPT CHECK!
Section 14.5 Calculating the pH of Weak Acid Solutions
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.
Copyright © Cengage Learning. All rights reserved 62
CONCEPT CHECK!
To make 4.00 M solution from 8.00 M solution, we must add water (dilute). With a
larger volume, Le Chatelier's principle tells us that equilibrium will shift to the right.
Thus, percent ionization will increase. However, this increase if offset by the
dilution factor, and the solution has an overall lower concentration of H+.
Section 14.5 Calculating the pH of Weak Acid Solutions
The pH of a 4.00 M formic acid solution should be:
higher than lower than the same as
the pH of an 8.00 M formic acid solution.
Explain.
Copyright © Cengage Learning. All rights reserved 63
CONCEPT CHECK!
Section 14.5 Calculating the pH of Weak Acid Solutions
The pH of a 4.00 M formic acid solution should be:
higher than lower than the same as
the pH of an 8.00 M formic acid solution.
Explain.
Copyright © Cengage Learning. All rights reserved 64
CONCEPT CHECK!
They should also understand that a lower concentration of the same weak acid will be
less acidic (and thus the pH will be higher).
Section 14.5 Calculating the pH of Weak Acid Solutions
Calculate the percent ionization of a 4.00 M formic acid solution in water.
% Ionization = 0.67%
Copyright © Cengage Learning. All rights reserved 65
EXERCISE!
Section 14.5 Calculating the pH of Weak Acid Solutions
Calculate the percent ionization of a 4.00 M formic acid solution in water.
% Ionization = 0.67%
Copyright © Cengage Learning. All rights reserved 66
EXERCISE!
% dissociation = 0.67%
Major species: HCHO2, H2O
Dominant reaction: HCHO2(aq) + H2O H3O+(aq) + CHO2-(aq)
HCHO2(aq) + H2O H3O+(aq) + CHO2-(aq)
I 4.00 0 0
C -x +x +x
E 4.00-x x x
Ka = 1.8 x 10-4 = (x)2/(4.00-x)
x = 0.027; (0.027/4.00) x 100 = 0.67%
Section 14.5 Calculating the pH of Weak Acid Solutions
Calculate the pH of a 4.00 M solution of formic acid.
pH = 1.57
Copyright © Cengage Learning. All rights reserved 67
EXERCISE!
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
Copyright © Cengage Learning. All rights reserved 68
Section 14.6 Bases
Calculate the pH of a 1.0 × 10–3 M solution of sodium hydroxide.
pH = 11.00
Copyright © Cengage Learning. All rights reserved 69
CONCEPT CHECK!
Section 14.6 Bases
Calculate the pH of a 1.0 × 10–3 M solution of calcium hydroxide.
pH = 11.30
Copyright © Cengage Learning. All rights reserved 70
CONCEPT CHECK!
Section 14.6 Bases
Equilibrium expression for weak bases uses Kb.
CN–(aq) + H2O(l) HCN(aq) + OH–(aq)
Copyright © Cengage Learning. All rights reserved 71
b
HCN OH =
CN
K
Section 14.6 Bases
pH calculations for solutions of weak bases are very similar to those for weak acids.
Kw = [H+][OH–] = 1.0 × 10–14
pOH = –log[OH–]
pH = 14.00 – pOH
Copyright © Cengage Learning. All rights reserved 72
Section 14.6 Bases
Calculate the pH of a 2.0 M solution of ammonia (NH3).
(Kb = 1.8 × 10–5)
pH = 11.78
Copyright © Cengage Learning. All rights reserved 73
CONCEPT CHECK!
Section 14.7 Polyprotic Acids
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:
Ka1 > Ka2 > Ka3
For a typical polyprotic acid in water, only the first dissociation step is important to pH.
Copyright © Cengage Learning. All rights reserved 74
Section 14.7 Polyprotic Acids
Calculate the pH of a 1.00 M solution of H3PO4.
Ka1 = 7.5 × 10-3
Ka2 = 6.2 × 10-8
Ka3 = 4.8 × 10-13
pH = 1.08
Copyright © Cengage Learning. All rights reserved 75
EXERCISE!
Section 14.7 Polyprotic Acids
Calculate the equilibrium concentration of PO43- in a
1.00 M solution of H3PO4.
Ka1 = 7.5 × 10-3
Ka2 = 6.2 × 10-8
Ka3 = 4.8 × 10-13
[PO43-] = 3.6 × 10-19 M
Copyright © Cengage Learning. All rights reserved 76
CONCEPT CHECK!
Section 14.8 Acid-Base Properties of Salts
Salts
Ionic compounds.
When dissolved in water, break up into its ions (which can behave as acids or bases).
Copyright © Cengage Learning. All rights reserved 77
Section 14.8 Acid-Base Properties of Salts
Salts
The salt of a strong acid and a strong base gives a neutral solution.
KCl, NaNO3
Copyright © Cengage Learning. All rights reserved 78
Section 14.8 Acid-Base Properties of Salts
Salts
A basic solution is formed if the anion of the salt is the conjugate base of a weak acid.
NaF, KC2H3O2
Kw = Ka × Kb
Use Kb when starting with base.
Copyright © Cengage Learning. All rights reserved 79
Section 14.8 Acid-Base Properties of Salts
Salts
An acidic solution is formed if the cation of the salt is the conjugate acid of a weak base.
NH4Cl
Kw = Ka × Kb
Use Ka when starting with acid.
Copyright © Cengage Learning. All rights reserved 80
Section 14.8 Acid-Base Properties of Salts
Copyright © Cengage Learning. All rights reserved 81
Cation
Anion
Acidic or Basic
Example
neutral neutral neutral NaCl
neutral conjugate base of
weak acid
basic NaF
conjugate acid of
weak base
neutral acidic NH4Cl
conjugate
acid of weak base
conjugate
base of weak acid
depends
on Ka & Kb values
Al2(SO4)3
Section 14.8 Acid-Base Properties of Salts
Qualitative Prediction of pH of Salt Solutions (from Weak Parents)
Copyright © Cengage Learning. All rights reserved 82
Section 14.8 Acid-Base Properties of Salts
HC2H3O2 Ka = 1.8 × 10-5
HCN Ka = 6.2 × 10-10
Calculate the Kb values for: C2H3O2− and CN−
Kb (C2H3O2-) = 5.6 × 10-10
Kb (CN-) = 1.6 × 10-5
Copyright © Cengage Learning. All rights reserved
EXERCISE!
Section 14.8 Acid-Base Properties of Salts
HC2H3O2 Ka = 1.8 × 10-5
HCN Ka = 6.2 × 10-10
Calculate the Kb values for: C2H3O2− and CN−
Kb (C2H3O2-) = 5.6 × 10-10
Kb (CN-) = 1.6 × 10-5
Copyright © Cengage Learning. All rights reserved
EXERCISE!
Kb (C2H3O2-) = 5.6 x 10-10
Kb (CN-) = 1.6 x 10-5
Section 14.8 Acid-Base Properties of Salts
Arrange the following 1.0 M solutions from lowest to highest pH.
HBr NaOH NH4Cl
NaCN NH3 HCN
NaCl HF
Justify your answer.
HBr, HF, HCN, NH4Cl, NaCl, NaCN, NH3, NaOH
Copyright © Cengage Learning. All rights reserved 85
CONCEPT CHECK!
Section 14.8 Acid-Base Properties of Salts
Arrange the following 1.0 M solutions from lowest to highest pH.
HBr NaOH NH4Cl
NaCN NH3 HCN
NaCl HF
Justify your answer.
HBr, HF, HCN, NH4Cl, NaCl, NaCN, NH3, NaOH
Copyright © Cengage Learning. All rights reserved 86
CONCEPT CHECK!
The order is: HBr (strong acid), HF (Ka = 7.2 x 10-4), HCN (Ka = 6.2
x 10-10), NH4Cl (Ka = 5.6 x 10-10), NaCl (neutral), NaCN (Kb = 1.6 x
10-5), NH3 (Kb = 1.8 x 10-5), NaOH (strong base). Have the students
use the Ka and Kb values to decide. They need not calculate the pH
values to answer this question.
Section 14.8 Acid-Base Properties of Salts
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
Copyright © Cengage Learning. All rights reserved 87
CONCEPT CHECK!
Section 14.8 Acid-Base Properties of Salts
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
Copyright © Cengage Learning. All rights reserved 88
CONCEPT CHECK!
Major Species: Na+, F-, H2O
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?
Copyright © Cengage Learning. All rights reserved 89
Section 14.8 Acid-Base Properties of Salts
Let’s Think About It…
The possibilities for the dominant reactions are:
1. F–(aq) + H2O(l) HF(aq) + OH–(aq)
2. H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
3. Na+(aq) + H2O(l) NaOH + H+(aq)
4. Na+(aq) + F–(aq) NaF
Copyright © Cengage Learning. All rights reserved 90
Section 14.8 Acid-Base Properties of Salts
Let’s Think About It…
How do we decide which reaction controls the pH?
F–(aq) + H2O(l) HF(aq) + OH–(aq)
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
Determine the equilibrium constant for each reaction.
Copyright © Cengage Learning. All rights reserved 91
Section 14.8 Acid-Base Properties of Salts
Let’s Think About It…
How do we decide which reaction controls the pH?
F–(aq) + H2O(l) HF(aq) + OH–(aq)
H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
Determine the equilibrium constant for each reaction.
Copyright © Cengage Learning. All rights reserved 92
The primary reaction is the first one, and the Kb value is 1.39 x 10-11.
This is over 1000 times larger than the K value for the second reaction
(Kw).
Section 14.8 Acid-Base Properties of Salts
Calculate the pH of a 0.75 M aqueous solution of NaCN.
Ka for HCN is 6.2 × 10–10.
Copyright © Cengage Learning. All rights reserved 93
EXERCISE!
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?
Copyright © Cengage Learning. All rights reserved 94
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?
Copyright © Cengage Learning. All rights reserved 95
Major Species: Na+, CN-, H2O
Possibilities:
CN-(aq) + H2O HCN(aq) + OH-(aq)
H2O + H2O H3O+(aq) + OH-(aq)
The primary reaction is the first one, and the Kb value is 1.6 x 10-5. This is many times larger than the K
value for the second reaction (Kw).
pH = 11.54
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:
1. CN–(aq) + H2O(l) HCN(aq) + OH–(aq)
2. H2O(l) + H2O(l) H3O+(aq) + OH–(aq)
3. Na+(aq) + H2O(l) NaOH + H+(aq)
4. Na+(aq) + CN–(aq) NaCN
Which of these reactions really occur?
Copyright © Cengage Learning. All rights reserved 96
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)
Copyright © Cengage Learning. All rights reserved 97
Section 14.8 Acid-Base Properties of Salts
Steps Toward Solving for pH
Kb = 1.6 × 10–5
pH = 11.54
CN–(aq) + H2O HCN(aq) + OH–(aq)
Initial 0.75 M 0 ~ 0
Change –x +x +x
Equilibrium 0.75–x x x
Section 14.9 The Effect of Structure on Acid-Base Properties
Models of Acids and Bases
Two factors for acidity in binary compounds:
Bond Polarity (high is good)
Bond Strength (low is good)
Copyright © Cengage Learning. All rights reserved 99
Section 14.9 The Effect of Structure on Acid-Base Properties
Bond Strengths and Acid Strengths for Hydrogen Halides
Copyright © Cengage Learning. All rights reserved 100
Section 14.9 The Effect of Structure on Acid-Base Properties
Oxyacids
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.
Copyright © Cengage Learning. All rights reserved 101
Section 14.9 The Effect of Structure on Acid-Base Properties
Several Series of Oxyacids and Their Ka Values
Copyright © Cengage Learning. All rights reserved 102
Section 14.9 The Effect of Structure on Acid-Base Properties
Comparison of Electronegativity of X and Ka Value
Copyright © Cengage Learning. All rights reserved 103
Section 14.10 Acid-Base Properties of Oxides
Oxides
Acidic Oxides (Acid Anhydrides):
O—X bond is strong and covalent.
SO2, NO2, CO2
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.
Copyright © Cengage Learning. All rights reserved 104
Section 14.10 Acid-Base Properties of Oxides
Oxides
Basic Oxides (Basic Anhydrides):
O—X bond is ionic.
K2O, 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.
Copyright © Cengage Learning. All rights reserved 105
Section 14.10 Acid-Base Properties of Oxides
Section 14.11 The Lewis Acid-Base Model
Lewis Acids and Bases
Lewis acid: electron pair acceptor
Lewis base: electron pair donor
Lewis acid Lewis base
Copyright © Cengage Learning. All rights reserved 107
OH
HO
H
H
AlAl3+ + 6
6
3+
Section 14.11 The Lewis Acid-Base Model
Three Models for Acids and Bases
Copyright © Cengage Learning. All rights reserved 108
Focus of AP Exam
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.
Copyright © Cengage Learning. All rights reserved 109