2. 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
Copyright © Cengage Learning. All rights reserved

3. Brønsted–Lowry Reaction
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
Copyright © Cengage Learning. All rights reserved

4. 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.
Copyright © Cengage Learning. All rights reserved

5. Acid Ionization Equilibrium
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
Copyright © Cengage Learning. All rights reserved

6. Ionization equilibrium lies far to the right.
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

7. Copyright © Cengage Learning. All rights reserved

8. Various Ways to Describe Acid Strength
Copyright © Cengage Learning. All rights reserved

9. 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

10. Three Possible Situations
[H+] = [OH–]; neutral solution
[H+] > [OH–]; acidic solution
[OH–] > [H+]; basic solution
Copyright © Cengage Learning. All rights reserved

11. Self-Ionization of Water
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
Copyright © Cengage Learning. All rights reserved

12. HA(aq) + H2O(l) H3O+(aq) + A-(aq)
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?
Use the following equation to write K:
HA(aq) + H2O  H3O+(aq) + A-(aq)
Although you can also write this as HA(aq)  H+(aq) + A-(aq), have the students use the form above until they are used to the fact that water is acting like a base.
K = [H3O+][A-] / HA
Copyright © Cengage Learning. All rights reserved

13. CONCEPT CHECK!
If the equilibrium lies to the right, the value for Ka is __________. If the equilibrium lies to the left, the value for Ka is ___________.
Large (or >1); small (or < 1)
The students should know this from Chapter 13, but it is worth re-emphasizing.
Copyright © Cengage Learning. All rights reserved

14. 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?
products; strong; greater than 1
One of the advantages in having students write out the weak acid reaction with water is that we can make comparisons between the relative strengths of the conjugate base and water acting as a base.
Copyright © Cengage Learning. All rights reserved

15. 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)
The bases from strongest to weakest are: A-, H2O, Cl-. Some students will try to claim that the Kb value for water is 1.0 x (or some will say 1.0 x 10-7) if they’ve read ahead in the chapter. Water does not have a Kb value; we must use reactions (like in the first Concept Check problem) to think about relative base strengths.
Copyright © Cengage Learning. All rights reserved

16. 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

17. Consider a solution of NaA where A– is the anion from weak acid HA:
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
Which way will equilibrium lie?
left
left
Copyright © Cengage Learning. All rights reserved

18. 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
less than 1
Copyright © Cengage Learning. All rights reserved

19. Consider a solution of NaA where A– is the anion from weak acid HA:
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
c) Does this mean A– is a strong or weak base?
weak base
weak base;
Students often try to memorize relationships. An easy (and incorrect one) is “The conjugate base of a weak acid is a strong base.” The terms “strong” and “weak” are relative terms. We reserve the term “strong base” for hydroxides.
Copyright © Cengage Learning. All rights reserved

20. 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 Cl– CN– C2H3O2–
Copyright © Cengage Learning. All rights reserved

21. 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–
Weakest to strongest: Cl-, H2O, C2H3O2-, CN-.
If HCN is weaker than HC2H3O2, the conjugate base will be stronger. Have the students see this by looking at the general equation
HA(aq) + H2O  H3O+(aq) + A-(aq)
They can also use this to see that water is a better base than the chloride ion.
Copyright © Cengage Learning. All rights reserved

22. 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.
The value for K is less than 1.
This reaction is neither a Kb nor a Ka reaction. To think about it, we need to know which is the stronger acid, HCN or HF. This can be decided by knowing the Ka values. HF is stronger, which means that it is more likely to donate a proton. Also, F- must be a weaker base than CN-. Thus, CN- has a greater affinity for a proton than F-. Because of all of this, equilibrium lies on the left, so K is less than 1.
Copyright © Cengage Learning. All rights reserved

23. 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 x 10-7
K = Ka(HCN)/Ka(HF) = [6.2 x 10-10]/[7.2 x 10-4]
The students should know how to manipulate the Ka expressions to derive this.
Copyright © Cengage Learning. All rights reserved

24. pH changes by 1 for every power of 10 change in [H+].
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

25. 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

26. The pH Scale and pH Values of Some Common Substances
Copyright © Cengage Learning. All rights reserved

27. Calculate the pH for each of the following solutions. 1.0 × 10–4 M H+
EXERCISE!
Calculate the pH for each of the following solutions.
1.0 × 10–4 M H+
b) M OH–
pH = –log[H+]
a) pH = –log[H+] = –log(1.0 × 10–4 M) = 4.00
b) Kw = [H+][OH–] = 1.00 × 10–14 = [H+](0.040 M) = 2.5 × 10–13 M H+
pH = –log[H+] = –log(2.5 × 10–13 M) = 12.60
Copyright © Cengage Learning. All rights reserved

28. The pH of a solution is 5.85. What is the [H+] for this solution?
EXERCISE!
The pH of a solution is What is the [H+] for this solution?
[H+] = 10^–5.85 = 1.4 × 10–6 M
Copyright © Cengage Learning. All rights reserved

29. 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

30. Calculate the pOH for each of the following solutions. 1.0 × 10–4 M H+
EXERCISE!
Calculate the pOH for each of the following solutions.
1.0 × 10–4 M H+
b) M OH–
pH = –log[H+] and pOH = –log[OH–] and = pH + pOH
a) pH = –log[H+] = –log(1.0 × 10–4 M) = 4.00; So, = pOH; pOH = 10.00
b) pOH = –log[OH–] = –log(0.040 M) = 1.40
Copyright © Cengage Learning. All rights reserved

31. The pH of a solution is 5.85. What is the [OH–] for this solution?
EXERCISE!
The pH of a solution is What is the [OH–] for this solution?
pH = –log[H+] and pOH = –log[OH–] and = pH + pOH
14.00 = pOH; pOH = 8.15
[OH–] = 10^–8.15 = 7.1 × 10–9 M
Copyright © Cengage Learning. All rights reserved

32. 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

33. CONCEPT CHECK!
Consider an aqueous solution of 2.0 × 10–3 M HCl. What are the major species in solution? What is the pH?
Major Species: H+, Cl-, H2O
pH = –log[H+] = –log(2.0 × 10–3 M) = 2.70
pH = 2.70
Copyright © Cengage Learning. All rights reserved

34. Calculate the pH of a 1.5 × 10–11 M solution of HCl.
CONCEPT CHECK!
Calculate the pH of a 1.5 × 10–11 M solution of HCl.
pH = 7.00
Watch for student to say pH = -log(1.5 x 10-11) =
This answers neglects that water is a major species and it turns out to be the major contributor of H+. If students think about this problem and consider major species, they will answer this correctly.
Major species are H+, Cl-, and H2O.
Dominant reaction is: H2O + H2O  H3O+ (aq) + OH- (aq)
Water is a major contributor. Thus , pH = 7.00.
Copyright © Cengage Learning. All rights reserved

35. 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

36. Let’s Think About It… Why is this reaction not likely?
NO3–(aq) + H2O(l) HNO3(aq) + OH–(aq)
Because the K value for HNO3 is very large and HNO3 virtually completely dissociates in solution.
Copyright © Cengage Learning. All rights reserved

37. 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

38. Solving Weak Acid Equilibrium Problems
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.
Copyright © Cengage Learning. All rights reserved

39. 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.
Copyright © Cengage Learning. All rights reserved

40. Solving Weak Acid Equilibrium Problems
Solve for x the “easy” way, that is, by assuming that [HA]0 – x about equals [HA]0.
Use the 5% rule to verify whether the approximation is valid.
Calculate [H+] and pH.
Copyright © Cengage Learning. All rights reserved

41. What are the major species in solution? HCN, H2O
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
Major Species: HCN, H2O. This is because HCN is a weak acid. Thus, while H+ and CN- are in solution, the concentrations are very small.
Primary reaction: HCN(aq) + H2O  H3O+(aq) + CN-(aq)
The primary reaction is chosen because the K value is significantly larger. Thus, it is the primary source of H3O+ (or H+).
Copyright © Cengage Learning. All rights reserved

42. Let’s Think About It… Why aren’t H+ or CN– major species?
See notes on slide 41.
Copyright © Cengage Learning. All rights reserved

43. 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.
See notes on slide 1.
Copyright © Cengage Learning. All rights reserved

44. Calculate the pH of a 0.50 M aqueous solution of the weak acid HF.
EXERCISE!
Calculate the pH of a 0.50 M aqueous solution of the weak acid HF.
(Ka = 7.2 × 10–4)
Major Species: HF, H2O
Possibilities for primary reaction:
HF(aq) + H2O  H3O+(aq) + F-(aq)
H2O + H2O  H3O+(aq) + OH-(aq)
The first reaction is the primary reaction because the K value is significantly larger. Thus, it is the primary source of H3O+ (or H+).
Use an ICE table, K expression, and pH = –log[H+] to solve for pH.
pH = 1.72
Copyright © Cengage Learning. All rights reserved

45. Let’s Think About It… What are the major species in solution? HF, H2O
Why aren’t H+ and F– major species?
See Slide 4.
Copyright © Cengage Learning. All rights reserved

46. HF(aq) + H2O(l) H3O+(aq) + F–(aq)
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?
See Slide 4.
Copyright © Cengage Learning. All rights reserved

47. Steps Toward Solving for pH
HF(aq) +
H2O
H3O+(aq) +
F–(aq)
Initial
0.50 M
~ 0
Change –x +x
Equilibrium
0.50–x x
See Slide 4.
Ka = 7.2 × 10–4 pH = 1.72
Copyright © Cengage Learning. All rights reserved

48. Percent Dissociation (Ionization)
For a given weak acid, the percent dissociation increases as the acid becomes more dilute.
Copyright © Cengage Learning. All rights reserved

49. A solution of 8.00 M formic acid (HCHO2) is 0.47% ionized in water.
EXERCISE!
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 x 10-4
If 8.00 M of the acid is 0.47% ionized, then M dissociates.
HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) I C E
Copyright © Cengage Learning. All rights reserved

50. EXERCISE!
Calculate the percent ionization of a 4.00 M formic acid solution in water.
% dissociation = 0.67%
Major species: HCHO2, H2O
Dominant reaction: HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq)
HCHO2(aq) + H2O  H3O+(aq) + CHO2-(aq) I
C -x x x
E 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%
Copyright © Cengage Learning. All rights reserved

51. 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