1. Chapter 16 Acids and Bases
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 16 Acids and Bases
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Acid-Base Definitions
Arrhenius (oldest, very narrow definition)
Acid: Substance that, when dissolved in water, increases the concentration of hydrogen (H+ or ions.
Base: Substance that, when dissolved in water, increases the concentration of hydroxide (OH-) ions.
Brønsted–Lowry
Acid: Proton (H+) donor
Base: Proton (H+) acceptor (water is the acid in this case, leaving OH- behind)
acids produce H+ ions in water The stronger the acid, the higher its concentration (molarity) of H+
bases produce OH- ions in water. The stronger the base, the higher it’s concentration (molarity) of OH-

3. Strong Acids and bases
You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.
Strong bases are the soluble hydroxides, which are the Group 1 and soluble group 2 hydroxides (Ca2+, Sr2+, and Ba2+).
These are, by definition, strong electrolytes and exist 100% as ions in aqueous solution. NO Equilibrum!!
For the monoprotic (all but H2SO4) strong acids, [H3O+] = [acid].
For alkali metal hydroxides (e.g. NaOH),
[OH-] = [base].
Remember that the alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+) give 2 OH- ions per unit of substance. So [OH-] = 2*[base].

4. Weak Acids and Bases All acids and bases that are not strong are weak.
In water there is an equilibrium between the molecular form and the ionized form of weak acids and bases.
GENERIC EXAMPLE for any weak acid HAA: HAA (aq) + H2O (l) H3O+ (aq) + AA- (aq)
GENERIC EXAMPLE for any weak base BB: BB (aq) + H2O (l) HO- (aq) + BBH+ (aq)
What is the equilibrium for the weak acid HNO2? For the weak base NH3?

5. Conjugate Acids and Bases:
Reactions between acids and bases always yield their conjugate bases and acids.
Notice that water can be an acid or base depending on what is added to it!

6. Acid and Base Strength
The strength of an acid is inversely related to the strength of its conjugate base
Strong acids are completely dissociated in water. HCl
Their conjugate bases are quite weak. Cl-
Weak acids only dissociate partially in water. HC2H3O2
Their conjugate bases are weak bases. C2H3O2-
Substances with no acidity do not give any hydrogen ions in water. CH4
Their conjugate bases are exceedingly strong. CH3-
What is the conjugate acid of HSO4-? What is the conjugate acid of H2SO4? What is the conjugate base of HSO4-? What is the conjugate acid of H2SO4?

7. What Happens When an Acid Dissolves in Water?
HCl(aq) + H2O(l)  H3O+(aq) + Cl−(aq)
Water acts as a Brønsted–Lowry base and abstracts (removes) a proton (H+) from the acid.
As a result, the conjugate base of the acid and a hydronium ion H3O+ are formed.

8. Chart of acids and bases p.674
You can use these examples to try to figure out if another substance is an acid, base or neither.
What do you think Br- is? C2H6? H2Se?

9. What is the conjugate base of HBr? Of H2O? Of H2SO4?
What is the conjugate acid of HS-? Of H2O? Of NH3?
Which is a stronger acid, HNO3 or HF?
Which is a stronger base, NO3- or F-?

10. What is the conjugate base of HBr? Of H2O? Of H2SO4? Br- HO- HSO4-
What is the conjugate acid of HS-? Of H2O? Of NH3? H2S H3O+NH4 +
Which is a stronger acid, HNO3 or HF?
Which is a stronger base, NO3- or F-?

11. A substance which can either accept or donate a proton (be an acid or a base) depending on what it is reacting with is amphiprotic EXAMPLES: HCO3−,HSO4− H2O Write a reaction where HCO3− reacts with a weak base. Write one where HCO3− reacts with an acid. Then determine whether each reaction favors reactants or products (HINT: The reaction always favors the side that has a weaker acid and base, so use the chart on p.674 to figure it out)

12. Autoionization of Water
water is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.
This is referred to as autoionization.
The equilibrium expression for this process is
Kc = [H3O+] [OH−]
This special equilibrium constant is referred to as the ion-product constant for water, Kw.
At 25C, Kw = 1.0  10−14
What are [H3O+] and [OH−] in pure water at 25C?
H2O(l) + H2O(l)
H3O+(aq) + OH−(aq)

13. Because in pure water [H3O+] = [OH−],pH
Kw = [H3O+] [OH−] = 1.0  10−14
Because in pure water [H3O+] = [OH−],
[H3O+] = (1.0  10−14)1/2 = 1.0  10−7
Kw is a CONSTANT! So Kw always has a value of 1.0  10−14. This means if [H3O+] increases (as in an acidic solution), what must happen to [OH−]? How about if [OH−] increases (as in an basic solution)?

14. pH
pH is defined as the negative base-10 logarithm of the hydronium ion concentration. pH = −log [H3O+]
What happens to pH as [H3O+] gets bigger? What happens to pH as gets [H3O+] smaller?
Turn to figure 16.5 and check it out!

15. pH
As [H3O+] rises, pH decreases.
As [H3O+] decreases, pH increases

16. pH Therefore, in pure water, pH = −log (1.0  10−7) = 7.00
An acid has a higher [H3O+] than pure water, so its pH is <7
A base has a lower [H3O+] than pure water, so its pH is >7.

17. −log [H3O+] + −log [OH−] = −log Kw = 14.00
Watch This!
Because
[H3O+] [OH−] = Kw = 1.0  10−14,
we know that [H3O+] and [OH−] are inversely related! One goes up the other goes down!
Mathmetically,
−log [H3O+] + −log [OH−] = −log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00

18. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25°C. Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is M,
(b) a solution in which [OH–] is 1.8  10–9 M.
Calculate the pH values for the two solutions described in Sample Exercise 16.5.

19. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25°C. Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is M,
(b) a solution in which [OH–] is 1.8  10–9 M.
Calculate the pH values for the two solutions described in Sample Exercise 16.5.

20. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25°C. Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is M,
(b) a solution in which [OH–] is 1.8  10–9 M.
Calculate the pH values for the two solutions described in Sample Exercise 16.5.

21. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25°C. Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is M,
(b) a solution in which [OH–] is 1.8  10–9 M.
Calculate the pH values for the two solutions described in Sample Exercise 16.5.
(a) pH = –log(1.0  10–12 ) = –(–12.00) = 12.00
(b) pH = –log (5.6  10–6 ) = 5.25 The rule for using significant figures with logs is that the number of decimal places in the log equals the number of significant figures in the original number

22. Other “p” Scales
The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).
Some similar examples are
pOH −log [OH−]
pKw −log Kw

23. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25°C. Calculate the concentration of H+ (aq) in (a) a solution in which [OH–] is M,
(b) a solution in which [OH–] is 1.8  10–9 M.
Calculate the pH values for the two solutions described in Sample Exercise 16.5.
(a) pH = –log(1.0  10–12 ) = –(–12.00) = 12.00
(b) pH = –log (5.6  10–6 ) = 5.25 The rule for using significant figures with logs is that the number of decimal places in the log equals the number of significant figures in the original number

24. Strong acid or base pH
1. What is the pH of a M solution of HClO4? 2. What is the pH of (a) a M solution of NaOH (b) a M solution of Ca(OH)2?

25. What is the pH of a 0. 040 M solution of HClO4
What is the pH of a M solution of HClO4? Because HClO4 is a strong acid, it is completely ionized, giving [H+] = [ClO4–] = M. pH = –log(0.040) = Because [H+] lies between benchmarks 1  10–2 and 1  10–1 in Figure 16.5, we estimate that the pH will be between 2.0 and 1.0 so this makes sense!

26. (a) a M solution of NaOH NaOH dissociates in water to give one OH– ion per formula unit. So the OH– concentration for the solution in (a) equals M.

27. What is the pH of (b) a 0. 0011 M solution of Ca(OH)2
What is the pH of (b) a M solution of Ca(OH)2? Ca(OH)2 is a strong base that dissociates in water to give two OH– ions per formula unit. Thus, the concentration of OH–for the solution is 2  (0.0011M) = M.

28. How Do We Measure pH?
For more accurate measurements, use a pH meter, which measures the voltage in the solution.
For less accurate measurements,
Litmus paper “Red” paper turns blue above ~pH = 8 Blue” paper turns red below ~pH = 5
An indicator like phenolphthalein

29. Dissociation Constants
For any weak acid dissociation,
the equilibrium expression is
This equilibrium constant for an acid is known as the acid-dissociation constant, Ka.
A larger value of Ka indicates a stronger acid (more product, H3O+)
What is the strongest acid in table 16.2?
HA(aq) + H2O(l)
A−(aq) + H3O+(aq)
[H3O+] [A−]
[HA]
Kc =

30. Dissociation Constants
Table 16.2

31. Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is Calculate Ka for formic acid at this temperature.
We know that
[H3O+] [HCOO−]
[HCOOH]
Ka =

32. Calculating Ka from pH Set up an ICE table… [HCOOH], M [H3O+], M
[HCOO−], M
Initially
0.10
Change
At Equilibrium

33. Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38.
We can find [H3O+] at equilibrium, which is the same as [HCOO−], from the pH.
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = 10log [H3O+] = [H3O+]
4.2  10−3 = [H3O+] = [HCOO−]

34. Calculating Ka from pH Now we can fill in our table… [HCOOH], M
[H3O+], M
[HCOO−], M
Initially
0.10
Change
−4.2  10-3
+4.2  10-3
+4.2  10−3
At Equilibrium
0.10 − 4.2  10−3
= = 0.10
4.2  10−3

35. Calculating Ka from pH [4.2  10−3] [4.2  10−3] Ka = [0.10]
= 1.8  10−4

36. Calculating Percent Ionization
[H3O+]eq
[HA]initial
Percent Ionization =  100

37. Calculating Percent Ionization
[H3O+]eq
[HA]initial
Percent Ionization =  100
In this example
[H3O+]eq = 4.2  10−3 M
[HCOOH]initial = 0.10 M

38. Calculating Percent Ionization
4.2  10−3
0.10
Percent Ionization =  100
= 4.2%

39. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25°C.
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Ka for acetic acid at 25°C is 1.8  10−5.

40. Calculating pH from Ka The equilibrium constant expression is
[H3O+] [C2H3O2−]
[HC2H3O2]
Ka =

41. Calculating pH from Ka We next set up a table… [HC2H3O2], M [H3O+], M
Initially
0.30
Change
At Equilibrium

42. Calculating pH from Ka [HC2H3O2], M [H3O+], M [C2H3O2−], M Initially
0.30
Change −x +x
At Equilibrium
0.30 − x  0.30 x
VERY IMPORTANT NOTE: If and only if the initial concentration of acid (0.30M in this example) is more than 100 times larger than Ka, ( in this example) then we can assume x will be very small compared to the initial concentration and can, therefore, be ignored.
This is why the “E” for HC2H3O2 is 0.30 − x  This is really helpful because it simplifies the algebra to solve for x

44. Calculating pH from Ka (x)2 (0.30) Ka =1.8  10−5 =
pH = −log [H3O+]
pH = −log (2.3  10−3)
pH = 2.64

45. Weak Bases [HB+] [OH−] Kb = [B]
Bases react with water to produce hydroxide ion.
The equilibrium constant expression for this reaction is
[HB+] [OH−]
[B]
Kb =
where Kb is the base-dissociation constant

46. Weak Bases
Kb can be used to find [OH−] and, through it, pH.

47. pH of Basic Solutions
Kb can be used to find [OH−] and, through it, pH.
What is the pH of a 0.15 M solution of NH3?
NH3(aq) + H2O(l)
NH4+(aq) + OH−(aq)
[NH4+] [OH−]
[NH3]
Kb =
= 1.8  10−5

48. pH of Basic Solutions Tabulate the data. [NH3], M [NH4+], M [OH−], M
Initially
Change
At Equilibrium

49. pH of Basic Solutions Tabulate the data. [NH3], M [NH4+], M [OH−], M
Initially
0.15
Change -x +x
At Equilibrium
x  0.15 x

50. pH of Basic Solutions [NH4+] [OH−] [NH3] Kb = = 1.8  10−5 (x)2 (0.15)
1.8  10−5 =

51. pH of Basic Solutions [NH4+] [OH−] [NH3] Kb = = 1.8  10−5 (x)2 (0.15)
1.8  10−5 =
(1.8  10−5) (0.15) = x2
2.7  10−6 = x2
1.6  10−3 = x2

52. pH of Basic Solutions [NH4+] [OH−] [NH3] Kb = = 1.8  10−5 (x)2 (0.15)
From our ICE table we know that x = [OH-] at equilibrium
Therefore,
[OH−] = 1.6  10−3 M
pOH = −log (1.6  10−3)
pOH = 2.80
pH = − 2.80
pH = 11.20
(x)2
(0.15)
1.8  10−5 =
(1.8  10−5) (0.15) = x2
2.7  10−6 = x2
1.6  10−3 = x2

53. Ka and Kb
Ka and Kb for a acid/conjugate base (or base/conjugate acid) pair are related in this way:
Ka  Kb = Kw
Therefore, if you know one of them, you can calculate the other.
For ethylamine, C2H5NH2, is Kb = 4.3 X 10-4 What is the conjugate acid? What is the Ka expression for the conjugate acid? What is the value of Ka for the conjugate acid?

54. Ka and Kb For ethylamine, C2H5NH2, is Kb = 6.4 X10-4.
What is the conjugate acid? C2H5NH3+
What is the Ka expression for the conjugate acid?
What is the value of Ka for the conjugate acid? Ka  Kb = Kw therefore Ka= Kw/Kb
Ka= 1.0 X / 4.3 X10-4 = 2.3 X
[H3O+] [C2H5NH2]
[C2H5NH3+ ]
Ka =

55. Ka and Kb
The stronger the acid (bigger Ka ) the weaker its conjugate base is (smaller Kb ).
What’s the strongest of the weak acids in the table? What’s the strongest of the weak bases?

56. Effect of Cations and Anions
Dissolving an ionic compound in water can raise or lower the pH, depending on the identity of the cation and anion in the compound.
An anion that is the conjugate base of a strong acid will not affect the pH.
An anion that is the conjugate base of a weak acid will increase the pH.
A cation that is the conjugate acid of a weak base will decrease the pH.

57. Effect of Cations and Anions
Cations of the strong Arrhenius bases will not affect the pH.
Other metal ions will cause a decrease in pH.
When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the Ka and Kb values.
Would you expect an aqueous solution of sodium nitrate to be acidic, basic or neutral? How about zinc nitrate? Sodium acetate? Zinc acetate?

59. Polyprotic Acids Have more than one acidic proton.
If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

60. Polyprotic Example First dissociation, Ka = 1.7 X 10-2
Second dissociation, Ka = 6.4 X 10-8
H2O(l) + H2SO3(aq)
HSO3-(aq) + H3O+(aq)
H2O(l) + HSO3-(aq)
SO32-(aq) + H3O+( aq)

61. Factors Affecting Acid Strength
In oxyacids, in which an OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

62. Factors Affecting Acid Strength
The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound.
Acidity increases from left to right across a row and from top to bottom down a group.

63. Factors Affecting Acid Strength
For a series of oxyacids, acidity increases with the number of oxygens.

64. Lewis Acids and Bases
Lewis acids are defined as electron-pair acceptors.
Lewis bases are defined as electron-pair donors.
Anything that could be a Brønsted–Lowry base is a Lewis base.
BUT Lewis bases can interact with things other than protons.

65. Lewis Bases Lewis bases are defined as electron-pair donors.
Anything that could be a Brønsted–Lowry base is a Lewis base.
Lewis bases can interact with things other than protons, however.

66. Reactions of Anions with Water
Anions are bases.
As such, they can react with water in a hydrolysis reaction to form OH− and the conjugate acid:
X−(aq) + H2O(l)
HX(aq) + OH−(aq)

67. Reactions of Cations with Water
Cations with acidic protons (like NH4+) will lower the pH of a solution.
Most metal cations that are hydrated in solution also lower the pH of the solution.

68. Reactions of Cations with Water
Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water.
This makes the O-H bond more polar and the water more acidic.
Greater charge and smaller size make a cation more acidic.

69. Factors Affecting Acid Strength
Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.