1. Acids and Bases
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2. Definitions Arrhenius Brønsted-Lowry
Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions.
Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.
Brønsted-Lowry
An acid is a proton donor.
A base is a proton acceptor.
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3. If it can be either… HCO3- HSO4- H2O …it is amphiprotic.
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4. What Happens When an Acid Dissolves in Water?
Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid.
As a result, the conjugate base of the acid and a hydronium ion are formed.
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5. Conjugate Acids and Bases
Reactions between acids and bases always yield their conjugate bases and acids.
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6. Sample Exercise
a) What is the conjugate base of each of the following acids:
HClO4
H2S
PH4+
HCO3-?
b) What is the conjugate acid of each of the following bases?
CN-
SO42-
H2O
HCO3-
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7. Acid and Base Strength
Strong acids are completely dissociated in water.
Their conjugate bases are quite weak.
Weak acids only dissociate partially in water.
Their conjugate bases are weak bases.
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8. Acid and Base Strength
Substances with negligible acidity do not dissociate in water.
Their conjugate bases are exceedingly strong.
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9. Autoionization of Water
As we have seen, water is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.
This is referred to as autoionization.
H2O (l) + H2O (l)
H3O+ (aq) + OH- (aq)
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10. Ion-Product Constant 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 25C, Kw = 1.0  10-14
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11. pH
pH is defined as the negative base-10 logarithm of the concentration of hydronium ion.
pH = -log [H3O+]
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.
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12. pH These are the pH values for several common substances.
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13. Practice Exercise 16.6
a) In a sample of lemon juice [H+] is 3.8 x 10-4 M. What is the pH?
(3.42)
b) A commonly available window-cleaning solution has a [OH-] of 1.9 x 10-6 M. What is the pH?
(8.28)
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14. Dissociation Constants
For a generalized acid dissociation,
the equilibrium expression would be
This equilibrium constant is called the acid-dissociation constant, Ka.
Note that H2O is omitted from Ka because it is a pure liquid
HA (aq) + H2O (l)
A- (aq) + H3O+ (aq)
[H3O+] [A-]
[HA]
Kc =
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15. Dissociation Constants
The larger the Ka, the stronger the acid
If Ka >> 1, then the acid is completely ionized  strong acid
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16. 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.
To calculate Ka, we need the equilibrium concentrations of all three things.
We can find [H3O+], which is the same as [HCOO-], from the pH.
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17. Calculating Ka from the pH
pH = -log [H3O+]
2.38 = -log [H3O+]
-2.38 = log [H3O+]
= 10log [H3O+] = [H3O+]
4.2  10-3 = [H3O+] = [HCOO-]
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18. Calculating Ka from pH Now we can set up a table… [HCOOH], M [H3O+], M
[HCOO-], M
Initially
0.10
Change
- 4.2  10-3
+ 4.2  10-3
At Equilibrium
 10-3
= = 0.10
4.2  10-3
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19. Calculating Ka from pH
Note that the [HCOOH] remains at 0.10 M at equilibrium
– sig figs. This will become very important later in this chapter.
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20. Calculating Ka from pH [4.2  10-3] [4.2  10-3] Ka = [0.10]
= 1.8  10-4
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21. Practice Exercise 16.10
Niacin, one of the B vitamins, has the following molecular structure:
A M solution of niacin has a pH of 3.26.
What is the acid-dissociation constant, Ka, for niacin?
(1.5 x 10-5)
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22. Calculating Percent Ionization Sample Exercise 16.11
A 0.10 M solution of formic acid (HCOOH) contains 4.2 x 10-3 M H+(aq). Calculate the percentage of the acid that is ionized.
(4.2%)
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23. Calculating Percent Ionization Sample Exercise 16.11
[H3O+]eq
[HA]initial
Percent Ionization =  100
In this example
[H3O+]eq = 4.2  10-3 M
[HCOOH]initial = 0.10 M
4.2  10-3
0.10
Percent Ionization =  100
= 4.2%
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24. Practice Exercise 16.11
A M solution of niacin has a pH of Calculate the percent ionization of the niacin.
(2.8%)
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25. Using Ka to Calculate pH
• Using Ka, we can calculate the concentration of H+ (and hence the pH).
Example: Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C.
Ka for acetic acid at 25C is 1.8  10-5.
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26. Using Ka to Calculate pH
1. Write the balanced chemical equation clearly showing the equilibrium.
CH3COOH(aq) D H+(aq) + CH3COO-(aq)
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27. Using Ka to Calculate pH
2. Write the equilibrium expression.
Look up the value for Ka (in a table).
Ka = [H+][ CH3COO-] = 1.8 x 10-5
[CH3COOH]
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28. Using Ka to Calculate pH
Write down the initial and equilibrium concentrations for everything except pure
water.
We usually assume that the equilibrium
concentration of H+ is x.
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29. Using Ka to Calculate pH
3. Write down the initial and equilibrium
concentrations for everything except pure
water.
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30. Using Ka to Calculate pH
Substitute into the equilibrium constant
expression and solve.
Ka = 1.8 x 10-5 = [H+][C2H3O2-] = (x)(x)
[HC2H3O2] – x
Note that Ka is very small (1.8 x 10-5) relative to [HC2H3O2] (0.30 M).
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31. Using Ka to Calculate pH
Substitute into the equilibrium constant expression and solve.
Keep this x
x2 = 1.8 x 10-5
0.30 – x
Neglect x in the denominator since it is extremely small relative to 0.30.
(Use ballpark figure of 103 X difference for neglecting x in the denominator )
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32. Using Ka to Calculate pH
Substitute into the equilibrium constant expression and solve.
 x2 = (1.8 x 10-5 )(0.30) = 5.4 x 10-6
 Ö x2 = Ö (5.4 x 10-6)
x = 2.3 x 10-3 M = [H+]
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33. Using Ka to Calculate pH
Substitute into the equilibrium constant expression and solve.
Check: Compare x with original [HC2H3O2] of 0.30 M:
2.3 x 10-3 M x 100% = 0.77%, which is < 5%
0.30 M
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34. Using Ka to Calculate pH
5. Convert x ([H+]) to pH.
pH = - log (2.3 x 10-3) = 2.64
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35. What do we do if we are faced with having to solve a quadratic equation in order to determine the value of x?
Often this cannot be avoided.
However, if the Ka value is quite small, we find that we can make a simplifying assumption.
• Assume that x is negligible compared to the initial concentration
of the acid.
• This will simplify the calculation.
• It is always necessary to check the validity of any assumption.
• Once we have the value of x, check to see how large it is compared to the
initial concentration.
• If x is <5% of the initial concentration, the assumption is probably a good
one.
• If x>5% of the initial concentration, then it may be best to solve the
quadratic equation or use successive approximations.
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36. Sample Exercise (p. 685)
Calculate the pH of a 0.20 M solution of HCN. Refer to Table 16.2 for Ka.
(5.00)
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37. Practice Exercise 16.12
The Ka for niacin is 1.6 x What is the pH of a M solution of niacin?
(3.41)
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38. Sample Exercise (p. 687)
Calculate the percentage of HF molecules ionized in
a) a 0.10 M HF solution (7.9%)
b) a M HF solution (23%)
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39. Practice Exercise 16.13
In Practice Exercise 16.11, we found that the percent ionization of niacin (Ka = 1.5 x 10-5) in a M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is
a) M (3.9%)
b) a 1.0 x 10-3 M (12%)
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40. 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.
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41. Sample Exercise (p. 689) HW
The solubility of CO2 in pure water at 25oC and 0.1 atm pressure is M. the common practice is to assume that all of the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced by reaction between the CO2 and H2O:
CO2(aq) + H2O(l) D H2CO3(aq)
What is the pH of a M solution of H2CO3?
(4.40)
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42. Practice Exercise 16.14 (p. 690) optional – requires quadratic
Calculate the pH of a M solution
of oxalic acid (H2C2O4).
(see Table 16.3 for Ka1 and Ka2).
b) Calculate the concentration of oxalate ion [C2O42-].
(pH = 1.80; [C2O42-] = 6.4 x 10-5 M)
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43. Weak Bases
Weak bases remove protons from substances. They react with water to produce hydroxide ion.
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44. Weak Bases The equilibrium constant expression for this reaction is
[HB] [OH-]
[B-]
Kb =
where Kb is the base-dissociation constant.
The larger Kb, the stronger the base.
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45. Weak Bases The equilibrium constant expression for this reaction is
[NH4+] [OH-]
[NH3]
Kb =
where Kb is the base-dissociation constant.
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46. Kb can be used to find [OH-] and, through it, pH.
Weak Bases
Kb can be used to find [OH-] and, through it, pH.
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47. pH of Basic Solutions Sample Exercise 16.15 (p. 691)
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
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48. pH of Basic Solutions Tabulate the data. [NH3], M [NH4+], M [OH-], M
Initially
0.15
At Equilibrium
x  0.15 x
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49. pH of Basic Solutions (x)2 (0.15) 1.8  10-5 =
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50. pH of Basic Solutions Check the assumption:
1.6  10-3 M x 100% = 1.1%, < 5%
0.15 M
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51. pH of Basic Solutions Therefore, [OH-] = 1.6  10-3 M
pOH = -log (1.6  10-3)
pOH = 2.80
pH =
pH = 11.20
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52. Practice Problem 16.15
Which of the following compounds should produce the highest pH as a 0.05 M solution: pyridine, methylamine, or nitrous acid?
(methylamine)
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53. Types of Weak Bases
Weak bases generally fall into one of two categories.
1. Neutral substances with a lone pair of electrons
that can accept protons.
Most neutral weak bases contain nitrogen.
Amines are related to ammonia and have one or more
N–H bonds replaced with N–C bonds
(e.g., CH3NH2 is methylamine).
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54. Types of Weak Bases
Like NH3, amines can abstract a proton from a water
molecule by forming an additional N-H bond,
as shown in this figure for methylamine.
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55. Types of Weak Bases 2. Anions of weak acids are also weak bases.
e.g.: ClO– is the conjugate base of HClO (weak acid):
ClO–(aq) + H2O(l) D HClO(aq) + OH–(aq) Kb = 3.3 x 10–7
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56. Sample Exercise (p. 692)
A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a pH of How many moles of NaClO were added to the water?
(See info immediately above.)
(0.60 mol)
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57. Practice Exercise 16.16
A solution of NH3 in water has a pH of What is the molarity of the solution?
(0.12 M)
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58. Ka and Kb Ka and Kb are related in this way: Ka  Kb = Kw
Therefore, if you know one of them, you can calculate the other.
Alternatively, pKa + pKb = pKw = (at 25oC)
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59. Sample Exercise 16.17 (p. 695) Calculate
the base-dissociation constant, Kb, for the fluoride ion (F-);
(1.5 x 10-11)
the acid-dissociation constant, Ka, for the ammonium ion (NH4+).
(5.6 x 10-10)
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60. Practice Exercise 16.17 (Use Appendix D for Ka)
a) Which of the following anions has the largest base-dissociation constant: NO2-, PO43-, or N3-?
(PO43-, Kb = 2.4 x 10-2)
b) The base quinoline has the following structure:
Its conjugate acid is listed in handbooks as having a pKa of 4.90.
What is the base-dissociation constant for quinoline?
(7.9 x 10-10)
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61. Acid/Base Properties of Salt Solutions
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62. 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)
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63. 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.
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64. 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.
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65. Effect of Cations and Anions
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.
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66. 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 effect on pH depends on the Ka and Kb values.
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67. Sample Exercise (p. 698)
Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral:
Hint: Analyze each compound – derived from strong or weak acids and bases?
Ba(CH3COO)2,
NH4Cl
CH3NH3Br
KNO3
Al(ClO4)3
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68. Sample Exercise (p. 698)
Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral:
Ba(CH3COO)2,
anion of weak acid  basic
NH4Cl
cation of weak base  acidic
CH3NH3Br
KNO3
neutral – derived from strong acid + strong base
Al(ClO4)3
cation of weak base (Al(OH)3) acidic,
Al3+ - increase the polarity of O-H in H2O
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69. Practice Exercise 16.18
In each of the following, indicate which salt in each of the following pairs will form the more acidic (or less basic) M solution:
a) NaNO3 or Fe(NO3)3
b) KBr, or KBrO
c) CH3NH3Cl or BaCl2
d) NH4NO2 or NH4NO3
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70. Sample Exercise (p. 698)
Predict whether the salt Na2HPO4 will form an acidic or a basic solution on dissolving in water.
(basic)
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71. Practice Exercise 16.19
Predict whether the dipotassium salt of citric acid (K2HC6H5O7) will form an acidic or basic solution in water.
(see Table 16.3 for data)
(acidic)
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72. Factors Affecting Acid Strength (Sec. 16.10-16.11)
The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound.
So acidity increases from left to right across a row and from top to bottom down a group.
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73. 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.
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74. Factors Affecting Acid Strength
For a series of oxyacids, acidity increases with the number of oxygens.
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75. Sample Exercise (p. 702)
Arrange the compounds in each of the following series in order of increasing acid strength:
a) AsH3, HI, NaH, H2O;
b) H2SO4, H2SeO3, H2SeO4.
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76. Sample Exercise (p. 702)
Arrange the compounds in each of the following series in order of increasing acid strength:
a) AsH3, HI, NaH, H2O
NaH < AsH3 < H2O < HI
b) H2SO4, H2SeO3, H2SeO4
H2SeO3 < H2SeO4 < H2SO4
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77. Practice 16.20
In each of the following pairs choose the compound that leads to the more acidic (or less basic) solution:
a) HBr, HF;
b) PH3, H2S;
c) HNO2, HNO3;
d) H2SO3, H2SeO3.
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78. Practice 16.20
In each of the following pairs choose the compound that leads to the more acidic (or less basic) solution:
a) HBr, HF;
b) PH3, H2S;
c) HNO2, HNO3;
d) H2SO3, H2SeO3.
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79. Factors Affecting Acid Strength
Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.
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80. Comparison of Different Types of Acids and Bases
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81. Arrhenius (traditional) acids and bases (C19th)
Acid: compound containing H that ionizes to yield H+ in solution
Base: compound containing OH that ionizes to yield OH- in solution
(Note: does not describe acid/base behavior in solvents other than water)
Note: Every Arrhenius acid/base is also a Brønsted-Lowry acid/base.
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82. Bronsted-Lowry Acids and Bases (1923)
Acid: H+ (proton) donor
Base: H+ (proton) acceptor
NH H2O D NH OH-
ammonia water ammonium ion hydroxide ion
(B-L base) (B-L acid) (B-L acid) (B-L base)
Note: Every Brønsted-Lowry acid/base is also a Lewis acid/base.
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83. Lewis Acids and Bases (1920’s)
Acid: accepts pair of e-‘s
Base: donates pair of e-‘s
H [:O:H]-  H:O:H
Lewis acid Lewis base .. .. .. ..
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84. Lewis Acids Lewis acids are defined as electron-pair acceptors.
Atoms with an empty valence orbital can be Lewis acids.
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85. 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.
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86. Hydrolysis of Metal Ions
The Lewis concept may be used to explain the acid properties of many metal ions.
Metal ions are positively charged and attract water molecules (via the lone pairs on the oxygen atom of water).
Hydrated metal ions act as acids.
• For example:
Fe(H2O)63+(aq) D Fe(H2O)5(OH)2+(aq) + H+(aq) Ka = 2 x 10–3
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87. Hydrolysis of Metal Ions
In general:
• The higher the charge, the stronger the M–OH2 interaction.
• Ka values generally increase with increasing charge
• The smaller the metal ion, the more acidic the ion.
• Ka values generally decrease with decreasing ionic radius
• Thus the pH of an aqueous solution increases as the size of the ion increases (e.g., Ca2+ vs. Zn2+) and as the charge increases (e.g., Na+ vs. Ca2+ and Zn2+ vs. Al3+).
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88. Sample Integrative Exercise 16 (p. 706)
Phosphorous acid (H3PO3) has the following Lewis structure:
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89. Sample Integrative Exercise 16 (p. 706)
a) Explain why phosphorous acid is diprotic and not triprotic.
b) A 25.0 mL sample of a solution of H3PO3 is titrated with M NaOH. It requires 23.3 mL of NaOH to neutralize both acidic protons. What is the molarity of the H3PO3 solution?
c) This solution has a pH of Calculate the percent ionization and Ka1 for H3PO3, assuming that Ka1 >>Ka2.
d) How does the osmotic pressure of a M solution of HCl compare qualitatively with that of a M solution of H3PO3? Explain.
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