1. Weak Bases NH3 + H2O NH4+ + OH-
in weak bases, only a small fraction of molecules accept H’s
weak electrolyte
most of the weak base molecules do not take H+ from water
much less than 1% ionization in water
[HO–] << [weak base]
finding the pH of a weak base solution is similar to finding the pH of a weak acid
NH3 + H2O NH4+ + OH-

3. Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5
Write the reaction for the base with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [OH-] from water is ≈ 0
NH3 + H2O NH4+ + OH-
[NH3]
[NH4+]
[OH]
initial
0.100
≈ 0
change
equilibrium
[NH3]
[NH4+]
[OH]
initial
change
equilibrium
since no products initially, Qc = 0, and the reaction is proceeding forward

4. Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[NH3]
[NH4+]
[OH]
initial
0.100
change
equilibrium -x +x +x x x x

5. Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5
determine the value of Kb from Table 15.8
since Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x
[NH3]
[NH4+]
[OH]
initial
0.100
≈ 0
change -x +x
equilibrium x x

6. Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5
check if the approximation is valid by seeing if x < 5% of [NH3]init
[NH3]
[NH4+]
[OH]
initial
0.100
≈ 0
change -x +x
equilibrium x
x = 1.33 x 10-3
the approximation is valid

7. Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5
substitute x into the equilibrium concentration definitions and solve
[NH3]
[NH4+]
[OH]
initial
0.100
≈ 0
change -x +x
equilibrium
0.099
1.33E-3
[NH3]
[NH4+]
[OH]
initial
0.100
≈ 0
change -x +x
equilibrium
0.100 x x
x = 1.33 x 10-3

8. Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5
use the [OH-] to find the [H3O+] using Kw
substitute [H3O+] into the formula for pH and solve
[NH3]
[NH4+]
[OH]
initial
0.100
≈ 0
change -x +x
equilibrium
0.099
1.33E-3

9. Find the pH of 0.100 M NH3(aq) solution given Kb for NH3 = 1.76 x 10-5
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
[NH3]
[NH4+]
[OH]
initial
0.100
≈ 0
change -x +x
equilibrium
0.099
1.33E-3
though not exact, the answer is reasonably close

10. Practice – Find the pH of a 0. 0015 M morphine solution, Kb = 1
Practice – Find the pH of a M morphine solution, Kb = 1.6 x 10-6

11. Practice – Find the pH of a 0. 0015 M morphine solution, Kb = 1
Practice – Find the pH of a M morphine solution, Kb = 1.6 x 10-6
Write the reaction for the base with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [OH-] from water is ≈ 0
B + H2O BH+ + OH-
[B]
[BH+]
[OH]
initial
0.0015
≈ 0
change
equilibrium
since no products initially, Qc = 0, and the reaction is proceeding forward

12. Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[B]
[BH+]
[OH]
initial
0.0015
change
equilibrium -x +x +x x x x

13. Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6
determine the value of Kb
since Kb is very small, approximate the [B]eq = [B]init and solve for x
[B]
[BH+]
[OH]
initial
0.0015
≈ 0
change -x +x
equilibrium x x

14. Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6
check if the approximation is valid by seeing if x < 5% of [B]init
[B]
[BH+]
[OH]
initial
0.0015
≈ 0
change -x +x
equilibrium x
x = 4.9 x 10-5
the approximation is valid

15. Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6
substitute x into the equilibrium concentration definitions and solve
[B]
[BH+]
[OH]
initial
0.0015
≈ 0
change -x +x
equilibrium
4.9E-5
[B]
[BH+]
[OH]
initial
0.0015
≈ 0
change -x +x
equilibrium x x
x = 4.9 x 10-5

16. Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6
use the [OH-] to find the [H3O+] using Kw
substitute [H3O+] into the formula for pH and solve
[B]
[BH+]
[OH]
initial
0.0015
≈ 0
change -x +x
equilibrium
4.9E-5

17. Find the pH of a 0.0015 M morphine solution, Kb = 1.6 x 10-6
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
[B]
[BH+]
[OH]
initial
0.0015
≈ 0
change -x +x
equilibrium
4.9E-5
the answer matches the given Kb

18. Acid-Base Properties of Salts
salts are water soluble ionic compounds
salts that contain the cation of a strong base and an anion that is the conjugate base of a weak acid are basic
Example: NaHCO3 solutions are basic
Na+ is the cation of the strong base NaOH
HCO3− is the conjugate base of the weak acid H2CO3
Conversely salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic
Example: NH4Cl solutions are acidic
NH4+ is the conjugate acid of the weak base NH3
Cl− is the anion of the strong acid HCl

19. Anions as Weak Bases
every anion can be thought of as the conjugate base of an acid
therefore, every anion can potentially be a base
A−(aq) + H2O(l) HA(aq) + OH−(aq)
the stronger the acid HA is, the weaker the conjugate base A- is
an anion that is the conjugate base of a strong acid is pH neutral
Cl−(aq) + H2O(l)  HCl(aq) + OH−(aq)
an anion that is the conjugate base of a weak acid is basic
F−(aq) + H2O(l) HF(aq) + OH−(aq)
since HF is a weak acid, the position of this equilibrium favors the right

20. Use the Table to Determine if the Given Anion Is Basic or Neutral
NO3−
the conjugate base of a strong acid, therefore neutral
HCO3−
the conjugate base of a weak acid, therefore basic
PO43−

21. Relationship between Ka of an Acid and Kb of its Conjugate Base
many reference books only give tables of Ka values because Kb values can be found from them
when you add equations, you multiply the K’s

22. Find the pH of 0. 100 M NaCHO2(aq) solution assuming Ka =1
Find the pH of M NaCHO2(aq) solution assuming Ka =1.8x10-4 for HCOOH
Na+ is the cation of a strong base – pH neutral. The CHO2− is the anion of a weak acid – pH basic
Write the reaction for the anion with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [OH-] from water is ≈ 0
HCO2− + H2O HCHO2 + OH-
[CHO2−]
[HCHO2]
[OH-]
initial
0.100
≈ 0
change
equilibrium

23. Find the pH of 0. 100 M NaCHO2(aq) solution assuming Ka =1
Find the pH of M NaCHO2(aq) solution assuming Ka =1.8x10-4 for HCOOH
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
Calculate the value of Kb from the value of Ka =1.8x10-4
substitute into the equilibrium constant expression
[CHO2−]
[HCHO2]
[OH-]
initial
0.100
≈ 0
change
equilibrium -x +x +x x x x

24. Find the pH of 0. 100 M NaCHO2(aq) solution assuming Ka =1
Find the pH of M NaCHO2(aq) solution assuming Ka =1.8x10-4 for HCOOH
since Kb is very small, approximate the [CHO2−]eq = [CHO2−]init and solve for x
[CHO2−]
[HCHO2]
[OH-]
initial
0.100
≈ 0
change -x +x
equilibrium x
Kb for CHO2− = 5.6 x 10-11 x

25. Find the pH of 0.100 M NaCHO2(aq) solution
check if the approximation is valid by seeing if x < 5% of [CHO2−]init
[CHO2−]
[HCHO2]
[OH-]
initial
0.100
≈ 0
change -x +x
equilibrium x
Kb for CHO2− = 5.6 x 10-11
x = 2.4 x 10-6
the approximation is valid

26. Find the pH of 0.100 M NaCHO2(aq) solution
substitute x into the equilibrium concentration definitions and solve
[CHO2−]
[HCHO2]
[OH-]
initial
0.100
≈ 0
change -x +x
equilibrium
0.100 −x x
[CHO2−]
[HCHO2]
[OH-]
initial
0.100
≈ 0
change -x +x
equilibrium
2.4E-6
Kb for CHO2− = 5.6 x 10-11
x = 2.4 x 10-6

27. Find the pH of 0.100 M NaCHO2(aq) solution
use the [OH-] to find the [H3O+] using Kw
substitute [H3O+] into the formula for pH and solve
[CHO2−]
[HCHO2]
[OH-]
initial
0.100
≈ 0
change -x +x
equilibrium
2.4E-6
Kb for CHO2− = 5.6 x 10-11

28. Find the pH of 0.100 M NaCHO2(aq) solution
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
[CHO2−]
[HCHO2]
[OH-]
initial
0.100
≈ 0
change -x +x
equilibrium
2.4E-6
though not exact, the answer is reasonably close
Kb for CHO2− = 5.6 x 10-11

29. Polyatomic Cations as Weak Acids
some polyatomic cations can be thought of as the conjugate acid of a base
therefore, some cations can potentially be an acid
MH+(aq) + H2O(l) MOH(aq) + H3O+(aq)
the stronger the base MOH is, the weaker the conjugate acid MH+ is
a cation that is the counterion of a strong base (Na+, K+ etc) is pH neutral
a cation that is the conjugate acid of a weak base is acidic
NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq)
since NH3 is a weak base, the position of this equilibrium favors the right

30. Metal Cations as Weak Acids
cations of small, highly charged metals are weakly acidic
alkali metal cations and alkali earth metal cations pH neutral
cations are hydrated
Al(H2O)63+(aq) + H2O(l) Al(H2O)5(OH)2+ (aq) + H3O+(aq)

31. Determine if the Given Cation is Acidic or Neutral
C2H5NH3+ (Kb of C2H5NH2 = 5.6 x 10-4)
the conjugate acid of a weak base, therefore acidic
Ca2+ (Kb of Ca(OH)2 = 3.74 x 10-3)
the counterion of a (not so) strong base, therefore neutral
Cr3+
a highly charged metal ion, therefore acidic

32. Classifying Salt Solutions as Acidic, Basic, or Neutral
if the salt cation is the counterion of a strong base and the anion is the conjugate base of a strong acid, it will form a neutral solution
NaCl Ca(NO3)2 KBr
if the salt cation is the counterion of a strong base and the anion is the conjugate base of a weak acid, it will form a basic solution
NaF Ca(C2H3O2) KNO2
if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a strong acid, it will form an acidic solution
NH4Cl
if the salt cation is a highly charged metal ion and the anion is the conjugate base of a strong acid, it will form an acidic solution
Al(NO3)3

33. Classifying Salt Solutions as Acidic, Basic, or Neutral
if the salt cation is the conjugate acid of a weak base and the anion is the conjugate base of a weak acid, the pH of the solution depends on the relative strengths of the acid and base
NH4F
HF is a stronger acid than NH4+
Ka of NH4+ is larger than Kb of the F−; therefore the solution will be acidic
Tro, Chemistry: A Molecular Approach

34. Determine whether a solution of the following salts is acidic, basic, or neutral
SrCl2
Sr2+ is the counterion of a strong base, pH neutral
Cl− is the conjugate base of a strong acid, pH neutral
solution will be pH neutral
AlBr3
Al3+ is a small, highly charged metal ion, weak acid
Br− is the conjugate base of a strong acid, pH neutral
solution will be acidic
CH3NH3NO3
CH3NH3+ is the conjugate acid of a weak base, acidic
NO3− is the conjugate base of a strong acid, pH neutral

35. Determine whether a solution of the following salts is acidic, basic, or neutral
NaCHO2
Na+ is the counterion of a strong base, pH neutral
CHO2− is the conjugate base of a weak acid, basic
solution will be basic
NH4F
NH4+ is the conjugate acid of a weak base, acidic
F− is the conjugate base of a weak acid, basic
Ka(NH4+) > Kb(F−); solution will be acidic

36. Polyprotic Acids
since polyprotic acids ionize in steps, each H has a separate Ka
Ka1 > Ka2 > Ka3
generally, the difference in Ka values is great enough so that the second ionization does not happen to a large enough extent to affect the pH
most pH problems just do first ionization
except H2SO4 use [H2SO4]o = [H3O+] for the second ionization
[A2-] = Ka2 as long as the second ionization is negligible

38. Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C
H2SO4 + H2O HSO4- + H3O+ Ka1= big!
Write the reactions for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations – assuming the [HSO4−] and [H3O+] is ≈ [H2SO4]
HSO4- + H2O SO42- + H3O Ka2= 0.012
[HSO4 -]
[SO42 -]
[H3O+]
initial
0.0100
change
equilibrium

39. Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
[HSO4 -]
[SO42 -]
[H3O+]
initial
0.0100
change −x +x
equilibrium −x x

40. Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C
expand and solve for x using the quadratic formula
Ka for HSO4− = 0.012

41. Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C
substitute x into the equilibrium concentration definitions and solve
[HSO4 -]
[SO42 -]
[H3O+]
initial
0.0100
change −x +x
equilibrium
0.0055
0.0045
0.0145
[HSO4 -]
[SO42 -]
[H3O+]
initial
0.0100
change −x +x
equilibrium −x x
x =

42. Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C
substitute [H3O+] into the formula for pH and solve
[HSO4 -]
[SO42 -]
[H3O+]
initial
0.0100
change −x +x
equilibrium
0.0055
0.0045
0.0145

43. Find the pH of 0.0100 M H2SO4(aq) solution @ 25°C
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HSO4 -]
[SO42 -]
[H3O+]
initial
0.0100
change −x +x
equilibrium
0.0055
0.0045
0.0145
the answer matches
Ka for HSO4− = 0.012

44. Strengths of Binary Acids
the more δ+ H-X δ- polarized the bond, the more acidic the bond
the stronger the H-X bond, the weaker the acid
binary acid strength increases to the right across a period
H-C < H-N < H-O < H-F
binary acid strength increases down the column
H-F < H-Cl < H-Br < H-I

45. Strengths of Oxyacids, H-O-Y
the more electronegative the Y atom, the stronger the acid
helps weakens the H-O bond
the more oxygens attached to Y, the stronger the acid
further weakens and polarizes the H-O bond

46. Lewis Acid - Base Theory
electron sharing
electron donor = Lewis Base = nucleophile
must have a lone pair of electrons
electron acceptor = Lewis Acid = electrophile
electron deficient
when Lewis Base gives electrons from lone pair to Lewis Acid, a covalent bond forms between the molecules
Nucleophile: + Electrophile  Nucleophile:Electrophile
product called an adduct
other acid-base reactions also Lewis
Tro, Chemistry: A Molecular Approach

47. Example - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and ElectrophileOH
H C H 
+ OH-1  OH
H C H 
+ OH-1 
Electrophile
Nucleophile •• •
Tro, Chemistry: A Molecular Approach

48. Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile
BF3 + HF 
CaO + SO3 
KI + I2 
Tro, Chemistry: A Molecular Approach

49. Practice - Complete the Following Lewis Acid-Base Reactions Label the Nucleophile and Electrophile
BF HF H+1BF4-1
CaO + SO Ca+2SO4-2
KI I KI3 F
B F
H F •• + -1
H+1
Nuc
Elec O
S O ••
Ca+2 O -2 + -2
Ca+2
Elec
Nuc
I I
K+1 I -1 •• +
Elec
Nuc
K+1 I I I -1

50. What is Acid Rain? natural rain water has a pH of 5.6
naturally slightly acidic due mainly to CO2
rain water with a pH lower than 5.6 is called acid rain
acid rain is linked to damage in ecosystems and structures

51. What Causes Acid Rain?
many natural and pollutant gases dissolved in the air are nonmetal oxides
CO2, SO2, NO2
nonmetal oxides are acidic
CO2 + H2O H2CO3
2 SO2 + O2 + 2 H2O H2SO4
processes that produce nonmetal oxide gases as waste increase the acidity of the rain
natural – volcanoes and some bacterial action
man-made – combustion of fuel
weather patterns may cause rain to be acidic in regions other than where the nonmetal oxide is produced

52. Damage from Acid Rain

53. pH of Rain in Different Regions

54. Sources of SO2 from Utilities

55. Damage from Acid Rain
acids react with metals, and materials that contain carbonates
acid rain damages bridges, cars, and other metallic structures
acid rain damages buildings and other structures made of limestone or cement
acidifying lakes affecting aquatic life
dissolving and leaching more minerals from soil
making it difficult for trees

56. Acid Rain Legislation 1990 Clean Air Act attacks acid rain
force utilities to reduce SO2
result is acid rain in northeast stabilized and beginning to be reduced