1. CHE 124: General Chemistry II
Chapter 15: Acids & Bases
CHE 124: General Chemistry II
Dr. Jerome Williams, Ph.D.
Saint Leo University

2. Overview Calculating pH Strong Bases Calculating pH Weak Bases
Polyprotic Acids
Determining pH of Salts

3. Strong Bases The stronger the base, the more willing it is to accept H
use water as the standard acid
For ionic bases, practically all units are dissociated into OH– or accept H’s
strong electrolyte
multi-OH strong bases completely dissociated
[HO–] = [strong base] x (# OH)
NaOH ® Na+ + OH−
Tro, Chemistry: A Molecular Approach, 2/e

4. Example 15. 11: Calculate the pH at 25 °C of a 0
Example 15.11: Calculate the pH at 25 °C of a M Sr(OH)2 solution and determine if the solution is acidic, basic, or neutral
Given:
Find:
[Sr(OH)2] = 1.5 x 10−3 M pH
Conceptual Plan:
Relationships:
[H3O+]
[OH] pH
[Sr(OH)2]
[OH]=2[Sr(OH)2]
Solution:
[OH]
= 2(0.0015)
= M
Check:
pH is unitless; the fact that the pH > 7 means the solution is basic
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5. Practice – Calculate the pH of the following strong acid or base solutions
M HCl
M Ca(OH)2
[H3O+] = [HCl] = 2.0 x 10−3 M
pH = −log(2.0 x 10−3) = 2.70
[OH−] = 2 x [Ca(OH)2] = 3.0 x 10−3 M
pOH = −log(3.0 x 10−3) = 2.52
pH = − pOH = − 2.52
pH = 11.48
Tro, Chemistry: A Molecular Approach, 2/e

6. Weak Bases
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 Û NH4+ + OH−
Tro, Chemistry: A Molecular Approach, 2/e

7. Base Ionization Constant, Kb
Base strength measured by the size of the equilibrium constant when react with H2O
:Base + H2O  OH− + H:Base+
The equilibrium constant is called the base ionization constant, Kb
larger Kb = stronger base
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8. Tro, Chemistry: A Molecular Approach, 2/e

9. Structure of Amines
Tro, Chemistry: A Molecular Approach, 2/e

10. Example 15.12:Find the pH of 0.100 M NH3(aq)
NH3 + H2O  NH4+ + OH
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]
[NH4+]
[OH]
initial
0.100
≈ 0
change
equilibrium
[NH3]
[NH4+]
[OH]
initial
change
equilibrium
because no products initially, Qc = 0,
and the reaction is proceeding forward
Tro, Chemistry: A Molecular Approach, 2/e

11. Example 15.12: Find the pH of 0.100 M NH3(aq)
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
0.100 x
Tro, Chemistry: A Molecular Approach, 2/e

12. Example 15.12: Find the pH of 0.100 M NH3(aq)
determine the value of Kb from Table 15.8
because Kb is very small, approximate the [NH3]eq = [NH3]init and solve for x
Kb for NH3 = 1.76 x 10−5
[NH3]
[NH4+]
[OH]
initial
0.100
≈ 0
change −x +x
equilibrium x
0.100 x
Tro, Chemistry: A Molecular Approach, 2/e

13. Example 15.12: Find the pH of 0.100 M NH3(aq)
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
Tro, Chemistry: A Molecular Approach, 2/e

14. Example 15.12: Find the pH of 0.100 M NH3(aq)
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
Tro, Chemistry: A Molecular Approach, 2/e

15. Example 15.12: Find the pH of 0.100 M NH3(aq)
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
Tro, Chemistry: A Molecular Approach, 2/e

16. Example 15.12: Find the pH of 0.100 M NH3(aq)
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
Tro, Chemistry: A Molecular Approach, 2/e

17. 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 Solve from first principles; write out all chemical equilibria. Assumptions must be stated and defended mathematically. Answer: pH = 8.69

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
NaHCO3 solutions are basic
Na+ is the cation of the strong base NaOH
HCO3− is the conjugate base of the weak acid H2CO3
Salts that contain cations that are the conjugate acid of a weak base and an anion of a strong acid are acidic
NH4Cl solutions are acidic
NH4+ is the conjugate acid of the weak base NH3
Cl− is the anion of the strong acid HCl
Tro, Chemistry: A Molecular Approach, 2/e

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 is, the weaker the conjugate base 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)
Tro, Chemistry: A Molecular Approach, 2/e

20. Example 15.13: Use the table to determine if the given anion is basic or neutral
NO3−
the conjugate base of a strong acid, therefore neutral
NO2−
the conjugate base of a weak acid, therefore basic
Tro, Chemistry: A Molecular Approach, 2/e

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
Tro, Chemistry: A Molecular Approach, 2/e

22. Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution
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
Example 15.14: Find the pH of M NaCHO2(aq) solution
CHO2− + H2O  HCHO2 + OH
[CHO2−]
[HCHO2]
[OH]
initial
0.100
≈ 0
change
equilibrium

23. Example 15.14: Find the pH of 0.100 M NaCHO2(aq) solution
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 of the weak acid from Table 15.5
substitute into the equilibrium constant expression
[CHO2−]
[HCHO2]
[OH]
initial
0.100
≈ 0
change
equilibrium x +x +x
0.100 x x x
Tro, Chemistry: A Molecular Approach, 2/e

24. Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
Kb for CHO2− = 5.6 x 10−11
because 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
0.100 x
Tro, Chemistry: A Molecular Approach, 2/e

25. Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
Kb for CHO2− = 5.6 x 10−11
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
x = 2.4 x 10−6
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e

26. Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
Kb for CHO2− = 5.6 x 10−11
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
x = 2.4 x 10−6
Tro, Chemistry: A Molecular Approach, 2/e

27. Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
Kb for CHO2− = 5.6 x 10−11
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
Tro, Chemistry: A Molecular Approach, 2/e

28. Example 15.14: Find the pH of 0.100 M NaCHO2(aq)
Kb for CHO2− = 5.6 x 10−11
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
Tro, Chemistry: A Molecular Approach, 2/e

29. Polyatomic Cations as Weak Acids
Some cations can be thought of as the conjugate acid of a weak base
others are the counter-ions of a strong base
Therefore, some cations can potentially be acidic
MH+(aq) + H2O(l)  MOH(aq) + H3O+(aq)
The stronger the base is, the weaker the conjugate acid is
a cation that is the counter-ion of a strong base is pH neutral
a cation that is the conjugate acid of a weak base is acidic
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
because NH3 is a weak base, the position of this equilibrium favors the right
Tro, Chemistry: A Molecular Approach, 2/e

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

31. Example 15.15: Determine if the given cation Is acidic or neutral
C5N5NH2+
the conjugate acid of the weak base pyridine, therefore acidic
Ca2+
the counter-ion of the strong base Ca(OH)2, therefore neutral
Cr3+
a highly charged metal ion, therefore acidic
Tro, Chemistry: A Molecular Approach, 2/e

32. Classifying Salt Solutions as Acidic, Basic, or Neutral
If the salt cation is the counter-ion 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 counter-ion 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
Tro, Chemistry: A Molecular Approach, 2/e

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 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
Tro, Chemistry: A Molecular Approach, 2/e

34. 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 because 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, 2/e

35. Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral
SrCl2
Sr2+ is the counter-ion 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
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
Tro, Chemistry: A Molecular Approach, 2/e

36. Example 15.16: Determine whether a solution of the following salts is acidic, basic, or neutral
NaCHO2
Na+ is the counter-ion 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
Tro, Chemistry: A Molecular Approach, 2/e

37. Practice – Determine whether a solution of the following salts is acidic, basic, or neutral
K+ is the counter-ion of a strong base, pH neutral
KNO3
CoCl3
Ba(HCO3)2
CH3NH3NO3
NO3− is the counter-ion of a strong acid, pH neutral
the solution is pH neutral
Co3+ is a highly charged cation, pH acidic
Cl− is the counter-ion of a strong acid, pH neutral
the solution is pH acidic
Ba2+ is the counter-ion of a strong base, pH neutral
the solution is pH basic
HCO3− is the conjugate of a weak acid, pH basic
CH3NH3+ is the conjugate of a weak base, pH acidic
the solution is pH acidic
NO3− is the counter-ion of a strong acid, pH neutral
Tro, Chemistry: A Molecular Approach, 2/e

38. Ionization in Polyprotic Acids
Because 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] as the [H3O+] for the second ionization
[A2−] = Ka2 as long as the second ionization is negligible
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39. Tro, Chemistry: A Molecular Approach, 2/e

40. Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3? (Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11)
Tro, Chemistry: A Molecular Approach, 2/e

41. Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3?
H2CO3 + H2O  HCO3 + H3O+
write the reactions for the acid with water one H at a time
construct an ICE table for the reaction
enter the initial concentrations – assuming the second ionization is negligible
HCO3− + H2O  CO32− + H3O+
[HA]
[A−]
[H3O+]
initial
0.12
≈ 0
change
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e

42. Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3?
H2CO3 + H2O  HCO3 + H3O+
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
[HA]
[A−]
[H3O+]
initial
0.12
change
equilibrium x +x +x x x
0.12 x
Tro, Chemistry: A Molecular Approach, 2/e

43. Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3?
determine the value of Ka1
because Ka1 is very small, approximate the [HA]eq = [HA]init and solve for x
Ka1 = 4.3 x 10−7, Ka2 = 5.6 x 10−11
[HA]
[A−]
[H3O+]
initial
0.12
≈ 0
change −x +x
equilibrium
0.012 x
0.12 x
Tro, Chemistry: A Molecular Approach, 2/e

44. the approximation is valid
Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3?
Ka1 for H2CO3 = 4.3 x 10−7
check if the approximation is valid by seeing if x < 5% of [H2CO3]init
[HA]
[A−]
[H3O+]
initial
0.12
≈ 0
change −x +x
equilibrium x
x = 2.27 x 10−4
the approximation is valid
Tro, Chemistry: A Molecular Approach, 2/e

45. Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3?
substitute x into the equilibrium concentration definitions and solve
[HA]
[A−]
[H3O+]
initial
0.12
≈ 0
change −x +x
equilibrium
0.12−x x
x = 2.3 x 10−4
Tro, Chemistry: A Molecular Approach, 2/e

46. Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3?
substitute [H3O+] into the formula for pH and solve
[HA]
[A−]
[H3O+]
initial
0.12
≈ 0
change −x +x
equilibrium
Tro, Chemistry: A Molecular Approach, 2/e

47. Practice – What is the pH of a 0.12 M solution of carbonic acid, H2CO3?
Ka1 for H2CO3 = 4.3 x 10−7
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
[HA]
[A−]
[H3O+]
initial
0.12
≈ 0
change −x +x
equilibrium
the values match
within sig figs
Tro, Chemistry: A Molecular Approach, 2/e