1. Acids and Bases
…and Airman 1st class HCl flew back over the front lines, confident that he had neutralized the enemy’s strongest base.

2. Arrhenius, BrØnsted-Lowry, and Lewis Please recall: The
definitions of an acid and a base

3. What is the conjugate base of…
HCl
CH3COOH
H2SO4
HSO4-
H2O
NH4+
NH3

4. What is the conjugate base of…
ACID (loses H+ to form) Conjugate base
HCl  Cl-
CH3COOH  CH3COO-
H2SO4  HSO4-
HSO4-  SO4-2
H2O  OH-
NH4+  NH3
NH3  NH2-

5. What is the conjugate acid of…
NO3-
C2O4-2
HPO4-2
HSO4-
H2O F-

6. What is the conjugate acid of…
Base (gains H+ to form) Conjugate acid
NO3-  HNO3
C2O  HC2O4-
HPO4-2  H2PO4-
HSO4-  H2SO4
H2O  H3O+
F-  HF

7. Show the conjugate acid/base pairs in the following reactions.
C2O4-2 + H3O+  HC2O H2O
CH3COOH + NH  NH3 + CH3COO-

8. Show the conjugate acid/base pairs in the following reactions.
C2O4-2 + H3O+  HC2O H2O
CH3COOH + NH  NH3 + CH3COO-
ConjugateBase
Conjugate Acid
Base

9. Show the conjugate acid/base pairs in the following reactions.
C2O4-2 + H3O+  HC2O H2O
CH3COOH + NH  NH3 + CH3COO-
ConjugateBase
Conjugate Acid
Base
Conjugate Acid
Base
Acid
ConjugateBase

10. Water dissociates H2O  H+ + OH- Kw=[H+][OH-]=1 x 10-14 at 25oC
(endothermic or exothermic?)
(Does Kw increase or decrease at higher T?)
(or 2H2O H3O+ + OH- )

11. [H+] is inversely related to [OH-]
When [H+] increases, [OH-] decreases in a water solution, and vice versa.
(Why?)

12. pH The basic (and acidic) definitions are: pH= -log [H+] [H+]= 10-pH
pOH= -log [OH-] [OH-]=10 -pOH
Kw=[H+][OH-]=1 x (at 25oC)
pH + pOH = 14 (at 25oC)

13. (as far as they dissolve.)
Please recall:
Strong acids and bases dissociate completely
in water.
Weak acids and bases do not.
Strong acids –HNO3, HClO4, H2SO4*, HI, HCl, HBr
Strong bases-Group 1 & 2 hydroxides
(as far as they dissolve.)

14. Contents pH
[H+] (M)
[OH-] (M)
pOH
Acidic or Basic 1
.023 mol HCl /L 2
1.5g NaOH /L 3
? mol LiOH/ 50ml
8.50 4
? mol KOH/25ml
2.50 5
? gHClO4 /150ml
.0200 6
? mol Ba(OH)2/L
.0070

15. Contents pH
[H+] (M)
[OH-] (M)
pOH
Acidic or Basic 1
.023 mol HCl /L
1.64
.023
4.3 x 10-13
12.36
Acidic 2
1.5g NaOH /L
12.57
2.7 x 10-13
.0375
1.43
Basic 3
1.6x10-7mol LiOH/ 50ml
8.50
3.2 x 10-9
3.2 x 10-6
5.50
basic 4
7.9x10-5mol KOH/25ml
11.50
3.2 x 10-12
3.2 x 10-3
2.50 5
.30 gHClO4 /150ml
1.70
.0200
5.00 x 10-13
12.30
acidic 6
.0035mol Ba(OH)2/L
11.85
1.4 x 10-12
.0070
2.15

16. Strength of acids and bases.
HCl: strong acid. Complete dissociation
H2CO3: weak acid. Partial dissociation
CH4: pathetic acid. No dissociation
What about their conjugates?

17. Strength of acids and bases.
Cl-: pathetic base. No association
HCO3-:weak base. Partial association
CH3-: so strong a base, complete association with water, leaving hydroxide.
What about their conjugates?

18. Strength of acids and bases.
The conjugate of a strong acid is a pathetic base
The conjugate of a weak acid is a weak base—the stronger the acid, the weaker the base and vice versa
The conjugate of a pathetic acid is a strong base

19. When comparing weak acids and bases…
HAH++A Ka= [H+][A-]
[HA]
Weak base,
B-+H2OHB+OH Kb=[HB][OH-]
[B-]
The position of the equilibrium is the strength of the acid or base.

20. For example:
Ka=1.8x10-5 for acetic acid
Ka=6.5x10-5 for benzoic acid

21. For example: Ka=1.8x10-5 for acetic acid Ka=6.5x10-5 for benzoic acid
Benzoic acid is a stronger acid.

22. For example: Ka=1.8x10-5 for acetic acid Ka=6.5x10-5 for benzoic acid
Benzoic acid is a stronger acid.
.10M solutions of each would have a lower pH for benzoic acid.

23. For example: Ka=1.8x10-5 for acetic acid Ka=6.5x10-5 for benzoic acid
Benzoic acid is a stronger acid.
.10M solutions of each would have a lower pH for benzoic acid.
An acetic acid solution could have a lower pH, but only at a higher concentration.

24. Write the reaction and eq. expression for:
Ammonia associating with water
Ammonium dissociating in water

25. Write the reaction and eq. expression for:
Ammonia associating with water
NH3(aq) + H2O (l)  NH4+ (aq) + OH- (aq)
Kb = [NH4+][OH-]
[NH3]
Ammonium dissociating in water
NH4+ (aq)  H+ (aq) + NH3 (aq)
Ka = [H+ ][NH3]
[NH4+]

26. Kb Ka = [NH4+][OH-][H+ ][NH3]
And…
Kb Ka = [NH4+][OH-][H+ ][NH3]
[NH4+] [NH3]

27. Kb Ka = [NH4+][OH-][H+ ][NH3]
And…
Kb Ka = [NH4+][OH-][H+ ][NH3]
[NH4+] [NH3]

28. Kb Ka = [NH4+][OH-][H+ ][NH3]
And…
Kb Ka = [NH4+][OH-][H+ ][NH3]
[NH4+] [NH3]
= [OH-][H+ ]=Kw
True for any conjugate pair.

29. For the following weak acids:
Hydrosulfuric acid
H2S
H2SH+ + HS-
Ka=[H+][HS-] =9.1 x 10-8
[H2S]
d) HS-+H2O OH- + H2S
e) Kb=[OH-][H2S] =1.1 x 10-7
[HS-]
You try it:
Chromic acid
Methanoic acid
Nitrous acid
Oxalic acid
Hydroselenic acid

30. The leveling effect
You can’t have a stronger acid than H+ or a stronger base than OH- in water.
In liquid ammonia, acids and bases are leveled to NH2- (a stronger base) and NH4+ (a weaker acid)
What about in CH3COOH (l)?

31. Where does the equilibrium lie?
HC2O H2O  C2O4-2 + H3O+
CH3COOH + NH2-  NH3 + CH3COO-

32. Where does the equilibrium lie
Where does the equilibrium lie? (on the side with the weaker acid and base!)
HC2O H2O  C2O4-2 + H3O+
CH3COOH + NH2-  NH3 + CH3COO-

33. Where does the equilibrium lie?
HC2O H2O  C2O4-2 + H3O+
(left, K<<1)
CH3COOH + NH2-  NH3 + CH3COO
(right, K>>1)

34. Where does the equilibrium lie?
F- + HNO  HF + NO2-
HCN + C6H5NH2  CN- + C6H5NH3+
HCO2H + NH3  HCO2- + NH4+

35. ?pH
What is the pH of a .25 M acetic acid solution?

36. ?pH What is the pH of a .25 M acetic acid solution?
Use the ICE method.
CH3COOH  CH3COO- + H+
I) .25 M 0M 0M
C) -x +x +x
E) .25-xM xM xM

37. Ka= [CH3COO-][H+] / [CH3COOH]
(This might be written Ka =[A-][H+] / [HA] or Ka =[Ac-][H+] / [HAc])
=(xM)(xM)/(.25-xM)=1.8 x 10-5
The two sides are equal where x=[H+]=.0021M, pH=2.68

38. For a weak acid solution…
x is very small, the .25 M doesn’t change very much. Try it.
Ka≈(x)(x)/(.25M)=1.8 x 10-5
The two sides are equal where x=[H+]=.0021M, pH=2.67

39. ?pH Various examples, using ICE method— Concentrate on changes moles
molarity&volumes
Henderson Hasselbach equation (later)

40. Titration--easy
32.8 mL of a .114 M sodium hydroxide solution is used to titrate 25.0 mL of an unknown nitric acid sample.
What is the concentration of the nitric acid solution?

41. Titration– mostly easy
48.6 mL of a sodium hydroxide solution of unknown concentration is used to titrate g KHP (Potassium hydrogen phthalate, a monoprotic acid, MW g/mol). What is the concentration of the unknown sodium hydroxide solution?

42. Titration– pretty easy
g of Mg(OH)2 dissolved in mL of H2O. This is titrated against a solution of HNO3. The initial volume of the burette is 2.35 ml. The final volume of the burette is ml. What can you determine?

43. Titration—not too easy
Start with ml of .100 M sulfuric acid. Add 5.00 ml .100 M KOH. What’s in the mixture? Add ml more. What’s in the mixture? Write the total, total ionic, and net ionic equations for these reactions.

44. Titration– you didn’t think it was easy, did you?
What is the pH of ml of a .115 M oxalic acid solution?
Add 4.00 ml of .170 NaOH. ?pH
Add another ml (19.00 ml total) of the NaOH. What is the new pH?

45. Diprotic acids Start with what you are given,
Conclude the direction of change based on K’s
Calculate things

46. Overtitration Once you’ve neutralized a weak acid or base, ignore it.
The pH is based on the excess of strong acid or base added after.
Subtract what was used in the neutralization, divide the excess by the total volume

47. Buffer solutions Resist changes in pH
Composed of significant amounts of a weak acid and its conjugate base
Can be a partly neutralized weak acid, or mixed with the sodium salt as the base
In the buffer range, the amount shifting is insignificant—ignore it. At the edges, use ICE

48. The Henderson-Hasselbach equation
Finding pH of a buffer solution
pH=pKa + log([base]/[acid])
pOH=pKb + log([acid]/[base])
--applies poorly when fraction is too large or small

49. Weak acid/base problems
4.66
9.26
3.15
3.55
2.83
.30
4.40
10.54
11.18
2.67
1.99
9.22
2.35
4.74
11.43
4.44
3.47
5.05
4.79
2.70 x 10-9
1.85 x 10-10

50. Salt hydrolysis
Q: What happens to the pH of a solution when you add a salt?

51. Salt hydrolysis
Q: What happens to the pH of a solution when you add a salt?
A: It depends if it is an acidic salt or a basic salt.

52. “I don’t know. Are you a good witch or a bad witch?”
Salt hydrolysis
Q: What happens to the pH of a solution when you add a salt?
A: It depends if it is an acidic salt or a basic salt.
“I don’t know. Are you a good witch or a bad witch?”

53. Neutral salts
If the cation is a group 1 metal, and the anion is the conjugate base of a strong acid, no effect.

54. If not…
The cation will show some tendency to associate itself with hydroxide—making more H+ in solution
The anion will show some tendency to associate itself with H+, leaving more OH- in solution

55. Generally-- Small, highly charged cations are more acidic
Kb’s are given for many anions
If Ka>Kb, the solution is acidic, If Ka<Kb, the solution is basic

56. Acidic or basic solutions? (Which ions have no effect?)
Aqueous:
Na2SO4
KBr
NH4Cl
Al(NO3)3
CuCl2
Li3PO4
NH4C2H3O2

57. Solubility Equilibria
When a minimally soluble salt dissolves in water,

58. Solubility Equilibria
Write the reaction for dissolving a minimally soluble ionic compound
Write the equilibrium expression
Given a Ksp—calculate solubility
Given solubility—calculate Ksp
Calculate solubility in a solution with one of the ions already.

60. Arrhenius acids and bases
Substances that ionize in water to form H+ ions are acids.
Substances that ionize in water to form OH- ions are bases.

61. BrØnsted-Lowry Definition
Substances that donate a proton (H+ ion) in a reaction are acids.
Substances that accept a proton (H+ ion) are bases.

62. Lewis Definition
Substances that accept an electron pair in a reaction are acids.
Substances that donate an electron pair are bases.

63. Conjugates
After an acid has donated a proton, the rest of the species is the conjugate base.
HAA- + H+
After a base has accepted a proton, the resulting species is the conjugate acid.
B- + H+ HB