1. Advanced Acid/Base Theory

2. pH of Strong Acids Strong acids completely dissociate in water.
Monoprotic strong acids release one mole of hydrogen ions for every mole of acid introduced into the water.
This makes pH calculations relatively easy.

3. Sample Question What is the pH of a 0.20 M solution of HCl(aq)?
[H+] = [HCl] because of complete dissociation.
pH = log [H+]
-log 0.20 = 0.70

4. Calculating pH of a strong bases
Strong bases also completely dissociate in water.
What is the pH of a M solution of NaOH?
What the pH of a M solution of Ba(OH)2?
[H+] = 3.57x10-13 M
pH =12.45
[H+] = 4.55x10-12 M
pH = 11.34

5. Another way of calculating it.
pOH = -log [OH-] ions
What would be the sum of pH and pOH of all aqueous solutions?
So what is the pOH of a M solution of NaOH?
What must it’s pH then be?
pOH 2.66
= 11.34

6. Another note about strong bases
Alkali oxides and alkaline earth metal oxides form strongly basic solutions in water.
Na2O + H2O  2 Na+ + 2 OH-
O2- + H2O  2 OH-
Note: These are not equilibrium reactions.

7. Weak Acids and pH
Write the equilibrium expression for the generic acid (HA) in water.
HA  H+ and A-

8. Ka – The acid dissociation constant.
How does Ka relate to acid strength?
Note Ka for weak acids are typically between 10-3 and
Because dissociation is incomplete is weak acids, [acid] cannot be directly used to calculate pH.
Ka is analogous to the equilibrium constant, so if it is less than 10-3, then it doesn’t dissociate very much.

9. Determining Ka from pH
A 0.10 M solution of methanoic acid (HCOOH) has a pH of What is the value of Ka?
HCOOH + H2O HCOO- + H3O+
Is there anything in this expression that we can determine from the information given?
Mol ratios, so every mol of HCOO- produced will = the amount of H3O+ produced, so whatever the [ ] of one will be the other. Also, the pH relates to the [H3O+].

10. pH = -log[H3O]+
10-pH = [H3O+]
[H3O+] = = 4.2x10-3 mol dm-3
Using this info, let’s build an ICE chart

11. ICE, ICE Baby HCOOH H3O+ HCOO- Initial 0.10 M Change - 4.2x10-3 M
Change
- 4.2x10-3 M
+ 4.2x10-3 M
Equilibrium
0.10 (What???)
4.2x10-3 M
4.2x10-3 M
Remember when we did equilibrium constants and we assumed that the amount of dissociation was so small, it really did not effect the initial concentration enough for it to change from it’s original concentration.

12. Percent Ionization Another less common measure of acid strength.
What is the % ionization in the previous problem?
4.2 %

13. Using Ka to calculate pH
Calculate the pH of a 0.20 M solution of HCN, Ka = 4.9x10-10.
Remember in weak acids, Δ[acid] is negligible.
Ka and Kb values for common weak acids and bases are the data booklet. Why aren’t Ka values for strong acids and bases included?

14. Solving Equation HCN H3O+ CN- Initial 0.20 M Change - x M + x M
Change
- x M
+ x M
Equilibrium
0.20 (Why???)

15. pH = -log [H+] pH = - log 9.9x10-6 = 5.00

16. Does concentration affect % ionization?
Calculate the percent ionization for HF(aq) solutions of 0.10 M and M?
Ka = 6.8x10-4.

17. Polyprotic Acids Acids with more than one dissociable H. Examples:
H2SO4
H3PO4
H2C6H6O6
H2CO3

18. Writing Ka Expressions
H3C6H5O H+ + H2C6H5O Ka1 = 7.4x10-4
H2C6H5O H+ + HC6H5O Ka2 = 1.7x10-5
HC6H5O H+ + C6H5O Ka3 = 4.0x10-7

19. Calculations with polyprotic acids
If the difference between successive Ka values is greater than 103, the pH of a solution can be estimated to be wholly due to the first Ka.
If the difference is less than 103, both values must be considered.

20. Sample Problem
What is the pH of a M solution of H2CO3? Ka1 = 4.3x10-7, Ka2 = 5.6x10-11
H2CO3 (aq) H+(aq) HCO3- (aq) M
- x
+ x
M - x +x

22. Weak Bases
Weak bases react with water to remove water and form hydroxide ions.
Write the reaction of ammonia with water in the space below.

23. Kb
Kb is the equilibrium expression for bases. Write the Kb expression for ammonia in water.

24. Sample Problem
Calculate the [OH-] of a 0.15 M solution of NH3 (aq). What is the pH of the solution? (Kb = 1.8x10-5)
X2/0.15=1.8x10-5
X2 = 2.7x10-6
X=1.6x10-3
pOH = 2.80
pH = 11.20

25. What are the weak bases? They are…
Molecules containing atoms with lone pairs of electrons,
e.g. NH3, any amine (N containing substance with only 3 bond pairs, leaving the 4th pair unattached.
Anions of weak acids
e.g. CH3COO-,

26. Ka and Kb for conjugate acid-base pairs
Write the reaction for ammonia going into aqueous solution with Kb expression.
Write the forward reaction for the conjugate acid of ammonia reacting with Ka expression .

27. Sum the two reactions.

28. Simplify the (Ka)(Kb) expression

29. Important
The product of Ka and Kb for any conjugate acid-base pair is always equal to Kw.
If Ka for HF is 6.8x104, what is the Kb for F-?

30. Buffer Solutions

31. Look at the following reaction
CH3COOH + H2O CH3COO- + H3O+
Ka = 1.8x10-5
What’s the acid?
What’s the base?
Conjugate acid?
Conjugate base?
Is the acid weak or strong?

32. Shifting the equilibrium
How would adding CH3COONa (sodium ethanoate) change the equilibrium?

33. Common ion effect
When a weak electrolyte and a strong electrolyte contain a common ion, the weak electrolyte ionizes less than it would if it were in solution alone.

34. What is the pH of a solution made by adding 0
What is the pH of a solution made by adding 0.30 mol of acetic acid to 0.30 mol of sodium acetate in enough water to make a 1.0 L solution?

35. Solving common ion problems
Identify strong and weak electrolytes.
Determine the source of H3O+ ions.
Determine the concentration of ions involved in the equilibrium.
Use equilibrium constant to make calculations

36. CH3COOH + H2O CH3COO- + H3O+ I
0.30 C -x +x E
(0.30-x)
(0.30+x)

37. Solve for hydronium ion concentration
X is small compare to and can be ignored
Ka = x(0.30)/0.30
Ka = x = 1.8x10-5
pH = -log 1.8e-5 = 4.74

38. Compare this to the pH of a solution of 0.30 M acetic acid

39. Calculate the fluoride ion concentration and pH of a solution that is 0.20 M HF and 0.10 M HCl. HF Ka = 6.8x10-4
Ka = 6.8e-4 = (H)(F)/HF = (0.10+x)(x)/(0.20-x) = (0.10)(x)/(0.20)
x = 0.2/0.1(6.8e-4) = 1.4e-3 = [F-]
[H+] = (0.10+x) = 0.10 = 1.00

40. Buffered Solutions
Solutions that resist drastic changes in pH upon the addition of a strong acid or base.
The solutions contain common ions as discussed previously.

41. Imagining a Buffered System
Must contain a weak acid (HX) and the ion that is its conjugate base (X-).
What would the equation and equilibrium expression for this reaction look like?

42. What are the major determinants of pH in this system?
Ka and ration of [HX]/[X-]

43. Adding a stress
How would adding NaOH to this solution affect the solution?
How would adding HCl to this solution affect the solution?

44. Determing the pH of a buffer system.
For the hypothetical acid HX, write the equilibrium expression and rearrange to solve for [H+]
Take the –log of both sides of the equation.
[H+] = Ka(HX/X-)
-log [H+] = -log Ka – log (HX/X-)
pH = pKa – log (HX/X-) or pH = pKa + log (X-/HX)

46. Henderson-Hasselbach Equation
Determines the pH of a buffer solution

47. Sample problem
What is the pH of a buffer solution that is 0.12 M lactic acid (HC3H5O3) and 0.10 M sodium lactate (NaC3H5O3)? Ka for lactic acid is 1.4x10-4.
Solve this problem two different ways.
ICE chart
Henderson-Hasselbach
1.7x10e-4

48. Calculating pH change when a strong acid or base is added to a buffered solution.
A 2.00 L solution containing mol of acetic acid and mol of sodium acetate has a pH of 4.74.
Calculate the pH of the solution after mol of NaOH is added.
What would the be the pH after mol of NaOH be if added to pure water?

49. Write out the reaction and set up an ICE chart
CH3COOH + OH-  H2O + CH3COO-
NOTE:
Strong bases react completely with weak acids
Strong acids react completely with weak bases.
pH = log (0.320/0.280)= 4.80
pH of water solution  14 = pH + pOH  pH = 14 – (-log [OH-]) = 14 – (-log 0.020) = 12.30

50. Acid-Base Titrations
A base of known concentration is added to an acid of unknown concentration OR
An acid of known concentration is added to a base of unknown concentration

51. What can we learn from acid-base titrations?
Concentration of an acid/base
Ka of a weak acid
Kb of a weak base

52. Strong Acid Strong Base Titrations
Curve shows how pH of a mL sample of M solution of HCl changes as M NaOH is added to the solution

53. Phases of a Titration (SA-SB)
Initial pH of the acid (or base)
The time between the initial pH and the equivalence point. pH first increases slowly and then very rapidly as the solution approaches the equivalence point.

54. Phases of a Titration (SA-SB)
The equivalence point. Equal numbers of moles of NaOH and HCl have reacted. pH is 7.
For strong acid-strong base titrations the equivalence point is at pH = 7.00 because Na+ and Cl- have no effect on pH.
How would the equivalence point change when titrating weak acids and bases?

55. Phases of Titrations
After the equivalence point. The pH of the solution is determine by the concentration of the excess NaOH.

56. Sample problem
Calculate the pH when 49.0 mL of M NaOH are added to 50.0 mL of M HCl.
Calculate the pH when 51.0 mL of M NaOH are added to 50.0 mL of M HCl.
pH = 3.00
pH = 11.00

57. Weak Acid-Strong Base Titrations
mL of M NaOH added to mL of M CH3COOH

58. Phases of Titration (WA-SB)
Initial pH of the acid. Ka can be used to determine this.
Between the initial pH and equivalence point. A buffer system is established at this point.

59. Phases of Titration (WA-SB)
Equivalence point- Equal number of moles of acid and base are present in the solution.
pH is not equal to Why?
Does Na+ have an effect on pH?
Does CH3COO- have an effect on pH?

60. Phases of Titration (SB-WA)
After the equivalence point: The OH- from NaOH has a dramatic effect the [OH-] of the solution.
The OH- from the reaction of acetate with water in negligable compared to the OH- from NaOH.

61. Sample Problem
Calculate the pH of the solution formed

62. Calculating pH at Equivalence Point
Calculate pH at the equivalence point of a titration of 50.0 mL of M acetic acid and M NaOH.

63. CH3COOH + OH-  CH3COO- + H2O
What is the original number of moles of acetic acid?
How many moles of OH- must be added?
How many mole acetate will form at equivalence point?

64. CH3COOH + OH-  CH3COO- + H2O
What is the concentration of acetate at the equivalence point?
Remember acetate is a weak base and the conjugate base of acetic acid.
For a conjugate acid/base pair Ka + Kb = Kw
What is the value of Kb?
Kb = 5.6e-10

65. CH3COOH + OH-  CH3COO- + H2O
What is the equilibrium expression for this Kb value.
Calculate [OH-] from this data.
Calculate pOH
Calculate pH
[OH-] = 5.3e-6
pOH = 5.28
pH = 8.72

66. Determining pKa from a titration
Write out Henderson-Hasselbach equation.
How could you make the
log [conj. base]/[acid] = 0?
At which point during a titration does this happen?

67. Determining pKa from a titration
Mathematically, how does pH relate to pKa at the half equivalence point?
How would you find Ka from pKa?