1. Applications of Aqueous Equilibria
McMurry & Fay ch. 15

2. Vocabulary: BUFFER SOLUTION an aqueous solution containing both an acid and its conjugate base

3. Calculate the pH in a solution prepared by dissolving 0
Calculate the pH in a solution prepared by dissolving 0.10 mol of solid NH4Cl in L of 0.40 M NH3 (Kb = 1.8 x 10-5). I C E
[NH4+][OH-]
[NH3]
Kb = = 1.8 x 10-5

4. [H+][A-] [HA] Ka = [H+][A-] [HA] - log Ka = - log [A-] [HA]
A shortcut
[H+][A-]
[HA]
Ka =
[H+][A-]
[HA]
- log Ka = - log
[A-]
[HA]
- log Ka = - log [H+] – log
= pKa
= pH

5. Henderson-Hasselbalch Equation
pH = pKa + log

6. Uses for Henderson-Hasselbalch Equation
Calculate pH of buffer solution
Prepare buffer solution of given pH

7. Calculating pH of buffer solution
Calculate the pH in a solution prepared by dissolving 0.10 mol of solid NH4Cl in L of 0.40 M NH3 (Kb = 1.8 x 10-5).
pH = pKa + log
[A-]
[HA]

8. Calculating pH of buffer solution
A buffer solution is created with 0.10 M HF (Ka = 6.8 x 10-4) and 0.15 M NaF. What is the pH?
pH = pKa + log
[A-]
[HA]

9. Buffer solutions resist changes in pH: adding acid
H2CO H+ + HCO3-

10. Buffer solutions resist changes in pH: adding base
H2CO3 + OH H2O + HCO3-

11. To calculate change in pH:
Choose appropriate reaction:
HA ⇌ A- + HA if adding acid
BH+ + OH- ⇌ B + H2O
Make an ICE table that shows number of moles of each species
Calculate change using # moles of base or acid added
Use final # of moles & Henderson-Hasselbalch eqn to calculate final pH

12. How much would the pH change if 100 mL of 1.0 HCl was added to water?
1.0 L of buffer solution has 0.30 M HNO2 (Ka=7.1 x 10-4) and 0.30 M NaNO mL of 1.0 M HCl are added to the solution.
How much would the pH change if 100 mL of 1.0 HCl was added to water?
How much does the pH change when the HCl is added to the buffer solution?
Choose appropriate reaction:
HA ⇌ A- + HA if adding acid
BH+ + OH- ⇌ B + H2O
Make an ICE table that shows number of moles of each species
Calculate change using # moles of base or acid added
Use final # of moles & Henderson-Hasselbalch eqn to calculate final pH

13. Preparing buffer solution of given pH
1. Pick acid with pKa within 1 pH unit of desired pH
2. Use Henderson-Hasselbalch equation to calculate ratio between acid & conjugate base
3. Multiply ratio by a factor appropriate for necessary
buffer strength

14. Pick acid with pKa near desired pHHF
3.17
CH3CO2H
4.76
Ethylenediamine
6.85
HOCl
7.53
Tris*
8.07
NH4+
9.25
*Tris(hydroxymethyl)aminomethane

15. Make a buffer with pH 7.8 [A-] pH = pKa + log [HA]
1. Pick acid with pKa within 1 pH unit of desired pH HF
3.17
CH3CO2H
4.76
Ethylenediamine
6.85
HOCl
7.53
Tris*
8.07
NH4+
9.25
2. Use Henderson-Hasselbalch equation to calculate ratio between acid & conjugate base
3. Multiply ratio by a factor appropriate for necessary
buffer strength
pH = pKa + log
[A-]
[HA]

16. Acid-Base Titrations

17. pH changes during Titration: Strong Acid-Strong Base
As you add base to an acid solution:
Some acid is neutralized; # of moles of H+ decreases
Volume of solution increases
Molarity =
so [H+] will decrease and pH will increase
Moles
Liters

18. Calculating pH changes during Strong Acid-Strong Base Titration: before equivalence point
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of H+ remaining:
# moles acid at start - # moles base added
Calculate volume of sample:
Initial volume + volume of base added
Calculate [H+] (moles H+/volume) and pH

19. 20. 00 mL of 0. 400 M HCl is titrated with 0
20.00 mL of M HCl is titrated with M NaOH Calculate pH when you have added: (a) 0 mL of acid (b) 5 mL of acid (c) 7.5 mL of acid (d) 10 mL of acid (e) 15 mL of acid (f) 19 mL of acid (g) 20 mL of acid
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of H+ remaining:
# moles acid at start - # moles base added
Calculate volume of sample:
Initial volume + volume of base added
Calculate [H+] (moles H+/volume) and pH

20. What happens after the equivalence point?
All acid has been used up; OH- dominates the pH
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate initial moles of acid :
# moles H+ = (Molarity acid) * (volume acid)
Calculate moles of OH-:
# moles of base – initial moles of acid
Calculate volume of sample:
Initial volume + volume of NaOH added
Calculate [OH-] (moles OH-/volume), pOH, and pH

21. Equivalence point # moles acid = # moles base

22. pH changes during Titration: Weak Acid-Strong Base
What’s different from strong acid titrations
Acid doesn’t completely dissociate so pH depends on Ka
Adding base produces the conjugate base of the weak acid
Acid + conjugate base both present = buffer solution
Use Henderson Hasselbalch to calculate pH

23. pH changes during Titration: Weak Acid-Strong Base (before equivalence point)
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of conjugate base formed:
# moles A- = # moles base added
Calculate moles of acid remaining:
# moles HA left =
initial # moles acid - # moles base added
Use Henderson Hasselbalch to calculate pH

24. Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH
What is [H+] at the start? Ka =
[H+][A-]
[HA]

25. Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH
Where do we expect our equivalence point to be?

26. Example: titration of 10 mL M benzoic acid (pKa = 4.202) with M NaOH What is pH after you have added:
5 mL 10 mL 12.5 mL 15 mL 20 mL 22.5 mL 24.5 mL
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of conjugate base formed:
# moles A- = # moles base added
pH = pKa + log
[A-]
[HA]
Calculate moles of acid remaining:
# moles HA left =
initial # moles acid - # moles base added
Use Henderson Hasselbalch to calculate pH

27. After the equivalence point: No acid left
A- + H2O HA + OH-

28. equivalence point # moles acid = # moles base
Half equivalence point
[A-] = [HA]

29. Half-equivalence point
Point at which [A-] = [HA] Henderson-Hasselbalch equation:
[A-]
[HA]
pKa = pH – log

30. At half-equivalence point,
pH = pKa

31. pH changes during Titration: Weak Acid-Strong Base (at equivalence point)
All acid has been turned into conjugate base Solution of basic salt: calculate pH as done in ch. 14

32. pH changes during Titration: Weak Acid-Strong Base (after equivalence point)
Most OH- comes from excess NaOH
Calculate moles of acid you start with:
# moles acid = (Molarity acid) * (volume acid)
Calculate moles of base added:
# moles base = (Molarity base) * (volume base)
Calculate moles of OH- remaining:
# moles base added - # moles acid
Calculate volume of sample:
Initial volume + volume of base added
Use [H+] = 10-14/[OH-] to calculate pH

33. Indicators Dyes that are weak acids or bases
Color of acid form different from conjugate base
HIn (aq)  H+ (aq) + In– (aq)
Acid form base form
(one color) (another color)

34. Choosing an indicator
Color change begins when two forms are equal:
Choose indicator that has pKa = pH of equiv. point!

36. Polyprotic acids

37. Carbonate chemistry
CO2 + H2O H2CO3 (carbonic acid) H2CO3 H+ + HCO3- (bicarbonate) pKa = 6.35 HCO3- H+ + CO32- (carbonate) pKa = 10.33

38. 2 Equilibrium constants for diprotic acids:
H2A H+ + HA-
HA H+ + A2-
[H+][HA-]
[H2A]
[H+][A2-]
[HA-]
Ka1 =
Ka2 =
Each contributes to [H+]

39. Calculating pH
[H+]equilibrium = [H+]1stIonization + [H+]2ndIonization

41. Calculating pH
[H+]equilibrium = [H+]1stIonization + [H+]2ndIonization [H+]1stIonization << [H+]2ndIonization

42. Simplification: [H+]equilibrium = [H+]1stIonization
In other words, treat the acid as a monoprotic acid!

43. Simplification # 2 H2A H+ + HA- HA- H+ + A2-
[H+]equilibrium = [H+]1stIonization
H2A H+ + HA-
[HA-]equilibrium = [HA-]1stIonization
HA H+ + A2-

44. Simplification #2 means
HA H+ + A2-
[H+][A2-]
[HA-]
Ka2 =
[H+][A2-]
[H+]
Ka2 =
[A2-] = Ka2

45. Calculating equilibrium concentrations for polyprotic acids:
1. Assume [A2-]≈ Ka2
2. Calculate remaining concentrations by treating as monoprotic acid

46. 2. Calculate remaining concentrations by treating as monoprotic acid
Calculate concentrations of solute species in 0.30 M oxalic acid (Ka1 = 5.9 x 10-2; Ka2 = 6.4 x 10-5)
1. Assume [A2-]≈ Ka2
2. Calculate remaining concentrations by treating as monoprotic acid

47. Solubility: An Equilibrium Point of View

48. Vocabulary
SATURATED SOLUTION A solution containing as much solute as can possibly dissolve
UNSATURATED SOLUTION
A solution containing less solute than can possibly dissolve
SUPERSATURATED SOLUTION
A solution containing more solute than is supposed to be able to dissolve

49. Solubility—Equilibrium Point of View
NaCl (s) Na+ (aq) + Cl- (aq)

50. Solubility—Equilibrium Point of View
NaCl (s) Na+ (aq) + Cl- (aq) Special equilibrium constant, Ksp:
Ksp= [Na+][Cl-]

51. Slightly soluble salts: How much can actually dissolve?
CaCO3 (s) Ca2+ (aq) + CO32- (aq) Ksp = 3.4 x 10-9

52. Vocabulary: Molar solubility
The number of moles of salt required to create one liter of a saturated solution

53. Calculating Ksp from solubility information
Write a balanced equation for ion dissociation.
Convert grams per liter to molarity if necessary
Use stoichiometry to convert molarity of salt to molarity of ions
Use stoichiometry to convert molarity of salt to ion concentrations
Use ion concentrations to calculate Ksp

54. The molar solubility of Ca(OH)2 is 1.06 x 10-2 M. What is the Ksp?
Write a balanced equation for ion dissociation.
Convert grams per liter to molarity if necessary
Use stoichiometry to convert molarity of salt to molarity of ions
Use stoichiometry to convert molarity of salt to ion concentrations
Use ion concentrations to calculate Ksp

55. The solubility of PbI2 is 4.34 g/L. What is the Ksp?
Write a balanced equation for ion dissociation.
Convert grams per liter to molarity if necessary
Use stoichiometry to convert molarity of salt to molarity of ions
Use stoichiometry to convert molarity of salt to ion concentrations
Use ion concentrations to calculate Ksp

56. Calculating solubility from Ksp
Write a balanced equation for ion dissociation.
Use Ksp to calculate ion concentrations
Use stoichiometry to convert ion concentrations to molarity of salt
Convert molarity to grams per liter if necessary

57. The Ksp for Ag2CO3 is 8. 5 x 10-12. What is the molar solubility
The Ksp for Ag2CO3 is 8.5 x What is the molar solubility? What is the solubility in grams per liter?
Write a balanced equation for ion dissociation.
Use Ksp to calculate ion concentrations
Use stoichiometry to convert ion concentrations to molarity of salt
Convert molarity to grams per liter if necessary

58. The Ksp for Cu3PO4 is 1. 4 x 10-37. What is the molar solubility
The Ksp for Cu3PO4 is 1.4 x What is the molar solubility? What is the solubility in grams per liter?
Write a balanced equation for ion dissociation.
Use Ksp to calculate ion concentrations
Use stoichiometry to convert ion concentrations to molarity of salt
Convert molarity to grams per liter if necessary

59. CaCO3 (s) Ca2+ (aq) + CO32- (aq)
Common Ion Effect
Application of Le Châtelier’s Principle:
CaCO3 (s) Ca2+ (aq) + CO32- (aq)

60. CaCO3 (s) Ca2+ (aq) + CO32- (aq)
Common Ion Effect
Application of Le Châtelier’s Principle: If we add more CO32-, equilibrium will shift towards reactants
CaCO3 (s) Ca2+ (aq) + CO32- (aq)

61. CaCO3 (s) Ca2+ (aq) + CO32- (aq)
Common Ion Effect
Application of Le Châtelier’s Principle: If we add more CO32-, equilibrium will shift towards reactants
CaCO3 (s) Ca2+ (aq) + CO32- (aq)
When common ions are present in solution, calculate solubilities using ICE tables

62. Calculating Solubility: Common Ion Effect
1. Write a balanced equation for ion dissociation.
2. Make ICE table. Add initial concentration of common ion
3. Use stoichiometry to calculate concentration changes in terms of x
4. Calculate equilibrium concentrations
5. Use Ksp expression to solve for x
6. Use value of x to calculate solubility

63. The Ksp of CaCO3 is 3.4 x 10-9. What is the molar solubility of CaCO3 in a 0.20 M Na2CO3 solution?
1. Write a balanced equation for ion dissociation.
2. Make ICE table. Add initial concentration of common ion
3. Use stoichiometry to calculate concentration changes in terms of x
4. Calculate equilibrium concentrations
5. Use Ksp expression to solve for x
6. Use value of x to calculate solubility

64. The Ksp of PbCrO4 is 1.8 x What is the molar solubility of PbCrO4 in a M K2CrO4 solution?
1. Write a balanced equation for ion dissociation.
2. Make ICE table. Add initial concentration of common ion
3. Use stoichiometry to calculate concentration changes in terms of x
4. Calculate equilibrium concentrations
5. Use Ksp expression to solve for x
6. Use value of x to calculate solubility

65. Will a precipitate form?
Only if solution is supersaturated
Supersaturation occurs if Q > Ksp

66. To determine if a precipitate will form…
1. Write the mass action expression for dissolution reaction
2. Use given concentrations to calculate Q
3. Compare Q to Ksp.
If Q > Ksp, than precipitate forms.
If Q < Ksp, then no precipitate forms.

67. A 0. 10 M solution of NaCl is mixed with a 0. 10 M solution of AgNO3
A 0.10 M solution of NaCl is mixed with a 0.10 M solution of AgNO3. Will a precipitate form? Ksp for AgCl is 1.8 x 10-10
1. Write the mass action expression for dissolution reaction
2. Use given concentrations to calculate Q
3. Compare Q to Ksp.
If Q > Ksp, than precipitate forms.
If Q < Ksp, then no precipitate forms.
Molar solubility = 1.3 x 10-5

68. A 0. 0010 M solution of CaCl2 is mixed with a 0
A M solution of CaCl2 is mixed with a M solution of K2SO4. Will a precipitate form? Ksp for CaSO4 is 2.4 x 10-5.
1. Write the mass action expression for dissolution reaction
2. Use given concentrations to calculate Q
Ksp = 4.93 x 10-5
3. Compare Q to Ksp.
If Q > Ksp, than precipitate forms.
If Q < Ksp, then no precipitate forms.