1. Additional Aspects of Aqueous Equilibria 熊同銘 tmhsiung@gmail.com
Chapter 17
Additional Aspects of Aqueous Equilibria
熊同銘

2. Contents
Common-Ion Effect
Buffers
Acid–Base Titrations
Solubility Equilibria
Factors That Affect Solubility
Precipitation and Separation of Ions
Qualitative Analysis for Metallic

3. 1. Common-Ion Effect
Common Ion: The ion which is formed in two different ionization processes.
Example, CH3COO– is the common ion, formed in ionization processes, for CH3COOH and CH3COONa.

4. Effect of Acetate on the Acetic Acid Equilibrium
Effect of Acetate on the Acetic Acid Equilibrium. Acetic acid is a weak acid:
CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)
Sodium acetate is a strong electrolyte:
NaCH3COO(aq) → Na+(aq) + CH3COO–(aq)
The presence of acetate from sodium acetate in an acetic acid solution will shift the equilibrium, according to LeChâtelier’s Principle:

5. The Common-Ion Effect
“Whenever a weak electrolyte and a strong electrolyte containing a common ion are together in solution, the weak electrolyte ionizes less than it would if it were alone in solution.”

6. Sample Exercise 17.1 Calculating the pH When a Common Ion Is Involved
What is the pH of a solution made by adding 0.30 mol of acetic acid (Ka = 1.8x10-5) and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?
Solution
x = 1.8 × 10–5 M = [H+]
pH = –log(1.8 × 10–5) = 4.74

7. Sample Exercise 17.2 Calculating Ion Concentrations When a Common Ion Is Involved
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. (Ka of HF is 6.8x10-4)
Solution
[H+] = ( x) M ≃ 0.10 M
pH = 1.00

8. 2. Buffers
Buffer solution: A solution that changes pH only slightly when small amounts of a strong acid or a strong base are added.
Components of buffer solution:
a weak acid and its conjugate base, e.g., CH3COOH + CH3COONa.
a weak base and its conjugate acid, e.g., NH3 + NH4Cl

9. Ways to Make a Buffer
Mix a weak acid and a salt of its conjugate base or a weak base and a salt of its conjugate acid.
Add strong acid and partially neutralize a weak base or add strong base and partially neutralize a weak acid.

10. Common ion effect for acid-base buffer HF/F– for example:
* Adding a small amount of acid or base only slightly neutralizes one component of the buffer, so the pH doesn’t change much.

11. An Equation for Buffer Solutions
HA(aq) + H2O  H3O+(aq) + A–(aq)
Therefore, (1)
Henderson-Hasselbalch equation: (2)
Either equation (1) or (2) available to estimate pH of the buffer
In general: [HA]  CHA, [A–]  CNaA
Effective buffer：
Buffer range: pH = pKa  1 (i.e., [HA]/[A–]: 0.1~ 10)
Molarity of each buffer component > Ka at least 100 fold
CHA: [HA]originally
CNaA: [NaA]originally

12. Sample Exercise 17.3 Calculating the pH of a Buffer
What is the pH of a buffer that is 0.12 M in lactic acid [CH3CH(OH)COOH, or HC3H5O3] and 0.10 M in sodium lactate [CH3CH(OH)COONa or NaC3H5O3]? For lactic acid, Ka = 1.4 × 10–4.
Solution

13. Buffer capacity and Buffer range
Buffer capacity: The amount of strong acid and/or strong base that a buffer solution can neutralize while maintaining an essentially constant pH.
* pH will be the same for a conjugate acid–base pair of 1 M each or 0.1 M each; however, the buffer which is 1 M can neutralize more acid or base.
Buffer range: The range of pH values over which a buffer solution can maintain a fairly constant pH. (pH = pKa  1)

14. Preparing Buffer Solutions
Making a buffer solution with a desired pH

15. New pH of a buffer after strong acid or base is added

16. (2.0 L)(0.18 mol NH4Cl/L) = 0.36 mol NH4Cl
Sample Exercise Calculating a Buffer
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.)
Solution
pOH = – pH = – 9.00 = 5.00
[OH–] = 1.0 × 10–5 M
[NH3] ≈ 0.10 M
(2.0 L)(0.18 mol NH4Cl/L) = 0.36 mol NH4Cl

17. Sample Exercise 17.6 Calculating pH Changes in Buffers
A buffer is made by adding mol CH3COOH and mol CH3COONa to enough water to make L of solution. The pH of the buffer is 4.74 (Ka of CH3COOH is 1.8x10-5 ). (a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH(aq) solution is added. (b) For comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M NaOH(aq) solution to L of pure water.
Solution
(a)
L × 4.0 mol/L = mol mol OH–

18. pOH = –log[OH–] = –log(0.020) = +1.70
Continued
(b)
[OH–] = mol/1.005 L = M
pOH = –log[OH–] = –log(0.020) = +1.70
pH = 14 – (+1.70) = 12.30

19. 3. Acid–Base Titrations
Acid–Base Titrations
An acid (or base) solution of known concentration is slowly added to a base (or acid) solution of unknown concentration.
A pH meter or indicators are used to determine when the solution has reached the equivalence point.
Application of titrations:
To estimate of the concentration of one of the reactants.
To estimate the equilibrium constant for the reaction.

20. Generally, the indicator range is expressed as pH = pKa  1:
Acid–Base Indicators
Acid-base indicators: A weak organic acid or a weak organic base whose undissociated form differs in color from its conjugate base or its conjugate acid.
HIn H2O  In– H3O+
Color in Color in
acidic solution basic solution
Generally, the indicator range is expressed as pH = pKa  1:
Litmus: 石蕊

22. Neutralization Reactions and Titration Curves
Equivalence point: The point in a titration in which the reactants are in stoichiometric proportions. They consume each other, and neither reactant is in excess.
Endpoint: The point in a titration at which indicator changes color.
Titration curve: The graph of pH versus volume of titrant.
To choose a proper indicator, the end point should as closely as possible to the equivalence point of the titration.
The millimole (mmol):
L/1000
mol/1000 =
M = L
mol mL
mmol

23. (Continuous)
Choosing a proper acid-base indicator-1

24. (Continuous)
Choosing a proper acid-base indicator-2

25. Strong Acid-Strong Base Titrations
Example: Titrating HCl with NaOH
(1) At beginning:
[H3O+] ≡ CHCl
(2) Before the equivalence point:
calculating net excess moles of HCl, then [H3O+]
(3) At the equivalence point:
[H3O+] = 1.0 x 10−7, pH = 7.00
(4) After the equivalence point:
calculating net excess moles of NaOH, then [OH–], then, [H3O+]

26. Sample Exercise 17.7 Calculations for a Strong Acid–Strong Base Titration
Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of M NaOH solution have been added to 50.0 mL of M HCl solution.
Solution
(a)
50.0 mL mL = 99.0 mL = L
–log(1.0 × 10–3) = 3.00

27. Continued (b) 50.0 mL + 51.0 mL = 101.0 mL = 0.1010 L
pOH = –log(1.0 × 10–3) = 3.00
pH = – pOH = – 3.00 = 11.00

28. Titration of 50.0 mL of a 0.100 M HCl with a 0.100 M NaOH

29. Weak Acid-Strong Base Titrations Example : Titrating CH3COOH with NaOH
At the begining:
HA H2O  A– + H3O+
CHA–x x x
x = [H3O+]
Before the equivalence point:
Calculating net excess moles of HA and the produced A–, then, [HA], [A–], then, or

30. (3) At the equivalence point:***
Calculating the expected moles of A–, then, [A–], then,
A– H2O  HA + OH–
[A–]–x x x
x = [OH–], then [H3O+], then pH
After the equivalence point:
Calculating the net excess moles of OH–, then, [OH–], then, [H3O+], then pH

31. Sample Exercise 17.8 Calculations for a Weak Acid–Strong Base Titration
Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH3COOH (Ka = 1.8 × 10–5).
Solution
45.0 mL mL = 95.0 mL = L

32. Continued

33. Sample Exercise 17.9 Calculating the pH at the Equivalence Point
Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH3COOH with M NaOH.
Solution
The number of moles of CH3COOH in the initial solution:
Moles = M × L = (0.100 mol/L)( L) = 5.00 × 10–3 mol CH3COOH
Total volume at the Equivalence Point :
50.0 mL mL = mL = L
The concentration of CH3COO– at the Equivalence Point :
CH3COO–(aq) + H2O(l) CH3COOH(aq) + OH–(aq)
Kb = Kw/Ka = (1.0 × 10–14)/(1.8 × 10–5) = 5.6 × 10–10 x x x
0.05-x

34. Continued 0.0500 – x ≃ 0.0500 x = [OH–] = 5.3 × 10–6 M pOH = 5.28
pH = 8.72

35. Titration of 50.0 mL of a 0.100 M CH3COOH with a 0.100 M NaOH
The best buffering occurred at pH ≈ pKa

36. pH values at Equivalence point
Type of titration
Example
pH value
Strong acid-Strong base
HCl/NaOH 7
Weak acid by Strong base
CH3COOH/NaOH
>7
Weak base by Strong acid
NH3/HCl
<7

37. Effect of acid strength
Each curve represents titration of 50.0 mL of 0.10 M weak acid with 0.10 M NaOH.

38. Titration of a Weak Polyprotic Acid
Example : Titrating phosphorous acid (H3PO3), a diprotic acid, with NaOH
H3PO3 + H2O  H2PO3– + H3O+
H2PO3– + H2O  HPO32– + H3O+

39. Titrating 50.0 mL of 0.10 M H3PO3 is titrated with 0.10 M NaOH
pH ≈
½(pKa1 + pKa2)
[H3PO3] = [H2PO3-] pH = pKa1
[H2PO3-] = [HPO32-]
pH = pKa2

40. 4. Solubility Equilibria The Solubility-Product Constant, Ksp
Mass solubility: The number of grams of solute in one liter of a saturated solution (g/L).
Molar solubility (s): The number of moles of solute in one liter of a saturated solution (M).
Solubility product constant (Ksp): The product of the molar concentrations of the constituent ions:
AxBy(s)  xAm+(aq) + yBn–(aq) Ksp = [Am+]x[Bn–]y
s x·s y·s
Solubility equilibria calculations:
• Determining a value of Ksp from experimental data
• Calculating equilibrium concentrations with known Ksp

41. Procedure for converting between solubility and Ksp

42. Some Ksp At 25 oC

43. CaF2(s) Ca2+(aq) + 2 F–(aq)
Sample Exercise Writing Solubility-Product (Ksp) Expressions
Write the expression for the solubility-product constant for CaF2,
Solution
CaF2(s) Ca2+(aq) + 2 F–(aq)
Ksp = [Ca2+][F–]2

44. Ksp = [Ag+]2[CrO42–] = (1.3 × 10–4)2(6.5 × 10–5) = 1.1 × 10–12
Sample Exercise Calculating Ksp from Solubility
Solid silver chromate is added to pure water at 25 ℃, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10–4 M. Assuming that the Ag2CrO4 solution is saturated and that there are no other important equilibria involving Ag+ or CrO42– ions in the solution, calculate Ksp for this compound.
Solution
Ag2CrO4(s) Ag+(aq) + CrO42–(aq)
2s s
Ksp = [Ag+]2[CrO42–]
Ksp = [Ag+]2[CrO42–] = (1.3 × 10–4)2(6.5 × 10–5) = 1.1 × 10–12

45. Sample Exercise 17.12 Calculating Solubility from Ksp
The Ksp for CaF2 is 3.9 × 10–11 at 25 ℃. Assuming equilibrium is established between solid and dissolved CaF2, and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.
Solution
Ksp = [Ca2+][F–]2 = (s)(2s)2 = 4s3 = 3.9 × 10–11

46. Factors That Affect Solubility
Presence of common ions
solution pH
presence of complexing agents

47. Common-Ion Effect
Followed Le Châtelier’s principle, i.e., the solubility of a slightly soluble ionic compound is lowered when a second solute that contains a common ion.
Example:
For equilibrated CaF2(s)  Ca2+(aq) + 2F–(aq)
What is happened once adding NaF.
What is happened once adding CaNO3.
What is happened once adding NaNO3
Ans:
Shift left
No effect

48. Ksp = 3.9 × 10–11 = [Ca2+][F–]2 = (0.010 + s)(2s)2
Sample Exercise Calculating the Effect of a Common Ion on Solubility
Calculate the molar solubility of CaF2 (Ksp = 3.1x10-11) at 25 ℃ in a solution that is (a) M in Ca(NO3)2 and (b) M in NaF.
Solution
(a)
CaF2(s)  Ca2+(aq) F2–(aq) I
C s s
E s s
Ksp = 3.9 × 10–11 = [Ca2+][F–]2 = ( s)(2s)2

49. Ksp = 3.9 × 10–11 = [Ca2+][F–]2 = (s)(0.010 + 2s)2
Continued
(b)
CaF2(s)  Ca2+(aq) F2–(aq) I
C s s
E s s
Ksp = 3.9 × 10–11 = [Ca2+][F–]2 = (s)( s)2

50. Solubility and pH Example 1: AgCl(s)  Ag+(aq) + Cl–(aq)
pH decrease ([H+] increase), no effect on solubility
Example 2:
PbF2(s)  Pb2+(aq) + 2F–(aq) Dissolution reaction
2H3O+(aq) + 2F–(aq)  2HF (aq) Acid-base reaction
PbF2(s) + 2H3O+(aq)  Pb2+(aq) + 2HF(aq) + 2H2O(l)
pH decrease ([H+] increase), solubility increase.
Example 3:
Mg(OH)2(s)  Mg2+(aq) + 2OH–(aq) Dissolution reaction
2H3O+(aq) + 2OH–(aq)  4H2O(l) Acid-base reaction
Mg(OH)2(s) + 2H3O+(aq)  Mg2+(aq) + 4H2O(l)
pH decrease ([H3O+] increase), solubility increase.

51. Formation of Complex Ion
Consists of a central metal atom or ion bonded with ligands.
Metal center, Lewis acid, accepts electron pairs
Ligands, Lewis bases, donate electron pairs.
At least one lone pair of electron in the Lewis structures of ligands (Lewis base).
The ligands and metal center are jointed by coordinate covalent (dative) bonds.
The constant of formation reaction of a complex ion is called a formation constant (or stability constant).

52. Common ligands examples
Complex ion formation example

53. Some Formation Constants

54. The Effect of Complex Ion Equilibria on Solubility
AgCl(s) in NH3(aq) solution for example:
AgCl(s)  Ag+(aq) + Cl–(aq) Ksp = 1.77 x 10–10
Ag+(aq) + 2NH3(aq)  [Ag(NH3)2]+(aq) Kf = 1.7 x 107
AgCl(s) + 2NH3(aq)  [Ag(NH3)2]+(aq) + Cl–(aq) K = Ksp x Kf = 3.0 x 10–3
In the presence of a high concentration of NH3, a significant amount of AgCl dissolves.

55. Examples of Amphoteric hidroxides: Al(OH)3, Zn(OH)2, and Cr(OH)3 etc.
*****
Amphoterism
Amphoteric oxides and hydroxides: Oxides and hydroxides that are only slightly soluble in water but that dissolve in either acidic or basic solutions.
Examples of Amphoteric hidroxides: Al(OH)3, Zn(OH)2, and Cr(OH)3 etc.
Amphoterism: 兩性

56. Amphoterism and Solubility

57. 6. Precipitation and Separation of Ions
Whether Precipitation Occurs
Calculating Q the reaction quotient:
Q > Ksp precipitation occur
Q = Ksp saturated, i.e., at equilibrium
Q < Ksp precipitation cannot occur
* The effect of dilution when solutions are mixed must be considered.
Quotient: 商值

58. Sample Exercise 17.16 Predicting Whether a Precipitate Forms
Does a precipitate form when 0.10 L of 8.0 × 10–3 M Pb(NO3)2 is added to 0.40 L of 5.0 × 10–3 M Na2SO4? (Ksp for PbSO4 is 6.3 × 10–7)
Solution.
PbSO4(s)  Pb2+(aq) + SO42–(aq) Ksp = 6.3 x 10–7
moles of Pb2+
concentration of Pb2+
moles of SO42–
concentration of SO42–
Q > Ksp, PbSO4 precipitates

59. Selective Precipitation of Ions
CuS (Ksp = 6x10-37), ZnS (Ksp = 2x10-25)

60. Sample Exercise 17.17 Selective Precipitation
A solution contains 1.0 × 10–2 M Ag+ and 2.0 × 10–2 M Pb2+. When Cl– is added, both AgCl(Ksp = 1.8 × 10–10) and PbCl2(Ksp = 1.7 × 10–5) can precipitate. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?
Solution
For AgCl:
For PbCl2:
AgCl precipitates first

61. 7. Qualitative Analysis for Metallic Elements
Classical qualitative inorganic analysis
Identify inorganic ions by acid-base chemistry, precipitation reactions, oxidation-reduction, and complex-ion formation etc.
Qualitative interest in what is present, not how much is present.
Quantitative analysis is concerned with quantity, or the amounts of substances in a solution or mixture.

62. Procedures of Qualitative Analysis

63. A General Qualitative Analysis Scheme

64. End of Chapter 17