1. CHAPTER 4
Reactions in Aqueous Solutions 1 1

2. Aqueous Solutions aqueous solutions -solute dissolved in water
nonelectrolytes - aqueous solutions do not conduct electricity
C2H5OH - ethanol 10 10

3. Aqueous Solutions
C6H12O6 - glucose (blood sugar) 11 11

4. Aqueous Solutions
C12H22O11 - sucrose (table sugar) 12 12

5. Properties of Aqueous Solution
Molecular compounds in water (e.g., CH3OH): no ions are formed.
If there are no ions in solution, there is nothing to transport electric charge.

6. Properties of Aqueous Solution
Ionic Compounds in Water
Ions dissociate in water.
In solution, each ion is surrounded by water molecules.
Transport of ions through solution causes flow of current.

7. Properties of Aqueous Solution

8. Properties of Aqueous Solution
Strong electrolytes: completely dissociate in solution.
For example:
Weak electrolytes: produce a small concentration of ions when they dissolve.
These ions exist in equilibrium with the unionized substance.

9. Properties of Aqueous Solution
strong electrolytes - extremely good conductors of electricity
HCl, HNO3, etc.
strong soluble acids
NaOH, KOH, etc.
strong soluble bases
NaCl, KBr, etc.
soluble ionic salts
ionize in water essentially 100% 13 13

10. Properties of Aqueous Solution
weak electrolytes
CH3COOH, (COOH)2
weak acids
NH3, Fe(OH)3
weak bases
some soluble covalent salts
ionize in water much less than 100% 14 14

11. Strong and Weak Acids acids generate H+ in aqueous solutions
strong acids ionize 100% in water 15 15

12. Strong and Weak Acids acids generate H+ in aqueous solutions
strong acids ionize 100% in water 15 15

13. entire list of strong water soluble acids
Strong and Weak Acids
entire list of strong water soluble acids
HCl - hydrochloric acid
HBr - hydrobromic acid
HI - hydroiodic acid
HNO3 - nitric acid
H2SO4 - sulfuric acid
HClO3 - chloric acid
HClO4 - perchloric acid 16 16

14. Strong and Weak Acids weak acids ionize less than 100% in water
10% or less!
some common weak acids
HF - hydrofluoric acid
CH3COOH - acetic acid (vinegar)
H2CO3 - carbonic acid (soda water)
H2SO3 - sulfurous acid
HNO2 - nitrous acid
H3PO4 - phosphoric acid 17 17

15. Strong and Weak Acids
weak acids ionize as reversible or equilibrium reactions
CH3COOH acetic acid
why they ionize less than 100% 18 18

16. Strong Soluble Bases bases produce OH- ions in solution
strong soluble bases ionize 100% in water 19 19

17. Strong Soluble Bases entire list of strong soluble bases
LiOH - lithium hydroxide
NaOH - sodium hydroxide
KOH - potassium hydroxide
RbOH - rubidium hydroxide
CsOH - cesium hydroxide
Ca(OH)2 - calcium hydroxide
Sr (OH)2 - strontium hydroxide
Ba (OH)2 - barium hydroxide 20 20

18. Insoluble Bases ionic but insoluble in water not very basic
Cu(OH)2 - copper (II) hydroxide
Fe(OH)2 - iron (II) hydroxide
Fe(OH)3 - iron (III) hydroxide
Zn(OH)2 - zinc (II) hydroxide
Mg(OH)2 - magnesium hydroxide 21 21

19. Weak Bases covalent compounds that ionize slightly in water
NH3 - ammonia is most common 22 22

20. Solubility Rules strong acids strong bases soluble ionic salts
completely water soluble
strong bases
soluble ionic salts
Figure 4.3, page 80 of textbook 23 23

21. Acids, Bases, and Salts
Dissociation = pre-formed ions in solid move apart in solution.
Ionization = neutral substance forms ions in solution.
Acid = substances that ionizes to form H+ in solution (e.g. HCl, HNO3, CH3CO2H, lemon, lime, vitamin C).
Bases = substances that react with the H+ ions formed by acids (e.g. NH3, Drano™, Milk of Magnesia™).

22. Acids, Bases, and Salts

23. Reactions in Aqueous Solutions
molecular equations
all reactants & products in molecular or ionic form
total ionic equation
ions as they exist in solution 24 24

24. Reactions in Aqueous Solutions
net ionic equation
shows ions that participate in reaction and removes spectator ions
spectator ions do not participate in the reaction 25 25

25. Displacement Reactions
Metathesis reactions involve swapping ions in solution:
AX + BY  AY + BX.
Metathesis reactions will lead to a change in solution if one of three things occurs:
an insoluble solid is formed (precipitate),
weak or nonelectrolytes are formed, or
an insoluble gas is formed.

26. Displacement Reactions

27. Displacement Reactions

28. Precipitation Reactions
form an insoluble compound
crystals
molecular equation 26 26

29. Precipitation Reactions
form an insoluble compound
crystals
molecular equation
total ionic reaction 26 26

30. Precipitation Reactions
total ionic reaction 26 26

31. Precipitation Reactions
total ionic reaction
net ionic reaction 26 26

32. Precipitation Reactions
net ionic reaction 28 28

33. HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Acids, Bases, and Salts
Neutralization occurs when a solution of an acid and a base are mixed:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Notice we form a salt (NaCl) and water.
Salt = ionic compound whose cation comes from a base and anion from an acid.
Neutralization between acid and metal hydroxide produces water and a salt.

34. Acid-Base Reactions acid + base salt + water
recall the acids and bases from before
molecular equation 30 30

35. Acid-Base Reactions acid + base salt + water
recall the acids and bases from before
molecular equation
total ionic equation 30 30

36. Acid-Base Reactions
total ionic equation
net ionic equation 31 31

37. Acid-Base Reactions
net ionic equation 32 32

38. Acid-Base Reactions
molecular equation 33 33

39. Acid-Base Reactions
molecular equation
total ionic equation 33 33

40. Acid-Base Reactions
total ionic equation
net ionic equation 34 34

41. Acid-Base Reactions
net ionic equation 35 35

42. Molarity: Moles of solute per liter of solution.
Solution Composition
Solution = solute dissolved in solvent.
Solute: present in smallest amount.
Water as solvent = aqueous solutions.
Change concentration by using different amounts of solute and solvent.
Molarity: Moles of solute per liter of solution.
If we know: molarity and liters of solution, we can calculate moles (and mass) of solute.

43. Solution Composition
Molarity: Moles of solute per liter of solution.

44. Concentration of Solutions
solution - one substance dissolved in another, commonly one is a liquid
concentration - amount of solute dissolved in a solvent
“sweet tea”
mixed drinks
units of concentration 20 14 27 17 3 55 55 3

45. Molarity (Molar Concentration)
molarity = mol solute/L of solution = M
Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. 22 16 29 19 5 62 62 5

46. Molarity (Molar Concentration)
molarity = mol solute/L of solution = M
Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. 22 16 29 19 5 63 63 5

47. Molarity (Molar Concentration)
Determine the mass of calcium nitrate required to prepare 3.50 L of M Ca(NO3)2 . 22 16 29 19 5 64 64 5

48. Molarity (Molar Concentration)
Determine the mass of calcium nitrate required to prepare 3.50 L of M Ca(NO3)2 . 22 16 29 19 5 65 65 5

49. Dilution of Solutions take a concentrated solution and add water to it
make tea “less sweet”
number of moles of solute remains constant
M1V1 = M2V2 works because # of moles is constant
If 10 mL of 12 M HCl is added to enough water to give 100 mL of solution, what is concentration of solution? 24 18 31 21 7 68 68 7

50. Dilution of Solutions take a concentrated solution and add water to it
make tea “less sweet”
number of moles of solute remains constant
M1V1 = M2V2 works because # of moles is constant
If 10 mL of 12 M HCl is added to enough water to give 100 mL of solution, what is concentration of solution?
(10 mL)(12 M) = M2(100 mL)
M2 = 1.2 M HCl 24 18 31 21 7 69 69 7

51. Dilution of Solutions
What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? 25 19 32 22 8 70 70 8

52. Dilution of Solutions
What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? 25 19 32 22 8 71 71 8

53. Stoichiometry with Solutions
There are two different types of units:
laboratory units (macroscopic units: measure in lab);
chemical units (microscopic units: relate to moles).
Always convert the laboratory units into chemical units first.
Grams are converted to moles using molar mass.
Volume or molarity are converted into moles using M = mol/L.
Use the stoichiometric coefficients to move between reactants and product.

54. Stoichiometry with Solutions

55. Stoichiometry with Solutions
What volume of M BaCl2 is required to completely react with 4.32 g of Na2SO4? 26 20 33 23 9 76 76 9

56. Stoichiometry with Solutions
What volume of M NaOH will react with 50.0 mL 0f M aluminum nitrate? What mass of aluminum hydroxide precipitates?
Al(NO3)3 + 3 NaOH ® Al(OH) NaNO3 27 21 34 24 10 77 77 10

57. Stoichiometry with Solutions
What volume of M NaOH will react with 50.0 mL 0f M aluminum nitrate? What mass of aluminum hydroxide precipitates?
Al(NO3)3 + 3 NaOH ® Al(OH) NaNO3 27 21 34 24 10 78 78 10

58. Stoichiometry in Solution
What mass of Al(OH)3 precipitates in (a)? 79 79

59. Stoichiometry in Solution
What mass of Al(OH)3 precipitates in (a)? 80 80

60. Titrations

61. Titrations
Suppose we know the molarity of a NaOH solution and we want to find the molarity of an HCl solution.
We know:
molarity of NaOH, volume of HCl.
What do we want?
Molarity of HCl.
What do we do?
Take a known volume of the HCl solution, measure the mL of NaOH required to react completely with the HCl.

62. Titrations What do we get? Next step? Can we finish?
Volume of NaOH. We know molarity of the NaOH, we can calculate moles of NaOH.
Next step?
We also know HCl + NaOH  NaCl + H2O. Therefore, we know moles of HCl.
Can we finish?
Knowing mol(HCl) and volume of HCl (20.0 mL above), we can calculate the molarity.

63. Titrations
What is the molarity of a KOH solution if 38.7 mL of KOH solution is required to react with 43.2 mL of M HCl? 29 23 36 26 12 84 84 12

64. Titrations
What is the molarity of a barium hydroxide solution if 44.1 mL of M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? 29 23 36 26 12 85 85 12

65. Titrations
What is the molarity of a barium hydroxide solution if 44.1 mL of M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? 29 23 36 26 12 88 88 12

66. Titrations
What is the molarity of a barium hydroxide solution if 44.1 mL of M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? 29 23 36 26 12 89 89 12

67. Oxidation Number Rules
The oxidation number of any free, uncombined element is zero.
The oxidation number of an element in a simple (monatomic) ion is the charge on the ion.
In the formula for any compound, the sum of the oxidation numbers of all elements in the compound is zero. In a polyatomic ion, the sum of the oxidation numbers of the constituent elements is equal to the charge on the ion.

68. Oxidation Number Rules
Hydrogen, H, in combined form, has the oxidation number +1, usually.
Oxygen, O, in combined form, has the oxidation number -2, usually.

69. Oxidation Numbers
assign oxidation numbers to each element in these compounds
NaNO3 K2Sn(OH)6 H3PO4 37 37

70. Oxidation Numbers
assign oxidation numbers to each element in these compounds
NaNO3 K2Sn(OH)6 H3PO4 37 37

71. Oxidation Numbers
assign oxidation numbers to each element in these compounds
NaNO3 K2Sn(OH)6 H3PO4 37 37

72. Oxidation Numbers
assign oxidation numbers to each element in these compounds
NaNO3 K2Sn(OH)6 H3PO4 37 37

73. Oxidation Numbers
assign oxidation numbers to each element in these polyatomic ions
SO32- HCO3- Cr2O72- 39 39

74. Oxidation Numbers
assign oxidation numbers to each element in these polyatomic ions
SO32- HCO3- Cr2O72-
+4 -2 39 39

75. Oxidation Numbers
assign oxidation numbers to each element in these polyatomic ions
SO32- HCO3- Cr2O72- 39 39

76. Oxidation Numbers
assign oxidation numbers to each element in these polyatomic ions
SO32- HCO3- Cr2O72- 39 39

77. Redox Reactions oxidation - loss of electrons
oxidation number increases
reduction - gain of electrons
oxidation number decreases or reduces!
oxidizing agents - chemical species that gain electrons and oxidize some other species
electron thieves
they are reduced 40 40

78. Redox Reactions
reducing agents - chemical species that lose electrons and reduce some other species
victims of electron thieves
they are oxidized
How do we balance redox reactions? 41 41

79. Balancing Redox Reactions
Half-Reaction Method 36 37 38 38 38

80. Half Reaction Method
Balancing redox reactions by half reaction method rules:
Write the unbalanced reaction
Break into 2 half reactions -
1 for oxidation
1 for reduction
Mass balance each half reaction by adding appropriate stoichiometric coefficents
in acidic solutions we can add H+ or H2O
in basic solutions we can add OH- or H2O 37 38 39 39 39

81. Half Reaction Method
Charge balance the half reactions by adding appropriate numbers of electrons
Multiply each half reaction by a number to make the number of electrons added equal to the number of electrons consumed
Add the two half reactions
Eliminate any common terms 38 39 40 40 40

82. Half Reaction Method
Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. 39 40 41 41 41

83. Half Reaction Method
Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. 40 41 42 42 42

84. Half Reaction Method
Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. 41 42 43 43 43

85. Half Reaction Method
Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. 42 43 44 44 44

86. Half Reaction Method
Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 43 44 45 45 45

87. Half Reaction Method
Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 43 44 46 46 45

88. Half Reaction Method
Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 44 45 47 47 46

89. Half Reaction Method
Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 45 46 48 48 47

90. Half Reaction Method
Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 46 47 49 49 48

91. Half Reaction Method
Example: In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH)4-, to chromate ions, CrO42-. Write and balance the net ionic equation for this reaction. 47 48 50 50 49

92. Half Reaction Method 48 49 51 51 50

93. Half Reaction Method
Example: When chlorine is bubbled into basic solution, it forms chlorate ions and chloride ions. Write and balance the net ionic equation. 49 50 52 52 51

94. Half Reaction Method 50 51 53 53 52

95. Synthesis Question
Nylon is made by the reaction of hexamethylene diammine
with adipic acid 90

96. Synthesis Question in a 1 to 1 mole ratio. The structure of nylon is:
where the value of n is typically 450,000. On a daily basis, a DuPont factory makes 1.5 million pounds of nylon. How many pounds of hexamethylene diamine and adipic acid must they have available in the plant each day? 91

97. Group Activity
Manganese dioxide, potassium hydroxide and oxygen react in the following fashion:
A mixture of g of MnO2, 26.6 L of M KOH, and g of O2 is allowed to react as shown above. After the reaction is finished, g of K2MnO4 is separated from the reaction mixture. What is the per cent yield of this reaction? 92