1. Reactions in Aqueous Solution
Chapter 5
Chapter 5

2. Compounds in Aqueous Solution
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
This is a reaction in which reactants are in solution
Solution – homogeneous mixture composed of two parts:
solute – the medium which is dissolved
solvent – the medium which dissolves the solute.
Chapter 5

3. Compounds in Aqueous Solution
Compounds in Water
Some compounds conduct electricity when dissolved in water – electrolytes
Those compounds which do not conduct electricity when dissolved in water are called – nonelectrolytes
Chapter 5

4. Compounds in Aqueous Solution
Ionic Compounds in Water (Electrolytes)
The conductivity of the solution is due to the formation of ions when the compound dissolves in water
These ions are not the result of a chemical reaction, they are the result of a dissociation of the molecule into ions that compose the solid.
Chapter 5

5. Compounds in Aqueous Solution
Ionic Compounds in Water
Chapter 5

6. Compounds in Aqueous Solution
Molecular Compounds in Water(Nonelectrolytes)
In this case no ions are formed, the molecules just disperse throughout the solvent.
Chapter 5

7. Compounds in Aqueous Solution
Molecular Compounds in Water(Nonelectrolytes)
There are exceptions to this, some molecules are strongly attracted to water and will react with it.
Chapter 5

8. Compounds in Aqueous Solution
Strong and Weak Electrolytes
Strong electrolytes – A substance which completely ionizes in water.
For example:
Chapter 5

9. Compounds in Aqueous Solution
Strong and Weak Electrolytes
Weak electrolyte: A substance which partially ionizes when dissolved in water.
For example:
Chapter 5

10. Compounds in Aqueous Solution
Strong and Weak Electrolytes
Notice that the arrow in this reaction has two heads, this indicates that two opposing reactions are occurring simultaneously.
Chapter 5

11. Compounds in Aqueous Solution
Strong and Weak Electrolytes
Since both reactions occur at the same time, this is called a chemical equilibrium.
Chapter 5

12. Precipitation Reaction
Chapter 5

13. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Precipitation Reaction
A reaction which forms a solid (precipitate)
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
AgCl is classified as an insoluble substance
Chapter 5

14. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Precipitation Reaction
Net Ionic Equation
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
AgNO3 and NaNO3 are electrolytes in solution so they actually occur as free ions.
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  AgCl(s) + Na+(aq) + NO3-(aq)
Chapter 5

15. Precipitation Reaction
Net Ionic Equation
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)  AgCl(s) + Na+(aq) + NO3-(aq)
Notice that NO3-(aq) and Na+(aq) occur in both the left and right side of the equation.
These are called spectator ions.
Chapter 5

16. Precipitation Reaction
Net Ionic Equation
Ag+(aq) + Cl-(aq) AgCl(s)
With the spectator ions removed, the resulting equation shows only the ions involved in the reaction remain.
This is a net ionic equation.
Chapter 5

17. Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Most nitrates (NO3-) and acetates (CH3CO2-) are soluble in water.
All chlorides are soluble except: Hg+, Ag+, Pb2+, Cu+
All sulfates are soluble except: Sr2+, Ba2+, Pb2+
Carbonates (CO32-), Phosphates (PO43-), Borates (BO33-),Arsenates (AsO43-), and Arsenites (AsO33-) are insoluble.
Hydroxides (OH-) of group Ia and Ba2+ and Sr2+ are soluble.
Most sulfides (S2-) are insoluble.
Chapter 5

18. Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4
AgCH3CO2
(NH4)3PO4
KClO4
Chapter 5

19. Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4 Insoluble
AgCH3CO2
(NH4)3PO4
KClO4
Chapter 5

20. Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4 Insoluble
AgCH3CO2 Soluble
(NH4)3PO4
KClO4
Chapter 5

21. Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4 Insoluble
AgCH3CO2 Soluble
(NH4)3PO4 Soluble
KClO4
Chapter 5

22. Compounds in Aqueous Solution
Solubility Guidelines for Ionic Compounds
Predict the solubility of the following compounds:
PbSO4 Insoluble
AgCH3CO2 Soluble
(NH4)3PO4 Soluble
KClO4 Soluble
Chapter 5

23. Acids and Bases
Acid - substance which ionizes to form hydrogen cations (H+) in solution
Examples:
Hydrochloric Acid HCl
Nitric Acid HNO3
Acetic Acid CH3CO2H
Sulfuric Acid H2SO4
Sulfuric acid can provide two H+’s - Diprotic acid,
The other acids can provide only one H Monoprotic acid.
Chapter 5

24. Acids and Bases Diprotic acid H2SO4  H+ + HSO4- HSO4- Û H+ + SO42-
Chapter 5

25. Acids and Bases Base - substance which reacts with H+ ions. Examples:
ammonia NH3
sodium hydroxide NaOH
Chapter 5

26. Acids and Bases Acid-Base Reaction H+ + OH-  H2O
It is clear that the metal hydroxides (NaOH for example) provide OH- by disassociation.
Bases like ammonia make OH- by reacting with water (ionization)
NH3 + H2O Û NH OH-
Chapter 5

27. Acids and Bases Strong and Weak Acids and Bases
Strong acids and bases are strong electrolytes.
Weak acids and bases are weak electrolytes.
Chapter 5

28. Acids and Bases Strong Acids
The strength of acids and bases are concerned with the ionization (or dissociation) of the substance, not its chemical reactivity.
Example:
Hydrofluoric acid (HF) is a weak acid, but it is very chemically reactive.
this substance can’t be stored in glass bottles because it reacts with glass (silicon dioxide).
Chapter 5

29. Acids and Bases Common Strong Acids and Bases Common Strong Acids
Hydrochloric Acid HCl
Hydrobromic Acid HBr
Hydroiodic acid HI
Nitric Acid HNO3
Perchloric Acid HClO4
Sulfuric Acid H2SO4
Chapter 5

30. Acids and Bases Common Strong Acids and Bases Common Strong Bases
Lithium Hydroxide LiOH
Sodium Hydroxide NaOH
Potassium Hydroxide KOH
Chapter 5

31. HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Acids, Bases, and Salts
Neutralization Reaction
Reaction between an acid and a base.
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
“The neutralization between acid and metal hydroxide produces water and a salt”
Salt – an ionic compound whose cation comes from a base and anion from an acid.
Chapter 5

32. HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Acids, Bases, and Salts
Neutralization Reaction
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
- despite the appearance of the equation, the reaction actually takes place between the ions.
Chapter 5

33. HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Acids, Bases, and Salts
Neutralization Reaction
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Total Ionic Equation
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)
Chapter 5

34. HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Acids, Bases, and Salts
Neutralization Reaction
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Total Ionic Equation
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)
Chapter 5

35. HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Acids, Bases, and Salts
Neutralization Reaction
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
Total Ionic Equation
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)
Net Ionic Equation
H+(aq) + OH-(aq)  H2O(l)
Chapter 5

36. Gas-Forming Reactions
Metal Carbonates and Acid
2 HCl(aq) + Na2CO3(aq)  2 NaCl(aq) + H2O(l) + CO2(g)
Metal carbonates (or bicarbonates) always form a salt, water and carbon dioxide
Chapter 5

37. Gas-Forming Reactions
Metal Sulfide and Acid
2 HCl(aq) + Na2S(s)  H2S(g) + 2 NaCl(aq)
Metal sulfides form a salt and hydrogen sulfide.
Chapter 5

38. Gas-Forming Reactions
Metal Sulfite and Acid
2 HCl(aq) + Na2SO3(s)  SO2(g) + 2 NaCl(aq) + H2O(l)
Metal sulfites form a salt, sulfur dioxide and water.
Chapter 5

39. Gas-Forming Reactions
Ammonium Salt and Strong Base
NH4Cl(s) + NaOH(aq)  NH3(g) + NaCl(aq) + H2O(l)
This reaction forms ammonia, salt and water
Chapter 5

40. Oxidation-Reduction Reactions
Reaction where electrons are exchanged.
2 Na(s) H2O(l)  2 NaOH(aq) + H2(g)
Na(s)  Na+(aq) e-
oxidation – loss of electrons
2 H+(g) e-  H2(g)
reduction – gain of electrons
Chapter 5

41. Oxidation-Reduction Reactions
Reaction where electrons are exchanged.
2 Na(s) H2O(l)  2 NaOH(aq) + H2(g)
An alternate approach is to describe how one reagent effects another.
Reducing Agent, a substance that causes another substance to be reduced. Na(s)  Na+(aq) e-
Oxidizing Agent, a substance that causes another substnace to be oxidized H+(g) e-  H2(g)
Chapter 5

42. Oxidation-Reduction Reactions
Oxidation Numbers
Each atom of a pure element has an oxidation number of zero(0).
For monatomic ions, the oxidation number equals the charge on the ion.
Fluorine always has an oxidation state of -1 in compounds.
Cl, Br, and I always have oxidation numbers of -1, except when combined with oxygen or fluorine.
The oxidation number of H is +1 and O is -2 in most compounds.
The sum of the oxidation numbers must equal the charge on the molecule or ion.
Chapter 5

43. Oxidation-Reduction Reactions
Oxidation Numbers
Examples
PCl5
P 1( ) =
Cl 5( ) = ________
Chapter 5

44. Oxidation-Reduction Reactions
Oxidation Numbers
Example
PCl5
P 1( ? ) = ?
Cl 5(-1) = __-5____
Chapter 5

45. Oxidation-Reduction Reactions
Oxidation Numbers
Example
PCl5
P 1(+5) =
Cl 5(-1) = __-5____
Chapter 5

46. Oxidation-Reduction Reactions
Oxidation Numbers
Example
CO32-
C 1( ) =
O 3( ) = _____ -2
Chapter 5

47. Oxidation-Reduction Reactions
Oxidation Numbers
Example
CO32-
C 1(?) = ?
O 3(-2) = __-6__ -2
Chapter 5

48. Oxidation-Reduction Reactions
Oxidation Numbers
Example
CO32-
C 1(+4) = +4
O 3(-2) = __-6__ -2
Chapter 5

49. Oxidation-Reduction Reactions
Oxidation Numbers
Example
K2CrO4
K 2( ) =
O 4( ) =
Cr 1( ) = ________
Chapter 5

50. Oxidation-Reduction Reactions
Oxidation Numbers
Example
K2CrO4
K 2(+1) =
O 4(-2) =
Cr 1( ? ) = ___?____
Chapter 5

51. Oxidation-Reduction Reactions
Oxidation Numbers
Example
K2CrO4
K 2(+1) =
O 4(-2) =
Cr 1(+6) = ___+6__
Chapter 5

52. Solutions Molarity(M)
Unit of concentration, moles of solute per liter of solution.
Chapter 5

53. Solutions
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): g/mol
Chapter 5

54. Solutions
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): g/mol
Chapter 5

55. Solutions
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): g/mol
Solution volume
Chapter 5

56. Solutions
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
F.W. (NaCl): g/mol
Solution volume
Chapter 5

57. Solutions
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
Chapter 5

58. Solutions
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
Chapter 5

59. Solutions
Example: What is the molarity(M) of a solution which contains 17.51g of NaCl in 751mL of solution?
Chapter 5

60. Solutions
Molarity
Chapter 5

61. MdiluteVdilute = MconcentratedVconcentrated
Solutions
Dilution
MdiluteVdilute = MconcentratedVconcentrated
Chapter 5

62. 0.100M(500mL) = 6.00M(Vconcentrated)
Solutions
Dilution
Example: What volume of 6.00M NaOH is required to make 500mL of 0.100M NaOH?
Mconcentrated = 6.00M Mdilute = 0.100M
Vconcentrated = ? Vdilute = 500mL
0.100M(500mL) = 6.00M(Vconcentrated)
Vconcentrated = 8.33mL
Chapter 5

63. pH Scale Concentration scale for acids and bases.
The square brackets around the H+ indicate that the concentration of H+ is in molarity.
So, a change of 1 pH unit indicates a 10X change in H+ concentration.
Chapter 5

64. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
We can now use molarity to determine stoichiometric quantities.
Example
How many grams of hydrogen gas are produced when 20.0 mL of 1.75M HCl is allowed to react with 15.0g of sodium metal?
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Chapter 5

65. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Convert quantities to moles
Chapter 5

66. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Convert quantities to moles
Chapter 5

67. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Convert quantities to moles
Determine limiting reagent
Chapter 5

68. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Convert quantities to moles
Determine limiting reagent
Chapter 5

69. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Calculate moles of H2
Chapter 5

70. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Calculate moles of H2
Chapter 5

71. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Calculate moles of H2
Calculate grams of H2
Chapter 5

72. 2 HCl(aq) + 2 Na(s)  H2(g) + 2 NaCl(aq)
Solution Stoichiometry
2 HCl(aq) Na(s)  H2(g) NaCl(aq)
Calculate moles of H2
Calculate grams of H2
Chapter 5

73. Solution Stoichiometry
Titrations
Chapter 5

74. Practice Problems
4,10, 16, 20, 24, 30, 36, 38, 44, 60, 62, 68
Chapter 5