1. Chapter 5: Reactions in Aqueous Solution
Chemistry Lecture 10
Chapter 5: Reactions in Aqueous Solution
Chapter Highlights
definitions of electrolytes & nonelectrolytes
recognize acids & bases
predict solubility of ionic compounds in water
determine net ionic equations
learn four basic reactions types & predict products
acid base
precipitation
gas-forming
oxidation -reduction

2. 100 % dissociation = strong electrolyte
Chemistry Lecture 10
Electrolytes
Definition: a substance whose aqueous solution conducts electricity is called an electrolyte
a substance can be a strong electrolyte, a weak electrolyte or a nonelectrolyte depending on the degree of dissociation (ionization) in solution
Example:
For sodium chloride, the ionic solid dissociates 100 % in water forming exclusively Na+ and Cl- ions in solution
100 % dissociation = strong electrolyte
NaCl(s) H2O(l) Na+(aq) Cl-(aq)

3. Electrolytes pure water acetic acid potassium dichromate solution
Chemistry Lecture 10
Electrolytes
pure water
acetic acid
solution
potassium dichromate
solution

4. Identifying Electrolytes
Chemistry Lecture 10
Identifying Electrolytes
Strong electrolytes : Substances that dissociate completely in water. Simple salts like NaCl that are combination of a metal and a nonmetal
Weak electrolytes : Substances that do not dissociate fully in water but do form some ions. Usually molecular compounds like acetic acid (CH3COOH) with ionizable groups (H+)
Nonelectrolytes : Substances that do not dissociate in water to form ions. Molecular compounds which are soluble but which remain intact as the molecule in solution

5. What type of electrolytes are these compounds?
Chemistry Lecture 10
Question:
What type of electrolytes are these compounds?
a) Epsom’s salt: MgSO4 . 7 H2O
b) Methanol: CH3OH
c) Acetic acid: CH3COOH

6. MgSO4 . 7 H2O(s) + H2O(l) Mg2+(aq) + SO42-(aq) strong electrolyte
Chemistry Lecture 10
Answer: a) b) c)
MgSO4 . 7 H2O(s) H2O(l) Mg2+(aq) SO42-(aq)
strong electrolyte
CH3OH(l) H2O(l) CH3OH(aq)
nonelectrolyte
CH3COOH(l) H2O(l) H+(aq) CH3COO-(aq)
weak electrolyte

7. Understanding & Predicting Reactions in Solution
Chemistry Lecture 10
Understanding & Predicting Reactions in Solution
Driving Force: a property of the reaction that can be identified as the reason for product formation
Examples:
NaCl(aq) KNO3(aq)
Na+(aq) Cl-(aq) K+(aq) NO3-(aq)
NaCl(aq) AgNO3(aq)
Na+(aq) NO3-(aq) AgCl(s)
solid formation

8. Chemistry Lecture 10
Types of Reactions
the reaction type depends on the driving force of the reaction. There are four basic types
Formation of an insoluble compound
Formation of a nonelectrolyte
Formation of a gas
Transfer of electrons

9. NaCl(aq) + AgNO3(aq) NaNO3(aq) + AgCl(s)
Chemistry Lecture 10
Solubility
certain combinations of cations and anions are soluble; that is they dissolve in water…..
if a compound will not dissolve in water it is insoluble
if a combination of anion and cation results in the formation of an insoluble solid, this is a precipitate
Example:
NaCl(aq) AgNO3(aq) NaNO3(aq) AgCl(s)
precipitate

10. Sulfates of Sr2+, Ba2+ & Pb2+
Chemistry Lecture 10
Soluble Compounds
Exceptions
Almost all salts of Na+,
K+ & NH4+
All salts of
Cl-, Br- & I-
Halides of Ag+, Hg22+ & Pb2+
Fluorides of Mg2+, Ca2+,
Sr2+ & Pb2+
Salts containing F-
Salts of
NO3-, ClO3-,
ClO4-, CH3CO2-
Salts of SO4-
Sulfates of Sr2+, Ba2+ & Pb2+

11. Most metal hydroxides (OH-) & oxides (O2-)
Chemistry Lecture 10
Insoluble Compounds
Exceptions
All salts of CO32-,
PO43-, C2O42-,
CrO42-, S2-
Most metal hydroxides (OH-) & oxides (O2-)
Salts of NH4+, and
alkali metal cations

12. Chemistry Lecture 10
Net Ionic Equations
the balanced equation that results from the omission of all spectator ions is the net ionic equation
spectator ions are the ions which do not participate in the reaction
Example:
Write a balanced net ionic equation for the reaction of AgNO3 with CaCl2 to produce AgCl and Ca(NO3)2.

13. 2 AgNO3 + CaCl2 2 AgCl + Ca(NO3)2
Chemistry Lecture 10
Step 1: Write the complete balanced equation with appropriate stoichiometry
2 AgNO CaCl AgCl Ca(NO3)2
Step 2: Decide on the physical state (eg solubility) of each compound.
2 AgNO3(aq) CaCl2(aq) AgCl(s) Ca(NO3)2(aq)

14. AgNO3(aq) Ag+(aq) + NO3-(aq)
Chemistry Lecture 10
Step 3: Recognize that all soluble ionic compounds dissociate to form ions in aqueous solution
AgNO3(aq) Ag+(aq) NO3-(aq)
CaCl2(aq) Ca2+(aq) Cl-(aq)
Ca(NO3)2(aq) Ca2+(aq) NO3-(aq)
2 Ag+(aq) NO3-(aq) Ca2+(aq) Cl-(aq)
2 AgCl(s) Ca2+(aq) NO3-(aq)

15. complete ionic equation
Chemistry Lecture 10
Step 4: Identify the spectator ions and remove them from the complete ionic equation to give the net ionic equation. Simplify the resulting equation in terms of stoichiometric coefficients.
2 Ag+(aq) NO3-(aq) Ca2+(aq) Cl-(aq)
2 AgCl(s) Ca2+(aq) NO3-(aq)
complete ionic equation
Ag+(aq) Cl-(aq) AgCl(s)
net ionic equation
The sum of ion charges is the same on
both sides of the net ionic equation

16. Precipitation Reactions
Chemistry Lecture 10
Precipitation Reactions
Write the net ionic equation for
the reaction of Pb(NO3)2 with KI.
Overall Reaction
Pb(NO3)2(aq) KI(aq) PbI2(s) KNO3(aq)
Pb2+(aq) NO3-(aq) K+(aq) I-(aq)
PbI2(s) K+(aq) NO3-(aq)
Net Ionic Equation
Pb2+(aq) I-(aq) PbI2(s)

17. HCl(g) + H2O(l) H+(aq) + Cl-(aq) NaOH(s) + H2O(l) Na+(aq) + OH-(aq)
Chemistry Lecture 10
Acids & Bases
Acid: any substance that , when dissolved in water, increases the concentration of hydrogen ions, H+, in the water
Base: any substance that, when dissolved in water, increases the concentration of hydroxide ions, OH-, in the water
HCl(g) H2O(l) H+(aq) Cl-(aq)
NaOH(s) H2O(l) Na+(aq) OH-(aq)

18. Chemistry Lecture 10
Strong Vs. Weak
A strong acid or strong base: an acid or base which ionizes completely in water; a strong electrolyte
A weak acid or base: an acid or base which does not ionize completely in water; a weak electrolyte
HCl(g) H2O(l) H+(aq) Cl-(aq)
CH3COOH(l) H2O(l) H+(aq) CH3COO-(aq)

19. Write the net ionic equation for the reaction of HNO3 with KOH.
Chemistry Lecture 10
Acid-Base Reactions I
Write the net ionic equation for
the reaction of HNO3 with KOH.
Overall Reaction
HNO3(aq) KOH(aq) KNO3(aq) HOH(l)
H+(aq) NO3-(aq) K+(aq) + OH-(aq)
K+(aq) + NO3-(aq) H2O(l)
Net Ionic Equation
H+(aq) OH-(aq) H2O(l)

20. Acid-Base Reactions II
Chemistry Lecture 10
Acid-Base Reactions II
Write the net ionic equation for
the reaction of CH3CO2H with Ca(OH)2.
Overall Reaction
2 CH3CO2H(aq) Ca(OH)2(s)
Ca(CH3CO2)2(aq) HOH(l)
Net Ionic Equation
2 CH3CO2H(aq) Ca(OH)2(s)
Ca2+(aq) (CH3CO2)- (aq) H2O(l)

21. Some Common Acids & Bases
Chemistry Lecture 10
Some Common Acids & Bases
Strong Acids
Strong Bases
HCl, HBr, HI, HNO3, H2SO4, HClO4
NaOH, KOH, Ca(OH)2
Weak Acids
Weak Bases
CH3CO2H, H3PO4, HF, H2CO3
NH3
H2SO4(l) H+(aq) HSO4-(aq)
Note:
HSO4-(aq) H+(aq) SO42-(aq)

22. Gas-Forming Reactions
Chemistry Lecture 10
Gas-Forming Reactions
The acids of some nonmetal ions are gases and a small number of aqueous acids easily decompose to form a gaseous product.
Examples
2 H+(aq) S2-(aq) H2S(g)
H+(aq) CN-(aq) HCN(g)
H2CO3(aq) H2O(l) CO2(g)
H2SO3(aq) H2O(l) SO2(g)

23. Gas-Forming Reactions
Chemistry Lecture 10
Gas-Forming Reactions
Write the net ionic equation for
the reaction of HNO3 with NiCO3.
Overall Reaction
2 HNO3(aq) NiCO3(s) Ni(NO3)2(aq) H2CO3(aq)
H2CO3(aq) H2O(l) CO2(g)
Net Ionic Equation
NiCO3(s) H+(aq) Ni2+(aq) H2O(l) CO2(g)

24. Properties of Compounds in Aqueous Solution
Chemistry Lecture 10
Properties of Compounds in Aqueous Solution
Aqueous solution: a solution of any substance or substances dissolved in water
Example:
Solid sodium chloride dissolves in water to give an aqueous solution of sodium cations and chloride anions
NaCl(s) H2O(l) Na+(aq) Cl-(aq)
aqueous solution
of sodium chloride

25. Oxides of Metals & Nonmetals
Chemistry Lecture 10
Oxides of Metals & Nonmetals
If a nonmetal oxide is dissolved in water an acidic solution results. This compounds is known as an acidic oxide
If a metal oxide is dissolved in water a basic solution results. This compounds is known as a basic oxide
SO3(g) H2O(l) H2SO4(aq)
H+(aq) + HSO42-(aq)
CaO(s) H2O(l) Ca(OH)2(aq)
Ca2+(aq) OH-(aq)

26. Summary: Types of Reactions
Chemistry Lecture 10
Summary: Types of Reactions
the reaction type depends on the driving force of the reaction. There are four basic types
Reaction Type Driving Force
Precipitation Reaction Formation of an insoluble compound
Acid-Base Neutralization Formation of a nonelectrolyte (water)
Gas-Forming Evolution of a water insoluble gas
Oxidation -reduction Transfer of electrons

27. Textbook Questions From Chapter #5
Chemistry Lecture 10
Textbook Questions From Chapter #5
Solubility: , 22, 24
Precipitation Reactions: 32
Net Ionic Equations: 36, 37
Reaction Types: 42, 49, 52, 57
Concentration/Stoichiometry: 70, 75, 76
Titration: , 86, 90

28. Chapter 5: Chemical Reactions
Chemistry Lecture 11
Chapter 5: Chemical Reactions
Chapter Highlights
definitions of electrolytes & nonelectrolytes
recognize acids & bases
predict solubility of ionic compounds in water
determine net ionic equations
learn four basic reactions types & predict products
acid base
precipitation
gas-forming
oxidation -reduction

29. Chemistry Lecture 11
Types of Reactions
the reaction type depends on the driving force of the reaction. There are four basic types
Reaction Type Driving Force
Precipitation Reaction Formation of an insoluble compound
Acid-Base Neutralization Formation of a nonelectrolyte (water)
Gas-Forming Evolution of a water insoluble gas
Oxidation -reduction Transfer of electrons

30. Oxidation-Reduction Reactions
Chemistry Lecture 11
Oxidation-Reduction Reactions
Write the net ionic equation for
the reaction of Cu with AgNO3.
Overall Reaction
2 AgNO3(aq) Cu(s) Cu(NO3)2(aq) Ag(s)
Net Ionic Equation
2 Ag+(aq) Cu(s) Cu2+(aq) Ag(s)

31. Oxidation-Reduction Reactions
Chemistry Lecture 11
Oxidation-Reduction Reactions
2 AgNO3(aq) Cu(s) Cu(NO3)2(aq) Ag(s)

32. Redox Reactions and Electron Transfer
Chemistry Lecture 11
Redox Reactions and Electron Transfer
All oxidation-reduction reactions involve the transfer of electrons between substances
2 Ag+(aq) Cu(s) Cu2+(aq) Ag(s)
2 Ag+(aq) e Ag(s)
Ag+ accepts electrons from Cu and is reduced to Ag
Ag+ is the oxidizing agent
Cu(s) Cu2+(aq) e-
Cu donates electrons to Ag+ and is oxidized to Cu2+
Cu is the reducing agent

33. Oxidation Numbers Question:
Chemistry Lecture 11
Oxidation Numbers
Question:
How can you tell an oxidation-reduction reaction when you see one ?
Answer:
Look for a change in oxidation number for an element(s)
Example
2 Ag+(aq) e Ag(s)
silver is reduced from oxidation state +1 to oxidation state 0

34. Guidelines For Determining Oxidation Numbers
Chemistry Lecture 11
Guidelines For Determining Oxidation Numbers
Each atom in a pure element has an oxidation number of 0.
The oxidation state for Cu in metallic Cu is 0 and is zero for each atom in I2 or S8.
For ions consisting of a single atom, the oxidation number is equal to the charge on the ion.
Al forms cation Al3+ which has oxidation number +3.
Cl forms anion Cl- which has oxidation number -1.

35. Guidelines For Determining Oxidation Numbers
Chemistry Lecture 11
Guidelines For Determining Oxidation Numbers
Fluorine is always -1 in compounds with other elements.
The oxidation state for F is -1 for the F- ion and -1 in compounds such as BrF5
Cl, Br, & I are always -1 in compounds except when combined with F or O.
The oxidation state for Cl is -1 for the Cl- ion and -1 in compounds such as CCl4, but is + 1 ClF.

36. Guidelines For Determining Oxidation Numbers
Chemistry Lecture 11
Guidelines For Determining Oxidation Numbers
The oxidation number of H is +1 and of O is -2 in most compounds.
Exceptions are very few BUT…. binary compounds between metals and hydrogen are hydrides (H-) and H has oxidation state -1.
Exceptions are very few BUT… peroxide O22- has oxygen in oxidation state -1.
The algebraic sum of the oxidation numbers in neutral compounds must be 0; in a polyatomic ion, the sum must be equal to the ion charge.

37. Assigning Oxidation Numbers
Chemistry Lecture 11
Assigning Oxidation Numbers
Question:
What are the oxidation numbers of:
a) Lithium and oxygen in Li2O
b) Manganese and oxygen in MnO4-

38. Assigning Oxidation Numbers
Chemistry Lecture 11
Assigning Oxidation Numbers
Answers:
a) Rule 1 states that the Li+ ion will have the oxidation state. This gives O2- an oxidation state of -2 which is consistent with Rule 5. Total is 0, correct for a neutral compound
b) Rule 5 states O should have oxidation state -2, for a contribution of -8. To have a total oxidation state sum of -1 for a monoanion Mn must be +7.

39. Common Oxidizing Agents
Chemistry Lecture 11
Common Oxidizing Agents
O2, oxygen O2-, oxide
X2, halogens X-, halide
HNO3, nitric acid NO, NO2 nitrogen oxides
Cr2O72-, dichromate Cr3+ ion (in acid)
MnO4-, permanganate Mn2+ ion (in acid)

40. Common Reducing Agents
Chemistry Lecture 11
Common Reducing Agents
H2, hydrogen H+, hyrogen ion
M, metals (Na, K, Al, Fe...) Mn+, metal cation
C, carbon (to reduce metal oxides) CO, CO2

41. Products of Simple Redox Reactions
Chemistry Lecture 11
Products of Simple Redox Reactions
Write the net ionic equation for the
reaction of Na with Cl2
Na is a good reducing agent
Na Na e-
Cl2 is a good oxidizing agent
Cl e Cl-
the overall redox equation would then be:
2 Na(s) Cl2(g) NaCl(s)

42. Products of Simple Redox Reactions
Chemistry Lecture 11
Products of Simple Redox Reactions
Write the net ionic equation for the
reaction of K with H2O
K is a good reducing agent
K K e-
H+ can be reduced to molecular hydrogen
2 H e H2
the overall redox equation would then be:
2 K(s) H2O(l) K+(aq) OH-(aq) H2(g)

43. Products of Simple Redox Reactions
Chemistry Lecture 11
Products of Simple Redox Reactions
Write the net ionic equation for the
reaction of Fe2O3 with Al
Al is a good reducing agent
Al Al e-
a metal oxide (Fe2O3) can be reduced back to the metal
Fe e Fe
the overall redox equation would then be:
Fe2O3(s) Al(s) Fe(s) Al2O3(s)

44. Summary: Types of Reactions
Chemistry Lecture 11
Summary: Types of Reactions
the reaction type depends on the driving force of the reaction. There are four basic types
Reaction Type Driving Force
Precipitation Reaction Formation of an insoluble compound
Acid-Base Neutralization Formation of a nonelectrolyte (water)
Gas-Forming Evolution of a water insoluble gas
Oxidation -reduction Transfer of electrons

45. Some Problems Question:
Chemistry Lecture 11
Some Problems
Question:
Write the net ionic equation for the reaction of aqueous solutions of K2CO3 and HClO4.

46. Chemistry Lecture 11
Some Problems
Answer:
Write a balanced equation with appropriate physical states indicated.
2 HClO4(aq) K2CO3(aq) KClO4(aq) H2CO3(aq)
and Remember
H2CO3(aq) H2O(l) CO2(g)

47. Some Problems Express the equation in a fully ionic form.
Chemistry Lecture 11
Some Problems
Express the equation in a fully ionic form.
2 H+(aq) ClO4-(aq) K+(aq) CO32-(aq)
2 K+(aq) ClO4-(aq) H2O(l) CO2(g)
Eliminate spectator ions and reduce to simplest stoichiometry.
2 H+(aq) CO32-(aq) H2O(l) CO2(g)

48. Some More Problems Question:
Chemistry Lecture 11
Some More Problems
Question:
Write the net ionic equation for the reaction of Ca with aqueous HCl.

49. Some More Problems Answer:
Chemistry Lecture 11
Some More Problems
Answer:
Write a balanced equation with appropriate physical states indicated.
Ca(s) HCl(aq) CaCl2(aq) H2(g)
because
Ca Ca e-
Ca is a good reducing agent
H+ can be reduced to molecular hydrogen
2 H e H2

50. Some More Problems Express the equation in a fully ionic form.
Chemistry Lecture 11
Some More Problems
Express the equation in a fully ionic form.
Ca(s) H+(aq) Cl-(aq)
Ca2+(aq) Cl-(aq) H2(g)
Eliminate spectator ions and reduce to simplest stoichiometry.
Ca(s) H+(aq) Ca2+(aq) H2(g)

51. Chapter 5: Chemical Reactions
Chemistry Lecture 12
Chapter 5: Chemical Reactions
Chapter Highlights
mass calculations related to stoichiometric reactions
limiting reagents & percent yields
empirical formula from analytical data
molarity
calculations from titration & standardization data

52. Solution Stoichiometry
Chemistry Lecture 12
Solution Stoichiometry
Molarity, M: An expression of concentration - the moles of solute per litre of solution
For example:
Suppose g of KMnO4 is dissolved in enough water to make a 250 mL solution. What is the concentration of the KMnO4(aq) solution ?

53. Chemistry Lecture 12
Making a Solution

54. Molarity of a Solution Step 1: Calculate the number of moles of KMnO4.
Chemistry Lecture 12
Molarity of a Solution
Step 1: Calculate the number of moles of KMnO4.
(0.435 g KMnO4) =
2.75 x 10-3 mol KMnO4
Step 2: Calculate the molarity of the KMnO4 solution.
molarity of KMnO4 = = M

55. stoichiometric factor
Chemistry Lecture 12
grams reactant A
stoichiometric factor x
direct calculation
not possible
grams product B
moles reactant A
moles reactant B
x (MW)
(molarity of A) x (volume A)
(molarity of B) x (volume B)

56. Textbook Questions From Chapter #5
Chemistry Lecture 12
Textbook Questions From Chapter #5
General Stoichiometry: 7
Limiting Reagents: 15, 17
Percent Yields: 21, 23
Chemical Analysis: 25, 27
Empirical Formula: 31, 33
Solution Stoichiometry: 45, 51
Titrations: , 61, 65
General & Review: 69, 73, 79, 85, 87