1. Types of Reactions & Solution Stoichiometry
Chapter 4
Types of Reactions & Solution Stoichiometry

2. Unit essential Question:
How do chemicals react with one another in aqueous solutions?

3. Lesson essential questions (4.1-4.4):
1) How do water molecules interact with chemicals?
2) How is the concentration of a solution measured?

4. Water, the Common Solvent
Section 4.1
Water, the Common Solvent

5. Aqueous solutions δ+ Dissolved in water.
Good solvent- polar molecules.
Hydration: ions in salts break apart due to attraction to polar water molecules.
Example:
NH4NO3 (s)  NH4+ (aq) + NO3- (aq) δ- δ+

6. Hydration H O H O H O H O H O H O H O H O H O

7. Solubility
Amount of substance that will dissolve in a given amount of water.
If they do dissolve, ions are separated, and can move around.
Water can also dissolve non-ionic compounds if they have polar bonds.

8. “Like dissolves like”
Polar substances generally dissolve other polar and ionic substances
Alcohol is slightly polar and dissolves (mixes) in water
Nonpolar substances dissolve other nonpolar substances
Fat will not dissolve in water

9. The Nature of Aqueous Solutions: Strong & Weak Electrolytes
Section 4.2
The Nature of Aqueous Solutions: Strong & Weak Electrolytes

10. Parts of Solutions Solution- homogeneous mixture.
Solute- what gets dissolved.
Solvent- what does the dissolving.
Soluble- Able to be dissolved.
Miscible- liquids dissolve in each other.

11. Electrolytes
Electrolytes- ionic compounds in solution that conduct electricity.
Strong electrolytes- completely dissociate (fall apart into ions).
Many ions = conduct electricity well.
Weak electrolytes- partially dissociate into ions.
Few ions = conduct electricity slightly.
Non-electrolytes- don’t dissociate at all.
No ions = don’t conduct electricity.

12. Acid/Base Electrolytes
Arrhenius acid- forms H+ ions when dissolved.
Strong acids dissociate completely.
Ex: H2SO4 HNO3 HCl HBr HI
Weak acids do not dissociate completely.
Ex: HC2H3O2
Arrhenius base - forms OH- ions when dissolved.
Strong bases also dissociate completely.
Ex: KOH NaOH (Groups 1 & 2 hydroxides)
Remember that dissociating completely means many ions are formed!

13. Sections 1&2 Homework
Pg #1,9,18,19

14. Warm-Up
HNO3 is a strong acid. Write the chemical equation for a solution of HNO3. Will it conduct electricity?

15. Composition of Solutions
Section 4.3
Composition of Solutions

16. Measuring Composition of Solutions
To do stoichiometry:
Need to know chemicals
Need to know amounts (concentrations)
Concentration- how much is dissolved.
Molarity = Moles of solute
Liters of solution
abbreviated M (molar)
1 M = 1mol solute / 1 liter solution
Chemicals & forms = what species are present and if they dissociate into ions or not

17. Molarity Calculations
Can solve for:
Amount or mass of solid to dissolve
Moles of solute
Volume of solution
Standard solution
Solution whose concentration is accurately known.

18. Examples
Calculate the molarity of a solution prepared by dissolving 11.5g of solid NaOH in water to make 1.50L of solution. (pg. 134)
Give the concentration of each ion in 0.50 M Co(NO3)2. (pg. 135)
#27 pg. 172

19. Dilutions Stock solution – a concentrated solution
Dilution – number of moles of solute stays the same, just adding more water
M1V1 = M2V2
Example: #30 (a) pg. 172
mol1 x V1 = mol2 x V2 L1 L2

20. Section 3 Homework
Pg #21-23,28,31

21. Lesson essential questions (4.5-4.7):
1) How do we identify and work with precipitation reactions?

22. Precipitation Reactions
Section 4.5
Precipitation Reactions

24. Precipitation Reactions
Solid forms when two solutions of ionic compounds are mixed.
Precipitate (ppt)
To help you remember: ‘If you’re not a part of the solution, your part of the precipitate!’

25. Precipitation reactions
NaOH(aq) + FeCl3(aq) ® NaCl(aq) + Fe(OH)3(s)
is really:
Na+(aq)+OH-(aq) + Fe+3 + Cl-(aq) ® Na+ (aq) + Cl- (aq) + Fe(OH)3(s)
So all that really happens is
OH- (aq) + Fe+3 (aq) ® Fe(OH)3 (s)
Also a double displacement reaction
net ionic equation!

26. Precipitation reaction
Can predict products, but can only be certain by experimenting.
The anion and cation switch partners.
Only occurs if a product is insoluble!
Otherwise all the ions stay in solution- nothing has happened (spectators)
Memorize solubility rules! Pg. 144

27. Solubility Rules All nitrates, Na+, K+, NH4+ are soluble.
You must know this for the AP exam!
Additional solubility rules on pg. 144.

28. Describing Reactions in Solutions
Section 4.6
Describing Reactions in Solutions

29. Three Types of Equations
1. Formula Equation- write formulas, not ions.
K2CrO4(aq) + Ba(NO3)2(aq) ®
2. Complete Ionic equation- show dissolved electrolytes as the ions.
2K+ + CrO4-2 + Ba NO3- ® BaCrO4(s) + 2K+ + 2 NO3-
Spectator ions are those that don’t react- appear as ions on both sides.

30. Three Type of Equations
3. Net Ionic equation- show only ions that react, not spectator ions
Ba+2 + CrO4-2 ® BaCrO4(s)
If all species in a reaction are aqueous (soluble), write NR!

31. Sections 5&6 Homework
Pg #36,42,44

32. AP Practice Question
How many moles of Na2SO4 must be added to 500 milliliters of water to create a solution that has a 2- molar concentration of the Na+ ion? (Assume the volume of the solution does not change.)
 Think about what you need to answer this!
 Need to find moles Na+. Then find moles Na2SO4
0.5 moles
1 mole
2 moles
5 moles

33. Stoichiometry of Precipitation Reactions
Section 4.7
Stoichiometry of Precipitation Reactions

34. Stoichiometry of Precipitation
Steps for reference: pg.148
Similar to other stoichiometry problems we’ve done!
Sample problem: What volume of 0.15M KCl is needed to precipitate out all of the lead ions from 100.mL of 0.20M Pb(NO3)2?
Not all of these steps may be necessary, depending upon what is being asked and what information you are given.
270mL KCl needed

35. Section 7 Homework
Pg. 173 #47,48,50,54

36. Section 4.8
Acid-Base Reactions

37. Lesson essential question (4.8):
How do we classify acids and bases?
What happens when acids and bases are mixed together?

38. Acid-Base Reactions
For our purposes an acid is a proton donor, H+ (BrØnsted-Lowry theory).
A base is a proton acceptor, usually OH-
acid + base ® salt + water
H+ + OH- ® H2O
Practice: Write the net ionic equation for the acid/base rxn. below:
HNO3(aq) + NaOH(aq)  ?
Note: H2CO3 always breaks down into CO2 & H2O

39. Acid-Base Reactions
Follow same steps as precipitation reactions for stoichiometry problems.
See p
Practice: What volume (in mL) of 0.100M HCl will react completely with 25.00mL of M NaOH?
(1) Write net ionic equation
(2) Find moles you’re starting with
(3) Find moles needed
(4) Find volume needed
Answer: 50.0mL HCl

40. Acid-Base Reactions Also called neutralization reactions.
Use titrations to determine concentrations.
Titrant: solution of known concentration
Analyte: solution of unknown concentration
Equivalence Point: when enough titrant has been added to exactly react with the analyte (neutralization is complete).
Stoichiometric amounts come from balanced equation!
Tells us how many moles of the titrant fully reacted with the analyte- then can solve for moles of analyte!

41. Titration
Solution of known concentration (titrant), is added to the unknown (analyte), until the equivalence point is reached.
How do we know when the equivalence point has been reached?
Add indicator to analyte at the beginning

42. Titration Where the indicator changes color is the endpoint.
Ex: phenolphthalein used often
Pink in base, colorless in acid
As close as we can get to the equivalence point; still assume complete neutralization.
The solution will not turn pink until one drop after the equivalence point (when the solution is more basic).
Can also use titration for non acid/base substances to find amounts/concentrations.

43. AP Practice Question
Which of the following best represents the balanced net ionic equation for the reaction of lead(II) carbonate & concentrated hydrochloric acid? (All lead compounds are insoluble.)
Pb2CO3 + 2H+ + Cl-  Pb2Cl + CO2 + H2O
PbCO3 + 2H+ + 2Cl-  PbCl2 + CO2 + H2O
PbCO3 + 2H+  Pb+2 + CO2 + H2O
PbCO3 + 2Cl-  PbCl2 + CO3-2

44. AP Practice Question
The conductivity of a solution of Ba(OH)2 is monitored as the solution is titrated with 0.10 M H2SO4. The original volume of the Ba(OH)2 solution is 25.0 mL. A precipitate of BaSO4 is formed during the titration. The data collected from the experiment is plotted in the graph above.

45. Question Continued
As the first 30.0 mL of 0.10 M H2SO4 are added to the Ba(OH)2 solution, two types of chemical reactions occur simultaneously. Write the balanced net-ionic equations for (i) the neutralization reaction and (ii) the precipitation reaction.
(i) Equation for neutralization reaction:
(ii) Equation for precipitation reaction:
OH- (aq) + 2H+ (aq)  H2O (l)
Ba+2 (aq) + SO4-2 (aq)  BaSO4 (s)

46. Question Continued
2) The conductivity of the Ba(OH)2 solution decreases as the volume of added 0.10 M H2SO4 changes from 0.0 mL to 30.0 mL.
Identify the chemical species that enable the solution to conduct electricity as the first 30.0 mL of 0.10 M H2SO4 are added.
(ii) On the basis of the equations you wrote in question 1, explain why the conductivity decreases.
OH- (aq) & Ba+2 (aq) (Can’t be anything from H2SO4 because the ions immediately react.)
[Ba+2] in sltn. decrease as they precipitate out, and [OH-] in sltn. decrease as they react to form H2O. Note: be specific in your answers!! Reference all species and reactions!

47. Question Continued
Think about what information can be determined from this point!
At equivalence point:
complete neutralization
3) Using the information in the graph, calculate the molarity of the original Ba(OH)2 solution.
0.12M Ba(OH)2

48. Section 8 Homework
Homework: pg #56, 58, 60, 64, 66

49. Oxidation – Reduction Reactions
Section 4.9
Oxidation – Reduction Reactions

50. Lesson essential questions (4.9-4.10):
How can we identify redox reactions?
How do we assign oxidation states?
Why is balancing different for redox reactions?

52. Redox Reactions
Ionic compounds are formed through the transfer of electrons.
An oxidation-reduction reaction involves the transfer of electrons.
One element gains, one loses
Non-ionic compounds can also undergo redox reactions.

53. Oxidation States = ‘charge’
A way of keeping track of the electrons.
Not necessarily true of what is in nature, but it works.
Need to memorize rules for assigning (pg.156):
The oxidation state of elements in their standard states is zero.
Oxidation state for monatomic ions are the same as their charge.

54. Oxidation states
Oxygen is assigned an oxidation state of -2 in its covalent compounds except in peroxide (-1).
In compounds with nonmetals hydrogen is assigned the oxidation state +1.
In its compounds fluorine is always –1.
The sum of the oxidation states must be zero in compounds or equal the charge of the ion.

55. Practicing Oxidation States
Determine the oxidation states in the following:
Cl2
SO4-2
CaBr2
C6H12O6
Cl: 0
S: O: -2
Ca: Br: -1
C: H: O: -2

56. Section 9 Homework
Pg. 174 #67(c-e),68(a-c),72

57. Balancing Redox Reactions
Section 4.10
Balancing Redox Reactions

58. Oxidation-Reduction e- transferred, so the oxidation states change.
Oxidation is the loss of electrons.
More positive oxidation state.
Reduction is the gain of electrons.
More negative oxidation state.
OIL RIG
LEO (the lion goes) GER

59. Agents
Oxidizing agent- substance that gets reduced (causes oxidation in another species).
Gains electrons.
More negative oxidation state.
Reducing agent- substance that gets oxidized (causes reduction in another species).
Loses electrons.
More positive oxidation state.

60. Identify the… Oxidizing agent Reducing agent Substance oxidized
Substance reduced
#1: 2Na + Cl2  2NaCl
#2: CH4 + 2O2  CO2 + 2H2O
reducing agent, substance oxidized
oxidizing agent, substance reduced
reducing agent, substance oxidized
oxidizing agent, substance reduced

61. Half-Reactions
All redox reactions can be thought of as happening in two halves.
One produces electrons - oxidation half.
The other requires electrons - reduction half.
Ex: Fe (s) + CuSO4 (aq)  Cu (s) + FeSO4 (aq)
Net Ionic: Fe (s) + Cu+2 (aq)  Cu (s) + Fe+2 (aq)
Oxidation: Fe (s)  Fe+2 (aq) + 2e-
Reduction: Cu+2 (aq) + 2e-  Cu (s)
Don’t necessarily need net ionic equation in the example, but it may make it more clear to you what’s being oxidized and reduced.

62. Balancing Redox Equations
Redox reactions may involve an acid or base as a reactant.
The number of electrons produced must be the same as those required.
8 step procedure for acidic solution, 10 step procedure for basic solution.
Called the half reaction method.
Balance each half reaction, then combine for total balanced reaction

63. Balancing in Acidic Solution
Write separate half reactions.
For each half reaction balance all species except H and O.
Balance O by adding H2O to one side.
Balance H by adding H+ to one side.
Balance charge by adding e- to the more positive side.

64. Balancing in Acidic Solution
Multiply equations by a number to make electrons equal.
Add equations together and cancel identical species. Reduce coefficients to smallest whole numbers.
Check that charges and elements are balanced.
Flow chart on pg. 167

65. Balancing in Acidic Solution
Ex: Balance the following equation:
H+ (aq) + Cr2O7-2 (aq) + C2H5OH (l) 
Cr+3 (aq) + CO2 (g) + H2O (l)
Reduction: 6e- + 14H+ + Cr2O7-2  2Cr+3 + 7H2O
Oxidation: C2H5OH + 3H2O  2CO2 + 12H+ + 12e-
Final: 16H+ + 2Cr2O7-2 + C2H5OH  4Cr H2O + 2CO2
*Note: there should NOT be any e- in the final balanced equation! If so, not balanced!

66. Basic Solution
Do everything you would with acid, but add one more step.
Add enough OH- to both sides to neutralize the H+.
Any H+ and OH- on the same side form water. Cancel out any H2O’s on both sides.
Simplify coefficients, if necessary.

67. Balancing in Basic Solution
Assume previous example in acidic solution was actually in a basic solution.
Had: 16H+ + 2Cr2O7-2 + C2H5OH  4Cr H2O + 2CO2
For any H+ ions, add same number of OH- ions to both sides. This forms water with H+. Cancel out waters on both sides.
Now: 16H2O + 2Cr2O7-2 + C2H5OH  4Cr H2O + 2CO2 + 16OH-
16H+, so add 16OH-
5 H2O

68. Practice Balancing Redox Rxns.
Pg. 174 #74(b)
Pg. 174 #75(b)
Answer: 6Cl- + Cr2O7 + 14H+  3Cl2 + 2Cr+3 + 7H2O
Answer: 2OH- + Cl2  OCl- + Cl- + H2O

69. Side Note: Redox Titrations
Same as titrations discussed before, just looking at redox reactions instead of acid/base reactions.
Permanganate ion is used often because it is its own indicator: MnO4- is purple, Mn+2 is colorless. When reaction solution remains clear, MnO4- is gone.
Chromate ion is also useful, but color change, orangish yellow to green, is harder to detect.

70. Pg. 174-175 #73-76 ONLY letter a for each
Section 10 Homework
Pg #73-76 ONLY letter a for each