1. Chemical Quantities and Aqueous Reactions
Chapter 4
Chemical Quantities and Aqueous Reactions
Chemistry: A Molecular Approach, 4nd Ed. Nivaldo Tro

2. Outline of Chapter 4 Climate Change and Combustion of Fossil Fuels
Ionic reaction
Acid-base Reactions
Reaction Stoichiometry
Titrations
Limiting Reactant, Theoretical Yield, and Percent Yield
- Acid-base titration
Gas-evolving reactions
Solution concentration and Stoichiometry
Acid + sulfide
Acid + sulfite via decomposition
Solute dissolving:
Acid + carbonate via decomposition
Electrolyte/non-electrolyte
Acid + metal
Molecular view of electrolyte/non-electrolyte
Ammonium salt + base via decomposition
Strong/Weak Electrolyte
Redox reactions
Solubility of ionic compounds
Combustion as redox
- Predicting when a salt will dissolve
Redox without combustion
Solubility rules – compounds in water
Reaction of metals with nonmetals
Precipitation/Precipitate
Oxidation states and rules for assigning oxidation state
- Determining products of precipitation reactions

3. Global Warming
Scientists have measured an average 0.7 °C rise in atmospheric temperature since 1860
During the same period atmospheric CO2 levels have risen 38%
Are the two trends causal?

4. The Sources of Increased CO2
One source of CO2 is the combustion reactions of fossil fuels we use to get energy
Another source of CO2 is volcanic action
How can we judge whether global warming is natural or due to our use of fossil fuels?

5. Reaction Stoichiometry
The amount of every substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction
Law of Conservation of Mass
Balancing equations by balancing atoms
The study of the numerical relationship between chemical quantities in a chemical reaction is called stoichiometry
Reaction Stoichiometry: The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 molecules of C8H18 react with 25 molecules of O2
to form 16 molecules of CO2 and 18 molecules of H2O
2 moles of C8H18 react with 25 moles of O2
to form 16 moles of CO2 and 18 moles of H2O
2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O

6. 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza
Making Pizza
The number of pizzas you can make depends on the amount of the ingredients you use
1 crust + 5 oz. tomato sauce + 2 cu cheese  1 pizza
This relationship can be expressed mathematically
1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza
If you want to make more or less than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make
assuming you have enough crusts and tomato sauce

7. Predicting Amounts from Stoichiometry
The amounts of any other substance in a chemical reaction can be determined from the amount of just one substance
How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18?
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
2 moles C8H18 : 16 moles CO2

8. because 6x moles of H2O as C6H12O6, the number makes sense
Practice  How many moles of water are made in the combustion of 0.10 moles of glucose?
Given:
Find:
0.10 moles C6H12O6
moles H2O
Conceptual Plan:
Relationships:
C6H12O O2 → 6 CO2 + 6 H2O
mol H2O
mol C6H12O6
1 mol C6H12O6 : 6 mol H2O
Solution:
Check:
0.6 mol H2O = 0.60 mol H2O
because 6x moles of H2O as C6H12O6, the number makes sense

9. 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasolne
Assuming that gasoline is octane, C8H18, the equation for the reaction is
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
The equation for the reaction gives the mole relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the conceptual plan will be
g C8H18
mol CO2
g CO2
mol C8H18

10. 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2
Example: Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 1015 g gasoline
Given:
Find:
3.4 x 1015 g C8H18
g CO2
Conceptual Plan:
Relationships:
1 mol C8H18 = g, 1 mol CO2 = 44.01g, 2 mol C8H18:16 mol CO2
g C8H18
mol CO2
g CO2
mol C8H18
Solution:
Check:
because 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

11. Which Produces More CO2; Volcanoes or Fossil Fuel Combustion?
Our calculation just showed that the world produced 1.1 x 1016 g of CO2 just from petroleum combustion in 2007
1.1 x 1013 kg CO2
Estimates of volcanic CO2 production are x 1011 kg/year
This means that volcanoes produce less than 2% of the CO2 added to the air annually

12. 1 mol C6H12O6 = 180.2g, 1 mol CO2 = 44.01g, 1 mol C6H12O6 : 6 mol CO2
Example 4.1: How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis?
Given:
Find:
37.8 g CO2, 6 CO2 + 6 H2O  C6H12O6+ 6 O2
g C6H12O6
Conceptual Plan:
Relationships:
1 mol C6H12O6 = 180.2g, 1 mol CO2 = 44.01g, 1 mol C6H12O6 : 6 mol CO2
g C6H12O6
mol CO2
g CO2
mol C6H12O6 g
44.01
mol 1
mol 1 g
180.2 2 6 12 CO
mol O H C 1
Solution: 6 12 2 O H C g
5.8
mol 1
180.2 CO
44.01
7.8 3 = 
Check:
because 6x moles of CO2 as C6H12O6, but the molar mass of C6H12O6 is 4x CO2, the number makes sense

13. 1 mol O2 = 32.00g, 1 mol PbO2 = 239.2g, 1 mol O2 : 2 mol PbO2
How many grams of O2 can be made from the decomposition of g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
Given:
Find:
100.0 g PbO2, 2 PbO2 → 2 PbO + O2
g O2
Conceptual Plan:
Relationships:
1 mol O2 = 32.00g, 1 mol PbO2 = 239.2g, 1 mol O2 : 2 mol PbO2
g O2
mol PbO2
g PbO2
mol O2
Solution:
Check:
because ½ moles of O2 as PbO2, and the molar mass of PbO2 is 7x O2, the number makes sense

14. More Making Pizzas We know that
1 crust + 5 oz. tomato sauce + 2 cu cheese  1 pizza
But what would happen if we had 4 crusts,
15 oz. tomato sauce, and 10 cu cheese?

15. More Making Pizzas, Continued
Each ingredient could potentially make a different number of pizzas
But all the ingredients have to work together!
We only have enough tomato sauce to make three pizzas, so once we make three pizzas, the tomato sauce runs out no matter how much of the other ingredients we have.
The tomato sauce limits the amount of pizzas we can make. In chemical reactions we call this the limiting reactant.
also known as the limiting reagent
The maximum number of pizzas we can make depends on this ingredient. In chemical reactions, we call this the theoretical yield.
it also determines the amounts of the other ingredients we will use!

16. Limiting and Excess Reactants in the Combustion of Methane
CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)
Our balanced equation for the combustion of methane implies that every one molecule of CH4 reacts with two molecules of O2

17. Limiting and Excess Reactants in the Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
If we have five molecules of CH4 and eight molecules of O2, which is the limiting reactant?
since less CO2 can be made from the O2 than the CH4, so the O2 is the limiting reactant

18. Practice — How many moles of Si3N4 can be made from 1
Practice — How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 moles of N2 in the reaction 3 Si + 2 N2  Si3N4?
Given:
Find:
1.20 mol Si, 1.00 mol N2
mol Si3N4
Conceptual Plan:
Relationships:
2 mol N2 : 1 Si3N4; 3 mol Si : 1 Si3N4
mol Si
mol Si3N4
Pick least
amount
Limiting
reactant and
theoretical
yield
mol N2
mol Si3N4
Solution:
Limiting
reactant
Theoretical
yield

19. More Making Pizzas
Let’s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield.
We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield.

20. Theoretical and Actual Yield
As we did with the pizzas, in order to determine the theoretical yield, we should use reaction stoichiometry to determine the amount of product each of our reactants could make
The theoretical yield will always be the maximum possible amount of product that can be made in a chemical reaction
the theoretical yield will always come from the limiting reactant
Because of both controllable and uncontrollable factors, the actual yield of product will always be less than the theoretical yield

21. Example:
When 28.6 kg of C are allowed to react with 88.2 kg of TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield.

22. Write down the given quantity and its units Given: 28.6 kg C
Example: When 28.6 kg of C reacts with kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Write down the given quantity and its units
Given: 28.6 kg C
88.2 kg TiO2
42.8 kg Ti produced

23. Write down the quantity to find and/or its units
Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: kg C, 88.2 kg TiO2, 42.8 kg Ti
Write down the quantity to find and/or its units
Find: limiting reactant
theoretical yield
percent yield

24. Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
Write a conceptual plan }
smallest
amount is
from
limiting
reactant kg
TiO2 C
smallest
mol Ti

25. Collect needed relationships
Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Collect needed relationships
1000 g = 1 kg
Molar Mass TiO2 = g/mol
Molar Mass Ti = g/mol
Molar Mass C = g/mol
1 mole TiO2 : 1 mol Ti (from the chem. equation)
2 mole C : 1 mol Ti (from the chem. equation)

26. Apply the conceptual plan
Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan
limiting reactant
smallest moles of Ti

27. Apply the conceptual plan
Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan
theoretical yield

28. Apply the conceptual plan
Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan

29. Check the solutions limiting reactant = TiO2
Example: Find the limiting reactant, theoretical yield, and percent yield TiO2(s) + 2 C(s)  Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: lim. rct., theor. yld., % yld.
CP: kg rct  g rct  mol rct  mol Ti
pick smallest mol Ti  TY kg Ti  %Y Ti
Rel: 1 mol C=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Check the solutions
limiting reactant = TiO2
theoretical yield = 52.9 kg
percent yield = 80.9%
Because Ti has lower molar mass than TiO2, the T.Y. makes
sense and the percent yield makes sense as it is less than 100%
Tro: Chemistry: A Molecular Approach, 2/e

30. Given: Find: 9.05 g NH3, 45.2 g CuO g N2 Conceptual Plan: g NH3 mol N2
Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield?
Given:
Find:
9.05 g NH3, 45.2 g CuO
g N2
Conceptual Plan:
Relationships:
1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = g
2 mol NH3 : 1 mol N2, 3 mol CuO : 1 mol N2
Choose
smallest
g NH3
mol N2
mol NH3
g N2
g CuO
mol N2
mol CuO

31. because the percent yield is less than 100, the answer makes sense
Practice — How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO? 2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield?
Solution:
Theoretical
yield
because the percent yield is less than 100, the answer makes sense
Check:

32. Solutions
When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous
the salt is still there, as you can tell from the taste, or simply boiling away the water
Homogeneous mixtures are called solutions
The component of the solution that changes state is called the solute
The component that keeps its state is called the solvent
if both components start in the same state, the major component is the solvent

33. Describing Solutions
Because solutions are mixtures, the composition can vary from one sample to another
pure substances have constant composition
saltwater samples from different seas or lakes have different amounts of salt
So to describe solutions accurately, we must describe how much of each component is present
we saw that with pure substances, we can describe them with a single name because all samples are identical

34. Solution Concentration
Qualitatively, solutions are often described as dilute or concentrated
Dilute solutions have a small amount of solute compared to solvent
Concentrated solutions have a large amount of solute compared to solvent

35. Concentrations—Quantitative Descriptions of Solutions
A more precise method for describing a solution is to quantify the amount of solute in a given amount of solution
Concentration = amount of solute in a given amount of solution
occasionally amount of solvent
Solution Concentration: Molarity
Moles of solute per 1 liter of solution
Used because it describes how many molecules of solute in each liter of solution

36. Preparing 1 L of a 1.00 M NaCl Solution

37. because most solutions are between 0 and 18 M, the answer makes sense
Example 4.5: Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution
Given:
Find:
25.5 g KBr, 1.75 L solution
molarity, M
Conceptual Plan:
Relationships:
1 mol KBr = g, M = moles/L
g KBr
mol KBr
L sol’n M
Solution:
Check:
because most solutions are between 0 and 18 M, the answer makes sense

38. Practice — What Is the molarity of a solution containing 3
Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in mL of solution?
Given:
Find:
3.4 g NH3, mL solution M
0.20 mol NH3, L solution M
Conceptual Plan:
Relationships:
M = mol/L, 1 mol NH3 = g, 1 mL = L
g NH3
mol NH3 M
mL sol’n
L sol’n
Solve:
Check:
the unit is correct, the number is reasonable because the fraction of moles is less than the fraction of liters

39. Using Molarity in Calculations
Molarity shows the relationship between the moles of solute and liters of solution
If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar
2 liters = 4.0 moles sugar
0.5 liters = 1.0 mole sugar
1 L solution : 2 moles sugar

40. Example 4.6: How many liters of 0.125 M NaOH contain 0.255 mol NaOH?
Given:
Find:
0.125 M NaOH, mol NaOH
liters, L
Conceptual Plan:
Relationships:
0.125 mol NaOH = 1 L solution
mol NaOH
L sol’n
Solution:
Check:
because each L has only mol NaOH, it makes sense that mol should require a little more than 2 L

41. Practice — Determine the mass of CaCl2 (MM = 110. 98) in 1. 75 L of 1
Practice — Determine the mass of CaCl2 (MM = ) in 1.75 L of 1.50 M solution
Given:
Find:
1.50 M CaCl2, 1.75 L
mass CaCl2, g
mol CaCl2
L sol’n
g CaCl2
Conceptual Plan:
Relationships:
1.50 mol CaCl2 = 1 L solution; g CaCl2 = 1 mol
Solution:
Check:
because each L has 1.50 mol CaCl2, it makes sense that 1.75 L should have almost 3 moles

42. Example: How would you prepare 250. 0 mL of a 1
Example: How would you prepare mL of a 1.00 M solution CuSO45 H2O(MM )?
Given:
Find:
250.0 mL solution
mass CuSO4 5 H2O, g
Conceptual Plan:
Relationships:
1.00 L sol’n = 1.00 mol; 1 mL = L; 1 mol = g
mL sol’n
L sol’n
g CuSO4
mol CuSO4
Solution:
Dissolve 62.4 g of CuSO4∙5H2O in enough water to total mL
Check:
the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter

43. Practice – How would you prepare 250.0 mL of 0.150 M CaCl2?
Given:
Find:
250.0 mL solution
mass CaCl2, g
Conceptual Plan:
Relationships:
1.00 L sol’n = mol; 1 mL = 0.001L; 1 mol = g
mL sol’n
L sol’n
g CaCl2
mol CaCl2
Solution:
Dissolve 4.16 g of CaCl2 in enough water to total mL
Check:
the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter

44. moles solute in solution 1 = moles solute in solution 2
Dilution
Often, solutions are stored as concentrated stock solutions
To make solutions of lower concentrations from these stock solutions, more solvent is added
the amount of solute doesn’t change, just the volume of solution
moles solute in solution 1 = moles solute in solution 2
The concentrations and volumes of the stock and new solutions are inversely proportional
M1∙V1 = M2∙V2

45. Example 4. 7: To what volume should you dilute 0. 200 L of 15
Example 4.7: To what volume should you dilute L of 15.0 M NaOH to make 3.00 M NaOH?
Given:
Find:
V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Conceptual Plan:
Relationships:
M1V1 = M2V2
V1, M1, M2 V2
Solution:
Check:
because the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does

46. Practice – What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO3 to mL?
Given:
Find:
V1 = 45.0 mL, M1 = 8.25 M, V2 = mL
M2, L
Conceptual Plan:
Relationships:
M1V1 = M2V2
V1, M1, V2 M2
Solution:
Check:
because the solution is diluted by a factor of 3, the molarity should decrease by a factor of 3, and it does

47. Practice – How would you prepare 200. 0 mL of 0
Practice – How would you prepare mL of 0.25 M NaCl solution from a 2.0 M solution?
Given:
Find:
M1 = 2.0 M, M2 = 0.25 M, V2 = mL
V1, L
Conceptual Plan:
Relationships:
M1V1 = M2V2
M1, M2, V2 V1
Solution:
Dilute 25 mL of 2.0 M solution up to mL
Check:
because the solution is diluted by a factor of 8, the volume should increase by a factor of 8, and it does

48. Solution Stoichiometry
Because molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

49. Example 4. 8: What volume of 0
Example 4.8: What volume of M KCl is required to completely react with L of M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq)?
Given:
Find:
0.150 M KCl, L of M Pb(NO3)2
L KCl
Conceptual Plan:
Relationships:
1 L Pb(NO3)2 = mol, 1 L KCl = mol, 1 mol Pb(NO3)2 : 2 mol KCl
L Pb(NO3)2
mol KCl
L KCl
mol Pb(NO3)2
Solution:
Check:
because you need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x the volume of Pb(NO3)2

50. Practice – 43. 8 mL of 0. 107 M HCl is needed to neutralize 37
Practice – 43.8 mL of M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the base? 2 HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2 H2O(aq)
Given:
Find:
L of M HCl, L Ba(OH)2
M Ba(OH)2
43.8 mL of M HCl, 37.6 mL Ba(OH)2
M Ba(OH)2
Conceptual Plan:
Relationships:
1 mL= L, 1 L HCl = mol, 1 mol Ba(OH)2 : 2 mol HCl
L HCl
mol Ba(OH)2
mol HCl
M Ba(OH)2
L Ba(OH)2
Solution:
Check:
the units are correct, the number makes sense because there are fewer moles than liters

51. What Happens When a Solute Dissolves?
There are attractive forces between the solute particles holding them together
There are also attractive forces between the solvent molecules
When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules
If the attractions between solute and solvent are strong enough, the solute will dissolve

52. Table Salt Dissolving in Water
Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal
When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions
The result is a solution with free moving charged particles able to conduct electricity

53. Electrolytes and Nonelectrolytes
Materials that dissolve in water to form a solution that will conduct electricity are called electrolytes
Materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes
Molecular View of Electrolytes and Nonelectrolytes
To conduct electricity, a material must have charged particles that are able to flow
Electrolyte solutions all contain ions dissolved in the water
ionic compounds are electrolytes because they dissociate into their ions when they dissolve
Non-electrolyte solutions contain whole molecules dissolved in the water
generally, molecular compounds do not ionize when they dissolve in water
the notable exception being molecular acids

54. Salt vs. Sugar Dissolved in Water
ionic compounds dissociate into ions when they dissolve
molecular compounds do not dissociate when they dissolve

55. Acids
Acids are molecular compounds that ionize when they dissolve in water
the molecules are pulled apart by their attraction for the water
when acids ionize, they form H+ cations and also anions
The percentage of molecules that ionize varies from one acid to another
Acids that ionize virtually 100% are called strong acids
HCl(aq)  H+(aq) + Cl−(aq)
Acids that only ionize a small percentage are called weak acids
HF(aq)  H+(aq) + F−(aq)

56. Dissociation and Ionization
When ionic compounds dissolve in water, the anions and cations are separated from each other. This is called dissociation.
Na2S(aq)  2 Na+(aq) + S2-(aq)
When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion
Na2SO4(aq)  2 Na+(aq) + SO42−(aq)
When strong acids dissolve in water, the molecule ionizes into H+ and anions
H2SO4(aq)  2 H+(aq) + SO42−(aq)

57. Strong and Weak Electrolytes
Strong electrolytes are materials that dissolve completely as ions
ionic compounds and strong acids
their solutions conduct electricity well
Weak electrolytes are materials that dissolve mostly as molecules, but partially as ions
weak acids
their solutions conduct electricity, but not well
When compounds containing a polyatomic ion dissolve, the polyatomic ion stays together
HC2H3O2(aq)  H+(aq) + C2H3O2−(aq)

58. Practice – Write the equation for the process that occurs when the following strong electrolytes dissolve in water
CaCl2(aq)  Ca2+(aq) + 2 Cl−(aq)
CaCl2
HNO3
(NH4)2CO3
HNO3(aq)  H+(aq) + NO3−(aq)
(NH4)2CO3(aq)  2 NH4+(aq) + CO32−(aq)

59. Solubility of Ionic Compounds
Some ionic compounds, such as NaCl, dissolve very well in water at room temperature
Other ionic compounds, such as AgCl, dissolve hardly at all in water at room temperature
Compounds that dissolve in a solvent are said to be soluble, where as those that do not are said to be insoluble
NaCl is soluble in water, AgCl is insoluble in water
the degree of solubility depends on the temperature
even insoluble compounds dissolve, just not enough to be meaningful
When Will a Salt Dissolve?
Predicting whether a compound will dissolve in water is not easy
The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results
we call this method the empirical method

60. Solubility Rules Compounds that Are Generally Soluble in Water

61. Solubility Rules Compounds that Are Generally Insoluble in Water

62. Solubility Rules Soluble Compounds Insoluble Exceptions
Soluble Compounds
Insoluble Exceptions
Insoluble Compounds
Soluble Exceptions
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

63. Practice – Determine if each of the following is soluble in water
KOH
AgBr
CaCl2
Pb(NO3)2
PbSO4
KOH is soluble because it contains K+
AgBr is insoluble; most bromides are soluble, but AgBr is an exception
CaCl2 is soluble; most chlorides are soluble, and CaCl2 is not an exception
Pb(NO3)2 is soluble because it contains NO3−
PbSO4 is insoluble; most sulfates are soluble, but PbSO4 is an exception

64. Precipitation Reactions
Precipitation reactions are reactions in which a solid forms when we mix two solutions
reactions between aqueous solutions of ionic compounds
produce an ionic compound that is insoluble in water
the insoluble product is called a precipitate

65. 2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2 KNO3(aq)

66. No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq)
all ions still present,  no reaction

67. Process for Predicting the Products of a Precipitation Reaction
1. Determine what ions each aqueous reactant has
2. Determine formulas of possible products
exchange ions
(+) ion from one reactant with (-) ion from other
balance charges of combined ions to get formula of each product
3. Determine solubility of each product in water
use the solubility rules
if product is insoluble or slightly soluble, it will precipitate
If neither product will precipitate, write no reaction after the arrow
5. If any of the possible products are insoluble, write their formulas as the products of the reaction using (s) after the formula to indicate solid. Write any soluble products with (aq) after the formula to indicate aqueous.
Balance the equation
- remember to only change coefficients, not subscripts

68. Example 4.10: Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride
1. Write the formulas of the reactants
K2CO3(aq) + NiCl2(aq) 
2. Determine the possible products
a) determine the ions present
(K+ + CO32−) + (Ni2+ + Cl−) 
b) exchange the Ions
(K+ + CO32−) + (Ni2+ + Cl−)  (K+ + Cl−) + (Ni2+ + CO32−)
c) write the formulas of the products
balance charges
K2CO3(aq) + NiCl2(aq)  KCl + NiCO3

69. 3. Determine the solubility of each product KCl is soluble
Example 4.10: Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride
3. Determine the solubility of each product
KCl is soluble
NiCO3 is insoluble
4. If both products are soluble, write no reaction
does not apply because NiCO3 is insoluble
5. Write (aq) next to soluble products and (s) next to insoluble products
K2CO3(aq) + NiCl2(aq)  KCl(aq) + NiCO3(s)
6. Balance the equation
K2CO3(aq) + NiCl2(aq)  2 KCl(aq) + NiCO3(s)

70. Practice – Predict the products and balance the equation
KCl(aq) + AgNO3(aq) 
Na2S(aq) + CaCl2(aq) 
(K+ + Cl−) + (Ag+ + NO3−)  (K+ + NO3−) + (Ag+ + Cl−)
KCl(aq) + AgNO3(aq)  KNO3 + AgCl
KCl(aq) + AgNO3(aq)  KNO3(aq) + AgCl(s)
(Na+ + S2−) + (Ca2+ + Cl−)  (Na+ + Cl−) + (Ca2+ + S2−)
Na2S(aq) + CaCl2(aq)  NaCl + CaS
Na2S(aq) + CaCl2(aq)  NaCl(aq) + CaS(aq)
No reaction

71. Practice – Write an equation for the reaction that takes place when an aqueous solution of (NH4)2SO4 is mixed with an aqueous solution of Pb(C2H3O2)2.
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) 
(NH4+ + SO42−) + (Pb2+ + C2H3O2−)  (NH4+ + C2H3O2−) + (Pb2+ + SO42−)
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq)  NH4C2H3O2 + PbSO4
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq)  NH4C2H3O2(aq) + PbSO4(s)
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq)  2 NH4C2H3O2(aq) + PbSO4(s)

72. 2 KOH(aq) + Mg(NO3)2(aq)  2 KNO3(aq) + Mg(OH)2(s)
Ionic Equations
Equations that describe the chemicals put into the water and the product molecules are called molecular equations
2 KOH(aq) + Mg(NO3)2(aq)  2 KNO3(aq) + Mg(OH)2(s)
Equations that describe the material’s structure when dissolved are called complete ionic equations
aqueous strong electrolytes are written as ions
soluble salts, strong acids, strong bases
insoluble substances, weak electrolytes, and nonelectrolytes are written in molecule form
solids, liquids, and gases are not dissolved, therefore molecule form
2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq)  2K+(aq) + 2NO3−(aq) + Mg(OH)2(s)

73. Ionic Equations
Ions that are both reactants and products are called spectator ions
2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq)  2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)
An ionic equation in which the spectator ions are removed is called a net ionic equation
2 OH−(aq) + Mg2+(aq)  Mg(OH)2(s)

74. Practice – Write the ionic and net ionic equation for each
K2SO4(aq) + 2 AgNO3(aq)  2 KNO3(aq) + Ag2SO4(s)
K2SO4(aq) + 2 AgNO3(aq)  2 KNO3(aq) + Ag2SO4(s)
2K+(aq) + SO42−(aq) + 2Ag+(aq) + 2NO3−(aq)
 2K+(aq) + 2NO3−(aq) + Ag2SO4(s)
2 Ag+(aq) + SO42−(aq)  Ag2SO4(s)
Na2CO3(aq) + 2 HCl(aq)  2 NaCl(aq) + CO2(g) + H2O(l)
2Na+(aq) + CO32−(aq) + 2H+(aq) + 2Cl−(aq)
 2Na+(aq) + 2Cl−(aq) + CO2(g) + H2O(l)
CO32−(aq) + 2 H+(aq)  CO2(g) + H2O(l)
Na2CO3(aq) + 2 HCl(aq)  2 NaCl(aq) + CO2(g) + H2O(l)

75. Acid-Base Reactions
Also called neutralization reactions because the acid and base neutralize each other’s properties
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)
The net ionic equation for an acid-base reaction is
H+(aq) + OH(aq)  H2O(l)
as long as the salt that forms is soluble in water

76. Acids and Bases in Solution acid + base  salt + water
Acids ionize in water to form H+ ions
more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+
most chemists use H+ and H3O+ interchangeably
Bases dissociate in water to form OH ions
bases, such as NH3, that do not contain OH ions, produce OH by pulling H off water molecules
In the reaction of an acid with a base, the H+ from the acid combines with the OH from the base to make water
The cation from the base combines with the anion from the acid to make the salt
acid + base  salt + water

77. Common Acids

78. Common Bases

79. HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

80. Example: Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
1. Write the formulas of the reactants
HNO3(aq) + Ca(OH)2(aq) 
2. Determine the possible products
a) determine the ions present when each reactant dissociates or ionizes
(H+ + NO3−) + (Ca2+ + OH−) 
b) exchange the ions, H+ combines with OH− to make H2O(l)
(H+ + NO3−) + (Ca2+ + OH−)  (Ca2+ + NO3−) + H2O(l)
c write the formula of the salt
(H+ + NO3−) + (Ca2+ + OH−)  Ca(NO3)2 + H2O(l)
3. Determine the solubility of the salt
Ca(NO3)2 is soluble
4. Write an (s) after the insoluble products and an (aq) after the soluble products
HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + H2O(l)
5. Balance the equation
2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l)

81. Example: Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide
6. Dissociate all aqueous strong electrolytes to get complete ionic equation
not H2O
2 H+(aq) + 2 NO3−(aq) + Ca2+(aq) + 2 OH−(aq) 
Ca2+(aq) + 2 NO3−(aq) + H2O(l)
7. Eliminate spectator ions to get net-ionic equation
2 H+(aq) + 2 OH−(aq)  2 H2O(l)
H+(aq) + OH−(aq)  H2O(l)

82. Practice – Predict the products and balance the equation
HCl(aq) + Ba(OH)2(aq) 
H2SO4(aq) + Sr(OH)2(aq) 
(H+ + Cl−) + (Ba2+ + OH−)  (H+ + OH−) + (Ba2+ + Cl−)
HCl(aq) + Ba(OH)2(aq)  H2O(l) + BaCl2
2 HCl(aq) + Ba(OH)2(aq)  2 H2O(l) + BaCl2(aq)
(H+ + SO42−) + (Sr2+ + OH−)  (H+ + OH−) + (Sr2+ + SO42−)
H2SO4(aq) + Sr(OH)2(aq)  H2O(l) + SrSO4
H2SO4(aq) + Sr(OH)2(aq)  2 H2O(l) + SrSO4
H2SO4(aq) + Sr(OH)2(aq)  2 H2O(l) + SrSO4(s)

83. Titration
Often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration
In the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed. At this point, called the endpoint, the reactants are in their stoichiometric ratio.
the unknown solution is added slowly from an instrument called a burette
a long glass tube with precise volume markings that allows small additions of solution

84. Acid-Base Titrations
The difficulty is determining when there has been just enough titrant added to complete the reaction
the titrant is the solution in the burette
In acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity
the chemical is called an indicator
At the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH
also known as the equivalence point

85. Titration

86. Titration The titrant is the base solution in the burette.
As the titrant is added to
the flask, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change.
At the titration’s endpoint,
just enough base has been added to neutralize all the acid. At this point the indicator changes color.

87. 12.54 mL of 0.100 M NaOH Write down the given quantity and its units
Example 4.14:
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Write down the given quantity and its units
Given: mL HCl
12.54 mL of M NaOH

88. Write down the quantity to find, and/or its units
Example 4.14:
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Write down the quantity to find, and/or its units
Find: concentration HCl, M

89. Collect needed equations and conversion factors
Example 4.14:
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Collect needed equations and conversion factors
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
 1 mole HCl = 1 mole NaOH
0.100 M NaOH 0.100 mol NaOH  1 L sol’n

90. Write a conceptual plan
Example 4.14:
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
Write a conceptual plan mL
NaOH L
NaOH
mol
NaOH
mol
HCl mL
HCl L
HCl

91. Apply the conceptual plan
Example 4.14:
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL of M NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the conceptual plan
= 1.25 x 103 mol HCl

92. Apply the conceptual plan
Example 4.14:
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Apply the conceptual plan

93. Check the solution HCl solution = 0.125 M Example 4.14:
The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?
Information
Given: mL HCl
12.54 mL NaOH
Find: M HCl
Rel: 1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
Check the solution
HCl solution = M
The units of the answer, M, are correct.
The magnitude of the answer makes sense because
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.

94. Gas-Evolving Reactions
Some reactions form a gas directly from the ion exchange
K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g)
Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water
K2SO3(aq) + H2SO4(aq)  K2SO4(aq) + H2SO3(aq)
H2SO3  H2O(l) + SO2(g)

95. NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)

96. Compounds that Undergo Gas-Evolving Reactions

97. Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves
1. Write the formulas of the reactants
Na2CO3(aq) + HNO3(aq) 
2. Determine the possible products
a) determine the ions present when each reactant dissociates or ionizes
(Na+ + CO32−) + (H+ + NO3−) 
b) exchange the anions
(Na+ + CO32−) + (H+ + NO3−)  (Na+ + NO3−) + (H+ + CO32−)
c) write the formula of compounds
balance the charges
Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
3. Check to see if either product is H2S - No
4. Check to see if either product decomposes – Yes
H2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)

98. Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l)
Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves
5. Determine the solubility of other product
NaNO3 is soluble
6. Write an (s) after the insoluble products and an (aq) after the soluble products
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l)
7. Balance the equation

99. Practice – Predict the products and balance the equation
HCl(aq) + Na2SO3(aq) 
H2SO4(aq) + CaS(aq) 
(H+ + Cl−) + (Na+ + SO32−) → (H+ + SO32−) + (Na+ + Cl−)
HCl(aq) + Na2SO3(aq) → H2SO3 + NaCl
HCl(aq) + Na2SO3(aq) → H2O(l) + SO2(g) + NaCl
2 HCl(aq) + Na2SO3(aq) ® H2O(l) + SO2(g) + 2 NaCl(aq)
(H+ + SO42−) + (Ca2+ + S2−) → (H+ + S2−) + (Ca2+ + SO42−)
H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4
H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4(s)

100. Other Patterns in Reactions
The precipitation, acid-base, and gas-evolving reactions all involve exchanging the ions in the solution
Other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions
also known as redox reactions
many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g)  2 Fe2O3(s)

101. Combustion as Redox 2 H2(g) + O2(g)  2 H2O(g)

102. Redox without Combustion 2 Na(s) + Cl2(g)  2 NaCl(s)
2 Na  2 Na+ + 2 e
Cl2 + 2 e  2 Cl

103. Oxidation and Reduction
Reactions of Metals with Nonmetals
Consider the following reactions:
4 Na(s) + O2(g) → 2 Na2O(s)
2 Na(s) + Cl2(g) → 2 NaCl(s)
The reactions involve a metal reacting with a nonmetal
In addition, both reactions involve the conversion of free elements into ions
4 Na(s) + O2(g) → 2 Na+2O2– (s)
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Oxidation and Reduction
To convert a free element into an ion, the atoms must gain or lose electrons
of course, if one atom loses electrons, another must accept them
Reactions where electrons are transferred from one atom to another are redox reactions
Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na → Na+ + 1 e– oxidation: Cl2 + 2 e– → 2 Cl– reduction

104. Electron Bookkeeping
For reactions that are not metal + nonmetal, or do not involve O2, we need a method for determining how the electrons are transferred
Chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction
even though they look like them, oxidation states are not ion charges!
oxidation states are imaginary charges assigned based on a set of rules
ion charges are real, measurable charges

105. Rules for Assigning Oxidation States
Rules are in order of priority
1. free elements have an oxidation state = 0
Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
2. monatomic ions have an oxidation state equal to their charge
Na = +1 and Cl = −1 in NaCl
3. (a) the sum of the oxidation states of all the atoms in a compound is 0
Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0
3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion
N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1
4. (a) Group I metals have an oxidation state of +1 in all their compounds
Na = +1 in NaCl
4. (b) Group II metals have an oxidation state of +2 in all their compounds
Mg = +2 in MgCl2

106. Rules for Assigning Oxidation States
5. in their compounds, nonmetals have oxidation states according to the table below
nonmetals higher on the table take priority

107. Example: Determine the oxidation states of all the atoms in a propanoate ion, C3H5O2–
There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b
(C3) + (H5) + (O2) = −1
Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order:
H = +1
O = −2
(C3) + 5(+1) + 2(−2) = −1
(C3) = −2
C = −⅔
Note: unlike charges, oxidation states can be fractions!

108. Practice – Assign an oxidation state to each element in the following
Br2 K+
LiF
CO2
SO42−
Na2O2
Br = 0, (Rule 1)
K = +1, (Rule 2)
Li = +1, (Rule 4a) & F = −1, (Rule 5)
O = −2, (Rule 5) & C = +4, (Rule 3a)
O = −2, (Rule 5) & S = +6, (Rule 3b)
Na = +1, (Rule 4a) & O = −1 , (Rule 3a)

109. Oxidation and Reduction: Another Definition
Oxidation occurs when an atom’s oxidation state increases during a reaction
Reduction occurs when an atom’s oxidation state decreases during a reaction
CH O2 → CO2 + 2 H2O
− – −2
oxidation
reduction

110. Oxidation–Reduction 2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Oxidation and reduction must occur simultaneously
if an atom loses electrons another atom must take them
The reactant that reduces an element in another reactant is called the reducing agent
the reducing agent contains the element that is oxidized
The reactant that oxidizes an element in another reactant is called the oxidizing agent
the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent

111. Example: Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions
Reducing
agent
Oxidizing
agent
Fe + MnO4− + 4 H+ → Fe3+ + MnO H2O +7 −2 +1 +3 +4 −2 +1 −2
Oxidation
Reduction

112. Sn4+ + Ca → Sn2+ + Ca2+ F2 + S → SF4
Practice – Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions
Sn4+ + Ca → Sn Ca2+
F2 + S → SF4
Ca is oxidized, Sn4+ is reduced
Ca is the reducing agent, Sn4+ is the oxidizing agent −1
S is oxidized, F is reduced
S is the reducing agent, F2 is the oxidizing agent

113. Balancing Redox Reactions
Some redox reactions can be balanced by the method we previously used, but many are hard to balance using that method
many are written as net ionic equations
many have elements in multiple compounds
The main principle is that electrons are transferred – so if we can find a method to keep track of the electrons it will allow us to balance the equation
Tro: Chemistry: A Molecular Approach, 2/e

114. Half-Reactions
We generally split the redox reaction into two separate half-reactions – a reaction just involving oxidation or reduction
the oxidation half-reaction has electrons as products
the reduction half-reaction has electrons as reactants
3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+
− − − −2 +1
oxidation: I− → IO3− + 6 e−
reduction: Cl2 + 2 e− → 2 Cl−
Tro: Chemistry: A Molecular Approach, 2/e

115. Balancing Redox Reactions by the Half-Reaction Method
In this method, the reaction is broken down into two half-reactions, one for oxidation and another for reduction
Each half-reaction includes electrons
electrons go on the product side of the oxidation half- reaction – loss of electrons
electrons go on the reactant side of the reduction half- reaction – gain of electrons
Each half-reaction is balanced for its atoms
Then the two half-reactions are adjusted so that the electrons lost and gained will be equal when combined
Tro: Chemistry: A Molecular Approach, 2/e

116. Example 18.2: Balancing redox reactions
in acidic solution
1. assign oxidation states and determine element oxidized and element reduced 2. separate into oxidation & reduction half- reactions
Fe2+ + MnO4– → Fe3+ + Mn2+ +2 +7 −2 +3 +2
oxidation
reduction
ox: Fe2+ → Fe3+
red: MnO4– → Mn2+
Tro: Chemistry: A Molecular Approach, 2/e

117. Example 18.2: Balancing redox reactions
in acidic solution
3. balance half-reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
Fe2+ → Fe3+
MnO4– → Mn2+
MnO4– → Mn2+ + 4H2O
MnO4– + 8H+ → Mn2+ + 4H2O
Tro: Chemistry: A Molecular Approach, 2/e

118. Example 18.2: Balancing redox reactions
in acidic solution
4. balance each half-reaction with respect to charge by adjusting the numbers of electrons
a) electrons on product side for oxidation
b) electrons on reactant side for reduction
Fe2+ → Fe e−
MnO4– + 8H+ → Mn2+ + 4H2O +7 +2
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
Tro: Chemistry: A Molecular Approach, 2/e

119. Example 18.2: Balancing redox reactions
in acidic solution
5. balance electrons between half-reactions 6. add half-reactions, canceling electrons and common species 7. Check that numbers of atoms and total charge are equal
Fe2+ → Fe e−
} x 5
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
5 Fe2+ → 5 Fe e−
MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O
5 Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5 Fe3+
reactant side
Element
product side 5 Fe 1 Mn 4 O 8 H
+17
charge
Tro: Chemistry: A Molecular Approach, 2/e

120. Balancing Redox Reactions
1. assign oxidation states
a) determine element oxidized and element reduced
2. write ox. & red. half-reactions, including electrons
a) ox. electrons on right, red. electrons on left of arrow
3. balance half-reactions by mass
a) first balance elements other than H and O
b) add H2O where need O
c) add H+ where need H
d) if reaction done in Base, neutralize H+ with OH−
4. balance half-reactions by charge
a) balance charge by adjusting electrons
5. balance electrons between half-reactions
6. add half-reactions
7. check by counting atoms and total charge
Tro: Chemistry: A Molecular Approach, 2/e

121. Practice – Balance the following equation in acidic solution I– + Cr2O72− → Cr3+ + I2
Tro: Chemistry: A Molecular Approach, 2/e

122. Practice – Balancing redox reactions
1. assign oxidation states and determine element oxidized and element reduced
2. separate into oxidation & reduction half- reactions
I− + Cr2O72– → I Cr3+ −1 +6 −2 +3
oxidation
reduction
ox: I− → I2
red: Cr2O72– → Cr3+
Tro: Chemistry: A Molecular Approach, 2/e
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123. Practice – Balancing redox reactions
3. balance half-reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
ox: I− → I2
ox: 2 I− → I2
red: Cr2O72– → Cr3+
red: Cr2O72– → 2 Cr3+
red: Cr2O72– → 2Cr3+ +7H2O
Cr2O72– +14H+ → 2Cr3+ +7H2O
Tro: Chemistry: A Molecular Approach, 2/e
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124. Practice – Balancing redox reactions
4. balance each half-reaction with respect to charge by adjusting the numbers of electrons
a) electrons on product side for oxidation
b) electrons on reactant side for reduction
2 I− → I2 + 2e−
2 I− → I2
Cr2O72– +14H+ → 2Cr3+ +7H2O
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O
Tro: Chemistry: A Molecular Approach, 2/e

125. Practice – Balancing redox reactions
5. balance electrons between half-reactions
6. add half-reactions, canceling electrons and common species
7. check
6 I− → 3 I2 + 6e−
2 I− → I2 + 2e−
} x 3
Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O
Cr2O72– +14H+ + 6 I−→ 2Cr3+ +7H2O + 3 I2
reactant side
Element
product side 2 Cr 6 I 7 O 14 H +6
charge
Tro: Chemistry: A Molecular Approach, 2/e
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126. assign oxidation states I(aq) + MnO4(aq)  I2(aq) + MnO2(s)
Example 18.3: Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
assign oxidation states
I(aq) + MnO4(aq)  I2(aq) + MnO2(s)
separate into half-reactions
ox:
red:
assign oxidation states
separate into half-reactions
ox: I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
− − −2
Reduction
Oxidation
Tro: Chemistry: A Molecular Approach, 2/e

127. Example 18.3: Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
balance half-reactions by mass
in base, neutralize the H+ with OH−
ox: 2 I(aq)  I2(aq)
red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l)
4 H+(aq) + 4 OH(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
4 H2O(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
2 H2O(l) + MnO4(aq)  MnO2(s) + 4 OH(aq)
balance half-reactions by mass
in base, neutralize the H+ with OH−
ox: 2 I(aq)  I2(aq)
red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l)
4 H+(aq) + 4 OH(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
4 H2O(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
balance half-reactions by mass
in base, neutralize the H+ with OH−
ox: 2 I(aq)  I2(aq)
red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l)
4 H+(aq) + 4 OH(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l) + 4 OH(aq)
balance half-reactions by mass
first, elements other than H and O
ox: 2 I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
balance half-reactions by mass
ox: I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s)
balance half-reactions by mass
then O by adding H2O
ox: 2 I(aq)  I2(aq)
red: MnO4(aq)  MnO2(s) + 2 H2O(l)
balance half-reactions by mass
then H by adding H+
ox: 2 I(aq)  I2(aq)
red: 4 H+(aq) + MnO4(aq)  MnO2(s) + 2 H2O(l)
Tro: Chemistry: A Molecular Approach, 2/e
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128. Example 18.3: Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
balance Half-reactions by charge
ox: 2 I(aq)  I2(aq) + 2 e
red: MnO4(aq) + 2 H2O(l) + 3 e− MnO2(s) + 4 OH(aq)
Balance electrons between half-reactions
ox: 2 I(aq)  I2(aq) + 2 e } x3
red: MnO4(aq) + 2 H2O(l) + 3 e−  MnO2(s) + 4 OH(aq) }x2
ox: 6 I(aq)  3 I2(aq) + 6 e
red: 2 MnO4(aq) + 4 H2O(l) + 6 e− 2 MnO2(s) + 8 OH(aq)
balance Half-reactions by charge
ox: 2 I(aq)  I2(aq)
red: MnO4(aq) + 2 H2O(l)  MnO2(s) + 4 OH(aq)
Balance electrons between half-reactions
balance Half-reactions by charge
ox: 2 I(aq)  I2(aq) + 2 e
red: MnO4(aq) + 2 H2O(l) + 3 e− MnO2(s) + 4 OH(aq)
Balance electrons between half-reactions
Tro: Chemistry: A Molecular Approach, 2/e

129. Example 18.3: Balance the equation: I(aq) + MnO4(aq)  I2(aq) + MnO2(s) in basic solution
add the half-rxns
ox: 6 I(aq)  3 I2(aq) + 6 e
red: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq)
tot: 6 I(aq)+ 2 MnO4(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH(aq)
check
add the half-rxns
ox: 6 I(aq)  3 I2(aq) + 6 e
red: 2 MnO4(aq) + 4 H2O(l) + 6 e  2 MnO2(s) + 8 OH(aq)
check
Reactant
Count
Element
Product 6 I 2 Mn 12 O 8 H 8
charge
Tro: Chemistry: A Molecular Approach, 2/e
129

130. Practice – Balance the following equation in basic solution I– + CrO42− → Cr3+ + I2
Tro: Chemistry: A Molecular Approach, 2/e

131. Practice – Balancing redox reactions
1. assign oxidation states and determine element oxidized and element reduced
2. separate into oxidation & reduction half- reactions
I− + CrO42– → I Cr3+ −1 +6 −2 +3
oxidation
reduction
ox: I− → I2
red: CrO42– → Cr3+
Tro: Chemistry: A Molecular Approach, 2/e
131

132. Practice – Balancing redox reactions
3. balance half-reactions by mass
a) first balance atoms other than O and H
b) balance O by adding H2O to side that lacks O
c) balance H by adding H+ to side that lacks H
d) if in basic solution, neutralize the H+ with OH−
ox: I− → I2
ox: 2 I− → I2
red: CrO42– → Cr3+
red: CrO42– → Cr3+ +4H2O
CrO42– +8H+ → Cr3+ +4H2O
CrO42– +8H2O → Cr3+ +4H2O +8OH−
CrO42– +8H+ +8OH− → Cr3+ +4H2O +8OH−
CrO42– +4H2O → Cr3+ +8OH−
Tro: Chemistry: A Molecular Approach, 2/e
132

133. Practice – Balancing redox reactions
4. balance each half-reaction with respect to charge by adjusting the numbers of electrons
a) electrons on product side for oxidation
b) electrons on reactant side for reduction
2 I− → I2 + 2e−
2 I− → I2
CrO42– +4H2O+ 3e− → Cr3+ +8OH−
CrO42– +4H2O → Cr3+ +8OH−
Tro: Chemistry: A Molecular Approach, 2/e

134. Practice – Balancing redox reactions
5. balance electrons between half-reactions
6. add half-reactions, canceling electrons and common species
7. check
6 I− → 3 I2 + 6e−
2 I− → I2 + 2e−
} x 3
2CrO42– +8H2O+ 6e− → 2Cr3+ +16OH−
CrO42– +4H2O+ 3e− → Cr3+ +8OH−
} x 2
2CrO42– +8H2O + 6 I− → 2Cr3+ +16OH− + 3 I2
reactant side
Element
product side 2 Cr 6 I 16 O H
−10
charge
Tro: Chemistry: A Molecular Approach, 2/e
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