1. Types of Chemical Reactions and Solution Stoichiometry
Chapter 4
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2. Types of Chemical Reactions and Solution Stoichiometry – ch 4
1. Distinguish between solute, solvent and solution.
2. What types of solutes are electrolytes?

3. Types of Chemical Reactions and Solution Stoichiometry – ch 4
Dissolved ionic (NaCl)
Dissolved molecules (Sugar)
Electrolyte
Solution
Non-electrolyte
Solution

4. Types of Chemical Reactions and Solution Stoichiometry – ch 4
3. Determine if the following substances are strong, weak or
non-electrolytes in water.
a. KC2H3O2
b. CH2O
c. HNO3
d. H2S

5. Types of Chemical Reactions and Solution Stoichiometry – ch 4
Lots of ions –
Conducts well
Few ions –
Conducts weakly
No ions –
Does not conduct

6. Types of Chemical Reactions and Solution Stoichiometry – ch 4
Type of compound
Strong
Weak
Non
Ionic
Only soluble
Slightly or insoluble
N/A
Acids
HCl, HBr, HI, HNO3, HClO3, HClO4, H2SO4
HBrO4, HIO4
All other acids
Molecular
NH3
All other

7. Types of Chemical Reactions and Solution Stoichiometry – ch 4

8. Types of Chemical Reactions and Solution Stoichiometry – ch 4
4. A solution is prepared by dissolving 50 g of barium fluoride in enough water to make 750 mL of solution.
a. What is the molarity of the barium fluoride?
b. What is the concentration for each of the ions in solution?
c. What is the concentration of the fluoride ion if 350 mL of water is added?

9. Types of Chemical Reactions and Solution Stoichiometry – ch 4
Molarity (M) ⇒ a unit of concentration
that measures the moles of solute
per liter of solution
M = nsolute Lsolution

10. Types of Chemical Reactions and Solution Stoichiometry – ch 4
5. What is the resulting concentration of bromide ion if 28 mL of 0.15 M sodium bromide solution were mixed with 45 mL of 0.12 M calcium bromide solution?

11. Types of Chemical Reactions and Solution Stoichiometry – ch 4
6. What volume of M Li3PO4 in milliliters is required to precipitate all of the mercury(II) ions from 250 mL of M Hg(NO3)2?

12. Types of Chemical Reactions and Solution Stoichiometry – ch 4
7. A precipitate will form when 10 mL of 0.25 M calcium chloride is mixed with 8 mL of 0.4 M silver nitrate.
a. Write the molecular and net ionic equation for this reaction
b. How many grams of precipitate form?
c. What is the concentration of the ions remaining in solution?

13. Types of Chemical Reactions and Solution Stoichiometry – ch 4
8. What type of solution results (acidic, basic or neutral) when 110 mL of 0.40 M HNO3 is mixed with 42 mL of 0.90 M Ba(OH)2? Write the molecular and net ionic equations.

14. Types of Chemical Reactions and Solution Stoichiometry – ch 4
9. What volume of 0.15 M H2SO4 is required to neutralize 60 mL of 0.80 M KOH?

15. Types of Chemical Reactions and Solution Stoichiometry – ch 4
10. Assign oxidation states for the following:
a. S in SO3
b. P in PF5
c. Cl in Cl2
d. C in C3H8
d. Mn in MnO4–
e. Cr in K2Cr2O7
f. N in NH4NO3

16. Types of Chemical Reactions and Solution Stoichiometry – ch 4
Rules and Hierarchy for Assigning Oxidation Numbers/States
1. What type of substance are you analyzing?
a. Element ⇒ The oxidation state for atoms in their elemental form are 0
b. Molecular Compound ⇒ the sum of the oxidation numbers in a compound
equals 0
c. Ionic compound ⇒ analyze the cation and anion individually
- Mono-atomic ions ⇒ The oxidation state for mono-atomic ions is the charge on the ion
- Polyatomic ions ⇒ The sum of the oxidation numbers in a polyatomic ion equals the net charge on the ion
2. The hierarchy tells which element gets assigned first:
Hierarchy of Substances
Oxidation State
Hydrogen in a molecular compound or polyatomic ion +1
Fluorine in a molecular compound -1
Oxygen in a molecular compound or polyatomic ion -2
Chlorine, bromine or iodine in a molecular compound
Nitrogen in a molecular compound or polyatomic ion -3

17. Types of Chemical Reactions and Solution Stoichiometry – ch 4
11. Identify the oxidizing and reducing agents for the following reactions and determine the moles of electrons transferred:
a. Ba Li  Ba Li+
b. 3 Cu NO3– H+  3 Cu NO H2O
c. Pb + PbO H2SO4  2 PbSO H2O

18. Types of Chemical Reactions and Solution Stoichiometry – ch 4
Oxidation-Reduction (Redox) Reactions 1. A reaction in which there is a transfer of electrons 2. Oxidation is when an element loses electrons and is noted by an increase in the oxidation number 3. Reduction is when an element gains electrons and is noted by a decrease in the oxidation number 4. The number of electrons gained must equal the number of electrons lost 5. An oxidizing agent is the substance getting reduced 6. A reducing agent is the substance getting oxidized

19. Types of Chemical Reactions and Solution Stoichiometry – ch 4
12. Balance the following red-ox reactions:
a. MnO NO3-  N2O MnO (acidic)
b. CNO As2O3  CN HAsO42- (basic)
c. Cr SO42–  Cr2O72– + H2SO (acidic)

20. Types of Chemical Reactions and Solution Stoichiometry – ch 4
Balancing Red-ox Reactions 1. Separate into two half reactions 2. Balance all elements except H and O 3. Balance the oxygen by adding H2O 4. Balance the hydrogen by adding H+ 5. Balance the net charges by adding electrons (e-) to the more positive side 6. Make the electrons lost equal to the electrons gained by multiplying the half reactions by the smallest common multiple 7. Add two half reactions back together canceling out electrons, H+ and H2O 8. If acidic stop. If basic add an OH- to each side for every H+. The OH- cancels out the H+ making water, which then needs to be adjusted for.

21. Types of Chemical Reactions and Solution Stoichiometry – ch 4
13. The reaction below can be used as a laboratory method of preparing small quantities of Cl2(g). If a 62.6 g sample that is 98.5% K2Cr2O7 by mass is allowed to react with 325 mL of HCl(aq) with a density of 1.15 g/mL and 30.1% HCl by mass, how many grams of Cl2(g) are produced?
Cr2O72- (aq) + Cl-(aq)  Cr3+(aq) + Cl2(g) (unbalanced)

22. Types of Chemical Reactions and Solution Stoichiometry – ch 4
You have completed ch. 4

23. Answer Key – Ch. 4
1. Distinguish between solute, solvent and solution.
solute – a substance dissolved (typically into a liquid) to form a solution
solvent – the dissolving medium in a solution
solution – homogenous mixture of solute and solvent or 2 solvents
2. What is an electrolyte?
electrolyte – a substance that produces ions upon dissolving in water thus causing a solution that can conduct electricity – all ionic compounds and acids are electrolytes – only some molecular are electrolytes

24. Answer Key – Ch. 4
3. Determine if the following substances are strong, weak or non electrolytes in water.
a. KC2H3O2 ⇒ soluble ionic ⇒ strong electrolyte
b. CH2O ⇒ molecular ⇒ non electrolyte
c. HNO3 ⇒ strong acid ⇒ strong electrolyte
d. H2S ⇒ weak acid ⇒ weak electrolyte

25. Answer Key – Ch. 4
4. A solution is prepared by dissolving 50 g of barium fluoride in enough water to make 750 mL of solution.
a. What is the molarity of the barium fluoride?
(50 g BaF2)/( g/mol) = mol BaF2
(750 mL)(1L/1000mL) = 0.75 L
M = (0.285 mol)/(0.75 L) = 0.38 M BaF2
b. What is the concentration for each of the ions in solution?
BaF2 (s)  Ba2+ (aq) + 2F– (aq)
0.38 mol/L BaF2 1 mol Ba2+ = 0.38 M Ba2+
0.38 mol/L BaF2 2 mol F– = 0.76 M F–
1 mol BaF2
1 mol BaF2

26. (0.76 M F–)(750 mL) = (M2)(750 mL + 350 mL)
Answer Key – Ch. 4
c. What is the concentration of the fluoride ion if 350 mL of water is added?
since the volume will go up the solution will get diluted – if you’re only adding solvent the moles of solute will remain constant ⇒ M1V1 = M2V2
(0.76 M F–)(750 mL) = (M2)(750 mL mL)
M2 = 0.52 M F–

27. Answer Key – Ch. 4
5. What is the resulting concentration of bromide ion if 28 mL of 0.15 M sodium bromide solution were mixed with 45 mL of 0.12 M calcium bromide solution?
NaBr (s)  Na+ (aq) + Br– (aq)
0.15 M NaBr ⇒ 0.15 M Br–
(0.15 mol/L)(28 mL) = 4.2 mmol Br– (before mixing)
CaBr2 (s)  Ca2+ (aq) + 2 Br– (aq)
0.12 M CaBr2 ⇒ 0.24 M Br–
(0.24 mol/L)(45 mL) = 10.8 mmol Br– (before mixing)
After mixing ⇒ 4.2 mmol Br– mmol = 15 mmol Br–
Final concentration ⇒ (15 mmol Br– )/(28 mL + 45 mL)
= 0.21 M Br–

28. Answer Key – Ch. 4
6. What volume of M Li3PO4 in milliliters is required to precipitate all of the mercury(II) ions from 250 mL of M Hg(NO3)2?
First write a balanced chemical reaction
2 Li3PO4 (aq) + 3 Hg(NO3)2 (aq)  6 LiNO3 (aq) + Hg3(PO4)2 (s)
M1V1 = M2V2
coeff
coeff
(0.15 M)(V1)/2 = (0.32 M)(250 mL)/3
V1 = 356 mL of Li3PO4

29. …continue to next slide
Answer Key – Ch. 4
7. A precipitate will form when 10 mL of 0.25 M calcium chloride is mixed with 8 mL of 0.4 M silver nitrate.
a. Write the molecular and net ionic equation for this reaction
Molecular ⇒
CaCl2 (aq) + 2 AgNO3 (aq)  Ca(NO3)2 (aq) + 2 AgCl (s)
Net ionic ⇒
Ca2+ and NO3– are the spectator ions
Ag+ (aq) + Cl– (aq)  AgCl (s)
b. How many grams of precipitate form?
First determine the LR
…continue to next slide

30. Answer Key – Ch. 4 7. …continued
CaCl2 ⇒ (10 mL)(0.25 M) = 2.5 mmols/1 = 2.5
AgNO3 ⇒ (8 mL)(0.4 M) = 3.2 mmols/2 = AgNO3 is the LR
3.2 mmols AgNO3 1mol AgCl g AgCl = 459 mg AgCl
c. What is the concentration of the ions remaining in solution?
Spectator ions don’t get consumed ⇒
[Ca2+] = (2.5 mmol)/(10 mL + 8 mL) = 0.14 M Ca2+
[NO3–] = (3.2 mmol)/(18 mL) = 0.18 M NO3–
Reacting ions get consumed ⇒
[Ag+] = (3.2 mmol – 3.2 mmol) = 0 M (consumed entirely b/c it’s the LR)
[Cl–] = (5 mmol – 3.2 mmol)/(18 mL) = 0.1 M Cl–
1 mol AgNO mol AgCl
or g AgCl

31. Answer Key – Ch. 4
8. What type of solution results (acidic, basic or neutral) when 110 mL of 0.40 M HNO3 is mixed with 42 mL of 0.90 M Ba(OH)2? Write the molecular and net ionic equations.
Molecular ⇒ 2 HNO3 (aq) + Ba(OH)2 (aq)  2 H2O (l) + Ba(NO3)2 (aq)
Net Ionic ⇒ H+ (aq) + OH– (aq) H2O (l)
H+ ⇒ (110 mL)(0.4 M) = 44 mmols
OH– ⇒ (42 mL)(1.8 M) = 75.6 mmols
The solution will be basic because the moles of OH – exceed the moles of H+

32. Answer Key – Ch. 4
9. What volume of 0.15 M H2SO4 is required to neutralize 60 mL of 0.80 M KOH?
H2SO4 + 2 KOH  2 H2O + K2SO4
M1V1 = M2V2
Coeff
Coeff
(0.15 M)(V1)/1 = (0.8 M)(60 mL)/2
V1 = 160 mL

33. Answer Key – Ch. 4 10. Assign oxidation states for the following:
a. S in SO3 ⇒ S + 3(-2) = 0 ⇒ S = +6
b. P in PF5 ⇒ P + 5(-1) = 0 ⇒ P = +5
c. Cl in Cl2 ⇒ Cl = 0
d. Mn in MnO4– ⇒ (+1) + Mn + 4(-2) = 0 ⇒ Mn = +7
e. Cr in K2Cr2O7 ⇒ 2(Cr) + 7(-2) = -2 ⇒ Cr = +6
f. N in NH4NO3 ⇒ In NH4+ ⇒ N + 4(+1) = +1 ⇒ N = -3
In NO3– ⇒ N + 3(-2) = -1 ⇒ N = +5

34. Answer Key – Ch. 4
11. Identify the oxidizing and reducing agents for the following reactions and determine the moles of electrons transferred:
a. Ba Li  Ba Li+
Ba2+ is getting reduced ⇒ oxidizing agent
Li is getting oxidized ⇒ reducing agent
b. 3 Cu NO3– H+  3 Cu NO H2O
Cu is getting oxidized reducing agent
N in NO3– is getting reduced oxidizing agent
c. Pb + PbO H2SO4  2 PbSO H2O
Pb is getting oxidized reducing agent
Pb in PbO2 is getting reduced oxidizing agent
2 moles of electrons
6 moles of electrons
2 moles of electrons

35. Answer Key – Ch. 4 …continue to next slide
12. Balance the following red-ox reactions:
a. MnO NO  N2O MnO4- (acidic)
Reduction ½ : 2x(3 e– + 2 H2O + MnO2  MnO4– + 4 H+)
Oxidation ½ : 3x(4 H+ + 2 NO3–  N2O4 + 2H2O + 2 e–)
2 MnO2 + 6 NO3– 4 H+  2 MnO4– + 3 N2O4 + 2 H2O
b. CNO As2O3  CN HAsO42- (basic)
Reduction ½ : 2x(2 e– + H2O + CNO–  CN– + 2 OH–)
Oxidation ½ : 8 OH– + As2O3  2 HAsO42– + 3 H2O + 4 e–
2 CNO– + As2O3 + 4 OH–  2 CN– + 2 HAsO42– + H2O
…continue to next slide

36. Answer Key – Ch. 4 12. …continued
c. Cr SO42–  Cr2O72– + H2SO (acidic)
Reduction ½ : 3x(2 e– + 4 H+SO42-  H2SO3 + H2O)
Oxidation ½ : 7 H2O + 2 Cr3+  6 e– + 14 H+ + Cr2O72–
3 SO42– + 2 Cr H2O  2 H+ + 3 H2SO3 + CrO72–

37. Answer Key – Ch. 4
13. The reaction below can be used as a laboratory method of preparing small quantities of Cl2(g). If a 62.6 g sample that is 98.5% K2Cr2O7 by mass is allowed to react with 325 mL of HCl(aq) with a density of 1.15 g/mL and 30.1% HCl by mass, how many grams of Cl2(g) are produced?
Cr2O72- (aq) + Cl-(aq)  Cr3+(aq) + Cl2(g) (unbalanced)
Oxidation ½ : 6 e– + 14 H+ + Cr2O72– 7 H2O + 2 Cr3+
Reductions ½ : 3x(2 Cl-  Cl2 + 2 e-)
14 H+ + Cr2O72– + 6 Cl-  7 H2O + 2 Cr Cl2
K2Cr2O7 ⇒ (0.985)(62.6g) = 61.7g 1mole g = mole
HCl ⇒ (325mL)(1.15g/mL)(0.301) = 112.5g 1mole g = 3.08 mole
…continue to next slide

38. Answer Key – Ch. 4
13. ….continued mole K2Cr2O7 1 = vs mole 6 = Limiting reagent⇒ K2Cr2O K2Cr2O7 1mol Cr2O72– 1 mol K2Cr2O7 3 mol Cl2 1mol Cr2O72– g Cl2 1 mol Cl2 = g Cl2