1. TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY
Chapter Four:
TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY

2. Water, the Common Solvent
One of the most important substances on earth
Can dissolve many different substances
A polar molecule
4.1
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3. Dissolution of a Solid in a Liquid
4.1
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4. Nature of Aqueous Solutions
Solute – substance being dissolved
Solvent – liquid water
Electrolyte – substance that when dissolved in water produces a solution that can conduct electricity
4.2
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5. Electrolytes
Strong Electrolytes – conduct current very efficiently (bulb shines brightly)
Weak Electrolytes – conduct only a small current (bulb glows dimly)
Nonelectrolytes – no current flows (bulb remains unlit)
4.2
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6. Electrolyte Behavior
4.2
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7. Chemical Reactions of Solutions
We must know:
Nature of the reaction
Amounts of chemicals present in the solutions
4.3
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8. Molarity
Molarity (M) = moles of solute per volume of solution in liters:
4.3
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9. Concept Check
Which of the following solutions contains the greatest number of ions?
400.0 mL of 0.10 M NaCl.
300.0 mL of 0.10 M CaCl2.
200.0 mL of 0.10 M FeCl3.
800.0 mL of 0.10 M sucrose.
a) contains mol of ions (0.400×0.10×2). b) contains mol of ions (0.300×0.10×3). c) contains mol of ions (0.200×0.10×4). d) does not contain any ions because sucrose does not break up into ions. Therefore, letter b) is correct.
4.3
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10. Let’s Think About It
Draw molecular level pictures showing each solution. Think about relative numbers of ions.
How many moles of each ion are in each solution?
4.3
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11. Notice
The solution with the greatest number of ions is not necessarily the one in which:
the volume of the solution is the largest.
the formula unit has the greatest number of ions.
4.3
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12. Dilution
The process of adding water to a stock solution to achieve the molarity desired for a particular solution.
Dilution with water does not alter the numbers of moles of solute present.
Moles of solute before dilution = moles of solute after dilution
M1V1 = M2V2
4.3
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13. Types of Chemical Reactions
Precipitation Reactions
Acid-Base Reactions
Oxidation-Reduction Reactions
4.4
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14. Precipitation Reaction
A double displacement reaction in which a solid forms and separates from the solution.
When ionic compounds dissolve in water, the resulting solution contains the separated ions
Precipitate – the solid that forms
4.5
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15. The Reaction of K2CrO4(aq) and Ba(NO3)2(aq)
Ba2+(aq) + CrO42–(aq) → BaCrO4(s)
solid BaCrO4 formed
4.5
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16. Precipitation of Silver Chloride
4.5
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17. Precipitates
Soluble – solid dissolves in solution; (aq) is used in reaction
Insoluble – solid does not dissolve in solution; (s) is used in reaction
Insoluble and slightly soluble are often used interchangeably
4.5
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18. Simple Rules for Solubility
Most nitrate (NO3) salts are soluble.
Most alkali (group 1A) salts and NH4+ are soluble.
Most Cl, Br, and I salts are soluble (except Ag+, Pb2+, Hg22+).
Most sulfate salts are soluble (except BaSO4, PbSO4, Hg2SO4, CaSO4).
Most OH salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally soluble).
Most S2, CO32, CrO42, PO43 salts are only slightly soluble.
4.5
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19. Describing Reactions in Solution
Formula Equation (Molecular Equation)
Reactants and products as compounds
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
Complete Ionic Equation
All strong electrolytes shown as ions
Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq) 
AgCl(s) + Na+(aq) + NO3(aq)
4.6
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20. Describing Reactions in Solution
Net Ionic Equation
Show only components that actually react
Ag+(aq) + Cl(aq)  AgCl(s)
Na+ and NO3 are spectator ions
4.6
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21. Concept Check
You have two separate beakers with aqueous solutions, one with 4 “units” of potassium sulfate and one with 3 “units” of barium nitrate.
Draw molecular level diagrams of both solutions.
The first beaker should have 8 K+ ions and 4 SO42- ions. The second beaker should have 3 Ba2+ ions and 6 NO3- ions.
4.6/4.7
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22. Concept Check
You have two separate beakers with aqueous solutions, one with 4 “units” of potassium sulfate and one with 3 “units” of barium nitrate.
b) Draw a molecular level diagram of the mixture of the two solutions before a reaction has taken place.
Now all of the ions should be mixed together in one beaker.
4.6/4.7
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23. Concept Check
You have two separate beakers with aqueous solutions, one with 4 “units” of potassium sulfate and one with 3 “units” of barium nitrate.
c) Draw a molecular level diagram of the product and solution formed after the reaction has taken place.
There should be 8 K+ ions, 6 NO3- ions, 1 SO42- ion, and 3 units of solid BaSO4 formed in the beaker.
4.6/4.7
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24. Solution Stoichiometry Problems
Identify the species present in the combined solution.
Write the balanced net ionic equation.
Calculate the moles of reactants.
Determine limiting reactant.
Calculate the moles of product(s).
Convert to grams or other units.
4.7
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25. Acid-Base Reactions (Brønsted-Lowry)
Acid – proton donor
Base – proton acceptor
For a strong acid and base reaction:
H+(aq) + OH–(aq)  H2O(l)
4.8
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26. Neutralization of a Strong Acid by a Strong Base
4.8
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27. Key Titration Terms
Titrant – solution of known concentration used in titration
Analyte – substance being analyzed
Equivalence point – enough titrant added to react exactly with the analyte
Endpoint – the indicator changes color so you can tell the equivalence point has been reached
4.8
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28. Performing Calculations for Acid-Base Reactions
List initial species and predict reaction.
Write balanced net ionic reaction.
Calculate moles of reactants.
Determine limiting reactant.
Calculate moles of required reactant or product.
Convert to grams or volume, as required.
4.8
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29. Oxidation-Reduction Reactions (Redox Reactions)
Reactions in which one or more electrons are transferred.
4.9
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30. Reaction of Sodium and Chlorine
4.9
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31. Rules for Assigning Oxidation States
Oxidation state of an atom in an element = 0
Oxidation state of monatomic element = charge
Oxygen = 2 in covalent compounds (except in peroxides where it = 1)
Hydrogen = +1 in covalent compounds
Fluorine = 1 in compounds
Sum of oxidation states = 0 in compounds
Sum of oxidation states = charge of the ion
4.9
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32. Exercise
Find the oxidation states for each of the elements in each of the following compounds:
K2Cr2O7
CO32-
MnO2
PCl5
SF4
K2Cr2O7; K = +1; Cr = +6; O = -2
CO32-; C = +4; O = -2
MnO2; Mn = +4; O = -2
PCl5; P = +5; Cl = -1
SF4; S = +4; F = -1
4.9
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33. Redox Characteristics
Transfer of Electrons
Transfer may occur to form ions
Oxidation – increase in oxidation state (loss of electrons); reducing agent
Reduction – decrease in oxidation state (gain of electrons); oxidizing agent
4.9
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34. Concept Check
Which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent.
a) Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
b) Cr2O72-(aq) + 2OH-(aq)  2CrO42-(aq) + H2O(l)
c) 2CuCl(aq)  CuCl2(aq) + Cu(s)
Zn – reducing agent; HCl – oxidizing agent
c) CuCl acts as the reducing and oxidizing agent
4.9
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35. Balancing Oxidation-Reduction Reactions
Cr2O72-(aq) + SO32-(aq)  Cr3+(aq) + SO42-(aq)
How can we balance this equation?
First Steps:
Separate into half-reactions
Balance elements except H and O
4.10
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36. Method of Half Reactions
Cr2O72-(aq)  2Cr3+(aq)  
SO32-(aq)  SO42-(aq)
How many electrons are involved in each half reaction?
4.10
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37. Method of Half Reactions (continued)
6e- + Cr2O72-(aq)  2Cr3+(aq) 
SO32-(aq)  + SO42-(aq) + 2e-
How can we balance the oxygen atoms?
4.10
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38. Method of Half Reactions (continued)
6e- + Cr2O72-(aq)  Cr3+(aq) + 7H2O 
H2O +SO32-(aq)  + SO42-(aq) + 2e-
How can we balance the hydrogen atoms?
4.10
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39. Method of Half Reactions (continued)
This reaction occurs in an acidic solution.
14H+ + 6e- + Cr2O72-  2Cr3+ + 7H2O 
H2O +SO32-  SO e- + 2H+
How can we balance the electrons?
4.10
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40. Method of Half Reactions (continued)
14H+ + 6e- + Cr2O72-  2Cr3+ + 7H2O
3[H2O +SO32-  SO e- + 2H+]
Final Balanced Equation:
Cr2O SO H+  2Cr3+ + 3SO H2O
4.10
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41. Exercise
Balance the following oxidation-reduction reaction that occurs in acidic solution.
Br-(aq) + MnO4-(aq)  Br2(l)+ Mn2+(aq)
10Br-(aq) + 16H+(aq) + 2MnO4-(aq)  5Br2(l)+ 2Mn2+(aq) + 8H2O(l)
4.10
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42. Half-Reaction Method – Balancing in Base
Balance as in acid.
Add OH that equals H+ ions (both sides!)
Form water by combining H+, OH. Cancel any water molecules that appear on both sides.
Check elements and charges for balance.
4.10
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