1. Chapter 4 Aqueous Reactions and Solution Stoichiometry
Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 4 Aqueous Reactions and Solution Stoichiometry
John D. Bookstaver
St. Charles Community College
Cottleville, MO
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2. Solutions
Solutions are defined as homogeneous mixtures of two or more pure substances.
The solvent is present in greatest abundance.
All other substances are solutes.
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3. Dissociation
When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them.
This process is called dissociation.
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4. Dissociation
An electrolyte is a substances that dissociates into ions when dissolved in water.
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5. Electrolytes
An electrolyte is a substances that dissociates into ions when dissolved in water.
A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.
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6. Electrolytes and Nonelectrolytes
Soluble ionic compounds tend to be electrolytes.
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7. Electrolytes and Nonelectrolytes
Molecular compounds tend to be nonelectrolytes, except for acids and bases.
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8. Electrolytes
A strong electrolyte dissociates completely when dissolved in water.
A weak electrolyte only dissociates partially when dissolved in water.
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9. Strong Electrolytes Are…
Strong acids
Strong bases
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10. Strong Electrolytes Are…
Strong acids
Strong bases
Soluble ionic salts
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11. Precipitation Reactions
When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.
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12. Metathesis (Exchange) Reactions
Metathesis comes from a Greek word that means “to transpose.”
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
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13. Metathesis (Exchange) Reactions
Metathesis comes from a Greek word that means “to transpose.”
It appears the ions in the reactant compounds exchange, or transpose, ions.
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
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14. Solution Chemistry
It is helpful to pay attention to exactly what species are present in a reaction mixture (i.e., solid, liquid, gas, aqueous solution).
If we are to understand reactivity, we must be aware of just what is changing during the course of a reaction.
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15. AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
Molecular Equation
The molecular equation lists the reactants and products in their molecular form.
AgNO3 (aq) + KCl (aq)  AgCl (s) + KNO3 (aq)
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16. Ionic Equation
In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions.
This more accurately reflects the species that are found in the reaction mixture.
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq) 
AgCl (s) + K+ (aq) + NO3- (aq)
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17. Net Ionic Equation
To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) 
AgCl (s) + K+(aq) + NO3-(aq)
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18. Ag+(aq) + Cl-(aq)  AgCl (s)
Net Ionic Equation
To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
The only things left in the equation are those things that change (i.e., react) during the course of the reaction.
Ag+(aq) + Cl-(aq)  AgCl (s)
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19. Net Ionic Equation
To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right.
The only things left in the equation are those things that change (i.e., react) during the course of the reaction.
Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions.
Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) 
AgCl (s) + K+(aq) + NO3-(aq)
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20. Writing Net Ionic Equations
Write a balanced molecular equation.
Dissociate all strong electrolytes.
Cross out anything that remains unchanged from the left side to the right side of the equation.
Write the net ionic equation with the species that remain.
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21. Solubility Rules for Ionic Salts
All sodium, potassium, nitrate (NO3-), and ammonium (NH4+) salts are soluble in water.

22. Electrochemical Reactions
In electrochemical reactions, electrons are transferred from one species to another.
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23. Oxidation Numbers
In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.
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24. Oxidation and Reduction
A species is oxidized when it loses electrons.
Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.
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25. Oxidation and Reduction
A species is reduced when it gains electrons.
Here, each of the H+ gains an electron, and they combine to form H2.
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26. Oxidation and Reduction
What is reduced is the oxidizing agent.
H+ oxidizes Zn by taking electrons from it.
What is oxidized is the reducing agent.
Zn reduces H+ by giving it electrons.
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27. Assigning Oxidation Numbers
Elements in their elemental form have an oxidation number of 0.
The oxidation number of a monatomic ion is the same as its charge.
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28. Assigning Oxidation Numbers
Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
Oxygen has an oxidation number of −2, except in the peroxide ion, which has an oxidation number of −1.
Hydrogen is −1 when bonded to a metal and +1 when bonded to a nonmetal.
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29. Assigning Oxidation Numbers
Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions.
Fluorine always has an oxidation number of −1.
The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.
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30. Assigning Oxidation Numbers
The sum of the oxidation numbers in a neutral compound is 0.
The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
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31. Balancing Oxidation-Reduction Equations
Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.
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32. Balancing Oxidation-Reduction Equations
This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.
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33. The Half-Reaction Method
Assign oxidation numbers to determine what is oxidized and what is reduced.
Write the oxidation and reduction half-reactions.
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34. The Half-Reaction Method
Balance each half-reaction.
Balance elements other than H and O.
Balance O by adding H2O.
Balance H by adding H+.
Balance charge by adding electrons.
Multiply the half-reactions by integers so that the electrons gained and lost are the same.
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35. The Half-Reaction Method
Add the half-reactions, subtracting things that appear on both sides.
Make sure the equation is balanced according to mass.
Make sure the equation is balanced according to charge.
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36. The Half-Reaction Method
Consider the reaction between MnO4− and C2O42− :
MnO4− (aq) + C2O42− (aq)  Mn2+ (aq) + CO2 (aq)
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37. The Half-Reaction Method
First, we assign oxidation numbers.
MnO4− + C2O42-  Mn2+ + CO2 +7 +3 +4 +2
Since the manganese goes from +7 to +2, it is reduced.
Since the carbon goes from +3 to +4, it is oxidized.
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38. Oxidation Half-Reaction
C2O42−  CO2
To balance the carbon, we add a coefficient of 2:
C2O42−  2 CO2
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39. Oxidation Half-Reaction
C2O42−  2 CO2
The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.
C2O42−  2 CO2 + 2 e−
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40. Reduction Half-Reaction
MnO4−  Mn2+
The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.
MnO4−  Mn H2O
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41. Reduction Half-Reaction
MnO4−  Mn H2O
To balance the hydrogen, we add 8 H+ to the left side.
8 H+ + MnO4−  Mn H2O
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42. Reduction Half-Reaction
8 H+ + MnO4−  Mn H2O
To balance the charge, we add 5 e− to the left side.
5 e− + 8 H+ + MnO4−  Mn H2O
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43. Combining the Half-Reactions
Now we evaluate the two half-reactions together:
C2O42−  2 CO2 + 2 e−
5 e− + 8 H+ + MnO4−  Mn H2O
To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.
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44. Combining the Half-Reactions
5 C2O42−  10 CO e−
10 e− + 16 H+ + 2 MnO4−  2 Mn H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn H2O + 10 CO2 +10 e−
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45. Combining the Half-Reactions
10 e− + 16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O42− 
2 Mn H2O + 10 CO2
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46. Balancing in Basic Solution
If a reaction occurs in basic solution, one can balance it as if it occurred in acid.
Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.
If this produces water on both sides, you might have to subtract water from each side.
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47. Displacement Reactions
In displacement reactions, ions oxidize an element.
The ions, then, are reduced.
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48. Displacement Reactions
In this reaction,
silver ions oxidize
copper metal.
Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)
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49. Displacement Reactions
The reverse reaction,
however, does not
occur.
Cu2+ (aq) + 2 Ag (s)  Cu (s) + 2 Ag+ (aq) x
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50. Activity Series
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51. volume of solution in liters
Molarity
Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different.
Molarity is one way to measure the concentration of a solution.
moles of solute
volume of solution in liters
Molarity (M) =
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52. Mixing a Solution
To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute.
The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.
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53. Dilution One can also dilute a more concentrated solution by
Using a pipet to deliver a volume of the solution to a new volumetric flask, and
Adding solvent to the line on the neck of the new flask.
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54. Dilution
The molarity of the new solution can be determined from the equation
Mc  Vc = Md  Vd,
where Mc and Md are the molarity of the concentrated and dilute solutions, respectively, and Vc and Vd are the volumes of the two solutions.
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55. Using Molarities in Stoichiometric Calculations
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56. Titration
Titration is an analytical technique in which one can calculate the concentration of a solute in a solution.
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