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Chapter 4
Reactions in Aqueous
Solutions
1
Solution
Terms
Homogenous mixture of 2 or more substances
Solvent:
Component with largest amount
Water is the universal solvent
Solute:
Remaining components: smaller amounts
Solvation/dissolving:
Water molecules surround & support solute molecules or ions
Water is NOT a part of the chemical reaction
2
Conductivity of Electrolytes in Aqueous Solutions
Non-electrolyte
No ionization
Ex: sugar
Weak electrolyte
Some ionization
Ex: acetic acid
Strong electrolyte
Full ionization
Ex: NaCl
3
Precipitation Reactions
4
Precipitation Terms
Precipitation: Ions in solution combine to form an insoluble solid salt
Precipitate: Solid salt that is formed
Spectator Ions: Ions that do not react in solution and remain as ions
Pb(N03 )2 (aq) +2KI (aq) Pb2+(aq)+ 2NO3- (aq)+ 2K+(aq)+2I -(aq)
Pb2+(aq)+ 2N03- (aq)+ 2K+(aq)+2I -(aq)  PbI2 (s, yellow)+2K+(aq)+ 2NO3 -(aq)
Net ionic equation: Pb2+ (aq) + 2I-(aq)  PbI2 (s, yellow)
5
Solubility Rules (must memorize)
Solubility rules classify compounds into those that usually are
soluble and those that usually are insoluble.
Soluble
Check first!
+
Ammonium (NH4 )
+
Hydrogen ion (H )
Alkali Metal (group 1A)
–
Nitrate (NO3 )
–
Perchlorate (ClO4 )
–
Acetate (C2H3O2 )
Always soluble
Always soluble
Always soluble
Always soluble
Always soluble
Always soluble
2+
2+
+
–
– –
–
Insoluble: Pb , Hg2 , Ag
Halide (Cl , Br , I & F )
2–
2+
2+
2+
2+
2+
Sulfate (SO4 )
Insoluble: Pb , Hg2 ,Ag+,Ba , Ca ,Sr
Sparingly
Soluble
(Insoluble)
2–
Sulfide (S )
–
Hydroxide (OH )
2Oxide (O )
2–
Carbonate (CO3 )
3–
Phosphate (PO4 )
Only if cation is soluble
Only if cation is soluble
Only if cation is soluble
Only if cation is soluble
See cations above
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Precipitation Reactions
1. Divide cations from anions in each reactant:
BaCI2 (aq) + Na2SO4 (aq) Ba2+(aq) + 2Cl-(aq) + 2Na+(aq)+ SO4 2-(aq)
2. Match cation from one salt with the anion from the other salt”
Ba2+(aq) + Cl-(aq) + Na+(aq)+ SO4 2-(aq) NaCl+ BaSO4
Note: Always keep the metal on the left in all salts!
3. Balance charges in salts and put in coefficients
Ba2+(aq) + 2Cl-(aq) + 2Na+(aq)+ SO4 2-(aq) 2 NaCl + BaSO4
4. Use solubility rules to predict solubility of products
BaCI2 (aq) + Na2SO4 (aq)  2 NaCl(aq) + BaSO4 (s)
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Acid-Base Reactions
8
Arrhenius Acids and Bases
Acid
Compound that ionizes in water to form a solution of H+ ions
(H3O+ ) and anions.
Base
Compound that ionizes in water to form a solution of OH- ions
and cations
Neutralization
Reaction between Arrhenius acid & base
H+ + OH- H2O and cation + anion  salt
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Strength of Acids and Bases
Strong acids and bases
Completely ionized in water to give either H3O+ or OHGood conductors of electricity.
Directional arrow () indicates dissociation is complete
Weak acids and bases
Partial ionization in water, most of original compound remains
Poor conductors
Double arrow () indicates dissociation is incomplete
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Common Acids and Bases(Memorize)
Strong Acids
Weak Acids
HCl, HI, HBr
HNO3
H2SO4
HClO4
HF
HNO2
Acetic acid, CH3COOH
H3PO4
Strong Bases
Weak Bases
NaOH, KOH, LiOH
Ba(OH)2
Other hydroxides (don’t dissolve)
Ammonia: NH3
11
Acid-Base Neutralizations
Strong acid with strong base
NaOH (aq) + HCl (aq) NaCI (aq) + H2O
Na+ +OH- + H+ + Cl-  Na+ + Cl- + H2O
Net ionic equation: OH- + H+H2O
Solid (weak) base with strong acid
Fe(OH)3 (s) + 3HCl(aq)FeCI3 (aq) + 3 H2O(l)
Fe(OH)3 (s) + 3H+ (aq) + 3Cl-(aq) Fe3+ (aq)+ 3Cl-+ 3H2O
Net ionic equation: Fe(OH)3 (s) + 3H+ Fe3+ + 3H2O
Weak acid with strong base
HCN (aq) + NaOH(aq)NaCN(aq) + 3 H2O(l)
HCN (aq) + Na+ (aq)+OH- (aq) Na+ (aq)+ CN-+ H2O
Net ionic equation: HCN (aq) +OH- (aq) CN-+ H2O
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Oxidation-Reduction
Reactions
13
Oxidation-Reduction Reactions
Oxidation-reduction reactions (REDOX reaction) occur when
electrons are transferred from one reactant to another during a
chemical reaction. There is a change in oxidation number for both
substances
Oxidation State/oxidation number: Theoretical charge on atom
Oxidation is the process where the oxidation number increases.
Electrons are lost from the substance
Reduction is the process where the oxidation number decreases.
Electrons are gained by the substance
Oxidation and reduction always accompany each other;
Neither can occur alone
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Redox Reaction: Half-reactions
Oxidation half-reaction:
Reduction half-reaction:
Sum of half-reactions:
Mg (s)  Mg2+ + 2e1/2O2(g) + 2e-  O2Mg (s) + 1/2O2(g)  MgO(s)
15
LEO the lion says GER
LEO
Lose
Electrons
Oxidation
GER
Gain
Electrons
Reduction
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Oxidation Number Rules
The rule earlier in the list always takes precedence.
1) ON = 0 for a compound or ionic charge for an ion
2) ON = +1 for IA elements and H
ON = +2 2A elements
3) ON= -2 for oxygen
4) ON= -1 for 7A elements
If both elements in 7A, then the one higher in the list is -1
5) ON = -2 for 6A elements other than oxygen
6) ON = -3 for 5A elements (very shaky!!!)
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Elemental Oxidation Numbers
18
Determining Oxidation Numbers
Determine the oxidation number of each element in:
NH3
CO32H202
NH4+
NO3-
H has ON = +1
0 has ON = -2
H has ON = +1
H has ON = +1
0 has ON = -2
N has ON = -3
C has ON = +4
0 has ON = -1
N has ON = -3
N has ON = +5
If composed of polyatomic ions, break down the
compound into ions before determining the oxidation
state of each element.
NH4NO3
N in NH4+ has ON = -3
N in NO3 – has ON=+5
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Oxidizing and Reducing Agents
Oxidizing agent: reactant that promotes oxidation
Electrons are gained so oxidizing agent is reduced.
Characteristic of nonmetals: ex: fluorine, oxygen.
High electron affinity: easily gains electrons
Within a group, ion with highest ON is better oxidizing agent.
Reducing agent: reactant that promotes reduction
Electrons are lost, so reducing agent is oxidized
Characteristic of an active metal, such as sodium.
Low ionization energy: easily loses electrons
Within a group, ion with lowest ON is better reducing agent.
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Types of Redox Reactions
Combination:
2Al(s)+ 3Br2(g)  2AlBr3(s)
0
Decomposition:
0
+3 -3
2KClO3(s)  2KCl(s) + 3O2(g)
+1 +5 -2
+1 -1
0
Disproportionation:Cl2(g)+ NaOH(aq)NaOCl(aq)+Cl- (aq)+ H2O(l)
0
Combustion:
+1 -2 +1
+1 -2 +1
-1
Bleach Production
CH4(g)+ 2O2(g)  CO2(g) + 2H2O(g)
-4 +1
0
+4 -2
+1 -2
+1 -2
Methane recovery from sewerage
Halogen Displacement: Cl2(g)+ 2KBr(s)  Br2(g)+ 2KCl(s)
F2>Cl2>Br2>I2
0
+1 -1
Production of Bromine gas
0
+1-1
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Displacement Reactions
A + BC  AB + C
Hydrogen Displacement
Ba(s)+ 2H2O(aq)  H2(g)+ Ba(OH)2(aq)
0
+1 -2
0
+2 -2 +1
Relative activity with water and acid
Metal Displacement
Cu(s)+ 2Ag +(aq)  2Ag(s)+ Cu2+(aq)
0
+1
0
+2
Silver Plating
Halogen Displacement: F2>Cl2>Br2>I2
Cl2(g)+ 2KBr()  Br2(g)+ 2KCl(s)
0
+1 -1
0
+1-1
Production of Bromine gas
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Other Displacement Reactions:
Activity Series
Reactivity of 2 metals
Higher metal becomes cation
Lower metal will be free metal
Pb(s)+2Cu+(aq) →Pb2 +(aq) + Cu(s)
Reactivity with water
Ca(s+ 2H2O(l) → Ca(OH)2(aq)+ H2(g)
Mg(s+ 2H2O(l) → Mg(OH)2(aq)+ H2(g)
Reactivity with acid
Pb(s)+ 2H+(aq) → Pb2 +(aq) + H2(g)
Pt(s)+ 2H+(aq) → NR
23
Concentration of Solutions
24
Molarity
Molarity (M)= moles solute/L solution
Units of mol/L
Conversion factor between moles solute & volume of solution.
Prepare 2 liters of a 1.0M solution of NaCl?
1. Calculate mass of Na Cl needed.
1molNaCl 58.5g NaCl 2LNaCl
 117g NaCl
x
x
1LNaCl
1molNaCl
1
2. Weigh out mass of NaCl.
3. Pour NaCl into volumetric flask.
4. Add water until the water reaches the 2L mark.
25
Dilution Of Solutions
Water is added to a small amount of stock solution to make a
less concentrated solution.
Addition of solvent does not change the mass of solute in a
solution but does change the solution concentration.
M1V1=M2V2
mol/L x L=mol/L x L
Calculate the volume of 1.0M stock solution needed to make
2.0L of a 0.12M solution of HCl.
M1= 0.12M M2=1.0M
V1= 2.0L
V2=?
M 1V1 0.12 Mx 2.0 L
V2 

 0.24 L  240mL
M2
1.0 M
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Conversion Relationships
Divide by density
Mass
g
Multiply by density
Divide by molar mass
Multiply by molar mass
Divide by molarity
Multiply by molarity
Moles
mol
Divide by Avagadro’s #
Volume
ml, L, cm3
Multiply by Avagadro’s #
Number of Particles
atoms, molecules, ions, etc.
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Calculating Ion Concentrations in Solution
What are the concentrations of aluminum ion, sulfate ion & nitrate
ion in a solution that is1.20 M aluminum sulfate and 1.0M
aluminum nitrate?
1. Write down how the salts break up in water.
Al2(SO4)3 2 Al3+(aq) + 3 SO42-(aq)
Al(NO3)3  Al3+(aq) + 3 NO3-(aq)
2. Add up all the concentrations and multiply by the number of ions
in the solution.
2 Al3+ + 3 SO42- 1.2M x 2 = 2.4M for Al3+
Al3+ + 3 NO3- 1.0M x 1 = 1.0M for Al3+
1.2M x 3 = 3.6M SO421.0M x 3 = 3.0M NO3-
3. Add up ions if there is more than 1 source.
2.4M Al3+ + 1.0M Al3+ = 3.4M Al3+
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Mathematical Solubility Problems
What is the molarity of a sodium phosphate solution if adding
AgNO3 (aq) to 75.0 mL produces 0.205 g Ag3PO4 (s)?
First produce the balanced chemical equation.
Na3PO4 (aq) +3 AgNO3 (aq)  Ag3PO4 (s) +3 NaNO3 (aq)
Then determine the amount of Na3PO4(s) in moles.
0 .205 g Ag 3 PO 4 x
1mol Ag 3 PO 4
418 g Ag 3 PO 4
x
1mol Na 3 PO 4
1mol Ag 3 PO 4
 4 .90 x10  4 Na 3 PO 4
Calculate the molarity of Na3PO4
4 . 90 x10  4 mol
Na 3 PO 4
75 . 0 ml Na 3 PO 4
1000 ml
x
 6 . 53 x10  3 M
1L
Na 3 PO 4
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Gravimetric Analysis
What is the % Cl in a sample of unknown composition?
1. Dissolve known mass of an unknown sample in water
0.5662g MxCly dissolved in waterxM+ + yCl-
React unknown with Ag+ to form a precipitate
Ag+(aq) + Cl-(aq) AgCl(s)
Filter and dry and weigh precipitate
1.0882g AgCl(s) recovered
Use stoichiometry to determine amount of unknown ion.
Mass Cl 
1 .0882 g AgCl
1
x
1mol AgCl
143 .3 g AgCl
x
1mol Cl
35 .45 g Cl
x
 0 .2692 g Cl
1mol AgCl
1mol Cl
%Cl= 0.2692g / 0.5662g x 100%= 47.54%Cl
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Titration
Titration reactions are used to determine acid/base concentration
1. React solution of known concentration with
measured volume of unknown solution
2. Reach endpoint of reaction
Ratio of reactants equals that in chemical
reaction.
For acid/base: moles H+ = moles OHUse an indicator to determine endpoint
chemical that changes color at endpoint
Record volume of second solution
3. Calculate molarity of unknown solution
based on molarity and volume of 1 solution
and volume recorded in titration.
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Acid/BaseTitration
25.00-mL of 0.200 M (H2SO4) is titrated with 12.32 ml of a
NaOH solution. What is the molarity of the NaOH solution?
1) Find the concentration of H+ and OH- in the chemicals
H2SO4(aq) + 2 NaOH (aq) Na2SO4(aq) + 2 H2O
2H+(aq) + 2 OH -(aq) 2 H2O
2 moles of H+ per mole of H2SO4(aq) = 0.400M H+
2) Solve for molarity OHmoles of OH- = moles of H3O+ at endpoint
M OH  
M H  xV H 
VOH 

0 .400 M H  x 25 .00 mL H 
12 .32 mL OH 
 0 .812 M OH 
2) Molarity of OH- = Molarity of NaOH
MNaOH = 0.812M
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Redox Titration
Determine the molarity of a potassium permanganate solution if
25.32 mLs are needed to react completely with 0.724 g Na2C2O4 (s).
2KMnO4 +5Na2C2O4 +8H2SO4 10CO2+ 8H2O +2MnSO4+ 5Na2SO4+ K2SO4
First determine amount in moles of KMnO4.
0.724g Na2C2O4 x
g Na2C2O4
1molNa2C2O4
134g Na2C2O4
>
x
2molKMnO4
5molNa2C2O4
mol Na2C2O
 2.16x103 molKMnO4
>
mol KMnO4
Determine concentration (Molarity!) of KMnO4
3
2.16 x10 molKMnO4
25.32mLKMnO4
X
1000mlKMnO4
1LKMnO4
 0.854M KMnO4
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