Chapter 5
Molecular View of Reactions
in Aqueous Solutions
Chemistry: The Molecular Nature
of Matter, 6E
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Definitions:
Solution : Homogeneous mixture
 2 or more components mix freely
 Contains at least 2 substances
Solvent : Medium that dissolves solutes
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Component present in largest amount
Can be gas, liquid, or solid
Liquids most common
Aqueous solution—water is solvent
Solute : Substance dissolved in solvent
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Solution is named by solute
Can be gas—CO2 in soda
Liquid—Ethylene glycol in antifreeze
Solid—Sugar in syrup
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Electrical Conductivity
Electrolyte
 Solutes that yield electrically conducting solutions
 Separate into ions when enter into solution
Strong electrolyte
 Electrolyte that dissociates 100% in water
 Yields aqueous solution that conducts electricity
 Good electrical conduction
 Ionic compounds
 Strong acids and bases
Ex. NaBr, KNO3, HClO4, HCl
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Electrical Conductivity
Weak electrolyte
 Aqueous solution that weakly conducts electricity
due to low ionization
 Weak acids and bases
Ex. Acetic acid (HC2H3O2), ammonia (NH3)
Non-electrolyte
 Aqueous solution that doesn’t conduct electricity
 Molecules remain intact in solution
Ex. Sugar, alcohol
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Your Turn
How many ions form on the dissociation of
Na3PO4?
A. 1
B. 2
C. 3
D. 4
E. 8
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Your Turn
How many ions form on the dissociation of
Al2(SO4)3?
A. 2
B. 3
C. 5
D. 9
E. 14
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Equations for Dissociation
Reactions
 Ionic compound dissolves to form hydrated ions
 Hydrated = surrounded by water molecules
 In chemical equations, hydrated ions are
indicated by
 Symbol (aq) after each ions
 Ions are written separately
KBr(s)  K+(aq) + Br(aq)
Mg(HCO3)2(s)  Mg2+(aq) + 2HCO3(aq)
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Learning Check
Write the equations that illustrate the dissociation
of the following salts:
Na3PO4(aq) →
3 Na+(aq) + PO43(aq)
Al2(SO4)3(aq) →
CaCl2(aq) →
2 Al3+(aq) + 3 SO42(aq)
Ca2+(aq) + 2 Cl(aq)
Ca(MnO4)2(aq) →
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Ca2+(aq) + 2 MnO4(aq)
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Equations of Ionic Reactions
 When two soluble ionic solutions are mixed,
sometimes an insoluble solid forms.
 Three types of equations used to describe
1. Molecular Equation
 Substances listed as complete formulas
2. Ionic Equation
 All soluble substances broken into ions
3. Net Ionic Equation
 Only lists ions that actually take part in
reaction
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Equations of Ionic Reactions
1. Molecular Equation
Ex.
Pb(NO3)2(aq) + 2KI(aq)  PbI2(s) + 2KNO3(aq)
2. Ionic Equation
Ex.
Pb2+(aq) + 2NO3(aq) + 2K+(aq) + 2I(aq) 
PbI2(s) + 2K+(aq) +
2NO3(aq)
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Equations of Ionic Reactions
3. Net Ionic Equation
Ex. Pb2+(aq) + 2I(aq)  PbI2(s)
Spectator Ions
 Ions that don’t take part in reaction
 They hang around and watch
 K+ & NO3 in our example
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Learning Check: Convert Molecular
to Ionic Equations:
Write the correct ionic equation for each:
Pb(NO3)2(aq) + 2NH4IO3(aq) → Pb(IO3)2(s) + 2NH4NO3(aq)
Pb2+(aq) + 2NO3–(aq) + 2NH4+(aq) + 2IO3–(aq) →
Pb(IO3)2(s) + 2NH4+(aq) + 2NO3–(aq)
2NaCl (aq) + Hg2(NO3)2 (aq) → 2NaNO3 (aq) + Hg2Cl2 (s)
2Na+(aq) + 2Cl–(aq) + Hg22+(aq) + 2NO3–(aq) → 2Na+(aq)
+ 2NO3–(aq) + Hg2Cl2(s)
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Your Turn
Consider the following reaction :
Na2SO4(aq) + BaCl2(aq) → 2NaCl(aq) + BaSO4(s)
Which is the correct ionic equation?
A. 2Na+(aq) + SO42–(aq) + Ba2+(aq) + Cl22–(aq) → 2Na+(aq)
+ 2Cl–(aq) + BaSO4(s)
B. 2Na+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) → 2Na+(aq)
+ 2Cl–(aq) + BaSO4(s)
C. 2Na+(aq) + SO42–(aq) + Ba2+(aq) + Cl22–(aq) → 2Na+(aq)
+ 2Cl–(aq) + Ba2+(s) + SO42–(s)
D. Ba2+(aq) + SO42–(aq) → BaSO4(s)
E. Ba2+(aq) + SO42–(aq) → Ba2+(s) + SO42–(s)
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Learning Check: Convert Ionic
Equation to Net Ionic Equation
Write the correct net ionic equation for each.
Pb2+(aq) + 2NO3–(aq) + 2K+(aq) + 2IO3–(aq) →Pb(IO3)2(s)
+ 2K+(aq) + 2NO3–(aq)
Pb2+(aq) + 2IO3–(aq) → Pb(IO3)2(s)
2Na+(aq) + 2Cl–(aq) + Hg22+(aq) + 2NO3–(aq) → 2Na+(aq)
+ 2NO3–(aq) + Hg2Cl2(s)
2Cl–(aq) + Hg22+(aq) → Hg2Cl2(s)
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Your Turn
Consider the following molecular equation:
(NH4)2SO4(aq) + Ba(CH3CO2)2(aq) →
2NH4CH3CO2(aq) + BaSO4(s)
Which is the correct net ionic equation?
A. Ba2+(aq) + SO42–(aq) → BaSO4(s)
B. 2NH4+(aq) + 2CH3CO2–(aq) → 2NH4CH3CO2(s)
C. Ba2+(aq) + SO42–(aq) → BaSO4(aq)
D. 2NH4+(aq) + Ba2+(aq) + SO42–(aq) + 2CH3CO2–(aq)
→ 2NH4+(aq) + 2CH3CO2–(aq) + BaSO4(s)
E. 2NH4+(aq) + 2CH3CO2–(aq) → 2NH4CH3CO2(aq)
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Acids & Bases as Electrolytes
Indicators
 Dye molecules that change color
in presence of acids or bases
Acids
 Turn blue litmus red
 Lemon juice, vinegar, H2SO4
Bases
 Turn red litmus blue
 Drano (lye, NaOH), ammonia (NH3)
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Neutralization Reaction
 Acid reacts with base to form water and salt
(ionic compound).
Acid + base  salt + H2O
Ex. HCl(aq) + NaOH(aq)  NaCl(aq) + H2O
HBr(aq) + LiOH(aq)  LiBr(aq) + H2O
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Acids Categorized by Number of H+s
Monoprotic Acids
 Furnish only one H+
HNO3(aq) + H2O  H3O+(aq) + NO3–(aq)
HC2H3O2(aq) + H2O  H3O+(aq) + C2H3O2–(aq)
Polyprotic acids
 Furnish more than one H+
Diprotic acids — furnish two H+
H2SO3(aq) + H2O  H3O+(aq) + HSO3–(aq)
HSO3–(aq) + H2O  H3O+(aq) + SO32–(aq)
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Acids Catagorized by Number of H+s
Polyprotic acids
 Triprotic acids — furnish three H+
–H+
–H+
–H+
H3PO4  H2PO4–  HPO42–  PO43–
 Stepwise equations
H3PO4(aq) + H2O  H3O+(aq) + H2PO4–(aq)
H2PO4–(aq) + H2O  H3O+(aq) + HPO42–(aq)
HPO42–(aq) + H2O  H3O+(aq) + PO43–(aq)
Net:
H3PO4(aq) + 3H2O  3H3O+(aq) + PO43–(aq)
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Strong Acids
HClO4(aq)
HClO3(aq)
HCl(aq)
HBr(aq)
HI(aq)
HNO3(aq)
H2SO4(aq)
perchloric acid
chloric acid
hydrochloric acid
hydrobromic acid
hydroiodic acid
nitric acid
sulfuric acid
 Dissociate completely when dissolved in water
Ex. HBr(g) + H2O  H3O+(aq) + Br–(aq)
 Good electrical conduction
 Any acid not on this list, assume weak
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Strong Bases
 Bases that dissociate completely in water
 Soluble metal hydroxides
 KOH(aq)  K+(aq) + OH–(aq)
 Good electrical conductors
 Behave as (aq) ionic compounds
 Common strong bases are:
 Group IA metal hydroxides
 LiOH, NaOH, KOH, RbOH, CsOH
 Group IIA metal hydroxides
 Ca(OH)2, Sr(OH)2, Ba(OH)2
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Dynamic Equilibrium
 2 opposing reactions occurring at same rate
 Also called Chemical equilibrium
Equilibrium
 Concentrations of substances present in solution do
not change with time
Dynamic
 Both opposing reactions occur continuously
 Represented by double arrow
HC2H3O2(aq) + H2O
H3O+(aq) + C2H3O2–(aq)
Forward reaction – Forms ions
Reverse reaction – Removes ions
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Weak Bases
 Molecular bases
 Do not dissociate
 Accept H+ from water inefficiently
 Accept H+ from acids preferentially
NH3(aq) + HCl(aq)  NH4Cl(aq)
Ex.
NH3(aq) + H2O  NH4+(aq) + OH(aq)
Or for general base
B(aq) + H2O  BH+(aq) + OH(aq)
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Naming of bases
NaOH
Sodium hydroxide
LiOH
Lithium hydroxide
Ca(OH)2
Calcium hydroxide
Mg(OH)2 Magnesium hydroxide
Ba(OH)2
Barium hydroxide
Equilibrium for Weak Base
Net is dynamic equilibrium
NH3(aq) + H2O
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NH4+(aq) + OH(aq)
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Learning Check
 Write the ionization equation for each of the
following with water:
1. Weak base methylamine, CH3NH2.
CH3NH2(aq) + H2O
CH3NH3+(aq) + OH–(aq)
2. Weak acid nitrous acid, HNO2.
HNO2(aq) + H2O
H3O+(aq) + NO2–(aq)
3. Strong acid chloric acid, HClO3.
HClO3(aq) + H2O  H3O+(aq) + ClO3–(aq)
4. Strong base strontium hydroxide, Sr(OH)2.
Sr(OH)2(aq)  Sr2+(aq) + 2 OH–(aq)
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Your Turn
Which of the following is a weak acid?
A. HCl
B. HNO3
C. HClO4
D. HC2H3O2
E. H2SO4
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MOLARITY
A measurement of the concentration of a solution
Molarity (M) is equal to the moles of solute (n) per liter of
solution M = n / V = mol / L
Calculate the molarity of a solution prepared by mixing 1.5 g
of NaCl in 500.0 mL of water.
First calculate the moles of solute:
1.5 g NaCl (1 mole NaCl) = 0.0257 moles of NaCl
58.45 g NaCl
Next convert mL to L:
0.500 L of solution
Last, plug the appropriate values into the correct
variables in the equation:
M = n / V = Chemistry:
0.0257
moles / 0.500 L = 0.051 mol/L
The Molecular Nature of Matter, 6E
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How many grams of LiOH is needed to prepare 250.0 mL of a
1.25 M solution?
First calculate the moles of solute needed:
M = n / V , now rearrange to solve for n:
n = MV
n = (1.25 mol / L) (0.2500 L)
= 0.3125 moles of solute needed
Next calculate the molar mass of LiOH:
23.95 g/mol
Last, use diminsional analysis to solve for mass:
0.3125 moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of
LiOH
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MOLARITY & DILUTION
M 1 V1 = M 2 V2
The act of diluting a solution is to simply add more water
(the solvent) thus leaving the amount of solute unchanged.
Since the amount or moles of solute before dilution (nb)
and the moles of solute after the dilution (na) are the same:
nb = na
And the moles for any solution can be calculated by n=MV
A relationship can be established such that
MbVb = nb = na = MaVa
Or simply :
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MbVb = MaVa
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Example:
Calculate the molarity of a solution prepared by diluting 25.0
mL of 0.05 M potassium iodide with 50.0 mL of water (the
densities are similar).
M1 = 0.05 mol/L
M2 = ?
V1 = 25.0 mL
V2 = 50.0 + 25.0 = 75.0 mL
M 1 V1 = M 2 V 2
M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI
V2
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75.0 mL
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Example:
Given a 6.00 M HCl solution, how would you prepare
250.0 mL of 0.150 M HCl?
M1 = 6.00 mol/L
M2 = 0.150
V1 = ? mL
V2 = 250.0 mL
M 1 V1 = M 2 V 2
M2 V2 = V1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl
M1
6.00 mol/L
You would need 6.25 mL of the 6.00 M HCl reagent which would be
added to about 100 mL of DI water in a 250.0 mL graduated cylinder
then more water would be added to the mixture until the bottom of the
menicus is at 250.0 mL. Mix well.
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Titration
• Titration is a method for determining the
concentration of a solution by reacting a
known volume of that solution with a solution
of known concentration.
• If you wish to find the concentration of an
acid solution, you would titrate the acid
solution with a solution of a base of known
concentration.
• You could also titrate a base of unknown
concentration with an acid of known
concentration.
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In Short Form...
MAVA = MBVB (mol acid/mol base)
This is the mole ratio
Does this make sense? Let’s find out using
the definition of molarity (mol/L) and
dimensional analysis...
MAVA = MBVB (mol acid/mol base)
(mol acid/L acid)(L acid) = (mol base/L base)(L base) (mol acid/mol
base)
mol acid = mol base (mol acid/mol base)
mol acid = mol acid
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Example:
A sample of 50 ml of NaOH solution required 40 ml of 0.3 M HCl
solution for complete neutralization. What is the molarity of
NaOH solution?
Solution:
NaOH + HCl ----> NaCl + H2O
M a ˣ Va = M b ˣ Vb ( mol acid/ mol base )
0.3 M ˣ 40 ml = M b ˣ 50ml
M b = 0.24 M
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Example:
A sample of 80ml of H2SO4 solution required 50ml of 1.25 M
LiOH solution for complete neutralization. What is the molarity
of H2SO4 solution?
Solution:
H2SO4 + 2LiOH ----> Li2SO4+ 2H2O
M a ˣ Va = M b ˣ Vb ( mol acid/ mol base )
M a ˣ 80 ml = 1.25 ˣ 50ml (1 / 2 )
M b = 0.39 M
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