10/16/2011
Chapter 4: Types of Chemical Reactions
and Solution Stoichiometry
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
4.11
4.12
Water, the Common Solvent
The Nature of Aqueous Solutions: Strong and Weak Electrolytes
The Composition of Solutions (MOLARITY!)
Types of Chemical Reactions
Precipitation Reactions
Describing Reactions in Solution
Selective Precipitation (limited coverage)
Stoichiometry of Precipitation Reactions
Acid-Base Reactions
Oxidation-Reduction Reactions
Balancing Oxidation-Reduction Equations (Exam 2)
Simple Oxidation-Reduction Titrations
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Definitions – Solutes, Solvents and Solutions
 Solute
– Substance being dissolved, mixed,
diluted.
– Example: compounds extracted
from coffee grounds, sugar, milk.
 Solvent
– Substance doing the dissolving,
mixing, dilution.
– Example: water
 Solution
– Final combination of dissolution,
mixing, and dilution.
– Example: morning coffee
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Water as a Solvent
• water is an important solvent – dissolves many substances
• “aqueous” means a solution in which water is the solvent
• water is a POLAR molecule
Red: more electron density
Blue: less electron density
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Polar and Nonpolar Solutes
• water also dissolves some nonionic substances if they are
polar (ethanol-water)
• ethanol molecules are polar (contain directional O-H bond)
• nonpolar substances (e.g., octane (C8H18), benzene (C6H6),
fats, oils) will not dissolve in water
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Ionic Solutes
• polar water molecules dissolve ionic compounds (salts)
• “hydration” breaks ionic compounds into anions and cations
• water dissolves different ionic compounds to different
degrees (more in Ch 8)
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The Role of Water as a Solvent:
Dissolution of Ionic Compounds
Electrical conductivity: the flow of electrical current in a solution
is an indicator of the presence of ions in solution and the
solubility of ionic compounds.
Electrolyte: a substance that conducts a current when
dissolved in water. Soluble, ionic compounds that dissociate
completely conduct a large current and are called strong
electrolytes.
NaCl(s) + H2O(l)
Na+(aq) + Cl -(aq)
Ions become solvated/hydrated – are surrounded by water
molecules.
These ions are labeled “aqueous”, are free to move throughout
the solution, and conduct electricity.
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Strong Electrolytes
• Produce ions in aqueous solution and conduct electricity well.
• Strong electrolytes are soluble salts (NaCl, NH4NO3), strong
acids, and strong bases.
• Strong acids produce H+ ions when they dissolve in water.
HCl, HNO3, and H2SO4 are strong acids:
HNO3(aq) → H+(aq) + NO3-(aq)
• Strong bases produce OH- ions when they dissolve in water:
NaOH and KOH are strong bases:
NaOH(s) → Na+(aq) + OH-(aq)
All of the above species are ionized nearly 100%.
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HCl (aq) is completely
ionized.
NaOH (aq) is completely
ionized.
Strong acids fully dissociate,
forming the anion and a
hydrated proton.
Strong bases fully dissociate,
forming a cation and the
hydroxide anion.
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Weak Electrolytes
• Produce relatively few ions in aqueous solutions
• The most common weak electrolytes are weak acids and
weak bases:
acetic acid is a typical weak acid:
HC2H3O2(aq) ⇌ H+(aq) + C2H3O2-(aq)
ammonia is a common weak base:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
Both of these species are ionized only ~1%
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Acetic acid (HC2H3O2) exists
in water mostly as
undissociated molecules.
Weak acids partially dissociate,
forming only a small number of
anions and hydrated protons.
The reaction of NH3 in water.
Weak bases partially dissociate
(or react with water to a limited
extent), forming only a small
number of cations and
hydroxide anions.
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Nonelectrolytes
• Dissolve in water but produce no ions in solution.
• Nonelectrolytes do not conduct electricity because when a
sample of the compound dissolves, the substance remains
intact as whole molecules and no ions are produced.
• Common nonelectrolytes include:
ethanol (CH3CH2OH)
table sugar (sucrose, C12H22O11)
• Insoluble ionic compounds are also nonelectrolytes:
Ca3(PO4)2, MgCO3
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Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
How many moles of each ion are formed when the following
compounds are dissolved in water:
a) 4.0 moles of sodium carbonate
b) 46.5 g of rubidium fluoride
c) 9.32 x 1021 formula units of iron(III) chloride
a) Na2CO3 (s)
H2O
moles of Na+ = 4.0 moles Na2CO3 x
2 Na+(aq) + CO3-2(aq)
2 mol Na+
1 mol Na2CO3
= 8.0 moles Na+ (and 4.0 moles of CO3-2)
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Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - II
b)
H2O
RbF(s)
Rb+(aq) + F -(aq)
1 mol RbF = 0.445 moles RbF
104.47 g RbF
thus, 0.445 mol Rb+ and 0.445 mol F -
moles of RbF = 46.5 g RbF x
c)
FeCl3 (s)
H2O
moles of FeCl3 =
9.32 x 1021 formula units x
= 0.0155 mol FeCl3
Fe+3(aq) + 3 Cl -(aq)
1 mol FeCl3
6.022 x 1023 formula units FeCl3
= 0.0155 mol Fe+3
moles of Cl - = 0.0155 mol FeCl3 x 3 mol Cl
= 0.0465 mol Cl 1 mol FeCl3
Molarity and Stoichiometry
• When we first met stoichiometry, we talked about
combining moles or masses of reactants.
• But, A LOT of interesting chemistry happens in aqueous
phase, where it’s more convenient to talk about molarity.
• The molarity of a solution gives you information about
the number of moles of solute available to participate in
a reaction.
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Describing Solutions: Molarity
Molarity (M) =
0.1 mol C6H12O6
moles solute
volume solution
=
n
V
0.1 mol 1000 mL = 0.2 mol
= 0.2 M
500. mL
1L
1L
500. mL
water
OR...
0.1 mol= 0.2 mol = 0.2 M
0.500 L
1L
[ C6H12O6 ] = 0.2 M
Using Molarity
Molarity (M) =
moles solute
volume solution
=
n
V
How many grams of glucose are in this solution?
(MM = 180.2 g/mol)
[C6H12O6 ] =
500. mL
water
1.5 M
n = MV = (1.5 mol/L)*(0.500 L) = 0.750 mol
? g = 0.75 mol 180.2 g
1 mol
= 135.15 g
= 140 g
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Molarity of Ionic Solutions
• Recall, ionic compounds separate into their component ions
when they dissolve in water.
MgCl2(s)
H2O(l)
For every MgCl2
formula unit that
dissolves...
Mg2+(aq)
+
...you get one
Mg2+ ion, and...
• This means... [MgCl2] = [Mg2+] =
1
2
2Cl-(aq)
... two Cl- ions
in solution.
[Cl-]
• Think about the solution in two ways:
– What is the molarity (M) with regards to the initial
composition of the solute?
– What is the M of each species given the electrolytic nature of
the solute?
Examples
Determine the concentrations of all the ions in each of the
following solutions:
1 M FeCl3:
[Fe3+] = 1 M
[Cl-] = 3 M
[ions] = 4 M
0.50 M Co(NO3)2:
[Co2+] = 0.50 M
[NO3-] = 1.0 M
[ions] = 1.50 M
How many moles of NO3- ions are there in 350. mL of a
solution that is 0.065 M in Co(NO3)2?
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Example
A water sample from the Great Salt Lake in Utah contains
83.6 mg of Na+ per 1.000 g of lake water. What is the
molarity of sodium ions in the lake?
(dlake water = 1.160 g/mL)
Mass of Na+
Mass of lake
water
Moles of Na+ in
1.000 g lake
water
Moles of Na+
L of lake
water
Vol of 1.000 g
lake water
Molarity
of Great
Salt Lake
Example (cont)
? mol Na+ =83.6 mg Na+
1g
1 mol Na+
1000 mg
22.99 g Na+
? L lake water = 1.000 g wtr
? [Na+] =
1 mL wtr
= 3.636 x 10-3 mol Na+
1L
1.160 g wtr 1000 mL
3.636 x 10-3 mol Na+
8.6207 x 10-4 L lake wtr
=
= 8.6207 x 10-4 L
4.218 mol/L
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Solution Preparation: Standard Solutions
Solution for which the concentration is accurately known.
Often prepared by starting with a solid solute weighed out
on a balance with several significant figures.
Solution Preparation: Standard Solutions
How would you prepare 1.00 L of a 0.375 M solution of
ammonium carbonate?
Determine the moles of ammonium carbonate required:
1.00 L x
0.375 mol (NH4)2CO3
L solution
= 0.375 mol (NH 4)2CO3
Convert to grams using the molar mass:
94.07 g (NH4)2CO3
0.375 mol (NH4)2CO3 x
= 35.276 g (NH4)2CO3
mol (NH4)2CO3
To make 1.00 L of solution, 35.3 g of (NH4)2CO3 are weighed out
and transferred to a 1.00 L volumetric flask. DI water is added to
dissolve the solid and then to dilute the solution to the mark on
the neck of the flask. Recalculate molarity based on specific mass.
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Solution Dilution
Dilution is the process of making a solution with a certain
molarity from a solution with a greater molarity.
Concentrated
acetic acid
Diluted
acetic acid
Volumetric
glassware
has very high
precision
500.0 mL
Dilution
• The number of moles (n) of solute stays the same…only the
volume of solution (V) changes.
• We can formalize this relationship:
M1V1 = M2V2
where…
M1, V1 = molarity and volume of concentrated solution
M2, V2 = molarity and volume of diluted solution
• Recall:
M = n/V  n = MV
• n remains unchanged:
M1V1 = n = M2V2
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Using M1V1 = M2V2
What volume of 5.00 M calcium nitrate solution is
needed to prepare 500.0 mL of a 0.250 M Ca(NO3)2
solution?
M1V1 = M2V2
V1 = ?
V2 = 500.0 mL
M1 = 5.00 M
M2 = 0.250 M
V1 =
M2V2
M1
=
(0.250 mol/L)*(0.5000 L)
5.00 mol/L
= 0.0250 L
= 25.0 mL
Stoichiometry Flowchart
Vol of
known soln
of known M
Convert
to moles
Vol of
desired soln
of known M
Convert
to volume
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