Chapter 4

Report
Chapter 4
Solution Chemistry:
The Hydrosphere
Chapter Outline
• 4.1 Solutions on Earth and Other Places
• Solutions: Solute vs. Solvent
•
•
•
•
•
•
•
•
4.2 Concentration Units
4.3 Dilutions
4.4 Electrolytes and Nonelectrolytes
4.5 Acid–Base Reactions: Proton Transfer
4.6 Titrations
4.7 Precipitation Reactions
4.8 Ion Exchange
4.9 Oxidation–Reduction Reactions: Electron
Transfer
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Earth: The Water Planet
• Earth’s surface:
• About 75% covered by water
• Depressions in Earth’s crust filled with
1.5 × 1021 L of H2O(ℓ)
• Properties of water are responsible for life
on Earth and many geographical features.
• All natural waters have ionic and
molecular compounds dissolved in them.
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Solutions
• Solutions:
• Homogeneous mixtures of two or more
substances
• solvent = a component of a solution that is
present in the greatest amount.
• solute = any component in a solution other than
the solvent (i.e., the other ingredients in the
mixture).
• Aqueous solutions → water is the solvent
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Solutions (cont.)
Water is the
solvent; sugar is
the solute.
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Chapter Outline
• 4.1 Solutions on Earth and Other Places
• 4.2 Concentration Units
• Defining the Amount of Solute in Solution
• Mass-based Units: ppm and ppb
• Mole-based Units: Molarity
•
•
•
•
•
•
•
4.3
4.4
4.5
4.6
4.7
4.8
4.9
Dilutions
Electrolytes and Nonelectrolytes
Acid–Base Reactions: Proton Transfer
Titrations
Precipitation Reactions
Ion Exchange
Oxidation–Reduction Reactions: Electron Transfer
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Concentration Units
• Concentration can be described
qualitatively or quantitatively, and define
the amount of solute in a solution:
amount of solute
amount of solvent
or
amount of solute
amount of solution
• Quantitative concentration units based on:
• Mass of solute
• Moles of solute
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Concentration Units (cont.)
• Parts per million (ppm):
 gram s o f so lu te   1 m g o f so lu te 
ppm = 
=

6
1
0
gram
s
o
f
so
lu
tio
n
1
kg
o
f
so
lu
tio
n

 

• Parts per billion (ppb):
 gram s of solute   1 μ g of solute 
ppb = 
=

9
1
0
gram
s
of
solution
1
kg
of
solution

 

• Useful for very small amounts of solute
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Molarity
• Molarity (M)
M o larity 
m o les o f so lu te
liters o f so lu tio n
• As a conversion factor  g of solute
M ass of solute   volum e  m olarity   M
m ol 
g

g = L 

L  m ol

• Small concentrations: mM (10–3 M), M
(10–6 M)
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Practice: Calculating Molarity
• What is the molarity of an aqueous
solution prepared by adding 36.5 g of
barium chloride to enough water to make
750.0 mL of solution?
- Collect and Organize: We have a mass of solute
(36.5 g BaCl2), and a final volume of solution
(750.0 mL). We can calculate the molar mass of
barium chloride from the chemical formula
(208.2 g/mol). We want to determine the
molarity of the BaCl2 solution.
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Practice: Calculating Molarity
• What is the molarity of an aqueous
solution prepared by adding 36.5 g of
barium chloride to enough water to make
750.0 mL of solution?
- Analyze: From the units for molarity, we need
moles of solute (BaCl2) per liter of solution.
With the mass of solute (36.5 g) and the molar
mass (208.2 g/mole) we can calculate the moles
of solute. The volume of solution will need to be
converted from mL to L.
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Practice: Calculating Molarity
• What is the molarity of an aqueous
solution prepared by adding 36.5 g of
barium chloride to enough water to make
750.0 mL of solution?
- Solve: (36.5 g/208.2 g/mol)/(0.7500L) = 0.234 M
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Practice: Calculating Molarity
• What is the molarity of an aqueous
solution prepared by adding 36.5 g of
barium chloride to enough water to make
750.0 mL of solution?
- Think About It: The mass of solute represents a
little less than 20% of one mole (.2 moles?).
Since the volume is less than 1 liter, the overall
molarity should be a little more than 0.2 M,
which matches our calculated result.
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Practice: Mass of Solute
• How many grams of aluminum nitrate are
required to make 500.0 mL of a 0.0525 M
aqueous solution?
- Collect and Organize: We know the molarity
(0.0525 M) and volume (500.0 mL) of the
solution. We want to calculate the mass of
solute needed to prepare the solution
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Practice: Mass of Solute
• How many grams of aluminum nitrate are
required to make 500.0 mL of a 0.0525 M
aqueous solution?
- Analyze: We can use molarity and volume to
calculate the number of moles of solute
required. We can then use the formula mass for
the solute, (Al(NO3)3 at 213.0 g/mol)) to convert
from moles to mass.
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Practice: Mass of Solute
• How many grams of aluminum nitrate are
required to make 500.0 mL of a 0.0525 M
aqueous solution?
-Solve:
(0.0525 mol/L) × (0.500 L) = 0.02625 mols
(0.02625 mol) × 213.0 g/mol) = 5.59 g
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Practice: Molarity from Density
• If the density of ocean water at a depth of
10,000 m is 1.071 g/mL and if 25.0 g of
water at that depth contains 190 mg of
KCl, what is the molarity of KCl?
- Collect and Organize: We know the mass of
solute (KCl, 190 mg), the mass of water (25.0 g),
and the density of water at the specified depth
(1.071 g/mL). We want to determine the molarity
of KCl in seawater at the specified depth.
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Practice: Molarity from Density
• If the density of ocean water at a depth of
10,000 m is 1.071 g/mL and if 25.0 g of
water at that depth contains 190 mg of
KCl, what is the molarity of KCl?
- Analyze: We can convert the mass of solute to
moles using the molar mass (74.55 g/mol). We
can convert the mass of water to a volume
using density. With the moles of solute and
volume of water (in L) we can calculate
molarity.
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Practice: Molarity from Density
• If the density of ocean water at a depth of
10,000 m is 1.071 g/mL and if 25.0 g of
water at that depth contains 190 mg of
KCl, what is the molarity of KCl?
-Solve:
(190 mg) × (1 g/1000 mg) × (1 mol/74.55 g) =
0.002549 mol KCl
(25.0 g) × ( 1 mL/1.071 g) × (1 L/1000 mL) =
0.0233 L
Molarity = (0.002549 mol/0.0233 L) = 0.109 M
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Chapter Outline
• 4.1 Solutions on Earth and Other Places
• 4.2 Concentration Units
• 4.3 Dilutions
• Concentrated (Stock) vs. Dilute Solutions
• Preparation of Dilute Solutions
• Measuring Concentration
•
•
•
•
•
•
4.4
4.5
4.6
4.7
4.8
4.9
Electrolytes and Nonelectrolytes
Acid–Base Reactions: Proton Transfer
Titrations
Precipitation Reactions
Ion Exchange
Oxidation–Reduction Reactions: Electron Transfer
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Dilutions
• Stock solution – a concentrated solution
of a substance used to prepare solutions
of lower concentration.
• Dilution – the process of lowering the
concentration of a solution by adding
more solvent.
• (mol solute)stock = (mol solute)dilute
• Minitial × Vinitial = Mdilute × Vdilute
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Dilutions (cont.)
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Practice: Diluting Stock
Solutions
•
Hydrochloric acid is obtained in 12.0 M
stock solution. What volume of stock
solution is required to make 500.0 mL of a
0.145 M dilute solution?
- Collect and Organize: We know the initial
molarity of the stock solution (12.0 M HCl), and
the desired molarity and volume of the dilute
solution (0.145 M and 500.0 mL, respectively).
We want to know the volume of stock solution
required to prepare the solution.
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Practice: Diluting Stock
Solutions
•
Hydrochloric acid is obtained in 12.0 M
stock solution. What volume of stock
solution is required to make 500.0 mL of a
0.145 M dilute solution?
- Analyze: Knowing three of the four variables in
the dilution equation, we can rearrange the
equation to solve for Vinitial.
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Practice: Diluting Stock
Solutions
•
Hydrochloric acid is obtained in 12.0 M
stock solution. What volume of stock
solution is required to make 500.0 mL of a
0.145 M dilute solution?
- Solve: Vinitial = (Mfinal × Vfinal)/(Minitial)  (0.145 M)(
500 mL)/(12.0 M) = 6.04 mL
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Practice: Diluting Stock
Solutions
•
Hydrochloric acid is obtained in 12.0 M
stock solution. What volume of stock
solution is required to make 500.0 mL of a
0.145 M dilute solution?
- Think About It: Since the final molarity is about
1/10th of the initial volume, we would
expect the initial volume to be about 1/10th of
the final volume, or about 5 mL. Our calculated
result is close and of the right order of
magnitude.
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Measuring Concentration
• Intensity of color can be used to measure
concentrations
• Beer’s Law: A = ε·b·c
• Where:
• A = absorbance (amount of light absorbed by
sample)
• ε = molar absorptivity
• b = path length
• c = concentration of absorbing species
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Measuring Concentration
(cont.)
(a)
(b)
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Chapter Outline
•
•
•
•
4.1
4.2
4.3
4.4
Solutions on Earth and Other Places
Concentration Units
Dilutions
Electrolytes and Nonelectrolytes
• Strong, Weak, and Nonelectrolytes
•
•
•
•
•
4.5
4.6
4.7
4.8
4.9
Acid–Base Reactions: Proton Transfer
Titrations
Precipitation Reactions
Ion Exchange
Oxidation–Reduction Reactions: Electron Transfer
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Electrolytes
• Strong electrolytes:
• Nearly 100% dissociated into ions
• Conducts current efficiently
• Examples: solutions of
NaCl, HNO3, HCl
• NaCl(s)
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

H O
Na+(aq) + Cl– (aq)
31
Electrolytes (cont.)
• Weak electrolytes:
• Only partially dissociate into
ions
• Slightly conductive
• Examples: vinegar (aqueous
solution of acetic acid); tap
water
•
CH3CO2H(aq) ⇄ CH3CO2–(aq) + H+(aq)
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Nonelectrolytes
• Substances in which no
ionization occurs; no
conduction of electrical
current
• Examples: aqueous
solutions of sugar, ethanol,
ethylene glycol
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Chapter Outline
•
•
•
•
•
4.1
4.2
4.3
4.4
4.5
Solutions on Earth and Other Places
Concentration Units
Dilutions
Electrolytes and Nonelectrolytes
Acid–Base Reactions: Proton Transfer
• Definition of Acids and Bases
• Neutralization Reactions
• Molecular vs. Ionic Equations: Net Ionic Equations
•
•
•
•
4.6
4.7
4.8
4.9
Titrations
Precipitation Reactions
Ion Exchange
Oxidation–Reduction Reactions: Electron Transfer
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Acid–Base Reactions
• Brønsted–Lowry definitions:
• Acids = proton (H+) donors
• Bases = proton acceptors
• HCl (aq)
Proton
donor (acid)
+ H2O(ℓ) → H3O+(aq) + Cl–(aq)
Proton
acceptor (base)
• H+ ions strongly associated with water
molecules  hydronium ions (H3O+)
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Acid–Base Reactions (cont.)
• Neutralization – reaction that takes place
when an acid reacts with a base and
produces a solution of a salt in water.
• Salt
• Product of a neutralization reaction
• Made up of the cation of the base plus the
anion of the acid.
• Example: HCl + NaOH → NaCl + H2O
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Types of Equations
• Molecular equations:
• Reactants/products are written as undissociated
molecules.
•
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(ℓ)
• Overall ionic equations:
• Distinguish between molecular and ionic substances
• Ionic species are represented as dissolved ions.
• H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → Na+(aq)
+ Cl–(aq) + H2O(ℓ)
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Types of Equations (cont.)
• Net ionic equations:
• Remove spectator ions (ions present in same
form on both reactants and products side of
chemical equation).
• H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) →
Na+(aq) + Cl–(aq) + H2O(ℓ)
• Net ionic equation: H+(aq) + OH–(aq) →
H2O(ℓ)
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Nonmetal Oxides and Acids
• Nonmetal oxides
• Form acids in hydrolysis reaction
• Example:
• SO3(g) + H2O(ℓ) 
H2SO4(aq)
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Strong and Weak Acids
• Strong acids/bases:
• Dissociate completely
in aqueous solution
(i.e., strong
electrolytes)
• Examples:
• H2SO4(aq) → H+(aq) +
HSO4–(aq)
• HSO4–(aq) ⇄ H+(aq) +
SO42–(aq)
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Strong and Weak Bases
• Strong bases
• 1A, 2A hydroxides
• NaOH(aq)  Na+(aq) +
OH–(aq)
• Weak bases
• NH3(aq) + H2O(ℓ) ⇄
NH4+(aq) + OH–(aq)
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Water: Acid or Base?
• Water as base:
• HCl(aq) +
H2O(ℓ) → H3O+(aq) + Cl– (aq)
Proton
Proton
donor (acid) acceptor (base)
• Water as acid:
• NH3(aq) + H2O(ℓ) ⇄ NH4+(aq) + OH–(aq)
Proton
Proton
acceptor (base) donor (acid)
• Amphiprotic: acts as acid or base
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Chapter Outline
•
•
•
•
•
•
4.1
4.2
4.3
4.4
4.5
4.6
Solutions on Earth and Other Places
Concentration Units
Dilutions
Electrolytes and Nonelectrolytes
Acid–Base Reactions: Proton Transfer
Titrations
• Terms: Standard Solutions (Titrants), Equivalence Points, and End
Points
• Using Titrations to Analyze Unknown Solutions
• 4.7 Precipitation Reactions
• 4.8 Ion Exchange
• 4.9 Oxidation–Reduction Reactions: Electron Transfer
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Key Titration Terms
• Titration – an analytical method to determine the
concentration of a solute in a sample by reacting it with
a standard solution.
• Standard solution – a solution of known concentration
(also called the titrant.)
• Equivalence point – point when moles of added titrant is
stoichiometrically equivalent to moles of substance
being analyzed.
• End point – point reached when the indicator changes
color.
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Titration Example
H2SO4 + 2 NaOH
Na2SO4 + 2 H2O
(titrant)
H2SO4
(unknown)
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End point
45
Stoichiometry Calculations
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
At the equivalence point:
# moles H2SO4 = (# moles NaOH)/2
Macid·Vacid = (Mbase·Vbase)/(2)
Rearrange to find Macid:
Macid = (Mbase·Vbase)/(2)(Vacid)
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Practice: Acid–Base Titration
#1
• What is the concentration of sulfuric acid
if 15.00 mL of it reacts with 18.45 mL of a
0.0973 M NaOH solution?
- Collect and Organize: We know Vacid = 15.00 mL,
Vbase= 18.45.mL, and Mbase= 0.0973 M. We need
to determine Macid for the original sulfuric acid
sample.
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Practice: Acid–Base Titration
#1
• What is the concentration of sulfuric acid
if 15.00 mL of it reacts with 18.45 mL of a
0.0973 M NaOH solution?
- Analyze: Knowing the stoichiometry of the
reaction, we can rearrange the equation to
solve for Macid.
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Practice: Acid–Base Titration
#1
• What is the concentration of sulfuric acid
if 15.00 mL of it reacts with 18.45 mL of a
0.0973 M NaOH solution?
- Solve: Macid = (Vbase)(Mbase)/(2)(Vacid) = 0.0598 M
H2SO4
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Practice: Acid–Base Titration
#1
• What is the concentration of sulfuric acid
if 15.00 mL of it reacts with 18.45 mL of a
0.0973 M NaOH solution?
- Think About It: The volume of base is only slightly more
than the volume of the original acid sample. If the
stoichiometry was 1:1, we would expect the molarity of
the acid to be similar to the molarity of the base.
However, the stoichiometry is 2:1—it takes twice as
much base to neutralize a given amount of the sulfuric
acid, so we would expect the molarity of acid to be a
little more than one half the molarity of the base. Our
calculated result is consistent with this estimate.
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Practice: Titration #2
• If 30.34 mL of a 0.135 M solution of
hydrochloric acid (HCl) were required to
neutralize 25.00 mL of a sodium hydroxide
(NaOH) solution, what is the molarity of the
sodium hydroxide solution?
- Collect and Organize: We know Vacid = 30.34 mL,
Macid = 0.135 M, Vbase= 25.00 mL. We need to
determine for the Mbase for the original sodium
hydroxide sample.
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Practice: Titration #2
• If 30.34 mL of a 0.135 M solution of
hydrochloric acid (HCl) were required to
neutralize 25.00 mL of a sodium hydroxide
(NaOH) solution, what is the molarity of the
sodium hydroxide solution?
- Analyze: Knowing the stoichiometry of the
reaction, we can rearrange the equation to
solve for Mbase. In this case, the stoichiometry is
1:1, so (# moles acid) = (# moles base).
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Practice: Titration #2
• If 30.34 mL of a 0.135 M solution of
hydrochloric acid (HCl) were required to
neutralize 25.00 mL of a sodium hydroxide
(NaOH) solution, what is the molarity of the
sodium hydroxide solution?
- Solve: Mbase = (Vacid)(Macid)/(Vbase) = 0.164 M
NaOH.
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Practice: Titration #2
• If 30.34 mL of a 0.135 M solution of
hydrochloric acid (HCl) were required to
neutralize 25.00 mL of a sodium hydroxide
(NaOH) solution, what is the molarity of the
sodium hydroxide solution?
- Think About It: The volume of acid is only slightly
more than the volume of the original base sample. The
stoichiometry in this case is 1:1. Since the volume of
base is slightly less than the volume of acid, we would
expect that the base is slightly more concentrated, i.e.,
has a slightly greater molarity than the base. Our
calculated result (0.164 M) is consistent with this
estimate.
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Chapter Outline
•
•
•
•
•
•
•
4.1
4.2
4.3
4.4
4.5
4.6
4.7
Solutions on Earth and Other Places
Concentration Units
Dilutions
Electrolytes and Nonelectrolytes
Acid–Base Reactions: Proton Transfer
Titrations
Precipitation Reactions
• Making Insoluble Salts
• Using Precipitation in Analysis
• Saturated Solutions and Supersaturation
• 4.8 Ion Exchange
• 4.9 Oxidation–Reduction Reactions: Electron Transfer
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Precipitation Reactions
• Precipitate:
• Solid product formed from a reaction in
solution
• AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)
• Can predict formation of precipitates
based on solubility “rules”
• Precipitation reactions can be written using
net ionic equations.
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Solubility Rules (Table 4.5)
• Soluble cations:
• Group I ions (alkali metals) and NH4+
• Soluble anions:
• NO3– and CH3COO– (acetate)
• Halides (Group 17): except Ag+, Cu+, Pb2+, Hg22+
• Sulfates (SO42–): except Pb2+, Hg22+, Ca2+, Ba2+, Sr2+
• Combining anions and cations not listed above
results in formation of an insoluble compound.
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Insoluble Compounds
• All hydroxides (OH–) except:
• Group IA (e.g, NaOH); Ca(OH)2, Sr(OH)2, and
Ba(OH)2
• All sulfides (S2–) except:
• Group IA and NH4+; CaS, SrS, and BaS
• All carbonates (CO32–) except IA, NH4+
• All phosphates (PO43–) except IA, NH4+
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Precipitation
Combining insoluble cation
with insoluble anion = ppt!
Example: PbNO3(aq) +
NaI(aq)
Cations: Pb2+
Na+
Anions: NO3–
I–
Soluble
Not soluble
Precipitate: PbI2(s)
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Net Ionic Equations
• Soluble ionic compounds
• Strong electrolytes!
Pb(NO3)2(aq) + 2 NaI(aq) →
2 NaNO3(aq) + PbI2(s)
• Total ionic equation:
Pb2+(aq) + 2 NO3–(aq) + 2 Na+(aq)
+ 2 I–(aq) → PbI2(s) + 2 Na+(aq) +
2 NO3 –(aq)
• Net ionic equation:
Pb2+(aq) + 2 I–(aq) →
PbI2(s)
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Practice: Will a Precipitate
Form?
Does a precipitate form when sodium chloride
is mixed with silver nitrate? If so, write the net
ionic equation for the formation of the
precipitate.
NaCl  Na+ + Cl–
AgNO3  Ag+ + NO3–
- Collect and Organize: We have two aqueous
solutions of sodium chloride and silver nitrate.
Upon mixing, we need to identify if any
combination of anion and cation will result in
formation of a precipitate.
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Practice: Will a Precipitate
Form?
Does a precipitate form when sodium chloride
is mixed with silver nitrate? If so, write the net
ionic equation for the formation of the
precipitate.
NaCl  Na+ + Cl–
AgNO3  Ag+ + NO3–
- Analyze: The ions in solution upon mixing are
Na+, Ag+, Cl–, and NO3–. We know that Na+ and
NO3– form soluble compounds, so the only
combination we need to evaluate is Ag+ with Cl–
.
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Practice: Will a Precipitate
Form?
Does a precipitate form when sodium chloride
is mixed with silver nitrate? If so, write the net
ionic equation for the formation of the
precipitate.
NaCl  Na+ + Cl–
AgNO3  Ag+ + NO3–
- Solve: Although halides generally form soluble
compounds, one of the exceptions is Ag+, so
we would expect that silver chloride (AgCl)
would precipitate.
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Practice: Will a Precipitate
Form?
Does a precipitate form when sodium chloride
is mixed with silver nitrate? If so, write the net
ionic equation for the formation of the
precipitate.
NaCl  Na+ + Cl–
AgNO3  Ag+ + NO3–
- Think About It: The result is consistent with the
solubility rules.
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Practice: Calculating the
Mass of Precipitate
• What mass of barium sulfate is produced
when 100.0 mL of a 0.100 M solution of
barium chloride is mixed with 100.0 mL of
a 0.100 M solution of iron(III) sulfate?
- Collect and Organize: We know that barium
sulfate precipitate forms when we mix solutions
of barium chloride and iron(III) sulfate. We
started with 100.0 mL of 0.100 M barium
chloride and 100.0 mL of 0.100 M iron(III)
sulfate. The net ionic reaction would be:
Ba2+(aq) + SO42–(aq)  BaSO4(s)
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Practice: Calculating the
Mass of Precipitate
• What mass of barium sulfate is produced
when 100.0 mL of a 0.100 M solution of
barium chloride is mixed with 100.0 mL of
a 0.100 M solution of iron(III) sulfate?
- Analyze: From the volumes and molarities, we
can calculate the number of moles of reactants
(barium and sulfate). Based on the moles of
reactants and the stoichiometry, we can
calculate the moles of product/precipitate.
Using the molar mass of barium sulfate (233.4
g/mol) we can calculate the mass of precipitate.
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Practice: Calculating the
Mass of Precipitate
• What mass of barium sulfate is produced
when 100.0 mL of a 0.100 M solution of
barium chloride is mixed with 100.0 mL of
a 0.100 M solution of iron(III) sulfate?
Solve:
Mol of Ba2+ = (0.1000 L) × (0.100 mol/L) = 0.0100 mol
Ba2+
Mol of SO42– = (0.1000 L) × (0.100 mol/L) = 0.0100 mol
SO42–
Since the stoichiometry is 1:1, we would obtain 0.0100
moles BaSO4(s) of
(0.0100 mol BaSO4) × (233.4 g/mol) = 2.334 g BaSO4
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Practice: Calculating the
Mass of Precipitate
• What mass of barium sulfate is produced
when 100.0 mL of a 0.100 M solution of
barium chloride is mixed with 100.0 mL of
a 0.100 M solution of iron(III) sulfate?
- Think About It: The formation of an insoluble
product, BaSO4, is consistent with the solubility
rules. The mass of BaSO4 is reasonable, given
that only 0.010 mol of product would be
obtained.
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Types of Solutions
• Saturated solution:
• a solution that contains the maximum
concentration of a solute possible at a given
temperature.
• Supersaturated solution:
• a solution containing more than the
maximum quantity of solute predicted to be
soluble in a given volume of solution at a
given temperature.
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A Saturated Solution Example
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Supersaturated Solution
Sodium acetate precipitates from a supersaturated solution.
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Chapter Outline
•
•
•
•
•
•
•
•
4.1 Solutions on Earth and Other Places
4.2 Concentration Units
4.3 Dilutions
4.4 Electrolytes and Nonelectrolytes
4.5 Acid–Base Reactions: Proton Transfer
4.6 Titrations
4.7 Precipitation Reactions
4.8 Ion Exchange
• Water Purification: Hard vs. Soft Water
• 4.9 Oxidation–Reduction Reactions: Electron
Transfer
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Ion Exchange
• Ion exchange – process by which one ion
is displaced by another
• Important in purification/softening of water
• “Soft” metal ions (Na+) exchange for
metals that contribute to “hard” water
(Ca2+, Mg2+)
• Uses ion exchange resin or zeolites:
• 2 (RCOO– Na+) + Ca2+ → (RCOO–)2Ca2+ + 2 Na+
Exchange resin
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Ion Exchange: Water Softeners
Ion exchange using resin
Hard water in—soft water out
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Zeolites: Natural Ion Exchangers
Zeolites
- Natural crystalline minerals or
synthetic materials consisting of
3D network of channels
containing Na+ or other +1 ion.
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Chapter Outline
•
•
•
•
•
•
•
•
•
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
Solutions on Earth and Other Places
Concentration Units
Dilutions
Electrolytes and Nonelectrolytes
Acid–Base Reactions: Proton Transfer
Titrations
Precipitation Reactions
Ion Exchange
Oxidation–Reduction Reactions: Electron Transfer
• Oxidation Numbers
• Electron Transfer in Redox Reactions
• Balancing Redox Reactions
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Oxidation–Reduction
Reactions
• Oxidation:
• Reaction that increases oxygen content of a
substance (loss of electrons)
• e.g. 4 Fe(s) + 3 O2(g)  2 Fe2O3(s)
• Reduction:
• Reaction involving loss of O2(gain of
electrons)
• e.g. Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
• Redox reactions = transfer of electrons
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Oxidation Number (O.N.) Rules
• O.N. of atoms in a neutral molecule sum to
zero; O.N. of atoms in an ion sum to charge on
the ion.
• Each atom in a pure element has O.N. = 0.
• In monatomic ion, O.N. = charge on the ion.
• Fluorine, O.N. = 1 in all compounds.
• Hydrogen = +1, oxygen = 2 in most
compounds (except in peroxides where oxygen
= 1).
• Br and Cl have 1 O.N., unless combined with
O or F.
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Practice: Assigning
Oxidation Numbers
Assign oxidation numbers to each element
in the following compounds. (The first two
are provided as examples.)
•
•
•
•
•
•
SO2
CrO42–
NH3
ClO3–
SF6
Cl2
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Oxygen is –2 and Sulfur is +4.
Oxygen is –2 and Chromium is +6.
For the first example:
- SO2 is a neutral molecule, so the sum
of O.N.s must equal zero. (rule 1)
- Oxygen is –2 (rule 5); there are two
oxygens, so (2) × ( –2) = –4
- Since the only other atom is S and the
sum of O.Ns must be zero, then the O.N
for S must be +4!
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Practice: Assigning
Oxidation Numbers
Assign oxidation numbers to each element
in the following compounds.
•
•
•
•
•
•
SO2
CrO42–
NH3
ClO3–
SF6
Cl2
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Answers:
-CrO42–: each O = –2 (total = –8), Cr = +6; the sum of
O.N. = charge on ion = –2.
-- ClO3–: each O = –2 (total = –6); since the sum of O.N.
= –1, the O.N. for Cl = +5.
-NH3: each H = +1; since the sum of O.N = O (neutral
molecule), the O.N. for N = +3.
-SF6: each F = –1 (total = –6); since the sum of O.N. = 0,
the O.N. for S = +6.
-- Cl2: the O.N. for any atom in the elemental state = 0.
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Redox Reactions: Electron
Transfer
• Change in oxidation state results from
gain or loss of electrons:
(+2 e–) × 4 = + 8e–
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
–8 e–
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Balancing Redox Reactions
• Copper wire immersed in silver nitrate
solution:
Cu(s) + 2 Ag+ (aq) →
Cu2+(aq) + 2 Ag(s)
Can divide overall redox
reaction into half-reactions:
Oxid: Cu(s) → Cu2+(aq) + 2 e–
Red: Ag+(aq) + e– → Ag(s)
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Half-Reaction Method
• Write separate reduction, oxidation halfreactions.
• Balance number of particles in each halfreaction.
• Balance charge by adding electrons to the
appropriate side.
• Multiply half-reactions by appropriate
whole number to balance electrons.
• Add half-reactions to generate a redox
equation.
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Practice: Balancing Redox
Reactions
A nail made of Fe(s) that is placed in a solution
of a soluble Pd2+ salt gradually disappears as
the iron enters the solution as Fe3+. Balance the
redox reaction.
- Collect and Organize: We know that the Fe(s)
converts to Fe3+ when placed in a solution of
Pd2+ ions.
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Practice: Balancing Redox
Reactions
A nail made of Fe(s) that is placed in a solution
of a soluble Pd2+ salt gradually disappears as
the iron enters the solution as Fe3+. Balance the
redox reaction.
- Analyze: The iron is oxidized to Fe3+ in the
presence of Pd2+. Since the Fe is oxidized, Pd2+
must be reduced to Pd(s). We can now write the
half-reactions and balance them using the steps
discussed previously.
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Practice: Balancing Redox
Reactions
A nail made of Fe(s) that is placed in a solution
of a soluble Pd2+ salt gradually disappears as
the iron enters the solution as Fe3+. Balance the
redox reaction.
-Solve:
Oxidation: [Fe(s)  Fe3+(aq) + 3 e–] × 2
Reduction: [Pd2+(aq) + 2 e–  Pd(s)] × 3
Net: 2 Fe(s) + 3 Pd2+(aq)  3 Pd(s) + 2
Fe3+(aq)
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Practice: Balancing Redox
Reactions
A nail made of Fe(s) that is placed in a solution
of a soluble Pd2+ salt gradually disappears as
the iron enters the solution as Fe3+. Balance the
redox reaction.
- Think About It: One species is oxidized while
another is reduced; the final reaction is
balanced both in terms of mass and charge,
since the number of electrons gained and lost
balance.
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Activity Series for Metals
• Metal cation will oxidize any metal
above it in the activity series (Table
4.6).
• Example:
• 2 Ag+(aq) + Cu(s)  2 Ag(s) + Cu2+(aq)
• (Ag+(aq) below Cu(s) in Table 4.6)
• Zn2+(aq) + Cu(s)  no reaction!
• (Zn2+(aq) above Cu(s) in Table 4.6)
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Redox in Nature
• Soil color influenced by mineral content
• Iron(III) – red-orange minerals (yellow-orange in
solution)
• Iron(II) – blue-gray minerals (pale green in solution)
• Analysis of iron in minerals  redox rxn
• Fe2+(aq) + MnO4–(aq)  Fe3+(aq) + Mn2+(aq)
(colorless)
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(deep purple) (yellow-orange) (pink)
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Redox in Nature (cont.)
• Balancing by half-reaction method:
• Oxidation: Fe2+(aq)  Fe3+(aq) + 1 e–
• Reduction: MnO4–(aq)  Mn2+(aq)
• Balance particles by adding H2O, H+:
• MnO4–(aq) + 8 H+(aq) + 5 e–  Mn2+(aq) + 4 H2O(ℓ)
• Balance e–: multiply Fe half-reaction × 5, add:
• 8 H+ + MnO4 – + 5 Fe2+  5 Fe3+ + Mn2+ + 4 H2O
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ChemTours: Chapter 4
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