Chapter 9
9.1
9.2
9.3
9.6
9.7
Solutions
9.1 Water
ƒ Water is the most common solvent.
ƒ The water molecule is polar.
ƒ Hydrogen bonds form between the hydrogen
atom in one molecule and the oxygen atom in a
different water molecule.
Properties of Water
Solutions
Electrolytes and Nonelectrolytes
Percent Concentration
Molarity
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Water
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Water in Foods
ƒ Water for the
body is
obtained from
fluids as well
as foods.
ƒ Some foods
have a high
percentage of
water.
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1
Surface Tension
9.2 Solutions: Solute and Solvent
Solutions
„ Are homogeneous
mixtures of two or
more substances.
„ Consist of a
solvent and one or
more solutes.
Water molecules
ƒ At the surface form hydrogen bonds with
molecules on or below the surface, which
pulls them closer.
ƒ At the surface behave like a thin, elastic
membrane, or “skin.”
ƒ Cannot hydrogen bond when compounds
called surfactants are added.
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Examples of Solutions
Nature of Solutes in Solutions
Solutes
ƒ Spread evenly
throughout the solution.
ƒ Cannot be separated by
filtration.
ƒ Can be separated by
evaporation.
ƒ Are not visible, but can
give a color to the
solution.
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„
7
The solute
and solvent
in a solution
can be a
solid, liquid,
and/or a gas.
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Learning Check
Solution
Identify the solute and the solvent in each.
A. brass: 20 g zinc + 50 g copper
solute =
1) zinc
solvent =
2) copper
Identify the solute and the solvent in each.
A. brass: 20 g zinc + 50 g copper
solute =
1) zinc
2) copper
solvent =
1) zinc
2) copper
B. 100 g H2O + 5 g KCl
solute =
1) KCl
solvent =
1) KCl
B. 100 g H2O + 5 g KCl
solute =
1) KCl
solvent =
2) H2O
2) H2O
2) H2O
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Learning Check
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Solution
Identify the solute in each of the following
solutions:
Identify the solute in each of the following solutions:
A. 2 g sugar (1)
A. 2 g sugar (1) and 100 mL water (2)
B. 60.0 mL of ethyl alcohol (1) and 30.0 mL of
methyl alcohol (2)
B. 30.0 mL of methyl alcohol (2)
C. 1.5 g NaCl (2)
C. 55.0 mL water (1) and 1.50 g NaCl (2)
D. 200 mL O2 (1)
D. Air: 200 mL O2 (1) and 800 mL N2 (2)
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3
Like Dissolves Like
„
„
„
Examples of Like Dissolves Like
A solution forms when there is an attraction
between the particles of the solute and solvent.
A polar solvent such as water dissolves polar
solutes such as sugar and ionic solutes such as
NaCl.
A nonpolar solvent such as hexane (C6H14)
dissolves nonpolar solutes such as oil or
grease.
Solvents
Solutes
Water (polar)
Ni(NO3)2
(ionic)
CH2Cl2 (nonpolar)
I2 (nonpolar)
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Learning Check
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Solution
1) Na2SO4
Which of the following solutes will dissolve
in water? Why?
1) Na2SO4
Yes, ionic
2) gasoline (nonpolar)
2) gasoline
No, nonnpolar
3) I2
3) I2
No, nonpolar
4) HCl
4) HCl
Yes, polar
Which of the following solutes will dissolve
in water? Why?
Most polar and ionic solutes dissolve in
water because water is a polar solvent.
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Formation of a Solution
„
„
Equations for Solution Formation
Na+ and Cl- ions on
the surface of a NaCl
crystal are attracted
to polar water
molecules.
In solution, the ions
are hydrated as
several H2O
molecules surround
each.
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„
When NaCl(s) dissolves in water, the
reaction can be written as
H2O
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Learning Check
NaCl(s)
Na+(aq) + Cl– (aq)
solid
separation of ions
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Solution
Solid LiCl is added to water. It dissolves because
A. The Li+ ions are attracted to the
1) oxygen atom (δ –) of water.
2) hydrogen atom (δ+) of water.
Solid LiCl is added to water. It dissolves because
A. The Li+ ions are attracted to the
B. The Cl- ions are attracted to the
1) oxygen atom (δ –) of water.
2) hydrogen atom (δ+) of water.
B. The Cl- ions are attracted to the
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1) oxygen atom (δ –) of water.
2) hydrogen atom (δ +) of water.
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Hydrates
„
„
Examples of Hydrates
Hydrates are solid
compounds that
contain water
molecules as part of
the crystal
structure.
Heating a hydrate
releases the water of
hydration to give
the anhydrate salt.
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Learning Check
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Solution
Write the equation for the dehydration of
AlCl3 • 6H2O.
Write the equation for the dehydration of
AlCl3 • 6H2O.
AlCl3 • 6H2O
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AlCl3 + 6H2O
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9.3 Electrolytes
Strong Electrolytes
Electrolytes
ƒ Produce positive (+) and negative (-) ions
when they dissolve in water.
ƒ In water conduct an electric current.
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„
„
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Learning Check
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Complete each of the following dissociation
equations for strong electrolytes dissolved in
water:
H2O
Ca2+ + 2Cl-
A. 3) CaCl2 (s)
H2O
H2O
1) 3K+ + PO432) K3PO4
3) K3+ + P3- + O4-
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Solution
Complete each of the following dissociation
equations for strong electrolytes dissolved in
H2O
water:
A. CaCl2 (s)
1) CaCl2
2) Ca2+ + Cl23) Ca2+ + 2ClB. K3PO4 (s)
Strong electrolytes ionize 100% in solution.
Equations for the dissociation of strong
electrolytes show the formation of ions in
aqueous (aq) solutions.
100% ions
H2O
+
NaCl(s)
Na (aq) + Cl-(aq)
H2O
CaBr2(s)
Ca2+(aq) + 2Br- (aq)
B. 1) K3PO4 (s)
27
3K+ + PO43-
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Weak Electrolytes
Nonelectrolytes
ƒ A weak electrolyte
ƒ Dissolves mostly as molecules in solution.
ƒ Produces only a few ions in aqueous
solutions.
ƒ Has an equilibrium that favors the
reactants.
HF + H2O
H3O+(aq) + F- (aq)
NH3 + H2O
ƒ Nonelectrolytes
ƒ Form only
molecules in water.
ƒ Do not produce ions
in water.
ƒ Do not conduct an
electric current.
NH4+(aq) + OH- (aq)
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9.6 Percent Concentration
„
„
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Mass Percent
The concentration of a solution is the amount
of solute dissolved in a specific amount of
solution.
amount of solute
amount of solution
The percent concentration describes the
amount of solute that is dissolved in 100 parts
of solution.
amount of solute
100 parts solution
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The mass percent (%m/m)
„ Concentration is the percent by mass of solute in a
solution.
mass percent = g of solute x 100%
g of solution
„ Is the g of solute in 100 g of solution.
mass percent =
g of solute
100 g of solution
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Mass of Solution
grams of solute
+
Calculating Mass Percent
grams of solvent
„
8.0 g KCl (solute) x 100 = 16% (m/m) KCl
50.0 g KCl solution
50.0 g KCl solution
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Mass percent (%m/m) is calculated from the grams
of solute (g KCl) and the grams of solution (g KCl
solution).
g of KCl
=
8.0 g
g of solvent (water) =
42.0 g
g of KCl solution
=
50.0 g
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Learning Check
34
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Solution
A solution is prepared by mixing 15 g Na2CO3
and 235 g of H2O. Calculate the mass percent
(%m/m) of the solution.
1) 15% (m/m) Na2CO3
2) 6.4% (m/m) Na2CO3
3) 6.0% (m/m) Na2CO3
3) 6.0% (m/m) Na2CO3
mass solute
=
15 g Na2CO3
mass solution
=
15 g + 235 g = 250 g
mass %(m/m) =
15 g Na2CO3
x
100
250 g solution
=
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6.0% Na2CO3 solution
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Preparing a Solution with a
Mass/Volume % Concentration
Mass/Volume Percent
The mass/volume percent (%m/v)
„ Concentration is the ratio of the mass in grams (g)
of solute in a volume (mL) of solution.
x 100%
mass/volume % = g of solute
mL of solution
„ Is the g of solute in 100 mL of solution.
mass/volume % =
g of solute
100 mL of solution
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Calculation of Mass/Volume Percent
„
Mass/volume percent (%m/v) is calculated from
the grams of solute (g KCl) and the volume of
solution (mL KCl solution).
g of KI
=
5.0 g KI
mL of KI solution
=
250.0 mL
5.0 g KI (solute)
x 100 = 2.0%(m/v) KI
250.0 mL KI solution
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„
A percent
mass/volume
solution is
prepared by
weighing out the
grams of solute
(g) and adding
water to give the
final volume of
the solution.
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Learning Check
A 500. mL samples of an IV
glucose solution contains 25 g
glucose (C6H12O6) in water.
What is the mass/volume %
(%m/v) of glucose of the IV
solution?
1) 5.0%
2) 20.% 3) 50.%
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Solution
Volume Percent
1) 5.0%
Mass/volume %(m/v)
x 100
=
25 g glucose
500. mL solution
The volume percent (%v/v)
„ Concentration is the percent volume (mL) of
solute (liquid) to volume (mL) of solution.
volume % (v/v) = mL of solute x 100%
mL of solution
„ Is the mL of solute in 100 mL of solution.
volume % (v/v) =
mL of solute
100 mL of solution
= 5.0 % (m/v) glucose solution
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Percent Conversion Factors
„
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Learning Check
Write two conversion factors for each solutions:
A. 8%(m/v) NaOH
Two conversion factors can be written for any
type of % value.
B. 12%(v/v) ethyl alcohol
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Solution
A. 8%(m/v) NaOH
and
8 g NaOH
100 mL solution
Using Percent Factors
How many grams of NaCl are needed to prepare
250 g of a 10.0% (m/m) NaCl solution?
1. Write the 10.0 % (m/m) as conversion factors.
and
100 g solution
10.0 g NaCl
100 g solution
10.0 g NaCl
2. Use the factor that cancels given (g solution).
250 g solution x 10.0 g NaCl = 25 g NaCl
100 g solution
100 mL solution
8 g NaOH
B. 12%(v/v) ethyl alcohol
12 mL alcohol and 100 mL solution
100 mL solution
12 mL alcohol
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Learning Check
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Solution
How many grams of NaOH are needed to
prepare 2.0 L of a 12%(m/v) NaOH solution?
1) 24 g NaOH
2) 240 g NaOH
3) 2400 g NaOH
2) 240 g NaOH
2.0 L x 1000 mL = 2000 mL
1L
2000 mL x 12 g NaOH
100 mL
= 240 g NaOH
12 % (m/v) factor
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Learning Check
Solution
2) 3000 mL
150 g glucose x
How many milliliters of 5% (m/v) glucose
solution are given if a patient receives 150 g of
glucose?
100 mL = 3000 mL
5 g glucose
5% m/v factor (inverted)
1) 30 mL
2) 3000 mL
3) 7500 mL
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9.7 Molarity (M)
„
Preparing a 1.0 Molar Solution
„
Molarity is a concentration unit for the moles
of solute in the liters (L) of solution.
Molarity (M)
Examples:
2.0 M HCl
6.0 M HCl
50
A 1.0 M NaCl solution is prepared by weighing
out 58.5 g NaCl ( 1.0 mole) and adding water to
make 1.0 liter of solution.
= moles of solute = moles
liter of solution
L
= 2.0 moles HCl
1L
= 6.0 moles HCl
1L
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Calculation of Molarity
Learning Check
What is the molarity of a NaOH solution prepared
by adding 4.0 g of solid NaOH to water to make
0.50 L of solution ?
1. Determine the moles of solute.
4.0 g NaOH x 1 mole NaOH = 0.10 mole
40.0 g NaOH
Calculate the molarity of an NaHCO3 solution
prepared by dissolving 36 g of solid NaHCO3 in
water to give a solution volume of 240 mL.
1) 0.43 M
2) 1.8 M
3) 15 M
2. Calculate molarity.
0.10 mole = 0.20 mole = 0.20 M NaOH
0.50 L
1L
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Solution
54
Learning Check
A glucose solution with a volume of 2.0 L contains
72 g glucose (C6H12O6). If glucose has a molar
mass of 180. g/mole, what is the molarity of the
glucose solution?
1) 0.20 M
2) 5.0 M
3) 36 M
2) 1.8 M
36 g x 1 mole NaHCO3 = 0.43 mole NaHCO3
84 g
0.43 mole NaHCO3 = 1.8 M NaHCO3
0.240 L
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Solution
1) 0.20 M
72 g x 1 mole x 1
180. g
2.0 L
Molarity Conversion Factors
=
0.20 moles
1L
=
0.20 M
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The units in molarity can be used to write
conversion factors.
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Learning Check
58
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Solution
Stomach acid is 0.10 M HCl solution.
How many moles of HCl are present in
1500 mL of stomach acid?
1) 15 moles HCl
2) 1.5 moles HCl
3) 0.15 mole HCl
3) 0.15 mole HCl
1500 mL x 1 L
1000 mL
1.5 L x 0.10 mole HCl
1L
=
1.5 L
= 0.15 mole HCl
Molarity factor
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Learning Check
Solution
3) 37 g KCl
Calculate the grams of KCl that must be
dissolved in water to prepare 0.25 L of a
2.0 M KCl solution.
1) 150 g KCl
2) 37 g KCl
3) 19 g KCl
Determine the number of moles of KCl.
0.25 L x 2.0 mole KCl = 0.50 moles KCl
1L
Convert the moles to grams of KCl.
0.50 moles KCl x 74.6 g KCl = 37 g KCl
1 mole KCl
molar mass of KCl
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Learning Check
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Solution
1) 25 mL
0.15 mole HNO3 x
How many milliliters of 6.0 M HNO3 contain
0.15 mole of HNO3?
1L
x 1000 mL
6.0 moles HNO3
1L
Molarity factor inverted
1) 25 mL
2) 90 mL
3) 400 mL
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= 25 mL HNO3
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