Chapter 14: Solutions
• A solution is a homogeneous mixture.
• A solution is composed of a solute dissolved in a
solvent.
• Solutions exist in all three physical states:
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Chapter 14
1
Gases in Solution
• Temperature affects the solubility of gases.
• The higher the temperature is, the lower the
solubility of a gas is in solution.
• An example is carbon dioxide in soda:
– Less CO2 escapes when you open a cold soda than
when you open a warm soda.
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Chapter 14
2
1
Pressure and Gas Solubility
• Pressure also influences the solubility of gases.
• According to Henry’s law, the solubility of a gas
is directly proportional to the partial pressure of
the gas above the liquid.
• If we double the partial pressure of a gas, we
double the solubility.
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Chapter 14
3
Henry’s Law
• We can calculate the solubility of a gas at a new
pressure using Henry’s law.
new pressure
solubility x
= new solubility
old pressure
• What is the solubility of oxygen gas at 25 °C and a
partial pressure of 1150 torr if the solubility of
oxygen is 0.00414 g/100 mL at 25 °C and 760 torr?
0.00414 g/100 mL x
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1150 torr
= 0.00626 g/100 mL
760 torr
Chapter 14
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2
Polar Molecules
• When two gases, liquids or solids make a solution,
the solute is the lesser quantity, and the solvent is
the greater quantity.
• Recall that a net dipole is present in a polar
molecule.
• Water is a polar molecule.
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Chapter 14
5
Polar and Nonpolar Solvents
• A liquid composed of polar molecules is a polar
solvent. Water and ethanol are polar solvents.
• A liquid composed of nonpolar molecules is a
nonpolar solvent. Hexane is a nonpolar solvent.
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Chapter 14
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3
Like Dissolves Like
• Polar solvents dissolve in one another.
• Nonpolar solvents dissolve in one another.
• This is the like dissolves like rule.
• Methanol dissolves in water, but hexane does not
dissolve in water.
• Hexane dissolves in toluene, but water does not
dissolve in toluene.
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Chapter 14
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Miscible and Immiscible
• Two liquids that completely
dissolve in each other are
miscible liquids.
• Two liquids that are not
miscible in each other are
immiscible liquids.
• Polar water and nonpolar oil
are immiscible liquids and
do not mix to form a
solution.
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Chapter 14
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4
Solids in Solution
• When a solid substance dissolves in a liquid, the
solute particles are attracted to the solvent
particles.
• When a solution forms, the solute particles are
more strongly attracted to the solvent particles
than other solute particles.
• We can also predict whether a solid will dissolve
in a liquid by applying the like dissolves like rule.
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Chapter 14
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Like Dissolves Like for Solids
• Ionic compounds, like sodium chloride, are
soluble in polar solvents and insoluble in nonpolar
solvents.
• Polar compounds, like table sugar (C12H22O11), are
soluble in polar solvents and insoluble in nonpolar
solvents.
• Nonpolar compounds, like naphthalene (C10H8),
are soluble in nonpolar solvents and insoluble in
polar solvents.
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Chapter 14
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5
The Dissolving Process
• When a soluble crystal is placed into a solvent, it
begins to dissolve.
• When a sugar crystal is
placed in water, the water
molecules attack the crystal
and begin pulling part of it
away and into solution.
• The sugar molecules are held
within a cluster of water
molecules called a solvent
cage.
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Chapter 14
11
Dissolving of Ionic Compounds
• When a sodium chloride crystal is placed in water,
the water molecules attack the edge of the crystal.
• In an ionic compound, the water
molecules pull individual ions
off of the crystal.
• The anions are surrounded by
the positively charged
hydrogens on water.
• The cations are surrounded by
the negatively charged oxygen
on water.
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Chapter 14
12
6
Rate of Dissolving
•
There are three ways we can speed up the rate of
dissolving for a solid compound:
1. Heating the solution
•
This increases the kinetic energy of the solvent, and the solute is
attacked faster by the solvent molecules.
2. Stirring the solution
•
This increases the interaction between solvent and solute
molecules.
3. Grinding the solid solute
•
There is more surface area for the solvent to attack.
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Chapter 14
13
Solubility of Solids and Temperature
• The solubility of a compound is the maximum
amount of solute that can dissolve in 100 g of
water at a given temperature.
• In general, a
compound
becomes more
soluble as the
temperature
increases.
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Chapter 14
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7
Saturated Solutions
• Saturated solution:
• Unsaturated solution:
• Supersaturated solution:
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Chapter 14
15
Supersaturated Solutions
• At 55 °C, the solubility of NaC2H3O2 is 100 g per
100 g water.
• If a saturated solution at 55 °C is cooled to 20 °C,
the solution is supersaturated.
• Supersaturated
solutions are
unstable. The excess
solute can readily be
precipitated.
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Chapter 14
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8
Concentration of Solutions
•
The concentration of a solution tells us how
much solute is dissolved in a given quantity of
solution.
•
We often hear imprecise terms such as a “dilute
solution” or a “concentrated solution.”
•
There are two precise ways to express the
concentration of a solution:
1. Mass/mass percent
2. Molarity
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Chapter 14
17
Mass Percent Concentration
• Mass percent concentration compares the mass of
solute to the mass of solvent.
• The mass/mass percent (m/m %) concentration is
the mass of solute dissolved in 100 g of solution.
mass of solute
x 100% = m/m %
mass of solution
g solute
x 100% = m/m %
g solute + g solvent
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Chapter 14
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9
Calculating Mass/Mass Percent
• A student prepares a solution from 5.00 g NaCl
dissolved in 97.0 g of water. What is the
concentration in m/m %?
5.50 g NaCl
x 100% = m/m %
5.00 g NaCl + 97.0 g H2O
5.00 g NaCl
102 g solution
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x
100% = 4.90 %
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Chapter 14
Mass Percent Unit Factors
• We can write several unit factors based on the
concentration 4.90 m/m % NaCl:
4.90 g NaCl
100 g solution
100 g solution
4.90 g NaCl
4.90 g NaCl
95.1 g water
95.1 g water
4.90 g NaCl
95.1 g water
100 g solution
100 g solution
95.1 g water
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Chapter 14
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10
Mass Percent Calculation
• What mass of a 5.00 m/m % solution of dextrose
contains 25.0 grams of dextrose?
• We want grams solution; we have grams dextrose.
25.0 g dextrose x
100 g solution
5.00 g dextrose
= 500 g solution
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Chapter 14
21
Molar Concentration
• The molar concentration, or molarity (M), is the
number of moles of solute per liter of solution, and
is expressed as moles/liter.
moles of solute
=M
liters of solution
• Molarity is the most commonly used unit of
concentration.
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Chapter 14
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11
Calculating Molarity
• What is the molarity of a solution containing 24.0
g of NaOH in 0.100 L of solution?
• We also need to convert grams NaOH to moles
NaOH (M = 40.00 g/mol).
24.0 g NaOH
0.100 L solution
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x
1 mol NaOH
= 6.00 M NaOH
40.00 g NaOH
Chapter 14
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Molarity Unit Factors
• We can write several unit factors based on the
concentration 6.00 M NaOH:
6.00 mol NaOH
1 L solution
1 L solution
6.00 mol NaOH
6.00 mol NaOH
1000 mL solution
1000 mL solution
6.00 mol NaOH
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Chapter 14
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12
Molar Concentration Problem
• How many grams of K2Cr2O7 are in 250.0 mL of
0.100 M K2Cr2O7?
• We want mass K2Cr2O7; we have mL solution.
250.0 mL solution x
0.100 mol K2Cr2O7 294.2 g K2Cr2O7
x
1 mol K2Cr2O7
1000 mL solution
= 7.36 g K2Cr2O7
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25
Chapter 14
Molar Concentration Problem,
Continued
• What volume of 12.0 M HCl contains 9.15 g of
HCl solute (M = 36.46 g/mol)?
• We want volume; we have grams HCl.
9.15 g HCl
x
1 mol HCl
36.46 g HCl
x
1000 mL solution
12.0 mol HCl
= 20.9 mL solution
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Chapter 14
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13
Dilution of a Solution
• Rather than prepare a solution by dissolving a
solid in water, we can prepare a solution by
diluting a more concentrated solution.
• When performing a dilution, the amount of solute
does not change, only the amount of solvent.
• The equation we use is: C1 x V1 = C2 x V2.
• This example C=M (molarity)
– M1 and V1 are the initial molarity and volume, and M2
and V2 are the new molarity and volume.
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Chapter 14
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Dilution Problem
• What volume of 6.0 M NaOH needs to be diluted
to prepare 5.00 L if 0.10 M NaOH?
• We want final volume and we have our final
volume and concentration.
M1 x V1 = M2 x V2
(6.0 M) x V1 = (0.10 M) x (5.00 L)
V1 =
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(0.10 M) x (5.00 L)
= 0.083 L
6.0 M
Chapter 14
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14
Solution Stoichiometry
• In Chapter 10, we performed mole calculations
involving chemical equations: stoichiometry
problems.
• We can also apply stoichiometry calculations to
solutions.
solution
concentration
balanced
equation
molarity known ⇒ moles known ⇒
moles unknown ⇒ mass unknown
molar mass
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29
Chapter 14
Solution Stoichiometry Problem
• What mass of silver bromide is produced from the
reaction of 37.5 mL of 0.100 M aluminum
bromide with excess silver nitrate solution?
AlBr3(aq) + 3 AgNO3(aq) → 3 AgBr(s) + Al(NO3)3(aq)
• We want g AgBr; we have volume of AlBr3.
37.5 mL soln
x
0.100 mol AlBr3
1000 mL soln
x
3 mol AgBr
1 mol AlBr3
x
187.77 g AgBr
1 mol AgBr
= 2.11 g AgBr
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Chapter 14
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15
Chapter Summary, Continued
• The mass/mass percent concentration is the mass
of solute per 100 grams of solution.
mass of solute
x 100% = m/m %
mass of solution
• The molarity of a solution is the moles of solute
per liter of solution.
moles of solute
=M
liters of solution
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Chapter 14
31
Chapter Summary, Continued
• You can make a solution by diluting a more
concentrated solution.
C1 x V1 = C2 x V2
• We can apply stoichiometry to reactions involving
solutions using the molarity as a unit factor to
convert between moles and volume.
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Chapter 14
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16