Transcript UNIT 6: Chemical Equilibrium Chapter 15 “Solutions”

Chapter 15
Solution Definitions
to Know & Use
Solution – homogenous mixture of two or
more substances in a single physical
state.
 Solute – the substance being dissolved.
 Solvent – the principal component that
dissolves another component of a
solution.
 Solubility – a quantifiable measure of the
degree to which a substance dissolves in
another substance.
 Soluble – a substance that can be
dissolved in another substance.
 Insoluble – a substance that cannot be
dissolved in another substance.

Chapter 15 “Solutions”

What are solutions?

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
Homogeneous mixtures of two or more
substances in a single physical state.
A solution consists of a solute dissolved in a
solvent.
Many examples exist.
What are the intrinsic properties of
solutions?
1.
Contain very small particles
(atoms, ions, molecules)
2.
Homogeneous throughout. (Particles are
evenly distributed on a molecular level).
3.
Particles do not separate with time under
constant conditions.
4.
Diverse physical states and chemical
compositions.
Types of Solutions

Solid Solutions


Gaseous Solutions


Air, scuba diving gases, vehicle
exhaust
Liquid Solutions


Alloys (14 carat gold, stainless steel,
brass)
Vinegar, antifreeze,
Aqueous Solutions


Solutions with water as solvent.
Seawater, soft drinks
15-2 Concentration of
Solutions
– the amount of solute
dissolved per unit of solvent.
 There are many ways to describe
concentration, but they are either
qualitative or quantitative.
 Qualitative – a representation of the
general nature of a solution.
 Quantitative – a measure of the amount of
a solute dissolved in the solution.
 Concentration
Qualitative Descriptions
of Solutions
Dilute – a solution containing very little
solute.
 Concentrated – a solution containing a large
amount of solute.

Qualitative Descriptions of
Solutions
Saturated – a solution containing the
maximum amount of solute that can be
dissolved at the current
temperature/pressure.
 Unsaturated – a solution containing less than
the maximum amount of solute that can be
dissolved at the current
temperature/pressure.
 Supersaturated – an unstable condition in
which a solution contains more than the
maximum amount of solute that can normally
be dissolved at the current
temperature/pressure.

Solution Stability


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Recall that a saturated solution contains the
maximum amount of solute that can be dissolved at
given conditions.
In a saturated solution, the rate of solute entering
into solution precisely balances the rate at which
solute comes out of solution (forming a solid
precipitate).
The saturated solution is stable and said to be in
dynamic equilibrium.
Solution Stability

In an unsaturated solution more solute can dissolve,
so it is not yet at equilibrium.

In a supersaturated solution more than the maximum
amount of solute that can normally be dissolved at
the current conditions is present. This is an
unstable situation that is resolved by precipitating
solute (solid). The end result is an equilibrium
condition.
Saturation:
Quantitative Descriptions of
Solutions

Quantitative methods are much more useful
than qualitative descriptions because they
specify the amounts of components in
solutions.
 The most common quantitative descriptions
include:
 Molarity, M = moles solute/L of solution

Molality, m = moles of solute/kg of solvent
Molarity, M

Molarity (M) – moles of solute per liter of
solution.
M = moles/L

What is the molarity of a solution made from 145 g
of NaCl in 2.75 L of solution?

Vinegar is a solution of acetic acid. What is
the molarity of the solution produced when
125g of acetic acid (C2H4O2) is dissolved in
sufficient water to prepare 1.50L of solution?
Molality, m

Molality (m) – moles of solute per kilogram of solvent.
m = moles/kg

What is the molality of a solution made from 20.4 g
KBr in 195 g of water?
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What is the molality of a solution containing 125g of
iodine (I2) and 750.g of carbon tetrachloride (CCl4)?
Mole Fraction, xsolute

Mole Fraction – moles of component per total moles of
solution.
Xsolute = moles solute
total moles
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Example: What is the mole fraction of sulfur dioxide in
an industrial exhaust gas containing 128.0 g of SO2
dissolved in every 1500. g of CO2?
Answer:
XSO2 = (mole fraction SO2)/(total moles of solution)
Moles SO2 = 128.0g • (1 mol SO2 /64.04g SO2) = 1.999 mol SO2
Moles CO2 = 1500.g CO2 • [1 mol CO2/44.01g CO2] = 34.08 mol CO2
XSO2 = 1.999 mol SO2 ___________ =
1.999 mol SO2 + 34.08 mol CO2
1.999
36.08
= 0.05540
Molarity and Dilution Factors
When diluting a solution into a less concentrated one, the total number
of moles of solute does not change.
(The compound didn’t go anywhere; it is still in the container!)
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Because Molarity = moles/volume,
the moles = Molarity * Volume, or M V.
Therefore we may write the following for the each of the two solutions:
moles1 = M1V1 and moles2 = M2V2

Since there are the same number of moles in the first solution as in
the second, we may let moles1 = moles2, or also
McVc = MdVd
Can use this simple equation to calculate the new molarity.
More Solution Definitions to
Know

Miscible – liquids that may be mixed together in any
amount.
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Immiscible – liquids that cannot be mixed.

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Acetic acid and water (vinegar).
Electrolyte – a substance that forms ions in solution,
enabling the solution to conduct electricity.

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Oil and water.
Aqueous Solution – liquid solutions for which the
solvent is water.

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Oil and gasoline.
NaCl in sea water, Gatorade.
Non-electrolyte – a substance that does not form ions
in solution, thus giving a non-conducting solution.

Sugar in tea.
15-3 Formation of Solutions

Dissolution - the complex interaction of two or more separate
substances (the solute and the solvent) to form a single system (the
solution).

Solvation – the process whereby solvent particles pull the solute
particles into solution and surround them; the interaction between
solute and solvent particles to form a solution.

Hydration - the process whereby water particles pull the solute
particles into solution and surround them to form a solution.

Solubility – a quantifiable measure of the degree to which a
substance dissolves in another substance; it is the amount of a
solute that will dissolve in a specific solvent under given conditions.
Expressed in gram of solute per 100 grams of solvent.
Saturation:
1.
Factors Affecting Solubility
Nature of Solute and Solvent
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Similar substances dissolve in one another. (“Likes
dissolve likes.”)
Polar substances dissolve in polar substances
•
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Water dissolves sugar & salt.
Water dissolves rubbing alcohol.
Nonpolar substances dissolve in nonpolar
substances.
•
•
Gasoline dissolves oil.
Dry cleaning fluids dissolve grease and oils.
Temperature – see next slides
Pressure
2.
3.

Gas solubility increases with pressure.
Solubility and Temperature
Solubility and Temperature
Temperature
Solubility of solids and liquids generally increases
with temperature.
Solubility of gases decreases with temperature.
Solubility of salts:
Solubility of gases:
http://www.elmhurst.edu/~chm/vchembook/174temppres.html
Energy Changes and the
Formation of Solutions
Solvation/Hydration may be
exothermic or endothermic.
ΔH depends on the balance of energy released by the
attraction of solute particles to the solvent versus
the energy consumed in breaking the attractions of
solute particles for each other (crystal lattice
energy).
Dissolving CaCl2 is very exothermic, but an ammonium
nitrate ‘cold pack’ works because the solvation is
endothermic.
Factors That Affect
Dissolution Rates
Surface area
1.

Increasing surface area (making smaller particles)
increases the rate of dissolution.
Stirring
2.

Stirring the solution increases the rate of
dissolution.
Temperature
3.

Increasing the temperature increases the rate of
dissolution.
15-4 Colligative Properties
 These
are properties that depend on
solution concentration rather than the
nature or type of solute.
 They are dependent on molality
(molsolute/kgsolvent)
 Examples include…
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Vapor Pressure Reduction
Boiling Point Elevation
Freezing Point Depression
Vapor Pressure Reduction
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Raoult’s Law – the magnitude of the vapor pressure reduction is
proportional to the solute concentration, regardless of the solute.
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Why? Nonvolatile solute molecules interfere with the solvent
molecules, preventing them from leaving the surface of the
solution, and thus decreasing the vapor pressure.
(Fig. 15-22, p 520)
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This results in an increase in the boiling point
of the solvent, and a decrease in its freezing point.
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Boiling point elevation
Freezing point depression
Applications?
“Freezing Point Depression”
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This is the ability of a dissolved
solute to lower the freezing point of
a solution.

Example: Antifreeze is added to a car’s
coolant system to prevent freezing of
the water in winter.

Decrease of freezing point is directly
proportional to the molality (m) of
the solute.
 Calculated from ΔTf = Kfm
…where ΔTf is the temperature
depression, m is molality and Kf is the
freezing point depression constant.
“Boiling Point Elevation”

This is the ability of a dissolved solute to raise
the boiling point of a solution.

Example: The antifreeze added to a car’s coolant
system also prevents overheating in summer!

Increase of boiling point is also directly
proportional to the molality (m) of the solute.
 Calculated from ΔTb = Kbm
…where ΔTb is the temperature depression, m is
molality and Kb is the freezing point depression
constant.
Post Lab Questions
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1. Which compound types (ionic or covalent)
produce more particles when dissolved in water and
why? Remember this is related to molality.
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2. Which type of compound (ionic or covalent) will
have a greater effect on the colligative properties of
a solution? Explain.
Determining Molar Mass
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A solution containing 16.9g of a nonvolatile molecular
compound in 250g of water has a freezing point of 0.744 °C. What is the molar mass of the compound?
Solution:
 First find the molality from ΔTf = Kfm.
0.744 °C = (1.86 °C/m) x m,
so the molality = 0.400 molal
 But molality = moles/kg, so
0.400 m = ??? moles/0.250 kg,
from which ??? moles = 0.100 mole
-Since 0.100 mole = 16.9g of solute, 1.00 mole of solute =
169 g, giving a molar mass of 169 g/mol.