Transcript Mixtures and Solutions

Mixtures and Solutions
Chemistry Matter and Change Chapter 14
Chapter 14 Main Idea
 Nearly
all of the gases,
liquids and solids that
make up our world are
mixtures.
Types of Mixtures
Chemistry- Matter and Change 14.4
14.1 Main Idea
 Mixtures
can be either
heterogeneous or
homogeneous.

14.1 Objectives




Compare the properties of
suspensions, colloids and solutions
Identify the types of colloids and
types of solutions
Describe the electrostatic forces in
colloids
Predict whether a solution is an
electrolyte or nonelectrolyte
Test your prediction
Review Vocabulary




Solute
Solvent
Molar mass
Concentration



New Vocabulary







Suspension
Colloid
Brownian motion
Tyndall effect
Soluble
Miscible
Insoluble
Immiscible
Electrolyte
Nonelectrolyte
Heterogeneous mixtures


Mixture- a combination of two or more
substances that each retain their individual
chemical properties
Not the same composition throughout the
substance
Suspensions

Type of heterogeneous mixture


Large particles settle if left undisturbed
Filter out
Colloids


Medium sized particles do not settle out, but
remain visibly different
Cannot filter out
Categories of Colloids
Particles Solid
Liquid
Gas
Medium
Solid
Solid sol
Solid emulsion or gel
Solid foam
Liquid
Sol
Emulsion
Foam
Gas
Solid aerosol
Liquid aerosol
Examples of Colloids
Category
Example
Dispersed
particles
Dispersion
medium
Solid sol
Ruby, sapphire
Solid
Solid
Sol
Blood, gelatin
Solid
Liquid
Solid emulsion
Butter, cheese
Liquid
Solid
Emulsion
Milk, mayonnaise
Liquid
Liquid
Solid foam
Marshmallow, Dove
Gas
Solid
Foam
Whipped cream
Gas
Liquid
Solid aerosol
Smoke, dust in air
Solid
Gas
Liquid aerosol
Fog, hairspray
Liquid
Gas
Brownian Motion



Movement of particles in a colloid
Erratic
Particles may repel preventing settling
Tyndall effect



Small particles that do not settle out
Too small to be seen unaided
Create a visible line when a laser is shined
through them (scatter light)
Classification of Matter
Solutions are
homogeneous mixtures
Homogeneous mixtures



Solvent- material in greater quantity
Solute- material in lesser quantity
Do not disperse laser light
Forming solutions


Not all materials will form solutions
Soluble- the two substances will form a
solution


Miscible- two liquids that will form a solution no
matter their proportions (tea and water)
Insoluble- two materials that never form a solution
(sand in water)

Immiscible- may be mixed, but will separate quickly
(oil and vinegar)
Electrolytes vs. Nonelectrolytes
The ammeter measures the flow of electrons (current) through the circuit.
If the ammeter measures a current, and the bulb glows, then the solution
conducts.
If the ammeter measures a current, and the bulb glows, then the solution
conducts.
If the ammeter fails to measure a current, and the bulb does not glow, the
solution is non-conducting.
Definition of Electrolytes and
Nonelectrolytes
An electrolyte is:

A substance whose
aqueous solution
conducts an electric
current.
A nonelectrolyte is:
A substance whose
aqueous solution
does not conduct an
electric current.
Electrolytes?
1. Pure water
2. Tap water
3. Sugar solution
4. Sodium chloride solution
5. Hydrochloric acid solution
6. Lactic acid solution
7. Ethyl alcohol solution
8. Pure sodium chloride
ELECTROLYTES:
NONELECTROLYTES:
Tap water (weak)
Pure water
NaCl solution
Sugar solution
HCl solution
Ethanol solution
Lactate solution
(weak)
Pure NaCl
But why do some compounds conduct electricity in solution while others do not…?
Answers to Electrolytes


Can you…



Compare the properties of
suspensions, colloids and solutions
Identify the types of colloids and
types of solutions
Describe the electrostatic forces in
colloids
Predict whether a solution is an
electrolyte or nonelectrolyte
Test your prediction
Solution Concentration
Chemistry- Matter and Change 14.2
14.2 Main Idea
 Concentration
can be
expressed in terms of
percent or in terms of
moles

14.2 Objectives



Describe concentration using
different units
Determine the concentration of
solutions
Calculate the molarity of a solution
Prepare different solutions with given
molarity, percent by mass, percent by
volume, molality, and mole ratio.
Review Vocabulary




Solvent
Ratio
Mass
Mole

New Vocabulary




Concentration
Molarity
Molality
Mole fraction
Dilution
Concentrated vs. Dilute
Expressing Concentration
Concentration Description
Ratio
Percent by mass
m assof solute
100
m assof solution
Percent by volume
volum eof solute
100
m assof solution
Molarity
m oles of solute
liter of solution
Molality
m oles of solute
kg of solution
Mole fraction
m oles of solute
m olesof solute  m oles of solution
Percent by mass

The mass of the solute divided by the mass of
the solution
In order to maintain a sodium chloride
concentration similar to ocean water, an
aquarium must contain 3.6 g sodium
chloride per 100.0 g of water. What is the
percent by mass of NaCl in the solution?



Mass of solute: 3.6 g
Mass of solution: 100.0 g + 3.6 g =103.6 g
Percent by mass = (3.6g/103.6g) x 100 = 3.5%
Percent by volume

The volume of the solute divided by the volume
of the solution
What is the percent by volume of
ethanol in a solution that contains
35mL of ethanol distributed in 155mL
of water?



Volume of solute: 35mL
Volume of solution: 155mL+35mL = 190mL
Percent by volume = (35mL/190mL) x100 = 18%
Molarity



Most commonly used
The moles of solute/liters of solution
Represented by M
n
M
V
Important Notes
Q: What do you call a tooth
in a glass of water?
A: One molar solution.


One way to measure the strength of a
solution
moles of solute per liter of solution.
mole solute
M
liter solvent
Important Notes
You must switch all units to moles and liters!

Recall 1000 mL = 1 L
Important Notes
You have 3.50 L of solution that contains
90.0 g of sodium chloride, NaCl. What is
the molarity of that solution?
90.0 g NaCl 1 mol NaCl
3.50 L Solution 58.44 g NaCl
=
0.440 M NaCl
A 100.5mL intravenous solution
contains 5.10g of glucose. What is the
molarity of this solution? (The molar
mass of glucose is 180.16 g/mol)
Moles of solute = 0.0283 mol C6H12O6
5.10 g glucose
1 mol glucose
180.16 g glucose
Liters of solution = 100.5 mL 1L
= 0.1005 L
1000mL
Molarity = 0.0283 mol/0.1005 L = 0.282M

Dilution of molar solutions

M1V1=M2V2

Important Notes
M is molarity; V is volume
What volume in milliliters, of a 2.00M
calcium chloride stock solution would
you use to make 0.50 L of 0.300M
calcium chloride solution?
M1=2.00M
 V1=?
 M2 = 0.300M
 V2=0.50L
 V1= 0.50L 0.300M
2.00M
V1=75mL

= 0.0750L 1000mL
1L
Molality



Used when volume of solution is highly
variable due to temperature differences
The moles of solute/kg of solvent
Represented by m
mol
m
kg
Important Notes
Molality
The concentration of a solution in moles of
solute per kilogram of solvent.
n
m
kg
Important Notes
You must switch all units to moles and
kilograms!
Recall 1000 g = 1 kg
Important Notes
In the lab, a student adds 4.5 g of
sodium chloride to 100.0 g of water.
Calculate the molality of the solution.



Mass of water = 100.0 g = 0.1000 kg
Mass of sodium chloride = 4.5g
Moles of NaCl
4.5g NaCl 1 mol NaCl
58.44g NaCl
= 0.077 mol NaCl
Molality =0.077 mol/0.1000 kg = 0.77m
Mole fraction

Represented by XA and XB and X…
Important Notes
nA
nB
XA 
; XB 
nA  nB
nA  nB
Mole Fraction
10% 9%
23%
58%
A
B
C
D
A hydrochloric acid solution contains
36g HCl and 64g H2O. What is the
mole fraction for the solution?




36g HCl = 1.0 mol HCl
64g H2O = 3.6 mol H2O
XHCl = 0.1mol /4.6 mol = 0.22
XH2O = 3.6 mol/4.6 mol = 0.78

14.2 Can you…



Describe concentration using different
units
Determine the concentration of
solutions
Calculate the molarity of a solution
Prepare different solutions with given
molarity, percent by mass, percent by
volume, molality, and mole ratio.
Factors Affecting Solvation
Chemistry- Matter and Change 14.3
14.3 Main Idea
 Factors
such as
temperature, pressure
and polarity affect the
formation of solutions

14.3 Objectives



Describe how intermolecular forces
affect solvation
Define solubility
Predict how different factors affect
solubility
Create solutions that are
unsaturated, saturated and supersaturated
Review Vocabulary







Exothermic
Endothermic
Solution
Polarity
Ionic bond
Covalent bond
Ionization


New Vocabulary





Solvation
Heat of solution
Unsaturated solution
Saturated solution
Supersaturated solution
Henry’s law
Dissociation
The solvation process

Like dissolves like



Only substances that can
become enmeshed in the
other substance will dissolve
Polar solutes require polar
solvents
Nonpolar solutes require
nonpolar solvents
Important Notes
Solvation of ionic compounds

Charged ends of the polar water molecules get
in between the ions in the crystal and force
them apart
Solvation of molecular compounds


Dissolve only when the attractive forces
between the molecules of the compound are
weaker than the attractive forces between the
polar water molecules and the polar solute
molecules
Water molecules form hydrogen bonds with the
polar portions of the solute
Heat of solution


Energy change during the formation of a
solution
Some solutions absorb energy during
formation while others release energy
Factors that affect the rate of solvation



Agitation
Surface area
Temperature
Degrees of solubility

Solubility of a substance can change with
temperature or pressure
Saturated


A solution that contains the exact amount of
solute at a specific temperature and pressure
The addition of more solute results in
undissolved particles
Unsaturated


A solution that contains less than the exact
amount of solute at a specific temperature and
pressure
The addition of more solute will result in a
greater concentration of the solution
Supersaturated



A solution that contains more than the exact
amount of solute at a specific temperature and
pressure
Highly unstable
Created at a higher temperature or pressure
Degrees of saturation
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED SUPERSATURATED
SOLUTION
SOLUTION
no more solutebecomes unstable,
dissolves
crystals form
concentration
Solubility of gases


As temperature increases, the solubility of
gases decreases
As pressure increases, the solubility of gases
also increases
Henry’s law

At a given temperature, the solubility (S) of a
gas in a liquid is directly proportional to the
pressure (P) of the gas above the liquid
S1 S 2

P1 P2

14.3 Can you…



Describe how intermolecular forces
affect solvation
Define solubility
Predict how different factors affect
solubility
Create solutions that are
unsaturated, saturated and supersaturated
Colligative Properties of Solutions
Chemistry- Matter and Change 14.4
14.4 Main Idea
 Colligative
properties
depend on the number
of solute particles in a
solution

14.4 Objectives


Describe colligative properties
Identify four colligative properties of
solutions
Determine the boiling point elevation
and freezing point depression of a
solution
14.4 Review Vocabulary



Ion
Boiling point
Freezing point
14.4 New Vocabulary





Colligative property
Vapor pressure lowering
Boiling point elevation
Freezing point depression
Osmosis
Electrolytes and Colligative Properties

Colligative properties depend on
the numbers of particles in a
solution rather than the identity
of the particles
Electrolytes

Allow electrons to flow

Can conduct electricity
Vapor Pressure Lowering

Reduced number of solvent molecules exposed
reduces the vapor pressure above a solution
Boiling point elevation

Addition of solutes requires more energy to
release molecules from a solution, thereby
elevating the boiling point.
Freezing Point Depression

Solute particles interfere with intermolecular
bonding, requiring more energy to return to a
solid state
Equations

Freezing point depression
Tf  K f m

Boiling point elevation
Tb  Kb m
Sodium chloride is often used to prevent
icy roads and to freeze ice cream. What are
the boiling and freezing points of a 0.029m
aqueous solution of sodium chloride?

Look up on tables





Kf (H2O) = 1.86 °C/m
Kb (H2O) = 0.512 °C/m
Tf = 0.00°C
Tb=100°C
Given in problem

m= 0.029m
Tf  K f m



ΔTf = (1.86 °C/m)(0.029m)=0.11°C
Tf =-0.11°C
Tb = 100.030°C
Osmosis


Diffusion of water through a semi-permeable
membrane
Osmotic pressure

Pressure resulting from osmosis

14.4 Can you…


Describe colligative properties
Identify four colligative properties of
solutions
Determine the boiling point elevation
and freezing point depression of a
solution