Solutions
Solutions
I. Classification of liquid mixtures.
II. Solution Concentration. Molarity.
III. Solubility.
-Dissolving process.
-Ionic Equations.
IV. Colligative properties.
-Types of solutions.
-Solubility curve.
-Molality.
-Freezing/Boiling Points.
Solutions are
homogeneous
mixtures
Classification
of Matter
Properties of Solutions,
Suspensions, and Colloids
Solutions
Solutions: a homogeneous
mixture of two or more
substances in a single
phase of matter.
 In a simple solution where,
for example, salt is
dissolved in water, the
particles of one substance
are randomly mixed with
the particles of another
substance.

SOLUTE – A solute is the dissolved
substance in a solution.
EX: CO2, KCl, Na2CO3
 SOLVENT – A solvent is the
dissolving medium in a solution.
EX: H2O, CCl4

The solute is generally designated as
that component of a solution that is of
lesser quantity.
If we had a mixture of 25 mL of
ethanol and 75 mL of water, the
ethanol would be the solute and water
would be the solvent.
 If we had a 50% to 50% ratio, it
would be unnecessary to designate
solvent or solution.

Heterogeneous Liquid Mixtures
There are two types:
1. Suspensions
2. Colloids
*Suspensions and colloids are not solutions.
Suspensions
If the particles in a solvent are so
large that they settle out due to
gravity unless the mixture is
constantly stirred or agitated, the
mixture is called a suspension.
 These particles (over 1000nm) can be
filtered out of the heterogeneous
mixture.

Colloids
Particles that are
intermediate in size between
those in solutions and
suspensions form mixtures
known as colloidal dispersions.
 Particles between 1nm and
1000 nm in diameter may
form colloids. After the
larger particles settle out
(suspensions), the water may
still be cloudy because
colloidal particles remain
dispersed in the water.
 Milk is an example of a colloid.

Suspensions
Colloids
Solutions
The Tyndall Effect
Many colloids appear homogeneous
because the individual particles
cannot be seen. The particles are,
however, large enough to scatter
light.
 Tyndall effect is a property that can
be used to distinguish between a
solution and a colloid.

When a laser is passed through a
solution and a colloid at the same time,
it is evident which glass contains the
colloid. (you can’t see the light in a colloid)
Colloid
Solution
Solutions
Colloids
Homogeneous
Particle size:
0.01-1 nm; can
be atoms, ions,
molecules
Suspensions
Heterogeneous
Particle size:
1-1000 nm,
dispersed; can
be large
molecules
Do not separate Do not separate
on standing
on standing
Heterogeneous
Particle size:
Over 1000 nm,
suspended;
can be large
particles
Particles settle
out
Cannot be
separated by
filtration
Most cannot be
separated by
filtration
All can be
separated by
filtration
Do not scatter
light
Scatter light
(Tyndall effect)
Not transparent
May scatter lite
Solution
Concentration
Solution Concentration
Molarity is simply a measure of the
"strength" of a solution. A solution
that we would call "strong" would have
a higher molarity than one that we
would call "weak".
 If you ever made or drank a liquid
made from a powdered mix, such as
Kool-Aid or hot cocoa, you probably
are familiar with the difference
between what is called a "weak"
solution or a "strong" solution.


To make Kool-Aid of "normal“ strength =
4 scoops of powder
----------------------2 quarts of water

To make Kool-Aid twice
the "normal" strength…
What could you do?
8 scoops of powder
-------------------or
2 quarts of water
4 scoops of powder
-------------------1 quart water
Solution Concentration


Molarity:
*One-molar (M) =
1 mole solute
1Liter solution
One mole of NaCl (molar mass of NaCl = 22.99 +
35.45 = 58.44 grams) is dissolved in enough
water (1 Liter) to make a 1M NaCl
solution.
Molarity Calculations

Calculate the molarity of:
*35.2 grams of CO2 in 500. mL.
Step 1: convert 35.2 g of CO2 into
moles
35.2g (1 mole ) = 0.800 mol
1
(44.01 g)
Step 2: divide moles by volume in liters
0.800 mol CO2 = 1.60 M CO2
0.500 L
Answers
WS: Molarity Problems
1)
2)
3)
4)
5)
6)
NaCl
Al2(SO4)3
HClO3
HCl
Ba(OH)2
Fe(NO3)2
1)
2)
3)
4)
5)
6)
7)
8)
1.22 M KClO3
0.77 M Na2SO4
1.00 M NaOH
0.66 M AlCl3
0.71 M HCl
0.82 M LiF
2.78 M KOH
0.10 M ZnCl2
Solutions and Solubility
Solubility
When we talk about the mixing of two
or more substances together in
solution was much consider solubility.
 Solubility is defined as the amount of
a substance that can be dissolved in a
given quantity of solvent.


When deciding what type of solvent to
use with a given solute it is important
to identify what types of substances
you have.
1. Polar substances (partial + or –
charges) tend to dissolve in polar
solvents
2. Nonpolar molecules (equal sharing of
e-) tend to dissolve in nonpolar solvents

Remember the rule:
L IK E D ISSO L V E S L IK E
Solvent-Solute Combinations
Solvent Type
Solute Type
Is solution likely?
Polar
Polar
Yes
Polar
Nonpolar
No
Nonpolar
Polar
No
Nonpolar
Nonpolar
Yes

Remember:
L IK E D ISSO L V E S L IK E
The Dissolving Process
Water is a polar solvent and is attracted
to polar solutes.
 Salt is polar (ionic).
 Water molecules surround and isolate
the surface ions. The ions become
hydrated and move away from each
other in a process called dissociation.

Insolubility
Any substance whose
solubility is less than 0.01
mol/L will be referred to
as insoluble.
 We can predict whether a
precipitate (insoluble
substance) will form when
solutions are mixed if we
know the solubilities of
different substances.

Experimental observations have led to
the development of a set of empirical
solubility rules for ionic compounds,
gases, and molecules.
 EX: Experiments demonstrate that
all ionic compounds that contain the
nitrate anion, NO3-, are soluble in
water.

*Refer to solubility table*
Ionic Equations
Referring to Solubility Tables & Writing Ionic Equations
are very useful tools when trying to determine if a reaction
will occur in an aqueous solution.
Ionic Equations:
An ionic equation is a chemical equation
in which electrolytes (soluble ions that
conduct electricity) are written as
dissociated ions.
 Ionic equations are used for single and
double replacement reactions which
occur in aqueous solutions.
 In an aqueous reaction ions that are
found as both reactants and products
are not part of a reaction. They are
termed spectator ions and essentially
cancel out of the ionic equation.

To write net ionic equations
follow these simple rules:
1.
2.
3.
4.
5.
Write a balanced equation.
Determine which substances are
soluble (refer to the solubility rules table)
Rewrite the equation in ionic form
by dissociating the soluble
reactants & products
Cancel the spectator ions
Write the net ionic equation
Hints… What will dissociate?
Refer to your Solubility Rules Table

Salts:
•
•

Write in ionic form if soluble.
EX: KCl  K+ + ClWrite in undissociated form if insoluble.
EX: AgCl  AgC
Acids:
•
•
STRONG ACIDS - Write in ionic form.
They are soluble. (listed on table)
EX: H2SO4  2H+ + SO4-2
WEAK ACIDS - Write in undissociated
form. They are insoluble. (not listed on table)
EX: H3PO4  H3PO4

Bases (OH-):
•
•
STRONG BASES - Write in ionic form.
They are soluble.
EX: Ca(OH)2  Ca+2 + 2OHWEAK BASES - Write in undissociated
form. They are insoluble.
EX: Mg(OH)2  Mg(OH)2
Oxides:
•
Always write in undissociated form.
EX: MgO, H2O
Gases:
•
Always write in undissociated form.
EX: SO2, NH3, H2, O2
Sample Problem
Step 1 & 2:
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
(s)
(s)
(s)
Step 3 & 4:
Ag+ + NO3- + Na++ Cl-  AgCl(s) + Na+ + NO3Step 5:
Ag+ + Cl-  AgCl(s)
Types of Solutions
Saturated Solutions




A solution at equilibrium with undissolved
solute is said to be saturated.
Additional solute will not dissolve if added
to this solution.
It is possible to dissolve less solute than
needed to form a saturated solution.
These solutions are unsaturated.
A supersaturated solution can be made by
dissolving the solute under high temps and
then carefully cooling them. These are
unstable solutions.
Degrees of Saturation
1. Saturated solution
 Solvent holds as much
solute as is possible at
that temperature.
 Undissolved solid
remains in flask.
 Dissolved solute is in
equilibrium with solid
solute particles.
2. Unsaturated Solution
Less than the
maximum amount of
solute for that
temperature is
dissolved in the
solvent.
 No solid remains in
flask.

3. Supersaturated
Solvent holds more solute than is
normally possible at that
temperature.
 These solutions are unstable;
crystallization can often be
stimulated by adding a “seed crystal”
or scratching the side of the flask.
 Solids form as solution cools.

Saturated Solution 

Supersaturated Solution
Factors Affecting Solubility
Solubility depends on the nature of
both the solvents and solutes,
temperature, and for gases, on
pressure.
 The solubility of most solid solutes in
water increases as the temp of the
solution increases.
 This means that more sucrose C12H22O11
can be dissolved in hot water than cold,
the basis for making “rock candy”.

Solubility Curves
The graph
represents the
solubility of
substances, including
NaCl, NaNO3, and
KNO3 at different
temps.
 Notice that when
temp increases,
solubility increases for
most substances.

Solubility of Gases- Temp
Solubility of gases is
dependent on both
temperature and
pressure.
 Based on the
solubility curve, the
solubility of NH3 and
SO2 (both gases)
decreases as
temperature
increases.

Solubility of Gases- Pressure





Gas Solubility increases when
pressure increases.
Carbonated beverages are
bottled with CO2 under
pressure to increase the
solubility of CO2 gas.
As bottle is opened, pressure
of CO2 decreases and
solubility of CO2 decreases.
Therefore, bubbles of CO2
escape from solution.
Temp also effects solubility,
colder sodas lose CO2 more
slowly than warm sodas.
Dissolving Chart
Solid
Gas
Increasing Solution
Temperature
Increases
solubility &
solubility rate
Decreases
solubility
Crushing Solute
Increases
solubility
rate
No Effect
Increases
solubility
rate
Increases
solubility
rate
No Effect
Increases
solubility &
solubility rate
Stirring Solution
Increasing
Atmospheric (air)
Pressure
Worksheet: Solubility Curves
Lab: Scooby-Doo Molarity Mayhem
Colligative Properties
Colligative comes from the Greek word
kolligativ meaning glue together.
 We use this term for the properties of
substances (solutes and solvents)
together.
 Colligative properties of solutions is
used to describe the effects of
antifreeze/summer coolant.

Molality
Recall the units for Molarity (M):
moles solute
L solution
 Molality (m) is the measure of the number
of moles of a solute per 1000g of solvent.
moles solute
1kg solvent
 Molality is best used to describe colligitive
properties and is represented by m.

Boiling Point and Freezing
Point
Review the phase
diagram of a pure
substance.
 How will the phase
diagram of a
solution (freezing
and boiling points)
differ from those
of a pure solvent?

The addition of a nonvolatile solute will
require a higher temperature in which
to reach boiling point, thus:
Boiling point elevation
 The addition of a nonvolatile solute will
require a lower temperature in which
to reach freezing point, thus
Freezing point depression

Pure water
Water with NaCl
The water with the solute of NaCl has fewer
liquid molecules becoming gases.
This will increase the temp needed to change
the state from (l)  (g)
Calculating Freezing and
Boiling Points

The following table contains the molal (K)
Boiling Point Elevations, Kb, and Freezing
Point Depressions, Kf.
Solvent
Water
Benzene
Ethanol
CCl4
Chloroform
Normal
boiling pt
(°C)
100.0
80.1
78.4
76.8
61.2
Kb
Normal
(°C/m) freezing pt
(°C)
0.52
2.53
1.22
5.02
3.63
0.0
5.5
-114.6
-22.3
-63.5
Kf
(°C/m)
1.86
5.12
1.99
29.8
4.68
The data for the table was found by
doing experiments.
 It has been found that 1 mole of a
nonvolatile solute particles will raise
the boiling temperatures of 1 kg of
water by 0.52 C°.
 The same concentration of solute will
lower the freezing point of 1 kg of
water by 1.86 C°.
 These two figures are the molal boiling
point constant (Kb) and the molal
freezing point constant (Kf).

A 1m solution of sugar in water
contains 1 mole of solute particles per
1 kg of solvent.
 A 1m solution of NaCl in water
contains 2 mole of solute (because
NaCl is an ion, it will dissociate in
water into Na+ and Cl- ions) per 1 kg
of solvent.
 How many mole solute would 1m
calcium nitrate, Ca(NO3)2, have per
1kg solvent?

Calculating Changes in Kb and Kf

Boiling point elevation is:
ΔTb = Kbmi
(moles)
(molality)
(change in boiling point) (boiling point
constant)
 Freezing point depression:
ΔTf = Kfmi
(moles)
(molality)
(change in freezing point) (freezing point
constant)
If 55.0 grams of glucose (C6H12O6) are
dissolved in 525 g of water, what will
be the change in boiling and freezing
points of the resulting solution?
Step 1: Calculate molality:
55.0 g ( 1 mol )
0.305 mol =
1
(180.18 g) =
0.525kg
0.581 m

Step 2: Obtain molal Kb from table.
 Step 3: Place values into equation
ΔTb = Kbmi
ΔTb = (0.52°C/m)(0.581m)(1) = 0.302 °C


This means that the boiling point will be elevated by
0.302 °C.
Normal Boiling Point + ΔTb = New Boiling Point
100 °C + 0.302 °C = 100.302 °C

This solution will reach boiling point at 100.302 °C.

Now let’s calculate the change in
freezing.
Calculate the change in freezing point
of 24.5g potassium bromide dissolved
in 445 g of water. (assume 100%
dissociation)
Step 1: Convert g of KBr into moles
24.5g ( 1 mol)
0.206mol =
1
(119.00 g) = 0.445kg
0.463m

•
•
•
Step 2: Obtain molal Kf from table.
Step 3: Place values into equation
KBr is ionic so the dissociation of KBr
makes 2 moles of ions (solute) per kg of
solvent:
KBr  K+ + Br –
ΔTf = Kfmi
ΔTf = (1.86°C/m)(0.926m)(2) = 1.72 °C
Normal Freezing Point - ΔTf = New Freezing Point
0 °C - 1.72 °C = -1.72 °C
*Freezing point has been depressed to -1.72 °C.
Coolant is used because it takes higher
temperatures to reach boiling point.
 Antifreeze needs lower temperatures in
order to freeze.
 This also why salt is used on frozen
roads and walkways. The salt dissolves
in the water and lowers the freezing
point of water. It now takes colder
temps to turn the water into ice.
 A 10-percent salt solution freezes at 20
F (-6 C), and a 20-percent solution
freezes at 2 F (-16 C).

Boiling point elevation is:
ΔTb = Kbmi
New Boiling Point = normal bp + ΔTb
Freezing point depression:
ΔTf = Kfmi
New Freezing Point = normal fp - ΔTf
Solvent
Normal
boiling pt
(°C)
Water
100.0
Benzene
80.1
Ethanol
78.4
CCl4
76.8
Chloroform 61.2
Normal
Kb
(°C/m) freezing
pt (°C)
0.52
0.0
2.53
5.5
1.22
-114.6
5.02
-22.3
3.63
-63.5
Kf
(°C/m)
1.86
5.12
1.99
29.8
4.68

1.
2.
3.
4.
5.
Practice problems: Compute both
boiling and freezing points of these
solutions: (assume 100% dissociation
of all ionic compounds)
27.6 g NaBr in 100.g of water.
100.0 g of C10H8 (naphthalene) in 250.
g of C6H6 (benzene).
25.9 g of C7H14BrNO4 (3-bromo-2nitrobenzoic acid) in 150. g of
benzene.
55.6 g of C12H22O11 in 500. g of water.
1500.g of NaCl in 4500. g of water.
Brief Summary
Heterogeneous liquid mixtures are
classified as suspensions (large particles
that settle out), or colloids (small
particles that stay dispersed).
 Homogeneous mixtures are solutions
made of a solute dissolved in a solvent.
 Solutes and solvents must be alike in
polarity in order to produce a solution.
 The concentration of a solution is
molarity (molar) and has the unit M,
which includes moles of solute per unit
volume of solvent.

When preparing a dilute solution from a
concentrated solution, use the formula:
M1V1 = M2V2
Where initial volume and molarity of
concentrated solution (EX: 12M HCl) is
compared to final volume and molarity of
diluted solution (EX: 6M HCl).
 Solubility of solutes can be reflected in
a solubility graph.
 Solubility of solid substances generally
increases as temperature increases.
 Solubility of gases decreases with
increased temperature.

Ionic equations can be written to
express the net reaction occurring in a
system after the spectator ions have
been removed.
 Solubility rules for substances have
been experimentally determined. They
indicate what substances are or are
not water soluble.
 Colligative properties demonstrate the
properties of the solution rather than
solute and solvent independently.

A solution with undissolved solute is
termed unsaturated.
 A solution with undissolved solute is
termed saturated.
 A solution that has more dissolved
solute at a particular temp due to
being dissolved at a higher temp is
termed supersaturated.
 Boiling points and Freezing points of
solutions can be calculated using
molality.

Molality (molal) is described by unit m and
expresses moles of solute per kg of solvent.
 When calculating BP and FP differences use
equation: Kfp or bp = Kb or f x m x moles
Kb or f is a standard and must be given
m must be calculated and adjusted to
express moles contributed.
-molecules contribute only 1 mol.
-ionic compounds contribute the number of
moles they dissociate into.
EX: KI  1 mole K+ + 1 mole I

After calculating difference, refer to
normal BP and FP of solvents and:
Boiling Point Elevation  add
difference to normal BP.
Freezing Point Depression  subtract
difference from normal FP.