Unit 11- Solubility &
Solutions, Ch. 17 & 18
I. Water
A. The Molecule
1. O—H bond is highly polar
2. Bond angle 105° making it Bent shaped
3. Water Molecule as a whole is polar
4. Attracted to each other by intermolecular
hydrogen bonds
Greater electronegativity
I. Water (cont.)
B. Important Properties
1. High surface tension
2. low vapor pressure

hydrogen bonds hold molecules to one another, tendency to
escape surface is low
3. high specific heat capacity

4.184 J/g×°C
4. high melting and boiling points

0°C and 100°C
I. Water (cont.)
C. Surface Tension – inward force, or pull, that
tends to minimize the surface area of a liquid

Surfactant – wetting agent such as soap or
detergent that decreases the surface tension by
interfering with hydrogen-bonding
Responsible for
high surface
tension
I. Water (cont.)
D. Atypical Ice
1. As a typical liquid cools, density increases b/c
Volume decreases as the mass stays constant
2. As water cools it first behaves like a typical
liquid until it reaches 4°C
3. Below 4°C the density of water starts to
decrease
**Ice is one of only a few solids that float in their own
liquid.
Atypical Ice
Do not need to write.
Density of Liquid Water and Ice
Temperature (°C)
Density (g/cm3)
100° (liquid water)
0.9584
50°
0.9881
25°
0.9971
10°
0.997
4°
1.000 (most dense)
0° (liquid water)
0° (ice)
0.9998
0.9168
Atypical
Why does ice behave
Ice so differently?
Open framework
arranged like a
honeycomb.
Framework collapses,
molecules packed closer
together, making it more dense
II. The Solution Process
A. Solution - homogeneous mixture
consisting of the same properties
throughout (exists in a single phase)
Solute - substance being
dissolved
- smaller quantity substance
Solvent - dissolving
medium
- larger quantity substance
- usually water
II. The Solution Process (cont.)

Aqueous Solution (aq) – a solution in
which the solvent is water
HCl(g) + H2O(l) → H3O(aq) + Cl-(aq)
potassium chromate,
cobalt(II) nitrate,
copper(II) sulfate,
K2CrO4 (yellow)
Co(NO3)2 (red)
CuSO4 (blue) potassium
potassium dichromate
nickel(II) chloride,
permanganate,
K2Cr2O7 (orange)
NiCl2 (green)
KMnO (purple)
4
II. The Solution
Process (cont.)
B. Solvation– the process
of dissolving
1st solute particles
(salt) are surrounded
by solvent particles
(water)
2nd solute particles
(salt) are separated
and pulled into
solution (salt water)
II. The Solution Process (cont.)
C. “Like Dissolves Like”
NONPOLAR
POLAR
NONPOLAR
Nonpolar solvents
dissolve nonpolar
compounds
POLAR
Polar solvents dissolve
polar molecules and
ionic compounds
“Polar Dissolves Polar”
Polar vs. Nonpolar

Polar Molecule – uneven electron forces acting
on the central atom.


Water
Nonpolar Covalent Bond – when 2 atoms
are joined by a covalent bond and the
bonding electrons are shared equally
III. Water of Hydration
Water molecules that are integral part of
the crystal structure
 Compound that contains water of
hydration is called a hydrate


(ex. CuSO4.5H2O)
Effloresce – losing the water of
hydration
 Deliquescent – removes water from air
to form a solution

III. Water of Hydration
Calculate the percent by mass of water in
sodium carbonate decahydrate
(Na2CO3 ·10H2O)
Percent H2O = mass of water
x 100%
mass of hydrate

Mass of 10H2O = 180 g
Mass of Na2CO3 ·10H2O = 286g
Percent H2O = 180 g
286 g
x 100% = 62.9% or
63%
IV. Electrolytes
A. Electrolytes – compounds that
conduct an electric current in solutions
 All ionic compounds are electrolytes
 Compounds that don’t conduct an
electric current are called
nonelectrolytes – not composed of
ions, includes many molecular
compounds (covalent compounds)
IV. Electrolytes (cont.)
-
-
+
salt
Strong
Electrolyte
solute exists as
ions only
+
-
water
+
sugar
Weak
Electrolyte
NonElectrolyte
solute exists as
ions and
molecules
solute exists as
molecules only
Some Examples of Strong Electrolytes, Weak Electrolytes
and Nonelectrolytes
Strong Electrolyte
Weak Electrolyte
Nonelectrolyte
Acids
(HCl, HBr, HI,
HNO3, HClO4)
Heavy metal
halides
HgCl2, PbCl2
Most organic
Compounds
Bases
(NaOH, KOH)
Water (very weak) Glucose
Ions, Ionic
compounds
(NaCl, KCl, CaCl2,
KClO3, MgSO4)
Also called salts
Organic (acids &
bases)
Acetic acid,
aniline
Glycerol
V. Heterogeneous Mixtures
A. Suspensions – mixtures from which particles
settle out upon standing and the average
particle size is greater than 100 nm in diameter.


Clearly identified as two substances
Gravity or filtration will separate the particles
B. Colloids – heterogeneous mixtures containing
particles that are between 1 nm and 100 nm in
diameter


Appear to be homogeneous but particles are
dispersed through medium
Ex: paint, aerosol spray, smoke, marshmallow,
whipped cream
V. Heterogeneous Mixtures
C. Tyndall Effect - phenomenon observed when
beam of light passes through a colloid or
suspension

Colloids exhibit the Tyndall effect
Colloid
Solution
VI. Solubility
• defined as the maximum grams of
solute that will dissolve in 100 g of
solvent at a given temperature
•based on a saturated solution
Solubility
SATURATED SUPERSATURATED
UNSATURATED
SOLUTION
SOLUTION
SOLUTION
max amount
over max amount
capable of dissolving no more solute becomes unstable,
more solute
dissolves
crystals form


Concentration Increasing
Supersaturated solutions are not in equilibrium with the solid
substance and can quickly release the dissolved solids.
Saturated solution is one that is in equilibrium with respect to
the dissolved substance. These conditions can quickly
change with temperature.
VI. Solubility (cont.)
B. Factors Affecting Solubility
1. Stirring (agitation)
 Increases
solubility b/c fresh solvent is
brought in contact with the surface of
the solute
 Surface phenomenon
VI. Solubility (cont.)
2. Temperature
 Increases solubility by increasing
kinetic energy, which increases the
collisions b/w molecules of solvent and
the surface of the solute
 Increases amount and rate of solute
dissolved
VI. Solubility (cont.)
3. Surface Area
 A smaller particle size dissolves more
rapidly than larger particles size
 Surface phenomenon
 More surface area exposed, faster
rate of dissolving
VI. Solubility (cont.)
A. Solubility Curve


shows the
dependence of
solubility on
temperature
Note: the solubility of
the gases are greater
in cold water than in
hot water
Out of the solids, which
has the lowest
solubility at 40°C?
KClO3
Which has the highest
solubility at 20°C?
KI
VI. Solubility (cont.)



How many grams of
KNO3 can be
dissolved in 100g of
water at 60°C?
110g
What is the solubility
of NH4Cl at 50°C?
50 g/L
Which salt shows the
least change in
solubility from 0°C to
100°C?
NaCl
Output side
1. Which of the
following compounds
dissolved the highest
at 20°C?
2. The lowest at 20°C?
3. Overall which
compound dissolved
the fastest from 0°C
to 100°C?
4. Name a compound
in the graph that is a
gas? How do you
know it’s a gas?
VII. Concentrations of Solutions




Concentration of a solution is a measure of
the amount of solute that is dissolved in a
given quantity of solution.
Dilute solution – contains a low
concentration of solute.
Concentrated solution – contains a high
concentration of solute.
Molarity (M) – number of moles of a solute
dissolved per liter of solution

a.k.a. molar concentration
A. Molarity
Calculates the number of moles in
1 L of the solution
Molarity (M) = moles of solute M = n
liters of solution
L

Ex 1
Calculate the molarity when 2 mol of glucose is
dissolved in 5 L of solution.
M=n
L
2 mol glucose
5 L solution
= 0.4 mol/L = 0.4 M
Ex 2
How many moles of solute are present
in 1.5 L of 0.24M Na2SO4?
M = n of solute
L of solution
1.5 × 0.24M = n
1.5 L
0.36 mol = n
“Triangle Trick”
× 1.5
Divide
n
0.24M 1.5L
Multiply
Ex 3
A saline solution contains 0.90g NaCl in
exactly 100.0 mL of solution. What is the
molarity of the solution?
M = n of solute
L of solution
What are you solving for?
Molarity (M)
1. CONVERT GRAMS TO MOLES!!
2. CONVERT mL to L.
0.90g NaCl 1 mol NaCl = 0.016 mol
58g NaCl
100.0 mL
1L
1000mL
KhDMdcm
L
= 0.1000 L
M = 0.016 mol
0.1000 L
Molarity = 0.16 M
Ex 4
How many grams of solute are in
2.40L of 0.650M HClO2?
M=n
2.40 L×0.650M = n
× 2.40 L
L
2.40 L
1.56 mol = n
1.56 mol HClO2 68g HClO2
1 mol HClO2
= 106.08g
HClO2
Title: Molarity Problems
You do not have to write the problem. You MUST show your work. BOX in your answer!
1. A solution has a volume of 2.0L and contains
36.0g of glucose. If the molar mass of glucose is
180g, what is the molarity of the solution?
2. A solution has a volume of 250 mL and contains
0.70 mol NaCl. What is its molarity?
3. How many moles of ammonium nitrate are in 335
mL of 0.425M NH4NO3?
4. How many moles of solute are in 250 mL of 2.0M
CaCl2? How many grams of CaCl2 is this?(molar
mass = 110g)
VII. Concentrations of Solutions
(cont.)
B. Making Dilutions


You can make a solution less concentrated by diluting
it with a solvent.
A dilution reduces the moles of solute per unit volume,
however, the total moles of solute in solution does not
change
M1 × V1 = M2 × V2

Volumes can be in L or mL, as long as the same units are
used for both V1 & V2
Ex 1
You need 150 mL of 0.40M NaCl and you
have a 1.0M of NaCl solution. Calculate
the volume of the NaCl solution.
V1 = 150 mL of 0.40 NaCl
M1 = 0.400M NaCl
M2 = 1.0M NaCl
Unknown = V2
M1 × V1 = M2 × V2
0.400M ×150 mL = 1.0M × V2
60. mL = V2
Ex 2
What volume of 5.00M sulfuric acid is
required to prepare 25.00L of 0.400M
sulfuric acid?
M1 = 5.00M
V2 = 25.00L
M2 = 0.400M
M1 × V1 = M2 × V2
5.0M × V1 = 0.40M ×25 L
V1 = 2.0 L
Ex 3
How many milliliters of a stock solution of
2.00M MgSO4 would you need to prepare
100.0 mL of 0.400M MgSO4?
M1 = 2.00M MgSO4
M2 = 0.400M MgSO4
V2 = 100 mL of 0.400M MgSO4
Unknown = V1
M1 × V1 = M2 × V2
2.00M × V1 = 0.400M ×100 mL
V1 = 20 mL
LEFT SIDE
Title: Dilution Problems
You do not have to write the problem. You MUST show your work. BOX in your answer.
5. In making a dilution how many mL of 4.0M HCl are
needed to make 250 mL of 0.30M HCl?
6. In making a dilution how many mL of a 3.00M
NaBr solution are needed to make 175 mL of
0.400M NaBr?
7. How many milliliters of a stock solution of 2.00M
MgSO4 would you need to prepare 100mL of
0.400M MgSO4?
8. How many milliliters of a stock solution of 4.00M
KI would you need to prepare 250.0mL of 0.760M
KI?
Adding to the mole road map!!
Solutions
Molarity:
5.0 M = 5.0 mol
1L
VIII. Using Molarity in
Stoichiometry
Zn + 2 HCl → ZnCl2 + H2
How many milliliters of 3.00M HCl are required to
react with 17.35 g of zinc?
17.35 g Zn
1 mol Zn
2 mol HCl
65 g Zn
1 mol Zn 3.00 mol HCl
1 L HCl
1000 mL HCl
1 L HCl
= 177.95 mL HCl
Using Molarity in Stoichiometry
Cu + 2 AgNO3 → 2 Ag + Cu(NO3)2
How many grams of copper are required to react
with 40.0 mL of 9.0M AgNO3?
40.0 mL AgNO3
1 L AgNO3
9.0 mol
AgNO3
1000 mL
1 L AgNO3
AgNO3
1 mol Cu
2 mol AgNO3
64 g Cu
1 mol Cu
= 11.52 g Cu
IX. Percent Solutions
If both solute and solvent are liquids the concentration of the
solute is expressed as a percent.
Percent by volume (%(v/v)) = volume of solute
solution volume
X 100
For solutions of solids dissolved in liquids is percent
(mass/volume).
Percent (mass/volume) (%(m/v)) = mass of solute (g) X 100
solution volume (mL)
IX. Percent Solutions
Ex 1
What is the percent by volume of ethanol
(C2H6O), or ethyl alcohol, in the final solution
when 85 mL of ethanol is diluted to a volume
of 250 mL with water?
% (v/v) = 85 mL ethanol X 100
250 mL solution

= 34% ethanol (v/v)
IX. Percent Solutions
Ex 2
A solution of glucose (C6H12O6) contains 2.8 g
in 20.0 mL of solution. What is the percent
(m/v) of the solution?
% (m/v) = 2.8g
X 100
20.0 mL

= 14% glucose (m/v)
Why do we get to make ice
cream?
X. Freezing-Point Depression
 the
difference in temperature
between the freezing point of a
solution and that of the pure solvent.
 The salt depresses the freezing point
of water because it disrupts the
crystal formation of the water
Why do we get to make ice
cream?
When you add salt to the ice, it lowers the freezing point of
the ice, so even more energy has to be absorbed from the
environment in order for the ice to melt. This makes the ice
colder than it was before, which is how your ice cream
freezes
Compounds that break into ions upon dissolving, like NaCl
breaks into Na+ and Cl-, are better at lowering the freezing
point than substances that don't separate into particles
because the added particles disrupt the ability of the water
to form crystalline ice.
The more particles there are, the greater the disruption and
the greater the impact on particle-dependent properties
(colligative properties) like freezing point depression.
Ice Cream Lab
1. Put all ingredients on my desk.
2. While I am mixing the ingredients. You need to
prepare your coffee can and freezer bag.
3. To prepare your coffee can, put a 5cm layer of ice then
sprinkle 1cm of rock salt, then add another layer of ice
then rock salt. (3/4th full, leave room to put in the ice cream
bag)
4. Then you will get your freezer zip lock bag and duct tape
the 3 side edges.
5. I will then put ice cream in your bag and you will seal the
opening with duct tape.
6. Place the taped bag of ice cream inside the middle of the
coffee can and duct tape the lid onto the coffee can.
7. Roll your coffee can outside for 20 minutes!
8. When complete, take bag out of can and rinse off bag.
9. Then enjoy ice cream!