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“Water, water
everywhere
The very boards did
shrink.
Water, water everywhere
Nor, any drop to drink.”
Rhyme of the Ancient Mariner
Samuel Coleridge
Heterogeneous matter:
Mixtures
Review of suspensions and
colloids with concentration on
solutions.
Introduction to Mixtures


A mixture is a combination of different
substances where each substance
retains its chemical properties.
Generally they can be separated by
non-chemical means such as filtration,
heating, or centrifugation.
3 Types of Mixtures

Heterogeneous



Suspensions
Colloid
Homogenous

solutions
Suspensions


A heterogeneous mixture in which some
of the particles will settle (separate)
upon standing.
 may be filtered
Example:
 oil/water
 clay/water
Colloids
A mixture containing particles that are
intermediate in size between those of
suspensions and true solutions.
Example: hair spray, cold cream, milk

Are usually cloudy or milky in
appearance, but look clear when very
dilute.
Particles CANNOT be separated by
filtering and do not settle.

Solutions



The same throughout.
Best mixed.
Examples




Kool-aide
iced tea
Air
metal alloys
®


gold jewelry
brass
Solutions
Introduction to solutions and
calculations for solutions
Introduction to solutions



A solution is a homogeneous mixture
where all particles exist as individual
molecules or ions.
Particles are so small (less than 1nm)
that the mixture never settles out.
Will not scatter light. Does not display
the Tyndall effect.
Tyndall effect

The Tyndall Effect is caused by reflection of
light by very small particles in suspension or
colloid. It is often seen from the dust in the
air when sunlight comes in through a
window, or comes down through holes in
clouds. It is seen when headlight beams are
visible on foggy nights, and in most X-File
episodes when Moulder and Sculley check out
some dark place with flashlights.
Tyndall effect

Components of a solution

Solvent




Solute
the dissolving medium

greater in amount
typically a liquid
usually water




aqueous
sometimes alcohol

tincture

Substance
dissolved
lesser in amount
frequently a solid
Solution Formation


If it will dissolve depends on the
solute and solvent.
Water is called the universal
solvent because it dissolves most
substances.
Liquid solvent and solute


Miscible – liquids that dissolve freely in
one another in any proportion are
completely miscible.
Immiscible – liquids solutes and
solvents that are not soluble in each
other.
Like dissolves like


A polar solvent (water) dissolves Salt
NaCl
(ionic)
most ionic solutes.
A polar solvent dissolves polar
Water(polar)
solutes.


A polar solvent does NOT
dissolve nonpolar solutes.
Nonpolar solvents (oil)
dissolve nonpolar solutes.
Solubility practice problem:


Determine whether the following
substances will dissolve in water
(polar), carbon tetrachloride (nonpolar),
and/or alcohol. (see worksheet 1)
MgCl2, F2, methanol (note: -ol ending
indicates and alcohol)
Solubility practice problem ans
Solutes
MgCl2
Water
alcohol
X
F2
Methanol
CH3OH
CCl4
X
X
X
X
X
Solubility practice problems

At this time complete the worksheet
solubility (polar vs. nonpolar)
Electrolytic Solutions


Electrolytic solutions
are composed of
substance that
dissolves in water to
give a solution that
conducts electric
current.
Ionic compounds
dissolved in water.
Nonelectrolytic solutions



substances dissolved in
water to give a solution that
does NOT conduct electricity
solutions containing neutral
solute molecules do not
conduct electric current
because there are no mobile
charged particles available
covalent compounds
dissolved in water
Electrolyte

At this time stop and complete the
electrolyte worksheet.
You are familiar with electrolytic
solutions. In the Ionic or Molecular lab
you tested conductivity of solutions .
(You dipped the wires in solutions and
the light lit up if the solution was ionic.)

What ionic solids dissolve?


To determine which ionic salts dissolve in
water you must consult a Table of
Solubility Rules.
Luckily for you one is included in you EOC
reference tables. Consult them now (page 6).
NCDPI Reference tables for Chemistry (adopted
2000)SOLUBILITY RULES:
SOLUBLE:

All Nitrates, Acetates, Ammonium and Group I salts

All Chlorides, Bromides, and Iodides, except Silver, Lead, and Mercury(I)

All Fluorides except Group II, Lead(II), and Iron(III)
 All Sulfates except Calcium, Strontium, Barium, Mercury, Lead(II), and
Silver
INSOLUBLE:

All Carbonates and Phosphates except Group I and Ammonium

All Hydroxides except Group I, Strontium, and Barium

All Sulfides except Group I, II, and Ammonium

All Oxides except Group I
INSOLUBLE means a precipitate forms when equal volumes of 0.10 M
solutions or greater are mixed
Solubility example problem #1

dentify the precipitate(s) and aqueous
component(s) that form when aqueous
solutions of zinc nitrate and ammonium
sulfide are combined.
Solubility example #1:
Plan

Write the possible double-replacement
reaction between reactants and use the
solubility rules to determine which
products are soluble and which will
precipitate (not soluble).
Solubility example #1:
solution




U:
K:
P:
S:

state of products
Zn(NO3)2 and (NH4)2S (I checked charges)
Write equation and consult rules.
Zn(NO3)2 + (NH4)2S  2NH4NO3 + ZnS
Consult chart
Solubility example #1:
solution

S: Zn(NO3)2 + (NH4)2S  2NH4NO3(aq) + ZnS(s)


Chart reveals that ammonium nitrate is soluble. Zinc sulfide is
not soluble and is therefore a precipitate according to the table.
All Nitrates, Acetates, Ammonium and Group I salts are soluble.
Solubility example #1:
solution

S: Zn(NO3)2 + (NH4)2S  2NH4NO3(aq) + ZnS(s)




Chart reveals that ammonium nitrate is soluble. Zinc sulfide is
not soluble and is therefore a precipitate according to the table.
All Nitrates, Acetates, Ammonium and Group I salts are soluble.
All Sulfides except Group I, II, and Ammonium are insoluble.
Note: You do not have to do a ukps for each of these problems.
You do need to justify your answers.
Solubility rules practice


At this time stop and work the solubility
rules practice problems.
You must complete the bold portion of
worksheet 5 (#1, 2, 4, 8)

Check your answers with your lab
partners.
Solubility

If you determine that the salt will
dissolve, how much of it will dissolve?
Solubility


The amount of a substance that
dissolves in a given quantity of a
solvent at a given temperature to
produce a saturated solution is its
solubility.
Unit = grams solute / 100 g solvent
NCDPI Reference tables for Chemistry (adopted 2000)
Solubility Curve
 Once you determine that
This is no longer
on the reference
a substance will dissolve
tables. It will beconsult
included
with a
the solubility
curve to determine how
question as needed.
much of it will dissolve
at a given temperature.

For example after adding
37 grams of NaCl to 100
ml of tap water no more
will dissolve. Any
additional salt settles to
the bottom.
Saturated


Same amount of solid
solidifying as dissolving.
Saturated solution – contains
the maximum amount of
solute for a given amount of
solvent at constant
temperature.
This is an equilibrium; the
same amount changes into a
solid as is dissolving in the
liquid.
Saturation


Unsaturated – a solution that contains less
solute than a saturated solution. (Less
than it could hold.)
Supersaturated – a solution which contains
more solute than it can theoretically hold
at a given temperature.


Additional solute will cause all the “extra”
solute to crystallize and precipitate.
this is how rock candy is made
Practice problem using
Solubility Curve

When 100 grams of water saturated
with KNO3 at 70oC are cooled to 25oC,
what is the total number of grams of
KNO3 that will precipitate?
Practice problem using
Solubility Curve


At 70oC 135g of KNO3
would be dissolved in
a saturated solution.
At 25oC only 40g of
KNO3 could be
dissolved.
Therefore 95grams of
KNO3 will precipitate.
Henry’s law


The solubility of a gas in a liquid is
directly proportional to the partial
pressure of that has on the surface of
the liquid.
Pressure has little effect on the
solubility of liquid or solid solutes.
Rate of Solution

How fast a solute will dissolve depends on:
 Size of solute particles


Agitation/stirring


Since dissolving occurs on the surface, increasing
surface area (power) allows more solvent to reach
more solute and speeds dissolving.
Moves solute away from solid so solvent can reach
more solute.
Temperature


For solids a warmer solvent moves faster and farther
apart thus increases rate of dissolving.
However, for gas solutes, as temperature of solvent
increases the solubility of a gas decreases.
Solubility Curve problems

At this time stop and complete the
solubility curve problems.
Concentration




Refers to how much solute is dissolved
Dilute means only a little solute is
dissolved
Concentrated means a lot is dissolved
These are NOT quantitative values just
words you should be familiar with.
Concentration Calculations

There are two major concentration
words that are quantitative in nature:
Molarity and Molality
Molarity



As is clear from its name, molarity
involves moles.
It is calculated by taking the moles of
solute and dividing by the liters of
solution.
Molarity = moles of solute
liters of solution
Molarity Example #1
Suppose we had 1.00 mole of sucrose (about
342.3 grams) and proceeded to mix it into
some water. It would dissolve and make
sugar water. We keep adding water,
dissolving and stirring until all the solid was
gone. We then make sure that when
everything is well-mixed there is exactly 1.00
liter of solution.
What would by the molarity of this solution?
Example #1 answer
U: molarity
K: 1.00 mol sucrose; 1.00 L solution
p: Molarity = moles of solute
liters of solution
s: M= 1.00 mol
1.00 liter solution
=1.00 mol/L = 1.00 M
Molarity Example #1 cont’d



Note, some textbooks make the M
using italics and some put in a dash,
like this: 1.00-M
When you say it out loud, say “one
point oh oh molar”
Never forget to replace M with mol/L
when you do calculations. “M” is just
shorthand.
Molarity Example #2

What is the molarity when 0.750
mol is dissolved in 2.50 L of
solution?
U: molarity
K: 0.750 mol; 2.50 L solution
p: Molarity = moles of solute
liters of solution
s: M= 0.750 mol
2.50 L solution
=0.300 mol/L = 0.300 M
Molarity Example #3 (using grams
which is much more common)

Suppose you had 58.44 g of NaCl and you dissolve it
in exactly 2.00L of solution. What would be the
molarity of the solution?
p: convert grams to moles
s:
Molarity = moles of solute
liters of solution
58.44 g NaCl 1 mol NaCl =1.00 mol NaCl
58.44 g NaCl
M=
1.00 mol NaCl
2.00 L solution
=0.500 mol/L = 0.500 M
Example #4

As a class calculate the molarity of 25.0g of
KBr dissolved to 750.0 mL solution.
Work the above problem.

Answer: 0.280M
Example #5


80.0 grams of glucose (C6H12O6) is
dissolved in 1.00L of solution. What is
the molarity?
Answer: 0.444M
Molarity

Notice how the phrase “of solution”
keeps showing up. The molarity
definition is based on the volume of
solution, NOT the volume of pure water
used. For example, to say: “A one
molar solution is prepared by adding
one mole of solute to one liter of water”
is totally INCORRECT.
Problems

At this time stop and complete the
Molarity Worksheet.
Dilution
Definition and Calculations
Dilution


To dilute a solution means to add more
solvent without the addition of more
solute.
Since the amount of solute stays the
same: moles before = moles after
Dilution equation



From the definition of molarity we get that
moles of solute equals molarity times the
volume.
If we are just diluting (adding water)
the moles of solute remains the
same.
Therefore: M1 V1 = M2 V2
Dilution example #1


53.4 mL of a 1.50M soln of NaCl is on hand,
but you need some 0.800M solution. How
many mL of 0.800M can you make?
S: (1.50mol/L)(53.4mL)= (0.800mol/L)(V2)
V2 = 100. mL
Dilution Practice problems

Complete the dilution practice problems
Boiling point elevation
Freezing point depression


When a solute is added to a solvent the
freezing point of the solvent decreases
and the boiling point increases.
For example, “salt” is added to the ice
in an ice cream freezer so that the
temperature of the ice (freezing point)
will decrease and the milk can freeze
(become solid)
Boiling point elevation
Freezing point depression


Likewise salt (any solute) added to a
liquid being heated raises the boiling
point.
If salt is added to water then the water
is heated, instead of the water boiling
at 100oC it would boil at a higher
temperature, maybe 104oC.