Solutions Powerpoint

 Solution Formation
 Solubility
 Solution concentration
 6. Solutions are homogeneous mixtures of two or more
substances. As a basis for understanding this concept:
 a. Students know the definition of solute and solvent
 b. Students know how to describe the dissolving process at the
molecular level by using the concept of random molecular
 c. Students know temperature, pressure, and surface area
affect the dissolving process
 d. Students know how to calculate the concentration of a solute
in terms of grams per liter, molarity, parts per million, and
percent composition.
 By the end of the lesson, the student will:
 Understand the dissolving process at the molecular level.
 Understand how the structure of the water molecule
allows it to dissolve many substances
 Know the factors which influence:
solubility and why
dissolving rate and why
 Know how to calculate the concentration of a solution
Solution: a homogeneous mixture of two or more substances
Solvent: a substance which dissolves another substance
Solute: the dissolved particles; whatever is being dissolved
Solution Formation Vocab:
 Soluble: a substance that dissolves in a solvent
 Insoluble: a substance that does not dissolve in
a solvent
Examples of Solutions
 Gas in Liquid:
Carbonic Acid (carbonation): CO2 in H2O
 Liquid in Liquid
Vinegar: Acetic Acid in water
 Solid in Solid
Brass: copper and zinc
Animation: Dissolving Process
(53 sec)
&feature=related (42 sec)
The dissolving process at the
molecular level:
Let’s look at the structure of
the water molecule
 The structure of the water molecule creates a
polar molecule due to the differences in
electronegativities of the oxygen and hydrogen
 Partial negative charge on the oxygen attracts the
partial positive charge on the neighboring
molecule. Molecules arrange themselves so as to
have opposite charges aligned, pulling it away
from its ‘structure’, “one ion or molecule at a time”
Dissolving Process
 Once released, the ions are surrounded by water molecules.
 The dissolving process is reversible
 The dissolved solute moves around in the solution and when it
comes into contact with un-dissolved solute particles it recrystallizes, meaning it returns to the solid state. And the
process repeats itself.
Na+(aq) + Cl- (aq)
 When the rates of the dissolving and
recrystallization become the same, the
solution is saturated at that temperature. A
dynamic equilibrium is reached,
meaning the rate of the forward reaction is the
same as the reverse reaction.
 Where does dissolving occur?
 On
the surface of a substance
To increase dissolving rate, contact between
solvent and solute must increase
 3 common ways to increase the collisions between solute and
Amount of Surface Area Exposed
 Temperature: raising the temperature increases
the kinetic energy of the particles, resulting in
more frequent and forceful collisions.
Amount of Surface Area Exposed
 Amount of Surface Area Exposed: breaking the
solute into smaller pieces increases its surface
area. A greater surface area allows more collisions
to occur and therefore, faster dissolving
Agitation (mixing, stirring, etc.)
 Agitation: stirring moves dissolved solute
particles away from the contact surfaces more
quickly and thereby allows new collisions
between solute and solvent particles to occur.
Without stirring, solvated particles move away
from the contact areas slowly.
 the maximum amount of solute that will dissolve in a
given amount of solvent at a specified temperature
and pressure
3 Types of Solutions
 Saturated solution
 Unsaturated solution
 Supersaturated solution
Saturated solution:
contains the
maximum amount of
solute. You cannot
dissolve any more
solute in the solvent.
Unsaturated solution: contains less than
the maximum amount of solute. You can
dissolve more solute in the solvent.
Supersaturated solution
contains more than the normal maximum amount of
solute. This is usually achieved by heating the
solution in order to dissolve more solute, then the
solution is cooled. This makes a supersaturated
solution. (ie.Rock candy; hot mineral springs)
Supersaturated Solutions
3 Factors Affecting Solubility
 1. Temperature:
many substances
are more soluble at
high temperatures
than low
Exception: Gases
Solubility of Gases with
Increased Temperature
As Temperature increases,
Solubility of gases decreases
Pressure affect on gases
2. Pressure:
 affects the solubility of
gaseous solutes. The
solubility of a gas in any
solvent increases as the
pressure above the
solution increases
3 Factors Affecting Solubility
 2. Pressure: affects the
solubility of gaseous
solutes. The solubility of
a gas in any solvent
increases as the
pressure above the
solution increases
3 Factors Affecting Solubility
 2. Pressure: affects
the solubility of
gaseous solutes.
The solubility of a
gas in any solvent
increases as the
pressure above the
solution increases
3 Factors Affecting Solubility
 3. Intermolecular forces:
“like dissolves like;”
meaning polar substances dissolve polar substances
and non-polar substances dissolve non-polar
 Miscibility: the ability to mix without separating into
two phases
Miscible: substances that mix together
ie. Vinegar and water
Immiscible: substances that do not mix together
ie. Oil and water
the amount of solute in a solution
 Dilute solution: contains small amounts
of solute (dissolved particles)
 Concentrated solution: contains large
amounts of solute
the number of grams of solute per 100 grams of solution
 Mass Percent = mass of solute x 100%
mass of solution
 Calculate the mass percent of NaCl in a solution
containing 15.3 g of NaCl and 155.0 g water. (8.98%NaCl)
 Calculate the mass percent of a solution containing 27.5 g of
ethanol and 175 mL of water. The density of water is 1.0 g/mL.
Example 3
 You have 1,500 g of a bleach solution. The percent by mass of
the solute sodium hypochlorite, NaOCl, is 3.62%. How many
grams of NaOCl are in the solution?
(54.3 g)
MOLARITY (M): the number of moles of
solute dissolved per liter of solution.
Note: Here, volume refers to the total volume of the
solution,not the volume of the solvent.
 Molarity (M) = moles of solute
liters of solution
 M is pronounced “molar”
Calculate the molarity if water is added to 2.0 mol of
glucose to give 5.0 L of solution.
(0.40 M)
 A solution contains 0.900 g of NaCl per 100 mL of solution.
What is the molarity? (0.154 M)
Example 3
 How many moles of Na2SO4 are needed to prepare 1.5L of 0.20
M Na2SO4?
(0.30 moles solute)
Example 4
 How many grams of solute are contained in 250 mL of 3.0
(30. g NaOH)
Example 5
 How would you prepare 250 mL of a 0.500 M
NaCl solution? (7.31 g NaCl)
Preparing a Solution from a
adding solvent to a concentrated (stock)
solution to make a less concentrated
Moles Before= Moles After
 Number moles before dilution = number moles after
 Use the dilution equation
M1 x
V1 = M2 x V2
 How many milliliters of 12.0 M HCl are needed
to prepare 750. ml of 0.250 M HCl?
(15.6 mL)
 How would you prepare 5.00 L of a 1.50 M
KCl from a 12.0 M stock solution? (0.625 L)
Making a dilute solution from
a concentrated solution
Example 3
 What is the molarity of the resulting solution when
the following mixture is prepared? 150.0 mL of
water is added to 55.0 mL of 6.50 M NaOH. (1.74 M)
Example 4
 How much water must be added to 125 mL of a
4.50 M NaCl solution to produce a 2.75 M
solution? (80. mL)
1. Construct a chart of values
2. Rearrange equation so that unknown is
3. Convert grams to moles, if necessary
4. Calculate and report to correct # of sig. figs!
 There are several different ways to solve solution
stoichiometry problems. Here are a few
guidelines/suggestions to get you started:
 Write a balanced chemical equation
 use Molarity as mol…
 ***allows conversion between moles and volume
or concentration and volume
 use mol-mol ratio if necessary
 there may be limiting reagent problems…in this
case determine how many moles there are for
each substance
 sample mini road map:
…from moles you can go to L (volume) or grams
(mass)…or even particles via Avogadro’s #!
Example 1
 1) Calculate the mass of solid NaCl that must be added
to 1.50 L of a 0.100 M AgNO3 solution to precipitate all
the Ag+ ions in the form of AgCl
Example 2
 2) How many milliliters of 0.0500M Pb(NO3)2 are
needed to react with 2.00L of a 0.0250M Na2SO4 solution
in order to produce a precipitate? How many grams of
precipitate are formed?

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