Solutions/
Concentrations
Georgia Performance Standard SC7:
Students will
characterize the properties that describe solutions and the
nature of acids and bases.
Review of Aqueous Solutions
• Ionic compounds most
readily dissolve in
water due to extreme
polarity
• Polar covalent
compounds also
dissolve in water
• Nonpolar compounds
don’t
Significant Vocabulary
•
•
•
•
•
•
•
Solution
Solute and Solvent
Solvation (n); Solvate (v)
Soluble and Insoluble
Solubility
Electrolyte and Nonelectrolyte
Saturate, Unsaturated, and Supersaturated
Determines IF Solute Will Dissolve...
• Like dissolves like:
Polar molecules will
solvate with polar molecules. Nonpolar
molecules will solvate with nonpolar molecules.
BUT, polar and nonpolar will not form solutions
together.
Determines Speed of Dissolving
BRING SOLUTE IN
CONTACT WITH SOLVENT
1. Agitation: Create more collisions mechanically
2. Temperature: More kinetic energy creates more
collisions
3. Surface Area: Dissolving process is a surface
phenomenon, the more surface of the solute that is
exposed the faster the solvation
Determines How Much Will Dissolve
SOLUBILITY:
HOW MUCH WILL DISSOLVE
1. Temperature: solubility of solid solute
increases as the temp. increases; solubility of
gaseous solute decreases as temp increases
ex. Hot water bubbles, thermal pollution
2. Pressure: solubility of gaseous solute
increases as the pressure increases
-Henry’s Law ex. Soft drinks
Temperature & Solubility
• Higher the
temperature, the
more solid will
dissolve in a liquid
• Higher the
temperature, the less
gas will dissolve in a
liquid
Henry’s Law:
Pressure & Solubility
• At a given temperature,
the solubility of a gas is
proportional to the
pressure of the gas
above the liquid.
The higher the pressure, the more carbon dioxide will
dissolve in the syrup giving a less “flat” taste.
Solubility Graph
Concentration of Solutions
The concentration of a solution is a measure of
how much solute is dissolved in a specific
amount of solvent or solution.
Molarity: most common units of solution
concentration; # of moles solute dissolved in
one liter of the solution
Molarity: moles of solute
liter of solution
Example 15.3 (page 531)
Calculate the molarity of a solution prepared
by dissolving 11.5 g of solid NaOH in enough
water to make 1.50L of solution.
Example 15.4 (page 532)
Calculate the molarity of a solution prepared
by dissolving 1.56 g of gaseous HCl into
enough water to make 26.8 mL of solution.
The most common concentration
term in chemistry is Molarity (M),
but chemists also report
concentration in molality (m)
sometimes.
molality: moles of solute
kg of solvent
If given grams, remember to
change it to moles.
Preparing a Solution in the Lab
• You’ve learned about concentration (Molarity and
molality). Therefore, you should be able to
create your own solutions for use in the lab from
now on.
– Example:
• 0.5 M HCl reacts with Mg(s)
• If I gave you HCl powder, what would you do to make the
solution?
• Watch these kids, and assess your plan.
• Choose a lab group and station. Draw an assignment from
the cup, and make the solution. Write your steps (including
materials) as you go.
• ALSO: Solve 15-2 Practice Problems WS 6, 8, 10-12, 14-16.
Diluting Solutions
If you already have a solution molarity, but you
want a different molarity:
Example: Your lab asks you to use 250 mL of
0.25M HCl, but you only find a jug of 6M HCl
in the stockroom.
USE THIS EQUATION:
M1V1 = M2V2
Example 15.8 (page 539)
What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of a 0.10 M H2SO4 solution?
*Be sure to solve the Dilution Worksheet in your
practice packet!
Using Molarity as a Conversion Factor
• You have learned to calculate molarity using the
equation:
– Molarity = moles of solute / liter of solution
• Did you realize that the calculation is simply a
ratio of solute to solution?
• Ratios (ie mole ratios, energy to mole ratios) can
be used as conversion factors in stoichiometry.
• THEREFORE, molarity can be written into our
dimensional analysis charts to solve
stoichiometry problems.
Solutions Stoichiometry
• What volume of 1.5 M HCl is needed to react with 21.5
grams of NaOH?
• What is the molarity of a solution of H2S if 48.5mL are
required to titrate 35.6mL of 0.35M Fe(OH)3 solution?
• A white precipitate forms when 200 mL of 0.200M
K3PO4 solution is mixed with 300 mL of 0.250 M CaCl2
solution. What mass of precipitate will form?
• BE SURE TO SOLVE THE PRACTICE PROBLEMS IN YOUR
PACKET!
Physical properties of solution are different
from the physical properties of the solvent.
Some properties are different simply
because there are “foreign” particles
(solute) in the solvent. Colligative properties
of solutions depend only on the number of
solute particles.
Boiling Point Elevation
• More solute particles means that vapor pressure
is lower which means that more kinetic energy is
needed to make vapor pressure equal
atmospheric pressure!
• Change in temperature is calculated:
– ∆Tb = Kb m i
• Since boiling point increases...ADD the
change.
Kb will be given.
Freezing Point Depression
• Liquid particles get into an orderly pattern to become
a solid. The solute particles disrupt the orderly
pattern causing more kinetic energy to be drawn
from the solution for it to freeze!
– ∆Tf = Kf m i
• Since freezing point decreases...SUBTRACT the
change.
• Kf will be given
Molality
• Another unit for concentration
• m=moles of solute per kilogram of solvent
• page 520
Solutions Stoichiometry
• A white precipitate forms when 200 mL of
0.200M K3PO4 solution is mixed with 300 mL
of 0.250 M CaCl2 solution. What mass of
precipitate will form?
• What volume of 1.5 M HCl is needed to react
with 21.5 grams of NaOH?
• What is the molarity of a solution of H2S if
48.5mL are required to titrate 35.6mL of
0.35M Fe(OH)3 solution?