Solutions
Ch 15
Mixtures (review)
• Mixture – has variable composition, different
types of parts. For example:
• air,
• water from the tap,
• hot chocolate,
Homogenized milk
• paint,
• 14k gold, stainless steel (metal mixtures are
called alloys)
• Heterogeneous “hetero” means different
• Homogenous “homo” means same
Non-homogenized milk
• Solution = homogenous mixture
Solutions
• The solvent is the substance that the
solute dissolves in
• So, if I stir sugar into my coffee, which
is which?
• Once the solute is dissolved, as long as
there’s none left at the bottom, what
do you call the new coffee?
• A homogenous mixture, or solution
• What if there’s sugar left at the
bottom? That would be called…?
• Saturated – there’s more solute than
the solvent can dissolve
Solutions
• Solution – mixture of two or more substances
in a single physical state.
• Characteristics:
• Homogeneous
• Does not settle out
• Cannot be separated by filtering
Solutions
• Solutions properties are:
• the same throughout the solution
• different from the individual properties of
their constituents (parts that make them up)
• i.e., salt and water have different individual
properties than salt water.
• Solubility refers to a solute’s ability to be
dissolved by the solvent.
• A substance is said to be insoluble if it stays in
its original state in the solvent
Properties of Solutions
Solutions can be made up of any combination of states of matter
for their solutes and solvents
• Can you think of an example • Gas in gas:
• Oxygen in nitrogen (71 % of
for each?
air)
• Liquid in liquid:
• Solid in liquid:
• concentrated syrup in water • Salt in water (= sea water)
to make soda mix
• Solid in solid:
• Gas in liquid:
• Copper in mercury to make
• CO2 in soda
dental amalgam
– (less gas can dissolve into a
• All metal solutions are
warm soda than a cold soda –
called ____
that’s why warm sodas loose
• Alloys
their fizz)
Properties of Solutions
• concentration = amt. of solute in a solvent. A
concentrated solution has a relatively high amount of
solute.
• There is a limit to the amount of solute that can be
dissolve into a solution.
• Eventually, the solution will contain so much solute that
no more solute can dissolve into it.
• At this point, the solution is considered to be
• Saturated - the solution is at its max. concentration for
the given temperature and pressure.
• You can tell a solution is saturated when the solute
precipitates out.
Properties of solutions
• Unsaturated solutions still have “room”
for more solute to dissolve into them.
• Anyone who has ever tried to dissolve
sugar into iced tea knows that it is much
harder to do than if the tea were hot.
• Iced tea simply has a lower saturation
point than hot tea does.
• So the saturation points can vary with
temperature and pressure.
• Usually, the hotter the solution, the
higher the saturation point. This is not
true for gases dissolved into liquids,
though – why?
Properties of solutions
• A supersaturated solution actually holds more
solute than its saturation point would dictate.
• A solution can become supersaturated when it
is brought to saturation, heated to raise the
saturation point, and then allowed to cool
carefully.
• At the cooler temperature, any slight
disturbance will trigger the precipitation of
the solid back out of the solution.
Solubility
• Water is the universal solvent because of its polarity
• Substances that are dissolved in water are said to
be aqueous.
• Most Ionic compounds are soluble in water
• Why?
• Ionic compounds are polar
• The polarity of the water molecules pulls the ionic
compounds apart.
• Some covalent compounds
solubility
dissolve in water because of
their polarity.
• Sucrose (table sugar) is polar,
and dissolves.
• Covalent substances that are
Like dissolves like. nonpolar are insoluble in water
• Oil and other fats are non-polar
•Polar dissolves polar.
and don’t dissolve in water.
•Nonpolar dissolves nonpolar
• Soap is non-polar, so dissolves
grease.
Rate of Dissolving
•
•
•
•
Affected by 3 factors:
Temperature
Agitation (mixing)
Surface area of solute
molarity
• Number of moles of solute dissolved in a liter of
solution
• Represented by M
• i.e. To make a 1M solution of aqueous NaCl,
measure a mole (58.44g) of NaCl and add water
to 1 Liter mark.
• Example problem: 5.7 g KNO3 in 233 mL solution.
What is the solution molarity?
molarity
• Example problem: 5.7 g KNO3 in 233 mL
solution. What is the solution molarity?
• Convert grams to mole of solute
• 5.7 g KNO3 mole KNO3
= 0.056 mol KNO3
•
101.11 g KNO3
• .056 mol KNO3 1000mL = 0.24 M KNO3
• 233mL
1L
molality
• Number of moles of solute dissolved in 1 kg of
solvent
• m = mol/kg
• Useful for studying colligative properties
• 1 liter of water = 1 kg
• All other solvents must be weighed, as they
have a different specific gravity and molality is
about the mass of solvent.
molality
• To make a 1m aqueous solution of NaCl,
measure 1 kg (=1 L) of water (solvent) and add
in 1 mole (58.44g) NaCl
molality
• What is the molality of a dentist’s amalgam?
Amalgam is made up of 70% mercury and 30%
copper by mass.
• Determine which is solute and which is solvent.
Mercury is solvent, since there is more mass of it.
• So convert the copper to moles
• 30 g Cu mol
= .47 mol Cu
•
63.55 g
• Now convert to molality units of mol/kg
• .47 mol Cu 1000g = 6.7 mol = 6.7 m
• 70 g Hg
kg
kg
Properties of Solutions
• Some physical properties of solutions are
dependent upon the solute to solvent ratio, or
the molality of the solution.
• Freezing point and melting point are examples of
these colligative properties
• Another factor affecting the freezing and boiling
points is the dissociation factor of the solute
• The dissociation factor is the number of particles
a solute breaks into when it dissolves.
• If the solute dissociates, it has a greater effect on
the temperature change.
• The freezing point is always lowered
• And the boiling point is always raised
Calculating changes in boiling and
freezing points
• The amount of change of the freezing point is Δ TF
• The amount of change of the boiling point is Δ TB
• The dissociation factor is d.f.
– i.e. NaCl’s d.f. is 2
– Covalent compounds’ d.f.’s are 1, since they stay whole in
solutions.
• In addition, every substance has a certain constant
which has been experimentally determined and is
always given, kF and kB
• The constant is related to the properties of the solvent
Calculating colligative property
• To calculate the change in boiling or freezing
point based on the molality:
• Δ TF = m d.f. kF
• Δ TB = m d.f. kB
• What is the new boiling point of water if 100.
g of glucose (C6H12O6) is dissolved into 750.
mL of H2O?
Calculating Colligative Property
• What is the new boiling point of water if 100. g of
glucose (C6H12O6) are dissolved into 750. mL of
H2O?
• Δ TB = m d.f. kB ,
kB of H2O = 0.52 °C/m
• First calculate the molality
• m = moles solute/mass solvent
• mol = 100 g glucose 1 mol
= 0.554 mol glucose
•
180.56 g
• d.f. of sugar is 1, as it is a covalent molecule
• Δ TB = 0.554 mol x 1 x 0.52 °C L = 0.384 °C
•
0.750 L
1 mol
Dilution
• Usually acids come from chemical supply
companies in highly concentrated forms.
• And you have to dilute them to a lower
molarity.
• You NEVER add the water to acid
• So you have to have a way to calculate how
much water to pour the concentrated acid into
to dilute it.
Dilution
• When we dilute a substance, it is always
assumed that we are adding water, the
universal solvent.
• The actual number of moles of the solute
never changes.
• Only the amount of water changes to reduce
the molarity of the solution.
• Extra dilution calculations:
– Text p. 556 #33 & 34
Dilution Calculations
• When you need to dilute a solution, you know
3 of the following 4 things:
• the molarity of what you started with (M1),
• the new molarity of the solution you want to
make (M2)
• The volume of the solution you start with (V1)
• The volume of the new solution (V2)
• Use the dilution equation M1V1= M2V2 to solve
for the last variable.
Dilution Calculations
• Remember that V1 is the volume of original
concentrated acid that must be added to
water.
• And that V2 is the amount of diluted solution.
• So the amount of water to add is:
• final volume (V2) – amt. of concentrated
solution (V1)
Dilution Calculations
• #31a in text p. 556
• Calculate the molarity that results when 250 mL
of water is added to 125 mL of .251 M HCl.
• First assign variables:
• V1 = 125 mL
• V2 = 125 mL + 250 mL = 375 mL
• M1 = .251 mol/liter
• M2 = ?
• Then plug into formula and solve for unknown
• M2 = M1V1/V2 = (.251 M)(125 mL)/375 mL
• M2 = .0837 M