Transcript Solutions

Topic 12
Solutions
Solutions
A solution is a homogeneous mixture of two or
more substances or components.
Solutions may exist as gases, liquids, or solids.
The solute is the substance being dissolved. It is the
gas or solid being dissolved in a liquid; if it is of the
same state, it is the component of lesser amount.
i.e. coffee: sugar, coffee / water
The solvent is the substance doing the dissolving. In
the case of a gas or solid being dissolved in a liquid,
the solvent is the liquid; if it is of the same state, it is
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the component of greater amount.
Solubility of Solutions
Fluids that dissolve in each other in all proportions are
said to be miscible fluids. Typically when discussing
solubility the phrase “like dissolves like” is used.
A more appropriate way of expressing this is to state that
two substances with intermolecular forces of about the
same type and magnitude are likely to be soluble in one
another. For example, CH3OH / H20 are both polar with
similar forces and magnitude; therefore, they are
miscible in each other.
If two fluids do not mix, they are said to be immiscible.
i.e. oil (nonpolar) / water (polar)
Layers separate with less dense
miscible
species on top (oil in this case). immiscible
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Solute-Solvent Interaction
In most cases, “like dissolves like.”
– This means that polar solvents dissolve polar (or
ionic) solutes and nonpolar solvents dissolve
nonpolar solutes.
– The relative force of attraction of the solute for
the solvent is a major factor in their solubility.
– i.e. the dipole-dipole interactions of water (polar
solvent) with a polar solute can be easily
explained as electrostatic attractions (d-, d+)
between the dipoles of the molecules.
H
d-O
d+
H
polar
d- solute d+
H
d- O
d+
H
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Ionic Solutions
Polar solvents, such as water, also interact well with
ionic solutes because of electrostatic attractions (d-, d+)
between the cation and anion with water.
H
d+
O
H
d-
d-
+ O
H
d+
H
H
d-O
d+
H
H
-
d+
O d-
H
Ionic compounds are the extreme in polarity. When water is
the solvent, the attraction of ions to water molecules is
referred to as hydration.
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Nonpolar Solutes
Nonpolar solutes interact with nonpolar solvents
primarily due to London forces.
Heptane, C7H16, and octane, C8H18, are both nonpolar
components of gasoline and are completely miscible
liquids.
However, for water (polar) to mix with gasoline
(nonpolar), hydrogen bonds must be broken and
replaced with weaker London forces between water
(A) and the gasoline (B).
Therefore, gasoline and water are immiscible because
water has a stronger attraction for itself (stronger
solvent-solvent interaction) causing the two substances
to separate into layers with the less dense gas on top.
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Solubility and the Solution Process
Solubility is the amount of solute that can dissolve in a
given amount of solvent at given conditions.
– Many factors affect solubility, such as
temperature (most solids T, solubility).
– There is a limit as to how much of a given solute
will dissolve at a given temperature.
– A saturated solution is one holding as much
solute as is possible at a stated temperature.
At 20oC, the solubility of NaCl in
water (saturation point) is
36 g NaCl / 100 mL of water.
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Solubility: Saturated Solutions
Sometimes it is possible to obtain a supersaturated
solution, that is, one that contains more solute than is
possible at a given temperature.
Supersaturated solutions are unstable and the slightest
disturbance will cause the excess solute to crystallize
out.
Crystallization from a
supersaturated solution of
sodium acetate.
HW 70
code: solubility
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Effects of Temperature and Pressure
on Solubility
The solubility of solutes is very
temperature dependent.
– For gases dissolved in liquids, as temperature
increases, solubility decreases.
– On the other hand, for most solids dissolved in
liquids, solubility increases as temperature
increases.
– Basically, an increase in temperature always shifts
the position of equilibrium towards the
endothermic process.
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Temperature Change
Usually dissolving a solid in a liquid is an endothermic process
because heat must be absorbed to break down the crystal lattice.
solid + liquid
solution
DH > 0 (endothermic)
Any additional heat would shift the equilibrium to the right (favor
forward endothermic reaction) causing more solid to dissolve. For
example, more sugar dissolves in hot coffee than cold coffee.
The process of a gas condensing to a liquid is always an
exothermic process.
gas + liquid
solution
DH < 0 (exothermic)
Any additional heat would shift the equilibrium to the left (favor
reverse endothermic reaction) causing less gas to dissolve. For
example, a warm beer goes flat faster than a cold beer.
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Effects of Pressure on Solubility
Henry’s Law states that the solubility of a gas in a liquid is directly
proportional to the partial pressure of the gas in direct contact with
the liquid.
– Expressed mathematically, the law is
S  k HP
where
S is the solubility of the gas,
kH is the Henry’s law constant
characteristic of the solution
P is the partial pressure of the gas.
Basically, the higher the pressure of a gas above the liquid, the more
soluble the gas is in the liquid. For example, if a piston pushes down on a
gas, more gas will dissolve in the liquid; P, solubility
For example, the fizz that occurs when a soda can is opened results from
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reduced pressure of carbon dioxide over the liquid; P,
solubility
Solution Concentration Expressions
Concentration expressions are a ratio
of the amount of solute to the amount of
solvent or solution.
– The quantity of solute, solvent, or solution can
be expressed in mass, volume, or moles.
– The common ways to express concentration
are molarity, molality, mass percent, and
mole fraction.
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Molarity
The molarity, M, of a solution is the moles of solute
in a liter of solution (volume of solute + solvent).
moles of solute
Molarity (M ) 
liters of solution
A solution can be prepared to a specific molarity by weighing
out the mass of the solute and dissolving in enough solvent
to obtain the needed volume of solution. Note: volumes are
temperature dependent.
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Mole Fraction
The mole fraction of a component “A”, A, in a
solution is defined as the moles of the component
substance divided by the total moles of solution
(moles of solute + solvent).
moles of substance A
A 
total moles of solution
Mole Fraction is a unitless quantity with the sum of mole
fractions of all components of the solution equaling to 1.
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Mass Percentage of Solute
The mass percentage of solute is defined as:
mass of solute
Mass percentage of solute 
 100%
mass of solution
Notes:
- the “%” in the formula is a unit like grams, etc. and not
the % key on the calculator.
- the unit of mass doesn’t matter as long as the same
unit is used for solute and solution.
- mass of solution = mass of solute + mass of solvent
For example, a 3.5% by mass solution contains
3.5 grams solute per 100 grams of solution.
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Mass Percentage of Solute
How many grams of water are needed to prepare
425.0 g of aqueous solution containing 2.40% by
mass of NaCl?
We know there is 2.40 g of NaCl needed per every 100 g of solution
by the mass percent; therefore, we can determine how much NaCl is
in 425.0 g of solution.
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙 = 425.0 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
2.40 𝑔 𝑁𝑎𝐶𝑙
100 𝑔 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 10.2 𝑔 𝑁𝑎𝐶𝑙
This is the amount of NaCl we need in a 2.40% by mass solution of
NaCl; however, we need to determine how much water must be added
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to obtain the 425.0 g of solution.
Mass Percentage of Solute
We know that of the 425.0 g of solution that 10.2 g consists of
NaCl and the remainder is the solvent water.
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒, 𝑁𝑎𝐶𝑙 + 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡, 𝐻2 𝑂
Rearranging,
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻2 𝑂 = 𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 − 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻2 𝑂 = 425.0 𝑔 − 10.2 𝑔 = 414.8 𝑔 𝐻2 𝑂
To prepare 425.0 g of 2.40% by mass NaCl, it would require
dissolving 10.2 g NaCl in 414.8 g of H2O.
Note: masses are additive and temperature independent while volumes are not additive
and are temperature dependent.
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Molality
The molality of a solution is the moles of solute
per kilogram of solvent.
moles of solute
molality ( m ) 
kilograms of solvent
This expression is useful in situations when concentrations must be
compared over a range of different temperatures because it is
based on mass which is temperature independent.
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A Problem to Consider
What is the molality of a solution containing 5.20 g
of glucose, C6H12O6, dissolved in 90.0 g of water?
molality ( m ) 
moles of solute
kilograms of solvent
First, convert the mass of glucose to moles and mass of water to kg
1 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶6 𝐻12 𝑂6 = 5.20 𝑔 𝐶6 𝐻12 𝑂6
180.2 𝑔 𝐶6 𝐻12 𝑂6
= 0.0289 𝑚𝑜𝑙𝑠 𝐶6 𝐻12 𝑂6
1 𝑘𝑔
𝑘𝑔 𝑜𝑓 𝐻2 𝑂 = 90.0 𝑔 𝐻2 𝑂
= 0.0900 𝑘𝑔 𝐻2 𝑂
1000 𝑔
Plugging these values into the definition of molality
HW 71
code: molality
0.0289 𝑚𝑜𝑙𝑠 𝐶6 𝐻12 𝑂6
𝑚=
= 0.321 𝑚 𝐶6 𝐻12 𝑂6
0.0900 𝑘𝑔 𝐻2 𝑂
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Colligative Properties
The properties of a solution differ from those of
the pure solvent.
The colligative properties of solutions are
those properties that depend on the number of
particles dissolved in solution rather than their
nature.
These properties include:
1. vapor pressure lowering
2. freezing point depression
3. boiling point elevation
4. Osmotic pressure
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Vapor Pressure Lowering (nonvolatile)
The vapor pressure of a liquid is the pressure of the gas above the liquid.
Vapor pressure is a colligative property that decreases by the addition of a
nonvolatile solute.
The vapor pressure of the solution (nonvolatile solute and nonelectrolyte
solvent) will be lower than the vapor pressure of the pure solvent.
Vapor pressure lowering is independent of the nature of the solute but
directly proportional to its concentration.
Any interference with the ability of solvent particles to vaporize results in a
decrease in gas molecules and hence a lower vapor pressure.
Adding a nonvolatile solute to the solution decreases the surface area
formerly occupied by just solvent (now mixture of solute and solvent
particles on surface) and diminishes the rate of vaporization of the solvent;
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hence, a lower vapor pressure compare to pure solvent.
Vapor Pressure Lowering (nonvolatile)
Raoult’s law states that the vapor pressure of a
solution containing a nonelectrolyte nonvolatile
solute is proportional to the mole fraction of the
solvent:
o
Psolution  ( Psolvent )( solvent )
where
Psolution is the vapor pressure of the solution
solvent is the mole fraction of the solvent
Posolvent is the pure vapor pressure of the solvent.
Basically, the vapor pressure of the solution is a fraction of that
of the pure solvent and depends on the percentage of solvent
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making up the solution (partial vapor pressure).
Vapor Pressure Lowering (volatile)
If a solution contains a volatile solute, the
vapor pressure of the solution will be a
combination of the partial vapor pressures of
each volatile component:
Psolution 
o
o
( Psolvent )( solvent ) + ( Psolute )( solute )
HW 72
code: vp
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van’t Hoff Factor (nonelectrolytes)
Colligative properties of solutions are directly proportional
to the concentration of solute particle; therefore, the effect
that solutes have on colligative properties depends on the
quantity of solute particles present in the solution.
When one mole of glucose, C6H12O6, dissolves in water, one
mole of solute molecules is obtained. Glucose is a molecular
compound composed of covalent bonding and is a
nonelectrolyte that does not ionize in water. For each mole
of nonelectrolyte solute, there is one mol of solute particles.
C6H12O6 (s) → C6H12O6 (aq)
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van’t Hoff Factor (electrolytes)
Electrolytes ionize in water and for each mole of solute, there could be
several moles of solute particles. For example, CaCl2 is an ionic
compound that is an electrolyte that dissociates into three solute
particles in water (1 mol of Ca2+ and 2 mol of Cl-).
CaCl2 (s) → Ca2+ (aq) + 2Cl- (aq)
Since calcium chloride has three times the number of solute particles as
glucose, we would expect CaCl2 to have approximately three times the
effect on the colligative properties as compared to glucose.
The ratio of moles of solute particles in solution to moles of formula
units dissolved is referred to as the van’t Hoff factor (i):
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑖 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑢𝑛𝑖𝑡𝑠 𝑑𝑖𝑠𝑠𝑜𝑙𝑣𝑒𝑑 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
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van’t Hoff Factor
We expect the moles of solute particles in solution and the van’t Hoff
factor to be the same, but this actually only occurs for very dilute
solutions.
For our purposes, we will assume that the ions of an electrolyte behave
independently meaning i should be equal to the number of moles of
ions per mole of electrolyte and 1 for nonelectrolytes.
For example, i should be 2 for NaCl [1Na+, 1Cl-], 5 for Al2(SO4)3
[2Al3+, 3SO42-], and 1 for C2H6O2 [1 C2H6O2].
In the previous section, we discussed the vapor pressure of
nonelectrolytes. When we calculate the vapor pressure of a solution
containing an ionic solute (electrolyte), we must account for the
number of solute particles when we calculate the mole fraction of the
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solvent.
van’t Hoff Factor
A solution contains 0.155 mol NaCl and 0.756 mol H2O.
Calculate the vapor pressure of the solution at 55oC given the
vapor pressure of pure water at 55oC is 118.1 mm Hg.
NaCl has 2 particles per every one mol of substance. This must
be accounted for in the mole fraction of water before we use
Raoult’s Law.
𝑛𝐻2𝑂
𝐻2𝑂 =
(2 × 𝑛𝑁𝑎𝐶𝑙 ) + 𝑛𝐻2𝑂
𝐻2𝑂 =
0.756 𝑚𝑜𝑙
= 0.709
2 × 0.155 𝑚𝑜𝑙 + 0.756 𝑚𝑜𝑙
𝑜
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = (𝑃𝑠𝑜𝑙𝑣𝑒𝑛𝑡
)(𝑠𝑜𝑙𝑣𝑒𝑛𝑡 )
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =
118.1 𝑚𝑚 𝐻𝑔 0.709 = 83.7 𝑚𝑚 𝐻𝑔
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Boiling Point Elevation and Freezing
Point Depression
For the same reason the vapor pressure reduces by the
addition of a nonvolatile solute, the boiling point of a solution
will be elevated.
The temperature at which the vapor pressure of a liquid
equals 1 atm is called the normal boiling point. Since the
addition of a nonvolatile solute will diminish the rate of
vaporization of a solvent, the temperature of the solution must
be increased to a value greater than the normal boiling point
of the pure solvent to achieve a vapor pressure of 1 atm.
The opposite effect occurs for freezing. When a solution is
cooled, it does not begin to freeze until a temperature is
achieved that is below the freezing point of the pure solvent.28
Boiling Point Elevation and Freezing
Point Depression
A good example of taking advantage of both properties is the
addition of antifreeze solution to the radiator of a car.
Typically, ethylene glycol is used as antifreeze and is a
nonvolatile solute. By adding antifreeze to our radiator, we
have raised the boiling point and lowered the freezing point of
our engine cooling system. It allows us to continue to use our
cars on hot summer days and cold winter days that otherwise
would not be possible with pure water.
The boiling point of a pure solvent remains constant while
changing from liquid to a gas; however, the boiling point of a
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solution continues to increase as solvent is vaporized.
Boiling Point Elevation and Freezing
Point Depression
The relevant equations associated with these
colligative properties are
DTb  ( K b )(m)(i )
DT f  ( K f )(m)(i)
where ∆𝑇𝑏 and ∆𝑇𝑓 are the change in temperature relative to
that of the pure solvent in oC, 𝐾𝑏 and 𝐾𝑓 are proportionality
constants called the molal boiling point elevation constant
and molal freezing depression point constant that are
dependent on the solvent, 𝑚 is the molality of the solution in
moles per kilogram solvent, and i is the number of moles of
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solute particles per mole of solute.
Boiling Point Elevation and Freezing
Point Depression
Once the change in temperature is calculated, the actual
boiling point or freezing point of a solution is calculated by
adding the change to the boiling point of the pure solvent or
subtracting the change from the freezing point of the pure
solvent.
Tb solution  Tb solvent + DTb
Tf
solution
 Tf
solvent
- DT f
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A Problem to Consider
Estimate the freezing point of 0.35 m solution of Cr(NO3)3.
Assume the value of 𝑖 is based on the number of solute
particles per mole of solute and the 𝐾𝑓 of water is 1.86oC/m.
Cr(NO3)3 has 4 particles per every one mol of substance
3+(aq) + 3NO -(aq)
Cr(NO
)
(s)

1Cr
which means 𝑖 = 4.
3 3
3
Plugging into the change in temperature equation we obtain
∆𝑇𝑓 = 𝐾𝑓 𝑚 𝑖
∆𝑇𝑓 = 1.86
o
C
𝑚
0.35 𝑚
4 = 2.6 oC
Tf solution = 𝑇𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 − ∆𝑇𝑓
HW 73
code: freezing
Tf solution = 0.00o𝐶 − 2.6o𝐶 = -2.6oC
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Molar Mass Calculation Problem
A 1.25 g sample of a nonelectrolyte compound was dissolved in
75.0 g of benzene. The solution froze 1.20oC below that of pure
benzene. Determine the molar mass of the compound if the 𝐾𝑓
of benzene is 5.12oC/m.
∆𝑇𝑓 = 𝐾𝑓 𝑚 𝑖
Re-arranging for the molality of the solution:
𝑚=
∆𝑇𝑓
𝐾𝑓 𝑖
1.20 𝑜𝐶
𝑚=
= 0.234 𝑚
𝑜
(5.12 𝐶/𝑚)(1)
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Molar Mass Calculation Problem
Knowing the relationship between moles and molar mass, we can
rewrite molality:
𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑔 𝑠𝑜𝑙𝑢𝑡𝑒 / 𝑀𝑚
𝑚 =
=
𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Re-arranging for molar mass:
𝑀𝑚 =
𝑀𝑚 =
𝑔 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚 (𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡)
1.25 𝑔 𝑠𝑜𝑙𝑢𝑡𝑒
= 71.2 𝑔/𝑚𝑜𝑙
0.234 𝑚𝑜𝑙 𝑠𝑜𝑙𝑢𝑡𝑒
(0.0750 𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡)
𝑘𝑔 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
HW 74
code: molar
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Osmotic Pressure
Osmosis is a process whereby a solvent passes from a dilute solution
into a more concentrated one through a membrane permeable only to the
solvent.
The equalization of concentration between the two solutions in contact
with one another across the membrane is what drives the process of
osmosis.
The osmotic pressure, 𝜋, is equal to the external pressure, P, just
sufficient to prevent osmosis. Osmotic pressure is a colligative property
and is directly proportional to the molar concentration of solute:
𝜋 = 𝑀𝑅𝑇
where M is the molarity of the solute, R is the ideal gas constant (0.0821
L . atm/mol . K), and T is absolute temperature (K).
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