Unit 10 Solutions
Solution: a homogenoueous mixture of 2 or more substances
2 parts to a solution
 The solvent does the dissolving; normally the component present in largest
amount
 The solute is dissolved; normaly the component present in smallest amount
Examples of all types of Solutions: Solids dissolve into solids, liquids can dissolve into
liquids, gases can dsissolve into liquids, etc
Salt in Water
Nitrogen
Alcohol in Water
Oxygen & Carbon Dioxide in
Silver in Gold
Oxygen in Water
Hydrogen gas in Platinum
Mercury in Silver
Descriptive terms
Miscible: when 2 or more liquids can dissolve into one another (water and alcohol)
Immiscible: when 2 or more cannot dissolve into one another (water & oil)
Soluble: when a solid or gas can dissolve into one another
Insoluble: when a solid or gas cannot dissolve into one another
We will focus on aqueous solutions-solutions where water is the solvent
Ways of Measuring the Concentration of Solutions
Qualitative terms:


Dilute : small amount of solute compared to the amount of solvent
Concentrated: large amount of solute compared to the amount of solvent
Quantitative terms:

Molarity = moles of solute
Liters of solution

abbreviated M; molarity changes with
temperature due to expansion/contraction
of solvent changing volume; most molar
solutions
are made at 25 C; 1.0 M means 1 molar
Mass Percent = Mass of solute x 100
often expressed as parts per
million(ppm)
Total Mass of solution
or (ppb) 1 mg/1L = 1ppm
 Can transfer easily to mole fraction from mass percent

Mole fraction – moles of solute over total moles of solution
Mole fraction of component χA =
NA_______
NA + NB….
 sum of all mole fractions equals 1


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Molality =
moles of solute
Kilograms of solvent
used in measuring colligative properties
does not change with
temperature / mass does not change
with temp; abbreviated m; 1 m
means 1 molal
Density (g/ml) can be used to convert between the different methods of
calculating concentration, especially between molarity and molality
Density of aqueous solution is usually identical to that of pure water (1g/1mL) at
normal temperatures
Example 1: A solution is prepared by mixing 1.00 g ethanol( C2H5OH) with 100.0 g
water to give a final volume of 101 ml. Calculate the molarity,
mass percent, mole fraction, and molality of ethanol in this solution.
Example 2: The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid
solution that has a density of 1.230 g/ml.
Calculate the mass percent, and molality of the sulfuric acid.
Energy of Making Solutions


Solutions form when attractive forces between solute and solvent particles are
comparable with those that exist between
the solute particles themselves or the solvent particles themselves.
Like Dissolves Like polar solvents dissolve ionic/polar solutes OR Nonpolar
solvents dissolve nonpolar solutes
Example : NaCl is soluble in water. the strong ion-dipole forces between the ions and
water are as strong as ionic bonds and Hydrogen Bonding
Example : Oil is immiscible in water. The weak dipole-induced dipole forces
between water & oil are not as strong as the weak LDF
and Hydrogen bonding of water

Interactions where the solvent completely envelops the solute is called solvation
or when water is the solvent, hydration.
Enthalpy(heat) of solution (ΔHsoln ) is the energy change for making a solution. ΔHsoln =
ΔH1 + ΔH2 + ΔH3


Most easily understood if broken into “3” steps. All 3 steps are added together to
calculate the heat of solution.
It can be positive(endothermic) or negative (exothermic)
1.Break apart solvent. ΔH1
 requires energy to overcome intermolecular forces. Expanding the solvent!
ΔH1 >0  Endothermic
 the values are large in polar solvents and low in nonpolar solvents
2. Break apart Solute. ΔH2
 Requires energy to overcome attractive forces of particles. Expanding the
solute! ΔH2 >0  Endothermic
 the values are large in ionic & polar solutes and low in nonpolar solutes
3. Mixing the solute and Solvent ΔH3 ΔH3 < 0  Exothermic
 requires the release of energy when solute & solvent interact to fomr a
solution
 the values are large in polar solutes-polar solvent and low in nonpolar
solutes-nonpolar solvent


Molecules can attract each other ; ΔH3 is large and negative.
Molecules can’t attract; outcome is ΔH3 is small and negative.
 All 3 terms can add together to get a positive or negative sum = ΔHsoln
ΔHsoln can be exothermic or slightly endothermic
Types of Solvent and Solutes
Case 1
 Oil and water do not mix
 Oil(large nonpolar with LDF)
 ΔH1 for solute is usually small & positive but large & positive due to size of oil
molecules
 ΔH2 for water is large & positive due to H- bonds
 ΔH3 is small and negative due to little to no interactions between polar and
nonpolar molecules
 So ΔHsoln is large and positive- Does not usually happen  too much energy
expended
Case 2
 Salt in water – mix
 ΔH1 for solute is usually large & positive due to electrostatic forces
 ΔH2 for water is large & positive due to H- bonds
 ΔH3 is large and negative due to ion-dipole forces
 ΔHsoln is small and positive in this case approximately 3 kJ/mol
 Remember, When heat of reaction is negative, reaction is spontaneous
 When ΔHsoln small and positive, what makes salt soluble?
 Entropy??? The increased disorder overrides the small cost of heat of solution.
 Of course, solution formation takes place based on 2 factors: tendency towards a
lower enthalpy(exothermic) and a higher entropy(favored)
Example 3: Calculate the enthalpy of solution for NaOH. Hf (s) = -425.6 kJ/mol and
Hf (aq) 1m = -470.1 kJ/mol
Structure and Solubility
 To be soluble in polar solvents, the molecules must be polar or ionic
 To be soluble in non-polar solvents the molecules must be non polar.
ΔH1solute
ΔH2solvent ΔH3interacti
ΔHsoln
Result
ons
Polar solvent,
polar solute
Polar solvent,
nonpolar solute
Nonpolar
solvent,
nonpolar solute
Nonpolar
solvent, polar
solute
Large
Large
Small
Large
Large,
negative
Small
Small
Small
Small
Large
Small
Small
Small
Large,
positive
Small
Large,
positive
Solution
forms*
No solution
forms
Solution
forms*
No solution
forms
Example 4: Decide whether liquid hexane or liquid methanol is the more appropriate
solvent for the substance grease (C20H42) and potassium iodide.
Solubility
Solution formation is a dynamic equilibrium process
Dissolution
Solute + Solvent
↔
Solution
Crystallization
 Defined as the maximum amount of solute that will dissolve in a sprcific amount
of solvent at a specific temperature
This max. amount forms a saturated solution.
Example: 35.7 g of NaCl dissolves per 100 ml at 0ºC



Saturated solution: a solution that is in a dynamic equilibrium with an undissolved
solute; think of it as being completely “full”,
no additional solute can fit into the solvent, it contains the maximum anout that
can dissolve
Unsaturated solution: is not “full”, additional solute can be added to solvent and
still be dissolved; it contains less than the maximum amount
Supersaturated solution: solutions that contain a greater amount of solute than
needed to form a saturated solution;
unstable; seed crystal disturbs system and crystallization takes place  needs to
be heated at high temp and cooled slowly
Pressure Effects
Changing the pressure doesn’t affect the amount of solid or liquid that dissolves
(solubility)
 They are incompressible.
Pressure DOES EFFECT the amount of gas that can dissolve in a liquid.
 The dissolved gas in solution is at equilibrium with the gas above the liquid.
 The equilibrium is dynamic- rate at which the gas molecules enter the solution
equals the rate at which they escape from solution and enter gas phase
 If you increase the pressure of the gas molecules, more will dissolve. The
equilibrium is disturbed.
 The system reaches a new equilibrium with more gas dissolved.
The solubility of a gas is directly proportional to its partial pressure above the solution
(assuming there is no reaction between the gas and the solvent)
Henry’s Law C = kP
C = solubility/concentration of gas
K = proportionality constant dependent on the gas-liquid mixture, varies with temperature
P = partial pressure of gas
Example 5: A certain soft drink is bottled so that a bottle at 25 °C contains CO2 gas at a
pressure of 5.0 atm over the liquid. Assuming that the
partial pressure of CO2 in the atmosphere is 4.0 x 10-4 atm, calculate the equilibrium
concentrations of CO2 in the bottle both before and
after the bottle is opened. The Henry’s law constant for CO2 in aqueous solution is 3.1 x
10-2 mol/atm•L at 25°C.
Temperature Effects
 Increased temperature usually increases the rate at which a solid dissolves. Most
solution foramtion is endothermic & decreases when the solution process is
exothermic
 It is difficult to predict. A graph of experimental data (Solubility Curves) will
show the relationship between temperature and the solubility of a solid in the
solvent.
 Gases are predictable: As temperature increases, the solubility of gases decreases

Environmental concern: Thermal pollution
Colligative Properties
 Physical properties that depend on the quantity of the particles in the solution, not
the kind of particles
 Properties of solution differ from properties of pure solvents
 These properties are: Vapor Pressure, Boiling Point, Freezing Point and Osmotic
Pressure
Example: Pure water freezes at 0ºC but a Salt Water solution would freeze at a lower
temperature
Vapor Pressure LOWERING of Solutions
 A volatile substance has a measurable vapor pressure, a nonvolatile substance has
no vapor pressure
 When comparing the vapor pressure of a pure solvent with those of their
solutions, addition of a nonvolatile solute to solvent will lowers the vapor
pressure
WHY? The molecules of the solvent must overcome the forces of both the solvent
molecules and the solute molecules. The more solute particles present the less the
solvent can evaporate AND fewer solvent molecules are at surface
Raoult’s Law (use with nonvolatile solutes)


Solutions with a nonvolatile solute -the solute doesn’t contribute to the vapor
pressure.
States that the vapor pressure of the solution is directly proportional to the mole
fraction of the solvent
Psoln = χsolvent x Pºsolvent




Psoln = Vapor pressure of the solution
χsolvent = mole fraction of solvent
Psolvent = vapor pressure of the pure solvent
i= van’t Hoff factor

Water has a
higher vapor
pressure than a
solution
 In its
linear
form,
Psoln = χsolvent x Pºsolvent
Y =m
x
+ b(0)
Example 6: Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/ml
at 25 °C.
Calculate the vapor pressure at 25C of a solution made by adding 50.0 ml of
glycerin to 500.0 ml of water.
The vapor pressure of pure water at 25 °C is 23.8 torr.
Can use this information to experimentally determine molar mass of a substance:
o If mass of substance is given and Raoult’s law determines moles of solute
present ,one can calculate molar mass M= mass/mole
Electrolytes in Solutions (NEED TO BE TAKEN INTO CONSIDERATION WHEN
CALCULATING CHANGES IN ANY COLLIGATIVE PROPERTY)
 Since colligative properties only depend on the number of molecules.
 Ionic compounds should have a bigger effect.; when they dissolve they dissociate.
 Individual Na and Cl ions fall apart.
1 mole of NaCl makes 2 moles of ions.
1mole Al(NO3)3 makes 4 moles ions.


Electrolytes have a bigger impact on vapor pressure lowering because they make
more pieces.
Relationship is expressed using the van’t Hoff factor ( i ) : factor equal to the
moles of ions present
i=


Moles of particles in solution
Moles of solute dissolved
The expected value can be determined from the formula.
The actual value is usually less because at any given instant some of the ions in
solution will be paired.
Ion pairing increases with concentration. ( the joining of oppositely charged ions due to
electrostatic attraction. The greater the charge on an ion the greater its tendency to pair in
solution)
 The more particles dissolve,d the more the property is affected. i=1 for a a
nonelectrolyte (molecular compound) i= number of particles fromed when one
formula unit of an ionic compound dissolves in the solvent
 NaCl (i= 2)
 Psoln = i (χsolvent ) x (Pºsolvent)
Example: Glucose is only1 molecule, NaCl has 2 ions that split apart, FeCl3 has 3 ions
that split apart vapor pressure will be lowered , lowered 2 times as expected, lowered 3
times as expected
What if the solute is volatile (NONIDEAL)?
 Must add together each substances vapor pressure
Modified Raoult’s Law
Ptotal soln = PA + PB
= χ AP0 A + χ BP0B
Ptotal = vapor pressure of mixture
χ A = mole fraction of A
χ B = mole fraction of B
P0A= vapor pressure of A
P0B= vapor pressure of B
Ideal solution- a liquid-liquid solution that obeys Raoult’s law
 Near ideal behavior is when 2 volatile liquids dissolve in one another and the
solute-solute, solvent-solvent, and solute-solvent interactions are very similar
 Can use Raoult’s law to see if the solution is ideal
 If it is ideal, (solute and solvent are alike) and predicted vapor pressure will be
correct
 Hexane and heptane or benzene and methylbenzene
Deviations: If it is not, the observed vapor pressure will be lower or higher than what
was predicted
Negative deviation from Raoult’s law.
 ΔHsoln is large & negative (exothermic).
 Vapor pressure of solution is lower than expected(calculated) the real measured
vapor pressure was lower than expected
 Interactions between solute-solvent bonds are greater than those in the solventsolvent bonds or solute- solute bonds
 Acetone and water
Positive Deviation
 ΔHsoln is large &positive(endothermic)
 Vapor pressure of solution is greater than expected(calculated) the real
measured vapor pressure is larger than what was calculated
 Interactions between the solute - solvent bonds are weaker than those of the
solvent-solvent bonds and solute- solute bonds
 Ethanol (polar) and hexane(nonpolar)
Example 7: At 40 C, the vapor pressure of pure heptane is 92.0 torr and the vapor
pressure of pure octane is 31.0 torr.
Consider a solution that contains 1.00 mol of heptane and 4.00 mol of octane.
Calculate the total vapor pressure of
each component and the total vapor pressure above the solution. Is this
considered an ideal solution?
Think of the type of substances in the solution.
Other Colligative Properties
 Dissolved particles affect vapor pressure so they affect phase changes as well.
 Electrolytes have a bigger impact on melting and freezing points per mole
because they make more pieces.
Boiling Point Elevation
 Because a non-volatile solute lowers the vapor pressure it raises the boiling point
proportionally to the amount of solute added.
 The equation is: ΔTb = iKbmsolute
ΔTb is the change in the boiling point
Kb is a molal boiling point constant specific to the solvent.
msolute is the molality of the solute
i = van’t Hoff Factor
One can calculate molar mass of an unknown compound if compound is soluble in a
solvent of a known Kb or Kf
Example 8: Antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile nonelectrolyte.
Calculate the boiling point of a 25.0 mass percent solution of ethylene glycol
in water. Kb = .52 °C/m
Freezing Point Depression
 Because a non-volatile solute lowers the vapor pressure of the solution it lowers
the freezing point.
 The equation is: ΔTf = Kfmsolute
ΔTf is the change in the freezing point
Kf is a molal freezing point constant specific to the solvent
msolute is the molality of the solute
i = van’t Hoff Factor
Example 9: When 15.0 g of ethyl alcohol, C2H5OH is dissolved in 750. g of formic acid,
the freezing point
of the solution is 7.20 °C. The freezing point of pure formic acid is 8.40 °C.
Evaluate the Kf for formic acid.
Example 10: A 1.20 g sample of an unknown compound is dissolved in 50.0 g of
benzene.
The solution freezes at 4.92 °C. Calculate the molecular mass of the
compound.
The freezing point of pure benzene is 5.48 °C and the Kf is 5.12 °C/m.
Example 11: Estimate the freezing point of .20m solution of Cr(NO3) 3 in water. Kf =
1.86 °C/m
Example 12: List the following solutions in order of their freezing points (from lowest to
highest)
.050 m CaCl2, 0.15 m NaCl, 0.10 m HCl, 0.050 m HC2H3O2, 0.10 m
C12H22O1
Osmotic Pressure
 Osmosis- selective passage of solvent molecules through a porous semi permeable
membrane from a dilute to more concentrated solution
 Osmotic pressure(π) –the pressure created by the movement of the solvent
through the membrane
 Equal to the pressure applied in order to prevent osmosis
 π = iMRT
M= molarity
R= universal gas constant(.08206 L·atm/mol ·K)
T= Kelvin temperature
i = van’t Hoff Factor
Colloids
 Heterogenous mixture
 Particles do not settle over time
 Particle size is intermediate to solutions and suspensions
 Exhibit the tyndall effect
 To destroy colloids, heat or add additional electrolytes
 Examples are: fog, aerosol sprays, smoke, whipped cream, soap, suds, milk,
mayo, paint, butter, and cheese