Solutions AP NOTES
2 parts to a solution (homogeneous mixture)

The solvent does the dissolving; normally the component present in largest amount

The solute is dissolved; normally the component present in smallest amount
There are examples of all types of solutes dissolving in all types of solvent.
Salt in water alcohol in water oxygen in air
Ag in Au
Hg in Ag oxygen in water H
2
in Pd
We will focus on aqueous solutions -solution where water is the solvent
Ways of Measuring
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 solution changing volume;
1M means 1 molar

% mass = 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 χ
A
= N
A_______
N
A
+ N
B….
 sum of all mole fractions equals 1

Molality = moles of solute used in measuring colligative properties
Kilograms of solvent 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 1-2
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.

Example : NaCl in water

Interactions where the solvent completely envelops the solute is called solvation or when water is the solvent, hydration.
Heat of solution ( ΔH soln
ΔH
2
+ ΔH
3
) is the energy change for making a solution. ΔH soln
= ΔH
1
+
 Most easily understood if broken into “3” steps.
1.Break apart solvent
 requires energy to overcome intermolecular forces. ΔH
1
>0

Endothermic
2. Break apart Solute.
 Requires energy to overcome attractive forces of particles. ΔH
2
>0

Endothermic
3. Mixing the solute and Solvent usually exothermic or slightly endothermic

 ∆
H3
depends on what you are mixing.
Molecules can attract each other ; ΔH
3
is large and negative.
 Molecules can’t attract; outcome is ΔH
3
is small and negative.

All 3 terms can add together to get a positive or negative sum. If exothermic or slightly endothermic, the solute dissolves into solvent
 This explains the rule “Like dissolves Like” - similar molecules with same type and magnitude of intermolecular forces attract one another
Types of Solvent and Solutes
Case 1

Oil and water do not mix

Oil(large nonpolar with LDF)
 Δ
H1 for solute is usually small & positive but large & positive due to size of oil molecules
 Δ
H2
for water is large & positive due to H- bonds

 Δ
H3
is small and negative due to little to no interactions between polar and nonpolar molecules
 So ΔH soln
is large and positive- Does not usually happen

too much energy expended
Case 2

Salt in water – mix
 Δ
H1 for solute is usually large & positive due to electrostatic forces
 Δ
H2
for water is large & positive due to H- bonds
 Δ
H3
is large and negative due to ion-dipole forces
 ΔH soln
is small and positive in this case approximately 3 kJ/mol

Remember, When heat of reaction is negative, reaction is spontaneous
 When ΔH soln
small and positive, what makes salt soluble?

Entropy??? Disorder

Of course, solution formation takes place based on 2 factors: tendency towards a lower enthalpy(exothermic) and a higher entropy(favored)
Example 4
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.
ΔH
1solute
ΔH
2solvent
ΔH
3interacti
ΔH soln ons
Polar solvent, polar solute
Large Large Large, negative
Small
Result
Solution forms*
Small Large Small Polar solvent, nonpolar solute
Nonpolar solvent, nonpolar solute
Small Small Small
Large, positive
Small
No solution forms
Solution forms*
Nonpolar solvent, polar solute
Large Small Small
Example 3
Solubility
Solution formation is a dynamic equilibrium process
Dissolution
Solute + Solvent ↔ Solution
Crystallization
Large, positive
No solution forms


Defined as the amount of solute needed to form a saturated solution in a given quantity of solvent at a certain temperature
 Example: 35.7 g of NaCl per 100 ml at 0ºC

Miscible: 2 liquids can dissolve in one another vs. Soluble –solid dissolves in liquid

Immiscible: 2 liquids cannot dissolve in one another vs. Insoluble- solids does not dissolve in liquid

Saturated solution: a solution that is in equilibrium with an undissolved solute; think of it as being completely “full”, no additional solute can fit into the solvent
 Unsaturated solution: is not “full”, additional solute can be added to solvent and still be dissolved

Supersaturated solution: solutions that contain a greater amount of solute than needed to form a saturated solution; like a super cooled liquids in the sense that the molecules could not arrange themselves in the ordered structure of a solid; unstable; seed crystal disturbs system and crystallization takes place
 needs to be heated at high temp and cooled quickly
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 the gas molecules, they dissolve faster. 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
Temperature Effects

Increased temperature usually increases the rate at which a solid dissolves.

A graph of experimental data (Solubility Curves)will show the relationship between temperature and the solubility of a solid in the solvent.
Example: See Solubility graph/ditto

Gases are predictable: As temperature increases, solubility decreases

Environmental concern: Thermal pollution

Colligative Properties

Properties of solution differ from properties of pure solvents
 Example: pure water freezes at 0ºC but a salt water solution would freeze at a lower temperature

Physical properties that depend on the quantity of the particles in the solution, not the kind of particles

Vapor Pressure, Boiling Point, Freezing Point and Osmotic Pressure
Vapor Pressure 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
P soln
= χ solvent
x Pº solvent

P soln
= Vapor pressure of the solution
 χ solvent
= mole fraction of solvent

P solvent
= vapor pressure of the pure solvent

In its linear form,
P soln
=

χ solvent
x Pº solvent
Y = m x + b(0)

Water has a higher vapor pressure than a solution

Can use this information to experimentally determine molar mass of a substance:
 If mass of substance is given and Raoult’s law determines moles of solute present ,one can calculate molar mass (mass/mole)

The more particles dissolve, the more the properties is affected
Example: glucose is only 1 molecule, NaCl has 2 ions that split apart, FeCl
3
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
P total soln
= P
A
+ P
B
= χ
A
P
0
A
+ χ
B
P
0
B
P total
= vapor pressure of mixture
χ
A
= mole fraction of A χ
B
= mole fraction of B
P
0
A
= vapor pressure of A P
0
B
= 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.
 ΔH soln
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 solvent- solvent bonds or solute- solute bonds

Acetone and water
Positive Deviation
 ΔH soln
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 6
Other Colligative Properties

Dissolved particles affect vapor pressure so they affect phase changes.
Boiling Point Elevation

Because a non-volatile solute lowers the vapor pressure it raises the boiling point proportionally to the amount of solute added.
ΔT b
 The equation is: ΔT b
= K b m solute
is the change in the boiling point
K b is a molal boiling point constant specific to the solvent. m solute
is the molality of the solute

One can calculate molar mass of an unknown compound if compound is soluble in a solvent of a known K b
or K f
Example 8
Freezing Point Depression

Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point.
 The equation is: ΔT f
= K f m solute
ΔT f
is the change in the freezing point
K f is a molal freezing point constant specific to the solvent m solute
is the molality of the solute
Example 9 -11
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 = MRT
M= molarity
R= universal gas constant(.08206 L·atm/mol ·K)
T= Kelvin temperature
Example

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(NO
3
)
3
makes 4 moles ions.

Electrolytes have a bigger impact on melting and freezing points per mole 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)

We can change our formulas to: π =iMRT ΔT b
= iK b m solute
ΔT f
= iK f m solute
Example 12
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