1. A solution is a homogeneous mixture of two or more substances.
A solution forms when a substance dissolves at the atomic level in a solvent.
If the substance being dissolved is ionic, the atomic level is ions.
If the substance being dissolved is covalent, the atomic level is molecules.
If the substance being dissolved is atomic (like a noble gas), the atomic level is atoms.
This is in contrast to a heterogeneous mixture where the solute will not break down to atomic level. I.E. the particles are larger than ions, atoms or molecules. Examples like this include suspensions and colloids.

2. A solution is a homogeneous mixture of two or more substances.
A solvent is a substance that is present in the largest amount in a solution. The solvent dissolves the other compound(s).
A solute is a substance that is present in a lesser amount in a solution. The solute dissolves in the solvent.
If a substance will dissolve in a solvent, then the substance is soluble.
If a substance will not dissolve in a solvent, then the substance is insoluble.
If a liquid substance will dissolve in a liquid solvent, the two substances are miscible.
If a liquid substance will not dissolve in a liquid solvent, the two substances are immiscible.

3. Solution formation (solid dissolving in water)
When the ions become surrounded by water molecules, they are said to be solvated.
Solvation is the process of surrounding a solute particle with solvent molecules. If the solvent is water, the process can also be called hydration.
If the intermolecular forces between the water molecules and the ions are stronger than the forces between the ions themselves, the ionic substance will dissolve.

4. Heat of solution (DHsoln): The heat of solution is overall energy change that occurs during solution formation.
If heat is released during solution formation, the heat of solution has a negative value and is exothermic (DHsoln is less than zero-negative)
If heat is absorbed during solution formation, the heat of solution has a positive value and is endothermic (DHsoln is greater than zero-positive)
The heat of solution is made up of three parts.
Part 1: the heat required to force solute particles apart from each other.
Part 2: the heat required to force solvent particles apart from each other.
Part 3: the heat released when the solute and solvent particles mix with each other.

5. Illustration of Energy Changes During Solution Formation
DH1 and DH2 are both endothermic steps (i.e. energy is required to separate the particles) DH3 is always an exothermic step (energy is released when the particles mix.

6. If DH3 > (DH1 + DH2), forming the solution will be an exothermic process (a).
If DH3 < (DH1 + DH2), forming the solution will be an endothermic process (b).
Notice that exothermic has a “down” arrow while endothermic has an “up” arrow.

7. When making a solution, there are two main concerns:
How much solute will dissolve in the solvent (Solubility)
How fast will the solute dissolve in the solvent (Rate of Solvation)
As we will see in the next few slides, these two things are not always impacted by the same factors.

8. Solubility
The amount of a solute that can dissolve in a solvent is the solubility of that substance in that solvent.
If a solvent contains the maximum amount of solute (at a given temperature), the solution is said to be saturated.
If a solvent contains less than the maximum amount of solute (at a given temperature), the solution is said to be unsaturated.
Factors impacting solubility:
Temperature
Pressure (mainly for gases dissolving in liquids)
Nature of the solute and solvent (mainly polarity issues)

9. Notice the units for solubility!
Notice that solubility increases with increasing temperature for most compounds (blue curves), but does decrease for others (red curves).
Notice the units for solubility!
What is the solubility of KBr at 20oC?
What is the solubility of KBr at 80oC?

10. Cross section of a water pipe.
When water is heated, some minerals have lower solubility in the hot water and start to crystallize onto the surface of the pipe. This is usually called scale and is a major problem for boilers that heat large amounts of water.

11. A supersaturated solution is one that holds more solute than it should at a given temperature.
If a solution is made at a high temperature (one where the solubility is higher at higher temperatures) and then the temperature is slowly lowered, the solution can end up with more solute dissolved in it than is normally possible.
Supersaturated solutions are unstable and if disturbed, solids will form rapidly in the solution-the solute crystallizes from the solution.
This type of crystallization event is often used to purify compounds because when done correctly, only solute particles will be incorporated into the crystal and impurities will be left behind in the solution.
After crystallization, the solid crystals are separated from the liquid solution by filtration.

12. Solutions form faster when the solute and solvent particles collide more often!
Therefore the rate of solvation can be increased by:
Increasing stirring rate
Increasing solvent temperature
Decreasing solute particle size
Decreasing solute particle size is the same as increasing surface area.
How does each of the above affect the number of collisions between
solute and solvent particles?

13. Solubility of gases in liquids
Notice that the solubility of gases in a liquid solvent always decreases as temperature is increased.
Why do you think gases behave this way?
What is the solubility of CO at 20oC?
Notice also that the solubility of gases in water are much lower than for the solids or liquids discussed so far.

14. Pressure has a large impact on the solubility of a gas in a liquid.
a) Initial situation b) Pressure is changed c) Final situation
What happens to the solubility of the gas when the pressure was increased?
Increasing the pressure, increases the solubility.
(P1)
(S1)
(P2)
(S2) =
Solubility of a gas is directly proportional to the pressure.
Henry’s Law:

15. Concentration: describing how much solute is dissolved in a solvent
A concentrated solution has a lot of solute dissolved.
While a dilute solution has only a little solute dissolved.
There are five ways that we will quantify concentration in this class.
Percent by Mass
(mass of solute/mass of solution)X100
Percent by Volume
(volume of solute/volume of solution)X100
Molarity
(moles of solute/liter of solution)
Molality
(moles of solute/kilogram of solvent)
Mole Fraction
(moles of solute)/(moles of solute + moles of solvent)

16. Here are five mathematical equations used to calculate concentration.
“m” is mass
Percent by Mass
Percent by Volume
“V” is volume
“n” is moles
Molarity
Molality
Mole Fraction

17. What is the percent by mass of NaHCO3 in a solution containing 20
What is the percent by mass of NaHCO3 in a solution containing 20.0 g of NaHCO3 dissolved in g of water?
If you wanted to make mL of a 30.0% aqueous solution of ethanol, what volume of ethanol would you need to use?
What is the molarity of a solution that was made by dissolving 50.0 g of KCl in enough water to make mL of solution?
What is the molality of a solution that was made by dissolving g of NaBr in g of water?

18. How many grams of a 35.5 % NaHCO3 aqueous solution would you need to use to have 50.0 grams of NaHCO3 ?
In a reaction, you use 64.5 mL of a 2.47 M solution of NaOH. How many grams of NaOH were used in the reaction?
What is the mole fraction of alcohol (C2H5OH) in an aqueous solution that contains 45.0 g of C2H5OH and 65.0 g of H2O?

19. Preparing solutions of a specific concentration:
Calculations Steps:
Step 1: identify the volume of solution to be made.
Step 2: identify the concentration of solution to be made.
Step 3: Use the concentration and volume information to determine the “amount” of solute needed.
Step 4: If the amount is in moles, convert it to grams using molar mass.
Step 5: If the solute is a liquid, convert the mass into volume using density.

20. Examples: a) How many grams of methanol (CH3OH) would be needed to make mL of a 1.50 molar solution?
Volume = mL = L
Concentration: molar (M) = 1.50 mol/L
M = n/V
So: n = MV
n = (1.50 mol/L)( L)
= mol CH3OH
Grams CH3OH = (0.750 mol CH3OH)(32.04 g/mol) = 24.0 g CH3OH
b) How many mL of CH3OH would be needed for this solution if the
density of CH3OH is g/mL?
density = mass/volume
So: volume = mass/density
volume = (24.0 g)/0.783 g/mL) = 30.7 mL

21. Steps Involved in the Preparation of a Standard Aqueous Solution using a solid compound.
Measure the mass of the correct number of grams of solid compound.
Add some water and shake until all solid is dissolved.
Carefully add enough water to fill the container to the volume marker.

22. Steps Involved in the Preparation of a Standard Aqueous Solution using a liquid compound.
Measure out the correct volume of liquid compound using a pipette or burette.
Add some water and shake until all liquid is dissolved.
Carefully add enough water to fill the container to the volume marker.

23. Dilution: If you take a small amount of a solution and add more solvent, then the concentration of the solute will be lower than it was before-this is dilution.
Dilution Problems: any time you increase the amount of solvent in a solution without changing the amount of solute, you are doing a dilution problem.
In a dilution problem the number of moles of solute stays the same while the volume and the concentration of the solution change.
Remember:
So: n = MV
Now in a dilution problem, there are two sets of conditions, the initial and the final (1 and 2 respectively).
M1V1 = mol of solute
and M2V2 = mol of solute
so M1V1 = M2V2
Use this form of the equation to solve dilution problems!

24. What will be the concentration of a solution that was made by using 50
What will be the concentration of a solution that was made by using 50.0 mL of a 6.50 M NaOH solution and diluting it to a final volume of 1.00 L?
What will be the concentration of NaOH in a solution that was made by mixing mL of a 4.20 M NaOH solution with mL of water?

25. Molality (m) is similar to Molarity (M) with one major difference
Molality (m) is similar to Molarity (M) with one major difference. The amount of solvent in kg is used instead of the amount of solution in L.
m = (mol solute)/(kg solvent)
M = (mol solute)/(L solution)
Molality is particularly useful for finding the molar mass of substances because the molality of a solution of unknown concentration can be determined by measuring Colligative properties.
Molality is really only used with Colligative properties while Molarity will be used in most other solution problems in the book.

26. (mol of A + mol of solvent)
Mole Fraction: like a percent by mole without the multiplication by 100.
Imagine dissolving solute A in a solvent. How would we fined the mole fraction of A in the solution?
Moles of solute
mol of A
Mole Fraction A =
(mol of A + mol of solvent)
Total moles in the solution.
The symbol for mole fraction is the Greek letter C
CA = (nA)/(nA + nB) for a two component mixture.
CA = (nA)/(nA + nB + nC) for a three component mixture.

27. Are the properties of a solution the same as the properties of the solute or of the solvent?
No, the properties of a solution are usually different than the properties of either the solute or the solvent.
One of the most important differences is that the vapor pressure of a solution is lower than the vapor pressure of the pure solvent.
Since vapor pressure is related to the boiling point and freezing point, a solution will have different boiling points and freezing points than the pure solvent has.
For example:
Why do we put salt on roads in the winter time?
Why do we add salt to water when we cook spaghetti?

28. WHY?
An experiment: a) a beaker of pure water and a beaker of aqueous solution in a closed container at the beginning of the experiment.
b) All of the solvent (water) ends up in the solution with a little water vapor in the gas phase.

29. When a non-volatile solute is dissolved in a solvent, the solution will have a lower vapor pressure than the pure solvent. The amount that it is lower depends upon the amount of solute particles-not the identity of the solute particles.
Why should the vapor pressure for the solution be lower than for the pure solvent?

30. Colligative Properties: Any property that depends upon the number of particles dissolved in a solvent.
Colligative properties exist because the presence of solute particles in the solution affects the vapor pressure of a solution.
Three common colligative properties:
Freezing point depression DTf
difference in the freezing point of a solution compared to pure solvent.
Remember that D means change, so DTf is the amount the freezing point changes.
Boiling point elevation DTb
difference in the boiling point of a solution compared to pure solvent.
Osmotic pressure P
the pressure produced by solvent molecules passing through a semi-permeable membrane into a solution.

31. Phase diagram for pure water (red lines) with the phase diagram for an aqueous solution superimposed on it (blue lines).
Since the vapor pressure of the solution is lower than the pure solvent, the temperatures for freezing and boiling change.

32. Freezing point depression DTf = Kfm
Kf = freezing point constant for a solvent
m = molality of particles in the solution
Notice that the new freezing point is not given by this equation, just the change in freezing point.
change in freezing point = DTf = Tinitial - Tfinal
so the new freezing point is: Tfinal = Tinitial - DTf
Tinitial will be the freezing point of the pure solvent.
Note: a 1.5 molal solution of NaCl will have make a 3.0 molal solution of particles because NaCl breaks apart into two particles (1.5 m)*(2) = 3.0 m
Note: a 1.5 molal solution of BaCl2 will have make a 4.5 molal solution of particles because BaCl2 breaks apart into three particles (1.5 m)*(3) = 3.0 m

33. Boiling point Elevation DTb = Kbm
Kb = boiling point constant for a solvent
m = molality of particles in the solution
Notice that the new boiling point is not given by this equation, just the change in boiling point.
change in boiling point = DTb = Tfinal - Tinitial
so the new boiling point is: Tfinal = Tinitial DTb
Tinitial will be the boiling point of the pure solvent.
Note: a 1.5 molal solution of NaCl will make a 3.0 molal solution of particles.
Note: a 1.5 molal solution of BaCl2 will make a 4.5 molal solution of particles.

34. What are the boiling point and freezing point of a 0
What are the boiling point and freezing point of a m aqueous solution of NaCl?
Aqueous means water is the solvent (use Kf (1.86 oC/m) and Kb (0.512 oC/m)).
NaCl has two particles so multiply the concentration by 2.
Boiling point Elevation DTb = Kbm = (0.512 oC/m)*(0.029 m)*(2) = oC
new boiling point is: Tfinal = Tinitial DTb = oC oC = oC
Freezing point depression DTf = Kfm = (1.86 oC/m)*(0.029 m)*(2) = oC
new freezing point is: Tfinal = Tinitial - DTf = 0.00 oC – 0.11 oC = oC

35. new boiling point is: Tfinal = Tinitial + DTb
new freezing point is: Tfinal = Tinitial - DTf
Notice that these two equations are different from each other.
Rather than worry about remembering the two different equations
remember that this is boiling point elevation (new boiling point must be higher than the original) and freezing point depression (new freezing point must be lower than the original).

37. Heterogeneous mixtures: mixtures that are not solutions
If you can see two phases in a mixture, the mixture is heterogeneous.
Sometimes you can not “see” through a mixture at all-this is also usually a heterogeneous mixture.
Example: muddy water-solid particles of “dirt” are suspended in the water so that you can not see through it.
Suspensions are a common class of heterogeneous mixtures-in theory, the phases can be forced to separate because the particle are large enough to either filter out or to settle apart from each other.
Colloids are like suspensions, but the particles are smaller and will not settle out.

38. How can we tell a suspension from a solution?
Suspended particles (even those that are typically to small to see with our eyes) will cause light to bounce off.
If light passes through a solution, you can not see the beam of light in the solution because there are no particles large enough to cause the light to bounce off of.
If light passes through a suspension, you can see the beam of light in the suspension because some of the light is scattered to the sides when it bounces off of the suspended particles. This is called the Tyndall effect.
Brownian motion explains why colloids do not settle out. The random motion of all the particles in the suspension keep the particles from settling to the bottom.

39. The following “slides” are related to osmotic pressure.
This concept is related to the colligative properties, but is not
really covered by the SOL, so these slides are for your information
only-the material will not be tested.

40. a) Solution on the right separated from pure solvent on the left by a semi-permeable membrane. Solute may not pass through the membrane by solvent may
b) Since the rate of solvent molecules passing through the membrane from right to left is greater than from left to right at the beginning, the level of liquid rises on the right and goes down on the left.

41. As the level of liquid in the flask increases, the pressure pushing solvent molecules from the solution into the pure solvent increases until the rate of solvent molecules traveling through the membrane becomes equal in both directions: the height of solution compared to the pure solvent is the osmotic pressure.
Osmotic pressure: P = MRT
M is molarity, R is gas constant, T is temperature in Kelvin

42. By choosing a membrane that lets impurities travel through but not blood, dialysis can be accomplished using osmosis.

43. If enough pressure is applied to overcome the osmotic pressure, reverse osmosis can take place. This allows you to obtain a pure solvent from a solution. An important application of this it to get fresh water from ocean water.