1. Chemistry 1 Chapter 18 - Solution Formation

2. Factors affecting rate of solution:
Agitation
brings fresh solvent in contact w/ solute
does not affect amount that dissolves

3. Factors affecting rate of solution:
Temperature
heat increases rate of solution by increasing kinetic energy and motion of particles

4. Factors affecting rate of solution:
Particle size
smaller particles dissolve faster because of greater surface area

5. Solubility - a measure of the amount of solute that will dissolve in a given amount of solvent.
we usually use g solute/100 g solvent
*See table of solubility on page 504
0ºC
20ºC
100ºC
Ba(OH)2
1.67
31.89
------
NaCl
35.7
36.0
39.2
Sucrose
179
230.9
487

6. Factors affecting solubility:
Nature of the solute and the solvent
like dissolves like

7. Factors affecting solubility:
Temperature
Most solids become more soluble at a higher temperature. A few become less soluble.

8. Factors affecting solubility:
solubility of gases decreases as temperature increases

9. Factors affecting solubility:
Pressure (for solutions of gases in liquids)
increasing pressure increases solubility of gas

10. Saturated solution -contains the maximum amount of solute that will dissolve in a given amount of solvent at a given temperature.
Unsaturated solution- more solute could be dissolved
-indicated by no solid present
Supersaturated solution- more solute is dissolved than is theoretically possible at a given temperature.

11. EXAMPLE:
How could you determine if a solution is saturated, unsaturated or supersaturated?
If solid is present – it is Saturated
If no solid is present – it is unsaturated or supersaturated
To determine which, add a crystal
Supersaturated will crystallize
Unsaturated will dissolve the crystal
DEMO

13. Cloud seeding is a practical example of supersaturation.

14. Miscible - liquids that are soluble in each other
Ex. ethanol and water
Immiscible- liquids that are not soluble in each other
Ex. oil and water

15. Henry’s Law- At a given temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.
S = S2
P P2

16. EXAMPLE:
At 25oC, the solubility of a certain gas in H2O is 0.75 g/L of H2O at a pressure of 1.0 atm. What is the solubility of the same gas at 10.0 atm pressure?
S1 = 0.75 g/L
P1 = 1.0 atm
S2 = ?
P2 = 10.0 atm
S S2
P P2 = x
x = g/L =

17. Concentration -a measure of the amount of solute per given amount of solvent or solution.
Dilute solution- small amount of solute
Concentrated solution- large amount of solute

18. Molarity(M) = moles of solute liters of solution
A 1 M solution contains 1 mole of solute per 1 L of solution. A 0.5 M NaCl solution has 0.5 mol NaCl in 1 L total solution. n V =

19. EXAMPLES:
What is the concentration in molarity of a solution made with 1.25 mol NaOH in 4.0 L of solution? n V
# mol
# L
M = =
M = ?
n = 1.25 mol
V = 4.0 L
M = 1.25 mol
4.0 L
= M
= M

20. How would you make 500. mL of 6.0 M NaOH?
How many grams of NaOH would be necessary? n V
V = 500. mL
= L
M =
M = 6.0 M
n =
M ∙ V
= 6.0 M ∙ L
n = ?
= mol
Na 23.0 O H
3 mol
40.0 g =
120 g NaOH
1 mol
40.0 g/mol

21. Dilutions
Diluting a solution won’t change the number of moles of solute. It only changes the concentration.
M1V1 = M2V2
This formula works only for dilution of a solution!!!!
Don’t use it for a reaction!

22. EXAMPLE:
For a lab we need 500. mL of 4.25 M HCl. Concentrated HCl is 12.0 M. How do we make up the solution?
M1V1 = M2V2
M1 = M
V1 = mL
M2 = M
V2 = ?
(4.25M)(500. mL) = (12.0 M)(V2)
V2 = 177 mL

23. Solution Stoichiometry
use Molarity as a conversion factor#
# mol
1 L
1 L
# mol

24. EXAMPLES:
What volume of M nitric acid is required to react completely with L of M Ba(OH)2?
2HNO3 + Ba(OH)2 → Ba(NO3) H2O
0.386 L Ba(OH)2
mol Ba(OH)2
2 mol HNO3
1 L HNO3
1 L Ba(OH)2
1 mol Ba(OH)2
0.246 mol HNO3
molarity
mol-mol ratio
molarity =
0.162 L HNO3

25. Calculate the number of grams of carbon dioxide that can react with 0
Calculate the number of grams of carbon dioxide that can react with L of a M solution of potassium hydroxide, according to the following reaction:
2KOH CO2  K2CO H2O
0.135 L KOH
0.357 mol KOH
1 mol CO2
44.0 g CO2
1 L KOH
2 mol KOH
1 mol CO2
= 1.06 g CO2
molarity
mol-mol ratio
GFM

26. % Solutions Percent by volume % v/v = volume solute (mL) x 100
volume solution (mL)
Percent mass/volume
% m/v = mass solute(g) x 100

27. EXAMPLES:
Rubbing alcohol is 70.0% isopropyl alcohol by volume. How many mL of isopropyl alcohol are in a 250 mL bottle of this solution?
% v/v = volume solute (mL) x 100
volume solution (mL)
% v/v = 70.0 %
v solution = 250 mL
v solute = X
70.0 % = __X__ x 100
250 mL
X = 175 mL

28. If the % mass/volume (m/v) for a solute is 6
If the % mass/volume (m/v) for a solute is 6.80 %, and the volume of the solution is 42.6 mL, what mass of the salt is present?
% m/v = mass solute (g) x 100
volume solution (mL)
% v/v = 6.80 %
v solution = 42.6 mL
g solute = X
6.80 % = __X__ x 100
42.6 mL
X = g
= g

29. Colligative Properties- properties that depend on the number of dissolved particles
freezing point depression
boiling point elevation
vapor pressure lowering
osmotic pressure

30. Most automobile engines use a water cooling system to dissipate engine heat. However, water freezes at 0 °C. With pure water as the coolant, automobiles could operate only in regions with temperatures above 0 °C . Instead, ethylene glycol or propylene glycol is mixed with the water as an "antifreeze." 

31. Ethylene glycol is non-corrosive to metal and rubber; mixes with water in all ratios, boils at 198 °C and freezes at -13 °C, has no odor, and is nonflammable and inexpensive. However, ethylene glycol is toxic, and its sweet taste is particularly attractive to pets and children. Because of its toxicity, propylene glycol is now used as an alternative to ethylene glycol. The antifreeze lowers the freezing point and raises the boiling point of the solution above that of pure water, permitting operation of the engine at a higher temperature.

32. An ice storm is coming. Your car’s radiator has no antifreeze in it; all the stores are closed, and you must use whatever you have around the house to save your car’s engine.
It would be ideal to have antifreeze (ethylene glycol) but it’s quantity not quality that counts here.

33. The next morning you wonder if what you did saved you thousands of dollars on a new engine.

34. Rule one: What you add has to dissolve in water.
Rule two: If you add a solid that dissolves in water, it doesn’t matter what it is, just the amount.
Rule three: Just like antifreeze, your goal is to replace about ½ of the water with a solid or liquid that is miscible with water.
Examples 1: Sugar, salt, baking soda, shampoo, laundry detergent, pancake syrup.
Examples 2: Rubbing alcohol, brake fluid

35. How does it work? Get in the way. Create Disorder
Freezing point depression
How does it work? Get in the way. Create Disorder
Reason 1: As the water tries to freeze, the other molecules get in the way.
Reason 2: By adding these other substances, you’ve added disorder to the mixture. Nature tends to favor disorder (entropy). When water tries to freeze, it has to get organized, which will take more energy.

36. Vapor Pressure- Pressure of a gas above a liquid in a closed system
The vapor pressure of a liquid decreases when solute particles are added because some of the H2O molecules at the surface are replaced by solute particles which don’t vaporize.

37. This decreases the number of H2O molecules that can escape and become gas. The more particles dissolved, the lower the vapor pressure becomes.

38. An ionic substance that ionized into 3 particles (Ex
An ionic substance that ionized into 3 particles (Ex. CaCl2) has 3 times the effect of a molecular substance that does not ionize.
CaCl2 (l) → Ca2+ (aq) + 2Cl- (aq)
C12H22O11 (l) → C12H22O11 (aq)

39. Boiling Point - temperature at which the vapor pressure of a liquid equals atmospheric pressure
The addition of a solute increases the boiling point of a solution because the vapor pressure of the solution is lower. It must reach a higher temperature before both vapor pressures are equal.
The boiling point goes up 0.512ºC for every mole of particles dissolved in a kilogram of water.

40. Freezing Point - temperature at which the vapor pressure of the solid and liquid are equal.
The addition of a solute lowers the freezing point of a solvent. Each mole of solute particles, per kg of water, decreases the freezing point by 1.86ºC.
This is caused by the disruption of intermolecular forces between solvent molecules

41. EXAMPLE: Molality (m)- moles solute Kg solvent
Calculate the molality of a solution formed by dissolving 20.0 g of NaOH in 2000.g water.
m = ?
mol Kg
m =
mol solute = 20.0 g
1 mol
0.500 mol =
40.0 g
0.5 mol
2.000 Kg
m =
Kg solvent = g
= Kg
= m

42. Calculating BP elevation & FP depression
Tb = (Kb)(m)(i)
Tf = (Kf)(m)(i)
Tb = boiling pt. elevation
Tf = freezing pt. depression
m = molality
Kb = molal boiling pt. elevation constant
-for water = 0.512
Kf = molal freezing pt. depression constant
-for water = 1.86
i = van’t Hoff factor (number of particles or ions)
- for NaCl, i = 2
- for MgCl2, i = 3
- for C12H22O11, i = 1
i is always 1 for nonelectrolytes (they don’t ionize)
After you find ΔT, add or subtract it to or from the normal BP or FP.

43. Calculate the i value for each of the following:
CaCl2
CaCl2 → Ca Cl-
Al2(SO4)3
NaCl
i = 3
i = 5
i = 2

44. EXAMPLE: Tf = (1.86)(0.855 m)(2) Tf = 3.1806 ºC
Calculate the freezing point of a solution that contains 100.g of NaCl in 2.00 kg H2O.
Tf = (Kf)(m)(i)
1 mol NaCl
100. g NaCl
= 1.71 mol NaCl
Kf =
m =
i =
1.86
0.855 m 2
58.5 g NaCl
Tf = (1.86)(0.855 m)(2)
Tf = ºC
Tf = 0 ºC – ºC
= ºC

45. EXAMPLE: Tb = (0.512)(0.855 m)(2) Tb = 0.8755 ºC
Calculate the boiling point.
Tb = (Kb)(m)(i)
Tb = (0.512)(0.855 m)(2)
Kb =
m =
i =
0.512
0.855 m 2
Tb = ºC
Tb = 100 ºC ºC
= ºC

46. We can use freezing or boiling points to calculate the molecular mass of an unknown solute.
Use the FP or BP to calculate molality
Use molality to find moles
Use g/moles to get molar mass

47. EXAMPLE:
Find the molecular mass (GFM) of a molecular compound that has a freezing point of ºC when 5.20 g is dissolved in 500. g water.
Tf = (Kf)(m)(i)
0.460 ºC = (1.86)(x)(1)
mol Kg
x = 0.86 m
m =
Tf =
Kf =
m =
i =
0.460 ºC
1.86 x 1
mol =
Kg∙m
mol =
(0.500 Kg)(0.86m)
= 0.43 mol g
mol
5.20 g
0.43 mol
GFM = = =
12 g/mol