1. Unit 8- Solutions Aqueous Boiling point Colligative property
Concentrated
Dilute
Molarity
Parts per million (ppm)
Percent by volume
Percent mass
Saturated
Solubility
Solute
Solution
Solvent
Supersaturated
Unsaturated
Vapor
Vapor pressure

2. Mixtures Homogeneous Heterogeneous Solutions
Can be in all phases of matter (s, l or g)
Solids- metals mixed form an alloy. Brass= Zn + Cu
Gases- ex: air
Liquids- usually a solid or liquid dissolved in a liquid
Heterogeneous
Temporary mixtures are suspensions
Ex: muddy water
Stable mixtures are colloids
Ex: milk
Particles have charge, they repel each other so they can’t clump and settle out

3. Characteristics and parts of solutions
Homogeneous mixture
Are clear and don’t disperse light
Can have color
Will not settle out
Will pass through a filter
Parts
Solute- substance that is being dissolved
present in a smaller amount
Solvent- substance that dissolves solute
Present in greater amount
Water- most common solvent
Water solutions = aqueous (aq)

4. Separating mixtures
Physically- decanting, centrifuge, filtration, evaporation
By solubility- chromatography
By boiling point- distillation

5. Solubility
How much solute will dissolve in a certain amount of solvent at a certain temperature
Materials with high solubility are soluble
Materials with low solubility are insoluble
“like dissolves like”
Liquids that are soluble in each other are miscible
If they aren’t they are immiscible

6. Factors that affect solubility
“like dissolves like”; degree of solubility (how much dissolves)
Soap* - long nonpolar chain with polar end
Grease= nonpolar, water= polar (soap connects the two)
Ionic substances- dissociate- ions separate
Takes E, entropy (randomness) increases as ions dissociate and then decreases as water molecules assemble around the ions
Solute type
Non polar solvent
Polar solvent
Nonpolar
Soluble
Insoluble
Polar
Ionic
soluble

7. Soap P O-
CH3
CH2 7

8. Soap P O-
CH3
CH2
Hydrophobic non-polar end 8

9. Soap P O-
CH3
CH2
Hydrophilic polar end 9

10. P O-
CH3
CH2 _ 10

11. A drop of grease in water Grease is non-polar Water is polar
Soap lets you dissolve the non-polar in the polar. 11

12. Hydrophobic ends dissolve in grease12

13. Hydrophilic ends dissolve in water13

14. Water molecules can surround and dissolve grease.14

15. Factors con’t Temperature Pressure
As temp of water increases, solids become more soluble (they separate), gases become less soluble (they come together)
Pressure
Has little or no effect on solubility in solids and liquids
Does effect the solubility of gases in liquids, as pressure increases, solubility of a gas in the liquid increases
**greater surface area = increased dissolving process

16. Factor review Factor Affect on Solid Solute Affect on Gaseous Solute
Particle Size
Stirring
Amount of
dissolved solute
Temperature
Reducing particle size by crushing increases the rate by increasing surface area.
Not applicable
Increases the rate by exposing fresh solvent to solute and increasing kinetic energy.
Decreases the rate by increasing kinetic energy, thereby reducing solubility.
As the amount of dissolved solute increases, the rate decreases.
Reducing particle size by crushing increases the rate by increasing surface area.
As the temperature increases, the rate increases.
As the temperature increases, the rate decreases.

17. Looking at solubility in the Reference Tables
Table F - Shows ionic compounds
Predicts whether a precipitate will form when 2 ionic compounds are mixed
A Rx will take place if 1 or both of the products are listed as insoluble

18. Looking at solubility in the Reference Tables
Table G-
Shows the amount of a substance that can be dissolved in 100g of water at different temperatures
Lines show the max amount of solute at a given temp

19. Solids- lines that increase as temp increases
3 lines- NH3, HCl, SO2- show decrease in solubility as temp increases- they are GASES

20. If given pt is below the line the solution is unsaturated
TEMPERATURE
MASS of SOLUTE
SATURATED
UNSATURATED
SUPERSATURATED
If given pt is below the line the solution is unsaturated
If given pt is above the line the solution is supersaturated
If given pt is on the line the solution is saturated- solubility equilibrium

21.    The morphology diagram tells us a great deal about what kinds of snow crystals form under what conditions.  For example, we see that thin plates and stars grow around -2 C (28 F), while columns and slender needles appear near -5 C (23 F).  Plates and stars again form near -15 C (5 F), and a combination of plates and columns are made around -30 C (-22 F).    Furthermore, we see from the diagram that snow crystals tend to form simpler shapes when the humidity (supersaturation) is low, while more complex shapes at higher humidities.  The most extreme shapes -- long needles around -5C and large, thin plates around -15C -- form when the humidity is especially high.     Why snow crystal shapes change so much with temperature remains something of a scientific mystery.  The growth depends on exactly how water vapor molecules are incorporated into the growing ice crystal, and the physics behind this is complex and not well understood.  It is the subject of current research in my lab and elsewhere.

22. Saturation- occurs at a pt of solubility equilibrium

23. Saturation Test 1- Add a crystal Saturated Solution
The crystal falls to the bottom.
Unsaturated Solution
The crystal dissolves.
Supersaturated Solution
Excess solute precipitates.

24. Interpreting Solubility Curves
Which compound which is the least soluble at 10°C?
How many grams of potassium nitrate needed to saturate 100 mL of water at 52°C?
One hundred mL of a potassium chloride solution is saturated at 10°C. How many additional grams are needed to saturate the solution at 50°C?
At what temperature do saturated solutions of sodium chloride and potassium chloride contain the same mass of solute per 100 mL of water?
How many more grams of sulfur dioxide can be dissolved in 100 mL of water at 20°C than at 90°C?
KClO3
90 g
12 g
37°C
7 g

25. Determining concentration
Concentration- ratio of solute to solvent- (g/L)
Dilute- solution with little solute
Concentrated- solution with a lot of solute
Molarity- M- moles/liter of soln
M = moles of solute ÷ liters of solution
*many times grams must be converted to moles before molarity is calculated
For dilute solutions- ppm- parts per million
ppm =
Mass (Solute)
Mass (Solution)
× 1,000,000 ppm
______________
Percent mass =
Mass (Solute)
Mass (Solution)
× 100%
______________
Percent volume =
Volume (Solute)
Volume (Solution)
× 100%
______________

26. Concentration Problems
What is the concentration of salt in a 20.0 mL solution containing 18.3 g of dissolved NaCl?
Procedure: Divide the mass of the solute by the volume of the solvent or the solution.
Concentration =
18.3 g
20.0 mL
= 0.92 g/mL
__________

27. Percent by Mass Problems
What is the percent by mass of ethanol in a solution containing 2.3 g of ethanol (C2H5OH) dissolved in 10.0 g of water?
Step 1: Find the mass of the solution
10.0 g g = 12.3 g
Step 2: Divide the mass of the solute by the mass of the solution and multiply by 100 %
Percent mass =
2.3 g
12.3 g
× 100% = 19 %
__________

28. Percent by Volume Problems
What is the percent by volume of glycerine in a solution containing 18.2 mL of glycerine (C3H6O3) dissolved in 85.0 mL of water?
Step 1: Find the volume of the solution
18.2 mL mL = mL
Step 2: Divide the volume of the solute by the volume of the solution and multiply by 100 %
Percent volume =
18.2 mL
103.2 mL
× 100% = 17.6 %
______________

29. Parts per Million Problems
About g of hydrogen sulfide are dissolved in
10.0 g of water. Express this in parts per million.
Step 1: Find the mass of the solution
10.0 g g = g
Step 2: Divide the mass of the solute by the mass of the solution and multiply by 1,000,000 ppm.
ppm = g g
× 1,000,000 ppm = 350 ppm
______________

30. Molarity Sample Problems
Find the molarity of 100. mL of a solution that contains 0.25 moles of dissolved solute.
Step 1: Convert all volumes to liters
Step 2: Substitute values into the definitional equation
100. mL ×
0.001 L
____________
1 mL
= L
M =
0.25 mol
____________
0.100 L
= 2.5 M

31. Sample Problem 2
Find the molarity of 250. mL of a solution that contains 4.0 g of dissolved sodium hydroxide (NaOH).
Step 1: Find the GFM
Na = 23 × 1 = 23
O = 16 × 1 = 16
H = 1 × 1 = 1 40
Step 2: Convert all volumes to liters
Step 3: Substitute values into the correct equation
250. mL ×
0.001 L
____________
1 mL
= L

32. Sample Problem 3 Step 1: Convert all volumes to liters
How many moles of solute are dissolved in 30 mL of a
2 M solution?
Step 1: Convert all volumes to liters
Step 2: Substitute values into the correct equation
30 mL ×
0.001 L
____________
1 mL
= 0.03 L
0.06 mol

33. Sample Problem 4
How many grams of silver nitrate (AgNO3) are needed to prepare 200. mL of a 0.10 M solution?
Step 1: Find the GFM
Ag = 108 × 1 = 108
N = 14 × 1 = 14
O = 16 × 3 = 48
170
Step 2: Convert all volumes to liters
Step 3: Substitute values into the correct equation
200. mL ×
0.001 L
____________
1 mL
= L
= 3.4 g

34. How to prepare a solution
Add desired amount of solute to volumetric flask
Add some DI water and mix until dissolved
Fill flask to the line on the neck, close and mix
What mass of sodium bicarbonate is required to prepare 2 L of a .250M sodium bicarbonate solution?
Knowns: concentration= .250M
vol of soln= 2L
- unknown: mass of Na2CO3 = ____g
- Find moles of using molarity equation then convert moles to grams

35. Preparing Solutions sample problems:
How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution?
How many grams of CaCl2 are needed to make 625 mL of a 2.0 M solution?

36. Physical Properties of Solutions
Conductivity-
Depends on solution having charged particles that can move
Electrolyte- substances that conduct electricity when melted or dissolved in water (ionic substances)
Sports drinks etc.
Strong electrolytes- completely dissociate
Weak electrolytes- only some ions dissociate (weak conductivity)
Nonelectrolyte- no ions; doesn’t conduct (molecular substances)

38. Colligative Properties
Dependent on # of dissolved particles
A physical effect of solute on solvent
Degree of effect determined by conc. of dissolved particles
Freezing pt depression, boiling pt elevation, lowering vapor pressure

39. Freezing Point depression
Crystallization process in water is disrupted by solute particles
Salt on sidewalks and roads
Making ice-cream
1 mole of any particles will have the same effect on freezing pt of water
1 mole of particles lowers freezing pt 1.86oC of 1000g of water

40. Vapor Pressure The pressure that a vapor exerts
Table H shows vapor pressure of liquids at different temperatures

41. BP elevation and VP lowering
Boiling pt- when VP = atmospheric pressure
As temp increases, VP increases until it reaches atmospheric pressure; then it starts to boil
Table H will show at what temp water will boil at different pressures
As solute is added to a solution, vapor pressure decreases; therefore it takes longer for the VP to reach the atmospheric pressure and the water boils at a higher temperature