1. I. The Nature of Mixtures

2. Mixtures
Heterogeneous Mixture – a non-uniform or unequal blend of two or more pure substances
Suspension – mixture containing particles that will settle out
Colloid – mixture containing particles with a size of 1nm – 1000nm, and do not separate – stay suspended

3. Mixtures
Tyndall Effect – Caused by dilute colloids, which appear to be homogeneous.
Is the scattering of light as it passes though a dilute colloid

4. Mixtures Solutions - homogeneous mixtures
Solute - substance being dissolved
Solvent - present in greater amount

5. Homogeneous Mixtures
A substance that dissolves in a solvent is soluble.
Two liquids that are soluble in each other in any proportion are miscible.
A substance that does not dissolve in a solvent is insoluble.
Two liquids that can be mixed but separate shortly after are immiscible.

6. Solutions
Solvation – the process of dissolving – sugar dissolves in to water
solute particles are surrounded by solvent particles
First...
solute particles are separated and pulled into solution
Then...

7. Solvation Non- Electrolyte Weak Electrolyte Strong Electrolyte+
sugar - +
acetic acid - +
salt
Non-
Electrolyte
Weak
Electrolyte
Strong
Electrolyte
solute exists as
molecules only
solute exists as
ions and
molecules
solute exists as
ions only

8. Solvation C6H12O6(s)  C6H12O6(aq) Molecular Solvation
molecules stay intact
C6H12O6(s)  C6H12O6(aq)

9. NaCl(s)  Na+(aq) + Cl–(aq)
Solvation
Dissociation
separation of an ionic solid into aqueous ions
NaCl(s)  Na+(aq) + Cl–(aq)

10. Solvation
“Like Dissolves Like”
NONPOLAR
POLAR

11. Solvation
Heat of Solution – energy change that occurs during the formation of the solution.
Exothermic – solvation produces a warm solution
Endothermic – solvation produces a cold solution

12. Factors Affecting Solvation (solubility)
Agitation – stirring or shaking – allows for more collisions (mixing) between solute and solvent
Surface Area – smaller pieces – more collisions
Temperature – in general solids dissolve faster in higher temps. – more collisions
Opposite for gases – colder the better

13. Solubility Solubility
maximum grams of solute that will dissolve in 100 g of solvent at a given temperature
varies with temp
based on a saturated soln
Book Reference: p 969 R-3 and p 974 R-8

14. Solubility Rules In general: Group I ions and Ammonium are soluble
Acetates and Nitrates are soluble
Cl, Br, I are soluble, except with Pb, Ag, Hg2+2
Sulfates are soluble, except with Ba, Sr, Pb, Ca, Ag, Hg2+2
Carbonates, Hydroxides, oxides, sulfides, phosphates are INsoluble

15. Solubility UNSATURATED SOLUTION more solute dissolves
no more solute dissolves
SUPERSATURATED SOLUTION
becomes unstable, crystals form
concentration

16. Solubility Solubility Curve
shows the dependence of solubility on temperature

17. Solubility Solids are more soluble at... high temperatures.
Gases are more soluble at...
low temperatures &
high pressures (Henry’s Law).

18. Solubility Types of Reactions in Aqueous Solutions
When two solutions that contain ions as solutes are combined, the ions might react.
If they react, it is always a double replacement reaction.
Three products can form: precipitates, water, or gases.

19. Solubility Replacement Reactions
Double Replacement Reactions occur when ions exchange between two compounds.
This figure shows a generic double replacement equation.

20. Solubility
Aqueous solutions of sodium hydroxide and copper(II) chloride react to form the precipitate copper(II) hydroxide.
2NaOH(aq) + CuCl2(aq) → 2NaCl(aq) + Cu(OH)2(s)
Ionic equations that show all of the particles in a solution as they actually exist are called complete ionic equations.
2Na+(aq) + 2OH–(aq) + Cu2+ (aq)+ 2Cl–(aq) → 2Na+(aq) + 2Cl–(aq) + Cu(OH)2(s)

21. Types of Reactions in Aqueous Solutions (cont.)
Ions that do not participate in a reaction are called spectator ions and are not usually written in ionic equations.
Formulas that include only the particles that participate in reactions are called net ionic equations.
2OH–(aq) + Cu2+(aq) → Cu(OH)2(s)

22. Solutions
II. Concentration

23. A. Concentration is...
a measure of the amount of solute dissolved in a given quantity of solvent
A concentrated solution has a large amount of solute
A dilute solution has a small amount of solute
thus, only qualitative descriptions

24. A. Concentration Describing Concentration Quantitatively % by mass
% by volume
Molarity
Molality
Mole Fraction

25. B. Percent

26. B. Percent
What is the percent solution if 25 mL of CH3OH is diluted to 150 mL with water?
17%
4.8 g of NaCl are dissolved in 82.0 mL of solution. What is the percent of the solution?
5.53%
How many grams of salt are there in 52 mL of a 6.3 % solution?
3.3g

27. C. Molarity
Volume of solvent only
1 kg water = 1 L water

28. C. Molarity
Find the molarity of a solution containing 75 g of MgCl2 in 250 mL of water.
75 g MgCl2
1 mol MgCl2
95.21 g MgCl2
0.25 L water
= 3.2 M MgCl2

29. C. Molarity
How many grams of NaCl are req’d to make a 1.54 M solution using L of water?
0.500 L water
1.54 mol NaCl
1 L water
58.44 g NaCl
1 mol NaCl
= 45.0 g NaCl

30. D. Molality 1 L water = 1 Kg water so …
Molality (m) = moles of solute / Kg of solvent
Basically the same calculations, but different units

31. E. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.

32. E. Dilution
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3

33. F. Preparing Solutions 1.54m NaCl in 0.500 kg of water
500 mL of 1.54M NaCl
mass 45.0 g of NaCl
add water until total volume is 500 mL
mass 45.0 g of NaCl
add kg of water
500 mL water
45.0 g NaCl
500 mL
mark
volumetric
flask

34. F. Preparing Solutions 250 mL of 6.0M HNO3 by dilution
measure 95 mL of 15.8M HNO3
95 mL of 15.8M HNO3
combine with water until total volume is 250 mL
250 mL mark
Safety: “Do as you oughtta, add the acid to the watta!”
water for safety

35. Solution Preparation Lab
Turn in one paper per team.
Complete the following steps:
A) Show the necessary calculations.
B) Write out directions for preparing the solution.
C) Prepare the solution.
For each of the following solutions:
1) mL of 0.500M NaCl
2) 0.250m NaCl in mL of water
3) mL of 1.00M Red Solution from 12.1M concentrate.

36. III. Colligative Properties
Solutions
III. Colligative Properties

37. A. Definition Colligative Property
property that depends on the concentration of solute particles, not their identity

38. A. Electrolytes
Dissolved Solute particles disrupt the normal Intermolecular Forces of the Solvent
Molecules – Count as one dissolved solute particle
Example: CH3OH = 1 solute particle
Ionic Compounds – number of solute particles is equal to the total number of ions in the neutral formula
Example: AlCl3 = 4 solute particles
1 Al and 3 Cl ions

39. B. Types Vapor Pressure Lowering
Solutions have a lower vapor pressure than the original pure solvent
Because the solute particles are attracted to solvent particles cause more IMF
So more IMF = less solvent particles becoming vapors

40. B. Types Freezing Point Depression (tf)
f.p. of a solution is lower than f.p. of the pure solvent
Boiling Point Elevation (tb)
b.p. of a solution is higher than b.p. of the pure solvent

41. Freezing Point Depression
B. Types
Freezing Point Depression

42. Boiling Point Elevation
B. Types
Boiling Point Elevation
Solute particles increase IMF of the solvent.

43. C. Phase Diagram

44. D. Applications salting icy roads making ice cream
Antifreeze - cars (-64°C to 136°C)

45. E. Calculations  t: change in temperature (°C)
t = k · m · n
 t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
m: molality (m)
n or i: # of particles

46. E. Calculations # of Particles Nonelectrolytes (covalent)
remain intact when dissolved
1 particle
Electrolytes (ionic)
dissociate into ions when dissolved
2 or more particles

47. E. Calculations
At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil?
GIVEN:
b.p. = ?
tb = ?
kb = 3.56°C·kg/mol
WORK:
m = 0.73mol ÷ 0.225kg
tb = (3.56°C·kg/mol)(3.2m)(1)
tb = 11°C
b.p. = 181.8°C + 11°C
b.p. = 193°C
m = 3.2m
n = 1
tb = kb · m · n

48. E. Calculations
Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
tf = ?
kf = 1.86°C·kg/mol
WORK:
m = 0.48mol ÷ 0.100kg
tf = (1.86°C·kg/mol)(4.8m)(2)
tf = 18°C
f.p. = 0.00°C - 18°C
f.p. = -18°C
m = 4.8m
n = 2
tf = kf · m · n