Notes One Unit Seven

1
Notes One Unit Seven Chapter 13 Solutions

• Definitions
• Types of Mixtures
• Example Solutions
• Factors Affecting Solubility
• Like Dissolves Like
• Solubility of Solids Changes with Temperature
• Solubility of Gases Changes with Temperature
• Pressure Factor
• Molar Concentration
• Finding Molarity From Mass and Volume
• Finding Mass from Molarity and Volume
• Finding Volume from Molarity and Mass

Pages 466-486
2
Definitions

• Solutions are homogeneous mixtures.
• Uniform throughout.
• Solvent.
• Determines the state of solution
• Is the largest component.
• Solute.
• Other components dissolved in solvent.

3
Common Mixtures
SOLUTE
SOLVENT
Type
EXAMPLE
liquid
liquid
emulsion
mayonnaise
gas
liquid
liquid foam
whipped cream
solid
dust in air
gas
aerosol
liquid
hair spray
gas
aerosol
solid
ruby glass
solid
solid
liquid
pearl
solid
emulsion
gas
Styrofoam
solid
solid foam
4
Solution Types
SOLUTE
SOLVENT
PHASE
EXAMPLE
gas
gas
gas
a i r
gas
liquid
liquid
soda pop
liquid
liquid
liquid
antifreeze
liquid
filling
solid
solid
solid
seawater
liquid
liquid
solid
brass
solid
solid
5
Factors Affecting Solubility

• 1. Nature of Solute / Solvent.
• 2. Temperature Increase
• i) Solid/Liquidincrease
• ii) gasdecreases
• 3. Pressure Factor -
• i) Solids/Liquids - Very little
• ii) gas - Increases.
• iii) squeezes gas into solution.

6
Like Dissolves Like

• Non-polar does not dissolve polar
• Oil in H2O
• Polar dissolves Polar
• C2H5OH in H2O
• Ionic compounds soluble in polar solvents
• NaCl in H2O

7
Solubility of solids Changes with Temperature

• Does Solubility always increase for solids or
  gases?
• How many grams potassium chromate will dissolve
  100g water at 70oC?
• 70g
• How many grams lead(II) nitrate will precipitate
  from 250g water as it cools from 70oC to 50oC?

solids
gases
101g
82g
250g
_____
19gx
48g
100g
8
Calculating Freezing Point Depression Mass

• A solution containing 1.89 g of methanol in 51.96
  g of water freezes at -3.4oC. Calculate the
  molecular weight of methanol .
• 1.)Calculate Temperature Change
• ?Tb
• 2.)Calculate moles per Kilograms
• ?Tf Kf x m ? m ?Tf /Kf
• m 0.500m/kg
• 3.)Calculate grams / kilograms
• g
• g 23.0g/kg
• fm
• 46.0g/m

0.000oC-
0.929oC
0.929oC
0.929
1.858oC/ m
m
7.67 g
0.3330kg
23.0 g/
0.500m
9
Solubility of Gases Changes with Temperature

• a) Why are fish stressed, if the temperature of
  the water increases?
• How much does the solubility of oxygen change,
  for a 20oC to 60oC change?
• 0.90-0.600.30mg

0.90mg
0.60mg
10
Pressure Factor
Greater pressure
more dissolved gas
11
Pressure Factor
12
Molar Concentration

• Mn/V
• MxVn
• Vn/M

13
Finding Molarity From Mass and Volume

• Calculate molarity for 25.5 g of NH3 in 600. mL
  solution.
• 1) Calculate Formula Mass
• 2) Calculate the moles of solute
• 3) Calculate the Moles/Liters Ratio
• M n / V
• M 1.50 moles / 0.600 L
• M 2.52 mol/L

E

Mass
14.0
N
1x
14.0
3.0
H
3x
1.0
17.0g/m
25.5g

17.0g/m
1.50m
14
Finding Volume from Molarity and Mass

• How many milliliters of 2.50M solution can be
  made using 25.5grams of NH3?
• 1)Calculate formula mass
• 2)Calculate the moles of solute
• 3)Calculate Volume from Moles/Liters Ratio
• V1.50m/2.50M0.600L
• 600.mL solution

E

Mass
14.0
N
1x
14.0
3.0
H
3x
1.0
17.0g/m
25.5g

17.0g/m
1.50m
15
Finding Mass from Molarity and Volume

• How many grams of NH3 are in 600. mL solution at
  2.50M?
• 1) Calculate formula mass
• 2) Calculate moles
• Mn/V ? nMxV
• n1.50m
• 3) Calculate mass
• ng/MW ? gMWxn
• g25.5g NH3

E

Mass
14.0
N
1x
14.0
3.0
H
3x
1.0
17.0g/m
n
2.50M
x
0.600L
x
g
(17.0g/m)
(1.50m)
16
Notes Two Unit Seven Chapter 13 Solutions

• Saturated versus Unsaturated
• Colligative properties of water
• Forming a Saturated Solution
• How Does a Solution Form?
• Colligative Properties
• Vapor Pressure
• Boiling and Freezing Point
• BP Elevation and Freezing FP Depression
• Calculating Freezing Point Depression Mass

Pages 487-501
17
Characteristics of Saturated Solutions
water
precipitate
precipitate
dissolve
dissolve
dissolve
Solid
Unsaturated
Unsaturated
Saturated
Dynamic
Equilibrium
Cooling causes precipitation.
Warming causes dissolving.
18
Solvation

• As a solution forms, the solvent pulls solute
  particles apart and surrounds, or solvates, them.

19
Colligative Properties

• Colligative properties depend on moles dissolved
  particles.
• Vapor pressure lowering
• Boiling point elevation
• Melting point depression
• Osmotic pressure

20
Vapor Pressure

• vapor pressure of a solvent.
• vapor pressure of a solution.

21
Phase Diagram
solid
Critical point?
melting?
?freezing
1 atm
Liquid
vaporizing?
Pressure
Triple point?
?condensing
gas
sublimation?
?depostion
NFP
NBP
Temperature
0.0oC
100.0oC
22
One Molal Solution of Water
solid
1 atm
Liquid
Pressure
gas
?Kf?
?Kb?
Temperature
0.512oC
1.858oC
23
BP Elevation Constants (Kb)FP Depression
Constants( Kf)

24
BP Elevation and FP Depression

• ?Tb Kb ? m
• Kb0.5120C/m
• ?Tf Kf ? m
• Kf1.8580C/m

25
Molarity versus Molality
moles of solute
________________
Molality (m)

kilograms solvent
moles of solute
________________
Molarity (M)

liters of solution
26
Calculating Tf andTb

• Calculate the freezing and boiling points of a
  solution made using 1000.g antifreeze (C2H6O2) in
  4450g water.
• 1) Calculate Moles
• 2) Calculate molality
• 3) Calculate Temperature Change
• ?tKxm
• ?Tf
• Tf
• ?Tb
• Tb

E

Mass
24.0
C
2x
12.0
1000.g
62.0g/mol
16.1 moles
32.0
O
2x
16.0
6.0
H
6x
1.0
16.1 mole
4.45 Kg water
3.62m
62.0g/m
(1.858oC/m)
(3.62 m)
6.73oC
0.000oC-
6.73oC
-6.73oC
(0.512oC/m)
(3.62 m)
1.96oC
100.000oC
1.96oC
101.96oC
27
Calculating Boiling Point Elevation Mass

• A solution containing 18.00 g of glucose in 150.0
  g of water boils at 100.34oC. Calculate the
  molecular weight of glucose.
• 1.)Calculate Temperature Change
• ?Tb
• 2.)Calculate moles per Kilograms
• ?Tb Kb x m ? m ?Tb /Kb
• m 0.67m/kg
• 3.)Calculate grams / kilograms
• g
• g 120 g/kg
• MW120 g/0.67m
• 180g/m

100.34oC-
100.00oC
0.34oC
0.34
0.512oC/ m
m
18.00 g
0.1500kg
28
Notes Three Unit Seven

• Ice-cream Lab A Calculating Freezing Point
• Depression Mass
• Colligative Properties of Electrolytes
• Distillation
• Osmotic Pressure
• Dialysis

Pages 487-501
29
Ice-cream
30
Calculating Freezing Point Depression Mass

• 1.)Calculate Temperature Change
• 2.)Calculate moles per Kilograms
• 3.)Calculate grams / kilograms

31
Colligative Properties of Electrolytes

• Colligative properties depend on the number of
  particles dissolved.
• NaCl?Na1Cl-1
  CH3OHAl2(SO4)3?2Al3 3SO4-2
  C6H12O6

32
Distillation
33
Distillation
34
Osmotic Pressure

• Hypertonic
• gt 0.92 (9.g/L)
• Crenation
• Isotonic Saline
• 0.92 (9.g/L)
• Hypotonic
• lt 0.92 (9.g/L)
• Rupture

35
Dialysis
36
Kidney
37
Dialysis