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![Page 1: Solutions (ch.16) n Solution – a homogeneous mixture of pure substances n The SOLVENT is the medium in which the SOLUTES are dissolved. (The solvent.](https://reader035.fdocuments.us/reader035/viewer/2022062423/56649e9d5503460f94b9eb3e/html5/thumbnails/1.jpg)
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Solutions (ch.16) Solution – a homogeneous mixture
of pure substances
The SOLVENT is the medium in which the SOLUTES are dissolved. (The solvent is usually the most abundant substance.)
– Example: • Solution: Salt Water• Solute: Salt• Solvent: Water
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READ ONLY SECTION:
HOW DO SUBSTANCES DISSOLVE?– “According to the kinetic theory, the water
molecules in each glass of tea are always moving. Some moving water molecules collide with sugar crystals. When this happens, energy is transferred to the sugar molecules at the surface of the crystal.” (Holt, p. 192)
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The process of dissolution is favored by:
1) A decrease in the energy of the system (exothermic)
2) An increase in the disorder of the system (entropy)
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Liquids Dissolving in Liquids
Liquids that are soluble in one another (“mix”) are MISCIBLE.– “LIKE dissolves LIKE”
POLAR liquids are generally soluble in other POLAR liquids.
NONPOLAR liquids are generally soluble in other NONPOLAR liquids.
LIKE DISSOLVES LIKE : demo
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Factors affecting rate of dissolution: think iced tea vs. hot tea &
the type of sugar you use: cubes or granulated
1) Surface area / particle size– Greater surface area, faster it dissolves
2) Temperature– Most solids dissolve faster @ higher temps
3) Agitation– Stirring/shaking will speed up dissolution
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Saturation: a solid solute dissolves in a solvent until the soln is SATURATED
Unsaturated solution – is able to dissolve more solute
Saturated solution – has dissolved the maximum amount of solute
Supersaturated solution – has dissolved excess solute (at a higher temperature). Solid crystals generally form when this solution is cooled.
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ROCK CANDY, YUM!!
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Applying Concepts QUESTION
When the crystallization has stopped, will the solution be saturated or unsaturated?
answer
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ANSWER: SATURATED
Solution has the maximum amount of solute for a given quantity of solvent at a constant temperature and pressure.
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SOLUBILITY
Solubility = the amount of solute that will dissolve in a given amount of solvent
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Factors Affecting Solubility The nature of the solute and solvent:
different substances have different solubilities Temperature: many solids substances
become more soluble as the temp of a solvent increases; however, gases are less soluble in liquids at higher temps.
Pressure: Only affects the solubility of gases. As pressure increases, the solubility of gases increases.
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Concentration of Solution
Concentration refers to the amount of solute dissolved in a solution.
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*MOLARITY
nsol' L
solute mol(M)Molarity
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Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) solid Na2SO4
b) 5.00 M Na2SO4
L 2.50
xM 665.0
nsol' L
solute mol(M)Molarity
mol 1.6625x
g236mol 1
g 1.142SONa mol 6625.1 42
Dissolve 236 g of Na2SO4 in enough water
to create 2.50 Lof solution.
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MOLARITY BY DILUTION
When you dilute a solution, you can use this equation:
2211 VMVM
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Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:a) 5.00 M Na2SO4
L) M)(2.50 665.0()M)(V (5.00 1
mL 333 L 333.0V1
2211 VMVM
Add 0.333 L of Na2SO4 to 2.17 L of water.
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Example: Describe how you would prepare 2.50 L of 0.665 M Na2SO4 solution starting with:b) solid Na2SO4
L 2.50
xM 665.0
nsol' L
solute mol(M)Molarity
mol 1.6625x
g236mol 1
g 04.142SONa mol 6625.1 42
Dissolve 236 g of Na2SO4 in enough water
to create 2.50 Lof solution.
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MASS PERCENT
100nsol' of mass total
solute mass % mass
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MASS PERCENT
Example: What is the percent of NaCl in a solution made by dissolving 24 g of NaCl in 152 g of water?
100g 176
g 24%14%6.13
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*MOLALITY
solvent kg
solute mol (m)molality
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MOLALITY Example: What is the molality of a
solution that contains 12.8 g of C6H12O6 in 187.5 g water?
solvent kg
solute molm
g 180.18
mol 1OHC g8.12 6126 mol 07104.0
kg 1875.0
mol 0.07104m m 379.0
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MOLALITY Example: How many grams of H2O must be
used to dissolve 50.0 g of sucrose to prepare a 1.25 m solution of sucrose, C12H22O11?
solvent kg
solute molm
g 342.34
mol 1OHC g 50.0 112212 mol 0.1461
x
mol 0.1461m 1.25
OH 117gkg 1168.0 2x
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Colligative Properties of Solutions (chapter 16)
Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.
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Colligative Properties
Lowering vapor pressure Raising boiling point Lowering freezing point Generating an osmotic pressure
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2 to focus on…
Lowering vapor pressure Raising boiling point Lowering freezing point Generating an osmotic pressure
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Boiling Point Elevation
a solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent.
Like when adding salt to a pot of boiling water to make pasta
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Boiling Point Elevation
Tb = kbmwhere: Tb = elevation of boiling pt
m = molality of solute (mol solute/kg solvent)
kb = the molal boiling pt elevation constant
kb values are constants; see table 16.3 pg. 495
kb for water = 0.52 °C/m
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Ex: What is the normal boiling pt of a 2.50 m glucose, C6H12O6, solution?
“normal” implies 1 atm of pressure Tb = kbm
Tb = (0.52 C/m)(2.50 m)
Tb = 1.3 C
Tb = 100.0 C + 1.3 C = 101.3 C
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Freezing/Melting Point Depression
The freezing point of a solution is always lower than that of the pure solvent.
Like when salting roads in snowy places so the roads don’t ice over or when making ice cream
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Freezing/Melting Point Depression
Tf = kfmwhere: Tf = lowering of freezing point
m = molality of solute
kf = the freezing pt depression constant
kf for water = 1.86 °C/m kf values are constants;
see table 16.2 pg. 494
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Ex: Calculate the freezing pt of a 2.50 m glucose solution.
Tf = kfm
Tf = (1.86 C/m)(2.50 m)
Tf = 4.65 C
Tf = 0.00C - 4.65 C = -4.65C
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Textbook pg. 495 #34
Calculate the freezing-point depression (ΔTf) of a benzene solution containing 400. g of benzene and 200. g of acetone, C3H6O (solute).
Kf for benzene is 5.12 °C/m
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Answer
ΔTf = Kf x m
ΔTf = (5.12 °C/m) x (m)
m =1
200. 58.09
8.61 0.400
molg x
mol solute gm
kg solvent kg
So, ΔTf = (5.12 °C/m) x (8.61 m) = 44.1 °C
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Do the following on your paper
1. What is the boiling point of each solution?a) 0.50 mol glucose in 1000. g water
b) 1.50 mol NaCl in 1000. g water 2. What is the freezing point of each solution?
a) 1.40 mol Na2SO4 in 1750 g water
b) 0.060 mol MgSO4 in 100. g water
answers
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ANSWERS
1a) 100.26 °C 1b) 100.78 °C 2a) -1.49 °C 2b) -1.1 °C
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More calcs.
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Ex: When 15.0 g of ethyl alcohol, C2H5OH, is dissolved in 750 grams of formic acid, the freezing pt of the solution is 7.20°C. The freezing pt of pure formic acid is 8.40°C. Determine Kf for formic acid.
Tf = kfm
1.20 C= (kf)( 0.4340 m)
kf = 2.76 C/m
mol 3255.0g 46.08
mol 1OHHC g 0.15 52
m 4340.0kg 75.0
mol 0.3255
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EXTRA NOTES
Important info for lab, etc.
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Electrolytes and Colligative Properties
• Colligative properties depend on the # of particles present in solution.• Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal conc.
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Electrolytes and Colligative Properties For example, the freezing pt of water is
lowered by 1.86°C with the addition of any molecular solute at a concentration of 1 m.– Such as C6H12O6, or any other covalent
compound
However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72°C…double that of a molecular solute.– NaCl Na+ + Cl- (2 particles)
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Electrolytes - Boiling Point Elevation and Freezing Point Depression
The relationships are given by the following equations:
Tf = kf ·m·n or Tb = kb·m·n
Tf/b = f.p. depression/elevation of b.p.m = molality of solute
kf/b = b.p. elevation/f.p depression constantn = # particles formed from the dissociation of
each formula unit of the solute
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Ex: What is the freezing pt of a 1.15 m sodium chloride solution?
NaCl Na+ + Cl- n=2
Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(2)
Tf = 4.28 C
Tf = 0.00C - 4.28 C = -4.28C