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Transcript of Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings.
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Chapter 12 Solutions
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Chapter 12 Slide 2 of 41
Solute and Solvent
Solutions• Are
homogeneous mixtures of two or more substances.
• Consist of a solvent and one or more solutes. Copyright © 2008 by Pearson Education, Inc.
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Chapter 12 Slide 3 of 41
Solutes• Spread evenly
throughout the solution.
• Cannot be separated by filtration.
• Can be separated by evaporation.
• Are not visible, but can give a color to the solution.
Nature of Solutes in Solutions
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Chapter 12 Slide 4 of 41
Examples of SolutionsThe solute and solvent can be a solid, liquid, and/ora gas.
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Chapter 12 Slide 5 of 41
Identify the solute in each of the following solutions:
1. _____ sugar (A) and _________ water (B)
2. _______ of ethyl alcohol (A) and ________ of methyl alcohol (B)
3. ________ water (A) and _______ NaCl (B)
4. Air: _________ (A) and __________ N2 (B)
Learning Check
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Chapter 12 Slide 6 of 41
WaterWater• Is the most common solvent.• Is a polar molecule.• Forms hydrogen bonds between the
hydrogen atom in one molecule and the oxygen atom in a different water molecule.
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Chapter 12 Slide 7 of 41
Formation of a Solution
Na+ and Cl- ions• On the surface of a
NaCl crystal are attracted to polar water molecules.
• In solution are hydrated as several H2O molecules surround each.
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Chapter 12 Slide 8 of 41
When NaCl(s) dissolves in water, the process can be written as:
H2O
NaCl(s) Na+(aq) + Cl- (aq)
solid separation of ions
The Na+ ions are attracted to the oxygen atom ( -) of water.
The Cl- ions are attracted to the hydrogen atom (+) of water.
Equations for Solution Formation
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Chapter 12 Slide 9 of 41
Two substances form a solution • when there is an attraction between the
particles of the solute and solvent.
• when a polar solvent such as water dissolves polar solutes such as sugar and ionic solutes such as NaCl.
• when a nonpolar solvent such as hexane (C6H14) dissolves nonpolar solutes such as oil or grease.
Like Dissolves Like
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Chapter 12 Slide 10 of 41
Water and a Polar Solute
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Chapter 12 Slide 11 of 41
Like Dissolves Like
Solvents Solutes
Water (polar) Ni(NO3)2
CH2Cl2(nonpolar) (polar)
I2 (nonpolar)Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Chapter 12 Slide 12 of 41
Electrolytes and Nonelectrolytes
Chapter 12 Solutions
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Chapter 12 Slide 13 of 41
In water, • Strong electrolytes produce ions and conduct an
electric current. • Weak electrolytes produce a few ions. • Nonelectrolytes do not produce ions.
Solutes and Ionic Charge
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Chapter 12 Slide 14 of 41
Strong electrolytes • Dissociate in water producing positive and negative
ions.• Conduct an electric current in water.• In equations show the formation of ions in aqueous (aq)
solutions.
H2O 100% ions
NaCl(s) Na+(aq) + Cl− (aq)
H2O
CaBr2(s) Ca2+(aq) + 2Br− (aq)
Strong Electrolytes
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Chapter 12 Slide 15 of 41
A weak electrolyte• Dissociates only slightly in water.• In water forms a solution of only a few ions and
mostly undissociated molecules.
HF(g) + H2O(l) H3O+(aq) + F- (aq)
NH3(g) + H2O(l) NH4+(aq) + OH- (aq)
Note: Unequal lengths of the arrows
Weak Electrolytes
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Chapter 12 Slide 16 of 41
Nonelectrolytes
Nonelectrolytes • Dissolve as
molecules in water. • Do not produce ions
in water.• Do not conduct an
electric current.
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Sucrose
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Chapter 12 Slide 17 of 41
Comparing Solutes in Solution
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Chapter 12 Slide 18 of 41
Chapter 12 Solutions
Solubility
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Chapter 12 Slide 19 of 41
Solubility • Is the maximum amount of solute that dissolves
in a specific amount of solvent. • Can be expressed as grams of solute in 100
grams of solvent, usually water.
g of solute
100 g water
Solubility
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Chapter 12 Slide 20 of 41
Effect of Temperature on Solubility
Solubility• Depends on
temperature.• Of most solids
increases as temperature increases.
• Of gases decreases as temperature increases.
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Chapter 12 Slide 21 of 4121
Solubility and Pressure
Henry’s law states • the solubility of a
gas in a liquid is directly related to the pressure of that gas above the liquid
• at higher pressures, more gas molecules dissolve in the liquid
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Chapter 12 Slide 22 of 41
Unsaturated Solutions
Unsaturated solutions • Contain less than the
maximum amount of solute.
• Can dissolve more solute. Dissolved
solute
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Chapter 12 Slide 23 of 41
Saturated Solutions
Saturated solutions • Contain the maximum amount
of solute that can dissolve. • Have undissolved solute at the
bottom of the container. • Have equal rates at which
solute dissolves and crystallizes
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Steady State
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Chapter 12 Slide 24 of 41
Comparing Unsaturated and Saturated Solutions
More solute can dissolve in an unsaturated solution but not in a saturated solution.
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Chapter 12 Slide 25 of 41
Soluble and Insoluble Salts
Ionic compounds that• dissolve in water are soluble salts• do not dissolve in water are
insoluble salts
Double Replacement AB + CD → AD + CB
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Chapter 12 Slide 26 of 41
Solubility Rules
Soluble salts • typically contain at least one ion from Groups
1A(1), NO3−, or C2H3O2
− (acetate)
• Most other combinations are insoluble.
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Chapter 12 Slide 27 of 41
Using the Solubility Rules
The solubility rules predict• if a salt is soluble in water • that a solid forms if ions of an insoluble
salt are present
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Chapter 12 Slide 28 of 41
Formation of a Solid
When solutions of salts are mixed, • A solid forms if ions of an insoluble salt
are present.
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Chapter 12 Slide 29 of 41
Equations for Forming Solids
A molecular equation shows the formulas of the
compounds.
Pb(NO3)2(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq)
An ionic equation shows the ions of the compounds.
Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq)
PbCl2(s) + 2Na+(aq) + 2NO3−(aq)
A net ionic equation shows only the ions that form asolid. Ions remaining in solution are spectator ions.Pb2+(aq) + 2Cl−(aq) PbCl2(s)
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Chapter 12 Slide 30 of 41
Equations for the Insoluble SaltSTEP 1 Observe the ions in the reactants.
Pb2+(aq) + 2NO3−(aq)
2Na+(aq) + 2Cl−(aq) STEP 2 Determine if any new ion combinations are insoluble salts. Yes. PbCl2(s)
STEP 3 Ionic equation with insoluble salt product.
Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq)
PbCl2(s) + 2Na+(aq) + 2NO3−(aq)
STEP 4 Net ionic equation.
Pb2+(aq) + 2Cl−(aq) PbCl2(s)
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Chapter 12 Slide 31 of 41
Molarity and Dilution
Chapter 12 Solutions
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Chapter 12 Slide 32 of 41
Molarity (M)Molarity (M) is
• A concentration term for solutions.
• The moles of solute in 1 L solution.
• moles of soluteliter of solution
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Chapter 12 Slide 33 of 41
Preparing a 1.0 Molar Solution
A 1.00 M NaCl solution is prepared• By weighing out 58.5 g NaCl (1.00 mol) and• Adding water to make 1.00 liter of solution.
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Chapter 12 Slide 34 of 41
What is the molarity of 0.500 L NaOH solution if itcontains 6.00 g NaOH?STEP 1 Given 6.00 g NaOH in 0.500 L solution
Need molarity (mol/L)
STEP 2 Plan g NaOH mol NaOH molarity
STEP 3 Conversion factors 1 mol NaOH = 40.00 g
1 mol NaOH and 40.00 g NaOH 40.00 g NaOH 1 mol NaOH
Calculation of Molarity
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Chapter 12 Slide 35 of 41
STEP 4 Calculate molarity.6.00 g NaOH x 1 mol NaOH = 0.150 mol
40.00 g NaOH
0.150 mol = 0.300 mol = 0.300 M NaOH 0.500 L 1 L
Calculation of Molarity (cont.)
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Chapter 12 Slide 36 of 41
Molarity Conversion Factors
The units of molarity are used as conversion factors in calculations with solutions.
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Chapter 12 Slide 37 of 41
Molarity in Calculations
How many grams of KCl are needed to prepare
125 mL of a 0.720 M KCl solution?
STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl
Need Grams of KCl
STEP 2 Plan L KCl mol KCl g KCl
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Chapter 12 Slide 38 of 41
Molarity in Calculations (cont.)
STEP 3 Conversion factors 1 mol KCl = 74.55 g
1 mol KCl and 74.55 g KCl 74.55 g KCl 1 mol KCl
1 L KCl = 0.720 mol KCl 1 L and 0.720 mol KCl
0.720 mol KCl 1 L
STEP 4 Calculate grams.0.125 L x 0.720 mol KCl x 74.55 g KCl = 6.71 g KCl 1 L 1 mol KCl
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Chapter 12 Slide 39 of 41
DilutionIn a dilution
• Water is added.
• Volume increases.
• Concentration decreases.
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Chapter 12 Slide 40 of 41
Comparing Initial and Diluted Solutions
In the initial and diluted solution
• The moles of solute are the same.
• The concentrations and volumes are related by the equation
M1V1 = M2V2
initial diluted
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Chapter 12 Slide 41 of 41
Dilution Calculations
What is the molarity if 0.180 L of 0.600 M KOH isdiluted to a final volume of 0.540 L?STEP 1 Prepare a table:
M1= 0.600 MV1 = 0.180 L
M2= ? V2 = 0.540 L
STEP 2 Solve dilution equation for unknown.
M1V1 = M2V2 M1V1/ V2 = M2
STEP 3 Set up and enter values:
M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M
V2 0.540 L