chapter 12 solutions copyright © 2008 by pearson education, inc. publishing as benjamin cummings
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Chapter 12 Solutions
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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.
Publishing as Benjamin Cummings
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
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chapter 12 Slide 4 of 41
Examples of SolutionsThe solute and solvent can be a solid, liquid, and/ora gas.
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
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
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.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
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
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
Chapter 12 Slide 12 of 41
Electrolytes and Nonelectrolytes
Chapter 12 Solutions
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
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
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
Chapter 12 Slide 17 of 41
Comparing Solutes in Solution
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
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.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
Chapter 12 Slide 22 of 41
Unsaturated Solutions
Unsaturated solutions • Contain less than the
maximum amount of solute.
• Can dissolve more solute. Dissolved
solute
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
Steady State
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.
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
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.
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
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.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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)
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)
Chapter 12 Slide 31 of 41
Molarity and Dilution
Chapter 12 Solutions
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
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.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
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.)
Chapter 12 Slide 36 of 41
Molarity Conversion Factors
The units of molarity are used as conversion factors in calculations with solutions.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
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
Chapter 12 Slide 39 of 41
DilutionIn a dilution
• Water is added.
• Volume increases.
• Concentration decreases.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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
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
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