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

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Page 1: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 12 Solutions

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

Page 2: 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

Page 3: Chapter 12 Solutions 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

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Page 4: Chapter 12 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.

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Page 5: Chapter 12 Solutions 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

Page 6: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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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Page 7: Chapter 12 Solutions 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

Page 8: Chapter 12 Solutions 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

Page 9: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Page 10: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 12 Slide 10 of 41

Water and a Polar Solute

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Page 11: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Page 12: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 12 Slide 12 of 41

Electrolytes and Nonelectrolytes

Chapter 12 Solutions

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Page 13: 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

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Page 14: Chapter 12 Solutions 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

Page 15: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Page 16: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Page 17: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 12 Slide 17 of 41

Comparing Solutes in Solution

Page 18: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 12 Slide 18 of 41

Chapter 12 Solutions

Solubility

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Page 19: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Page 20: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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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Page 21: Chapter 12 Solutions 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

Page 22: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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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Page 23: Chapter 12 Solutions 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

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Steady State

Page 24: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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.

Page 25: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Page 26: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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.

Page 27: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Page 28: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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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Page 29: Chapter 12 Solutions 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)

Page 30: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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)

Page 31: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 12 Slide 31 of 41

Molarity and Dilution

Chapter 12 Solutions

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Page 32: 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

Page 33: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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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Page 34: Chapter 12 Solutions 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

Page 35: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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.)

Page 36: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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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Page 37: Chapter 12 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

Page 38: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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

Page 39: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 12 Slide 39 of 41

DilutionIn a dilution

• Water is added.

• Volume increases.

• Concentration decreases.

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Page 40: Chapter 12 Solutions 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

Page 41: Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

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