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Transcript of Chapter 15 Solutions. Chapter 15 Table of Contents Copyright © Cengage Learning. All rights...
![Page 1: Chapter 15 Solutions. Chapter 15 Table of Contents Copyright © Cengage Learning. All rights reserved 2 15.1 Solubility 15.2 Solution Composition: An Introduction.](https://reader035.fdocuments.us/reader035/viewer/2022062421/56649cdc5503460f949a74ca/html5/thumbnails/1.jpg)
Chapter 15
Solutions
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Chapter 15
Table of Contents
Copyright © Cengage Learning. All rights reserved 2
15.1 Solubility
15.2 Solution Composition: An Introduction
15.3 Solution Composition: Mass Percent
15.4 Solution Composition: Molarity
15.5 Dilution
15.6 Stoichiometry of Solution Reactions
15.7 Neutralization Reactions
15.8 Solution Composition: Normality
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Chapter 15
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What is a Solution?
• Solution – homogeneous mixture Solvent – substance present in largest
amount Solutes – other substances in the solution Aqueous solution – solution with water as the
solvent
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Section 15.1
Solubility
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Various Types of Solutions
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Section 15.1
Solubility
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• Ionic substances breakup into individual cations and anions.
Solubility of Ionic Substances
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Section 15.1
Solubility
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• Polar water molecules interact with the positive and negative ions of a salt.
Solubility of Ionic Substances
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Section 15.1
Solubility
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• Ethanol is soluble in water because of the polar OH bond.
Solubility of Polar Substances
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Section 15.1
Solubility
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• Why is solid sugar soluble in water?
Solubility of Polar Substances
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Section 15.1
Solubility
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• Nonpolar oil does not interact with polar water.• Water-water hydrogen bonds keep the water
from mixing with the nonpolar molecules.
Substances Insoluble in Water
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Section 15.1
Solubility
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• A “hole” must be made in the water structure for each solute particle.
• The lost water-water interactions must be replaced by water-solute interactions.
• “like dissolves like”
How Substances Dissolve
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Section 15.1
Solubility
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Concept Check
Which of the following solutes will generally not dissolve in the specified solvent? Choose the best answer. (Assume all of the compounds are in the liquid state.)
a) CCl4 mixed with water (H2O)
b) NH3 mixed with water (H2O)
c) CH3OH mixed with water (H2O)
d) N2 mixed with methane (CH4)
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Section 15.2
Solution Composition: An Introduction
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• The solubility of a solute is limited. Saturated solution – contains as much
solute as will dissolve at that temperature.
Unsaturated solution – has not reached the limit of solute that will dissolve.
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Section 15.2
Solution Composition: An Introduction
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• Supersaturated solution – occurs when a solution is saturated at an elevated temperature and then allowed to cool but all of the solid remains dissolved. Contains more dissolved solid than a
saturated solution at that temperature. Unstable – adding a crystal causes
precipitation.
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Section 15.2
Solution Composition: An Introduction
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• Solutions are mixtures. • Amounts of substances can vary in different
solutions. Specify the amounts of solvent and
solutes. Qualitative measures of concentration
concentrated – relatively large amount of solute
dilute – relatively small amount of solute
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Section 15.3
Solution Composition: Mass Percent
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mass of soluteMass percent = 100%
mass of solution
grams of soluteMass percent = 100%
grams of solute + grams of solvent
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Section 15.3
Solution Composition: Mass Percent
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Exercise
What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?
6.6%
[5.5 g / (5.5 g + 78.2 g)] × 100 = 6.6% glucose
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Section 15.4
Solution Composition: Molarity
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• Molarity (M) = moles of solute per volume of solution in liters:
moles of solute = Molarity = liters of solution
M
6 moles of HCl3 HCl = 2 liters of solution
M
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Section 15.4
Solution Composition: Molarity
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Exercise
You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.
8.00 M
1.00 mol / (125.0 / 1000) = 8.00 M
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Section 15.4
Solution Composition: Molarity
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Exercise
A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution?
1.57 M
500.0 g is equivalent to 2.355 mol K3PO4 (500.0 g / 212.27 g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L).
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Section 15.4
Solution Composition: Molarity
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Exercise
You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?
0.200 L
2.00 mol / 10.0 M = 0.200 L
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Section 15.4
Solution Composition: Molarity
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Exercise
Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity.
10.0 M NaOH[100.0 g NaOH / 39.998 g/mol] / [250.0 / 1000] = 10.0 M NaOH
5.37 M KCl[100.0 g KCl / 74.55 g/mol] / [250.0 / 1000] = 5.37 M KCl
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Section 15.4
Solution Composition: Molarity
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Concept Check
You have two HCl solutions, labeled Solution A and Solution B. Solution A has a greater concentration than Solution B. Which of the following statements are true?
a) If you have equal volumes of both solutions, Solution B must contain more moles of HCl.
b) If you have equal moles of HCl in both solutions, Solution B must have a greater volume.c) To obtain equal concentrations of both solutions, you must add a certain amount of water to
Solution B.d) Adding more moles of HCl to both solutions will make them less concentrated.
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Section 15.4
Solution Composition: Molarity
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• For a 0.25 M CaCl2 solution:
CaCl2 → Ca2+ + 2Cl–
Ca2+: 1 × 0.25 M = 0.25 M Ca2+
Cl–: 2 × 0.25 M = 0.50 M Cl–.
Concentration of Ions
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Section 15.4
Solution Composition: Molarity
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Concept Check
Which of the following solutions containsthe greatest number of ions?
a) 400.0 mL of 0.10 M NaCl.
b) 300.0 mL of 0.10 M CaCl2.
c) 200.0 mL of 0.10 M FeCl3.
d) 800.0 mL of 0.10 M sucrose.
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Section 15.4
Solution Composition: Molarity
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• Where are we going? To find the solution that contains the greatest
number of moles of ions.
• How do we get there? Draw molecular level pictures showing each
solution. Think about relative numbers of ions. How many moles of each ion are in each
solution?
Let’s Think About It
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Section 15.4
Solution Composition: Molarity
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• The solution with the greatest number of ions is not necessarily the one in which: the volume of the solution is the
largest. the formula unit has the greatest
number of ions.
Notice
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Section 15.4
Solution Composition: Molarity
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• A solution whose concentration is accurately known.
Standard Solution
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Section 15.4
Solution Composition: Molarity
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• Weigh out a sample of solute.• Transfer to a volumetric flask.• Add enough solvent to mark on flask.
To Make a Standard Solution
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Section 15.5
Dilution
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• The process of adding water to a concentrated or stock solution to achieve the molarity desired for a particular solution.
• Dilution with water does not alter the numbers of moles of solute present.
• Moles of solute before dilution = moles of solute after dilution
M1V1 = M2V2
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Section 15.5
Dilution
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• Transfer a measured amount of original solution to a flask containing some water.
• Add water to the flask to the mark (with swirling) and mix by inverting the flask.
Diluting a Solution
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Section 15.5
Dilution
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Concept Check
A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution?
a) Add water to the solution.
b) Pour some of the solution down the sink drain.
c) Add more sodium chloride to the solution.
d) Let the solution sit out in the open air for a couple of days.
e) At least two of the above would decrease the concentration of the salt solution.
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Section 15.5
Dilution
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Exercise
What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution?
60.0 mL
M1V1 = M2V2
(2.00 M)(V1) = (0.800 M)(150.0 mL)
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Section 15.6
Stoichiometry of Solution Reactions
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1. Write the balanced equation for the reaction. For reactions involving ions, it is best to write the net ionic equation.
2. Calculate the moles of reactants.
3. Determine which reactant is limiting.
4. Calculate the moles of other reactants or products, as required.
5. Convert to grams or other units, if required.
Steps for Solving Stoichiometric Problems Involving Solutions
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Section 15.6
Stoichiometry of Solution Reactions
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Concept Check (Part I)
10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).
What precipitate will form?
lead(II) phosphate, Pb3(PO4)2
What mass of precipitate will form?
1.1 g Pb3(PO4)2
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Section 15.6
Stoichiometry of Solution Reactions
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• Where are we going? To find the mass of solid Pb3(PO4)2 formed.
• How do we get there? What are the ions present in the combined solution? What is the balanced net ionic equation for the
reaction? What are the moles of reactants present in the
solution? Which reactant is limiting? What moles of Pb3(PO4)2 will be formed?
What mass of Pb3(PO4)2 will be formed?
Let’s Think About It
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Section 15.6
Stoichiometry of Solution Reactions
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Concept Check (Part II)
10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).
What is the concentration of nitrate ions left in solution after the reaction is complete?
0.27 M
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Section 15.6
Stoichiometry of Solution Reactions
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• Where are we going? To find the concentration of nitrate ions left in
solution after the reaction is complete.
• How do we get there? What are the moles of nitrate ions present in the
combined solution? What is the total volume of the combined
solution?
Let’s Think About It
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Section 15.6
Stoichiometry of Solution Reactions
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Concept Check (Part III)
10.0 mL of a 0.30 M sodium phosphate solution reacts with 20.0 mL of a 0.20 M lead(II) nitrate solution (assume no volume change).
What is the concentration of phosphate ions left in solution after the reaction is complete?
0.011 M
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Section 15.6
Stoichiometry of Solution Reactions
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• Where are we going? To find the concentration of phosphate ions left in
solution after the reaction is complete.
• How do we get there? What are the moles of phosphate ions present in
the solution at the start of the reaction? How many moles of phosphate ions were used
up in the reaction to make the solid Pb3(PO4)2?
How many moles of phosphate ions are left over after the reaction is complete?
What is the total volume of the combined solution?
Let’s Think About It
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Section 15.7
Neutralization Reactions
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• An acid-base reaction is called a neutralization reaction.
• Steps to solve these problems are the same as before.
• For a strong acid and base reaction:
H+(aq) + OH–(aq) H2O(l)
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Section 15.7
Neutralization Reactions
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Concept Check
For the titration of sulfuric acid (H2SO4) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of 0.500 M sulfuric acid?
1.00 mol NaOH
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Section 15.7
Neutralization Reactions
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• Where are we going? To find the moles of NaOH required for the
reaction.
• How do we get there? What are the ions present in the combined
solution? What is the reaction? What is the balanced net ionic equation for the
reaction? What are the moles of H+ present in the solution? How much OH– is required to react with all of the
H+ present?
Let’s Think About It
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Section 15.8
Solution Composition: Normality
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• One equivalent of acid – amount of acid that furnishes 1 mol of H+ ions.
• One equivalent of base – amount of base that furnishes 1 mol of OH ions
• Equivalent weight – mass in grams of 1 equivalent of acid or base.
Unit of Concentration
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Section 15.8
Solution Composition: Normality
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Section 15.8
Solution Composition: Normality
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Section 15.8
Solution Composition: Normality
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• To find number of equivalents:
number of equivalents equivalents equivNormality = = = =
1 liter of solution liter LN
equiv V = L = equiv
LN
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Section 15.8
Solution Composition: Normality
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Concept Check
If Ba(OH)2 is used as a base, how many equivalents of Ba(OH)2 are there in 4 mol Ba(OH)2?
a) 2b) 4c) 8d) 16
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Section 15.8
Solution Composition: Normality
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Homework
• Reading assignment– Pages 475 through 502
• Homework Problems– Questions and problems 3, 5, 7, 11, 17, 19,
21, 23, 25, 27, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 57, 59, 61, 71, 73, 85, 87.
• Due on
Chapter 15 Homework