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Chapter 15 Solutions

Chapter 15 Return to TOC Copyright © Cengage Learning. All rights reserved 3 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

Section 15.1 Solubility Return to TOC Copyright © Cengage Learning. All rights reserved 11 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)CCl 4 mixed with water (H 2 O) b)NH 3 mixed with water (H 2 O) c)CH 3 OH mixed with water (H 2 O) d)N 2 mixed with methane (CH 4 )

Section 15.2 Solution Composition: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 12 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.

Section 15.2 Solution Composition: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 13 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.

Section 15.2 Solution Composition: An Introduction Return to TOC Copyright © Cengage Learning. All rights reserved 14 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

Section 15.3 Solution Composition: Mass Percent Return to TOC Copyright © Cengage Learning. All rights reserved 16 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

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 18 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

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 19 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 K 3 PO 4 (500.0 g / 212.27 g/mol). The molarity is therefore 1.57 M (2.355 mol/1.50 L).

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 20 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

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 21 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

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 22 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.

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 23 For a 0.25 M CaCl 2 solution: CaCl 2 → Ca 2+ + 2Cl –  Ca 2+ : 1 × 0.25 M = 0.25 M Ca 2+  Cl – : 2 × 0.25 M = 0.50 M Cl –. Concentration of Ions

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 24 Concept Check Which of the following solutions contains the greatest number of ions? a) 400.0 mL of 0.10 M NaCl. b) 300.0 mL of 0.10 M CaCl 2. c) 200.0 mL of 0.10 M FeCl 3. d) 800.0 mL of 0.10 M sucrose.

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 25 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

Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 26 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

Section 15.5 Dilution Return to TOC Copyright © Cengage Learning. All rights reserved 29 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 M 1 V 1 = M 2 V 2

Section 15.5 Dilution Return to TOC Copyright © Cengage Learning. All rights reserved 31 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.

Section 15.5 Dilution Return to TOC Copyright © Cengage Learning. All rights reserved 32 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 M 1 V 1 = M 2 V 2 (2.00 M)(V 1 ) = (0.800 M)(150.0 mL)

Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 33 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

Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 34 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, Pb 3 (PO 4 ) 2  What mass of precipitate will form? 1.1 g Pb 3 (PO 4 ) 2

Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 35 Where are we going?  To find the mass of solid Pb 3 (PO 4 ) 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 Pb 3 (PO 4 ) 2 will be formed?  What mass of Pb 3 (PO 4 ) 2 will be formed? Let’s Think About It

Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 36 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

Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 37 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

Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 38 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

Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 39 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 Pb 3 (PO 4 ) 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

Section 15.7 Neutralization Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 40 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)  H 2 O(l)

Section 15.7 Neutralization Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 41 Concept Check For the titration of sulfuric acid (H 2 SO 4 ) 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

Section 15.7 Neutralization Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 42 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

Section 15.8 Solution Composition: Normality Return to TOC Copyright © Cengage Learning. All rights reserved 43 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