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Chapter 15a Solutions

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Chapter 15 Table of Contents Solubility 15.2 Solution Composition: An Introduction 15.3 Solution Composition: Mass Percent 15.4 Solution Composition: Molarity 15.5 Dilution

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Chapter 15 Return to TOC 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

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Section 15.1 Solubility Return to TOC 4 Various Types of Solutions

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Section 15.1 Solubility Return to TOC Copyright © Cengage Learning. All rights reserved 5 Ionic substances breakup into individual cations and anions. Solubility of Ionic Substances

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Section 15.1 Solubility Return to TOC Copyright © Cengage Learning. All rights reserved 6 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 Return to TOC Copyright © Cengage Learning. All rights reserved 7 Ethanol is soluble in water because of the polar OH bond. Solubility of Polar Substances

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Section 15.1 Solubility Return to TOC 8 Why is solid sugar soluble in water? Solubility of Polar Substances Why is isopropyl alcohol soluble in water? O δ- H δ+ H δ+

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Section 15.1 Solubility Return to TOC 9 Why is blue food coloring soluble in water? Solubility of Polar Substances Blue food dye mixes with alcohol and water unless a dissolved salt dominates the water as a solvent.

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Section 15.1 Solubility Return to TOC Copyright © Cengage Learning. All rights reserved 10 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 Return to TOC 11 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 Return to TOC Copyright © Cengage Learning. All rights reserved 12 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 )

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Section 15.1 Solubility Return to TOC Copyright © Cengage Learning. All rights reserved 13 Factors that affect Solubility Molecular structure-like dissolves like –ionic nature –polarity Temperature –Increased temperatures increase solubility of solids dissolved in liquids. –Increased temperatures decrease solubility of gasses dissolved in liquids. Pressure –Increased pressures increase solubility of gasses dissolved in liquids.

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Section 15.1 Solubility Return to TOC 14

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Section 15.2 Solution Composition: An Introduction Return to TOC 15 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 Return to TOC 16 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 Return to TOC Copyright © Cengage Learning. All rights reserved 17 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 Return to TOC Copyright © Cengage Learning. All rights reserved 18

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Section 15.3 Solution Composition: Mass Percent Return to TOC Copyright © Cengage Learning. All rights reserved 19 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 g)] × 100 = 6.6% glucose

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Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 20 Molarity (M) = moles of solute per volume of solution in liters:

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Section 15.4 Solution Composition: Molarity Return to TOC 21 Exercise You have 1.00 mol of sugar in mL of solution. Calculate the concentration in units of molarity M 1.00 mol / (125.0 / 1000) = 8.00 M

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Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 22 Exercise A 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 g is equivalent to mol K 3 PO 4 (500.0 g / 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 Return to TOC 23 Exercise You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? L 2.00 mol / 10.0 M = L

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Section 15.4 Solution Composition: Molarity Return to TOC 24 Exercise Consider separate solutions of NaOH and KCl made by dissolving g of each solute in mL of solution. Calculate the concentration of each solution in units of molarity M NaOH [100.0 g NaOH / g/mol] / [250.0 / 1000] = M NaOH M KCl [100.0 g KCl / g/mol] / [250.0 / 1000] = M KCl

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Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 25 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 Return to TOC 26 For a 0.25 M CaCl 2 solution: CaCl 2 → Ca Cl – Ca 2+ : 1 × 0.25 M = 0.25 M Ca 2+ Cl – : 2 × 0.25 M = 0.50 M Cl –. Concentration of Ions

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Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 27 Concept Check Which of the following solutions contains the greatest number of ions? a) mL of 0.10 M NaCl. b) mL of 0.10 M CaCl 2. c) mL of 0.10 M FeCl 3. d) mL of 0.10 M sucrose.

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Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 28 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 Return to TOC 29 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 Return to TOC 30 A solution whose concentration is accurately known. Standard Solution

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Section 15.4 Solution Composition: Molarity Return to TOC Copyright © Cengage Learning. All rights reserved 31 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 Return to TOC 32 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

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Section 15.5 Dilution Return to TOC Copyright © Cengage Learning. All rights reserved 33 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 Return to TOC Copyright © Cengage Learning. All rights reserved 34 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 Return to TOC 35 Exercise What is the minimum volume of a 2.00 M NaOH solution needed to make mL of a 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)

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Section 15.5 Dilution Return to TOC 36 W Chapter 15b Solutions

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Section 15.5 Dilution Return to TOC Stoichiometry of Solution Reactions 15.7Neutralization Reactions 15.8Solution Composition: Normality

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Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 38 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 Return to TOC Copyright © Cengage Learning. All rights reserved 39 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? 0.91 g Pb 3 (PO 4 ) 2 see next 2 slides

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Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 40 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

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Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved L P(0.30 moles P/L)(1 mole LP/2mole P) (812g LP/mole) = 1.1g LP L Lead(0.20 moles Lead/L)(1 mole LP/3mole L)(812g LP/mole) = 0.91g LP 2PO Pb 2+ Pb 3 (PO 4 ) 2 (s)

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Section 15.6 Stoichiometry of Solution Reactions Return to TOC 42 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 see next two slides

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Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 43 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 Return to TOC Copyright © Cengage Learning. All rights reserved L LN(0.20 moles LN/L)(2 mole N/1mole LN)= moles Nitrate ion moles N/(0.0100L L) = 0.27 M Nitrate 2NO Pb 2+ Pb(NO 3 ) 2 (aq)

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Section 15.6 Stoichiometry of Solution Reactions Return to TOC 45 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? 0.02 M see next two slides

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Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 46 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

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Section 15.6 Stoichiometry of Solution Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 47 (1.1g LP-0.91g LP)(1mole LP/812g)(2mole P/1mole LP)/ (0.0100L L) = 0.02 M Phosphate Ion 2PO Pb 2+ Pb 3 (PO 4 ) 2 (s)

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Section 15.7 Neutralization Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 48 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)

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Section 15.7 Neutralization Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 49 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 M sulfuric acid? 1.00 mol NaOH see next 2 slides

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Section 15.7 Neutralization Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 50 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 15.7 Neutralization Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 51 H 2 SO 4 + 2NaOH Na 2 SO 4 + 2H 2 O 1.00L SA(0.500 moles SA/L)(2 mole SH/1mole SA)= 1.00 mol NaOH

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Section 15.7 Neutralization Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 52 HA + NaOH Salt + H 2 O If I use 8.34 ml of a M solution of NaOH to titrate or neutralize ml of a monoprotic acid (one H), what is the concentration of the acid? L SH (0.1562mol SH/L)(1 mole HA/1 mole SH) / L HA = M HA

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Section 15.8 Solution Composition: Normality Return to TOC Copyright © Cengage Learning. All rights reserved 53 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 Return to TOC Copyright © Cengage Learning. All rights reserved 54

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Section 15.8 Solution Composition: Normality Return to TOC Copyright © Cengage Learning. All rights reserved 55

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Section 15.8 Solution Composition: Normality Return to TOC Copyright © Cengage Learning. All rights reserved 56 To find number of equivalents:

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Section 15.8 Solution Composition: Normality Return to TOC Copyright © Cengage Learning. All rights reserved 57 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)2 b)4 c)8 d)16

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Section 15.8 Solution Composition: Normality Return to TOC 58 Colligative Properties (not in textbook) A property that does not depend on the identity of a solute in solution Vary only with the number of solute particles present in a specific quantity of solvent 4 colligative properties: –Osmotic pressure –Vapor pressure lowering –Boiling point elevation –Freezing point depression 11-

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Section 15.8 Solution Composition: Normality Return to TOC 59 Osmotic Pressure Osmosis –A process in which solvent molecules diffuse through a barrier that does not allow the passage of solute particles The barrier is called a semipermeable membrane. –A membrane that allows the passage of some substances but not others Osmotic Pressure –Pressure that can be exerted on the solution to prevent osmosis 11-

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Section 15.8 Solution Composition: Normality Return to TOC 60 Osmotic Pressure Reverse Osmosis- Ultra Filtration –A process in which pressure is applied to reverse the osmosis flow. –A form of molecular filtration used to remove salt from water. 11-

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Section 15.8 Solution Composition: Normality Return to TOC p. 479

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Section 15.8 Solution Composition: Normality Return to TOC 62 Vapor Pressure Lowering Solutes come in 2 forms: Volatile –Solutes that readily form a gas Nonvolatile –Solutes that DO NOT readily form a gas Generally, the addition of a solute lowers the vapor pressure of a solution when compared to the pure solvent. 11-

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Section 15.8 Solution Composition: Normality Return to TOC 63 Solutions under sealed container to start with. Solutions under sealed container at later time. Vapor Pressure Lowering

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Section 15.8 Solution Composition: Normality Return to TOC 64 Phase Diagram of Water 11-

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Section 15.8 Solution Composition: Normality Return to TOC 65 Supercritical Extraction Removal of caffeine from coffee using supercritical CO 2 or natural effervescence. Removal of oils (fats) from potato chips. Dry Cleaning Cloths 31.1 °C 72.9 atm

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Section 15.8 Solution Composition: Normality Return to TOC 66 Boiling Point Elevation The addition of solute affects the boiling point because it affects the vapor pressure. The boiling point is raised with the addition of solute in comparison to the pure solvent. An equation that gives the increase in boiling point: ∆T b = K b m Where ∆T b is the increase in temperature from the pure solvent’s boiling point, K b is the boiling point constant, which is characteristic of a particular solvent, and m is the molality (moles of solute per kg of solvent) 11-

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Section 15.8 Solution Composition: Normality Return to TOC 67 Freezing Point Depression The freezing point is lowered with the addition of solute in comparison to the pure solvent. An equation that gives the decrease in freezing point: ∆T f = K f m Where ∆T f is the decrease in temperature from the pure solvent’s freezing point, K f is the freezing point constant, which is characteristic of a particular solvent, and m is the molality (moles of solute per kg of solvent) 11-

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Section 15.8 Solution Composition: Normality Return to TOC 68 Examples of Colligative Properties Antifreeze lowers the freezing point of your radiator fluid and raises the boiling point. Salting roads melts the ice.

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Section 15.8 Solution Composition: Normality Return to TOC 69 Example of Freezing Point Depression Making Ice Cream You must add salt so as to lower the freezing point of the ice water cold enough to freeze the ice cream.

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Section 15.8 Solution Composition: Normality Return to TOC 70 ΔT f = K f mol solute/Kg solvent = (78)/5.0= 29 o C 0.0 o C – 29 o C = - 29 o C freezing point Calculations of Freezing Point Lowering and Boiling Point Elevation What will be the coldest temperature of an ice water solution with 5.0 lbs of salt in 3.0 gallon container with about 5.0 Kg of water? What will be the boiling Point? 5.0 lbs NaCl(454g/lb)(1mole NaCl/58 g) = 39 moles NaCl ΔT b = K b mol solute/Kg solvent = 0.52 (78)/5.0= 8.1 o C o C o C = o C boiling point 39 moles NaCl x 2 (two particles or ions are formed) = 78

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Section 15.8 Solution Composition: Normality Return to TOC 71 Colligative Properties and Strong Electrolytes Strong electrolytes dissociate most of the time into their constituent ions. Therefore, the number of particles (in this case ions) increases with the number of ions. Colligative properties are proportional to number of particles in solution. Example: MgCl 2 (s) Mg 2+ (aq) + 2 Cl - (aq) In this case the number of particles increases to 3 particles. Therefore, we would multiply the colligative property amount by

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Section 15.8 Solution Composition: Normality Return to TOC 72 Practice – Strong vs. Weak Electrolytes Which of the following aqueous solutions is expected to have the lowest freezing point? –0.5 m CH 3 CH 2 OH –0.5 m Ca(NO 3 ) 2 –0.5 m KBr 11- Forms 1 particle per molecule or formula unit: 1 x 0.5 m particles = 0.5 m particles Forms 3 particle per molecule or formula unit: 3 x 0.5 m particles = 1.5 m particles Forms 2 particle per molecule or formula unit: 2 x 0.5 m particles = 1.0 m particles Thus Ca(NO 3 ) 2 should the greatest freezing point lowering or the lowest freezing point.

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