Chapter 13 Solutions. Homework Assigned Problems (odd numbers only) Assigned Problems (odd numbers only) “Problems” 25 to 59 (begins on page 478) “Problems”

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Chapter 13 Solutions

Homework Assigned Problems (odd numbers only) Assigned Problems (odd numbers only) “Problems” 25 to 59 (begins on page 478) “Problems” 25 to 59 (begins on page 478) “Cumulative Problems” 109-129 (begins on page 482) “Cumulative Problems” 109-129 (begins on page 482) “Highlight Problems” 131, 133, page 484-485 “Highlight Problems” 131, 133, page 484-485

Solutions A solution is a homogeneous mixture of two or more substances A solution is a homogeneous mixture of two or more substances It is uniform, same composition throughout It is uniform, same composition throughout One substance is dissolved into another One substance is dissolved into another Requires the interaction of particles of atomic or molecular size Requires the interaction of particles of atomic or molecular size

Solutions Two parts: Solvent and solute Two parts: Solvent and solute Solute: Substance being dissolved (present in a smaller amount relative to the solvent) Solute: Substance being dissolved (present in a smaller amount relative to the solvent) Solvent: Substance that dissolves the solvent (present in the greatest amount) Solvent: Substance that dissolves the solvent (present in the greatest amount) Most solutions are liquid but can be gaseous or solid Most solutions are liquid but can be gaseous or solid

Solutions “Like dissolves like” “Like dissolves like” Substances that are similar should form a solution Substances that are similar should form a solution Refers to the overall polarity of the solvent (polar and nonpolar) and the solute (polar, nonpolar, and ionic) Refers to the overall polarity of the solvent (polar and nonpolar) and the solute (polar, nonpolar, and ionic) Must be an attraction between the solute and solvent for a solution to form Must be an attraction between the solute and solvent for a solution to form

Solutions Nonpolar molecules (no dipole, cannot hydrogen bond) Nonpolar molecules (no dipole, cannot hydrogen bond) Examples: oil, iodine Examples: oil, iodine Both do not dissolve well in water because it (water) is polarBoth do not dissolve well in water because it (water) is polar Both dissolve well in nonpolar solvents such as carbon tetrachlorideBoth dissolve well in nonpolar solvents such as carbon tetrachloride CCl 4 H2OH2O I 2 in CCl 4 Ni(NO 3 ) 2 in H 2 O

Solutions of Solids Dissolved in Water (Water as a Solvent) Most common solutions (with a solid, liquid, or gas) contain water as the solvent Most common solutions (with a solid, liquid, or gas) contain water as the solvent Water is a polar molecule due to its bent shape Water is a polar molecule due to its bent shape It also has the ability to hydrogen bond It also has the ability to hydrogen bond Dissolves many polar and ionic substances Dissolves many polar and ionic substances Due to intermolecular interactions (dipole-dipole or H-bonding) upon mixing Due to intermolecular interactions (dipole-dipole or H-bonding) upon mixing

Solutions of Solids Dissolved in Water (Water as a Solvent) Polar compounds (a permanent dipole, can H- bond) and ionic compounds dissolve into polar solvents Polar compounds (a permanent dipole, can H- bond) and ionic compounds dissolve into polar solvents A polar molecule (with ionic bonding) dissolves into water if attractions for water overcome the attractions between the ions A polar molecule (with ionic bonding) dissolves into water if attractions for water overcome the attractions between the ions As each ion enters the solution, it is immediately surrounded by water molecules: hydration (solvation) As each ion enters the solution, it is immediately surrounded by water molecules: hydration (solvation)

Solutions of Solids Dissolved in Water (Water as a Solvent) When sodium chloride crystals are placed in water, they begin to dissolve When sodium chloride crystals are placed in water, they begin to dissolve The attractive forces between the ions and water are stronger than forces between the ions in the crystal The attractive forces between the ions and water are stronger than forces between the ions in the crystal Water molecules surround each ion, keeping them apart Water molecules surround each ion, keeping them apart

Solutions of Solids Dissolved in Water (Water as a Solvent) When sodium chloride crystals are placed in water, they begin to dissolve When sodium chloride crystals are placed in water, they begin to dissolve The attractive forces between the ions and water are stronger than forces between the ions in the crystal The attractive forces between the ions and water are stronger than forces between the ions in the crystal Water molecules surround each ion, keeping them apart Water molecules surround each ion, keeping them apart

Electrolyte Solutions: Dissolved Ionic Solids Compounds that ionize in water are called electrolytes Compounds that ionize in water are called electrolytes Electrolytes are solutes that exist as ions in solutionElectrolytes are solutes that exist as ions in solution Formed from an ionic compound that dissociates in water forming an electroyte solution with cations and anionsFormed from an ionic compound that dissociates in water forming an electroyte solution with cations and anions These solutions conduct electricityThese solutions conduct electricity

Ions In Solution (in Water) When ionic compounds dissolve in water the ions dissociate When ionic compounds dissolve in water the ions dissociate Separate into the ions floating in waterSeparate into the ions floating in water Potassium chloride dissociates in water into potassium cations and chloride anionsPotassium chloride dissociates in water into potassium cations and chloride anions KCl(aq) = K + (aq) + Cl - (aq) K+K+ Cl - K Cl

Ions In Solution (in Water) Copper(II) sulfate dissociates in water into copper(II) cations and sulfate anions Copper(II) sulfate dissociates in water into copper(II) cations and sulfate anions CuSO 4 (aq) = Cu +2 (aq) + SO 4 2- (aq) Cu +2 SO 4 2- Cu SO 4

Ions In Solution (in Water) Potassium sulfate dissociates in water into potassium cations and sulfate anions Potassium sulfate dissociates in water into potassium cations and sulfate anions K 2 SO 4 (aq) = 2 K + (aq) + SO 4 2- (aq) K+K+ SO 4 2- K+K+ KK SO 4

Nonelectrolyte Solutions Compounds that do not ionize in water are called nonelectrolytes Compounds that do not ionize in water are called nonelectrolytes Solute is a molecular substanceSolute is a molecular substance Substance dispersed throughout the solvent as individual moleculesSubstance dispersed throughout the solvent as individual molecules Each molecule is separated (dissolved) by molecules of the solvent forming a nonelectrolyte solutionEach molecule is separated (dissolved) by molecules of the solvent forming a nonelectrolyte solution These solutions do not conduct electricityThese solutions do not conduct electricity

Electrolytes and Nonelectrolytes Strong electrolyte: Strong electrolyte: Dissociates completely into ionsDissociates completely into ions Conduct electricityConduct electricity Weak electrolyte: Weak electrolyte: Mainly whole moleculesMainly whole molecules Very few separate (into ions)Very few separate (into ions) Conduct electricity less than strong electrolytesConduct electricity less than strong electrolytes Nonelectrolyte: Nonelectrolyte: No dissociation into ionsNo dissociation into ions Do not conduct electricityDo not conduct electricity

Electrolytes and Nonelectrolytes Strong electrolytes are completely ionized when dissolved in water Strong electrolytes are completely ionized when dissolved in water Sodium chloride dissociates to form Na + and Cl - Sodium chloride dissociates to form Na + and Cl - Good conductor of electricity Good conductor of electricity

Electrolytes and Nonelectrolytes Weak electrolytes are only partially ionized when dissolved in water Weak electrolytes are only partially ionized when dissolved in water Hydrofluoric acid only partially dissociates to form H + and F - Hydrofluoric acid only partially dissociates to form H + and F - Poor conductor of electricity Poor conductor of electricity

Electrolytes and Nonelectrolytes Nonelectrolyes are not ionized when dissolved in water Nonelectrolyes are not ionized when dissolved in water e.g. sugar and ethanol do not dissociate into ions in water e.g. sugar and ethanol do not dissociate into ions in water Do not conduct electricity Do not conduct electricity

Solubility and Saturation Solubility is the maximum amount of solute that will dissolve into a given amount of solvent Solubility is the maximum amount of solute that will dissolve into a given amount of solvent It is affected by It is affected by Type of solute (solid, liquid, or gas)Type of solute (solid, liquid, or gas) Type of solvent (and the solute interaction)Type of solvent (and the solute interaction) TemperatureTemperature

Solubility and Saturation Unsaturated: Less solute than the maximum amount possible is dissolved into the solution Unsaturated: Less solute than the maximum amount possible is dissolved into the solution Saturated: Contains the maximum amount of solute that can be dissolved Saturated: Contains the maximum amount of solute that can be dissolved saturated

Solubility and Saturation A supersaturated solution contains more of the dissolved particles than could be dissolved by the solvent under normal circumstances A supersaturated solution contains more of the dissolved particles than could be dissolved by the solvent under normal circumstances These solutions result from altering a condition of the saturated solution such as T, V, or P These solutions result from altering a condition of the saturated solution such as T, V, or P

Solutions of Solids in Water: Effect of Temperature on Solubility For most solids (solute), solubility increases with an increase in temperature For most solids (solute), solubility increases with an increase in temperature More sugar will dissolve in hot water than in cold water More sugar will dissolve in hot water than in cold water

Solutions of Gases in Water: Effect of Temperature on Solubility For gases, solubility decreases with an increase in temperature For gases, solubility decreases with an increase in temperature Henry’s Law: The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution Henry’s Law: The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution

Solution Concentration Solution concentration Solution concentration The amount of solute (mass or moles) dissolved into a certain amount of a solution or solventThe amount of solute (mass or moles) dissolved into a certain amount of a solution or solvent Qualitative Qualitative Dilute, concentrated, saturated, unsaturatedDilute, concentrated, saturated, unsaturated Quantitative Quantitative Mass to mass, volume to volume, and molarityMass to mass, volume to volume, and molarity

Solution Concentration Mass PercentMass Percent Mass of the solute divided by the total mass of solution multiplied by 100 The mass of the solute and solution must be in the same units Volume Percent The volume of solute divided by the total volume of solution multiplied by 100 The solute and solution volumes must be in the same units

Mass Percent Grams of solute per grams of solution Grams of solute per grams of solution Remember that the mass of solution is grams of solute + grams of solventRemember that the mass of solution is grams of solute + grams of solvent

Calculating Mass Percent: Example 1 A 135 g sample of seawater is evaporated to dryness, leaving 4.73 g of solid residue. What is the mass percent of the solute in the original sea water? A 135 g sample of seawater is evaporated to dryness, leaving 4.73 g of solid residue. What is the mass percent of the solute in the original sea water?

Mass Percent in Calculations: Example 2 What mass of water must be added to 425 g of formaldehyde to prepare a 40.0% (by mass) solution of formaldehyde? What mass of water must be added to 425 g of formaldehyde to prepare a 40.0% (by mass) solution of formaldehyde?

Mass Percent Conc. Example 2

Solution Concentration: Molarity Molarity is the concentration expression most commonly used in the laboratory Molarity is the concentration expression most commonly used in the laboratory The amount of solute is expressed in moles The amount of solute is expressed in moles To obtain the molarity, we need to know the solution volume in liters and the number of moles of solute present To obtain the molarity, we need to know the solution volume in liters and the number of moles of solute present

Solution Concentration: Molarity Moles of solute per liters of solution Moles of solute per liters of solution More useful than mass percent More useful than mass percent More common to measure liquids by volume, not massMore common to measure liquids by volume, not mass Amount of solute expressed in moles (quantity of particles)Amount of solute expressed in moles (quantity of particles) Chemical reactions occur between molecules and atomsChemical reactions occur between molecules and atoms Since it expressed in moles, you can do chemical calculation (stoichiometry) problemsSince it expressed in moles, you can do chemical calculation (stoichiometry) problems

Making Solutions of a Specific Molarity Make up in a volumetric flask Make up in a volumetric flask Flask with a long, narrow neck that is marked with a line indicating an exact volumeFlask with a long, narrow neck that is marked with a line indicating an exact volume Method Method Add measured amount of solid (mass in grams)Add measured amount of solid (mass in grams) Add some water to dissolve the solidAdd some water to dissolve the solid Fill with water up to the line (volume in mL or L)Fill with water up to the line (volume in mL or L)

Using Molarity in Calculations: Example 1 Calculate the molarity of a solution made by dissolving 15.0 g of NaOH (sodium hydroxide) in enough water to give a final volume of 100. mL. Calculate the molarity of a solution made by dissolving 15.0 g of NaOH (sodium hydroxide) in enough water to give a final volume of 100. mL. Convert volume to liters

Using Molarity in Calculations: Example 1 Convert mass to moles

Using Molarity in Calculations: Example 2 Formalin is an aqueous solution of formaldehyde (H 2 CO). How many grams of formaldehyde must be used to prepare 2.50 L of 12.3 M formalin? Formalin is an aqueous solution of formaldehyde (H 2 CO). How many grams of formaldehyde must be used to prepare 2.50 L of 12.3 M formalin? molarity × volume = moles 30.8 mol formaldehyde

Standard Solutions A solution whose concentration is exactly known A solution whose concentration is exactly known A std. solution can be diluted to make up less concentrated solutionsA std. solution can be diluted to make up less concentrated solutions A std. solution is like concentrated orange juice. For example, one can of orange juice concentrate is diluted with three cans of water.A std. solution is like concentrated orange juice. For example, one can of orange juice concentrate is diluted with three cans of water.

Solution Dilution Dilution is the process in which more solvent is added to a solution in order to lower its concentration Dilution is the process in which more solvent is added to a solution in order to lower its concentration A common laboratory routine is diluting a solution of known concentration (stock solution) to a lower concentration A common laboratory routine is diluting a solution of known concentration (stock solution) to a lower concentration A dilution always lowers the concentration because the same amount of solute is present in a larger amount of solvent A dilution always lowers the concentration because the same amount of solute is present in a larger amount of solvent

Dilution Most often a solution of a specific molarity must be prepared by adding a predetermined volume of solvent to a specific volume of stock solutionMost often a solution of a specific molarity must be prepared by adding a predetermined volume of solvent to a specific volume of stock solution When solvent is added to dilute a solution, the number of moles remains unchangedWhen solvent is added to dilute a solution, the number of moles remains unchanged A relationship exists between the volumes and molarities of the diluted and stock solutionsA relationship exists between the volumes and molarities of the diluted and stock solutions Moles of solute = (initial solution) M 1 V 1 = Moles of solute (diluted solution) M 2 V 2

Solution Dilution: Example 1 Determine the volume required to prepare 0.75L of 0.10 M HCl from a 12 M HCl stock solution. Determine the volume required to prepare 0.75L of 0.10 M HCl from a 12 M HCl stock solution. How many moles of HCl do we eventually want?How many moles of HCl do we eventually want? Initially we have 12 M HCl. Calculate what volume of the stock solution will contain the number of moles needed.Initially we have 12 M HCl. Calculate what volume of the stock solution will contain the number of moles needed.

Solution Dilution: Example 1 How many liters of 12 M HCl contains 0.075 mol of HCl ?How many liters of 12 M HCl contains 0.075 mol of HCl ? 6.25 mL of 12 M HCl are needed

Solution Dilution: Example 2 What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO 4 is diluted to a final volume of 250 mL.What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO 4 is diluted to a final volume of 250 mL.

Solution Dilution: Example 2 What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO 4 is diluted to a final volume of 250 mL.What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO 4 is diluted to a final volume of 250 mL. Initial M 1 = 1.0 M V 1 = 0.025 L Final M 2 = ? V 2 = 0.250 L Calculate the unknown molarity using the relationship Moles of solute = (initial solution) M 1 V 1 = Moles of solute (diluted solution) M 2 V 2

Solution Dilution: Example 2 What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO 4 is diluted to a final volume of 250 mL.What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO 4 is diluted to a final volume of 250 mL. Set Up Problem by solving for M 2 1)Moles of solute before dilution equals moles of solute after dilution 2) Calculate the unknown molarity by solving for M 2 3) Set up problem

Solutions in Chemical Reactions: Solution Stoichiometry Stoichiometry is the calculation of quantitative relationships between reactants and products Stoichiometry is the calculation of quantitative relationships between reactants and products Calculations are based on balanced chemical equations Calculations are based on balanced chemical equations The coefficients in the balanced equation indicate the moles of products and reactants The coefficients in the balanced equation indicate the moles of products and reactants

Solutions in Chemical Reactions: Solution Stoichiometry Many reactions take place in solution and the solution concentration (molarity) directly relates the solution volume and moles of solute present Many reactions take place in solution and the solution concentration (molarity) directly relates the solution volume and moles of solute present Stoichiometric calculations are the same as in chapter 8, but with the addition of some molarity calculations Stoichiometric calculations are the same as in chapter 8, but with the addition of some molarity calculations

Quantitative Relationships Needed for Solving Chemical Formula Based Problems Grams A Grams B P A, T A, V A P B, T B, V B Liters A Liters B Moles AMoles B pV = nRT Mole-mole Factor molarity Molar mass molarity 1 mol = 22.4 L at STP M × V

Solutions in Chemical Reactions: Solution Stoichiometry When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates.

Solutions in Chemical Reactions: Solution Stoichiometry Calculate the mass of PbSO 4 that will be formed when 1.25 L of 0.050 M Pb(NO 3 ) 2 reacts with 2.00 L of 0.025 M Na 2 SO 4. Calculate the mass of PbSO 4 that will be formed when 1.25 L of 0.050 M Pb(NO 3 ) 2 reacts with 2.00 L of 0.025 M Na 2 SO 4. Pb(NO 3 ) 2 and 2.00 L of 0.025 M Na 2 SO 4 Given: 1.25 L of 0.050 M Pb(NO 3 ) 2 and 2.00 L of 0.025 M Na 2 SO 4 Need: Mass (g) of PbSO 4 Plan Na 2 SO 4 : M sodium sulfate × V sodium sulfate = mol Na 2 SO 4 Plan: Use volume and molarity to determine the moles of each reactant Plan Pb(NO 3 ) 2 : M lead (II) nitrate × V lead (II) nitrate = mol Pb(NO 3 ) 2

Solutions in Chemical Reactions: Solution Stoichiometry 2 Determine the moles of reactants

Solutions in Chemical Reactions: Solution Stoichiometry Determine the limiting reactant Limiting reactant 2

Solutions in Chemical Reactions: Solution Stoichiometry Calculate the mass of lead (II) sulfate that forms

Freezing Pt. Depression and Boiling Point Elevation There are some physical properties of solutions that are different from those of the pure solvent There are some physical properties of solutions that are different from those of the pure solvent The ability to lower a freezing point or raise a boiling point of a solution is called a colligative property. The ability to lower a freezing point or raise a boiling point of a solution is called a colligative property. For example: Pure water freezes at T = 0 °C Aqueous solutions freeze at lower temperatures For example: The addition of antifreeze in water makes a solution with a lower f.p. and higher b.p. than that of pure water

Freezing Pt. Depression and Boiling Point Elevation A colligative property of a solution is a physical property that depends on the quantity of solute particles present. A colligative property of a solution is a physical property that depends on the quantity of solute particles present. The ability to lower a freezing point or raise a boiling point depends on the quantity of solute particles present, not the kind of particles. The ability to lower a freezing point or raise a boiling point depends on the quantity of solute particles present, not the kind of particles. For f.p. depression and b.p. elevation, the concentration of the solution is expressed in molality. For f.p. depression and b.p. elevation, the concentration of the solution is expressed in molality.

Freezing Pt. Depression and Boiling Point Elevation The molality of a solution (m) is the ratio that gives the number of moles of solute per kilogram of solvent The molality of a solution (m) is the ratio that gives the number of moles of solute per kilogram of solvent

Freezing Pt. Depression and Boiling Point Elevation The decrease in the freezing point relative to the pure solvent is directly proportional to the number of solute particles per mole of solvent molecules The decrease in the freezing point relative to the pure solvent is directly proportional to the number of solute particles per mole of solvent molecules The increase in the boiling point relative to the pure solvent is directly proportional to the number of solute particles per mole of solvent molecules The increase in the boiling point relative to the pure solvent is directly proportional to the number of solute particles per mole of solvent molecules K f is the freezing point depression constant for the solvent K b is the boiling point elevation constant for the solvent

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