Chapter 12 Solutions

Homework Assigned Problems (odd numbers only) Assigned Problems (odd numbers only) “Questions and Problems” 12.1 to (begins on page 379) “Questions and Problems” 12.1 to (begins on page 379) “Additional Questions and Problems” to (page 407 & 408) “Additional Questions and Problems” to (page 407 & 408) “Challenge Questions” and 12.95, page 408 “Challenge Questions” and 12.95, page 408

Solutions A homogeneous mixture of two or more substances A homogeneous mixture of two or more substances Same composition throughout Same composition throughout One substance is dissolved into another One substance is dissolved into another

Solutions Two parts: Solvent and Solute Two parts: Solvent and Solute Solute: Substance being dissolved Solute: Substance being dissolved Solvent: Substance that dissolves the solvent Solvent: Substance that dissolves the solvent Most solutions are liquid but can be gaseous or solid Most solutions are liquid but can be gaseous or solid

Water as a 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

Formation of Solutions 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)

Formation of Solutions 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

Formation of Solutions 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 “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 Do not dissolve well in water because it is polar Do not dissolve well in water because it is polar Dissolve well in nonpolar solvents such as carbon tetrachloride Dissolve well in nonpolar solvents such as carbon tetrachloride CH 2 Cl 2 H2OH2O I 2 in CH 2 Cl 2 Ni(NO 3 ) 2 in H 2 O

Electrolytes and Nonelectrolytes Electrolytes Electrolytes Solutes that exist as ions in solutionSolutes that exist as ions in solution Formed from an ionic compound that dissociates in water forming a solution with cations and anionsFormed from an ionic compound that dissociates in water forming a solution with cations and anions These substances conduct electricityThese substances 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

Electrolytes and Nonelectrolytes Nonelectrolytes 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 by molecules of the solventEach molecule is separated by molecules of the solvent These substances do not conduct electricityThese substances 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 i.e. sugar and ethanol do not dissociate into ions in water i.e. sugar and ethanol do not dissociate into ions in water Do not conduct electricity Do not conduct electricity

Solubility The maximum amount of solvent that will dissolve into a given amount of solvent The maximum amount of solvent 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 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

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

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

Soluble and Insoluble Salts Not all ionic compounds are soluble in water Not all ionic compounds are soluble in water If water cannot overcome the ionic forces to separate the ions in solution: insoluble If water cannot overcome the ionic forces to separate the ions in solution: insoluble The factors to determine solubility are complex which make predictions difficult The factors to determine solubility are complex which make predictions difficult A series of statements or rules have developed to guide predictions called the solubility rules A series of statements or rules have developed to guide predictions called the solubility rules

Solubility Rules 1) Most compounds that contain NO 3 - and C 2 H 3 O 2 - ions are soluble in water 2) Most compounds that contain Na +, K +, or NH 4 + ions are soluble in water 3) Most compounds that contain Cl - ions are soluble, except AgCl, PbCl 2, and Hg 2 Cl 2 4) Most compounds that contain SO 4 2- ions are soluble, except BaSO 4, PbSO 4, CaSO 4

Solubility Rules 5) Most compounds that contain OH - ions are slightly soluble (will precipitate). Exceptions: NaOH, KOH, are soluble and Ba(OH) 2, Ca(OH) 2 are moderately soluble 6) Most compounds that contain S 2-, CO 3 2-, or PO 4 3- ions are slightly soluble (will precipitate)

Predicting Reactions When chemicals (dissolved in water) are mixed and one of these four events can occur, the reaction will generally happen When chemicals (dissolved in water) are mixed and one of these four events can occur, the reaction will generally happen “Forces” that drive a reaction “Forces” that drive a reaction formation of a solidformation of a solid formation of waterformation of water transfer of electronstransfer of electrons formation of a gasformation of a gas

Formation of a Solid Upon mixing two solutions, if two ions of an insoluble salt come into contact, a solid forms Upon mixing two solutions, if two ions of an insoluble salt come into contact, a solid forms This is a precipitation reaction This is a precipitation reaction The solid that forms is called a precipitate The solid that forms is called a precipitate Can use the solubility rules to predict whether a solid will form when two ionic solutions are mixed Can use the solubility rules to predict whether a solid will form when two ionic solutions are mixed

Formation of a Solid Two aqueous clear solutions are mixed together: lead (II) nitrate and potassium iodide Two aqueous clear solutions are mixed together: lead (II) nitrate and potassium iodide The reaction between two solutes produces an insoluble product: lead (II) iodide The reaction between two solutes produces an insoluble product: lead (II) iodide

Predicting Products To predict the identities of the products: AB + CD (Reactants) 1)Separate all soluble ionic compounds into ions A + B - C + D -

Predicting Products 2) Write the new combination of the ions Exchange pairs Exchange pairs Charge balance each compound Charge balance each compound Determine if any are insoluble Determine if any are insoluble A + B - C + D - A + D - C + B - Products

Predicting Products 3)Write the ionic equation including any solid 4) Write the net ionic equation to remove any spectator ions

Predicting Reactions Determine if product(s) are: Determine if product(s) are: Insoluble (precipitation reaction)Insoluble (precipitation reaction) Water (acid/base reaction)Water (acid/base reaction) Electron transfer (redox reaction)Electron transfer (redox reaction) Gas (gas-forming reaction)Gas (gas-forming reaction)

Types of Equations Molecular Equations Molecular Equations Equations which show the complete chemical formulas for all reactants and productsEquations which show the complete chemical formulas for all reactants and products Uses neutral formulas or symbols without indicating ionic characterUses neutral formulas or symbols without indicating ionic character

Types of Equations Ionic Equations Ionic Equations Separates the neutral formulas into the ions they will actually form in solutionSeparates the neutral formulas into the ions they will actually form in solution Equations which describe the actual ions and molecules in the solutions as well as the molecules of solid, liquid and gas not dissolvedEquations which describe the actual ions and molecules in the solutions as well as the molecules of solid, liquid and gas not dissolved

Types of Equations Spectator Ions Spectator Ions Ions that appear in identical forms on both sides of the equationIons that appear in identical forms on both sides of the equation Although present, they play no role in the reactionAlthough present, they play no role in the reaction Net Ionic Equation Net Ionic Equation An ionic equation in which the spectator ions are omittedAn ionic equation in which the spectator ions are omitted It shows only the ions that are directly involved in a chemical reactionIt shows only the ions that are directly involved in a chemical reaction

Percent 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

Percent Concentrations 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

Mass Percent Conc. 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 Conc. 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

Molarity and Dilutions 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

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 stoichiometry problemsSince it expressed in moles, you can do stoichiometry problems

Solutions 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 solidAdd measured amount of solid Add some water to dissolve the solidAdd some water to dissolve the solid Fill with water up to the lineFill with water up to the line

Molarity Example Calculate the molarity of a solution prepared by dissolving 1.0 g of ethanol in enough water to give a final volume of 101 mL. Calculate the molarity of a solution prepared by dissolving 1.0 g of ethanol in enough water to give a final volume of 101 mL.

Molarity Example

Formalin is an aqueous solution of formaldehyde (HCHO). How many grams of formaldehyde must be used to prepare 2.5 L of 12.3 M formalin? Formalin is an aqueous solution of formaldehyde (HCHO). How many grams of formaldehyde must be used to prepare 2.5 L of 12.3 M formalin?

Standard Solutions A solution whose concentration is exactly known A solution whose concentration is exactly known Dilute to make up less concentrated solutionsDilute to make up less concentrated solutions Like concentrated orange juice: One can of orange juice concentrate is diluted with three cans of water.Like concentrated orange juice: One can of orange juice concentrate is diluted with three cans of water.

Molarity and Dilutions 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

Dilution Example 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.

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

Dilution Example II 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.

Dilution Example II 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 = L Final M 2 = ? V 2 = 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

Dilution Example II 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 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 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

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

Stoichiometry Example 2 Determine the moles of reactants

Stoichiometry Example Determine the limiting reactant Limiting reactant 2

Stoichiometry Example Calculate the mass of lead (II) sulfate that forms

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