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

Chapter 13Chapter 13SolutionsSolutions

HomeworkHomework

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 Cumulative Problems” 109-129 (begins on page

482)482) ““Highlight Problems” 131, 133, page 484-485Highlight Problems” 131, 133, page 484-485

SolutionsSolutions

A A solutionsolution is a homogeneous mixture is a homogeneous mixture of two or more substancesof two or more substances

It is uniform, same composition It is uniform, same composition throughoutthroughout

One substance is dissolved into One substance is dissolved into anotheranother

Requires the interaction of particles Requires the interaction of particles of atomic or molecular sizeof atomic or molecular size

SolutionsSolutions Two parts: Solvent and Two parts: Solvent and solutesolute Solute: Substance being dissolved (present in a smaller Solute: Substance being dissolved (present in a smaller

amount relative to the solvent)amount relative to the solvent) Solvent: Substance that dissolves the solvent (present in Solvent: Substance that dissolves the solvent (present in

the greatest amount)the greatest amount) Most solutions are liquid but can be gaseous or solidMost solutions are liquid but can be gaseous or solid

SolutionsSolutions

““Like dissolves like” Like dissolves like” Substances that are similar should Substances that are similar should

form a solutionform a solution Refers to the overall polarity of the Refers to the overall polarity of the

solvent (polar and nonpolar) and the solvent (polar and nonpolar) and the solute (polar, nonpolar, and ionic)solute (polar, nonpolar, and ionic)

Must be an attraction between the Must be an attraction between the solute and solvent for a solution to solute and solvent for a solution to form form

SolutionsSolutions Nonpolar molecules Nonpolar molecules

(no dipole, cannot (no dipole, cannot hydrogen bond)hydrogen bond)

Examples: oil, iodineExamples: oil, iodine• Both do not dissolve Both do not dissolve

well in water well in water because it (water) because it (water) is polaris polar

• Both dissolve well Both dissolve well in nonpolar solvents in nonpolar solvents such as carbon such as carbon tetrachloridetetrachloride

CCl4

H2O

I2 in

CCl4

Ni(NO3)2

in H2O

Solutions of Solids Dissolved in WaterSolutions of Solids Dissolved in Water(Water as a Solvent)(Water as a Solvent)

Most common solutions (with a solid, Most common solutions (with a solid, liquid, or gas) contain water as the liquid, or gas) contain water as the solventsolvent

Water is a polar molecule due to its bent Water is a polar molecule due to its bent shapeshape

It also has the ability to hydrogen bond It also has the ability to hydrogen bond Dissolves many polar and ionic Dissolves many polar and ionic

substancessubstances Due to intermolecular interactions Due to intermolecular interactions

(dipole-dipole or H-bonding) upon mixing(dipole-dipole or H-bonding) upon mixing


Polar compounds (a permanent dipole, can H-Polar compounds (a permanent dipole, can H-bond) and ionic compounds dissolve into polar bond) and ionic compounds dissolve into polar solventssolvents

A polar molecule (with ionic bonding) dissolves A polar molecule (with ionic bonding) dissolves into water if attractions for water overcome the into water if attractions for water overcome the attractions between the ionsattractions between the ions

As each ion enters the solution, it is immediately As each ion enters the solution, it is immediately surrounded by water molecules: surrounded by water molecules: hydration hydration (solvation)(solvation)


When sodium chloride crystals are placed in water, they When sodium chloride crystals are placed in water, they begin to dissolve begin to dissolve

The attractive forces between the ions and water are The attractive forces between the ions and water are stronger than forces between the ions in the crystalstronger than forces between the ions in the crystal

Water molecules surround each ion, keeping them apart Water molecules surround each ion, keeping them apart


When sodium chloride When sodium chloride crystals are placed in crystals are placed in water, they begin to water, they begin to dissolve dissolve

The attractive forces The attractive forces between the ions and between the ions and water are stronger water are stronger than forces between than forces between the ions in the crystalthe ions in the crystal

Water molecules Water molecules surround each ion, surround each ion, keeping them apart keeping them apart

Electrolyte Solutions: Electrolyte Solutions: Dissolved Ionic SolidsDissolved Ionic Solids

Compounds that ionize in water are Compounds that ionize in water are called called electrolyteselectrolytes• Electrolytes are Electrolytes are solutessolutes that exist as that exist as

ions in solutionions in solution• Formed from an ionic compound that Formed from an ionic compound that

dissociates in water forming an dissociates in water forming an electroyte solutionelectroyte solution with cations and with cations and anionsanions

• These solutions conduct electricityThese solutions conduct electricity

Ions In Solution (in Water)Ions In Solution (in Water)

When ionic compounds dissolve in When ionic compounds dissolve in water the ions dissociatewater the ions dissociate• Separate into the ions floating in waterSeparate into the ions floating in water• Potassium chloride dissociates in water into Potassium chloride dissociates in water into

potassium cations and chloride anionspotassium cations and chloride anions

KCl(aq) = KKCl(aq) = K+ + (aq) + Cl(aq) + Cl- - (aq)(aq)

K+ Cl-K Cl


Copper(II) sulfate dissociates in water Copper(II) sulfate dissociates in water into copper(II) cations and sulfate into copper(II) cations and sulfate anionsanions

CuSOCuSO44(aq) = Cu(aq) = Cu+2+2(aq) + SO(aq) + SO442-2-(aq)(aq)

Cu+2 SO42-Cu SO4


Potassium sulfate dissociates in water Potassium sulfate dissociates in water into potassium cations and sulfate into potassium cations and sulfate anionsanionsKK22SOSO44(aq) = 2 K(aq) = 2 K+ + (aq) + SO(aq) + SO44

2-2-(aq)(aq)

K+

SO42-

K+

KK SO4

Nonelectrolyte SolutionsNonelectrolyte Solutions

Compounds that do not ionize in Compounds that do not ionize in water are called water are called nonelectrolytesnonelectrolytes• Solute is a molecular substanceSolute is a molecular substance• Substance dispersed throughout the Substance dispersed throughout the

solvent as individual moleculessolvent as individual molecules• Each molecule is separated (dissolved) Each molecule is separated (dissolved)

by molecules of the solvent forming a by molecules of the solvent forming a nonelectrolyte solutionnonelectrolyte solution

• These solutions do not conduct These solutions do not conduct electricityelectricity

Electrolytes and NonelectrolytesElectrolytes and Nonelectrolytes Strong electrolyteStrong electrolyte::

• Dissociates completely into ionsDissociates completely into ions• Conduct electricityConduct electricity

Weak electrolyteWeak electrolyte::• Mainly whole moleculesMainly whole molecules• Very few separate (into ions)Very few separate (into ions)• Conduct electricity less than strong Conduct electricity less than strong

electrolyteselectrolytes NonelectrolyteNonelectrolyte::

• No dissociation into ionsNo dissociation into ions• Do not conduct electricityDo not conduct electricity

Electrolytes and NonelectrolytesElectrolytes and Nonelectrolytes

Strong electrolytes Strong electrolytes are completely are completely ionized when ionized when dissolved in waterdissolved in water

Sodium chloride Sodium chloride dissociates to form dissociates to form NaNa+ + and Cland Cl- -

Good conductor of Good conductor of electricityelectricity


Weak electrolytes Weak electrolytes are only partially are only partially ionized when ionized when dissolved in waterdissolved in water

Hydrofluoric acid Hydrofluoric acid only partially only partially dissociates to form dissociates to form HH+ + and Fand F- -

Poor conductor of Poor conductor of electricityelectricity


Nonelectrolyes are Nonelectrolyes are not ionized when not ionized when dissolved in waterdissolved in water

e.g. sugar and e.g. sugar and ethanol do not ethanol do not dissociate into ions dissociate into ions in waterin water

Do not conduct Do not conduct electricityelectricity

Solubility and SaturationSolubility and Saturation

SolubilitySolubility is the maximum amount of is the maximum amount of solute that will dissolve into a given solute that will dissolve into a given amount of solventamount of solvent

It is affected byIt is affected by• Type of solute (solid, liquid, or gas)Type of solute (solid, liquid, or gas)• Type of solvent (and the solute Type of solvent (and the solute

interaction)interaction)• TemperatureTemperature

Solubility and SaturationSolubility and Saturation UnsaturatedUnsaturated: Less solute than the : Less solute than the

maximum amount possible is dissolved maximum amount possible is dissolved into the solutioninto the solution

SaturatedSaturated: Contains the maximum amount : Contains the maximum amount of solute that can be dissolved of solute that can be dissolved

saturated

Solubility and SaturationSolubility and Saturation

A A supersaturatedsupersaturated solution contains more of the solution contains more of the dissolved particles than could be dissolved by the dissolved particles than could be dissolved by the solvent under normal circumstancessolvent under normal circumstances

These solutions result from altering a condition of These solutions result from altering a condition of the saturated solution such as T, V, or Pthe saturated solution such as T, V, or P

Solutions of Solids in Water:Solutions of Solids in Water: Effect of Temperature on Solubility Effect of Temperature on Solubility

For most solids For most solids (solute), solubility (solute), solubility increases with an increases with an increase in increase in temperaturetemperature

More sugar will More sugar will dissolve in hot dissolve in hot water than in cold water than in cold waterwater

Solutions of Gases in Water:Solutions of Gases in Water:Effect of Temperature on SolubilityEffect of Temperature on Solubility

For gases, solubility For gases, solubility decreases with an decreases with an increase in increase in temperaturetemperature

Henry’s Law: The Henry’s Law: The amount of a gas amount of a gas dissolved in a solution dissolved in a solution is directly proportional is directly proportional to the pressure of the to the pressure of the gas above the solutiongas above the solution

Solution ConcentrationSolution Concentration

Solution concentrationSolution concentration• The amount of solute (mass or moles) dissolved into a The amount of solute (mass or moles) dissolved into a

certain amount of a solution or solventcertain amount of a solution or solvent

QualitativeQualitative• Dilute, concentrated, saturated, unsaturatedDilute, concentrated, saturated, unsaturated

QuantitativeQuantitative• Mass to mass, volume to volume, and molarityMass to mass, volume to volume, and molarity

%100 solutionof amount

soluteof amountsolutionaofononcentratic

Solution ConcentrationSolution Concentration

%100)/( solutionofvolumesoluteofvolumevvpercentvolume %

•Mass PercentMass PercentMass of the solute divided by the total mass of solution Mass of the solute divided by the total mass of solution multiplied by 100multiplied by 100

The mass of the solute and solution must be in the same units

•Volume PercentThe volume of solute divided by the total volume of solution multiplied by 100

The solute and solution volumes must be in the same unitsThe solute and solution volumes must be in the same units

100%g solventof massg soluteof mass

g soluteof massm/m%percentmass )( )(

)()(

Mass PercentMass Percent

Grams of solute per grams of Grams of solute per grams of solutionsolution• Remember that the mass of solution Remember that the mass of solution

is is grams of solute grams of solute ++ grams of solvent grams of solvent

100solution of mass

solute of massm/m)(%percent mass

Calculating Mass Percent: Example 1Calculating Mass Percent: Example 1

A 135 g sample of seawater is A 135 g sample of seawater is evaporated to dryness, leaving 4.73 g of evaporated to dryness, leaving 4.73 g of solid residue. What is the solid residue. What is the mass percentmass percent of the solute in the original sea water?of the solute in the original sea water?

%50.3%100solution g 135

solid g 4.73

100solution of mass

solute of massm/m)(%percent mass

Mass Percent in Calculations: Example 2Mass Percent in Calculations: Example 2

What mass of water must be added to 425 g What mass of water must be added to 425 g of formaldehyde to prepare a 40.0% (by of formaldehyde to prepare a 40.0% (by mass) solution of formaldehyde?mass) solution of formaldehyde?

40% 425 g formaldehyde

? g total100%


100 g total

Mass Percent Conc. Example 2Mass Percent Conc. Example 2


100 g total

1062.5 g total

1062.5 g total - 425 g formaldehyde = 637.5 g water

425 g formaldehyde 100 g total 40 g formaldehyde

Solution Concentration: MolaritySolution Concentration: Molarity

Molarity is the concentration Molarity is the concentration expression most commonly used in expression most commonly used in the laboratorythe laboratory

The amount of solute is expressed in The amount of solute is expressed in molesmoles

To obtain the molarity, we need to To obtain the molarity, we need to know the solution volume in liters know the solution volume in liters and the number of moles of solute and the number of moles of solute presentpresent

Solution Concentration: Molarity Solution Concentration: Molarity

MolesMoles of solute of solute per litersper liters of solution of solution More useful than mass percentMore 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 Amount of solute expressed in moles (quantity of particles)particles)

• Chemical reactions occur between molecules and atomsChemical reactions occur between molecules and atoms

• Since it expressed in moles, you can do chemical Since it expressed in moles, you can do chemical calculation (calculation (stoichiometrystoichiometry) problems) problems

solution liters

solute moles(M)Molarity

Making Solutions of a Specific MolarityMaking Solutions of a Specific Molarity

Make up in a volumetric flaskMake up in a volumetric flask• Flask with a long, narrow neck that is marked Flask with a long, narrow neck that is marked

with a line indicating an exact volumewith a line indicating an exact volume MethodMethod

• 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 1Using Molarity in Calculations: Example 1

Calculate the molarity of a solution made Calculate the molarity of a solution made by dissolving 15.0 g of NaOH (sodium by dissolving 15.0 g of NaOH (sodium hydroxide) in enough water to give a final hydroxide) in enough water to give a final volume of 100. mL.volume of 100. mL.

g/mol 40.00NaOH of massMolar

L100.0mL1000

L1mL.100

Convert volume to liters


NaOH g 40.00

NaOH mol 1 NaOH g 15.0NaOH mol 0.375

Morliter

mol75.375.3

L100.0

mol375.0

Convert mass to moles

solution liters

solute moles(M)Molarity


Formalin is an aqueous solution of Formalin is an aqueous solution of formaldehyde (Hformaldehyde (H22CO). How many grams of CO). How many grams of formaldehyde must be used to prepare 2.50 L formaldehyde must be used to prepare 2.50 L of 12.3 M formalin?of 12.3 M formalin?

2.5 L 12.3 mol formaldehyde1 L

molarity × volume = moles

30.8 mol formaldehyde

COH mol 1

COH g 30.03 COH mol 30.8

2

22 COHg925 2

Standard SolutionsStandard Solutions A solution whose concentration is A solution whose concentration is exactlyexactly known known

• A std. solution can be diluted to make up A std. solution can be diluted to make up less less concentratedconcentrated solutions solutions

• A std. solution is like concentrated orange juice. A std. solution is like concentrated orange juice. For example, one can of orange juice For example, one can of orange juice concentrate is diluted with three cans of water.concentrate is diluted with three cans of water.

Solution DilutionSolution Dilution

Dilution is the process in which more Dilution is the process in which more solvent is added to a solution in order to solvent is added to a solution in order to lower its concentrationlower its concentration

A common laboratory routine is diluting a A common laboratory routine is diluting a solution of known concentration (solution of known concentration (stock stock solutionsolution) to a lower concentration) to a lower concentration

A dilution always lowers the concentration A dilution always lowers the concentration because the same amount of solute is because the same amount of solute is present in a larger amount of solventpresent in a larger amount of solvent

solutiondilutedsolutionstock soluteMolessolute Moles

DilutionDilution

• Most often a solution of a specific molarity must be Most often a solution of a specific molarity must be prepared by adding a predetermined volume of solvent prepared by adding a predetermined volume of solvent to a specific volume of to a specific volume of stock solutionstock solution

• When solvent is added to dilute a solutionWhen solvent is added to dilute a solution, the number , the number of moles remains unchangedof moles remains unchanged

• A relationship exists between the volumes and A relationship exists between the volumes and molarities of the diluted and stock solutionsmolarities of the diluted and stock solutions

solution L

solute molM

Moles of solute =

(initial solution)

M1V1 =

Moles of solute

(diluted solution)

M2V2

solute molMsolution L

Solution Dilution: Example 1Solution Dilution: Example 1 Determine the volume required to prepare Determine the volume required to prepare

0.75L of 0.10 0.75L of 0.10 MM HCl from a HCl from a 12 12 MM HCl stock HCl stock solutionsolution..

• How many moles of HCl do we How many moles of HCl do we eventually want?eventually want?

0.075 mol HCl0.75 L 0.10 mol HCl

L

• Initially we have 12 M HCl. Calculate Initially we have 12 M HCl. Calculate what volume of the what volume of the stock solutionstock solution will will contain the number of moles needed.contain the number of moles needed.

Solution Dilution: Example 1Solution Dilution: Example 1

• How many liters of How many liters of 12 M HCl12 M HCl contains contains 0.075 mol of HCl ?0.075 mol of HCl ?

6.2510-3 L 12 M HCl0.075 mol HCl 1 L

12 mol HCl

6.25 mL of 12 M HCl are needed


• What is the molarity of a solution prepared What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSOwhen 25.0 mL of a 1.0 M CuSO4 4 is diluted to a is diluted to a final volume of 250 mL. final volume of 250 mL.


• What is the molarity of a solution prepared What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSOwhen 25.0 mL of a 1.0 M CuSO4 4 is diluted to is diluted to

a final volume of 250 mL.a final volume of 250 mL.

Initial M1 = 1.0 M V1 = 0.025 L Final M2 = ? V2 = 0.250 L

Calculate the unknown molarity using the relationship

Moles of solute =

(initial solution)

M1V1 =

Moles of solute

(diluted solution)

M2V2


2211 VMVM

2250.0

025.00.1M

L

LL

mol

• What is the molarity of a solution prepared What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSOwhen 25.0 mL of a 1.0 M CuSO4 4 is diluted to is diluted to

a final volume of 250 mL.a final volume of 250 mL.

Set Up Problem by solving for M2

1) Moles of solute before dilution equals moles of solute after dilution

2) Calculate the unknown molarity by solving for M2

3) Set up problem

22

11 MV

VM

Solutions in Chemical Reactions:Solutions in Chemical Reactions:Solution StoichiometrySolution Stoichiometry

Stoichiometry is the calculation of Stoichiometry is the calculation of quantitative relationships between quantitative relationships between reactants and productsreactants and products

Calculations are based on balanced Calculations are based on balanced chemical equationschemical equations

The coefficients in the balanced equation The coefficients in the balanced equation indicate the moles of products and indicate the moles of products and reactants reactants


Many reactions take place in solution and Many reactions take place in solution and the solution concentration (molarity) the solution concentration (molarity) directly relates the solution volume and directly relates the solution volume and moles of solute presentmoles of solute present

Stoichiometric calculations are the same Stoichiometric calculations are the same as in chapter 8, but with the addition of as in chapter 8, but with the addition of some molarity calculationssome molarity calculations

Quantitative Relationships Needed for Quantitative Relationships Needed for Solving Chemical Formula Based ProblemsSolving Chemical Formula Based Problems

Grams A Grams B

PA , TA, VA PB , TB, VB

Liters A Liters BMoles A Moles B

pV = nRT

Mole-moleFactormolarity

Molar mass

molarity

1 mol = 22.4 Lat STP

M × V M × V


When aqueous solutions of NaWhen aqueous solutions of Na22SOSO44 and and Pb(NOPb(NO33))22 are mixed, PbSO are mixed, PbSO44 precipitates. precipitates.


Calculate the mass of PbSOCalculate the mass of PbSO44 that will be that will be formed when 1.25 L of 0.050 formed when 1.25 L of 0.050 MM Pb(NO Pb(NO33))22 reacts with 2.00 L of 0.025 reacts with 2.00 L of 0.025 MM Na Na22SOSO44..

(aq)NaNO2 (s)PbSO (aq))Pb(NO (aq)SONa 342342

Given: 1.25 L of 0.050 M Pb(NOPb(NO33))2 2 and 2.00 L of 0.025 M Naand 2.00 L of 0.025 M Na22SOSO4 4

Need: Mass (g) of PbSONeed: Mass (g) of PbSO44

Plan Na2SO4 : Msodium sulfate× Vsodium sulfate = mol Na2SO4

Plan: Use volume and molarity to determine the moles of each reactant

Plan Pb(NO3)2 : Mlead (II) nitrate× Vlead (II) nitrate = mol Pb(NO3)2


Na2SO4 (aq) Pb(NO3)2 (aq) PbSO 4(s) NaNO3(aq)2

23 )Pb(NO mol 0.063

42SONa mol 0.050

1.25 L 0.050 mol Pb(NO 3)2

1L

2.00 L 0.025 mol Na 2SO4

1L

Determine the moles of reactants


0.0625 mol PbSO 4

4PbSO mol 0.05042

442

SONa mol 1

PbSO mol 1SONa mol 0.050

Determine the limiting reactant

Limiting reactant

Na2SO4 (aq) Pb(NO3)2 (aq) PbSO4(s) NaNO3(aq)2

0.0625 mol Pb(NO 3)2 1 mol PbSO 4

1 mol Pb(NO3 )2


14.96 g PbSO4

4

44

PbSO mol 1

PbSO g 299.27PbSO mol 0.05

Calculate the mass of lead (II) sulfate that forms

Freezing Pt. Depression and Freezing Pt. Depression and Boiling Point ElevationBoiling Point Elevation

There are some physical properties of There are some physical properties of solutions that are different from those of solutions that are different from those of the pure solventthe pure solvent

The ability to lower a freezing point The ability to lower a freezing point or raise a boiling point of a solution is or raise a boiling point of a solution is called a called a colligative propertycolligative property..

For example: Pure water freezes at T = 0 °CAqueous 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


A colligative property of a solution is a A colligative property of a solution is a physical property that depends on the physical property that depends on the quantity of solute particles present.quantity of solute particles present.

The ability to lower a freezing point or The ability to lower a freezing point or raise a boiling point depends on the raise a boiling point depends on the quantity of solute particles present, not quantity of solute particles present, not the kind of particles.the kind of particles.

For f.p. depression and b.p. elevation, the For f.p. depression and b.p. elevation, the concentration of the solution is expressed concentration of the solution is expressed in in molalitymolality. .


The molality of a solution (m) is the The molality of a solution (m) is the ratio that gives the number of moles ratio that gives the number of moles of solute per kilogram of solventof solute per kilogram of solvent

solventKilograms

soluteMolesMolality


The decrease in the freezing point relative to the The decrease in the freezing point relative to the pure solvent is directly proportional to the pure solvent is directly proportional to the number of solute particles per mole of solvent number of solute particles per mole of solvent moleculesmolecules

The increase in the boiling point relative to the The increase in the boiling point relative to the pure solvent is directly proportional to the pure solvent is directly proportional to the number of solute particles per mole of solvent number of solute particles per mole of solvent moleculesmolecules

bb KmT

ff KmT Kf is the freezing point depression constant for the solvent

Kb is the boiling point elevation constant for the solvent

endend

Documents

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