Introduction

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CHE102/112/114

Dr. Stacey Gulde

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ChapterTopic(s)13Solution chemistry14Kinetics15Equilibrium16Acids and bases17Other aqueous equilibriaSpring Break19Thermodynamics20Electrochemistry21Nuclear Chemistry23Transition Metals & Coordination Chemistry18Environmental chemistry

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We expect you to try problems before asking questions, show us your work and how far youve gotten in the problem, BRING YOUR BOOKYou should expect to get correct help in timely & courteous mannerIf not TELL an instructor! Give day & time of incident13Chapter 13:Properties of SolutionsSuggested review: Chap 4 (Concentration of solutions)Chap 11 (11.1 11.5)

Skip section 13.6

14SolutionsSolutions are homogeneous (uniform) mixturesSolute usually present in smaller amounts, its dissolvingSolvent usually present in larger amounts, causes dissolvingUsually water but not always15

Dynamic EquilibriumDissolution (dissolves) solute breaks apart (usually into ions)

Crystallize - when dissolved solutecomes out of soln

Dynamic Equilibrium when the ratesof these processes are equal16

dissolvecrystallize60

Types of SolutionsSupersaturated solutions contains more than the max amount of solute, at a specific tempTo make: heat saturated soln, add solute then cool to original tempNot stableIf add more solute, it crystallizes18Link:NaC2H3O2

Why do Solutions form?

Existing forces & bonds are broken then new ones created

Intermolecular forces(IMF) Chap. 11IntraTypes of IMFWhy do Solutions form?Solvation when ions or molecules are surrounded by solvent molecules in a specific wayHydration special case b/c water is the solvent

20NaCl in waterWater orientsto NaClWater pullsNaCl apartNaClMovie

Na+Cl-A Closer Look atNaCl in WaterMust break solute-solute attraction (Ionic bond)Breaking requires energy, endothermic

Must break solvent-solvent attraction (IMF: H-bonds)Breaking requires energy, endothermic

Must form IMF solute-solvent interactions (Solvation/Hydration)Forming gives off energy, exothermicInteractions increase: w/inc. _______& dec. _____ sizeNaCl (s) Na+ (aq) + Cl- (aq)

chargeion

Why do Solutions form?Each step is associated with an energy change (H)

Hsoln can beendothermic (+)or exothermic (-)Depends on size of IMF

Soln formation will always occur when Hsoln (-)Solute-solvent particles are attracted to each other, they want to be together22

Why do Solutions formif Hsoln > 0Ex: NH4NO3 in H2O, DHsoln = +26 kJ/mol, Endothermic

Solution formation is favored by anincrease in disorder (randomness)Aka: Entropy!! (S)

Spontaneous happens automaticallyNo input of energy

Ex: NH4NO3 forms a soln Enthalpy (H) : disfavoredEntropy (S) : favored23Entropically disfavoredEntropically favoredWhy do Solutions form?ReviewEnthalpy (H): Solutions always form when Hsoln < 0 (exo)Strong solvent-solute interactions

Entropy (S): Increased disorderHelps compensate for Hsoln > 0 (endo)24Factors Affecting SolubilitySolute-Solvent Interactions:Substances likely to dissolve if they have similar forces

LIKE dissolves LIKE!

Polar solvents dissolve: polar or ionic solutesNonpolar solvents dissolve: nonpolar solutes

Liquid solubility terminology:Miscible mixes completelyImmiscible doesnt mix significantlyEx: Oil (nonpolar hydrocarbon) & Water (polar)25Sample ProblemsIs ethanol (C2H6O) miscible in water?

Dominate IMF involved:Water =Ethanol =26

YESH-bondingH-bonding

Sample ProblemsIs hexane (C6H14) miscible in water?27

hexanewater

Factors Affecting Solubility(in water)Structural Factors: Like Dissolves Like

Solubility of liquids decreaseas carbon chain lengthincrease

Solubility of gases increase w/Increased MmIncreased polarity28

2828324084Mm(g/mol)

Factors Affecting SolubilityPressure:Strongly affects gasses in liquids, no effect on solids or liquids in liquidsGas solubility increases as thepressure of the gas (partial pressure)over the solvent increases

Henrys Law29

Sg = solubility of gas (M)Pg = partial pressure of gas above solutionk = Henrys law constant (varies w/temp & solute-solvent pair)

Factors Affecting SolubilityTemperature:Depends on phase:As temperature increases:Solubility of most solid solutes in waterincreases

Solubility of most gas solutes in waterdecreases30

Factors Affecting SolubilityTemperature:Depends on solution process:Endothermic: Added heat overcomes solute-soluteSolubility increases w/increased tempAdditional heat pushes for more soln to form

Exothermic: Added heat overcomes solute-solvent forcesSolubility decreases w/increased tempAdditional heat counteracts more soln to form31A + B solnA + B solnHEAT ++ HEATConcentration of SolutionsQualitatively: relative termInsoluble slightly soluble soluble very soluble

Dilute vs. concentratedThe more solute you have the more concentrated the soln32< 0.1g/100g> 2g/100g

Concentration of SolutionsQuantitatively: numerically calculated

Mass percent, parts per million & parts per billionUnits = None

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x 100x 106x 109Sample ProblemExpress the concentration of 1.0g glucose (C6H12O6) in 100.kg of water, in mass% & ppm

34Mass%=1.0g100,001gx100 =0.0010%ppm=1.0g100,001gx106 =10.ppm

Soln = 100,000g1.0g+ 100,000gConcentration of SolutionsMole fraction, X: Subscript indicates the component of interestUnits = none

Sample: 2.0 mols of glucose dissolved in 12 mols water35

Concentration of SolutionsMolarity, M: (Chap4)

Most widely used Units = mol/L

Easy to measure volume, but it changes with temp36

Concentration of Solutionsmolality, m:Units = mol/kgIndependent of temp

Must know density of soln to calculate molalityAlso allows you to convert b/t M & m37

Sample ProblemsWhat is the m & M concentration of a dilute solution that has 1.00g glucose (C6H12O6) dissolved in 100.kg of water?glucose: Mm = 180.16 g/mol 381.00g C6H12O6 100.kg H2Om =?

1.00g C6H12O6

180.16g C6H12O6 1mol C6H12O6= 0.00555mol C6H12O6

0.00555mol C6H12O6100.kg H2O

(solute)(solvent)Sample ProblemsWhat is the m & M concentration of a dilute solution that has 1.00g glucose (C6H12O6) dissolved in 100.kg of water?glucose: Mm = 180.16 g/mol 391.00g C6H12O6 100.kg H2OM =?

1.00g C6H12O6

180.16g C6H12O6 1mol C6H12O6= 0.00555mol C6H12O6100.kg H2O

1 mL1.00g H2O

Dwater

0.00555mol C6H12O6100.L soln

= 100. L H2O (solute)(solvent)To get volume,need Density

Vsolvent = VsolnSoln = 1g+100,000g=100,001gSoln =100kg Concentration of SolutionsNOTE:If water is solvent: M & m are ~ equal for dilute solnsEx: 1.0g glucose (C6H12O6) in 100.kg of water (See above problem)

1 kg solvent ~ 1 kg solution

Density of water = 1.00 g/mL or 1.00kg/L40Sample ProblemsA solution is prepared by adding 50.0g of glycerol, C3H8O3, to 60.0mL of water. The density of the solution is 1.10 g/mL. Calculate the molality and molarity of the solution.Glycerol: Mm = 92.11 g/mol4150.0g C3H8O3 60.0mL H2ODsoln=1.10g/mLm =?

50.0g C3H8O3

92.11g C3H8O3 1mol C3H8O3 = 0.543mol C3H8O3 60.0mL H2O

1 mL1.00g H2O = 0.0600kg H2O

Dwater

0.543mol C3H8O3 0.0600kg H2O

= 60.0g H2O (solute)(solvent)To get mass,need DensityDo we use1.10g/mL?YesNoSample ProblemsA solution is prepared by adding 50.0g of glycerol, C3H8O3, to 60.0mL of water. The density of the solution is 1.10 g/mL. Calculate the molality and molarity of the solution.Glycerol: Mm = 92.11 g/mol4250.0g C3H8O3 60.0mL H2ODsoln=1.10g/mL110.0g Soln

1.10 g Soln1 mL= 0.100L Soln

0.543mol C3H8O3

= 0.100L Soln

DSoln(solute)(solvent)M =?

DoesVsolvent=Vsoln?YesNoColligative PropertiesColligative property depends on quantity but not the kind of particles (concentration important)

Vapor PressureBoiling PointFreezing PointOsmotic Pressure43Must consider what the solute dissolves into?Non-Electrolytes: dissolve as entire moleculesMolecular compoundsEx. Glucose: C6H12O6(s) C6H12O6(aq)

Strong Electrolytes: break up into many IONSIonic, Strong Acid or Strong BaseAl2(SO4)3(s)

vant Hoff factor, i:compensates for # ofparticles dissolvedESTIMATIONS

442Al3+(aq)+3SO42-(aq)

ChemicalParticles dissolved (i)Glucose:C6H12O61NaCl2MgCl23FeCl34Al2(SO4)35MolecularIonic1. Vapor Pressure LoweringVapor pressure, VP pressure exerted by a vapor at equilibriumA high vapor pressure means many gas molecules presentVolatile exhibits VP (acetone)Nonvolatile no VP (glycerin)

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1. Vapor Pressure LoweringAdding a nonvolatile solute to a solvent ALWAYS LOWERS the vapor pressure of the solvent

Solvent becomes more stabilized in liquid state, thus producing fewer gas particles46

1. Vapor Pressure LoweringRaoults Law:

Psoln depends on the total conc. of soluteparticles added1mol CaCl2 lowers VP more than 1mol NaClCaCl2 = ____ ionsNaCl = ____ ionsIdeal solutions obey Raoults LawSolute conc. is lowSolute & solvent have similar size & IMF47

Psoln = VP of solution (with nonvolatile solute)Xsolvent = mole fractionP solvent = VP of pure solvent

32Sample ProblemPredict the vapor pressure of a solution prepared by mixing 42.9g glycerin (C3H8O3) with 175g of water at 25C. The vapor pressure of pure water at 25C is 23.76 torr. 4842.9g C3H8O3175g H2OPsoln=?Psolvent=23.76 torr

42.9g C3H8O3

92.09 g C3H8O31mol C3H8O3= 0.466 mol C3H8O3175g H2O

18.02g H2O 1mol H2O = 9.71 mol H2O Total = 10.18 mol

9.71 mol10.18 mol

= 22.7 torr

2. Boiling pointBoiling point temp at which liquidVP equals external (gas) pressure

What does curve look like foraqueous solns?

Adding a nonvolatile solute to asolvent increases (ELEVATES)the boiling point relative to theof a pure solvent49

Soln.WaterAqueous Boil pt?IncreaseDecreaseStay Same

3. Freezing Point(Exp. 13!)Freezing point = melting point temp at which VP of solid equals VP of liquidStarts at the Triple point

What happens for aqueous solns?Adding a nonvolatile solute to a solvent decreases the freezing point relative to that of a pure solventHOW?50

SolidLiquidGas51Freezing point depressionDTf = iKfmBoiling point elevationDTb = iKbmComparing:Phase DiagramsBlack = Pure solventBlue = SolnBp: increasedFp: decreasedShifted down

i = # of solute particles dissolved m = molality of solute (mol/kg solvent)

Careful with T: (always want to be positive)Tf= Tpure Tsoln Tb= Tsoln Tpure

Kb=molal boiling point elevation constantKf=molal freezing point depression constantBoth are dependent on solvent52

Sample problemEthylene glycol [EG, CH2(OH)CH2(OH)] is a common car antifreeze. It is water soluble and is a nonvolatile nonelectrolyte (bp=197C, d=1.11kg/L). Calculate the freezing point of a 30.0% (by volume) solution.53

Solute=EGSolvent=water30.0 L70.0 L

1L1.11kg EG

1kg1000 g

62.07g EG1mol EG

1kg H2O1L H2O= 7.66mKf = Tpure =

= 14.2C

Bp=197Cd=1.11kg/L14.2C = 0.0C Tsoln Tsoln = -14.2C1.86C/mi = 1table 13.30.0C4. OsmosisOsmosis net movement of solvent molecules across a semipermeable membraneSemipermeable membrane allows passageof certain small molecules (water)but not large solute (moleculesor ions) due to tiny pores

Movement of solvent always occursfrom less solute more solute As if to equalize the concentrations

Note: Bio uses high solvent low solvent

Pure H2OSugar SolnSugar conc.decreasesafter osmosis(solute = 0)(solute > 0)As solvent moves, liquid levels become unequal resulting in different pressureson each end of the tube

Osmotic pressure () pressure required to stoposmosisSimilar to ideal gas law55

= Osmotic pressure (atm)i = # solute particles in solutionV = volume (L)R = ideal gas constant (0.08206 Latm/molK)T = temperature (K)M = molarity of particles (molecules or ions)4. OsmosisIf osmotic pressures are equal for 2 solns, they are called isotonicOsmosis wont occur through a semipermeable membrane

Hypotonic soln of lower concentrationIf place red blood cells in Hypotonic soln:Hemolyisis water enters the cell = ruptures

Hypertonic soln of higher concentrationIf place red blood cells in Hypertonic soln:Crenation water leaves the cell = shrivel56

Sample ProblemsWhat is the osmotic pressure of a 0.010M aqueous solution of CaCl2 at 25C?57 = ?M = 0.010 mol/LT = 25C

Predicted i?1234i = 3

298K

Sample Problems(try on your own)The boiling point of ethanol (C2H5OH) is 78.4C & it has a boiling point elevation constant of 1.22C/m. How many grams of glycerin (C3H8O3) must be added to 200.0g of ethanol to produce a solution temp of 81.1C.58T = 78.4CKb=1.22C/mgC3H8O3 = ?

Predicted i?1234Tsoln = 81.1C

gC2H5OH = 200.0gTb = Tsoln - TPure= 81.1C- 78.4CTb = 2.70Ci = 1

(solute)(solvent)

0.2000kg

92.09g C3H8O31mol C3H8O3= 40.7g C3H8O3Next slideChap. 13 SummarySolution formationEntropy increased disorderEnthalpy favorable intermolecular interactionsSolubilityInfluence of pressure Henrys LawInfluence of molecular structure like dissolves likeInfluence of temperatureConcentrationMass %, ppm, ppbMole fractionMolarityMolalityColligative propertiesVapor Pressure Raoults LawBoiling pt. elevation & Freezing pt. depressionOsmosisvant Hoff factor, i59