2 Overview Types of Mixtures Concentration Dilution SuspensionColloidSolutionConcentrationMass percentMole fractionMolalityMolarityDilutionSolution StoichiometrySolubility“Like dissolves like”Factors affectingHenry’s LawSolubility product (intro)Common ion effect (intro)SaturationElectrolytesColligative PropertiesVapor Pressure LoweringFreezing Point DepressionBoiling Point ElevationOsmotic Pressure
3 Vocabulary Solute – substance being dissolved Solvent – the dissolving medium
4 Types of MixturesColloid – heterogeneous mixture with intermediate sized solute particles that do not settle out of the mixtureSuspension – heterogeneous mixture with large solute particles that can settle out of the mixtureSolution – homogeneous mixture of two or more components
5 ColloidsHeterogeneous mixture with intermediate sized solute particles that do not settle out of the mixtureWhen filtered, particles do not separate outParticles do not separate when left to standColloid particles make up the dispersed phaseExampleWhen large soil particles settle out of muddy water, the water is still cloudyEmulsion and foam are used to classify colloidsEmulsifying agents help keep colloid particles dispersed
6 Examples of Colloids Class of Colloid Phases Example Sol Solid dispersed in liquidPaints, mudGelSolid network extending throughout liquidGelatin, jell-oLiquid emulsionLiquid dispersed in liquidMilk, mayonnaiseFoamGas dispersed in liquidShaving cream, whipped creamSolid aerosolSolid dispersed in gasSmoke, auto exhaustLiquid aerosolLiquid dispersed in gasFog, mist, clouds, aerosol spraySolid emulsionLiquid dispersed in solidCheese, butterSolid foamGas dispersed in solidMarshmallow, styrofoamSolid solSolid dispersed in solidRuby glass
7 Tyndall EffectColloids scatter light, making a beam visible. Solutions do not scatter light.
8 SuspensionHeterogeneous mixture with large solute particles that can settle out of the mixtureParticles are so large that they settle out unless the mixture is constantly stirredParticles can be separated from the mixture through filtrationExampleMuddy water
9 Solution Solvation: interactions between solute and solvent In a solution…The solute can’t be filtered outThe solute always stays mixedParticles are always in motionA solution will have different properties than the solventSolvation: interactions between solute and solventHydration: when the solvent is water
10 Other Solution Examples SoluteSolventExampleGasAirLiquidSodaAlcohol in waterSolidSalt in waterHydrogen in palladiumMercury in silverSilver in gold
12 Comparison Colloid Suspension Solution Heterogeneous Homogeneous Particle size = nm, dispersed; can be aggregates or small moleculesParticle size = over 1000 nm, suspended; can be large particles or aggregatesParticle size = nm, can be atoms, ions, moleculesDo not separate on standingParticles settle outCannot be separated by filtrationCan be separated by filtrationScatter light (Tyndall effect)May scatter light but are not transparentDo not scatter light
13 Concentration Concentrated solution – strong solution Dilute solution – “Watered-down” solutionLow concentrationsPollutants often found in air and water are typically found at very low concentrations. Two common units are used to express these trace amounts.Parts per million (ppm) = (volume solute/volume solution) × 106Parts per billion (ppb) = (volume solute/volume solution) × 109Example. One cm3 of SO2 in one m3 of air would be expressed as 1 ppm or 1000 ppb.
14 Solution Concentration Mass percent: the ratio of mass units of solute to mass units of solution, expressed as a percentMole Fraction (X): the ratio of moles of solute to total moles of solution
15 Solution Concentration Molality (m): moles of solute per kilogram of solvent
16 AssumptionsAssume that solutions with water as the solvent have the density of pure water (1 mL = 1 gram)1 ml of water = 1 gram of water1000 ml of water = 1 liter = 1000 grams
17 Solution Concentration Molarity (M): the ratio of moles of solute to liters of solutionRecognizes that compounds have different formula weightsA 1 M solution of glucose contains the same number of molecules as 1 M ethanol.[ ] - special symbol which means molar (mol/L )M =moles solute molliters of solution L=
18 Molarity ExamplesCalculate the molarity of a 2.0 L solution that contains 10 moles of NaOH MNaOH = 10 molNaOH / 2.0 L = 5.0 MWhat’s the molarity of a solution that has g HCl in 2.0 liters?First, you need the FM of HCl.MassHCl = g/molNext, find the number of moles.molesHCl = gHCl / g/mol= 0.50 molFinally, divide by the volume.MHCl = mol / 2.0 L= M
19 Solution Preparation Solutions are typically prepared by: Dissolving the proper amount of solute and diluting to volume.Dilution of a concentrated solution.
20 Dissolving Solute in Proper Volume How do you prepare ml of a M solution of sodium chloride.First, you need to know how many moles of NaCl are in ml of a 0.5 M solution.M = mol/liters and mL = liters0.5 M = (X mol)/( liters)X = mols NaClNext, we need to know how many grams of NaCl to weigh out.mol NaCl ÷ mol = gramsFinally, you’re ready to make the solution.Weigh out exactly grams of dry, pure NaCl and transfer it to a volumetric flask.Fill the flask about 1/3 of the way with pure water and gently swirl until the salt dissolves.Now, dilute exactly to the mark, cap and mix.
21 Diluting Solutions to Proper Concentration Once you have a solution, it can be diluted by adding more solvent. This is also important for materials only available as solutionsM1V1 = M2V21 = initial 2 = finalAny volume or concentration unit can be used as long as you use the same units on both sides of the equation.
22 Diluting Solutions to Proper Concentration What is the concentration of a solution produced by diluting ml of 1.5 MNaOH to liters?M1V1 = M2V2M1 = 1.5 M M2 = ???V1 = ml V2 = 2000 ml(1.5)(100.0) = (M2 )(2000)M2 = M
23 Solution Stoichiometry Extension of earlier stoichiometry problems.First step is to determine the number of moles based on solution concentration and volume.Final step is to convert back to volume or concentration as required by the problem.You still need a balanced equation and must use the coefficients for working the problem.
24 Stoichiometry Example Determine the volume of M HCl that must be added to completely react with 250 ml of 2.50 M NaOHHCl(aq) + NaOH(aq) →NaCl(aq) + H2O (l)We have 250 ml of a 2.50 M solution2.50 mol/L = (molNaOH)(0.250 L)molNaOH = 0.6250.625 molNaOH = × 1 molHCl/1 molNaOH = molHCl
25 Stoichiometry Example (cont…) Now we can determine what volume of our M HCl solution is required.M = molHCl /literHCl0.100 = (0.625 mol)/(X liters)= L of HCl
26 Solubility of Solutions “Like” dissolves “like”Nonpolar solutes dissolve best in nonpolar solventsExamplesFats in benzene, steroids in hexane, waxes in toluenePolar and ionic solutes dissolve best in polar solventsInorganic salts in water, sugars in small alcoholsMiscible – pairs of liquids that dissolve in one anotherImmiscible – pairs of liquids that do not dissolve in one another
27 Factors Affecting Dissolution Factors that increase the rate of dissolving a soluteIncreasing surface area of soluteMore contact surfaces for solvent to dissolve soluteAgitating (stirring/shaking) solutionIncrease surfaces of solute exposed to solventHeating the solventSolvent molecules move fasterAverage kinetic energy increasesAt higher temperatures, collisions between solvent molecules and solute are more frequent
28 Henry’s LawSolubility of a gas is directly proportional to the partial pressure of that gas on the surface of the liquidIn carbonated beverages, solubility of CO2 is increased by increasing the pressureCarbonation escapes when the pressure is reduced to atmospheric pressure so CO2 escapesEffervescence – the rapid escape of gas from a liquid in which it is dissolvedSg = kPgSg = solubility of gasK = Henry’s law constant (table p 536)Pg = partial pressure of gas over solution
29 Effects on Solubility of Gases To increase the solubility of gasesIncrease pressureGas is forced into solution under pressureSoda goes “flat” when exposed to atmospheric pressureDecrease temperatureIncreased temperature means increased kinetic energyWith increased kinetic energy, more gas molecules can escape from the liquidSoda goes “flat” faster in warm temperatures than cold
31 Solubility ProductSalts are considered “soluble” if more than 1 gram can be dissolved in 100 mL of water.Salts that are “slightly soluble” and “insoluble” still dissociate to some extentSolubility product (Ksp) is the extent to which a salt dissociates in solutionGreater the value of Ksp, the more soluble the salt
33 Ksp Values for Salts at 25°C NameFormulaKsp Barium carbonate BaCO3 2.6 x 10-9 Barium chromate BaCrO4 1.2 x 10-10 Barium sulfate BaSO4 1.1 x 10-10 Calcium carbonate CaCO3 5.0 x 10-9 Calcium oxalate CaC2O4 2.3 x 10-9 Calcium sulfate CaSO4 7.1 x 10-5 Copper(I) iodide CuI 1.3 x 10-12 Copper(II) iodate Cu(IO3)2 6.9 x 10-8 Copper(II) sulfide CuS 6.0 x 10-37 Iron(II) hydroxide Fe(OH)2 4.9 x 10-17 Iron(II) sulfide FeS 6.0 x 10-19 Iron(III) hydroxide Fe(OH)3 2.6 x 10-39 Lead(II) bromide PbBr2 6.6 x 10-6 Lead(II) chloride PbCl2 1.2 x 10-5 Lead(II) iodate Pb(IO3)2 3.7 x 10-13 Lead(II) iodide PbI2 8.5 x 10-9 Lead(II) sulfate PbSO4 1.8 x 10-8 NameFormulaKsp Lead(II) bromide PbBr2 6.6 x 10-6 Lead(II) chloride PbCl2 1.2 x 10-5 Lead(II) iodate Pb(IO3)2 3.7 x 10-13 Lead(II) iodide PbI2 8.5 x 10-9 Lead(II) sulfate PbSO4 1.8 x 10-8 Magnesium carbonate MgCO3 6.8 x 10-6 Magnesium hydroxide Mg(OH)2 5.6 x 10-12 Silver bromate AgBrO3 5.3 x 10-5 Silver bromide AgBr 5.4 x 10-13 Silver carbonate Ag2CO3 8.5 x 10-12 Silver chloride AgCl 1.8 x 10-10 Silver chromate Ag2CrO4 1.1 x 10-12 Silver iodate AgIO3 3.2 x 10-8 Silver iodide AgI 8.5 x 10-17 Strontium carbonate SrCO3 5.6 x 10-10 Strontium fluoride SrF2 4.3 x 10-9 Strontium sulfate SrSO4 3.4 x 10-7 Zinc sulfide ZnS 2.0 x 10-25
34 Common Ion Effect AgCl(s) is added to water The Ag+ ions and the Cl- ions completely dissociateNaCl(aq) is mixed with AgCl(aq)The Na+ ions do not have an effect on the Ag+ or the Cl- ionsThere are already Cl- ions in solutionThe addition of a “common ion” can cause some of the salt to precipitate out of solutionThe salt with the smaller Ksp value will precipitate first
35 Saturation of Solutions Supersaturated solution – contains more solute than can dissolve in the solventSaturated solution – a solution that cannot dissolve any more solute under the given conditionsUnsaturated solution – a solution that is able to dissolve additional solute
36 Electrolytes Electrolyte – solution that conducts electricity ionic compounds in polar solvents dissociate (break apart) in solution to make ionsmay be strong (100% dissociation) or weak (less than 100%)Strong Electrolyte – all or almost all of compound dissociatesExample: strong acids (H2SO4, HNO3, HClO4, HCl, HBr, HI)Weak Electrolyte – Small amount of compound dissociatesExample – weak acids (HF, acetic acid)Nonelectrolyte – solution that does not conduct electricitysolute is dispersed but does not dissociateExample: sugar (dissolves but does not dissociate), organic acids (contain carboxyl groups)
38 Concentration of Ions Strong electrolytes – ions completely dissociate Relative concentration of ions depends on chemical formulaExample 1: In a solution of 0.1 M NaClConcentration of Na+ = 0.1 MConcentraion of Cl - = 0.1 MExample 2: In a solution of 0.1 M Na2SO4Concentration of Na+ = 0.1 M × 2 = 0.2 M (account for 2 sodium ions)Concentraion of SO4-2 = 0.1 M
39 Colligative Properties Properties that depend on the concentration of solute particles but not on their identityVapor pressure loweringFreezing point depressionBoiling point elevationOsmotic Pressure
40 Vapor Pressure Lowering Increased number of solute particles means fewer water moleculesFewer molecules exist to escape from the liquidMolecules have lower tendency to leave the liquid phaseWhen a solute is added to a solution, the vapor pressure of the solution decreases (Raoult’s Law)
41 Vapor Pressure Lowering Raoult’s Law:P = XP°P = vapor pressure of the solutionP° = vapor pressure of the pure solventX = mole fraction of the solvent
42 Vapor Pressure Lowering (Example) Calculate the vapor pressure of the solution when 235 grams of sugar (C12H22O11) are added to 650 mL of water at 40°C. The vapor pressure of water at 40°C is 55 mm Hg. 235 grams C12H22O11 = mol C12H22O mL H2O= 650 g H2O= mol H2O X= /( ) = P = XP° P = (0.981)(55mmHg) = 54.0 mmHg
43 Freezing Point Depression When solute is added to a solution, freezing point decreasesParticles in first beaker can easily shift into solid phase from liquid phase at normal freezing temperatureParticles in second beaker are blocked by solute particles and cannot as easily move into the solid phase from the liquid phase so a colder temperature is needed to shift into the solid phase
44 Freezing Point Depression ∆tf = iKfm∆tf = change in freezing temperaturei = van’t Hoff factor (number of particles into which the added solute dissociates)Kf = molal freezing-point depression constantm = molality
45 Boiling Point Elevation When a solute is added to a solution, the boiling point of the substance increasesFewer water molecules exist due to addition of solute particlesFewer particles leave liquid phaseHigher temperature needed to increase kinetic energy and allow liquid particles to escape
46 Boiling Point Elevation ∆tb= iKbm∆tb = change in boiling temperaturei = van’t Hoff factor (number of particles into which the added solute dissociates)Kb = molal boiling-point elevation constantm = molality
47 Osmotic PressureExternal pressure that must be applied to stop osmosisOsmosis – movement of solvent through a semi-permeable membrane (allows passage of some particles but not others) from the side of lower solute concentration to the side of higher solute concentrationSemi-permeable membrane - allows passage of some particles but not others
48 Osmotic PressureSemi-permeable membrane allows passage of water molecules but not solute particlesSolute particles prevent water molecules from striking surface of semi-permeable membraneOn pure water side, water molecules free to hit surface
49 Osmotic Pressure (cont…) Rate of water molecules leaving pure water side is greater than rate of water molecules leaving solution sideHeight of solution rises until pressure exerted by the height of solution is large enough to forces water molecules back through the membrane at equal rate to which they enter the solution side
50 Osmotic Pressure П = RTi = MRTi λ = osmotic pressure (atm) n = moles of soluteR = gas constantT = temperature (K)V = volume of solutioni = van’t Hoff factor (number of particles into which the added solute dissociates)M = molarity of solution
51 Van’t Hoff FactorSalt lowers the freezing point of water nearly twice as much as sugar does. Why is this?Nonelectrolyte solutions – particles do not dissociate into ionsStrong electrolyte solutions – particles completely dissociate into ionsSugar is a nonelectrolyte so sugar dissolves to produce only one particle in solution. Salt is a strong electrolyte so it dissociates completely to produce two ions in solution (Na+ and Cl-). Salt has twice as many particles to lower the freezing point as sugar.This is accounted for with the van’t Hoff factor (i)What is the van’t Hoff factor for Ba(NO3)2? _________
52 Forces of AttractionA 0.1 m solution of KCl lowers the freezing point more than a 0.1 m solution of MgSO4. Why is that?Forces of attraction (charges of ions) interfere with movements of aqueous ionsOnly in very dilute solutions is the forces of attraction between ions negligibleIons of higher charge attract each other more strongly and cluster together more acting as a single unit more than multiple particlesMgSO4 has ions of +2 and -2. Ions are more attracted to each other more than in KCl (ions are +1 and -1). MgSO4 particles cluster together more than KCl particles so they have less effect on colligative properties.