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  • 1. Solutions1Chemistry I SAVE PAPER AND INK!!! When youChapters 15 & 16 print out the notes on PowerPoint, print "Handouts" instead ofChemistry I HD "Slides" in the print setup. Also,Chapter 15turn off the backgrounds(Tools>Options>Print>UNcheckICP Chapter 22 "Background Printing")! Why does a raw egg swell or shrink when placed in different solutions?

2. Some Definitions2A solution is a _______________ mixture of 2 or more substances in a single phase.One constituent is usually regarded as the SOLVENT and the others as SOLUTES. 3. 3Parts of a Solution SOLUTE thepart of a solution Solute SolventExamplethat is beingdissolved (usually solidsolidthe lesseramount)solidliquid SOLVENT thegassolidpart of a solutionthat dissolves the liquid liquidsolute (usuallythe greatergasliquidamount) gasgas Solute + Solvent =Solution 4. 4DefinitionsSolutions can be classified as saturated or unsaturated.A saturated solution contains the maximum quantity of solute that dissolves at that temperature.An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature 5. 5Example: Saturated and Unsaturated FatsSaturated fats arecalled saturatedbecause all of thebonds between thecarbon atoms in a fatare single bonds.Thus, all the bondson the carbon areoccupied orsaturated withhydrogen. These arestable and hard todecompose. Thebody can only usethese for energy, andso the excess is Unsaturated fats have at least one double bondstored. Thus, thesebetween carbon atoms; monounsaturated meansshould be avoided in there is one double bond, polysaturated meansdiets. These are there are more than one double bond. Thus, thereusually obtained fromare some bonds that can be broken, chemicallysheep and cattle fats. changed, and used for a variety of purposes.Butter and coconut These are REQUIRED to carry out many functionsoil are mostly in the body. Fish oils (fats) are usuallysaturated fats.unsaturated. Game animals (chicken, deer) are usually less saturated, but not as much as fish. Olive and canola oil are monounsaturated. 6. 6 DefinitionsSUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolvedSupersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways:1. Warm the solvent so that it will dissolve more, then cool the solution2. Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution. 7. 7 SupersaturatedSodium Acetate One applicationof asupersaturatedsolution is thesodium acetateheat pack. 8. 8 IONIC COMPOUNDS Compounds in Aqueous SolutionMany reactions involve ionic compounds, especially reactions in water aqueous solutions.KMnO4 in water K+(aq) + MnO4-(aq) 9. Aqueous Solutions 9How do we know ions are present in aqueous solutions?The solutions _______________ __________They are called ELECTROLYTESHCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. 10. Aqueous 10SolutionsSome compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugar ethanol ethylene glycol 11. Its Time to Play Everyones 11Favorite Game Show Electrolyte or Nonelectrolyte! 12. 12Electrolytes in the BodyMake your own Carry messages to50-70 g sugarand from the brainOne liter of warm waterPinch of saltas electrical signals 200ml of sugar free fruitsquash Maintain cellular Mix, cool and drinkfunction with thecorrectconcentrationselectrolytes 13. 13Concentration of SoluteThe amount of solute in a solution is given by its concentration.moles soluteMolarity (M) = liters of solution 14. 141.0 L ofwater wasused tomake 1.0 Lof solution.Notice the water leftover. 15. 15 PROBLEM: Dissolve 5.00 g of NiCl26 H2O in enough water to make 250 mL of solution. Calculate the Molarity.Step 1: Calculate molesof NiCl26H2O1 mol5.00 g = 0.0210 mol 237.7 gStep 2: Calculate Molarity 0.0210 mol= 0.0841 M 0.250 L[NiCl26 H2O ] = 0.0841 M 16. 16 USING MOLARITYWhat mass of oxalic acid, H2C2O4, isrequired to make 250. mL of a 0.0500 Msolution?moles = MVStep 1: Change mL to L.250 mL * 1L/1000mL = 0.250 LStep 2: Calculate.Moles = (0.0500 mol/L) (0.250 L) = 0.0125 molesStep 3: Convert moles to grams.(0.0125 mol)(90.00 g/mol) = 1.13 g 17. 17 Learning CheckHow many grams of NaOH are requiredto prepare 400. mL of 3.0 M NaOHsolution?1) 12 g2) 48 g3) 300 g 18. 18 Concentration UnitsAn IDEAL SOLUTION is one where the properties depend only on the concentration of solute.Need conc. units to tell us the number of solute particles per solvent particle.The unit molarity does not do this! 19. 19Two Other Concentration Units MOLALITY, mmol solute m of solution = kilograms solvent% by mass% by mass =grams solute grams solution 20. 20 Calculating ConcentrationsDissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate molality and % by mass of ethylene glycol. 21. Calculating Concentrations 21 Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate m & % of ethylene glycol (by mass). Calculate molality1.00 mol glycolconc (molality) = 4.00 molal 0.250 kg H2O Calculate weight % 62.1 g%glycol = x 100% = 19.9%62.1 g + 250. g 22. 22 Learning CheckA solution contains 15 g Na2CO3 and 235 g ofH2O? What is the mass % of the solution?1) 15% Na2CO32) 6.4% Na2CO33) 6.0% Na2CO3 23. 23Using mass %How many grams of NaCl are needed toprepare 250 g of a 10.0% (by mass) NaClsolution? 24. 24Try this molality problem 25.0 g of NaCl is dissolved in 5000. mL ofwater. Find the molality (m) of the resultingsolution.m = mol solute / kg solvent25 g NaCl 1 mol NaCl = 0.427 mol NaCl58.5 g NaClSince the density of water is 1 g/mL,5000 mL = 5000 g, which is 5 kg 0.427 mol NaCl= 0.0854 m salt water 5 kg water 25. Colligative Properties25On adding a solute to a solvent, the propertiesof the solvent are modified. Vapor pressuredecreases Melting point decreases Boiling point increases Osmosis is possible (osmotic pressure)These changes are called COLLIGATIVEPROPERTIES.They depend only on the NUMBER of soluteparticles relative to solvent particles, not onthe KIND of solute particles. 26. 26Change in Freezing Point Ethylene glycol/waterPure water solutionThe freezing point of a solution is LOWERthan that of the pure solvent 27. Change in Freezing Point 27Common Applications of Freezing Point DepressionEthyleneglycol deadly toPropylene glycolsmallanimals 28. Change in Freezing Point 28 Common Applicationsof Freezing PointDepressionWhich would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision?a) sand, SiO2b) Rock salt, NaClc) Ice Melt, CaCl2 29. Change in Boiling Point 29Common Applications of Boiling Point Elevation 30. 30 Boiling Point Elevation and Freezing Point Depression T = Kmii = vant Hoff factor = number of particlesproduced per molecule/formula unit. Forcovalent compounds, i = 1. For ioniccompounds, i = the number of ionspresent (both + and -)CompoundTheoretical Value of iglycol1NaCl2CaCl2 3Ca3(PO4)2 5 31. 31Boiling Point Elevation andFreezing Point Depression T = Kmim = molalityK = molal freezing point/boiling point constantSubstance KfSubstanceKbbenzene5.12 benzene 2.53camphor40.camphor 5.95carboncarbon 30.5.03tetrachloride tetrachlorideethyl ether1.79 ethyl ether 2.02water1.86 water 0.52 32. Change in Boiling Point 32Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution?Kb = 0.52 oC/molal for water (see Kb table).SolutionTBP = Kb m i1.Calculate solution molality = 4.00 m2.TBP = Kb m iTBP = 0.52 oC/molal (4.00 molal) (1)TBP = 2.08 oCBP = 100 + 2.08 = 102.08 oC (water normally boils at 100) 33. 33Freezing Point DepressionCalculate the Freezing Point of a 4.00 molal glycol/water solution.Kf = 1.86 oC/molal (See Kf table)SolutionTFP = Kf m i= (1.86 oC/molal)(4.00 m)(1)TFP = 7.44FP = 0 7.44 = -7.44 oC (because water normally freezes at 0) 34. Freezing Point Depression34At what temperature will a 5.4 molal solution of NaCl freeze?Solution TFP = Kf m i TFP = (1.86 oC/molal) 5.4 m 2 TFP = 20.1 oC FP = 0 20.1 = -20.1 oC 35. 35Preparing Solutions Weigh out a solidsolute and dissolve in agiven quantity ofsolvent. Dilute a concentratedsolution to give onethat is lessconcentrated. 36. 36ACID-BASE REACTIONSTitrationsH2C2O4(aq) + 2 NaOH(aq) --->acidbase Na2C2O4(aq) + 2 H2O(liq)Carry out this reaction using a TITRATION. Oxalic acid, H2C2O4 37. Setup for titrating an acid with a base37 38. 38Titration1. Add solution from the buret.2. Reagent (base) reacts with compound (acid) in solution in the flask.3. Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base)This is calledNEUTRALIZATION.