- 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.