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Student Ch 15 Solutions - Chemistry I – Chapters 15&...

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Unformatted text preview: Chemistry I – Chapters 15 & 16 Chemistry I HD – Chapter 15 ICP – Chapter 22 Solutions 1 SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")! Why does a raw egg swell or shrink when Why placed in different solutions? placed Some Definitions A solution is a solution 2 _______________ _______________ mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and SOLVENT the others as SOLUTES. SOLUTES Parts of a Solution • SOLUTE – the part of a solution that is being dissolved (usually the lesser amount) • SOLVENT – the part of a solution that dissolves the solute (usually the greater amount) • Solute + Solvent = Solution Solute solid solid gas liquid gas gas Solvent solid liquid solid liquid liquid gas Example 3 4 Definitions Solutions can be classified as Solutions saturated or unsaturated. saturated un A saturated solution contains saturated the maximum quantity of solute that dissolves at that temperature. temperature. An unsaturated solution An unsaturated contains less than the maximum amount of solute that can dissolve at a particular temperature particular Example: Saturated and Unsaturated Fats Saturated fats are called saturated because all of the bonds between the carbon atoms in a fat are single bonds. Thus, all the bonds on the carbon are occupied or “saturated” with hydrogen. These are stable and hard to decompose. The body can only use these for energy, and so the excess is stored. Thus, these should be avoided in diets. These are usually obtained from sheep and cattle fats. Butter and coconut oil are mostly saturated fats. 5 Unsaturated fats have at least one double bond between carbon atoms; monounsaturated means there is one double bond, polysaturated means there are more than one double bond. Thus, there are some bonds that can be broken, chemically changed, and used for a variety of purposes. These are REQUIRED to carry out many functions in the body. Fish oils (fats) are usually unsaturated. Game animals (chicken, deer) are usually less saturated, but not as much as fish. Olive and canola oil are monounsaturated. 6 Definitions SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved possible Supersaturated solutions are Supersaturated unstable. The supersaturation is only temporary, and usually accomplished in one of two ways: ways: 1. Warm the solvent so that it will Warm dissolve more, then cool the solution 2. Evaporate some of the solvent carefully so that the solute does not solidify and come out of 7 Supersaturated Supersaturated Sodium Acetate • One application One of a supersaturated solution is the sodium acetate “heat pack.” “heat 8 Compounds in Aqueous Solution Many reactions involve ionic Many compounds, especially reactions in water — aqueous solutions. aqueous KMnO in water 4 IONIC COMPOUNDS K (aq) + MnO (aq) + 4- Aqueous Solutions How do we know ions are How present in aqueous solutions? solutions? The solutions The 9 _________________ _________________ ________ ELECTROLYTES ELECTROLYTES They are called They HCl, MgCl2, and NaCl are and strong electrolytes. strong They dissociate completely (or nearly so) into ions. into Aqueous Aqueous Solutions Solutions Some compounds Some dissolve in water but do not conduct electricity. They are called nonelectrolytes. nonelectrolytes. Examples include: sugar ethanol ethylene glycol 10 It’s Time to Play Everyone’s Favorite Game Show… Electrolyte or Nonelectrolyte! 11 12 Electrolytes in the Body Carry messages to Carry and from the brain as electrical signals as Maintain cellular Maintain function with the correct concentrations electrolytes electrolytes Make your own 50­70 g sugar One liter of warm water Pinch of salt 200ml of sugar free fruit squash Mix, cool and drink 13 Concentration of Solute The amount of solute in a solution The is given by its concentration. concentration Molarity (M) = moles solute liters of solution 14 1.0 L of 1.0 water was used to make 1.0 L of solution. Notice the water left over. over. 15 PROBLEM: Dissolve 5.00 g of NiCl2•6 •6 H2O in enough water to make 250 mL of in solution. Calculate the Molarity. solution. Step 1: Calculate moles Step of NiCl2•6H2O of 1 mol 5.00 g • = 0.0210 mol 237.7 g Step 2: Calculate Molarity Step Calculate 0.0210 mol = 0.0841 M 0.250 L [NiCl2•6 H2O ] = 0.0841 M 16 USING MOLARITY What mass of oxalic acid, H2C2O4, is What required to make 250. mL of a 0.0500 M solution? moles = M•V Step 1: Change mL to L. Step Change 250 mL * 1L/1000mL = 0.250 L Step 2: Calculate. Step Calculate. Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles Step 3: Convert moles to grams. Step Convert (0.0125 mol)(90.00 g/mol) = (0.0125 1.13 g 1.13 17 Learning Check How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 1) 12 g 2) 48 g 3) 300 g 18 Concentration Units An IDEAL SOLUTION is An IDEAL one where the properties depend only on the concentration of solute. concentration Need conc. units to tell us the Need number of solute particles per solvent particle. per The unit “molarity” does not The do this! do 19 Two Other Concentration Units MOLALITY, m mol solute m of solution = kilograms solvent % by mass % by mass = grams solute grams solution 20 Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol Dissolve in 250. g of H2O. Calculate molality and % in O. by mass of ethylene glycol. by Calculating Concentrations 21 Calculate molality 1.00 mol glycol conc (molality) = = 4.00 molal 0.250 kg H2O Calculate weight % Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate m & % of ethylene glycol (by of O. mass). mass). 62.1 g %glycol = x 100% = 19.9% 62.1 g + 250. g 22 Learning Check A solution contains 15 g Na2CO3 and 235 g of H2O? What is the mass % of the solution? 1) 15% Na2CO3 2) 6.4% Na2CO3 3) 6.0% Na2CO3 23 Using mass % How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution? 24 Try this molality problem • 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution. m = mol solute / kg solvent 25 g NaCl 1 mol NaCl 58.5 g NaCl = 0.427 mol NaCl Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg 0.427 mol NaCl 5 kg water = 0.0854 m salt water Colligative Properties On adding a solute to a solvent, the properties On of the solvent are modified. of • Vapor pressure decreases Vapor decreases • Melting point decreases Melting decreases • Boiling point increases Boiling increases • Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE These PROPERTIES. PROPERTIES They depend only on the NUMBER of solute NUMBER particles relative to solvent particles, not on the KIND of solute particles. KIND 25 26 Change in Freezing Point Change Pure water Ethylene glycol/water Ethylene solution solution The freezing point of a solution is LOWER than LOWER that of the pure solvent that Change in Freezing Point Change Common Applications of Freezing Point Depression Depression 27 Propylene glycol Ethylene glycol – deadly to small animals Change in Freezing Point Change Common Applications Common of Freezing Point Depression Depression Which 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) b) c) sand, SiO2 Rock salt, NaCl Ice Melt, CaCl2 28 Change in Boiling Point Change Common Applications of Boiling Point Elevation Elevation 29 30 Boiling Point Elevation and Boiling Freezing Point Depression Freezing ∆T = K•m•i i = van’t Hoff factor = number of particles van’t produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -) present Compound Theoretical Value of i glycol 1 NaCl 2 31 Boiling Point Elevation and Boiling Freezing Point Depression Freezing m = molality molality K = molal freezing molal point/boiling point constant Substance benzene camphor ethyl ether water Kf 5.12 40. 1.79 1.86 ∆T = K•m•i Substance benzene camphor ethyl ether water Kb 2.53 5.95 2.02 0.52 carbon tetrachloride 30. carbon tetrachloride 5.03 Change in Boiling Point Change Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution? solution? b o b 32 K = 0.52 C/molal for water (see K table). Solution ∆TBP = Kb • m • i 1. 2. Calculate solution molality = 4.00 m ∆TBP = Kb • m • i ∆TBP = 0.52 oC/molal (4.00 molal) (1) ∆TBP = 2.08 oC BP = 100 + 2.08 = 102.08 oC (water normally boils at 100) (water 33 Freezing Point Depression Calculate the Freezing Point of a 4.00 molal Calculate glycol/water solution. glycol/water Kf = 1.86 oC/molal (See Kf table) Solution ∆TFP = Kf • m • i = (1.86 oC/molal)(4.00 m)(1) ∆TFP = 7.44 FP FP = 0 – 7.44 = -7.44 oC (because water normally freezes at 0) Freezing Point Depression At what temperature will a 5.4 molal solution At of NaCl freeze? of Solution ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = 20.1 oC 20.1 FP o 34 FP = 0 – 20.1 = -20.1 C 35 Preparing Solutions • Weigh out a solid Weigh solute and dissolve in a given quantity of solvent. solvent. • Dilute a concentrated Dilute solution to give one that is less concentrated. concentrated. 36 ACID-BASE REACTIONS Titrations H2C2O4(aq) + 2 NaOH(aq) ---> acid acid base base Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a TITRATION. Carry TITRATION Oxalic acid, Oxalic H2C2O4 Setup for titrating an acid with a base 37 38 Titration 1. Add solution from the buret. 2. Reagent (base) reacts with 2. compound (acid) in solution in the flask. in 1. Indicator shows when exact Indicator stoichiometric reaction has occurred. (Acid = Base) occurred. This is called NEUTRALIZATION. NEUTRALIZATION. ...
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