无机化学课件：10 Ch13 Solution

1、Chemistry: Atoms FirstJulia Burdge & Jason OverbyChapter 13Physical Properties of SolutionsPhysical Properties of Solutions1313.1 Types of Solutions13.2 A Molecular View of the Solution ProcessThe Importance of Intermolecular ForcesEnergy and Entropy in Solution Formation13.3 Concentration UnitsMola

2、lityPercent by MassComparison of Concentration Units13.4 Factors that Affect SolubilityTemperaturePressure13.5 Colligative PropertiesVapor-Pressure LoweringBoiling-Point ElevationFreezing-Point DepressionOsmotic PressureElectrolyte Solutions13.6 Calculations Using Colligative Properties13.7 Colloids

3、Types of SolutionsA solution is a homogeneous mixture of two or more substances.A solution consists of a solvent and one or more solutes.13.1Solutions can be classified by the amount of solute dissolved.An unsaturated solution is one that contains less solute than the solvent has the capacity to dis

4、solve at a specific temperature.Types of SolutionsSolutions can be classified by the amount of solute dissolved.A saturated solution is one that contains the maximum amount of solute that will dissolve in a solvent at a specific temperature.Types of SolutionsSupersaturated solutions are generally un

5、stable.Types of SolutionsA Molecular View of the Solution ProcessSolvation occurs when solute molecules are separated from one another and surrounded by solvent molecules.Solvation depends on three types of interactions:1) Solute-solute interactions2) Solvent-solvent interactions3) Solute-solvent in

6、teractions13.2A Molecular View of the Solution ProcessSoluteSolventSolventSeparated soluteSeparated soluteSeparated solventSolutionEnergyStep 2H2 0Step 1H1 0Step 3H3 0Hsoln = H1 + H2 + H3 “Like dissolves like”Two substances with similar type and magnitude of intermolecular forces are likely to be so

7、luble in each other.The Importance of Intermolecular ForcesToluene, C7H8Octane, C8H18Both non-polar liquids,solution forms when mixedTwo liquids are said to be miscible if they are completely soluble in each other in all proportions.“Like dissolves like”Two substances with similar type and magnitude

8、 of intermolecular forces are likely to be soluble in each other.The Importance of Intermolecular ForcesPolar and non-polar liquids,solution does not form when mixedWater, H2OOctane, C8H18“Like dissolves like”Two substances with similar type and magnitude of intermolecular forces are likely to be so

9、luble in each other.The Importance of Intermolecular ForcesEthanol, C2H6OBoth polar liquids, solution forms when mixedWater, H2O Worked Example 13.1Strategy Consider the structure of each solute to determine whether or not it is polar. For molecular solutes, start with a Lewis structure and apply th

10、e VSEPR theory. We expect polar solutes, including ionic compounds, to be more soluble in water. Nonpolar solutes will be more soluble in benzene.Determine for each solute whether the solubility will be greater in water, which is polar, or in benzene (C6H6), which is nonpolar: (a) Br2, (b) sodium io

11、dide (NaI), (c) carbon tetrachloride, and (d) formaldehyde (CH2O).Solution (a) Bromine is a homonuclear diatomic molecule and is nonpolar. Bromine is more soluble in benzene.(b) Sodium iodide is ionic and more soluble in water. Worked Example 13.1 (cont.)Solution (c) Carbon tetrachloride has the fol

12、lowing Lewis structure:With four electron domains around the central atom, we expect a tetrahedral arrangement. A symmetrical arrangement of identical bonds results in a nonpolar molecule. Carbon tetrachloride is more soluble in benzene.(d) Formaldehyde has the following Lewis structure:Crossed arro

13、ws represent individual bond dipoles. This molecule is polar and can form hydrogen bonds in water. Formaldehyde is more soluble in water.Think About It Remember that molecular formula alone is not sufficient to determine the shape or polarity of a polyatomic molecule. It must be determined by starti

14、ng with a correct Lewis structure and applying VSEPR theory.Concentration UnitsThe amount of solute relative to the volume of a solution or to the amount of solvent in a solution is called concentration.Molarity:Mole fraction:13.3Concentration UnitsMolality (m) is the number of moles of solute disso

15、lved in 1 kg (1000 g) solvent:Percent by Mass: Worked Example 13.2Strategy Use the molar mass of glucose to determine the number of moles of glucose in a liter of solution. Use the density (in g/L) to calculate the mass of a liter of solution. Subtract the mass of glucose from the mass of solution t

16、o determine the mass of water. The molar mass of glucose is 180.2 g/mol.A solution is made by dissolving 170.1 g of glucose (C6H12O6) in enough water to make a liter of solution. The density of the solution is 1.062 g/mL. Express the concentration in (a) molality, (b) percent by mass, and (c) parts

17、per million.Solution (a)1 liter of solution 1062 g 170.1 g = 892 g water = 0.892 kg water170.1 g180.2 g/mol= 0.9440 mol glucose per liter of solution1062 gL= 1062 g0.9440 mol glucose0.892 kg water= 1.06 m Worked Example 13.2 (cont.)Solution (b)(c)170.1 g1062 g solution 100% = 16.02% glucose by mass1

18、70.1 g1062 g solution 1,000,000 = 1.602105 ppm glucoseThink About It Pay careful attention to units in problems such as this. Most require conversions between grams and kilograms and/or liters and milliliters. Worked Example 13.3Strategy (a) Use density to determine the total mass of a liter of solu

19、tion, and use percent by mass to determine the mass of isopropyl alcohol in a liter of solution. Convert the mass of isopropyl alcohol to moles, and divide moles by liters of solution to get molarity.(b) Subtract the mass of C3H7OH from the mass of solution to get the mass of water. Divide moles of

20、C3H7OH by the mass of water (in kg) to get molality.The mass of a liter of rubbing alcohol is 790 g, and the molar mass of isopropyl alcohol is 60.09 g/mol.“Rubbing alcohol” is a mixture of isopropyl alcohol (C3H7OH) and water that is 70 percent isopropyl alcohol by mass (density = 0.79 g/mL at 20C)

21、. Express the concentration of rubbing alcohol in (a) molarity and (b) molality. Worked Example 13.3 (cont.)Solution (a)(b) 790 g solution 553 g C3H7OH = 237 g water = 0.237 kg waterRubbing alcohol is 9.2 M and 39 m in isopropyl alcohol.790 g solutionL solutionThink About It Note the large differenc

22、e between molarity and molality in this case. Molarity and molality are the same (or similar) only for very dilute aqueous solutions.70 g C3H7OH100 g solution=553 g C3H7OHL solution553 g C3H7OHL solution1 mol60.09 g C3H7OH=9.20 mol C3H7OHL solution= 9.2 M9.20 mol C3H7OH0.237 kg water= 39 mFactors Th

23、at Affect SolubilityTemperature affects the solubility of most substances.13.4Pressure greatly influences the solubility of a gas.Henrys law states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution.cmolar concentration (mol/L)P pressure (atm)kpropo

24、rtionality constant called Henrys law constantFactors That Affect Solubilityc = kP Worked Example 13.4Strategy Use c = kP and the given Henrys law constant to solve for the molar concentration (mol/L) of CO2 at 25C and the two pressures given.Calculate the concentration of carbon dioxide in a soft d

25、rink that was bottled under a partial pressure of 5.0 atm CO2 at 25C (a) before the bottle is opened and (b) after the soda has gone “flat” at 25C . The Henrys law constant for CO2 in water at this temperature is 3.110-2 mol/Latm. Assume that the partial pressure of CO2 in air is 0.0003 atm and that

26、 Henrys law constant for the soft drink is the same as that for water.Solution (a) c = (3.110-2 mol/Latm)(5.0 atm) = 1.610-1 mol/L(b) c = (3.110-2 mol/Latm)(0.0003 atm) = 910-6 mol/LThink About It With a pressure approximately 15,000 smaller in part (b) than in part (a), we expect the concentration

27、of CO2 to be approximately 15,000 times smallerand it is.Colligative PropertiesColligative properties are properties that depend on the number of solute particles in solution.Colligative properties do not depend on the nature of the solute particles.The colligative properties are:vapor-pressure lowe

28、ringboiling-point elevationfreezing-point depressionosmotic pressure13.5Raoults law states that the partial pressure of a solvent over a solution is given by the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution.P1partial pressure ofsolvent over solutionP vapo

29、r pressure of pure solvent1mole fraction of solventPvapor pressure lowering2mole fraction of soluteColligative Properties Worked Example 13.5Strategy Convert the masses of glucose and water to moles, determine the mole fraction of water, and use P1 = 1P1 to find the vapor pressure over the solution.

30、 The molar masses of glucose and water are 180.2 and 18.02 g/mol, respectively.Calculate the vapor pressure of water over a solution made by dissolving 225 g of glucose in 575 g of water at 35C. (At 35C, P = 42.2 mmHg.)Solution P = waterP = 0.962 42.4 mmHg = 40.6 mmHgThe vapor pressure of water over

31、 the solution is 40.6 mmHg.H2O225 g glucose180.2 g/mol= 1.25 mol glucose575 g water18.02 g/mol= 31.9 mol waterand31.9 mol water1.25 mol glucose + 31.9 mol water= 0.962H2OH2Owater =Think About It This problem can also be solved using Equation 13.5 to calculate the vapor-pressure lowering, P.If both c

32、omponents of a solution are volatile, the vapor pressure of the solution is the sum of the individual partial pressures. Colligative PropertiesBenzeneTolueneAn ideal solution obeys Raoults law.Colligative PropertiesBenzeneTolueneSolutions boil at a higher temperature than the pure solvent.Tb boiling

33、 point elevationKbboiling point elevation constant (C/m)mmolalityColligative PropertiesSolutions freeze at a lower temperature than the pure solvent.Tffreezing point depressionKffreezing point depression constant (C/m)mmolalityColligative PropertiesColligative Properties Worked Example 13.6Strategy

34、Convert grams of ethylene glycol to moles, and divide by the mass of water in kilograms to get molal concentration. Use molal concentrations and Tb = Kbm and Tf = Kfm, respectively. The molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol. Kf and Kb for water are 1.86C/m and 0.52C/m, respectively.E

35、thylene glycol CH2(OH)CH2(OH) is a common automobile antifreeze. It is water soluble and fairly nonvolatile (b.p. 197C). Calculate (a) the freezing point and (b) the boiling point of a solution containing 685 g of ethylene glycol in 2075 g of water.Solution (a) Tf = Kfm = (1.86C/m)(5.32 m) = 9.89CTh

36、e freezing point of the solution is (0 9.89)C = 9.89C(b) Tb = Kbm = (0.52C/m)(5.32 m) = 2.8CThe boiling point of the solution is (100.0 + 2.8)C = 102.8C685 g C2H6O262.07 g/mol= 11.04 mol C2H6O211.04 mol C2H6O22.075 kg water= 5.32 m C2H6O2andThink About It Because it both lowers the freezing point an

37、d raises the boiling point, antifreeze is useful at both temperature extremes.Osmosis is the selective passage of solvent molecules through a porous membrane from a more dilute solution to a more concentrated one.Colligative PropertiesOsmotic pressure () of a solution is the pressure required to sto

38、p osmosis. Osmotic pressure (atm)Mmolarity (moles/L)Rgas constant (0.08206 Latm/molK)Tabsolute temperature (kelvins)Colligative PropertiesElectrolytes undergo dissociation when dissolved in water.The vant Hoff factor (i) accounts for this effect.Colligative PropertiesThe vant Hoff factor (i) is 1 fo

39、r all nonelectrolytes:For strong electrolytes i should be equal to the number of ions:1 particle dissolved, i = 1Colligative PropertiesNaCl(s)Na+(aq) + Cl(aq)H2OC12H22O11(s)C12H22O11(aq)H2ONa2SO4(s)2Na+(aq) + SO42(aq)H2O2 particles dissolved, i = 23 particles dissolved, i = 3The vant Hoff factor (i)

40、 is usually smaller than predicted due to the formation of ion pairs.An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces.Colligative Propertiesion pairThe vant Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs.A

41、n ion pair is made up of one or more cations and one or more anions held together by electrostatic forces.Colligative PropertiesConcentration has an effect on experimentally measured vant Hoff factors (i).Colligative Properties Worked Example 13.7Strategy Use osmotic pressure to calculate the molar

42、concentration of KI, and divide by the nominal concentration of 0.01000 M. R = 0.08206 Latm/Kmol, and T = 298 K.The osmotic pressure of a 0.0100 M potassium iodide (KI) solution at 25C is 0.465 atm. Determine the experiment vant Hoff factor for KI at this concentration.Solution Solving = MRT for M,M

43、 = =i =The experimental vant Hoff factor for KI at this concentration is 1.90.RT= 0.0190 M0.465 atm(0.08206 Latm/Kmol)(298 K)0.0190 M0.0100 M= 1.90Think About It The calculated vant Hoff factor for KI is 2. The experimental vant Hoff factor must be less than or equal to the calculated value. Worked

44、Example 13.8Strategy Use Tf = Kfm to determine the molal concentration of the solution. Use the density of ethanol to determine the mass of the solvent. The molal concentration of quinine multiplied by the mass of ethanol (in kg) gives moles of quinine. The mass of quinine (in grams) divided by mole

45、s of quinine gives the molar mass. Kf for ethanol is 1.99C/m.Quinine was the first drug widely used to treat malaria, and it remains the treatment of choice for severe cases. A solution prepared by dissolving 10.0 g of quinine in 50.0 mL of ethanol has a freezing point 1.55C below that of pure ethan

46、ol. Determine the molar mass of quinine. (The density of ethanol is 0.789 g/mL.) Assume that quinine is a nonelectrolyte.Solution mass of ethanol = 50.0 mL 0.789 g/mL = 39.5 g or 3.9510-2 kgSolving Tf = Kfm for molal concentrationm = =Tf Kf= 0.779 m1.55C1.99C/m Worked Example 13.8 (cont.)Solution Th

47、e solution is 0.779 m in quinine (i.e., 0.779 mol of quinine/kg ethanol solvent.)molar mass of quinine = = 325 g/mol(3.9510-2 kg ethanol)0.779 mol quininekg ethanol10.0 g quinine0.00308 mol quinine = 0.00308 mol quinineThink About It Check the result using the molecular formula of quinine: C20H24N2O

48、2 (324.4 g/mol). Multistep problems such as this one require careful tracking of units at each step. Worked Example 13.9Strategy Use = MRT to calculate the molarity of the solution. Because the solution volume is 1 L, the molarity is equal to the number of moles of hemoglobin. Dividing the given mas

49、s of hemoglobin by the number of moles gives the molar mass. R = 0.08206 Latm/Kmol, T = 298 K, and = 14.3 mmHg/(760 mmHg/atm) = 1.8810-2 atm.A solution is prepared by dissolving 50.0 g of hemoglobin (Hb) in enough water to make 1.00 L of solution. The osmotic pressure of the solution is measured and

50、 found to be 14.3 mmHg at 25C. Calculate the molar mass of hemoglobin. (Assume that there is no change in volume when the hemoglobin is added to water.) Worked Example 13.9 (cont.)Solution Rearranging = MRT to solve for molarity we get,M = =Thus, the solution contains 7.6910-4 moles of hemoglobin.mo

51、lar mass of hemoglobin =RT= 7.6910-4 M1.8810-2 atm(0.08206 Latm/Kmol)(298 K)50.0 g7.6910-4 mol= 6.50104 g/molThink About It Biological molecules can have very high molar masses.Percent dissociation is the percentage of dissolved molecules (or formula units, in the case of an ionic compound) that sep

52、arate into ions in a solution.Strong electrolytes should have complete, or 100%, dissociation, however, experimentally determined vant Hoff factors indicate that this is not the case.Percent dissociation of a strong electrolyte is more complete at lower concentration.Percent ionization of weak elect

53、rolytes is also dependent on concentration.Calculations Using Colligative Properties13.6 Worked Example 13.10Strategy Use the osmotic pressure and = MRT to determine the molar concentration of the particles in solution. Compare the concentration of particles to the nominal concentration (0.100 M) to

54、 determine what percentage of the original HF molecules are ionized. R = 0.08206 Latm/Kmol, and T = 298 K.A solution that is 0.100 M in hydrofluoric acid (HF) has an osmotic pressure of 2.64 atm at 25C. Calculate the percent ionization of HF at this concentration.Solution Rearranging = MRT to solve

55、for molarity,M = =RT= 0.108 M2.64 atm(0.08206 Latm/Kmol)(298 K) Worked Example 13.10 (cont.)Solution The concentration of dissolved particles is 0.108 M. Consider the ionization of HF:HF(aq) H+(aq) + F-(aq)According to this equation, if x HF molecules ionize, we get x H+ ions and x F- ions. Thus, the total concentration of particles in solution will be the original concentration of HF minus x, which gives the concentration of intact HF molecules, plus 2x, which is the concentration of ions (H+ and F-):(0.100 x) + 2x = 0.100 + xTherefore, 0.108 = 0.100 + x and x = 0.008. Because we earlie