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CHM 1046 - General Chemistry 2 Summer B 2012 Dr. Jeff Joens

CHM 1046 - General Chemistry 2 Summer B 2012 Dr. Jeff Joens Classroom: CP 145 Time: M, T, W, R 9:30am to 10:45am Office: CP 331 phone 348-3121 web page: www.joenschem.com email: joensj@fiu.edu. Homework Problems

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CHM 1046 - General Chemistry 2 Summer B 2012 Dr. Jeff Joens

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  1. CHM 1046 - General Chemistry 2 Summer B 2012 Dr. Jeff Joens Classroom: CP 145 Time: M, T, W, R 9:30am to 10:45am Office: CP 331 phone 348-3121 web page: www.joenschem.com email: joensj@fiu.edu

  2. Homework Problems Chapter 12 Homework Problems: 2, 5, 8, 12, 20, 23, 32, 34, 36, 42, 52, 64, 72, 74, 78, 80, 84, 89, 98 Note: Homework is not to be turned in. My solutions to the homework problems are available at my website. Answers (but not detailed solutions) to the in chapter practice problems and the odd numbered end of chapter problems are given at the end of the book. There are also several copies of the textbook and solution manual (Solutions Manual for Chemistry: A Molecular Approach) available at the Reserve Desk (2nd Floor of the Green Library) under Dr. Graves name. The solution manual gives detailed solutions to all of the problems in the book.

  3. CHAPTER 12 Solutions

  4. Solutions A solution is a homogeneous mixture. The major component of a solution is called the solvent, while the minor components are called solutes. There are several common kinds of solutions kind of solutionexample gas in gas air (78% N2, 21 % O2, 0.9 % Ar + others) gas in liquid O2 in water liquid in liquid ethanol in water solid in liquid NaCl in water solid in solid alloys (bronze - copper + tin) We will usually focus on solutions where the solvent is a liquid, but all of the above combinations represent possible types of solutions.

  5. Solubility The term solubility means the amount of a particular solute that will dissolve in a given amount of solvent (usually a liquid). There are two common ways in which solubility is given. 1) Mass solute per mass solvent. Indicates the maximum mass of solute that will dissolve in a given mass of solvent. Example: At 25 C the solubility of sodium chloride in water is 36. g NaCl per 100. g water. 2) Molar solubility. The maximum number of moles of solute per liter of solution. Example: The molar solubility of silver chloride (AgCl) in pure water at 25 C is 1.3 x 10-5 mol/L. (Equivalent to 1.9 x 10-4 g per 100 g water). There are other ways of expressing solubility, including mass per volume of solvent, and solubility product.

  6. Soluble and Insoluble There are several common terms that refer to the ability of a solid solute to dissolve in a liquid solvent. Soluble - Indicates that a large amount of solute will dissolve in a given amount of solvent. Insoluble - Indicates that a small amount, or no, solute will dissolve in a given amount of solvent. In our above examples sodium chloride is soluble in water and silver chloride is insoluble in water. In fact, many substances classified as “insoluble” will dissolve to a small extent in water or other solvents. We will discuss this in more detail later in the semester.

  7. Miscible and Immiscible Liquids We can classify liquids in terms of their ability to form solutions when mixed. Two liquids are miscible if any amounts of the two liquids will combine to form a solution (water + ethyl alcohol). Two liquids are immiscible if they do not mix (water + gasoline).

  8. Thermodynamics of Solution Formation Consider the following general process for the formation of a solution by dissolving a liquid or solid solute in a liquid solvent. M(s)  M(soln) There are two factors that favor formation of a solution: 1) An increase in the randomness of the system. 2) A decrease in the energy of the system. For now, we will take energy (E) and enthalpy (H) as meaning the same thing, as they are usually close in value.

  9. Entropy of Solution (Ssoln) 1) An increase in the randomness or disorder of the system. In thermodynamics randomness is measured in terms of entropy. For solutions the change in entropy is called Ssoln. Consider the dissolution of a solid in a liquid. We would expect that the solution is less ordered than the starting pure solid and pure liquid. Therefore, we expect Ssoln > 0 (surprisingly, there are a few cases where Ssoln < 0, but they are extremely rare).

  10. Enthalpy of Solution (Hsoln) 2) A decrease in the energy of the system. If mixing lowers the energy of a system then it is more likely to occur. Since solutions are usually prepared under conditions of constant pressure, we look at the enthalpy of solution, Hsoln, defined as the change in enthalpy for the process M(s)  M(soln) Hsoln To decide whether Hsoln will be positive, negative, or approximately zero we need to consider three type of attractive forces. solvent-solvent - Attractive forces acting between solvent molecules. solute-solute - Attractive forces acting between solute molecules. solvent-solute - Attractive forces acting between a solvent molecule and a solute molecule.

  11. Steps In Solution Formation We can think solution formation as a three step process: 1) Separate the solute particles Hsolute > 0 2) Separate the solvent particles Hsolvent > 0 3) Mix solute and solvent particles Hmixing < 0 ____________________________________________ Formation of solution Hsolution From Hess’ law Hsolution = Hsolute + Hsolvent + Hmixing

  12. Examples of Solvent-Solute Attractive Forces

  13. Hsoln For Various Solvent + Solute Combinations solvent polar polar nonpolar nonpolar solute polar nonpolar polar nonpolar solvent-solvent attraction strong strong weak weak solute-solute attraction strong weak strong weak solvent-solute attraction strong weak weak weak enthalpy change (Hsoln) ~ 0 positive positive ~ 0

  14. Solution Formation Hsoln Ssoln Will a solution form? negative positive Yes. ~ zero positive Yes. positive positive Maybe. A general consequence of the above is the general rule in mixing liquids that “like dissolves like”. Two polar liquids or two nonpolar liquids will usually mix (because Hsoln  0), but a polar and a nonpolar liquid will not usually mix (because Hsoln >> 0).

  15. Solutions of Ionic Compounds in Water For solutions of ionic compounds in water we may use Hess’ law to understand the factors involved in the change in energy for solution formation. The change in energy for solution formation, Hsoln, is Hsoln = |Hlattice| - |Hhydration|

  16. Example - KF(s) in water Since |Hlattice|  |Hhydration|, then Hsoln  0. (Actually endo-thermic by ~ 2 kJ/mol).

  17. Classification of Solutions Solutions can be classified according to the relative amount of dissolved solute. Unsaturated solution - Contains less than the maximum equilibrium amount of dissolved solute. Saturated solution - Contains the max-imum equilibrium amount of dissolved solute. Supersaturated solution - Contains more than the maximum equilibrium amount of dissolved solute. Note that a supersaturated solution is unstable. It will spontaneously form a preci-pitate until it becomes a saturated solution.

  18. Effect of Temperature On Solubility Note that the solubility of most solids in water or other liquids increases with increasing temperature.

  19. Recrystallization The increase in the solubility of a solid in a liquid solvent is the basis of a purification technique called recrystallization. The solid is dissolved in a minimum amount of hot solvent. When the solvent cools, the solute precipitates to form pure crystals.

  20. Gas Solubility Gases can also dissolve in liquids. Gas solubility is described by Henry’s law [A] = kH pA [A] = concentration of dissolved A (in mol/L) pA = partial pressure of A above the solution kH = Henry’s law constant The value for the Henry’s law constant depends on the solvent, the gas, and temperature. For most gases solubil-ity is low (exceptions are gases that can chemically interact with the solvent) and solubility decreases as temperature in-creases.

  21. Example: The Henry’s law constant for water is kH = 2.02 x 10-3 mol/L.atm at T = 20 C. What is the concentration of dissolved oxygen in water in equilibrium with the Earth’s atmosphere (pO2 = 0.21 atm)?

  22. Example: The Henry’s law constant for water is kH = 2.02 x 10-3 mol/L.atm at T = 20 C. What is the concentration of dissolved oxygen in water in equilibrium with the Earth’s atmosphere (pO2 = 0.21 atm)? [O2] = kH pO2 = (2.02 x 10-3 mol/L.atm) (0.21 atm) = 4.2 x 10-4 mol/L (or 0.0135 g/L) While small, this is enough dissolved oxygen to support fish life.

  23. Concentration Units - Molarity There are several common units for solutions. Molarity (M) = moles solute liters solution Molarity is the most common concentration unit used for solutions because it is easy to prepare solutions with know values of molarity. Note that molarity depends on temperature, because the volume occupied by a solution changes when temperature changes.

  24. Concentration Units - Molality and Mole Fraction Solution concentrations can also be given in terms of molality and mole fraction. Molality (m) = moles solute kg solvent Mole fraction (Xi) = ni = moles of ith substance ntotal total number of moles Molality and mole fraction are used in some specialized applications, such as in equations for some of the colligative properties of solutions and in Dalton’s law of partial pressure. Note that molality and mole fraction are temperature independent.

  25. Concentration Units - Percent By Mass, Parts By Mass or Volume, and Related Units We can also give solution concentrations in terms of the masses of substances present in the solution. Percent by mass (mass %) = mass of solute x 100 % mass of solution Parts per million by mass (ppm) = mass of solute x 106 mass of solution Parts per million by volume (ppm) = volume of solute x 106 volume of solution By analogy with ppm (parts per million) we can also define ppb (parts per billion; multiply by 109) or parts per trillion (ppt; multiply by 1012). We always should indicate if parts per… is by mass or by volume. We can also define percent by volume, or mole percent (percent by moles).

  26. Solution Calculations There are two general types of problems you should be able to do in connection with solutions. 1) Given information on how a solution is prepared, you should be able to find the solute concentration (molarity, molality, mole fraction, mass percent, and so forth). 2) Conversion from one concentration unit to a different concentration unit. Example: A solution is prepared by mixing 40.00 g ethylene glycol (C2H6O2, M = 62.07 g/mol) and 60.00 g water (H2O, M = 18.02 g/mol). The density of the solution is D = 1.0514 g/mL. What are the molarity, molality, mole fraction, and percent by mass ethylene glycol in the solution?

  27. Example: A solution is prepared by mixing 40.00 g ethylene glycol (C2H6O2, MW = 62.07 g/mol) and 60.00 g water (H2O. MW = 18.02 g/mol). The density of the solution is D = 1.0514 g/mL. What are the molarity, molality, mole fraction, and percent by mass ethylene glycol in the solution? nE = 40.00 g 1 mol = 0.6444 mol E 62.07 g nW = 60.00 g 1 mol = 3.3296 mol W 18.02 g XE = nE = (0.6444 mol) = 0.1622 nE + nW (0.6444 + 3.3296)mol mE = mol E = 0.6444 mol = 10.74 mol/kg kg H2O 0.06000 kg

  28. The total mass of solution is m = 40.00 g + 60.00 g = 100.00 g. Since the density of the solution is D = 1.0514 g/mL, the volume of the solution is V = 100.00 g 1 mL = 95.11 mL 1.0514 g So ME = 0.6444 mol = 6.775 mol/L 0.09511 L Finally, the mass percent ethylene glycol is mass % = mE. 100 % = 40.00 g . 100 % = 40.00 % mE + mW (40.00 + 60.00)g

  29. Volatile Liquids and Vapor Pressure A volatile liquid is a liquid that has a significant vapor pressure. A volatile liquid will, if left exposed in the atmosphere, evaporate. Liquids that do not have a significant vapor pressure are called nonvolatile. Vapor pressure is defined as the equilibrium pressure of gas above a sample of pure liquid. The value for the vapor pressure depends on temperature, and increases when temperature increases.

  30. Solutions of Volatile Liquids - Raoult’s Law Raoult’s law is a relationship that applies to some solutions of volatile liquids. Consider a solution consisting of two volatile liquids A and B. Raoult’s law predicts that the partial pressure of vapor from each liquid above the solution is given by the relationships pA = XA pA pB = XB pB where pA, pB are the partial pressures of A and B above the solution XA, XB are the mole fraction of A and B in the solution pA, pB are the vapor pressures of pure A and pure B

  31. Example: At 20C the vapor pressure of pure benzene is 24. torr and the vapor pressure of pure toluene is 76. torr. Consider a solution with XB = 0.40. Assuming that benzene and toluene form an ideal solution, find the partial pressure of benzene and toluene and the total pressure above the solution.

  32. Example: At 20C the vapor pressure of pure benzene (B) is 24. torr and the vapor pressure of pure toluene (T) is 76. torr. Consider a solution with XB = 0.40. Assuming that benzene and toluene form an ideal solution, find the partial pressure of benzene and toluene and the total pressure above the solution. From Raoult’s law, pA = XApA XB + XT = 1, so XT = 1 – XB = 1 – 0.40 = 0.60 So pB = (0.40)(24. torr) = 9.6 torr pT = (0.60)(76. torr) = 45.6 torr ptotal = psoln = pB + pT = 9.6 torr + 45.6 torr = 55.2 torr

  33. Ideal Solution An ideal solution is a solution whose components obey Raoult’s law. The solvent in a solution will always obey Raoult’s law in the limit XA 1. In a mixture of two liquids A and B there are three types of inter-molecular forces A --- A molecules of A with A B --- B molecules of B with B A --- B molecules of A with B Solutions will generally be close to ideal if they are composed of similar molecules with similar intermolecular attractive forces (so usually molecules that are both nonpolar).

  34. Ideal and Nonideal Solution For an ideal solution a plot of pressure vs mole fraction will be linear. If such a plot is not linear, then we have a nonideal solution. A = C6H6; B = C6H5CH3 A = CH3COCH3; B = CS2 Note that since XA + XB = 1, XB = 1 - XA.

  35. Solutions With a Nonvolative Ionic Solute For a solution containing a nonvolatile ionic solute (or any nonvolatile solute that can dissociate into smaller particles) there is one additional complicating factor. As before, there will only be solvent molecules above the solution. If the solvent obeys Raoult’s law (true if dilute solution), then pA = psoln = XA pA However, we must take into account the ionization or dissociation of the solute in calculating the mole fraction of solvent. XA = moles of solvent particles total moles of particles XB = moles of solute particles total moles of particles

  36. Finding the Moles of Particles (Examples) Consider the following three substances: C6H12O6 (sugar) NaCl (sodium chloride) K2SO4 (potassium sulfate) When the above substances are dissolved in water how many moles of particles will form per mole of dissolved substance?

  37. Finding the Moles of Particles (Examples) C6H12O6 (sugar) C6H12O6(s)  C6H12O6(aq) mol particles = 1.00 mole sugar 1 mole particle = 1.00 mol particles 1 mol sugar NaCl (sodium chloride) NaCl(s)  Na+(aq) + Cl-(aq) mol particles = 1.00 mol NaCl 2 mol particles = 2.00 mol particles 1 mol NaCl K2SO4 (potassium sulfate) K2SO4(s)  2 K+(aq) + SO42-(aq) mol particles = 1.00 mol K2SO43 mol particles = 3.00 mol particles 1 mol K2SO4

  38. Van’t Hoff Factor (i) The van’t Hoff factor, i, is defined as i = moles of particles moles of solute So moles of particles = i (moles of solute) For the above examples: C6H12O6(s)  C6H12O6(aq) i = 1 NaCl(s)  Na+(aq) + Cl-(aq) i = 2 K2SO4(s)  2 K+(aq) + SO42-(aq) i = 3

  39. Experimental Values for theVan’t Hoff Factor The value for the van't Hoff factor can be found experimentally. For dilute solutions the experimental value for the van’t Hoff factor is close to the value predicted from theory. However, for more concentrated solutions the experimental value is lower than that expected. For example, for NaCl NaCl(s)  Na+(aq) + Cl-(aq) Molality NaCl itheory iexperimental 0.0001 2.0 1.99 0.001 2.0 1.97 0.01 2.0 1.94 0.1 2.0 1.87 Unless otherwise stated we will use the theoretical value for i.

  40. Example The vapor pressure of pure water is pW = 149.4 torr at T = 60. C. Find the partial pressure of water above each of the following solutions at T = 60. C. You may assume the solvent (water) behaves ideally. a) 1.00 mol sugar (C6H12O6) and 1000. g water. b) 1.00 mol calcium nitrate (Ca(NO3)2) and 1000. g water.

  41. The vapor pressure of pure water is pW = 149.4 torr at T = 60. C. Find the partial pressure of water above each of the following solutions at T = 60. C. a) 1.00 mol sugar (C6H12O6) and 1000. g water. moles water = 1000. g water 1 mol water = 55.56 mol 18.0 g water moles solute particles (sugar) = 1.00 mol (i = 1) XW = 55.56 mol = 0.982 (55.56 + 1.00)mol pW = (0.982) (149.4 torr) = 146.7 torr

  42. The vapor pressure of pure water is pW = 149.4 torr at T = 60. C. Find the partial pressure of water above each of the following solutions at T = 60. C. b) 1.00 mol calcium nitrate (Ca(NO3)2) and 1000. g water. moles water = 1000. g water 1 mol water = 55.56 mol 18.0 g water Ca(NO3)2(s)  Ca2+(aq) + 2 NO3-(aq) (i = 3) moles solute particles (Ca(NO3)2) = 1.00 mol Ca(NO3)23 mol particle = 3 mol particles 1 mol Ca(NO3)2 XW = 55.56 mol = 0.949 (55.56 + 3.00)mol pW = (0.949) (149.4 torr) = 141.8 torr

  43. Colligative Property A colligative property is any property of a solution consisting of a volatile solvent and a nonvolatile solute that depends at most on two things: 1) The physical properties of the solvent 2) The number or concentration of solute particles There are four colligative properties - vapor pressure lowering - boiling point elevation - freezing point depression - osmotic pressure The equations given for colligative properties apply to ideal conditions and are therefore expected to be correct in the limit Xsolvent 1.

  44. Vapor Pressure Lowering Consider the situation below. The left side contains pure solvent A, while the right side contains a solution of A and a nonvolatile solute. pleft = pA pright = XApA We may define the vapor pressure lowering p = pleft - pright, and so p = pA - XApA = (1 - XA) pA But XA + XB = 1, so 1 - XA = XB, and so p = XBpA , the final expression for vapor pressure lowering.

  45. Measurement of Vapor Pressure Lowering The apparatus shown below can be used to accurately measure the vapor pressure of a solution relative to that of a pure liquid. Note that the working fluid in the manometer is often chosen to be a non-volatile liquid of lower density than mercury, such as sulfuric acid or mineral oil.

  46. Boiling Point Elevation and Freezing Point Depression Consider the situation below. The left side contains pure solvent A, while the right side contains a solution of A and a nonvolatile solute. The boiling point of the solution will be higher than the boiling point of the pure liquid, and the freezing point of the solution will be lower than the freezing point of the pure liquid. Tb = Tb - Tb° = Kb mB  boiling point elevation Tf = Tf° - Tf = Kf mB  freezing point depression Kb, Kf = boiling point elevation and freezing point depression constants Tb, Tf = boiling and freezing point for the solution Tb°, Tf° = boiling and freez-ing point for the pure liquid mB = molality of solute par-ticles

  47. Example A solution is prepared by dissolving 1.00 mol sodium chloride (NaCl) in 1000. g of water. What are the normal freezing point and normal boiling point for this solution. Tb = 100.00 C Tf = 0.00 C Kb = 0.512 C/m Kf = 1.86 C/m

  48. A solution is prepared by dissolving 1.00 mol sodium chloride (NaCl) in 1000. g of water. What are the normal freezing point and normal boiling point for this solution. NaCl(s)  Na+(aq) + Cl-(aq) (i = 2) Tb = 100.00 C Tf = 0.00 C Kb = 0.512 C/m Kf = 1.86 C/m molality NaCl = 1.00 mol NaCl 2 mol particles = 2.00 mol particles 1.000 kg water 1 mol NaCl kg water So Tb = (0.512 C/m) (2.00 m) = 1.02 C Tf = (1.86 C/m) (2.00 m) = 3.72 C Tb = 100.00 C + 1.02 C = 101.02 C Tf = 0.00 C - 3.72 C = - 3.72 C

  49. Phase Diagram For Solutions We can picture the differences between the behavior of a pure liquid and a solution of the liquid and a nonvolative solute in terms of the corresponding phase diagram. In a sense all that has changed for the solution is the location of the triple point, which affects the locations of the phase boundaries, as indicated in the phase diagram at right.

  50. Osmotic Pressure Consider the apparatus below. The right chamber contains a pure liquid, while the left chamber contains a solution of the liquid and a non-volatile solute. The two chambers are separated by a semipermeable membrane that allows solvent molecules to pass but prevents the passage of solute molecules. The osmotic pressure, , is defined as the additional pressure that must be applied to the solution to keep the level of liquid in both cham-bers the same.

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