This document discusses properties of solutions, including: - Solutions are homogeneous mixtures of two or more substances, with the solute dispersed uniformly throughout the solvent. - A solute dissolves as the solvent molecules interact with and surround the solute particles or ions, changing the enthalpy of the system. - The entropy of the system typically increases during dissolution, making dissolution spontaneous even for endothermic processes. - Concentrations of solutions can be expressed using various units including molarity, molality, mass percent, and parts per million or billion.

1. Chapter 13 Properties of Solutions Adapted by SA Green from: John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc. Chemistry, The Central Science , 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Solutions Solutions are homogeneous mixtures of two or more pure substances. In a solution, the solute is dispersed uniformly throughout the solvent .

3. Solutions How does a solid dissolve into a liquid? What ‘drives’ the dissolution process? What are the energetics of dissolution?

4. How Does a Solution Form? Solvent molecules attracted to surface ions. Each ion is surrounded by solvent molecules. Enthalpy (  H) changes with each interaction broken or formed. Ionic solid dissolving in water

5. How Does a Solution Form? Solvent molecules attracted to surface ions. Each ion is surrounded by solvent molecules. Enthalpy (  H) changes with each interaction broken or formed.

6. How Does a Solution Form The ions are solvated (surrounded by solvent). If the solvent is water, the ions are hydrated . The intermolecular force here is ion-dipole.

7. Energy Changes in Solution To determine the enthalpy change, we divide the process into 3 steps. Separation of solute particles. Separation of solvent particles to make ‘holes’. Formation of new interactions between solute and solvent.

8. Enthalpy Changes in Solution The enthalpy change of the overall process depends on  H for each of these steps. Start End End Start

9. Enthalpy changes during dissolution The enthalpy of solution,  H soln , can be either positive or negative.  H soln =  H 1 +  H 2 +  H 3  H soln (MgSO 4 )= -91.2 kJ/mol --> exothermic  H soln (NH 4 NO 3 )= 26.4 kJ/mol --> endothermic

10. Why do endothermic processes sometimes occur spontaneously? Some processes, like the dissolution of NH 4 NO 3 in water, are spontaneous at room temperature even though heat is absorbed, not released.

11. Enthalpy Is Only Part of the Picture Entropy is a measure of: Dispersal of energy in the system. Number of microstates (arrangements) in the system. b. has greater entropy,  is the favored state (more on this in chap 19)

12. Entropy changes during dissolution Each step also involves a change in entropy. Separation of solute particles. Separation of solvent particles to make ‘holes’. Formation of new interactions between solute and solvent.

13. SAMPLE EXERCISE 13.1 Assessing Entropy Change In the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Does the entropy of the system increase or decrease?

14. Dissolution vs reaction Dissolution is a physical change — you can get back the original solute by evaporating the solvent. If you can’t, the substance didn’t dissolve, it reacted. Ni(s) + HCl(aq) NiCl 2 (aq) + H 2 (g) NiCl 2 (s) dry

15. Degree of saturation Saturated solution Solvent holds as much solute as is possible at that temperature. Undissolved solid remains in flask. Dissolved solute is in dynamic equilibrium with solid solute particles.

16. Degree of saturation Unsaturated Solution Less than the maximum amount of solute for that temperature is dissolved in the solvent. No solid remains in flask.

17. Degree of saturation Supersaturated Solvent holds more solute than is normally possible at that temperature. These solutions are unstable; crystallization can often be stimulated by adding a “seed crystal” or scratching the side of the flask.

18. Degree of saturation Unsaturated, Saturated or Supersaturated?  How much solute can be dissolved in a solution? More on this in Chap 17 (solubility products, p 739)

19. Factors Affecting Solubility Chemists use the axiom “like dissolves like”: Polar substances tend to dissolve in polar solvents. Nonpolar substances tend to dissolve in nonpolar solvents.

20. Factors Affecting Solubility The stronger the intermolecular attractions between solute and solvent, the more likely the solute will dissolve. Example: ethanol in water Ethanol = CH 3 CH 2 OH Intermolecular forces = H-bonds; dipole-dipole; dispersion Ions in water also have ion-dipole forces.

21. Factors Affecting Solubility Glucose (which has hydrogen bonding) is very soluble in water. Cyclohexane (which only has dispersion forces) is not water-soluble.

22. Factors Affecting Solubility Vitamin A is soluble in nonpolar compounds (like fats). Vitamin C is soluble in water.

23. Which vitamin is water-soluble and which is fat-soluble?

24. Gases in Solution In general, the solubility of gases in water increases with increasing mass. Why? Larger molecules have stronger dispersion forces.

25. Gases in Solution The solubility of liquids and solids does not change appreciably with pressure. But, the solubility of a gas in a liquid is directly proportional to its pressure. Increasing pressure above solution forces more gas to dissolve.

26. Henry’s Law S g = kP g where S g is the solubility of the gas; k is the Henry’s law constant for that gas in that solvent; P g is the partial pressure of the gas above the liquid.

27. Temperature Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.

28. Temperature The opposite is true of gases. Higher temperature drives gases out of solution. Carbonated soft drinks are more “bubbly” if stored in the refrigerator. Warm lakes have less O 2 dissolved in them than cool lakes.

29. Chap 13: Ways of Expressing Concentrations of Solutions

30. Mass Percentage Mass % of A =  100 mass of A in solution total mass of solution

31. Parts per Million and Parts per Billion ppm =  10 6 Parts per Million (ppm) Parts per Billion (ppb) ppb =  10 9 mass of A in solution total mass of solution mass of A in solution total mass of solution

32. Mole Fraction ( X ) In some applications, one needs the mole fraction of solvent , not solute — make sure you find the quantity you need! moles of A total moles in solution X A =

33. Molarity ( M ) Because volume is temperature dependent, molarity can change with temperature. mol of solute L of solution M =

34. Molality ( m ) Because neither moles nor mass change with temperature, molality (unlike molarity) is not temperature dependent. mol of solute kg of solvent m =

35. Moles/Mass Mass/Mass Moles/Moles Moles/L

36. Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

37. PRACTICE EXERCISE (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? PRACTICE EXERCISE A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the molality and (b) the mole fraction of NaOCl in the solution. SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations (a) A solution is made by dissolving 13.5 g of glucose (C 6 H 12 O 6 ) in 0.100 kg of water. What is the mass percentage of solute in this solution? ( b) A 2.5-g sample of groundwater was found to contain 5.4   g of Zn 2+ What is the concentration of Zn 2+ in parts per million?

38. PRACTICE EXERCISE (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? Answers: (a) 2.91%, (b) 90.5 g of NaOCl Answers: (a) 0.505 m , (b) 9.00  10 –3 SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations (a) A solution is made by dissolving 13.5 g of glucose (C 6 H 12 O 6 ) in 0.100 kg of water. What is the mass percentage of solute in this solution? ( b) A 2.5-g sample of groundwater was found to contain 5.4   g of Zn 2+ What is the concentration of Zn 2+ in parts per million? Solution Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute. Plan: We can calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water). Solve: Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%. (b) Analyze: In this case we are given the number of micrograms of solute. Because 1   g is 1  10 –6 g, 5.4   g = 5.4  10 –6 g. Plan: We calculate the parts per million using Equation 13.6. Solve:

39. Colligative Properties Colligative properties depend only on the number of solute particles present, not on the identity of the solute particles. Among colligative properties are Vapor pressure lowering Boiling point elevation Melting point depression Osmotic pressure

40. Vapor Pressure As solute molecules are added to a solution, the solvent become less volatile (=decreased vapor pressure). Solute-solvent interactions contribute to this effect.

41. Vapor Pressure Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

42. Raoult’s Law P A = X A P  A where X A is the mole fraction of compound A P  A is the normal vapor pressure of A at that temperature NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent .

43. PRACTICE EXERCISE The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr (Appendix B).

44. Solution Analyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute. Plan: We can use Raoult’s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, X A , is the ratio of the number of moles of solvent (H 2 O) to total solution (moles C 3 H 8 O 3 + moles H 2 O). SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr (Appendix B). Solve: To calculate the mole fraction of water in the solution, we must determine the number of moles of C 3 H 8 O 3 and H 2 O:

45. PRACTICE EXERCISE The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: 0.290 SAMPLE EXERCISE 13.8 continued The vapor pressure of the solution has been lowered by 0.6 torr relative to that of pure water. We now use Raoult’s law to calculate the vapor pressure of water for the solution:

46. Boiling Point Elevation and Freezing Point Depression Solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

47. Boiling Point Elevation The change in boiling point is proportional to the molality of the solution:  T b = K b  m where K b is the molal boiling point elevation constant, a property of the solvent.  T b is added to the normal boiling point of the solvent.

48. Freezing Point Depression The change in freezing point can be found similarly:  T f = K f  m Here K f is the molal freezing point depression constant of the solvent.  T f is subtracted from the normal freezing point of the solvent.

49. Boiling Point Elevation and Freezing Point Depression In both equations,  T does not depend on what the solute is , but only on how many particles are dissolved.  T b = K b  m  T f = K f  m

50. Colligative Properties of Electrolytes Because these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) show greater changes than those of nonelectrolytes. e.g. NaCl dissociates to form 2 ion particles; its limiting van’t Hoff factor is 2.

51. Colligative Properties of Electrolytes However, a 1 M solution of NaCl does not show twice the change in freezing point that a 1 M solution of methanol does. It doesn’t act like there are really 2 particles.

52. van’t Hoff Factor One mole of NaCl in water does not really give rise to two moles of ions.

53. van’t Hoff Factor Some Na + and Cl − reassociate as hydrated ion pairs, so the true concentration of particles is somewhat less than two times the concentration of NaCl.

54. The van’t Hoff Factor Reassociation is more likely at higher concentration. Therefore, the number of particles present is concentration dependent.

55. The van’t Hoff Factor We modify the previous equations by multiplying by the van’t Hoff factor, i  T f = K f  m  i i = 1 for non-elecrtolytes

56. Osmosis Semipermeable membranes allow some particles to pass through while blocking others. In biological systems, most semipermeable membranes (such as cell walls) allow water to pass through, but block solutes.

57. Osmosis In osmosis, there is net movement of solvent from the area of higher solvent concentration ( lower solute concentration ) to the are of lower solvent concentration ( higher solute concentration ). Water tries to equalize the concentration on both sides until pressure is too high.

58. Osmotic Pressure The pressure required to stop osmosis, known as osmotic pressure ,  , is where M is the molarity of the solution If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic . n V  = ( ) RT = MRT

59. Osmosis in Blood Cells If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic . Water will flow out of the cell, and crenation results.

60. Osmosis in Cells If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic . Water will flow into the cell, and hemolysis results.

62. Molar Mass from Colligative Properties We can use the effects of a colligative property such as osmotic pressure to determine the molar mass of a compound.

63. Colloids: Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity.

64. Tyndall Effect Colloidal suspensions can scatter rays of light. This phenomenon is known as the Tyndall effect.

65. Colloids in Biological Systems Some molecules have a polar, hydrophilic ( water-loving ) end and a nonpolar, hydrophobic ( water-hating ) end.

66. Colloids in Biological Systems Sodium stearate is one example of such a molecule.

67. Colloids in Biological Systems These molecules can aid in the emulsification of fats and oils in aqueous solutions.

68. END Chap 13

69. Why Do Endothermic Processes Occur? Things do not tend to occur spontaneously (i.e., without outside intervention) unless the energy of the system is lowered.

70. Enthalpy Is Only Part of the Picture So even though enthalpy may increase, the overall energy of the system can still decrease if the system becomes more disordered.

71. Student, Beware! Just because a substance disappears when it comes in contact with a solvent, it doesn’t mean the substance dissolved.