1. Chapter 14 Solutions

2. What are solutions? A Solution is…

3. What are solutions? Solution: a uniform mixture that contains 2 or more substances (solids, liquids, or gases); also called a homogeneous mixture

4. What are solutions? Solution: a uniform mixture that contains 2 or more substances (solids, liquids, or gases); also called a homogeneous mixture Solute: Solvent:

5. What are solutions? Solution: a uniform mixture that contains 2 or more substances (solids, liquids, or gases); also called a homogeneous mixture Solute: substance that gets dissolved Solvent: dissolving medium Example: salt water - salt is solute - water is solvent

6. What are solutions? What are some solutions that you use in everyday life?

7. Characteristics of Solutions Soluble vs Insoluble

8. Characteristics of Solutions Soluble vs Insoluble dissolves in solventdoes NOT dissolve in solvent *** Like Dissolves Like ***

9. Soluble OR Insoluble 1)Sand in water 2)Oil in vinegar 3)Vinegar in water 4)Melted butter in oil 5)Carbon dioxide (g) in water 6)Chocolate syrup in milk

10. Solvation in aqueous solutions Why are some substances soluble in one solvent but not another?

11. Solvation in aqueous solutions Why are some substances soluble in one solvent but not another? * It has to do with attractive forces between particles: 1. forces between pure solute particles 2. forces between pure solvent particles 3. forces between solute and solvent particles

12. Attractive Forces & Solvation To dissolve a solute in a solvent: If the attractive forces between solute & solvent are greater than forces holding solute together → the solvent will pull apart the solute, and dissolve it (“solvation”) If the attractive forces between solute & solvent are smaller than attractive forces holding solute together → the solvent will NOT pull apart the solute.

13. Aq. Solutions of ionic compounds Dissolving a solute (NaCl) in water: Charged ends of water attract + and - ions in crystal. Attraction between dipoles (in water) & ions is greater than attraction of ions within the crystal So, crystal dissolves in water

14. Aq. Solutions of ionic compounds Dissolving CaSO 4 in water: Attractive forces between Ca 2+ and SO 4 2- are so strong compared to attractive forces between water & ions. Calcium sulfate does NOT dissolve in water “Hard Water” stains

15. Aq. solutions of molecular compounds Dissolving sugar in water: O-H bond in sugar becomes site for hydrogen-bonding with water molecules Strong attractive forces between water & sugar are stronger than those in sugar alone; sugar dissolves.

16. Factors that affect rate of solvation Solvation can only occur when solute and solvent are in contact with each other. Three common ways to increase collisions between solute & solvent: 1.Agitating mixture 2.Increasing surface area of solute (smaller particles) 3.Increasing temperature of solvent

17. Heat of Solution Energy is released or absorbed during the solvation process. Energy is required to overcome attractive forces within solute and within solvent…endothermic. Energy is released when solute and solvent particles mix (particles attract each other)…exothermic. Overall energy change during the formation of a solution is call the heat of solution.

18. Solubility Every solute has a characteristic solubility in a given amount of solvent at a specified temperature and pressure. High solubility = more solute can be dissolved Low solubility = less solute can be dissolved

19. Factors that affect Solubility Temperature: most solutes are more soluble at higher temperatures. Pressure (gaseous solutes and solutions) Nature of solute and solvent (high or low attractive forces between solute and solvent; compared to attractive forces with solute alone)

20. Solubility & Temperature Substance0 °C20 °C60 °C100 °C Sucrose179.2203.9287.3487.2 NaCl35.735.937.139.2 Ca(OH) 2 0.1890.1730.1210.076 Solubility of some solutes in water (g/100 g H 2 O) *Most soluble? *Least soluble? *Solubilities affected by temperature?

21. Saturated, unsaturated, supersaturated 1.Saturated solution 2.Unsaturated solution 3.Supersaturated solution A.Contains more solute that a saturated solution at the same temperature B.Contains less dissolved solute for a given T & P. C.Contains the maximum amount of dissolved solute for a given T & P.

22. Solubility & Pressure Solubility of gases increases as external pressure (pressure above solution) increases. Carbonated beverages contain carbon dioxide dissolved in aqueous solution; results in the “fizz”. If container is left open, the CO 2 will escape, resulting in “flat” soda.

23. Henry’s Law Henry’s Law: solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid, at a given T. S 1 = S 2 P 1 P 2 S 1 : solubility of gas at pressure P 1 S 2 : solubility of gas at new pressure P 2

24. Practice Problem If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25 °C, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature?

25. Practice Problem If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25 °C, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature? Hint: Solubility has units of g solute/L solvent

26. Practice Problem If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25 °C, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature? S 1 = 0.85 g/1.0 L = 0.85 g/L

27. Practice Problem If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25 °C, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature? S 1 = 0.85 g/LS 2 = ? P 1 = 4.0 atmP 2 = 1.0 atm

28. If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25 °C, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature? Known: S 1 = S 2 P 1 P 2 Unknown: Practice Problem

29. If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25 °C, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature? S 2 = (S 1 )(P 2 ) (P 1 ) Practice Problem

30. If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 25 °C, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature? S 2 = (S 1 )(P 2 ) (P 1 ) S 2 = (0.85 g/L)(1.0 atm) = 0.21 g/L (4.0 atm) Practice Problem

31. 15.2 Solution Concentration Concentration: a measure of how much solute is dissolved in a specific amount of solvent or solution. Qualitatively: think in terms of “concentrated” or “dilute”. Which one is dilute? Concentrated?

32. Expressing Solution Concentration As a chemist, we want quantitative descriptions! (Want an actual number…How much?) Concentration can be represented by: 1.percent by mass or volume 2.Molarity 3.Molality 4.Mole fraction

33. Concentration Ratios Concentration Description Ratio Percent by Mass Percent by Volume **Molarity Molality Mole Fraction

34. Percent by Mass & Concentration

35. Example: Percent by mass What is the percent by mass of NaCl, if 3.6 g NaCl is added to 100.0 g water?

36. Example: Percent by mass

40. What is the percent by mass of NaHCO 3 in a solution containing 20 g NaHCO 3 dissolved in 600 mL H 2 O?

41. Example: Percent by mass What is the percent by mass of NaHCO 3 in a solution containing 20 g NaHCO 3 dissolved in 600 mL H 2 O?

42. Example: Percent by mass

43. What is the percent by mass of NaHCO 3 in a solution containing 20 g NaHCO 3 dissolved in 600 mL H 2 O? ** We need to convert 600 mL water to grams water

44. Example: Percent by mass What is the percent by mass of NaHCO 3 in a solution containing 20 g NaHCO 3 dissolved in 600 mL H 2 O? ** We need to convert 600 mL water to grams water ** 1 mL H 2 O = 1 g H 2 O ** So… 600 mL = 600 g

45. Example: Percent by mass

48. Percent by Volume & Concentration

49. Example: Percent by Volume What is the percent by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL water?

50. Example: Percent by Volume What is the percent by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL water?

51. Example: Percent by Volume What is the percent by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL water?

52. Example: Percent by Volume What is the percent by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL water?

53. Example: Percent by Volume What is the percent by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL water?

54. Molarity

55. Examples of Molarity For example: 1)0.1M solution: contains 0.1 mol solute in 1 L solution 2)3M solution: contains 3 mol solute in 1 L solution 3)0.5M solution: contains 0.5 mol solute in 1 L solution

56. Practice Problem: Molarity Calculate the molarity of 100.0 mL of an aqueous solution containing 0.85 mol potassium chloride.

57. Practice Problem: Molarity

59. Calculate the molarity of 100.0 mL of an aqueous solution containing 0.85 mol potassium chloride. 100.0 mL x 1 L = 0.1 L 1000 mL

60. Practice Problem: Molarity

61. What is the molarity of a solution that contains 5.10-g glucose in 100.5 mL solution. The molar mass of glucose is 180.16 g/mol.

62. Practice Problem: Molarity What is the molarity of a solution that contains 5.10-g glucose in 100.5 mL solution. The molar mass of glucose is 180.16 g/mol. Known:mass solute = 5.10 g molar mass = 180.16 g/mol volume = 100.5 mL Unknown:solution concentration = ? M

63. Practice Problem: Molarity

64. What is the molarity of a solution that contains 5.10-g glucose in 100.5 mL solution. The molar mass of glucose is 180.16 g/mol. Convert mass to moles solute:

65. Practice Problem: Molarity What is the molarity of a solution that contains 5.10-g glucose in 100.5 mL solution. The molar mass of glucose is 180.16 g/mol. Convert mL to L:

66. Practice Problem: Molarity

68. What is the molar concentration of a solution containing 1.0 mol of KCl in 0.750 L solution?

69. Making Molar Solutions How do you prepare 1 Liter of a 1.50 M aqueous solution of NaCl for an experiment?

70. Making Molar Solutions How do you prepare 1 Liter of a 1.50 M aqueous solution of NaCl for an experiment? What this is really asking is… What is the mass of NaCl needed to prepare 1 Liter solution of 1.50 M NaCl?

71. Making Molar Solutions How do you prepare 1 Liter of a 1.50 M aqueous solution of NaCl for an experiment? Step 1: determine the number of moles in 1 L of a 1.50 M solution of NaCl.

72. Making Molar Solutions

74. How do you prepare 1 Liter of a 1.50 M aqueous solution of NaCl for an experiment? Step 2: convert mol NaCl to mass NaCl

75. Making Molar Solutions

76. How do you prepare 1 Liter of a 1.50 M aqueous solution of NaCl for an experiment? Step 3: Make solution Add 87.66 g NaCl into 1-L container first, then add enough water to make final volume equal to 1 liter. Remember: Volume solution = volume solute + volume solvent

77. Example: Making Molar Solutions How many grams of CaCl 2 should be dissolved in 1.75 L of a 0.10 M solution of CaCl 2 ?

78. Review: Molarity 1)Determine the molarity of 1.60 L of a solution containing 1.55 g of dissolved KBr. 2)What volume of 0.125 M NiCl 2 contains 3.25 mol? 3)How many moles of NaOH are needed to make 100 mL of a 2 M solution of NaOH?

79. Diluting Solutions In lab, chemists start with concentrated solutions of standard molarities, called stock solutions. Example: concentrated HCL is 12 M. We can dilute solutions to any concentration we want.

80. Dilution Equation The total number of moles of solute does NOT change when we dilute…so we can use the following relationship: M 1 V 1 = M 2 V 2 M 1 = molarity of stock solution V 1 = volume of stock solution M 2 = molarity of dilute solution V 2 = volume of dilute solution

81. Practice Problem What volume of 2.00 M calcium chloride stock is needed to make 0.50 L of 0.300 M calcium chloride?

82. Practice Problem What volume of 2.00 M calcium chloride stock is needed to make 0.50 L of 0.300 M calcium chloride? M 1 V 1 = M 2 V 2

83. Practice Problem What volume of 2.00 M calcium chloride stock is needed to make 0.50 L of 0.300 M calcium chloride? M 1 V 1 = M 2 V 2 Known: Unknown:

84. Practice Problem What volume of 2.00 M calcium chloride stock is needed to make 0.50 L of 0.300 M calcium chloride? M 1 V 1 = M 2 V 2 Known:M 1 = 2.00 M M 2 = 0.300 M V 2 = 0.50 L Unknown:V 1 = ?

85. Practice Problem

86. If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of the dilute solution?

87. Practice Problem If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of the dilute solution? M 1 V 1 = M 2 V 2

88. Practice Problem If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of the dilute solution? M 1 V 1 = M 2 V 2 Known: Unknown:

89. Practice Problem If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of the dilute solution? M 1 V 1 = M 2 V 2 Known:M 1 = 3.5 M V 1 = 20.0 mL V 2 = 100.0 mL Unknown:M 2 = ?

90. Practice Problem

91. Molality

92. Practice Problem: Molality In lab, a student adds 4.5 g of NaCl to 100.0 g of water. Calculate the molality of the solution.

93. Practice Problem: Molality

96. To calculate molality: 1) Convert mass solute to moles solute

97. Practice Problem: Molality To calculate molality: 2) Convert grams solvent to kilogram solvent

98. Practice Problem: Molality

99. Mole Fraction

100. Practice Problem: Mole Fraction What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by mass?

101. Practice Problem: Mole Fraction What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by mass? Step 1: Determine mass of solute and solvent

102. Practice Problem: Mole Fraction What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by mass? Step 1: Determine mass of solute and solvent. Assume a 100 gram sample: 22.8 % NaOH = 22.8 g NaOH 77.2% H 2 O = 77.2 g H 2 O

103. Practice Problem: Mole Fraction What is the mole fraction of NaOH in an aqueous solution that contains 22.8% NaOH by mass? Step 2: Convert mass to moles, using Molar Mass. 22.8 g NaOH x 77.2 g H 2 O x

104. Practice Problem: Mole Fraction

105. Colligative Properties Colligative Properties: -Means “depending on the collection” -physical properties of solutions are affected by number of particles but not the identity of dissolved solute particles. -Includes: -Vapor pressure lowering -Boiling point elevation -Freezing point depression -Osmostic pressure

106. Electrolytes/Nonelectrolytes Electrolytes: ionic compounds that produce many ions when dissolved in water and conduct electricity; ex. NaCl Weak electrolytes: ionic compounds that produce a few ions when dissolved in water Nonelectrolytes: molecular compounds that dissolve in water but do not produce ions in water and don’t conduct electricity ex. sugar

107. Vapor Pressure Lowering Vapor pressure: pressure exerted in a closed container by liquid particles that have escaped liquid’s surface. Equilibrium in a closed container: rate of escaping liquid = rate of entering liquid

108. Vapor Pressure Lowering The addition of nonvolatile solutes to solutions lowers the vapor pressure.

109. Boiling Point Elevation Addition of nonvolatile solute lowers vapor pressure, solution must be heated to higher temperature to boil. Boiling Point (b.p.) Elevation: temperature difference between a solution’s boiling point and a pure solvent’s boiling point. A greater number of solute particles results in greater b.p. elevation.

110. Boiling Point Elevation For nonelectrolytes, the b.p. elevation (ΔT b ) is proportional to the solution’s molality: ΔT b = K b m ΔT b = value of the b.p. elevation K b = molal b.p. elevation constant; difference in b.p. between 1 m nonvolatile and pure solvent m = molality

111. Boiling Point Elevation SolventBoiling Point (°C)Kb (°C/m) Water100.00.512 Benzene80.12.53 CCl 4 76.75.03 Ethanol78.51.22 1m nonelectrolyte solution in Water: ΔT b = K b m = (0.512°C/m)(1m) = 0.512 °C New B.P. = 100.0 °C + 0.512 °C = 100.512°C

112. Freezing Point Depression At lower temps, solvent particles don’t have enough kinetic energy to overcome intermolecular forces. Particles form more organized structure in solid state. Solute particles interfere with attractive forces of solvent particles and prevents solvent from entering solid state at its normal freezing point. *Freezing point of solution is always lower than that of pure solvent.

113. Freezing Point Depression Freezing Point (f.p.) Depression: difference in temp between f.p. and f.p. of pure solvent. - ΔT f is directly proportional to the solution’s molality: ΔT f = K f m ΔT f = value of the b.p. elevation K f = molal f.p. depression constant for specific solvent m = molality

114. Phase Diagram The addition of a solute changes the freezing point and the boiling point of the solvent.

115. Practice Problem What is the boiling point and freezing point of a 0.029m aqueous solution of sodium chloride? K b = 0.512 °C/m, K f = 1.86 °C/m

116. Practice Problem What is the boiling point and freezing point of a 0.029m aqueous solution of sodium chloride? K b = 0.512 °C/m, K f = 1.86 °C/m Known: Unknown:

117. Practice Problem What is the boiling point and freezing point of a 0.029m aqueous solution of sodium chloride? K b = 0.512 °C/m, K f = 1.86 °C/m Known: solute = NaCl molality (m) of NaCl solution= 0.029 m Unknown: boiling point = ? freezing point = ?

118. Practice Problem What is the boiling point and freezing point of a 0.029m aqueous solution of sodium chloride? K b = 0.512 °C/m, K f = 1.86 °C/m Step 1: Calculate particle molarity. Even though NaCl Molality = 0.029 m, the actual Particle Molality is equal to … Particle Molality = 2 x 0.029 m = 0.058 m

119. Practice Problem What is the boiling point and freezing point of a 0.029m aqueous solution of sodium chloride? K b = 0.512 °C/m, K f = 1.86 °C/m Step 2: Calculate boiling point elevation and freezing point depression. ΔT b = K b m = (0.512 °C/m)(0.058m) = 0.030 °C ΔT f = K f m= (1.86 °C/m)(0.058m) = 0.11 °C

120. Practice Problem What is the boiling point and freezing point of a 0.029m aqueous solution of sodium chloride? K b = 0.512 °C/m, K f = 1.86 °C/m Step 3: Calculate the new boiling point and new freezing point after the addition of solute. T b = 100.0 °C + 0.030 °C = 100.030 °C T f = 0.0 °C - 0.11 °C = -0.11 °C

121. Osmosis Osmosis: diffusion of solvent particles across a semipermeable membrane from an area of higher solvent concentration to an area of lower solvent concentration.

122. Osmosis Osmotic Pressure: amount of additional pressure caused by water molecules that moved into the solution.