1. Solutions

2. The Nature of Aqueous Solutions A solution is a homogeneous mixture. A solution is a homogeneous mixture. The substance that is dissolved is the solute. The substance that is dissolved is the solute. The substance that does the dissolving is the solvent. (Water is known as the universal solvent due to its ability to dissolve a wide range of solutes) The substance that does the dissolving is the solvent. (Water is known as the universal solvent due to its ability to dissolve a wide range of solutes)

3. How do solutes dissolve? Ionic solids are all soluble in water. Ionic solids are all soluble in water. The “positive” ends of a water molecule are attracted to the anion of the solid. The “positive” ends of a water molecule are attracted to the anion of the solid. The “negative” ends of a water molecule are attracted to the cation of the solid. The “negative” ends of a water molecule are attracted to the cation of the solid. This process is called hydration and results in the ionic solid being pulled apart, or dissolved. This process is called hydration and results in the ionic solid being pulled apart, or dissolved.

4. What is an electrolyte? Solutions that conduct an electric current efficiently contain strong electrolytes. Solutions that conduct an electric current efficiently contain strong electrolytes. Strong electrolytes are substances that completely ionize in water. Strong electrolytes are substances that completely ionize in water. Three types of compounds are classified as strong electrolytes. Three types of compounds are classified as strong electrolytes.

5. Strong Electrolytes The following are classified as strong electrolytes: The following are classified as strong electrolytes: –Soluble salts (see the handout of the solubility rules) –Strong acids (HCl, HBr, HI, H 2 SO 4, HNO 3, HClO 3, HClO 4 ) –Strong bases (group 1A hydroxides, and Ba(OH) 2

6. What is an acid? According to the Arrhenius definition, an acid is any substance that produces H + ions when it is dissolved in water. According to the Arrhenius definition, an acid is any substance that produces H + ions when it is dissolved in water. Strong acids are those acids in which every molecule ionizes into H + ions and anions. Strong acids are those acids in which every molecule ionizes into H + ions and anions.

7. What is a strong base? According to the Arrhenius definition, a strong base is any soluble ionic compound containing the hydroxide ion (OH - ). According to the Arrhenius definition, a strong base is any soluble ionic compound containing the hydroxide ion (OH - ). When these compounds dissolve in water, the cations and OH - ions separate and move independently. When these compounds dissolve in water, the cations and OH - ions separate and move independently.

8. What is a weak electrolyte? Weak electrolytes are substances that exhibit a small degree of ionization in water and therefore conduct an electric current inefficiently. Weak electrolytes are substances that exhibit a small degree of ionization in water and therefore conduct an electric current inefficiently. The most common weak electrolytes are weak acids and weak bases. The most common weak electrolytes are weak acids and weak bases. An example of a weak acid is acetic acid (HC 2 H 3 O 2 ). An example of a weak base is ammonium hydroxide (NH 4 OH) An example of a weak acid is acetic acid (HC 2 H 3 O 2 ). An example of a weak base is ammonium hydroxide (NH 4 OH)

9. What is a nonelectrolyte? Nonelectrolytes are substances that dissolve in water but do not produce any ions. Nonelectrolytes are substances that dissolve in water but do not produce any ions. Organic/covalent molecules are nonelectrolytes. Table sugar (C 12 H 22 O 11 ) is an example of a soluble compound that is a nonelectrolyte. This is because, when dissolved, the sugar molecule remains intact. Organic/covalent molecules are nonelectrolytes. Table sugar (C 12 H 22 O 11 ) is an example of a soluble compound that is a nonelectrolyte. This is because, when dissolved, the sugar molecule remains intact.

10. Concentration Concentrations of solutions are expressed qualitatively and quantitatively Concentrations of solutions are expressed qualitatively and quantitatively Qualitative expressions include the terms saturated, unsaturated, and supersaturated. Qualitative expressions include the terms saturated, unsaturated, and supersaturated.

11. Concentration (continued) Saturated solution-a solution in which no more solute will dissolve at a given temperature (the maximum has been dissolved) Saturated solution-a solution in which no more solute will dissolve at a given temperature (the maximum has been dissolved) Unsaturated solution-a solution in which more solute will dissolve at a given temperature (less than the maximum has been dissolved) Unsaturated solution-a solution in which more solute will dissolve at a given temperature (less than the maximum has been dissolved) Super-saturated solution-a solution that contains higher-than-saturation concentration of solute (a slight disturbance will cause crystallization) Super-saturated solution-a solution that contains higher-than-saturation concentration of solute (a slight disturbance will cause crystallization)

12. Quantitative Expressions of Concentration Percent by mass Percent by mass Molarity Molarity Molality Molality Dilution of solutions Dilution of solutions

13. Percent by Mass Percent by mass gives the mass of solute per 100 mass units of solution Percent by mass gives the mass of solute per 100 mass units of solution Percent solute=mass of solute x 100% Percent solute=mass of solute x 100% mass of solution mass of solution Mass of solution = mass of solute + mass of solvent Mass of solution = mass of solute + mass of solvent

14. Percent by Mass Problems Calculate the mass of NiSO 4 contained in 200. g of a 6.00% solution of NiSO 4 Calculate the mass of NiSO 4 contained in 200. g of a 6.00% solution of NiSO 4 A 6.00% solution of NiSO 4 contains 40.0 g of NiSO 4. Calculate the mass of the solution. A 6.00% solution of NiSO 4 contains 40.0 g of NiSO 4. Calculate the mass of the solution. Calculate the mass of NiSO 4 present in 200. mL of a 6.00% solution of NiSO 4. The density of the solution is 1.06 g/mL at 25 o C. Calculate the mass of NiSO 4 present in 200. mL of a 6.00% solution of NiSO 4. The density of the solution is 1.06 g/mL at 25 o C. What is the volume of a solution that is 15.0% Fe(NO 3 ) 3 and contains 30.0 g of the solute? The density of the solution is 1.16 g/mL at 25 o C. What is the volume of a solution that is 15.0% Fe(NO 3 ) 3 and contains 30.0 g of the solute? The density of the solution is 1.16 g/mL at 25 o C.

15. Molarity Molarity-the moles of solute pervolume of solution in liters. Molarity-the moles of solute pervolume of solution in liters. M = molarity = moles of solute M = molarity = moles of solute liters of solution liters of solution (solution = solvent + solute) (solution = solvent + solute)

16. Molarity Problems Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. Calculate the mass of Ba(OH) 2 required to prepare 2.50 L of a Calculate the mass of Ba(OH) 2 required to prepare 2.50 L of a 0.0600 M solution of Ba(OH) 2.

17. Molality Molality-the number of moles of solute per kilogram of solvent Molality-the number of moles of solute per kilogram of solvent Molality = moles of solute Molality = moles of solute kilogram of solvent kilogram of solvent

18. Molality Problems A solution is prepared by mixing 1.00 g of ethanol with 100.0 g of water to give a final volume of 101 mL. What is the molality? What is the molarity? What is the mass percent? What is the mole fraction? A solution is prepared by mixing 1.00 g of ethanol with 100.0 g of water to give a final volume of 101 mL. What is the molality? What is the molarity? What is the mass percent? What is the mole fraction?

19. Dilution of Solutions Dilution occurs when more solvent is added to a solution without changing the amount of solute that is present. Dilution occurs when more solvent is added to a solution without changing the amount of solute that is present. V 1 M 1 = V 2 M 2 V 1 M 1 = V 2 M 2 For safety purposes when in a lab setting, concentrated solutions of acids and bases are always added slowly to water. For safety purposes when in a lab setting, concentrated solutions of acids and bases are always added slowly to water.

20. Dilution Problems How many mL of 18.0 M H 2 SO 4 are required to prepare 1.00 L of a How many mL of 18.0 M H 2 SO 4 are required to prepare 1.00 L of a 0.900 M solution of H 2 SO 4 ?

21. Review of Solution Concentration The electrolyte in car batteries is a 3.75 M H 2 SO 4 solution that has a density of 1.230 g/mL. What is the mass %, molality, and mole fraction of the sulfuric acid? The electrolyte in car batteries is a 3.75 M H 2 SO 4 solution that has a density of 1.230 g/mL. What is the mass %, molality, and mole fraction of the sulfuric acid?

22. Factors Affecting Solubility Solubility is the maximum amount of solute that will dissolve in a given amount of solvent at a particular temperature.

23. Molecular Structure The statement “like dissolves like” summarizes the formation of solutions based on the structure of the molecules. The statement “like dissolves like” summarizes the formation of solutions based on the structure of the molecules. “Like dissolves like” means that polar solvents dissolve ionic and polar solutes while nonpolar solvents dissolve nonpolar solutes. “Like dissolves like” means that polar solvents dissolve ionic and polar solutes while nonpolar solvents dissolve nonpolar solutes. The term miscibility is used to describe the ability of one liquid to dissolve in another. (The observation of “like dissolves like” still applies). The term miscibility is used to describe the ability of one liquid to dissolve in another. (The observation of “like dissolves like” still applies).

24. Example of Differentiating Solvent Properties Based on Structure Decide whether liquid hexane or liquid methanol is the more appropriate solvent for the substances grease (C 20 H 42 ) and potassium iodide. Decide whether liquid hexane or liquid methanol is the more appropriate solvent for the substances grease (C 20 H 42 ) and potassium iodide.

25. Effect of Temperature on Solubility The solubility of most solid solutes increases with increasing temperature. The solubility of most solid solutes increases with increasing temperature. (exceptions include: Na 2 SO 4 and Ce 2 (SO 4 ) 3 ) (exceptions include: Na 2 SO 4 and Ce 2 (SO 4 ) 3 ) The solubility of gases decreases with increasing temperature. The solubility of gases decreases with increasing temperature. (example of this includes thermal pollution) (example of this includes thermal pollution)

26. Solubility Graphs Solubility curves are graphs that illustrate the effect of temperature on the solubilities of various salts. Solubility curves are graphs that illustrate the effect of temperature on the solubilities of various salts. Using the graph on p.496, answer the following questions: Using the graph on p.496, answer the following questions: *How many grams of KNO 3 will dissolve in 200 g of water at 70 o C? *At what temperature will 80 g of KBr dissolve in 100 g of water? *Which solute’s solubility is least affected by temperature?

27. Effect of Pressure on Solubility Changing the pressure has no effect on the solubilities of either solids or liquids in liquids. Changing the pressure has no effect on the solubilities of either solids or liquids in liquids. The solubilities of gases in all solvents will increase as the partial pressure of the gases increase. The solubilities of gases in all solvents will increase as the partial pressure of the gases increase.

28. Henry’s Law Henry’s law applies to gases that do not react with the solvent in which they dissolve. Henry’s law applies to gases that do not react with the solvent in which they dissolve. The law states that the pressure of a gas above the surface of a solution is proportional to the concentration of the gas in the solution. The law states that the pressure of a gas above the surface of a solution is proportional to the concentration of the gas in the solution.

29. Henry’s Law (continued) The equation for Henry’s law is: The equation for Henry’s law is: P gas = kC gas P gas is the pressure of the gas above the solution k is a constant for a particular gas and solvent at a particular temperature C gas represents the concentration of the dissolved gas; usually expressed as molarity or mole fraction.

30. Calculations Using Henry’s Law A certain soft drink is bottled so that at 25 o C a bottle contains CO 2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO 2 in the atmosphere is 0.0004 atm, calculate the equilibrium concentration of CO 2 in the soda both before and after the bottle is opened. The Henry’s law constant for CO 2 at 25 o C is A certain soft drink is bottled so that at 25 o C a bottle contains CO 2 gas at a pressure of 5.0 atm over the liquid. Assuming that the partial pressure of CO 2 in the atmosphere is 0.0004 atm, calculate the equilibrium concentration of CO 2 in the soda both before and after the bottle is opened. The Henry’s law constant for CO 2 at 25 o C is 32 L. atm/mol at 25 o C. How does your answer explain why soft drinks go flat after being opened?

31. Colligative Properties Colligative properties are physical properties that depend on the number, not the kind, of solute particles in a given amount of solvent. Colligative properties are physical properties that depend on the number, not the kind, of solute particles in a given amount of solvent. There are four important colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, osmotic pressure There are four important colligative properties: vapor pressure lowering, boiling point elevation, freezing point depression, osmotic pressure

32. Lowering of Vapor Pressure Solutions containing a nonvolatile liquid or a solid as a solute always has a lower vapor pressure than the pure solvent. Solutions containing a nonvolatile liquid or a solid as a solute always has a lower vapor pressure than the pure solvent. When a solute is dissolved, there are fewer solvent molecules at the surface of the liquid. When a solute is dissolved, there are fewer solvent molecules at the surface of the liquid. Therefore, the solvent molecules vaporize at a slower rate than if no solute were present. Therefore, the solvent molecules vaporize at a slower rate than if no solute were present. The lowering of the vapor pressure of a solvent due to the presence of a nonvolatile solute is summarized by Raoult’s Law. The lowering of the vapor pressure of a solvent due to the presence of a nonvolatile solute is summarized by Raoult’s Law.

33. Raoult’s Law The vapor pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent in the solution. The vapor pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent in the solution. P solvent = X solvent P 0 solvent P solvent = X solvent P 0 solvent The lowering of the vapor pressure can be expressed as: The lowering of the vapor pressure can be expressed as: ΔP solvent = X solute P 0 solvent

34. Sample Problem Determine the vapor pressure lowering of a sucrose solution made by dissolving 50.0 g of sucrose (C 12 H 22 O 11 ) in 117 g of water. The vapor pressure of pure water at 25 o C is 23.8 torr. Determine the vapor pressure lowering of a sucrose solution made by dissolving 50.0 g of sucrose (C 12 H 22 O 11 ) in 117 g of water. The vapor pressure of pure water at 25 o C is 23.8 torr.

35. Sample Problem #2 Predict the vapor pressure of a solution made by mixing 35.0 g of solid Na 2 SO 4 with 175 g water at 25 o C. The vapor pressure of pure water at 25 o C is 23.76 torr. Predict the vapor pressure of a solution made by mixing 35.0 g of solid Na 2 SO 4 with 175 g water at 25 o C. The vapor pressure of pure water at 25 o C is 23.76 torr.

36. Sample Problem #3 At 40 o C, the vapor pressure of pure heptane is 92.0 torr and the vapor pressure of pure octane is 31.0 torr. Consider a solution that contains 1.00 mole of heptane and 4.00 moles of octane. Calculate the vapor pressure of each component and the total vapor pressure above the solution. At 40 o C, the vapor pressure of pure heptane is 92.0 torr and the vapor pressure of pure octane is 31.0 torr. Consider a solution that contains 1.00 mole of heptane and 4.00 moles of octane. Calculate the vapor pressure of each component and the total vapor pressure above the solution.

37. Boiling Point Elevation A nonvolatile solute elevates the boiling point of the solvent. Boiling occurs at a temperature where the vapor pressure is equal to 1 atm. Boiling occurs at a temperature where the vapor pressure is equal to 1 atm. Addition of a nonvolatile solute lowers vapor pressure. Addition of a nonvolatile solute lowers vapor pressure. Therefore, a solution must be heated to a higher temperature than the boiling point of the pure solvent to reach a vapor pressure of 1 atm. Therefore, a solution must be heated to a higher temperature than the boiling point of the pure solvent to reach a vapor pressure of 1 atm.

38. Equation for Boiling Point Elevation ∆T = K b m solute ∆T is the boiling point elevation K b is the boiling point elevation constant for the solvent m solute is the molality of the solute

39. Calculating Boiling Point Elevation Find the boiling point of a 1.25 m sucrose solution. Find the boiling point of a 1.25 m sucrose solution. A solution was prepared by dissolving 18.00 g glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34 o C. Calculate the molar mass of glucose. A solution was prepared by dissolving 18.00 g glucose in 150.0 g of water. The resulting solution was found to have a boiling point of 100.34 o C. Calculate the molar mass of glucose.

40. Freezing Point Depression When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. This is the reason that compounds such as sodium chloride and calcium chloride are spread on roadways to prevent ice from forming in freezing weather. This is the reason that compounds such as sodium chloride and calcium chloride are spread on roadways to prevent ice from forming in freezing weather.

41. Equation for Freezing Point Depression ∆T = K f m solute ∆T = K f m solute ∆T is the freezing point depression ∆T is the freezing point depression K f is the freezing point depression constant K f is the freezing point depression constant m solute is the molality of the solute m solute is the molality of the solute

42. Calculating Freezing Point Depression Calculate the freezing point of a 1.25 m sucrose solution. Calculate the freezing point of a 1.25 m sucrose solution. What mass of ethylene glycol (C 2 H 6 O 2 = 62.1 g/mol), the main component of antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at -10.0 o F What mass of ethylene glycol (C 2 H 6 O 2 = 62.1 g/mol), the main component of antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at -10.0 o F (-23.3 o C)? (-23.3 o C)? A sample weighing 0.546 g was dissolved in 15.0 g of benzene, and the freezing point depression was determined to be 0.240 o C. Calculate the molar mass of the sample. A sample weighing 0.546 g was dissolved in 15.0 g of benzene, and the freezing point depression was determined to be 0.240 o C. Calculate the molar mass of the sample.

43. The van’t Hoff Factor Colligative properties depend on the number of solute particles in a given mass of solvent. Colligative properties depend on the number of solute particles in a given mass of solvent. The dissociation of electrolytes in aqueous solutions causes a greater change in these properties than a solution containing a nonelectrolyte (as in the previous examples). The dissociation of electrolytes in aqueous solutions causes a greater change in these properties than a solution containing a nonelectrolyte (as in the previous examples). One measure of the extent of dissociation of an electrolyte in water the van’t Hoff factor, i, for the solution. One measure of the extent of dissociation of an electrolyte in water the van’t Hoff factor, i, for the solution. i = moles of particles/moles of solute ( or the number of ions per formula unit) i = moles of particles/moles of solute ( or the number of ions per formula unit) The observed value of i is always slightly lower than the predicted value due to ion pairing. (see page 513). The observed value of i is always slightly lower than the predicted value due to ion pairing. (see page 513). For problem solving purposes, use the ideal value of i unless the problem indicates otherwise. For problem solving purposes, use the ideal value of i unless the problem indicates otherwise.

44. Determining the van’t Hoff factor What is the value of the van’t Hoff factor, i, for the following strong electrolytes? What is the value of the van’t Hoff factor, i, for the following strong electrolytes? a) Na 2 SO 4 b) KOH b) KOH c) Al 2 (SO 4 ) 3 c) Al 2 (SO 4 ) 3 d) SrSO 4 d) SrSO 4

45. Properties of Electrolyte Solutions From the following solutions: From the following solutions: 0.010 m Na 3 PO 4 0.020 m CaBr 2 0.020 m KCl 0.020 m HF in water (HF is a weak acid) a) Which would have the same boiling point as 0.040 m C 6 H 12 O 6 in water? 0.040 m C 6 H 12 O 6 in water? a) Which solution would have the highest vapor pressure at 28 o C? b) Which solution would have the largest freezing point depression?

46. Properties of Electrolyte Solutions (continued) From the following: From the following: pure water 0.01 m C 12 H 22 O 11 in water 0.01 m NaCl in water 0.01 m CaCl 2 in water Choose the one with the a) Highest freezing point b) Lowest freezing point c) Highest boiling point d) Lowest boiling point e) Highest vapor pressure

47. Solving Problems Using the van’t Hoff Factor? Calculate the freezing point and the boiling point of each of the following aqueous solutions. Calculate the freezing point and the boiling point of each of the following aqueous solutions. a) 0.050 m MgCl 2 b) 0.050 m FeCl 3

48. Osmotic Pressure Osmosis is the passage of a solvent through a semi-permeable membrane. Osmosis is the passage of a solvent through a semi-permeable membrane. Osmotic Pressure is the minimum pressure that stops osmosis. Osmotic Pressure is the minimum pressure that stops osmosis. The equation for osmotic pressure is The equation for osmotic pressure is π = iMRT π is the pressure in atmM is the molarity of the solution π is the pressure in atmM is the molarity of the solution R = 0.08206 L. atm/K. molT is the Kelvin temperature R = 0.08206 L. atm/K. molT is the Kelvin temperature

49. Calculating Osmotic Pressure To determine the molar mass of a certain protein,.00100 g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure was found to be 1.12 torr at 25.0 o C. Calculate the molar mass of the protein. To determine the molar mass of a certain protein,.00100 g of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure was found to be 1.12 torr at 25.0 o C. Calculate the molar mass of the protein. What concentration of NaCl in water is needed to produce an aqueous solution isotonic with blood ( π = 7.70 atm at 25 o C)? What concentration of NaCl in water is needed to produce an aqueous solution isotonic with blood ( π = 7.70 atm at 25 o C)?

50. Calculating Osmotic Pressure What osmotic pressure would 50.0 g of sucrose in 117 g of water exhibit at 25 o C? The density of this solution is 1.34 g/mL. What osmotic pressure would 50.0 g of sucrose in 117 g of water exhibit at 25 o C? The density of this solution is 1.34 g/mL.