1. CHAPTER 13 SOLUTIONS

2. BASIC DEFINITIONS Solution Solution – a homogeneous mixture of 2 or more substances in a single phase Solute – The dissolved substance usually in smaller amount Solvent – the dissolving medium usually in larger amount

3. Visual Concepts Solutes, Solvents, and Solutions Chapter 13 Solvent - present in greater amount Solute - substance being dissolved

4. Visual Concepts Solutions Chapter 13

5. MORE DEFINITIONS Saturated – The maximum amount of solute is dissolved in the solvent Unsaturated – Less than the maximum amount of solute is dissolved in the solvent Dilute – The amount of solvent is much greater than the amount of solute dissolved in it.

6. MORE DEFINITIONS Electrolyte – A substance that dissolves in water to give a solution that will conduct electricity Nonelectrolyte – A substance that dissolves in water to give a solution that will not conduct electricity

7. TYPES OF SOLUTIONS Gases (Air is a mixture of O 2, N 2, CO 2, H 2 O, etc.) Solids (White gold is a mixture of gold & palladium) Liquids (salt dissolved in water; carbon dioxide in soda)

8. Particle Model for a Suspension Chapter 13 Section 1 What Is a Solution?

9. SUSPENSION A heterogeneous mixture in which particles in a solvent are so large, they will settle out unless it is constantly stirred Particles can be filtered out due to their size (>1000 nm in diameter)

10. Visual Concepts Suspensions Chapter 13

11. COLLOIDS A heterogeneous mixture in which particles in a solvent remain suspended by the movement of surrounding molecules Think of fruit suspended in jello Particles are not easily filtered due to their size (<1000 nm in diameter)

12. Visual Concepts Factors Affecting the Rate of Dissolution Chapter 13

13. FACTORS AFFECTING SOLUBILITY 1.Surface Area – increasing the surface area gives molecules more places to interact causing the solute to dissolve faster 2.Agitation/Stirring – stirring gives molecules more opportunities to interact – faster dissolving 3.Temperature – Adding heat adds energy to molecules = faster dissolving 4.“Like dissolves like” – polar solutes dissolve in polar solvents or Nonpolar solutes dissolve in Nonpolar solvents

14. Visual Concepts Like Dissolves Like Chapter 13

15. CONCENTRATION CALCULATIONS

16. Visual Concepts Concentration Chapter 13

17. Concentration, continued Calculating Concentration, continued Concentrations can be expressed in many forms. Section 2 Concentration and Molarity Chapter 13

18. MOLARITY (M)

19. MOLARITY (m) Moles of solute per Liter of solution

20. SAMPLE PROBLEM A You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution?

21. SAMPLE PROBLEM A You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution? Step 1: Outline what you know. M = ? Mol / L mol = 90.0 g L = 3.50 L

22. SAMPLE PROBLEM A You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution? Step 2: Convert any units necessary. mol = 90.0 g NaCl 90.0 g NaClx 1 mol NaCl 58 g NaCl = 1.55 mol NaCl

23. SAMPLE PROBLEM A You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution? Step 3: Plug into the equation and solve. M = ? Mol / L mol = 1.55 mol L = 3.50 L

24. SAMPLE PROBLEM A You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of that solution? Step 3: Plug into the equation and solve. M=M= 1.55 mol 3.50 L =0.44 mol / L

25. SAMPLE PROBLEM B You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain?

26. SAMPLE PROBLEM B You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? Step 1: Outline what you know. M = 0.5 Mol / L mol = ? mol L = 0.8 L

27. SAMPLE PROBLEM B You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? Step 2: Convert any units necessary. All units correct! Moving on…

28. SAMPLE PROBLEM B You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? Step 3: Plug into the equation and solve. M = 0.5 Mol / L mol = ? mol L = 0.8 L

29. SAMPLE PROBLEM B You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? Step 3: Plug into the equation and solve. 0.5 M= mol 0.8 L (0.8 L)

30. SAMPLE PROBLEM B You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? Step 3: Plug into the equation and solve. mol = 0.4

31. MOLALITY (m)

32. Moles of solute per kilogram of solvent

33. SAMPLE PROBLEM C A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C 12 H 22 O 11 ) in 125 g of water. Find the molal concentration of this solution.

34. SAMPLE PROBLEM C A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C 12 H 22 O 11 ) in 125 g of water. Find the molal concentration of this solution. Step 1: Outline what you know. m = ? Mol / kg mol = 17.1 g Kg = 125 g

35. 342.34 g C 12 H 22 O 11 SAMPLE PROBLEM C A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C 12 H 22 O 11 ) in 125 g of water. Find the molal concentration of this solution. Step 2a: Convert any units necessary. mol = 17.1 g C 12 H 22 O 11 17.1 g C 12 H 22 O 11 x 1 mol C 12 H 22 O 11 =0.05 mol C 12 H 22 O 11

36. 1000 g SAMPLE PROBLEM C A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C 12 H 22 O 11 ) in 125 g of water. Find the molal concentration of this solution. Step 2b: Convert any units necessary. kg = 125 g H 2 O 125 g H 2 O x 1 kg =0.125 kg H 2 O

37. SAMPLE PROBLEM C A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C 12 H 22 O 11 ) in 125 g of water. Find the molal concentration of this solution. Step 3: Plug into the equation and solve. m = ? Mol / kg mol = 0.05 mol kg = 0.125 kg

38. SAMPLE PROBLEM C A solution was prepared by dissolving 17.1 g of sucrose (table sugar, C 12 H 22 O 11 ) in 125 g of water. Find the molal concentration of this solution. Step 3: Plug into the equation and solve. m=m= 0.05 mol 0.125 kg =0.400 mol / kg

39. SAMPLE PROBLEM D How much iodine must be added to prepare a 0.480 m solution of iodine in CCl 4 if 100.0 g of CCl 4 is used?

40. SAMPLE PROBLEM D How much iodine must be added to prepare a 0.480 m solution of iodine in CCl 4 if 100.0 g of CCl 4 is used? Step 1: Outline what you know. m = 0.480 m mol = ? mol kg = 100.0 g

41. SAMPLE PROBLEM D How much iodine must be added to prepare a 0.480 m solution of iodine in CCl 4 if 100.0 g of CCl 4 is used? Step 2: Convert any units necessary. kg = 100.0 g Kg = 0.100 kg

42. SAMPLE PROBLEM D How much iodine must be added to prepare a 0.480 m solution of iodine in CCl 4 if 100.0 g of CCl 4 is used? Step 3: Plug into the equation and solve. m = 0.480 Mol / kg mol = ? mol kg = 0.100 kg

43. SAMPLE PROBLEM D How much iodine must be added to prepare a 0.480 m solution of iodine in CCl 4 if 100.0 g of CCl 4 is used? Step 3: Plug into the equation and solve. 0.480 m = mol 0.100 kg (0.100 kg)

44. SAMPLE PROBLEM D How much iodine must be added to prepare a 0.480 m solution of iodine in CCl 4 if 100.0 g of CCl 4 is used? Step 3: Plug into the equation and solve. mol = 0.048

45. PARTS PER MILLION (ppm) ppm = grams of solute Liters of solution X 1000 Grams of solute in 1 million grams of solution. Useful in very dilute samples, but for the sake of simplicity, your worksheet will not work with very small exponents.