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- 1. Chapter 13 The Properties of Mixtures: Solutions and Colloids 13-1
- 2. SOLUTIONS I Unit 13 • Concentrations of solutions • Chapter 11.1-11.4 McM • Chapter 13.5 Silberberg 13-2
- 3. The Properties of Mixtures: Solutions and Colloids 13.1 Types of Solutions: Intermolecular Forces and Solubility 13.3 Why Substances Dissolve: Understanding the Solution Process 13.5 Concentration Terms 13-3
- 4. Goals & Objectives • See the following Learning Objectives on pages 512. • Understand these Concepts: • 13.14 • Master these Skills: • 13.3-4. 13-4
- 5. Solutions and Colloids A solution is a homogeneous mixture and exists as a single phase. The particles in a solution are individual atoms, ions, or small molecules. A colloid is a heterogeneous mixture and exists as two or more phases, which may be visibly distinct. The particles in a colloid are typically macromolecules or aggregations of small molecules. 13-5
- 6. Solutions and Solubility A solute dissolves in a solvent to form a solution. Usually, the solvent is the most abundant component. The solubility (S) of a solute is the maximum amount that dissolves in a fixed quantity of solvent at a given temperature. Substances that exhibit similar types of intermolecular force dissolve in each other. This is often expressed by saying “like dissolves in like.” 13-6
- 7. A Sodium Chloride Solution • Sodium chloride is the solute • water is the solvent (understood) 13-7
- 8. Figure 13.1 Ion-dipole (40-600) H bond (10-40) Dipole-dipole (5-25) 13-8 Types of intermolecular forces in solutions. Ion-induced dipole (3-15) Dipole-induced dipole (2-10) Dispersion (0.05-40)
- 9. Concentration • A measure of the amount of solute dissolved per given amount of solvent or solution. • Typical units – Percent by mass (% w/w) – Molarity (M, mol/L) – Molality (m) – Proof 13-9
- 10. Proof • Approximately twice the percent by volume of alcohol • • • • 13-10 Proof 26 80 190 %Alcohol by Volume 13% 40% 95%
- 11. Percent by Mass of Solute • %mass of solute = g solute x 100 • g solution • A 25.9% solution of glucose contains: • 35.0g of glucose /35.g glucose + 100 g of water. The density of water is 1.00 g/ml 13-11
- 12. 13-12
- 13. Calculate the concentration in %(w/w) of the following solutions. Assume that water has a density of 1.00 g/mL. A 3.50 g sample of morphine hydrochloride is dissolved in 750 mL of water. 13-13
- 14. 13-14
- 15. Calculate the concentration in %(w/w) of the following solutions. Assume that water has a density of 1.00 g/mL. A 3.50 g sample of morphine hydrochloride is dissolved in 750 mL of water. 13-15
- 16. 13-16
- 17. Calculate the concentration in %(w/w) of the following solutions. Assume that water has a density of 1.00 g/mL. 30.0 mL of alcohol (density = 0.789 g/mL) is mixed into 370 mL of water 13-17
- 18. 13-18
- 19. 13-19
- 20. 13-20
- 21. Examples • What volume of 12.9% KOH solution contains 40.0 grams of KOH? The density of the solution is 1.11g/mL. 13-21
- 22. 13-22
- 23. 13-23
- 24. Solutions and Intermolecular Forces When a solution forms, solute-solute attractions and solvent-solvent attractions are replaced by solutesolvent attractions. This can only occur if the forces within the solute and solvent are similar to the forces that replace them. 13-24
- 25. Figure 13.2 Hydration shells around an Na+ ion. Ion-dipole forces orient water molecules around an ion. In the innermost shell here, six water molecules surround the cation octahedrally. 13-25
- 26. Dual Polarity and Effects on Solubility • Alcohols are organic compounds that have dual polarity. – The general formula for an alcohol is CH3(CH2)nOH. • The –OH group of an alcohol is polar. – It interacts with water through H bonds and – with hexane by through weak dipole-induced dipole forces. • The hydrocarbon portion is nonpolar. – It interacts through weak dipole-induced dipole forces with water – and by dispersion forces with hexane. 13-26
- 27. Figure 13.3 H 2O Like dissolves like: solubility of methanol in water. Methanol CH3OH Both H2O and methanol have H bonds between their molecules. A solution of water and methanol. 13-27
- 28. Figure 13.18 Sodium acetate crystallizing from a supersaturated solution. A supersaturated solution contains more than the equilibrium concentration of solute. It is unstable and any disturbance will cause excess solute to crystallize immediately. 13-28
- 29. Factors that affect Solubility Temperature affects solubility. Most solids are more soluble at higher temperatures. Gases become less soluble as temperature increases. Pressure affects the solubility of gases – they become more soluble at higher pressure. 13-29
- 30. Figure 13.19 13-30 Relation between solubility and temperature for several ionic compounds.
- 31. Table 13.5 Concentration Definitions Concentration Term Molarity (M) Molality (m) Ratio amount (mol) of solute volume (L) of solution amount (mol) of solute mass (kg) of solvent Parts by mass Parts by volume mass of solute mass of solution volume of solute volume of solution Mole fraction (X) amount (mol) of solute amount (mol) of solute + amount (mol) of solvent 13-31
- 32. Molarity • Moles of solute per liter of solution • M = moles of solute liters of solution • Abbreviate: mol/L • A 2.0M (molar) solution of NaCl contains 2.0 moles of NaCl in 1.0 liter of solution. • Or 4.0 moles of NaCl in 2.0 liters of solution etc. 13-32
- 33. 13-33
- 34. Examples • Calculate the molarity of a solution prepared by dissolving 12.5 g of sulfuric acid (H2SO4) in sufficient water to give 1.75L of solution. • Determine the mass of Ca(NO3)2 required to prepare 3.50 L of a solution that is 0.800M in Ca(NO3)2. 13-34
- 35. 13-35
- 36. 13-36
- 37. Examples • The density of commercial hydrochloric acid is 1.185g/mL. It is 36.31% HCl by mass. Determine its molarity. 13-37
- 38. 13-38
- 39. 13-39
- 40. Using Solutions in Chemical Reactions • Na2SO4 + BaCl2 BaSO4 + 2NaCl • Determine the volume of 0.500M BaCl2 solution required for complete reaction with 4.32g of Na2SO4 according to the reaction 13-40
- 41. 13-41
- 42. Using Solutions in Chemical Reactions • H2SO4 + 2NaOH Na2SO4 + 2H2O • Determine the molarity of the H2SO4 solution required for complete reaction with 15.14 ml of 0.1022M NaOH solution, if you started with 10.00ml of H2SO4 solution in the erlenmeyer flask. 13-42
- 43. Using Solutions in Chemical Reactions: Titration 13-43 The process, operation, or method of determining the concentration of a substance in solution by adding to it a standard reagent of known concentration in carefully measured amounts until a reaction of definite and known proportion is completed, as shown by a color change or by electrical measurement, and then calculating the unknown concentration.
- 44. Dilution of Solutions • A more concentrated solution can be diluted with solvent to produce a solution that is more dilute. The moles of solute will be the same in both solutions. However, since the volumes will be different the molarity will change. • M1V1 = M2V2 13-44
- 45. 13-45
- 46. 13-46
- 47. Examples • Determine the volume of 18.0M H2SO4 required to prepare 2.50 liters of 2.40M H2SO4 solution. • If 50.0 ml of 12.0M HCl is diluted to a final total volume of 200.0 ml, determine the final concentration of the solution. 13-47
- 48. 13-48
- 49. 13-49
- 50. Molality • Moles of solute per kilogram of solvent • m = moles of solute • kg of solvent • A 1.0m (molal) solution of NaCl contains 1.0 moles of NaCl in 1.0 kg of water. 13-50
- 51. Sample Problem 13.3 Calculating Molality PROBLEM: What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water? PLAN: Molality is defined as moles of solute (CaCl2) divided by kg of solvent (H2O). We convert the mass of CaCl2 to moles using the molar mass, and then divide by the mass of H2O, being sure to convert from grams to kilograms. mass (g) of CaCl2 divide by M of (g/mol) amount (mol) of CaCl2 divide by kg of water molality (m) of CaCl2 solution 13-51
- 52. Sample Problem 13.3 SOLUTION: 32.0 g CaCl2 molality = 13-52 1 mol CaCl2 x 110.98 g CaCl2 0.288 mol CaCl2 1 kg 271 g H2O x 103 g = 0.288 mol CaCl2 = 1.06 m CaCl2
- 53. Examples 13-53 • Determine the molality of a solution prepared by dissolving 400 g of NaOH in 500mL of water. The density of water is 1.00g/mL. M.W. of NaOH is 40.0 amu. • Calculate the molarity and molality of an aqueous solution that is 10.0% glucose. The density of the solution is 1.04g/mL. M.W. of glucose = 180.0amu.
- 54. 13-54
- 55. 13-55
- 56. Examples • Determine the molality of a solution that contains 7.25 g of benzoic acid, C6H5COOH, in 200 mL of benzene, C6H6. The density of benzene is 0.879g/mL and the molar mass of benzoic acid = 122 amu. 13-56
- 57. 13-57
- 58. 13-58
- 59. Mole Fraction • Moles of component A per total moles of all components in the solution. • XA = molesA • molesA + molesB + molesC • XB = • XA + XB + XC + ……… = 1 13-59
- 60. 13-60
- 61. Sample Problem 13.4 Expressing Concentrations in Parts by Mass, Parts by Volume, and Mole Fraction PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50-g pill that contains 40.5 mg of Ca. (b) The label on a 0.750-L bottle of Italian chianti indicates “11.5% alcohol by volume.” How many liters of alcohol does the wine contain? (c) A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water? PLAN: (a) We convert mg to g of Ca2+, find the mass ratio of Ca2+ to pill and multiply by 106. (b) We know the volume % of the alcohol and the total volume, so we can find the volume of alcohol. (c) We convert g of solute and solvent to moles in order to calculate the mole fractions. 13-61
- 62. Sample Problem 13.4 SOLUTION: (a) 40.5 mg Ca2+ x 1g 103 mg 3.5 g (b) 0.750 L chianti x x 106 = 1.16x104 ppm Ca2+ 11.5 L alcohol 100. L chianti = 0.0862 L alcohol (c) moles isopropyl alcohol = moles water = 13-62 58.0 g x 142 g x 1 mole = 2.36 mol C H OH 3 7 60.09 g 1 mole = 3.22 mol H2O 18.02 g
- 63. Sample Problem 13.4 X C H OH = 3 7 XH O = 2 13-63 2.36 mol C2H8O2 2.36 mol C3H7OH + 3.22 mol H2O = 0.423 3.22 mol H2O = 0.577 2.36 mol C3H7OH + 3.22 mol H2O
- 64. Interconverting Concentration Terms • To convert a term based on amount (mol) to one based on mass, you need the molar mass. • To convert a term based on mass to one based on volume, you need the solution density. • Molality involves quantity of solvent, whereas the other concentration terms involve quantity of solution. 13-64
- 65. Sample Problem 13.5 Interconverting Concentration Terms PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution as a hair bleach. An aqueous solution H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its (a) Molality (b) Mole fraction of H2O2 (c) Molarity PLAN: (a) To find the mass of solvent we assume the % is per 100 g of solution. Take the difference in the mass of the solute and solution for the mass of peroxide. (b) Convert g of solute and solvent to moles before finding X. (c) Use the density to find the volume of the solution. 13-65
- 66. Sample Problem 13.5 SOLUTION: (a) From mass % to molality: g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O 1 mol H2O2 30.0 g H2O2 x = 0.882 mol H2O2 34.02 g H2O2 0.882 mol H2O2 molality = 70.0 g x 1 kg 103 g 13-66 = 12.6 m H2O2
- 67. Sample Problem 13.5 (b) From mass % to mole fraction: 1 mol H2O 70.0 g H2O x = 3.88 mol H2O 18.02 g H2O XH O = 2 2 0.882 mol H2O2 3.88 mol H2O + 0.882 mol H2O2 = 0.185 (c) From mass % and density to molarity: 1 mL volume (mL) of solution = 100.0 g x 1.11 g mol H2O2 0.882 mol H2O2 molarity = = L soln 90.1 mL x 1 L soln 103 mL 13-67 = 90.1 mL = 9.79 M H2O2
- 68. Examples • What are the mole fractions of glucose and water in 10.0% glucose solution. 13-68
- 69. 13-69
- 70. 13-70

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