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Chapter 9

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Chapter 9

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  1. Chapter 9 Solutions

  2. Water • Water is the most common solvent. • The water molecule is polar. • Hydrogen bonds form between the hydrogen atom in one molecule and the oxygen atom in a different water molecule.

  3. Water

  4. Water in Foods • Water for the body is obtained from fluids as well as foods. • Some foods have a high percentage of water.

  5. Solution Formation 02

  6. Solution Formation 02

  7. How Does a Solution Form? As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

  8. 9.2 The Solution Process • Solubility depends primarily on the strength of the attractions between solute and solvent particles relative to the strengths of the attractions within the pure substances.

  9. Like Dissolves Like • A solution forms when there is an attraction between the particles of the solute and solvent. • A polar solvent such as water dissolves polar solutes such as sugar and ionic solutes such as NaCl. • A nonpolar solvent such as hexane (C6H14)dissolves nonpolar solutes such as oil or grease.

  10. Examples of Like Dissolves Like Solvents Solutes Water (polar) Ni(NO3)2 (ionic) CH2Cl2 (nonpolar) I2 (nonpolar)

  11. Learning Check Which of the following solutes will dissolve in water? Why? 1) Na2SO4 2) gasoline (nonpolar) 3) I2 4) HCl

  12. Solution Which of the following solutes will dissolve in water? Why? 1) Na2SO4Yes, ionic 2) gasoline No, nonpolar 3) I2 No, nonpolar 4) HCl Yes, polar Most polar and ionic solutes dissolve in water because water is a polar solvent.

  13. Formation of a Solution • Na+ and Cl- ions on the surface of a NaCl crystal are attracted to polar water molecules. • In solution, the ions are hydrated as several H2O molecules surround each.

  14. Equations for Solution Formation • When NaCl(s) dissolves in water, the reaction can be written as H2O NaCl(s) Na+(aq) + Cl– (aq) solid separation of ions

  15. Learning Check Solid LiCl is added to water. It dissolves because A. The Li+ ions are attracted to the 1) oxygen atom ( –) of water. 2) hydrogen atom (+) of water. B. The Cl- ions are attracted to the 1) oxygen atom ( –) of water. 2) hydrogen atom (+) of water.

  16. Solution Solid LiCl is added to water. It dissolves because A. The Li+ ions are attracted to the 1) oxygen atom ( –) of water. B. The Cl- ions are attracted to the 2) hydrogen atom ( +) of water.

  17. 9.4 Solubility

  18. Solubility • Solubility states the maximum amount of solute that dissolves in a specific amount of solvent at a particular temperature. • Typically, solubility is expressed as the grams of solute that dissolves in 100 g of solvent, usually water. g of solute 100 g water

  19. Saturated Solutions • A saturated solution • Contains the maximum amount of solute that can dissolve. • Has some undissolved solute at the bottom of the container.

  20. Unsaturated Solutions • An unsaturated solution • Contains less than the maximum amount of solute. • Can dissolve more solute.

  21. Learning Check At 40C, the solubility of KBr is 80 g/100 g H2O. Identify the following solutions as either (1) saturated or (2) unsaturated. Explain. A. 60 g KBr added to 100 g of water at 40C. B. 200 g KBr added to 200 g of water at 40C. C. 25 g KBr added to 50 g of water at 40C.

  22. Solution A. 2 Amount is less than the solubility. B. 1 In 100 g of water, 100 g KBr exceeds the solubility at 40C. C. 2 This would be 50 g KBr in 100 g of water, which is less than the solubility at 40C.

  23. 9.5 The Effects of Temperature on Solubility • The solubility of most solids increases with an increase in temperature.

  24. Solubility of Gases and Temperature • The solubility of gases decreases with an increase in temperature.

  25. Learning Check A. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun? B.Why do fish die in water that is too warm?

  26. Solution A. The pressure in a bottle increases as the gas leaves solution as it becomes less soluble at high temperatures. As pressure increases, the bottle could burst. B. Because O2 gas is less soluble in warm water, fish cannot obtain the amount of O2 required for their survival.

  27. 9.6 The Effect of Pressure on Solubility: Henry’s Law • According to Henry’s Law, the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid.

  28. The Effect of Pressure on Solubility: Henry’s Law

  29. 9.7 Units of Concentration

  30. Molarity (M) • Molarity is a concentration unit for the moles of solute in the liters (L) of solution. Molarity (M) = moles of solute = moles liter of solution L Examples: 2.0 M HCl = 2.0 moles HCl 1 L 6.0 M HCl = 6.0 moles HCl 1 L

  31. Preparing a 1.0 Molar Solution • A 1.0 M NaCl solution is prepared by weighing out 58.5 g NaCl ( 1.0 mole) and adding water to make 1.0 liter of solution.

  32. Calculation of Molarity What is the molarity of a NaOH solution prepared by adding 4.0 g of solid NaOH to water to make 0.50 L of solution ? 1. Determine the moles of solute. 4.0 g NaOH x 1 mole NaOH = 0.10 mole 40.0 g NaOH 2. Calculate molarity. 0.10 mole = 0.20 M NaOH 0.50 L

  33. Learning Check Calculate the molarity of an NaHCO3 solution prepared by dissolving 36 g of solid NaHCO3 in water to give a solution volume of 240 mL. 1) 0.43 M 2) 1.8 M 3) 15 M

  34. Solution 2) 1.8 M 36 g x 1 mole NaHCO3 = 0.43 mole NaHCO384 g 0.43 mole NaHCO3 = 1.8 M NaHCO3 0.240 L

  35. Learning Check A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 0.20 M 2) 5.0 M 3) 36 M

  36. Solution 1) 0.20 M 72 g x 1 mole = 0.40 moles 180. g 0.40 moles / 2L = 0.20 M

  37. Molarity Conversion Factors The units in molarity can be used to write conversion factors.

  38. Learning Check Stomach acid is 0.10 M HCl solution. How many moles of HCl are present in 1500 mL of stomach acid? 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 mole HCl

  39. Solution 3) 0.15 mole HCl 1500 mL x 1 L = 1.5 L 1000 mL 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L Molarity factor

  40. Learning Check Calculate the grams of KCl that must be dissolved in water to prepare 0.25 L of a 2.0 M KCl solution. 1) 150 g KCl 2) 37 g KCl 3) 19 g KCl

  41. Solution 2) 37 g KCl Determine the number of moles of KCl. 0.25 L x 2.0 mole KCl = 0.50 moles KCl 1 L Convert the moles to grams of KCl. 0.50 moles KCl x 74.6 g KCl = 37 g KCl 1 mole KCl molar mass of KCl

  42. Learning Check How many milliliters of 6.0 M HNO3 contain 0.15 mole of HNO3? 1) 25 mL 2) 90 mL 3) 400 mL

  43. Solution 1) 25 mL 0.15 mole HNO3 x 1 L x 1000 mL 6.0 moles HNO3 1 L Molarity factor inverted = 25 mL HNO3

  44. Percent Concentration • The concentration of a solution is the amount of solute dissolved in a specific amount of solution. amount of solute amount of solution • The percent concentration describes the amount of solute that is dissolved in 100 parts of solution. amount of solute 100 parts solution

  45. Mass Percent The mass percent (%m/m) • Concentration is the percent by mass of solute in a solution.mass percent = g of solute x 100 g of solution • Is the g of solute in 100 g of solution.mass percent = g of solute 100 g of solution

  46. Mass of Solution grams of solute + grams of solvent 50.0 g KCl solution

  47. Calculating Mass Percent • Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.0 g g of solvent (water) = 42.0 g g of KCl solution = 50.0 g 8.0 g KCl (solute) x 100 = 16% (m/m) KCl 50.0 g KCl solution

  48. Learning Check A solution is prepared by mixing 15 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution. 1) 15% (m/m) Na2CO3 2) 6.4% (m/m) Na2CO3 3) 6.0% (m/m) Na2CO3

  49. Solution 3) 6.0% (m/m) Na2CO3 mass solute = 15 g Na2CO3 mass solution = 15 g + 235 g = 250 g mass %(m/m) = 15 g Na2CO3 x 100 250 g solution = 6.0% Na2CO3 solution

  50. Mass/Volume Percent The mass/volume percent (%m/v) • Concentration is the ratio of the mass in grams (g) of solute in a volume (mL) of solution. mass/volume % = g of solute x 100 mL of solution • Is the g of solute in 100 mL of solution. mass/volume % = g of solute 100 mL of solution