1. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 7 Solutions 7.1 Solutions 1

2. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Solutions: Solute and Solvent Solutions are homogeneous mixtures of two or more substances form when there is sufficient attraction between solute and solvent molecules have two components: the solvent, present in a much larger amount, and the solute, present in a smaller amount 2

3. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Water as a Solvent Water is one of the most common solvents in nature is a polar molecule due to polar O–H bonds molecules form hydrogen bonds important in many biological compounds 3

4. Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. “Like Dissolves Like” “Like dissolves like”: polarity of solute and solvent must be similar to form a solution. 4

5. Chapter 7 Solutions 7.2 Electrolytes and Nonelectrolytes 5

6. Solutes and Ionic Charge In water, strong electrolytes produce ions and conduct an electric current weak electrolytes produce a few ions nonelectrolytes do not produce ions 6

7. Strong Electrolytes Strong electrolytes dissociate in water, producing positive and negative ions conduct an electric current in water in equations show the formation of ions in aqueous (aq) solutions 100% ions 7

8. Weak Electrolytes A weak electrolyte dissociates only slightly in water in water forms a solution of a few ions and mostly undissociated molecules 8

9. Nonelectrolytes dissolve as molecules in water do not produce ions in water do not conduct an electric current 9

10. Chapter 7 Solutions 7.3 Solubility 10

11. Solubility Solubility is the maximum amount of solute that dissolves in a specific amount of solvent temperature sensitive for solutes expressed as grams of solute in 100 grams of solvent, usually water g of solute 100 g water 11

12. Unsaturated Solution Unsaturated solutions contain less than the maximum amount of solute can dissolve more solute 12

13. Saturated Solution Saturated solutions contain the maximum amount of solute that can dissolve have undissolved solute at the bottom of the container contain solute that dissolves as well as solute that recrystallizes in an equilibrium process 13

14. Effect of Temperature on Solubility Solubility depends on temperature of most solids increases as temperature increases of gases decreases as temperature increases 14 In water, most common solids are more soluble as the temperature increases.

15. Learning Check 1. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun? 2. Why do fish die in water that is too warm? 15

16. Solution 1. The pressure in a bottle increases as the gas leaves solution when it becomes less soluble at higher temperatures. As pressure increases, the bottle could burst. 2. Because O 2 gas is less soluble in warm water, fish cannot obtain the amount of O 2 required for their survival. 16

17. Soluble vs. Insoluble Salts Not all ionic compounds, salts, are soluble in water. 17

18. Soluble vs. Insoluble Salts Mixing certain aqueous solutions produces insoluble salts. Barium sulfate, BaSO 4, an insoluble salt, is used to enhance X-rays. 18

19. Chapter 7 Solutions 7.4 Concentration of a Solution 19

20. Solution Concentration The concentration of a solution is expressed as amount of solute x amount of solution Units of concentration include: Mass percent (m/m) Volume percent (v/v) Mass/volume percent (m/v) Molarity (moles solute/liters solution) 20

21. Mass Percent Mass percent (% m/m) is the concentration by mass of solute in a solution mass percent = g of solute  100 g of solute + g of solvent amount in g of solute in 100 g of solution (conversion factor for mass percent) mass percent = g of solute x 100 g of solution 21

22. Mass of Solute: Mass Solution 22

23. Calculating Mass Percent The calculation of mass percent (% m/m) requires the grams of solute (g KCl) and grams of solution (g KCl + g water) or g of KCl 8.00 g g of solvent (water) 42.00 g g of KCl solution = 50.00 g 8.00 g KCl (solute)  100 = 16.0% (m/m) 50.00 g KCl solution 23

24. Learning Check A solution is prepared by mixing 15.0 g of Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (% m/m) of the solution. A. 15.0% (m/m) Na 2 CO 3 B. 6.38% (m/m) Na 2 CO 3 C. 6.00% (m/m) Na 2 CO 3 24

25. Molarity Molarity (moles of solute/liter of solution) is moles of solute per volume (L) of solution. M = moles solute = 1.0 mole NaCl liter of solution 1 L solution = 1.0 M NaCl solution 25

26. Preparing a 1.0 M Solution A 1.0 M NaCl solution is prepared by weighing out 58.5 g of NaCl (1.0 mole) and adding water to make 1.00 liter of solution 26

27. Molarity Calculations What is the molarity of 0.500 L of NaOH solution if it contains 6.00 g of NaOH? Step 1 Determine the quantities of solute and solution. Solute: 1 mole of NaOH = 40.0 grams of NaOH 1 mole NaOH and 40.0 g NaOH x 40.0 g NaOH1 mole NaOH 6.00 g NaOH  1 mole NaOH = 0.150 mole of NaOH 40.0 g NaOH The volume of the solution is 0.500 L. 27

28. Molarity Calculations What is the molarity of 0.500 L of NaOH solution if it contains 6.00 g of NaOH? Step 2 Write the concentration expression. M = moles solute L solution Step 3 Substitute solute and solution quantities into the expression. M = moles solute = 0.150 mole NaOH = 0.300 M NaOH L solution 0.500 L solution 28

29. Learning Check What is the molarity of 0.225 L of a KNO 3 solution containing 34.8 g of KNO 3 ? A. 0.344 M B. 1.53 M C. 15.5 M 29

30. Solution What is the molarity of 0.225 L of a KNO 3 solution containing 34.8 g of KNO 3 ? Step 1 Determine the quantities of solute and solution. Solute: 1 mole of KNO 3 = 101.1 grams of KNO 3 1 mole KNO 3 and 101.1 g KNO 3 101.1 g KNO 3 1 mole KNO 3 34.8 g KNO 3  1 mole KNO 3 = 0.344 mole of KNO 3 101.1 g KNO 3 The volume of the solution is 0.225 L. 30

31. Solution Step 2 Write the concentration expression. M = moles solute L solution Step 3 Substitute solute and solution quantities into the expression. M = moles solute = 0.344 mole of KNO 3 = 1.53 M KNO 3 L solution 0.225 L solution The answer is B, 1.53 M KNO 3. 31

32. Molarity as a Conversion Factor The units of molarity are used as conversion factors in calculations with solutions. MolarityEquality 3.5 M HCl1 L = 3.5 moles of HCl Written as Conversion Factors 3.5 moles HCl and 1 L 1 L 3.5 moles HCl 32

33. Learning Check How many grams of AlCl 3 are needed to prepare 125 mL of a 0.150 M solution? A. 20.0 g of AlCl 3 B. 16.7 g of AlCl 3 C. 2.50 g of AlCl 3 33

34. Solution How many grams of AlCl 3 are needed to prepare 125 mL of a 0.150 M solution? 0.125 L  0.150 mole AlCl 3  133.5 g AlCl 3 = 2.50 g of AlCl 3 1 L 1 mole AlCl 3 molarity molar mass conversion factor The answer is C, 2.50 g of AlCl 3. 34

35. Chapter 7 Solutions 7.5 Dilution of Solutions 35 When water is added to a concentrated solution, there is no change in the number of particles. However, the solute particles spread out as the volume of the diluted solution increases.

36. Dilution In a dilution, water is added volume increases concentration decreases mass of solute remains the same 36

37. Dilution 37

38. Solute Concentration in Initial Diluted Solutions In the initial and diluted solution, the moles of solute are the same the concentrations and volumes are related by the following equations: for percent concentration: C 1 V 1 = C 2 V 2 initial diluted for molarity: M 1 V 1 = M 2 V 2 initial diluted 38

39. Guide to Calculating Dilution Quantities 39

40. Dilution: Molarity What is the final concentration when 0.50 L of 6.0 M HCl solution is diluted to a final volume of 1.0 L? Step 1 Prepare a table of the concentrations and volumes of the solutions. M 1 = 6.0 MV 1 = 0.50 L M 2 = ? MV 2 = 1.0 L 40

41. Dilution: Molarity What is the final concentration when 0.50 L of 6.0 M HCl solution is diluted to a final volume of 1.0 L? Step 2 Rearrange the dilution expression to solve for the unknown quantity. M 1  V 1 = M 2 V 2 Step 3 Substitute the known quantities into the dilution expression and solve. 6.0 M  0.50 L = 3.0 M HCl 1.0 L 41

42. Learning Check What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution? 42

43. Solution What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution? V 2 = (25.0 mL)(14.0%) = 175 mL of solution 2.00% 43

44. Learning Check What is the molarity (M) of a solution prepared by diluting 0.180 L of 0.600 M HNO 3 to 0.540 L? 44

45. Solution What is the molarity (M) of a solution prepared by diluting 0.180 L of 0.600 M HNO 3 to 0.540 L? Step 1 Prepare a table of the concentrations and volumes of the solutions. M 1 = 0.600 MV 1 = 0.180 L M 2 = ?V 2 = 0.540 L Step 2 Rearrange the dilution expression to solve for the unknown quantity. M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 V 2 45

46. Solution What is the molarity (M) of a solution prepared by diluting 0.180 L of 0.600 M HNO 3 to 0.540 L? Step 3 Substitute the known quantities into the dilution expression and solve. M 2 = (0.600 M)(0.180 L) = 0.200 M solution 0.540 L 46

47. Chapter 7 Solutions 7.6 Properties of Solutions 47 Ethylene glycol is added to a radiator to form an aqueous solution that has a lower freezing point and a higher boiling point than water.

48. Solutions are transparent do not separate contain small particles, ions or molecules that cannot be filtered and pass through semipermeable membranes 48

49. Colloids have medium-size particles cannot be filtered can be separated by semipermeable membranes 49

50. Suspensions have very large particles settle out can be filtered must be stirred to stay suspended Examples include blood platelets, muddy water, and calamine lotion. 50

51. Solutions, Colloids, and Suspensions 51 Properties of different types of mixtures: (a) suspensions settle out; (b) suspensions are separated by a filter; (c) solution particles go through a semipermeable membrane, but colloids and suspensions do not.

52. Colligative Properties Colligative properties depend on the number of solute particles in solution include freezing point lowering include boiling point elevation The solute particles disrupt the formation of solid crystals, lowering the freezing point of the solvent. 52 Ethylene glycol is added to a radiator to form an aqueous solution that has a lower freezing point and a higher boiling point than water.

53. Colligative Properties and Electrolytes Electrolytes break up into ions in solution, increasing the number of particles include freezing point lowering include boiling point elevation 1 mole of CaCl 2 = 3 moles of ions 53

54. Osmosis In osmosis, water (solvent) flows from the lower solute concentration into the higher solute concentration the level of the solution with the higher solute concentration rises the concentrations of the two solutions become equal with time 54 Water flows into the solution with a higher solute concentration until the flow of water becomes equal in both directions.

55. Osmotic Pressure Osmotic pressure is produced by the solute particles dissolved in a solution equal to the pressure that would prevent the flow of additional water into the more concentrated solution greater as the number of dissolved particles in the solution increases 55

56. Osmotic Pressure in Blood Red blood cells have cell walls that are semipermeable membranes maintain an osmotic pressure that cannot change or damage occurs must maintain an equal flow of water between the red blood cell and its surrounding environment 56

57. Isotonic Solutions An isotonic solution exerts the same osmotic pressure as body fluids such as red blood cells, RBCs of 5.0% (m/v) glucose or 0.90% (m/v) NaCl are typical isotonic solutions 57 (a) In an isotonic solution, a red blood cell retains its normal volume.

58. Hypotonic Solution A hypotonic solution has a lower solute concentration than red blood cells means water flows into cells by osmosis The increase in fluid causes the cells to swell and burst, a condition called hemolysis. 58 (b) Hemolysis: In a hypotonic solution, water flows into a red blood cell, causing it to swell and burst.

59. Hypertonic Solution A hypertonic solution has a higher solute concentration than RBCs water goes out of the cells by osmosis causes crenation: RBCs shrink in size 59 (c) Crenation: In a hypertonic solution, water leaves the red blood cell, causing it to shrink.

60. Dialysis In dialysis, solvent and small solute particles pass through an artificial membrane large particles are retained inside waste particles such as urea from blood are removed using hemodialysis (artificial kidney) 60 (a) In an isotonic solution, a red blood cell retains its normal volume. (b) Hemolysis: In a hypotonic solution, water flows into a red blood cell, causing it to swell and burst. (c) Crenation: In a hypertonic solution, water leaves the red blood cell, causing it to shrink.