1. II III I C. Johannesson I. The Nature of Solutions (p. 401 - 410, 425 - 433) Ch. 13 & 14 - Solutions

2. C. Johannesson A. Definitions  Solution -  Solution - homogeneous mixture Solvent Solvent - present in greater amount Solute Solute - substance being dissolved

3. C. Johannesson A. Definitions Solute Solute - KMnO 4 Solvent Solvent - H 2 O

4. C. Johannesson

5. Terms C. Johannesson Suspensions – a mixture when particles in a solvent are so large that they settle out unless constantly stirred or agitated. (Ex: muddy water) Colloids – when particles that are intermediate in size between those in solutions and suspensions form mixtures

6. C. Johannesson B. Solvation  Solvation –  Solvation – the process of dissolving solute particles are separated and pulled into solution solute particles are surrounded by solvent particles

7. C. Johannesson B. Solvation Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules DISSOCIATIONIONIZATION View animation online.animation

8. C. Johannesson B. Solvation  Dissociation separation of an ionic solid into aqueous ions NaCl(s)  Na + (aq) + Cl – (aq)

9. C. Johannesson B. Solvation  Ionization breaking apart of some polar molecules into aqueous ions HNO 3 (aq) + H 2 O(l)  H 3 O + (aq) + NO 3 – (aq)

10. C. Johannesson B. Solvation  Molecular Solvation molecules stay intact C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq)

11. C. Johannesson B. Solvation NONPOLAR POLAR “Like Dissolves Like”

12. C. Johannesson B. Solvation  Soap/Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water

13. C. Johannesson C. Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

14. C. Johannesson C. Solubility  Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated soln

15. C. Johannesson C. Solubility  Solubility Curve shows the dependence of solubility on temperature

16. C. Johannesson C. Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

17. II III I C. Johannesson II. Concentration (p. 412 - 418) Ch. 13 & 14 - Solutions

18. C. Johannesson A. Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

19. C. Johannesson A. Concentration SAWS Water Quality Report - June 2000

20. C. Johannesson B. Molality mass of solvent only 1 kg water = 1 L water

21. C. Johannesson B. Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 m MgCl 2 0.25 kg water

22. C. Johannesson B. Molality  How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl 58.44 g NaCl 1 mol NaCl

23. C. Johannesson B. Molarity L of solution

24. Practice  You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? C. Johannesson

25. Practice  What volume of 3.00 M NaCl is needed for a reaction that requires 146.3g of NaCl? C. Johannesson

26. C. Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

27. C. Johannesson C. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

28. C. Johannesson D. Preparing Solutions  500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL mark 500 mL volumetric flask  1.54m NaCl in 0.500 kg of water

29. C. Johannesson D. Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Filling the volumetric flask”)

30. C. Johannesson D. Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Using your hand as a stopper”)

31. C. Johannesson D. Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!”

32. C. Johannesson Solution Preparation Lab  Turn in one paper per team.  Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution.  For each of the following solutions: 1) 1 L of 0.21M NaCl 2) 0.65m NaCl in 200.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.1M concentrate.

33. II III I C. Johannesson III. Colligative Properties (p. 436 - 446) Ch. 13 & 14 - Solutions

34. C. Johannesson A. Definition  Colligative Property property that depends on the concentration of solute particles, not their identity

35. C. Johannesson B. Types  Freezing Point Depression  Freezing Point Depression (  t f ) f.p. of a solution is lower than f.p. of the pure solvent  Boiling Point Elevation  Boiling Point Elevation (  t b ) b.p. of a solution is higher than b.p. of the pure solvent

36. C. Johannesson B. Types View Flash animation.Flash animation Freezing Point Depression

37. C. Johannesson B. Types Solute particles weaken IMF in the solvent. Boiling Point Elevation

38. C. Johannesson B. Types  Applications salting icy roads making ice cream antifreeze cars (-64°C to 136°C) fish & insects

39. C. Johannesson C. Calculations  t :change in temperature (° C ) k :constant based on the solvent (° C·kg/mol ) m :molality ( m ) n :# of particles  t = k · m · n

40. C. Johannesson C. Calculations  # of Particles Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles

41. C. Johannesson C. Calculations  At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? m = 3.2m n = 1  t b = k b · m · n WORK: m = 0.73mol ÷ 0.225kg GIVEN: b.p. = ?  t b = ? k b = 3.60°C·kg/mol  t b = (3.60°C·kg/mol)(3.2m)(1)  t b = 12°C b.p. = 181.8°C + 12°C b.p. = 194°C

42. C. Johannesson C. Calculations  Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. m = 4.8m n = 2  t f = k f · m · n WORK: m = 0.48mol ÷ 0.100kg GIVEN: f.p. = ?  t f = ? k f = 1.86°C·kg/mol  t f = (1.86°C·kg/mol)(4.8m)(2)  t f = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C

43. II III I C. Johannesson I. Introduction to Acids & Bases (p. 453 - 473) Ch. 15 & 16 - Acids & Bases

44. C. Johannesson A. Properties  electrolytes  turn litmus red  sour taste  react with metals to form H 2 gas  slippery feel  turn litmus blue  bitter taste ChemASAP  vinegar, milk, soda, apples, citrus fruits  ammonia, lye, antacid, baking soda

45. C. Johannesson B. Definitions  Arrhenius - In aqueous solution… HCl + H 2 O  H 3 O + + Cl – AcidsAcids form hydronium ions (H 3 O + ) H HHHH H Cl OO – + acid

46. C. Johannesson B. Definitions  Arrhenius - In aqueous solution… BasesBases form hydroxide ions (OH - ) NH 3 + H 2 O  NH 4 + + OH - H H H H H H N NO O – + H H H H base

47. C. Johannesson B. Definitions  Brønsted-Lowry HCl + H 2 O  Cl – + H 3 O + AcidsAcids are proton (H + ) donors. BasesBases are proton (H + ) acceptors. conjugate acid conjugate base baseacid

48. C. Johannesson B. Definitions H 2 O + HNO 3  H 3 O + + NO 3 – CBCAAB

49. C. Johannesson B. Definitions - can be an acid or a base.  Amphoteric - can be an acid or a base. NH 3 + H 2 O  NH 4 + + OH - CACBBA

50. C. Johannesson B. Definitions F - H 2 PO 4 - H2OH2O HF H 3 PO 4 H 3 O +  Give the conjugate base for each of the following: - an acid with more than one H +  Polyprotic - an acid with more than one H +

51. C. Johannesson B. Definitions Br - HSO 4 - CO 3 2- HBr H 2 SO 4 HCO 3 -  Give the conjugate acid for each of the following:

52. C. Johannesson B. Definitions  Lewis AcidsAcids are electron pair acceptors. BasesBases are electron pair donors. Lewis base Lewis acid

53. C. Johannesson C. Strength  Strong Acid/Base 100% ionized in water strong electrolyte - + HCl HNO 3 H 2 SO 4 HBr HI HClO 4 NaOH KOH Ca(OH) 2 Ba(OH) 2

54. C. Johannesson C. Strength  Weak Acid/Base does not ionize completely weak electrolyte - + HF CH 3 COOH H 3 PO 4 H 2 CO 3 HCN NH 3

55. II III I C. Johannesson Ch. 15 & 16 - Acids & Bases II. pH (p. 481 - 491)

56. C. Johannesson A. Ionization of Water H 2 O + H 2 O H 3 O + + OH - K w = [H 3 O + ][OH - ] = 1.0  10 -14

57. C. Johannesson A. Ionization of Water  Find the hydroxide ion concentration of 3.0  10 -2 M HCl. [H 3 O + ][OH - ] = 1.0  10 -14 [3.0  10 -2 ][OH - ] = 1.0  10 -14 [OH - ] = 3.3  10 -13 M Acidic or basic? Acidic

58. C. Johannesson pH = -log[H 3 O + ] B. pH Scale 0 7 INCREASING ACIDITY NEUTRAL INCREASING BASICITY 14 pouvoir hydrogène (Fr.) “hydrogen power”

59. C. Johannesson B. pH Scale pH of Common Substances

60. C. Johannesson B. pH Scale pH = -log[H 3 O + ] pOH = -log[OH - ] pH + pOH = 14

61. C. Johannesson B. pH Scale  What is the pH of 0.050 M nitric acid (HNO 3) ? pH = -log[H 3 O + ] pH = -log[0.050] pH = 1.3 Acidic or basic? Acidic

62. C. Johannesson B. pH Scale  What is the molarity of hydrobromic acid (HBr) in a solution that has a pOH of 9.6? pH + pOH = 14 pH + 9.6 = 14 pH = 4.4 Acidic pH = -log[H 3 O + ] 4.4 = -log[H 3 O + ] 10 -4.4 = [H 3 O + ] [H 3 O + ] = 4.0  10 -5 M HBr

63. II III I C. Johannesson III. Titration (p. 493 - 503) Ch. 15 & 16 - Acids & Bases

64. C. Johannesson A. Neutralization  Chemical reaction between an acid and a base.  Products are a salt (ionic compound) and water.

65. C. Johannesson A. Neutralization ACID + BASE  SALT + WATER HCl + NaOH  NaCl + H 2 O HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O Salts can be neutral, acidic, or basic. Neutralization does not mean pH = 7. weak strong neutral basic

66. C. Johannesson B. Titration  Titration Analytical method in which a standard solution is used to determine the concentration of an unknown solution. standard solution unknown solution

67. C. Johannesson  Equivalence point (endpoint) Point at which equal amounts of H 3 O + and OH - have been added. Determined by… indicator color change B. Titration dramatic change in pH

68. C. Johannesson B. Titration moles H 3 O + = moles OH - M  V  n = M  V  n M:Molarity V:volume n:# of H + ions in the acid or OH - ions in the base

69. C. Johannesson B. Titration  42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H 2 SO 4. Find the molarity of H 2 SO 4. H3O+H3O+ M = ? V = 50.0 mL n = 2 OH - M = 1.3M V = 42.5 mL n = 1 MV# = MV# M(50.0mL)(2) =(1.3M)(42.5mL)(1) M = 0.55M H 2 SO 4