1. Molarity, Dilution, and pH Main Idea: Solution concentrations are measured in molarity. Dilution is a useful technique for creating a new solution from a stock solution. pH is a measure of the concentration of hydronium ions in a solution. 1

2. Properties of Aqueous Solutions Solution- a homogeneous mixture of two or more substances. Solute- a substance in a solution that is present in the smallest amount. Solvent- a substance in a solution that is present in the largest amount. In an aqueous solution, the solute is a liquid or solid and the solvent is always water.

3. Molarity Review One of the most common units of solution concentration is molarity. Molarity (M) is the number of moles of solute per liter of solution. Molarity is also known as molar concentration, and the unit M is read as “molar.” A liter of solution containing 1 mol of solute is a 1M solution, which is read as a “one-molar” solution. A liter of solution containing 0.1 mol of solute is a 0.1 M solution. 3

4. Molarity Equation To calculate a solution’s molarity, you must know the volume of the solution in liters and the amount of dissolved solute in moles. Molarity (M) = moles of solute liters of solution 4

5. Molarity Example A 100.5-mL intravenous (IV) solution contains 5.10 g of glucose (C 6 H 12 O 6 ). What is the molarity of the solution? The molar mass of glucose is 180.16 g/mol. SOLUTION: 1)Calculate the number of moles of C 6 H 12 O 6 by dividing mass over molar mass = 0.0283 mol C 6 H 12 O 6 2)Convert the volume of H 2 O to liters by dividing volume by 1000 = 0.1005 L 3)Solve for molarity by dividing moles by liters = 0.282 M 5

6. Preparing Molar Solutions Now that you know how to calculate the molarity of a solution, how would you prepare one in the laboratory? STEP 1: Calculate the mass of the solute needed using the molarity definition and accounting for the desired concentration and volume. STEP 2: The mass of the solute is measured on a balance. STEP 3: The solute is placed in a volumetric flask of the correct volume. STEP 4: Distilled water is added to the flask to bring the solution level up to the calibration mark. 6

7. http://www.ltcconline.net/stevenson/2008CHM101Fall/CHM101LectureNotes20081022.htm 7

8. Properties of Aqueous Solutions All solutes that dissolve in water fit into one of two categories: electrolyte or non-electrolyte. Electrolyte- a substance that when dissolved in water conducts electricity Non-electrolyte- a substance that when dissolved in water does not conduct electricity. To have an electrolyte, ions must be present in water.

9. Electrolytic Properties of Aqueous Solutions NaCl in water. – What happens? – NaCl (s) → Na + (aq) + Cl – (aq) – Completely dissociates

10. Strong vs. Weak Electrolytes How do you know when an electrolyte is strong or weak? Take a look at how HCl dissociates in water. – HCl (s) → H + (aq) + Cl – (aq)

11. Electrolytic Properties of Aqueous Solutions

13. Hydrated Ions

14. Electrolytic Properties of Aqueous Solutions What about weak electrolytes? What makes them weak? – Ionization of acetic acid CH 3 COOH (aq) ↔ CH 3 COO – (aq) + H + (aq)

15. Electrolytic Solutions

16. Precipitation Reactions Precipitation Reaction- a reaction that results in the formation of an insoluble product. These reactions usually involve ionic compounds. Formation of PbI 2 : – Pb(NO 3 ) 2 (aq) + 2KI (aq) → PbI 2 (s) + 2KNO 3 (aq)

17. Preciptate

18. Precipitate

19. Precipitation Reactions How do you know whether or not a precipitate will form when a compound is added to a solution? By knowing the solubility of the solute! Solubility- The maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature. Three levels of solubility: Soluble, slightly soluble or insoluble.

20. Precipitation Reactions

21. Determining Solubility Determine the solubility for the following: (1) Ag 2 SO 4 (2) CaCO 3 (3) Na 3 PO 4

22. Diluting Molar Solutions In the laboratory, you might use concentrated solutions of standard molarities, called stock solutions. – For example, concentrated hydrochloric acid (HCl) is 12 M. You can prepare a less-concentrated solution by diluting the stock solution with additional solvent. – Dilution is used when a specific concentration is needed and the starting material is already in the form of a solution (i.e., acids). 22

23. Dilution of Solutions When you want to dilute a solution, what happens to the number of moles present in the solution? – Do they increase? – Decrease? – Stay the same?

24. Dilution of Solutions

25. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution! 25

26. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add? 26

27. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do ? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution 27

28. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = = M V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH) / (0.50 M) = 0.30 L or or 300 mL 28

29. PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH. 29

30. A shortcut A shortcut M 1 V 1 = M 2 V 2 Where M represents molarity and V represents volume. The 1s are for the stock solution and the 2s are for the solution you are trying to create. Preparing Solutions by Dilution 30

31. What factors affect the rate of dissolving? 31

32. What factors affect the rate of dissolving? Temperature You can dissolve more into a warm liquid than you can into a cold liquid 32

33. What factors affect the rate of dissolving? Temperature You can dissolve more into a warm liquid than you can into a cold liquid Surface area Which dissolves faster, a cube of sugar or grains of sugar? Why? 33

34. What factors affect the rate of dissolving? Temperature You can dissolve more into a warm liquid than you can into a cold liquid Surface area Which dissolves faster, a cube of sugar or grains of sugar? Why? Concentration The more solute already dissolved in a solvent, the slower the rate of dissolving. 34

35. What factors affect the rate of dissolving? Pressure What affect would increasing pressure have on the rate of dissolving? Why? 35

36. What factors affect the rate of dissolving? Pressure What affect would increasing pressure have on the rate of dissolving? Why? Mixing Describe what happens when you mix a solution. 36

37. How do you get from this…

38. …to this?

39. Add an ionic compound!

40. Colligative Properties Properties that depend only on the number of solute particles and not on their identity.

41. Some Colligative Properties are: Vapor pressure lowering Boiling point elevation Freezing Point depression

42. Vapor Pressure

43. Vapor Pressure Lowering The particles of solute are surrounded by and attracted to particles of solvent. Now the solvent particles have less kinetic energy and tend less to escape into the space above the liquid. So the vapor pressure is less.

44. Ionic vs Molecular Solutes Ionic solutes produce two or more ion particles in solution. They affect the colligative properties proportionately more than molecular solutes (that do not ionize). The effect is proportional to the number of particles of the solute in the solution.

45. How many particles do each of the following give upon solvation? NaCl CaCl 2 Glucose

46. Freezing Point Depression

47. Example Salt is added to melt ice by reducing the freezing point of water.

48. Boiling Point Elevation

49. Example Addition of ethylene glycol C 2 H 6 O 2 (antifreeze) to car radiators.

50. Ready for a test soon?!?!?! It’s review time! In this unit we studied SOLUTIONS! 50

51. Freezing Point Depression and Boiling Point Elevation Boiling Point Elevation ∆T b =mk b (for water k b =0.51 o C/m) Freezing Point Depression ∆T f =mk f (for water k f =1.86 o C/m) Note: m is the molality of the particles, so if the solute is ionic, multiply by the #of particles it dissociates to.

52. Which is more effective for lowering the freezing point of water? NaCl or CaCl 2

53. Example 1: Find the new freezing point of 3m NaCl in water.

54. Example 2: Find the new boiling point of 3m NaCl in water.

55. The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H + (or OH - ) ion. Under 7 = acid 7 = neutral Over 7 = base 55

56. Acid-Base Reactions Acids- generally have a sour taste, change litmus from blue to red, can react with certain metals to produce gas, conduct electricity. Bases- generally have a bitter taste, change litmus from red to blue, feel slippery, conduct electricity. Br Ø nstead Acid- proton donor Br Ø nstead Base- proton acceptor

57. Acid-Base Reactions Acid or Base? – HCl (aq) + H 2 O (l) → H 3 O + (aq) + Cl – (aq) – NH 3 (aq) + H 2 O (l) → NH 4 + (aq) + OH – (aq)

58. Acid-Base Reactions Look at the following compounds and decide whether they are a Br Ø nstead Acid or a Br Ø nstead Base. – HBr – NO 2 – – HCO 3 –

59. 59 Arrhenius definition of acids and bases: Acids are compounds that give off H + ions (also called hydronium ions, H 3 O + ions, or simply “protons”) when you dissolve them in water. Bases are compounds that give off OH - (hydroxide) ions when you dissolve them in water.

60. Arrhenius acids almost always start with the letter “H” in their formulas – this is the source of the H + ion that comes off when you dissolve the compound. Common examples include the following: HNO 3(l)  H + (aq) + NO 3(aq) - HCl HBr H 2 SO 4 Arrhenius bases always have “OH” in their formulas, indicating the presence of the hydroxide ion. Common examples include: NaOH (s)  Na +1 (aq) + OH -1 (aq) KOH Mg(OH) 2 60

61. Acid-Base Reactions

62. pH of Common Substances 62

63. pH Value H + Concentration Relative to Pure Water Example 010 000 000battery acid 11 000 000concentrated sulfuric acid 2100 000lemon juice, vinegar 310 000orange juice, soda 41 000tomato juice, acid rain 5100black coffee, bananas 610urine, milk 71pure water 80.1sea water, eggs 90.01baking soda 100.001Great Salt Lake, milk of magnesia 110.000 1ammonia solution 120.000 01soapy water 130.000 001bleach, oven cleaner 140.000 000 1liquid drain cleaner 63

64. Put the following substances in order, from lowest pH to highest pH: Drain Cleaner Water Vinegar Soap Orange Juice Baking Soda Battery Acid 64

65. Put the following substances in order, from lowest pH to highest pH: Battery AcidLowest pH (most Acidic) Vinegar Orange Juice Water Baking Soda Soap Drain CleanerHighest pH (most Basic) 65

66. Calculating the pH pH = - log [H+] (The [ ] means Molarity) Example: If [H + ] = 1 X 10 -10 pH = - log 1 X 10 -10 pH = - (- 10) pH = 10 Example: If [H + ] = 1.8 X 10 -5 pH = - log 1.8 X 10 -5 pH = - (- 4.74) pH = 4.74 66

67. pH calculations – Solving for H+ If the pH of Coke is 3.12, [H + ] = ??? Because pH = - log [H + ] then - pH = log [H + ] - pH = log [H + ] Take antilog (10 x ) of both sides and get 10 -pH = [H + ] [H + ] = 10 -3.12 = 7.6 x 10 -4 M *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button 67

68. pH calculations – Solving for H+ A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution? A solution has a pH of 8.5. What is the Molarity of hydrogen ions in the solution? pH = - log [H + ] 8.5 = - log [H + ] -8.5 = log [H + ] Antilog -8.5 = antilog (log [H + ]) 10 -8.5 = [H + ] 3.16 X 10 -9 = [H + ] pH = - log [H + ] 8.5 = - log [H + ] -8.5 = log [H + ] Antilog -8.5 = antilog (log [H + ]) 10 -8.5 = [H + ] 3.16 X 10 -9 = [H + ] 68

69. More About Water H 2 O can function as both an ACID and a BASE…..water is AMPHOTERIC In pure water there can be AUTOIONIZATION Equilibrium constant for water = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C

70. More About Water K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so K w = [H 3 O + ] 2 = [OH - ] 2 and so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M Autoionization 70

71. pOH Since acids and bases are opposites, pH and pOH are opposites! Since acids and bases are opposites, pH and pOH are opposites! pOH does not really exist, but it is useful for changing bases to pH. pOH does not really exist, but it is useful for changing bases to pH. pOH looks at the perspective of a base pOH looks at the perspective of a base pOH = - log [OH - ] Since pH and pOH are on opposite ends, pH + pOH = 14 71

72. pH [H + ] [OH - ] pOH 72

73. [H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10 -3 M) pOH = - log 0.0010 pOH = - log 0.0010 pOH = 3 pOH = 3 pH = 14 – 3 = 11 OR K w = [H 3 O + ] [OH - ] [HO + ] = 1.0 x 10 -11 M [H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00 73

74. [OH - ] [H + ] pOH pH 10 -pOH 10 -pH -Log[H + ] Log[OH - ] -Log[OH - ] 14 - pOH 14 - pH 1.0 x 10 -14 [OH - ] [OH - ] 1.0 x 10 -14 [H + ] [H + ] 74

75. Oxidation Reduction Reactions Oxidation Reaction- refers to the half-reaction that involves the loss of electrons. Reduction Reaction- refers to the half-reaction that involves the gain of electrons. OILRIG Oxidizing agent- the compound or ion in a redox reaction that donates electrons. Reducing agent- the compound or ion in a redox reaction that accepts electrons.

76. Oxidation-Reduction Reactions

77. HOMEWORK 1)How much calcium hydroxide [Ca(OH) 2 ], in grams, is needed to produce 1.5 L of a 0.25 M solution? 2)What volume of a 3.00M KI stock solution would you use to make 0.300 L of a 1.25 M KI solution? 3)How many mL of a 5.0 M H 2 SO 4 stock solution would you need to prepare 100.0 mL of 0.25 M H 2 SO 4 ? 4)If 0.50 L of 5.00 M stock solution is diluted to make 2.0 L of solution, how much HCl, in grams, is in the solution? 77

78. HOMEWORK 5) Calculate the pH of solutions having the following ion concentrations at 298 K. a) [H + ] = 1.0 x 10 -2 Mb) [H + ] = 3.0 x 10 -6 M 6) Calculate the pH of a solution having [OH - ] = 8.2 x 10 -6 M. 7) Calculate pH and pOH for an aqueous solution containing 1.0 x 10 -3 mol of HCl dissolved in 5.0 L of solution. 8) Calculate the [H + ] and [OH - ] in a sample of seawater with a pOH = 5.60. 78