2. 1 IONIC COMPOUNDS Compounds in Aqueous Solution Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO 4 in water K + (aq) + MnO 4 - (aq)

3. 2 An Ionic Compound, CuCl 2, in Water CCR, page 149

4. 3 How do we know ions are present in aqueous solutions? The solutions conduct electricity! They are called ELECTROLYTES HCl, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Aqueous Solutions

5. 4 HCl, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Aqueous Solutions

6. 5 Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH 3 CO 2 H(aq) ---> CH 3 CO 2 - (aq) + H + (aq)

7. 6 Aqueous Solutions Acetic acid ionizes only to a small extent, so it is a weak electrolyte. CH 3 CO 2 H(aq) ---> CH 3 CO 2 - (aq) + H + (aq)

8. 7 Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugarethanol ethylene glycol Examples include: sugarethanol ethylene glycol

9. 8 If one ion from the “Soluble Compd.” list is present in a compound, the compound is water soluble. Water Solubility of Ionic Compounds Screen 5.4 & Figure 5.1

10. 9 Water Solubility of Ionic Compounds Common minerals are often formed with anions that lead to insolubility: sulfidefluoride sulfidefluoride carbonateoxide carbonateoxide Azurite, a copper carbonate Iron pyrite, a sulfide Orpiment, arsenic sulfide

11. 10 An acid -------> H + in water ACIDSACIDS Some strong acids are HClhydrochloric H 2 SO 4 sulfuric HClO 4 perchloric HNO 3 nitric HNO 3

12. 11 An acid -------> H + in water ACIDSACIDS HCl(aq) ---> H + (aq) + Cl - (aq)

13. 12 The Nature of Acids HCl H 2 O Cl - H 3 O + hydronium ion

14. 13 Weak Acids WEAK ACIDS = weak electrolytes CH 3 CO 2 H acetic acid H 2 CO 3 carbonic acid H 3 PO 4 phosphoric acid HF hydrofluoric acid Acetic acid

15. 14 ACIDSACIDS Nonmetal oxides can be acids CO 2 (aq) + H 2 O(liq) ---> H 2 CO 3 (aq) SO 3 (aq) + H 2 O(liq) ---> H 2 SO 4 (aq) and can come from burning coal and oil.

16. 15 Base ---> OH - in water BASES see Screen 5.9 and Table 5.2 NaOH(aq) ---> Na + (aq) + OH - (aq) NaOH is a strong base

17. 16 Ammonia, NH 3 Ammonia, NH 3 An Important Base

18. 17 BASESBASES Metal oxides are bases CaO(s) + H 2 O(liq) --> Ca(OH) 2 (aq) --> Ca(OH) 2 (aq) CaO in water. Indicator shows solution is basic.

19. 18 Know the strong acids & bases!

20. 19 Net Ionic Equations Mg(s) + 2 HCl(aq) --> H 2 (g) + MgCl 2 (aq) We really should write Mg(s) + 2 H + (aq) + 2 Cl - (aq) ---> H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) The two Cl - ions are SPECTATOR IONS — they do not participate. Could have used NO 3 -.

21. 20 Net Ionic Equations Mg(s) + 2 HCl(aq) --> H 2 (g) + MgCl 2 (aq) Mg(s) + 2 H + (aq) + 2 Cl - (aq) ---> H 2 (g) + Mg 2+ (aq) + 2 Cl - (aq) We leave the spectator ions out — Mg(s) + 2 H + (aq) ---> H 2 (g) + Mg 2+ (aq) NET IONIC EQUATION to give the NET IONIC EQUATION

22. 21 Chemical Reactions in Water Sections 5.2 & 5.4-5.6—CD-ROM Ch. 5 We will look at EXCHANGE REACTIONS Pb(NO 3 ) 2 (aq) + 2 KI(aq) ----> PbI 2 (s) + 2 KNO 3 (aq) The anions exchange places between cations.

23. 22 Precipitation Reactions The “driving force” is the formation of an insoluble compound — a precipitate. Pb(NO 3 ) 2 (aq) + 2 KI(aq) -----> 2 KNO 3 (aq) + PbI 2 (s) Net ionic equation Pb 2+ (aq) + 2 I - (aq) ---> PbI 2 (s)

24. 23 Acid-Base Reactions The “driving force” is the formation of water.The “driving force” is the formation of water. NaOH(aq) + HCl(aq) ---> NaCl(aq) + H 2 O(liq) Net ionic equationNet ionic equation OH - (aq) + H + (aq) ---> H 2 O(liq) OH - (aq) + H + (aq) ---> H 2 O(liq) This applies to ALL reactions of STRONG acids and bases.This applies to ALL reactions of STRONG acids and bases.

25. 24 Acid-Base Reactions CCR, page 162

26. 25 Acid-Base Reactions A-B reactions are sometimes called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end.A-B reactions are sometimes called NEUTRALIZATIONS because the solution is neither acidic nor basic at the end. The other product of the A-B reaction is a SALT, MX.The other product of the A-B reaction is a SALT, MX. HX + MOH ---> MX + H 2 O HX + MOH ---> MX + H 2 O M n+ comes from base & X n- comes from acid M n+ comes from base & X n- comes from acid This is one way to make compounds! This is one way to make compounds!

27. 26 Gas-Forming Reactions This is primarily the chemistry of metal carbonates. CO 2 and water ---> H 2 CO 3 H 2 CO 3 (aq) + Ca 2+ ---> H 2 CO 3 (aq) + Ca 2+ ---> 2 H + (aq) + CaCO 3 (s) (limestone) 2 H + (aq) + CaCO 3 (s) (limestone) Adding acid reverses this reaction. MCO 3 + acid ---> CO 2 + salt

28. 27 Gas-Forming Reactions CaCO 3 (s) + H 2 SO 4 (aq) ---> 2 CaSO 4 (s) + H 2 CO 3 (aq) 2 CaSO 4 (s) + H 2 CO 3 (aq) Carbonic acid is unstable and forms CO 2 & H 2 O H 2 CO 3 (aq) ---> CO 2 + water (Antacid tablet has citric acid + NaHCO 3 )

29. 28

30. 29 Quantitative Aspects of Reactions in Solution Sections 5.8-5.10

31. 30 TerminologyTerminology In solution we need to define the - SOLVENTSOLVENT the component whose physical state is preserved when solution forms SOLUTESOLUTE the other solution component

32. 31 Concentration of Solute The amount of solute in a solution is given by its concentration The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution Concentration (M) = [ …]

33. 32 1.0 L of water was used to make 1.0 L of solution. Notice the water left over. CCR, page 177

34. 33 PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = 0.0841 M

35. 34 The Nature of a CuCl 2 Solution Ion Concentrations CuCl 2 (aq) --> Cu 2+ (aq) + 2 Cl - (aq) If [CuCl 2 ] = 0.30 M, then [Cu 2+ ] = 0.30 M [Cl - ] = 2 x 0.30 M

36. 35 USING MOLARITY What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution? Because Conc (M) = moles/volume = mol/V this means that moles = MV

37. 36 Step 1: Calculate moles of acid required. (0.0500 mol/L)(0.250 L) = 0.0125 mol Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = 1.13 g USING MOLARITY moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution?

38. 37 Preparing Solutions Weigh out a solid solute and dissolve in a given quantity of solvent.Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.Dilute a concentrated solution to give one that is less concentrated.

39. 38 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!

40. 39 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?

41. 40 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

42. 41 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 = 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)(1 L/0.50 mol) = 0.30 L or 300 mL

43. 42 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.

44. 43 A shortcut A shortcut C initial V initial = C final V final Preparing Solutions by Dilution

45. 44 pH, a Concentration Scale pH: a way to express acidity -- the concentration of H + in solution. Low pH: high [H + ] High pH: low [H + ] Acidic solutionpH < 7 Neutral pH = 7 Neutral pH = 7 Basic solution pH > 7 Basic solution pH > 7 Acidic solutionpH < 7 Neutral pH = 7 Neutral pH = 7 Basic solution pH > 7 Basic solution pH > 7

46. 45 The pH Scale pH = log (1/ [H + ]) = - log [H + ] In a neutral solution, [H + ] = [OH - ] = 1.00 x 10 -7 M at 25 o C pH = - log [H + ] = -log (1.00 x 10 -7 ) = - (-7) = 7 See CD Screen 5.17 for a tutorial

47. 46 [H + ] and pH If the [H + ] of soda is 1.6 x 10 -3 M, the pH is ____? Because pH = - log [H + ] then pH= - log (1.6 x 10 -3 ) pH= - log (1.6 x 10 -3 ) pH = - (-2.80) pH = 2.80

48. 47 pH and [H + ] If the pH of Coke is 3.12, it is ____________. Because pH = - log [H + ] then log [H + ] = - pH log [H + ] = - pH Take antilog and get [H + ] = 10 -pH [H + ] = 10 -3.12 = 7.6 x 10 -4 M

49. 48 Zinc reacts with acids to produce H 2 gas.Zinc reacts with acids to produce H 2 gas. Have 10.0 g of ZnHave 10.0 g of Zn What volume of 2.50 M HCl is needed to convert the Zn completely? What volume of 2.50 M HCl is needed to convert the Zn completely? SOLUTION STOICHIOMETRY Section 5.10

50. 49 GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS Mass zinc Stoichiometric factor Moles zinc Moles HCl Mass HCl Volume HCl

51. 50 Step 1: Write the balanced equation Zn(s) + 2 HCl(aq) --> ZnCl 2 (aq) + H 2 (g) Step 2: Calculate amount of Zn Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 3: Use the stoichiometric factor

52. 51 Step 3: Use the stoichiometric factor Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 4: Calculate volume of HCl req’d

53. 52 ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

54. 53 Setup for titrating an acid with a base CCR, page 186

55. 54 TitrationTitration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H + + OH - --> H 2 O H + + OH - --> H 2 O 5. At equivalence point moles H + = moles OH - moles H + = moles OH -

56. 55 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentra-tion of the NaOH? 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentra-tion of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

57. 56 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amount of H 2 C 2 O 4 Step 2: Calculate amount of NaOH req’d

58. 57 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amount of H 2 C 2 O 4 = 0.0118 mol acid = 0.0118 mol acid Step 2: Calculate amount of NaOH req’d = 0.0236 mol NaOH = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH [NaOH] = 0.663 M

59. 58 LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown. Apples contain malic acid, C 4 H 6 O 5. C 4 H 6 O 5 (aq) + 2 NaOH(aq) ---> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(liq) Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(liq) 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

60. 59 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1: Calculate amount of NaOH used. C V = (0.663 M)(0.03456 L) = 0.0229 mol NaOH = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated. = 0.0115 mol acid

61. 60 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 3: Calculate mass of acid titrated. Step 1: Calculate amount of NaOH used. = 0.0229 mol NaOH = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated = 0.0115 mol acid = 0.0115 mol acid

62. 61 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1: Calculate amount of NaOH used. = 0.0229 mol NaOH = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated = 0.0115 mol acid = 0.0115 mol acid Step 3: Calculate mass of acid titrated. = 1.54 g acid = 1.54 g acid Step 4: Calculate % malic acid.

63. 62 Oxidation-Reduction Reactions Section 5.7 Thermite reaction Fe 2 O 3 (s) + 2 Al(s) ----> ----> 2 Fe(s) + Al 2 O 3 (s)

64. 63 REDOX REACTIONS EXCHANGEAcid-BaseReactionsEXCHANGEGas-FormingReactions EXCHANGE: Precipitation Reactions REACTIONS

65. 64 REDOX REACTIONS Oxidation— 2 H 2 (g) + O 2 (g) ---> 2 H 2 O(liq) Mg(s) + 2 HCl(aq) --> MgCl 2 (aq) + H 2 (g) All corrosion reactions are oxidations. 2 Al(s) + 3 Cu 2+ (aq) ---> 2 Al 3+ (aq) + 3 Cu(s) Reduction— Fe 2 O 3 (s) + 2 Al(s) --> 2 Fe(s) + Al 2 O 3 (s)

66. 65 Cu(s) + 2 Ag + (aq) ---> Cu 2+ (aq) + 2 Ag(s) In all reactions if something has been oxidized then something has also been reduced REDOX REACTIONS

67. 66 Why Study Redox Reactions Manufacturing metals FuelsFuels CorrosionCorrosion BatteriesBatteries

68. 67 Redox reactions are characterized by ELECTRON TRANSFER between an electron donor and electron acceptor. Transfer leads to— 1. increase in oxidation number of some element = OXIDATION 2.decrease in oxidation number of some element = REDUCTION REDOX REACTIONS

69. 68 OXIDATION NUMBERS The electric charge an element APPEARS to have when electrons are counted by some arbitrary rules: 1.Each atom in free element has ox. no. = 0. Zn O 2 I 2 S 8 Zn O 2 I 2 S 8 2.In simple ions, ox. no. = charge on ion. -1 for Cl - +2 for Mg 2+ -1 for Cl - +2 for Mg 2+

70. 69 OXIDATION NUMBERS 3.O has ox. no. = -2 (except in peroxides: in H 2 O 2, O = -1) 4.Ox. no. of H = +1 (except when H is associated with a metal as in NaH where it is -1) 5.Algebraic sum of oxidation numbers = 0 for a compound = 0 for a compound = overall charge for an ion = overall charge for an ion

71. 70 OXIDATION NUMBERS NH 3 N = ClO - Cl = H 3 PO 4 P = MnO 4 - Mn = Cr 2 O 7 2- Cr = C 3 H 8 C = Oxidation number of F in HF?

72. 71 Recognizing a Redox Reaction Corrosion of aluminum 2 Al(s) + 3 Cu 2+ (aq) --> 2 Al 3+ (aq) + 3 Cu(s) Al(s) --> Al 3+ (aq) + 3 e - Ox. no. of Al increases as e - are donated by the metal.Ox. no. of Al increases as e - are donated by the metal. Therefore, Al is OXIDIZEDTherefore, Al is OXIDIZED Al is the REDUCING AGENT in this balanced half- reaction.Al is the REDUCING AGENT in this balanced half- reaction.

73. 72 Recognizing a Redox Reaction Corrosion of aluminum 2 Al(s) + 3 Cu 2+ (aq) --> 2 Al 3+ (aq) + 3 Cu(s) Cu 2+ (aq) + 2 e - --> Cu(s) Ox. no. of Cu decreases as e - are accepted by the ion.Ox. no. of Cu decreases as e - are accepted by the ion. Therefore, Cu is REDUCEDTherefore, Cu is REDUCED Cu is the OXIDIZING AGENT in this balanced half- reaction.Cu is the OXIDIZING AGENT in this balanced half- reaction.

74. 73 Recognizing a Redox Reaction Notice that the 2 half-reactions add up to give the overall reaction —if we use 2 mol of Al and 3 mol of Cu 2+. 2 Al(s) --> 2 Al 3+ (aq) + 6 e - 2 Al(s) --> 2 Al 3+ (aq) + 6 e - 3 Cu 2+ (aq) + 6 e - --> 3 Cu(s) 3 Cu 2+ (aq) + 6 e - --> 3 Cu(s)----------------------------------------------------------- 2 Al(s) + 3 Cu 2+ (aq) ---> 2 Al 3+ (aq) + 3 Cu(s) Final eqn. is balanced for mass and charge.

75. 74 Common Oxidizing and Reducing Agents See Table 5.4 HNO 3 is an oxidizing agent 2 K + 2 H 2 O --> 2 KOH + H 2 Metals (Na, K, Mg, Fe) are reducing agents Metals (Cu) are reducing agents Cu + HNO 3 --> Cu 2+ + NO 2

76. 75 Recognizing a Redox Reaction See Table 5.4 In terms of oxygengainloss In terms of halogengainloss In terms of electronslossgain Reaction Type OxidationReduction

77. 76 Examples of Redox Reactions Metal + halogen 2 Al + 3 Br 2 ---> Al 2 Br 6

78. 77 Examples of Redox Reactions Metal (Mg) + Oxygen Nonmetal (P) + Oxygen

79. 78 Examples of Redox Reactions Metal + acid Mg + HCl Mg = reducing agent H + = oxidizing agent Metal + acid Cu + HNO 3 Cu = reducing agent HNO 3 = oxidizing agent