2. 1.Definitions Solution - homogeneous mixture of two or more substances Aqueous - dissolved in water (aq) Anion - negatively charged ion Cation - positively charged ion

3. Electrode - conductor of electricity - electrical conductors (wires, plates) partially immersed in a solution and connected to a source of electricity (e - ) Cathode- connected to the negative pole Anode- connected to the positive pole

4. 2. Arrhenius’ theory of electrolytic dissolution –a. strong electrolyte- solute that is present in solution almost exclusively as ions; good electrical conductor (ex. KCl, HBr, ) –b. weak electrolyte- solute that is only partly ionized in solution; poor conductor of electricity (ex. Acetic Acid (vinegar)) –c. Nonelectrolyte - Solute that is present in solution almost exclusively as molecules; does not conduct electricity (ex. Sugar)

5. 3.Solubility Rules Memorize; refer to back of stations lab

6. 4. Dissolution equations Ex5.1 Write dissolution equations for: NaCl(s), Al(NO 3 ) 3 (s), Na 2 SO 4 (s), and MgCl 2 (s )

7. 1.Definitions –Double Replacement Reaction: reaction between two ionic compounds in which the cations and anions are exchanged to form two new ionic compounds –Precipitate: Insoluble ionic solid Symbolized by: (s) or 

8. – Full Ionic Equation Equation in which all compounds are written as ions REMEMBER Coefficients, Subscripts, etc.! – Net Ionic Equation “net result” of the chemical reaction Formation of solid, gas, H 2 O (l), etc. – Spectator Ions Ions that do not take part in the reaction; present before and after

9. Double Replacement Rxns The positive ions from one solution interact with the negative ions from the other solution to form: 1.A precipitate that will settle out Pb(NO 3 ) 2(aq) + 2 NaCl (aq) ---> PbCl 2(s) + 2 NaNO 3(aq) 2.A gas that will bubble out FeS (s) + 2HCl (aq) ---> H 2 S (g) + FeCl 2 (aq) 3.A molecular compound, usually water HCl (aq) + NaOH (aq) ---> NaCl (aq) + H 2 O (l)

10. What is happening in the solution? –The reactants are aqueous, which means that ionic compounds are ionized. –Individual positive and negative ions are present in the aqueous solutions KCl (s)  K + (aq) + Cl - (aq)

11. Formula Eq: AgNO 3 (aq) + NaCl (aq) ---> AgCl (s) + NaNO 3(aq) **Write the Full (Complete) Ionic Equation by separating all aqueous compounds Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq) ---> AgCl (s) + Na + (aq) + NO 3 - (aq) Net Ionic:

12. Spectator Ions – Ions that do not take part in the reaction and are present before and after the reaction

13. 2.Ex5.2 Predict whether or not the following pairs of reactants will form precipitates. Then write the double replacement reaction, the full ionic equation, and the net ionic equation and identify spectator ions. a. CuCl 2 and (NH 4 ) 2 SO 4

14. – b. Ba(NO 3 ) 2 and Na 2 CO 3

15. – c. MgCl 2 and AgNO 3

16. Ex5.3 What two aqueous solutions of soluble compounds could be mixed to produce a precipitate of CuCO 3 ?

17. 1.Arrhenius Definition of acids and bases – Acids Produces H + in soln – Bases Produces OH - in soln

18. 2.Strong acids – strong electrolytes HCl (g)  H + (aq) + Cl - (aq) – Strong Acids to Memorize: “Br I Cl SO NO ClO 4-3-4”

19. 3.Weak acids – weak electrolytes Any acid that isn’t “strong” Majority of acids HC 2 H 3 O 2 (aq)   H + (aq) + C 2 H 3 O 2 -1 (aq)

20. 4.Strong bases – strong electrolytes NaOH (s) --------> Na + (aq) + OH - (aq) Strong Bases to Memorize: (Group I, II Hydroxides, except the 1 st one in group I and the 1 st two in group II) –LiOH, NaOH, KOH, RbOH, CsOH –Ca(OH) 2, Sr(OH) 2, Ba(OH) 2

21. 5.Weak bases – weak electrolytes Any base that isn’t strong Watch for Nitrogen-containing compounds NH 3 (aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq)

22. 6.Oxides of metals and nonmetals –Oxides of nonmetals form acids Ex. CO 2 + H 2 O  H 2 CO 3 –Oxides of metals form bases Ex. CaO + H 2 O  Ca(OH) 2

24. 1.Neutralization reaction and net ionic equation –Neutralization Rxn: Reaction between an acid and a base –**Most Rxns: Acid + Base  Water + Salt Net Ionic Eqn: H + (aq) + OH - (aq)  H 2 O (l)

25. Ex5.5 Write the full neutralization reaction and net ionic equations for the interactions between HNO 3 and Ba(OH) 2, and between H 2 SO 4 and KOH.

26. 3. Acid Base Titration –Titration Lab procedure where acid and base are made to combine in their stoichiometric proportions Follow the same stoich process as in Chap 4

27. 1.Definition –Reaction in which a gas is formed, sometimes as the result of a decomposition of one of the products

28. 2. Metal and acid PURE metals react with acid to produce H 2 gas Mg (s) + HCl (aq)  MgCl 2(aq) + H 2(g)

29. 3.Metal carbonate & acid  metal salt+carbon dioxide+water MgCO 3(aq) + HCl (aq)  MgCl 2 (aq) + H 2 CO 3 (aq) H 2 CO 3 (aq)  H 2 O (ℓ) + CO 2 (g)

30. 4.Metal sulfide & acid  metal salt & hydrogen sulfide gas Na 2 S (aq) + 2HCl (aq)  2NaCl (aq) + H 2 S (g)

31. 5.Metal sulfite & acid  metal salt+sulfur dioxide gas+water Na 2 SO 3(aq) + 2HCl (aq)  2NaCl (aq) SO 2(g)+ H 2 O (ℓ)

32. 6.Ammonium salt&strng. base  metal salt+ammonia+water NH 4 Cl (aq) + NaOH (aq)  NaCl (aq) +NH 3(g) + H 2 O (ℓ)

33. 1. General Classification: S, D, SR, DR, and C 2. Redox Reaction Species lose/gain electrons as reaction occurs

34. 3. Product favored reactions – All of the preceding have been product-favored reactions Removing ions from solution to form a new product (solid, gas, water, etc.)

35. 1.Examples and categories – Uncombined Elements Fe (s) + Cu +2 (aq)  Fe +2 (aq) + Cu (s) – Combined Elements (Pb and V) PbO + V +3 + H 2 O  PbO 2 + VO + H +

36. 2. Species which lose electrons are said to be: – Oxidized 3. Species which gain electrons are said to be: – Reduced “LEO” the Lion GER!!!!!! “OIL RIG”

37. 4. Half-reaction examples Fe (s) + Cu +2 (aq)  Fe +2 (aq) + Cu (s) Fe  Fe +2 + 2e - OXIDATION Cu +2 + 2e -  Cu REDUCTION

38. 5. Rules for determining oxidation numbers a. Neutral species (isolated atom, molecule, formula unit) – The total charge is 0. – Ex. Fe = 0 Cl 2 = 0 C 6 H 12 O 6 = 0 NaCl = 0 b. Monatomic ions, Grp 1A, 2A … – +1, +2, +3, skip -3, -2, -1 (usually) c. Oxygen and Hydrogen – Oxygen most of the time -2 – Exception-Peroxides Oxidation # = -1 (O 2 -2 ) – Hydrogen most of the time +1 5.7OXIDATION-REDUCTION REACTIONS

39. – d. Sum of oxidation numbers for neutral and charged species: Neutral Species: total charge = 0 Charged Species: total charge = overall ion charge – Ex. NH 4 + = +1 NO 3 -1 = -1 PO 4 -3 = -3 Cr +3 = +3

40. Ex5.6 Assign oxidation numbers for each element in the following species: CaC 2 O 4 Cr 2 O 7 2- N 2 ON 2 O 4

41. – ClO 1- ClO 4 1- – HAsO 4 2- K 2 O – HCO 3 1- MgSO 4

42. – HIO 3 Na 2 MoO 4 – NO 2 1- NH 4 1+ – S 2 O 3 2- TeF 8 2-

43. 1. Write all reactants and products EXCEPT H + and H 2 O in the form of separate oxidation and reduction half-reactions. 2. For each half-rxn, adjust coefficients for all atoms except H and O. 3. For each half-rxn, if the number of oxygen atoms differs between the left and right sides of the equation, add H+ to the side with more oxygen atoms. 4. For each half-rxn, add water to the opposite side of the H+ to balance it and provide additional Oxygens.

44. 5. For each half-rxn, add the number of e- need to balance the charge. (Oxidation rxn will have e- on the right and the Reduction will have e- on the left). 6. Multiply each half-rxn by the minimum factor required to equalize the number of e- in each half-rxn. 7.Add the equations for the half-rxns together, canceling e- and excess water molecules or hydrogen ions. 8. IF DONE IN BASIC SOLUTION: – Add OH - ions to the side with less Oxygen. – Add H 2 O to the opposite side to balance “H’s & O’s”

45. See Steps on Previous Slide for Explanations!! Ex. PbO + V +3  PbO 2 + VO (in acid soln.) Step 1: – PbO  PbO 2 – V +3  VO Step 2: (all atoms except O & H are already balanced) – PbO  PbO 2 – V +3  VO Step 3 &4: – H 2 O + PbO  PbO 2 +2H + – H 2 O + V +3  VO +2H +

46. Step 5: – H 2 O + PbO  PbO 2 + 2H + + 2e- – 1 e- + H 2 O + V +3  VO + 2H + Step 6: – H 2 O + PbO  PbO 2 + 2H + + 2e- – 2e- + 2H 2 O + 2V +3  2VO + 4H + Step 7: – H 2 O + PbO + 2e- + 2H 2 O + 2V +3  PbO 2 + 2H + + 2e- + 2VO + 4H + FINAL: PbO + 3H 2 O + 2V +3  PbO 2 + 2VO + 6H +

47. 9. Ex5.7 Balance in acidic solution: MnO 4 1- + Fe 2+  Mn 2+ + Fe 3+

48. Cr 2 O 7 2- + H 2 SO 3  Cr 3+ + SO 4 2-

49. 10. Ex5.8 Balance in Basic Solution I 1- + MnO 4 1-  I 2 + MnO 2

50. S 2 O 3 2- + I 2  SO 4 2- + I 1-

51. 1.Molarity – definition, equation, and notation (M and [ ]) – Unit of concentration for solutions – Molarity = M = mol-solute L-soln – [ ] = molar concentration Ex. [Mg +2 ] [NaCl]

52. Ex5.9 What is the molarity of a solution made by dissolving 20.0g of NaCl(s) in enough water to make 1.00L of solution? In enough water to make 300.mL of solution?

53. Ex5.10 How many liters of 0.430M solution of Na 2 SO 4 can be made starting with 10.0g of solid?

54. Ex5.11 How many grams of solid solute can be recrystallized from 10.0mL of 0.020M CuCl 2 ?

55. 5.Dissolution equations and molarity – Concentration of compound is related to concentration of ions via equation stoich! – Ex. 0.500 M MgCl 2

56. Ex5.12 State the concentration of each ion in a. 0.250M Na 2 CO 3 b. 0.023M barium phosphate c. 0.380M glucose

57. 7. Dilution problems and equation – Start with conc. soln.  dilute with H 2 O to lower conc. – M 1 V 1 = M 2 V 2

58. 8. Ex5.13 What is the new concentration of a solution made by adding 150.mL of water to 23.0mL of a 2.50M solution of NaCl?

59. 9.Ex5.14 A student starts with100mL of a 0.500M solution of HCl. How much water needs to be added to obtain a solution of HCl with a concentration of 0.0250M. What is the final volume of the solution?

60. 1.pH scale – definitions and diagram pH scale: ranges from 0 (acid) to 14 (base); measures “ the Power of Hydrogen ” pH = -log [H + ] [H 3 O + ] = 10 -pH

61. Ex5.15 State the pH of the following solutions with given hydrogen ion concentrations: a. [H + ] = 0.0230Mb. [H + ] = 0.000560Mc. [H + ] = 3.00M

62. Ex5.16 What is the hydrogen ion concentration for each of the following solutions with given pH? a. pH = 4.30b. pH = 11.2c. pH = 6.30

63. Precipitation Reaction Ex5.17 A precipitate forms when solutions of silver nitrate and scandium (III) chloride are mixed. What volume of 0.0385M scandium (III) chloride is needed to react completely with 22.00mL of 0.130M silver nitrate? What is the mass of the precipitate formed?

64. Precipitation Reaction Ex5.18 What is the mass percent of NaCl in a mixture of sodium chloride and sodium nitrate if a 0.9056g sample of the mixture yields 0.9372g of AgCl(s) when allowed to react with excess AgNO 3 (aq)?

65. Acid Base Reaction Ex5.19 How many milliliters of 0.0195M HCl are required to titrate 10.00mL of 0.0116M Ca(OH) 2 ?

66. Acid Base Reaction Ex5.20 A Vitamin C capsule (ascorbic acid) is analyzed by titrating is with 0.250M sodium hydroxide. It is found that 10.30mL of the base is required to react with a capsule with a mass of 0.518g. What is the percentage of vitamin C, C 8 H 8 O 6, in the capsule if the acid and base react in a 1 to 1 molar ratio?

67. Gas Forming Reaction Ex5.21 If 38.55mL of HCl are used to react with 2.150g of Na 2 CO 3, what is the molarity of the HCl solution?

68. Redox Reaction Ex5.22 To analyze an iron containing compound, you convert all the iron to Fe 2+ in aqueous solution and then titrate the solution with a known concentration of KMnO 4 according to the following balanced equation: – MnO 4 1- (aq) + 5Fe 2+ (aq) + 8H 1+ (aq)  Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O(l) A 0.598g sample of the iron containing compound requires 22.25mL of 0.0123M KMnO 4 for titration to the equivalence point. What is the mass percent of iron in the sample?

69. #1 – 21, 23, 25, 27, 29, 105, 107, 117 – Predicting Solubilities #2 – 31, 33, 35, 37, 39, 41, 43 – Net Ionic Equations & Acid-Base #3 – 45, 47, 49, 95, 97, 53, 55, 57 – Gas forming and Predicting Products #4 – Worksheet – Balancing Redox Equations #5 – 59, 61, 63, 64, 67, 69, 71, 113 – Molar concentrations, preparing solutions #6 – 73, 75, 77, 109, 87, 89, 91, 93, 101 – pH & Acid-Base Titration #7 – 79, 83, 99, 1-3, 111, 115, 125 – Reaction Stoichiometry