1. AP CHEMISTRY CHAPTER 4 SOLUTION STOICHIOMETRY

2. -Covalent bonds -Electrons aren’t shared evenly (oxygen is more electronegative) -Electrons spend more time close to O than to H

3. This uneven distribution of charge makes water polar. Because of this, water is a good solvent.

4. When water surrounds an ionic crystal, the H end attracts the anion and the O end attracts the cation. This process is called hydration. Hydration #- The number of H 2 O molecules associated with a particular ion, usually 4 or 6.

5. Hydration causes salts to dissolve. H 2 O also dissolves polar covalent substances such as C 2 H 5 OH. H 2 O doesn’t dissolve nonpolar covalent substances.

6. Hydration

7. Review: Solute Solvent Electrical conductivity Strong electrolyte Weak electrolyte

8. Arrhenius determined that the extent to which a solution can conduct an electrical current depends directly on the number of ions present.

9. Solubility- g/given volume solvent or moles/given volume solution

10. Strong electrolytes 1.soluble salts 2.strong acids –completely ionize HCl(aq), HNO 3 (aq), H 2 SO 4 (aq)

11. Ex. Show how HCl dissociates when dissolved in water. HCl  H + + Cl -

12. Acid (Arrhenius) – a substance that produces H + ions in water solution

13. 3.strong bases- completely ionize -contain OH - -bitter taste and slippery feel -NaOH, KOH

14. Weak electrolytes -only ionize slightly (weak acids and bases) HC 2 H 3 O 2  H + + C 2 H 3 O 2 - 99% 1%

16. Ammonia (NH 3 ) -weak base NH 3 + H 2 O  NH 4 + + OH -

17. Molarity (M) = moles of solute liters of solution

18. Ex. Calculate the molarity of a solution made by dissolving 23.4g of sodium sulfate in enough water to form 125 mL of solution. 23.4 g Na 2 SO 4 1 mol Na 2 SO 4 = 0.165 mol Na 2 SO 4 142.06g Na 2 SO 4 0.165 mol = 1.32 M 0.125 L

19. Ex. How many grams of Na 2 SO 4 are required to make 350 mL of 0.50 M Na 2 SO 4 ? 0.350L 0.50 mol Na 2 SO 4 142.06g Na 2 SO 4 = 24.9g 1 L 1 mol Na 2 SO 4

20. Ex. What volume of 1.000 M KNO 3 must be diluted with water to prepare 500.0 mL of 0.250 M KNO 3 ? Dilution problem (M 1 V 1 = M 2 V 2 ) (1.000M)(V 1 ) = (0.250M)(500.0mL) V 1 = 125 mL

21. Read procedure for using volumetric flasks and types of pipets. We will be using both in several labs this year.

23. Writing net-ionic equations 1. molecular equation -overall reaction stoichiometry 2. complete ionic equation -all strong electrolytes are represented as ions 3. net ionic equation -spectator ions are not included

24. 1. NaCl(aq) + AgNO 3 (aq)  NaNO 3 (aq) + AgCl(s) 2. Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq)  Na + (aq) + NO 3 - (aq) + AgCl(s) 3. Cl - (aq) + Ag + (aq)  AgCl(s)

25. Precipitation Reaction- A reaction in which two solutions are mixed and an insoluble solid forms.

26. Review solubility rules Use solubility rules to determine which product, if any, will ppt in a double replacement rxn.

27. Simple Rules for Solubility 1.Most nitrate (NO 3  ) salts are soluble. 2.Most alkali (group 1A) salts and NH 4 + are soluble. 3.Most Cl , Br , and I  salts are soluble (NOT Ag +, Pb 2+, Hg 2 2+ ) 4.Most sulfate salts are soluble (NOT BaSO 4, PbSO 4, HgSO 4, CaSO 4 ) 5.Most OH  salts are only slightly soluble (NaOH, KOH are soluble, Ba(OH) 2, Ca(OH) 2 are marginally soluble) 6.Most S 2 , CO 3 2 , CrO 4 2 , PO 4 3  salts are only slightly soluble.

28. SOLUBILITY SONG To the tune of “ My Favorite Things” from “The Sound of Music” Nitrates and Group One and Ammonium, These are all soluble, a rule of thumb. Then you have chlorides, they’re soluble fun, All except Silver, Lead, Mercury I. Then you have sulfates, except for these three: Barium, Calcium and Lead, you see. Worry not only few left to go still. We will do fine on this test. Yes, we will! Then you have the--- Insolubles…. Hydroxide, Sulfide and Carbonate and Phosphate, And all of these can be dried! By Kimberly Chin and Jeannie Chen, class of 2001

29. Selective precipitation- process by which ions are caused to ppt one by one in sequence to separate mixtures of ions. Qualitative analysis- process of separating and identifying ions

30. Ex. Separate Ag +, Ba 2+, Fe 3+ 1.Add Cl - to remove Ag + as AgCl. 2.Add SO 4 2- to remove Ba 2+ as BaSO 4. 3.Add OH - or S 2- to remove Fe 3+ as Fe(OH) 3 or Fe 2 S 3.

31. Ex. Separate Pb 2+, Ba 2+, Ni 2+ 1.Add Cl - to remove Pb 2+ as PbCl 2. 2.Add SO 4 2- to remove Ba 2+ as BaSO 4. 3.Add OH - or S 2- to remove Ni 2+ as Ni(OH) 2 or NiS.

32. Quantitative analysis- determines how much of a component is present. Gravimetric analysis- quantitative procedure where a ppt containing the substance is formed, filtered, dried & weighed.

33. Ex. The zinc in a 1.2000g sample of foot powder was precipitated as ZnNH 4 PO 4. Strong heating of the ppt yielded 0.4089 g of Zn 2 P 2 O 7. Calculate the mass percent of zinc in the sample of the foot powder. 0.4089gZn 2 P 2 O 7 1 mol Zn 2 P 2 O 7 2 mol Zn 65.37g = 304.7 g 1 mol Zn 2 P 2 O 7 1 mol Zn 0.1754g Zn x 100 = 14.62% Zn 1.200g sample

34. Ex. A mixture contains only NaCl and Fe(NO 3 ) 3. A 0.456g sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of Fe(OH) 3. The ppt is filtered, dried, & weighed. Its mass is 0.128g. Calculate: a. the mass of the iron b.the mass of Fe(NO 3 ) 3 c.the mass percent of Fe(NO 3 ) 3 in the sample 0.128g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe 55.85g Fe= 0.0669g Fe 106.9g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe 0.0669g Fe 1 mol Fe 1 mol Fe(NO 3 ) 3 241.9g Fe(NO 3 ) 3 = 0.290g Fe(NO 3 ) 3 55.85g Fe 1 mol Fe 1 mol Fe(NO 3 ) 3 0.290g x 100 = 63.6% Fe(NO 3 ) 3 0.456g

35. Acid-Base Reactions

36. Bronsted-Lowry acid-base definitions: acid- proton donor base- proton acceptor

37. When a strong acid reacts with a strong base the net ionic rxn is: H + (aq) + OH - (aq)  H 2 O(l) When a strong acid reacts with a weak base or a weak acid reacts with a strong base, the reaction is complete (the weak substance ionizes completely.) HC 2 H 3 O 2 (aq) + OH - (aq)  H 2 O(l) + C 2 H 3 O 2 - (aq)

38. neutralization reaction - acid-base rxn When just enough base is added to react exactly with the acid in a solution, the acid is said to be neutralized.

39. Volumetric Analysis titration- process in which a solution of known concentration (standard solution) is added to analyze another solution.

40. titrant- solution of known concentration (in buret) equivalence point or stoichiometric point- point where just enough titrant has been added to react with the substance being analyzed

41. Indicator - chemical which changes color at or near the equivalence point End point- point at which the indicator changes color

42. Ex. 54.6 mL of 0.100 M HClO 4 solution is required to neutralize 25.0 mL of an NaOH solution of unknown molarity. What is the concentration of the NaOH solution? HClO 4 + NaOH  H 2 O + NaClO 4 0.0546 L HClO 4 0.100 mol HClO 4 1 mol NaOH = 1 L HClO 4 1 mol HClO 4 0.00546 mol NaOH 0.00546 mol NaOH = 0.218 M NaOH 0.025L

43. Oxidation-Reduction Reactions Redox Rxns - reactions in which one or more electrons are transferred.

44. Electronegativity - attraction for shared electrons most electronegative  F>O>N=Cl elements “Phone Call” These are most likely to have negative oxidation numbers.

45. Rules for Assigning Oxidation States 1. Oxidation state of an atom in an element = 0 2. Oxidation state of monatomic element = charge 3. Oxygen =  2 in covalent compounds (except in peroxides where it =  1) 4. H = +1 in covalent compounds 5. Fluorine =  1 in compounds 6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

46. Review oxidation state rules on page 167. N 2 O PBr 3 HPO 3 2- P 4 O 6 NH 2 - +1-2 +3-1 +1+3-2 +3 -2 -3 +1

47. Noninteger states are rare, but possible. Fe 3 O 4 8/3 -2 O = 4(-2) = -8 Fe = 8/3 = 2 2/3 or Fe 2+, Fe 3+, Fe 3+

48. Oxidation - loss of electrons - increase in oxidation number Reduction - gain of electrons - decrease in oxidation number

49. OIL RIG Oxidation Is Loss (of e-), Reduction Is Gain (of e-)

50. LEO the lion goes GER Lose Electrons = Oxidation, Gain Electrons = Reduction

51. Oxidizing agent - electron acceptor - substance that is reduced Reducing agent - electron donor - substance that is oxidized

52. 2KI + F 2  2KF + I 2 +1-1 0 +1-1 0 oxidized I reduced F OA F2F2 RA KI

53. 2PbO 2  2PbO + O 2 +4 -2 +2 -2 0 oxidized: O reduced: Pb OA: PbO 2 RA: PbO 2

54. Balancing redox reactions by the half-reaction method 1.Write skeleton half-reactions. 2.Balance all elements other than O and H. 3.Balance O by adding H 2 O. 4.Balance H by adding H +.

55. 5.Balance charge by adding e - to the more positive side. 6.Make the # of e - lost = # of e - gained by multiplying each half-rxn by a factor. 7.Add half-reactions together. 8.Cancel out anything that is the same on both sides.

56. 9.If the reaction occurs in basic solution, add an equal number of hydroxide ions to both sides to cancel out the hydrogen ions. Make water on the side with the hydrogen ions. Cancel water if necessary. 10.Check to see that charge and mass are both balanced.

57. Practice: Sn 2+ + Cr 2 O 7 2-  Sn 4+ + Cr 3+ (acidic solution) Sn 2+  Sn 4+ Cr 2 O 7 2-  Cr 3+ Sn 2+  Sn 4+ Cr 2 O 7 2-  2Cr 3+ Sn 2+  Sn 4+ Cr 2 O 7 2-  2Cr 3+ +7H 2 O Sn 2+  Sn 4+ Cr 2 O 7 2- + 14H +  2Cr 3+ +7H 2 O Sn 2+  Sn 4+ +2e - Cr 2 O 7 2- + 14H + +6e -  2Cr 3+ +7H 2 O 3(Sn 2+  Sn 4+ +2e - ) Cr 2 O 7 2- + 14H + +6e -  2Cr 3+ +7H 2 O 3Sn 2+ + Cr 2 O 7 2- + 14H + +6e -  3Sn 4+ +6e - + 2Cr 3+ +7H 2 O 3Sn 2+ + Cr 2 O 7 2- + 14H +  3Sn 4+ + 2Cr 3+ +7H 2 O

58. MnO 4 2- + I -  MnO 2 + I 2 (basic solution) MnO 4 2-  MnO 2 I -  I 2 MnO 4 2-  MnO 2 2I -  I 2 MnO 4 2-  MnO 2 + 2H 2 O 2I -  I 2 MnO 4 2- +4H +  MnO 2 + 2H 2 O 2I -  I 2 MnO 4 2- +4H + + 2e -  MnO 2 + 2H 2 O 2I -  I 2 +2e - MnO 4 2- +4H + + 2e - +2I -  MnO 2 + 2H 2 O +I 2 +2e - MnO 4 2- +4H + +2I -  MnO 2 + 2H 2 O +I 2 MnO 4 2- +4H + + 4OH - +2I -  MnO 2 + 2H 2 O +I 2 + 4OH - MnO 4 2- + 4H 2 O +2I -  MnO 2 + 2H 2 O +I 2 + 4OH - MnO 4 2- + 2H 2 O +2I -  MnO 2 + I 2 + 4OH -

59. OXIDATION-REDUCTION TITRATIONS Most common oxidizing agents: KMnO 4 & K 2 Cr 2 O 7

60. Potassium permanganate used to disinfect ponds and fish in Egypt. Photo by Will Rooney (AP 2008)

61. MnO 4 - in acidic solution: MnO 4 - + 8H + + 5e -  Mn 2+ + 4H 2 O Purple colorless When you titrate with MnO 4 -, the solution is colorless until you use up all of the reducing agent (substance being oxidized).

62. In calculations, work redox titrations like acid-base titrations. You must have a balanced reaction to know the mole ratio.