2. Types of Chemical Reactions and Solution Stoichiometry

3. Classification of Matter Solutions are homogeneous mixtures

4. Solute A solute is the dissolved substance in a solution. A solvent is the dissolving medium in a solution. Solvent Salt in salt water Sugar in soda drinks Carbon dioxide in soda drinks Water in salt waterWater in soda

5. Saturation of Solutions  A solution that contains the maximum amount of solute that may be dissolved under existing conditions is saturated.  A solution that contains less solute than a saturated solution under existing conditions is unsaturated.  A solution that contains more dissolved solute than a saturated solution under the same conditions is supersaturated.

6. The ammeter measures the flow of electrons (current) through the circuit.  If the ammeter measures a current, and the bulb glows, then the solution conducts.  If the ammeter fails to measure a current, and the bulb does not glow, the solution is non-conducting. Electrolytes vs. Nonelectrolytes

7. An electrolyte is:  A substance whose aqueous solution conducts an electric current. A nonelectrolyte is:  A substance whose aqueous solution does not conduct an electric current. Try to classify the following substances as electrolytes or nonelectrolytes… Definition of Electrolytes and Nonelectrolytes

8. 1.Pure water 2.Tap water 3.Sugar solution 4.Sodium chloride solution 5.Hydrochloric acid solution 6.Lactic acid solution 7.Ethyl alcohol solution 8.Pure, solid sodium chloride Electrolytes?

9. ELECTROLYTES:NONELECTROLYTES: Tap water (weak) NaCl solution HCl solution Lactate solution (weak) Pure water Sugar solution Ethanol solution Pure, solid NaCl But why do some compounds conduct electricity in solution while others do not…? We must understand the nature of water Answers…

10. The Nature of Water Water molecules are BENT (105 ⁰ <) H & O share electrons UNEVENLY O pulls harder on e-s, so the O end of the molecule is slightly negative H pulls less hard on e-s, so the H end of the molecule is slightly positive This makes H2O a POLAR molecule (oppositely charged ends), like a little magnet AKA “a dipole”

11. The Nature of Water The polarity of water gives it great ability to dissolve compounds Imany ionic compounds Some acids: HCl, NO3,

12. 1. Ionic Compounds Ionize in Solution + ions associate with the - (oxygen) end of the water dipole. - ions associate with the + (hydrogen) end of the water dipole. Ions tend to stay in solution where they can conduct a current rather than reforming a solid. IONIC CPDS ARE ELECTROLYTES! Dissociation of sodium chloride Dissociation of sodium chloride

13. Many Ionic Compounds Dissociate (break apart) NaCl(s)  AgNO 3 (s)  MgCl 2 (s)  Na 2 SO 4 (s)  AlCl 3 (s)  Na + (aq) + Cl - (aq) Ag + (aq) + NO 3 - (aq) Mg 2+ (aq) + 2 Cl - (aq) 2 Na + (aq) + SO 4 2- (aq) Al 3+ (aq) + 3 Cl - (aq) Note: write “H 2 O” above each arrow to show that the ionic compound is placed in water, NOT reacting with water. H2O

14. Practice Dissociation Equations Study Guide p 106 #1, 2

15. Again, b/c water is polar…and also b/c covalent acids are also polar! For instance, hydrogen chloride molecules, which are polar, give up their hydrogens to water, 2. Covalent (molecular) acids IONIZE in solution forming chloride ions (Cl - ) and hydronium ions (H 3 O + ).

16. Examples of strong acids include:  Hydrochloric acid, HCl  Sulfuric acid, H 2 SO 4  Nitric acid, HNO 3  Hydroiodic acid, HI  Perchloric acid, HClO 4 Strong acids are completely ionized in solution (Strong Electrolytes) In general, we can assume nearly all other acids are weak

17. Many of these weaker acids are “organic” acids that contain a “carboxyl” group. The carboxyl group does not easily give up its hydrogen. Weak acids ionize only slightly (Weak Electrolytes) HC 2 H 3 O 2 (acetic acid or vinegar)is a weak acid

18. Other organic acids and their sources include: o Citric acid – citrus fruit o Malic acid – apples o Butyric acid – rancid butter o Amino acids – protein o Nucleic acids – DNA and RNA o Ascorbic acid – Vitamin C This is an enormous group of compounds; these are only a few examples. FYI: Because of the carboxyl group, organic acids are sometimes called “carboxylic acids”.

19. 3.Some Other Polar Covalent Compounds are Weak Electrolytes They ionize very slightly in water Ex: Ammonia, NH 3 Only about 1% of the molecules dissociate!

20. Sugar (sucrose – C 12 H 22 O 11 ), and ethanol (ethyl alcohol – C 2 H 5 OH) 4. MOST covalent compounds do not ionize at all in solution. (Nonelectrolyes) NOTE: These molecular compounds DISSOLVE (the water pulls the molecules away from each other, but do not IONIZE (the molecules are not charged.) DISSOLVE

21. Practice Identifying Electrolytes Study Guide, p 106 # 4 & 5

22. Molarity The concentration of a solution measured in moles of solute per liter of solution. mol = M L

23. Preparation of Molar Solutions Problem: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?  Step #1: Ask “How Much?” (What volume to prepare?) 1.500 L  Step #2: Ask “How Strong?” (What molarity?) 0.500 mol 1 L  Step #3: Ask “What does it weigh?” (Molar mass is?) 58.44 g 1 mol = 43.8 g

24. Practice calculating mass needed to make a solution How many grams of NaOH are needed to make 3.5 L of a 2.5 molar solution? Given 3.5 L x Unknown =?g NaOH 2.5 mol NaoH x 1 L of soln 39.9g NaoH 1 mol NaOH

25. Serial Dilution It’s not practical to keep solutions of many different concentrations on hand, so chemists prepare more dilute solutions from a more concentrated “stock” solution. Problem: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? (Note: must convert volumes to L) M stock V stock = M dilute V dilute OR M 1 V 1 =M 2 V 2 (11.6 M)(x Liters) = (3.0 M)(0.250 Liters) x Liters = (3.0 M)(0.250 Liters) 11.6 M = 0.065 L

26. Molarity of Ions in Solution When an ionic compound ionizes in solution, the number of ions formed may be different than the moles of compound. H 2 O(l) Ex: CrCl 3 (s)  Cr 3+ (aq) + 3Cl - (aq) Say “1 mole of CrCl 3 ionizes to form 1 mole of Cr 3+ ions and 3 moles Cl - ions”

27. Calculating Molarity of Ions in Solution 0.25 M CrCl 3 What is the molarity of Cr 3+ ions? Cl- ions? CrCl 3 (s)  Cr 3+ (aq) + 3Cl - (aq) ANSWER Molarity of Cr 3+ = molarity of CrCl 3 = 0.25 M Molarity of Cl- = 3 x molarity of CrCl 3 = 3 x 0.25M = 0.75M

28. Practice Dilutions & Molarity Calcs Study Guide, p 106 # 7, 9, 11, 13,

29. Types of Reactions 5 basic Precipitation Net ionic Oxidation-reduction (redox) Acid-Base

30. 1. Single Replacement Reactions Replacement of:  Metals by another metal  Hydrogen in water by a metal  Hydrogen in an acid by a metal  Halogens by more active halogens A + BX  AX + B BX + Y  BY + X

31. The Activity Series of the Metals Lithium Potassium Calcium Sodium* Magnesium Aluminum Zinc Chromium Iron Nickel Lead Hydrogen Bismuth Copper Mercury Silver Platinum Gold Metals can replace other metals provided that they are above the metal that they are trying to replace. Metals above hydrogen can replace hydrogen in acids. *Metals from sodium upward can replace hydrogen in water

32. The Activity Series of the Halogens Fluorine Chlorine Bromine Iodine Halogens can replace other halogens in compounds, provided that they are above the halogen that they are trying to replace. 2NaCl(s) + F 2 (g)  2NaF(s) + Cl 2 (g) MgCl 2 (s) + Br 2 (g)  ???No Reaction ???

33. 2.Double Replacement Reactions The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY  AY + BX One of the compounds formed is usually a precipitate (an insoluble solid), an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

34. ormation of a Precipitate Formation of a Precipitate

35. Highly Soluble Ionic Compounds (& their exceptions) MEMORIZE THESE RULES! IonSolubilityExceptions NO 3 - SolubleNone ClO 4 - SolubleNone Na + SolubleNone K+K+K+K+SolubleNone NH 4 + SolubleNone Cl -,I -, Br - Soluble Pb 2+, Ag +, Hg 2 2+ SO 4 2- Soluble Ca 2+, Ba 2+, Sr 2+, Pb 2+, Ag +, Hg 2+

36. Slightly Soluble (usually considered insoluble) Ionic Compounds MEMORIZE THESE RULES! IonSolubilityExceptions CO 3 2- Slightly soluble Group 1 (IA) and NH 4 + PO 4 3- “ Group 1 (IA) and NH 4 + OH - “ Group 1 (IA) and Ca 2+, Ba 2+, Sr 2+ S 2- “ Groups 1 (IA), 2 (IIA), and NH 4 + CrO 4 2- “none

37. Completing a Double replacement Rxn: Ex: Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 + 2KNO 3 1. Determine the products formed when the ions are exchanged, & balance 2. Decide if the products are soluble (aq) or insoluble (s) PbI 2 Insoluble, so it will be a solid KNO 3 Soluble, so it will be in aqueous solution (s) (aq)

38. Complete Ionic Equation Shows all soluble compounds as aqueous ions Shows all insoluble compounds as a unit, use the symbol (s) to show it is a solid. Pb 2+ (aq) + 2 NO 3 - (aq) + 2 K + (aq) +2 I - (aq)  2K + (aq) + 2 NO 3 - (aq)PbI 2 (s) + Pb(NO 3 ) 2 (aq) + 2KI2KNO 3 (aq)  PbI 2 (s) +

39. Net Ionic Equation Eliminates all “spectator” ions (ions that appear identically on both sides of equation) Pb 2+ (aq) + 2NO 3 - (aq) + 2K + (aq) +2I - (aq)  2K + (aq) + 2NO 3 - (aq)PbI 2 (s) + Pb 2+ (aq) + 2 I - (aq)  PbI 2 (s)

40. Practice Complete Ionic Equation & Net Ionic Equation CuSO 4 + Na 2 S  CuS (insoluble) + Na 2 SO 4 (soluble) Complete Ionic Net Ionic Cu 2+ (aq) +SO 4 2- (aq) +2Na + (aq) +S 2- (aq)  CuS(s) + 2Na + (aq)+SO 4 2- (aq) Cu 2+ (aq) + S 2- (aq)  CuS(s)

41. Practice Problems Study Guide: Please Note: show all work in your notebook, not your study guide! Pp 108 #28 Net Ionic Equation Worksheet Q#1-10

42. Stoichiometry of Precipitation Rxns It is helpful to be able to predict the amount of precipitant formed b/c it is often collected & used. Ex: Calculate the mass of solid NaCl that must be added to 1.50L of a 0.100 M AgNO 3 solution to precipitate all the Ag+ ions in the form of AgCl. First, we need a balanced equation: NaCl(s) + AgNO3(aq)  AgCl(s) + NaNO3(aq) Given 1.5 L AgNO3 x 0.100 mol AgNO3 x 1 L AgNO3 1 mol NaCl x 1 mol AgNO3 58.45 g NaCl = 1 mol NaCl ___g NaCl8.77

43. Stoichiometry of Precipitation Rxns, cont. Please note: you will also encounter limiting reactant problems with precip. Rxns! Remember, set up 2 separate equations to see which reactant forms the LEAST product. This is your limiting reactant. The equations will be identical to those on the prior page.

44. Stoichiometry of Precipitation Rxns, cont. HINT: label all equations with your givens & unknown. Example: When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. Calculate the mass of PbSO 4 formed when 1.25 L of 0.0500 M Pb(NO 3 ) 2 and 2.00 L of 0.0250 M Na 2 SO 4 are mixed. Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq)  PbSO 4 (s) + 2NaNO 3 (aq) V=2.00 L M=0.0250mol/L V=1.25 L M=0.0500mol/L m=? g NOTE: set up 2 separate equations & solve for limiting reactant. See prior slide. Equations are set up same way.

45. Practice Problems: Study Guide, pp __________ Q # _________ Please show all your work in your notebook, not your study guide.

46. Oxidation and Reduction Reactions (Redox) Def: Rxns in which electrons are transferred Ex: 2Na(s) + Cl 2 (g)  2 NaCl(s) An electron transfers from the Na atom to the Cl atom. 

47. What about molecular compounds? Non-ionic compounds can also be formed from redox reactions. Even though e-s aren’t FULLY transferred, they can be assumed to involve a transfer…let’s see how!

48. Oxidation and Reduction Reactions (Redox) Ex:CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O + energy Carbon is less EN than oxygen, so we assume there is a transfer of e-s from C to O. NOTE: We will study electronegativity (EN) in Chapter 8. EN is the strength with which an atom in a bond pulls on electrons. (Check Figure 8.3 on p 353.) This shows us the value for EN of O is 3.5 and the EN value for C is 2.5.

49. Oxidation and Reduction Rxns, cont. Oxidation States (Oxidation #s) provide a way to track e-s in redox reactions, especially in molecular substances.

50. Rules for Assigning Oxidation Numbers Rules 1 & 2 1.The oxidation number of any uncombined element is zero 2. The oxidation number of a monatomic ion equals its charge

51. Rules for Assigning Oxidation Numbers Rules 3 & 4 3. The oxidation number of oxygen in compounds is -2 (except in peroxides, in which the oxidation number is -1) 4. The oxidation number of hydrogen in compounds is +1

52. Rules for Assigning Oxidation Number Rule 5 5. The sum of the oxidation numbers in the formula of a compound is 0 2(+1) + (-2) = 0 H O (+2) + 2(-2) + 2(+1) = 0 Ca O H

53. Rules for Assigning Oxidation Numbers Rule 6 6. The sum of the oxidation numbers in the formula of a polyatomic ion is equal to its charge X + 3(-2) = -1 N O  X = +5  X = +6 X + 4(-2) = -2 S O  X = +6

54. Practice Together Assign oxidation states to all atoms in the following: 1. CO 2 2. SF 6 3. NO 2 - 1. C +4 O -2 2. S +6 F -1 3. N +3 O -2 Check the math on #3 (NO 2 - ) +3 + 2(-2) = -1

55. Practice Assigning Oxidation Numbers Study Guide, p 110 Q# 50-52, Please Note: show all work in your notebook, not your study guide!

56. Oxidation and Reduction G ain E lectrons = R eduction An old memory device for oxidation and reduction goes like this… LEO says GER L ose E lectrons = O xidation

57. Using Half Reactions to understand Redox Reactions Each sodium atom loses one electron: Each chlorine atom gains one electron:

58. LEO says GER : LEO says GER : Lose Electrons = Oxidation Sodium is oxidized Gain Electrons = Reduction Chlorine is reduced

59. Reducing Agents and Oxidizing Agents The substance reduced is the oxidizing agent The substance oxidized is the reducing agent Sodium is oxidized – it is the reducing agent Chlorine is reduced – it is the oxidizing agent

60. Using Half-reactions to Balance Redox Equations Sometimes it is helpful to separate the rxn into two half-reactions: 1. Oxidation 2. Reduction Example: Ce 4+ (aq) + Sn 2+ (aq)  Ce 3+ (aq) + Sn 4+ (aq) 1.Oxidation: Sn 2+ (aq)  Sn 4+ (aq) + 2 e- 2. Reduction: Ce 4+ (aq) + e-  Ce 3+ (aq) NOTE: If you combine these equations, the e-s and charges don’t balance. So, multiply the 2 nd equation by 2.

61. Using Half-reactions to Balance Redox Equations 1.Oxidation: Sn 2+ (aq)  Sn 4+ (aq) + 2 e- 2.Reduction: 2(Ce 4+ (aq) + e-  Ce 3+ (aq)) Now, combine (add) the two equations, cancelling items that appear identically on both sides of the arrow. 2. Reduction:2Ce 4+ (aq) + 2e-  2Ce 3+ (aq)= Sn 2+ (aq) + 2Ce 4+ (aq) + 2 e-  + Sn 4+ (aq) +2Ce 3+ (aq) + 2 e- EXTRA: Which is the oxidizing agent? Reducing agent? cesium tin

62. Practice Problems text book: p 183-184, practice problems #57, 61, 62, 64 Show all your work in your notebook. Let’s do # 64a together Zn(s) + HCl(aq)  Zn 2+ (aq) + H 2 (g) + Cl- (aq)

63. Let’s try 53 c Cu + H+ + NO3-  Cu2+ + NO2 + H2O Cu  Cu2+ + 2e- 2 (NO3- + 1e-  NO2 ) Cu + 2 NO3- + 2e-  2NO2 + 2e-

64. Trends in Oxidation and Reduction Active metals: Lose electrons easily Are easily oxidized Are strong reducing agents Active nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

65. Redox Reaction Prediction #1 Important OxidizersFormed in reaction MnO 4 - (acid solution) MnO 4 - (basic solution) MnO 2 (acid solution) Cr 2 O 7 2- (acid) CrO 4 2- HNO 3, concentrated HNO 3, dilute H 2 SO 4, hot conc Metallic Ions Free Halogens HClO 4 Na 2 O 2 H 2 O 2 Mn(II) MnO 2 Mn(II) Cr(III) NO 2 NO SO 2 Metallous Ions Halide ions Cl - OH - O 2

66. Redox Reaction Prediction #2 Important ReducersFormed in reaction Halide Ions Free Metals Metalous Ions Nitrite Ions Sulfite Ions Free Halogens (dil, basic sol) Free Halogens (conc, basic sol) C 2 O 4 2- Halogens Metal Ions Metallic ions Nitrate Ions SO 4 2- Hypohalite ions Halate ions CO 2

67. Not All Reactions are Redox Reactions Reactions in which there has been no change in oxidation number are not redox rxns. Examples:

68. Acid-Base Reactions Arrhenius definition: Acid is a proton (hydrogen ion) donor Base is a hydroxide (OH-) donor Bronsted-Lowry Definition (applies to more compounds) Acid is a proton (hydrogen ion) donor Base is a proton (hydrogen ion) acceptor

69. Strong Acids: there are 7 Some oxyacids: HClO 4, HClO 3, H 2 SO 4, HNO 3 Most halide acids are strong:HI, HCl, HBr Except HF, which is weak

70. Strong Bases: there are 8 Hydroxides of Group 1 metals are all strong bases: LiOH, NaOH, KOH, RbOH, CsOH 3 Hydroxides of Group 2 metals are strong bases Ca(OH) 2 Sr(OH) 2 Ba(OH) 2

71. Acid-Base Reactions: Strong Acid/Strong Base To predict results of rxn, focus on the species present in mixed solution. Ex#1 : HCl(aq) + NaOH(aq)  ??? Species Present before rxn occurs: H + (aq) + Cl- (aq) + Na + (aq) + OH- (aq) (b/c HCl is a strong acid and NaOH is a strong base=complete dissociation of both acid & base) NaCl is soluble in water (check your table) so Na+ & Cl- are spectator ions.

72. Acid-Base Reactions: Strong Acid/Strong Base Ex#1(continued): HCl(aq) + NaOH(aq)  ??? H + (aq) + OH- (aq) are the only other species present. Water is a nonelectrolyte, so H + (aq) + OH- will NOT exist in solution together. Therefore, they must combine to form H 2 O. Our net ionic equation for this reaction is therefore: H + (aq) + OH- (aq)  H 2 O (l)

73. Acid-Base Reactions: Weak Acid/Strong Base Again, to predict results of rxn, focus on the species present in mixed solution. Ex#2 : HC 2 H 3 O 2 (aq) + KOH(aq)  ??? Species Present before rxn occurs: HC 2 H 3 O 2 (aq) + K + (aq) + OH- (aq) b/c HC 2 H 3 O 2 (aq) is a weak acid (almost no dissociation = 1%) & b/c KOH is a strong base (complete dissociation)

74. Acid-Base Reactions: Weak Acid/Strong Base Note: OH- is such a strong base that it will strip H+ ions from the HC 2 H 3 O 2 (acetic acid)  we can assume it reacts completely with any weak acid it will be combined with HC 2 H 3 O 2 (aq) + K + (aq) + OH- (aq)  HOH + C 2 H 3 O 2 - (aq) + K+(aq) Net Ionic: HC 2 H 3 O 2 (aq) + OH- (aq)  HOH + C 2 H 3 O 2 - (aq) But the net ionic equation is quite different than for a strong acid/strong base rxn : Remember, acetic acid is NOT dissociated before the reaction!

75. Acid-Base Reactions Arrhenius Acids & Bases Often called “neutralization reactions” Complete equation shows formation of a salt and water Ex: HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)

76. Acids that Undergo Multi-step Rxn H 2 CO 3 (aq) breaks into water and carbon dioxide H 2 CO 3  H 2 O + CO 2

77. Stoichiometry of Neutralization Reactions 1. List the species present in the combined solution before any reaction occurs & decide what will occur. 2. Write balanced net ionic equation. 3. Set up your calculation, identifying Given, Unknown, and conversion factors that cancel the necessary units. Or identify the proper formula and solve for unknown.

78. Stoichiometry of Neutralization Reactions: Example 1 (p159 of text) Example: What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? 1. List the species present in the combined solution before any reaction occurs & decide what will occur. H + (aq) + Cl- (aq) + Na + (aq) + OH- (aq) (HCl is a strong acid and NaOH is a strong base, so both completely ionize in aqueous solution) NaCl is soluble, so the Na + and Cl- ions are spectators. (This reaction :Na + (aq) + Cl- (aq)  NaCl will not occur.) H+ and OH- do not coexist, so H 2 O will form

79. Stoichiometry of Acid-Base Rxns 2. Write the balanced net ionic equation for this reaction. H + (aq) + OH- (aq)  H 2 O (l)

80. Stoichiometry of Acid-Base Rxns 3. Set up your calculation, identifying Given, Unknown, and conversion factors that cancel the necessary units. Or identify the proper formula and solve for the unknown. Analysis: M A V A = M B V B M A =0.100 mol/L V A = ??? L M B =0.350 mol/L V B = 0.025 L H + (aq) + OH- (aq)  H 2 O (l) Given 0.025L OH- Unknown = L H+ x 0.350 mol OH- 1 L OH- x 1 mol H+ 1 mol OH- x 1L H+ 0.100 mol H+ 8.75 x 10 -3 NOTE: Our usual analysis will also work! Solution: V A =(0.350M) (0.025L) (0.100 M) VA = 8.75 x 10 -3 L H+

81. Stoichiometry of Acid-Base Rxns: Example 2 (p160 of text) Example2 : 28.0 mL of 0.250M HNO3 and 53.0 mL of 0.320 M KOH are mixed. Calculate the amount of water formed in the resulting reaction. What is the concentration of H+ or OH- ions in excess after the rxn goes to completion? 1. List the species present in the combined solution before any reaction occurs & decide what will occur. H + (aq) + NO 3 - (aq) + K + (aq) + OH- (aq) (HNO 3 is a strong acid and KOH is a strong base, so both completely ionize in aqueous solution) KNO 3 is soluble, so the K + and NO 3 - ions are spectators. (This reaction :K + (aq) + NO 3 - (aq)  KNO 3 will not occur.) H+ and OH- do not coexist, so H 2 O will form

82. Stoichiometry Acid-Base Rxns: Example 2 (p160 of text) 2. Write the balanced net ionic equation for this reaction. H + (aq) + OH- (aq)  H 2 O (l)

83. Practice Acid-Base Stoichiometry Problems Text Book, p 183, Q # 45, 47, 49, 50, 51, 53 Show all your work in your notebook

84. Stoichiometry of Neutralization Reactions 3. Set up your calculation, identifying Given, Unknown, and conversion factors that cancel the necessary units. Or identify the proper formula and solve for the unknown. Analysis: M A V A = M B V B Will NOT work for this problem. All your variables have values. This is a limiting reactant problem M A =0.250 mol/L V A = 0.0280 L M B =0.320 mol/L V B = 0.0530 L H + (aq) + OH- (aq)  H 2 O (l) Unknown 1: L H 2 O Unknown 2: M H+ or M OH- in excess

85. Stoichiometry of Neutralization Reactions H + (aq) + OH- (aq)  H 2 O (l) Given 0.0530L OH- Unknown = mol OH- x 0.350 mol OH- 1 L OH- 0.0170 Unknown 1: L H 2 O Unknown 2: M H+ or M OH- in excess Solution Given 0.0280L H+ x 0.250 mol H+ 1 L H+ 0.00700 Unknown = __molH+

86. Titration