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This exam will take 2 hours. Please be ready. The grade has to be above 80%. I will be very appreciate your help. These are all the materials you need to know for the exam. This exam will only cover Ch4-6.

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Lecture PowerPoint Chemistry The Molecular Nature of Matter and Change Sixth Edition Martin S. Silberberg 4-1 Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 4 Three Major Classes of Chemical Reactions 4-2 The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions 4.4 Acid-Base Reactions 4.5 Oxidation-Reduction (Redox) Reactions 4.6 Elements in Redox Reactions 4.7 Reaction Reversibility and the Equilibrium State 4-3 Water as a Solvent • Water is a polar molecule – since it has uneven electron distribution – and a bent molecular shape. • Water readily dissolves a variety of substances. • Water interacts strongly with its solutes and often plays an active role in aqueous reactions. 4-4 Figure 4.1 Electron distribution in molecules of H2 and H2O. A. Electron charge distribution in H2 is symmetrical. B. Electron charge distribution in H2O is asymmetrical. C. Each bond in H2O is polar. D. The whole H2O molecule is polar. 4-5 Figure 4.2 4-6 An ionic compound dissolving in water. Figure 4.3 4-7 The electrical conductivity of ionic solutions. Sample Problem 4.1 Using Molecular Scenes to Depict an Ionic Compound in Aqueous Solution PROBLEM: The beakers shown below contain aqueous solutions of the strong electrolyte potassium sulfate. (a) Which beaker best represents the compound in solution? (H2O molecules are not shown). (b) If each particle represents 0.10 mol, what is the total number of particles in solution? 4-8 Sample Problem 4.1 PLAN: (a) Determine the formula and write and equation for the dissociation of 1 mol of compound. Potassium sulfate is a strong electrolyte; it therefore dissociates completely in solution. Remember that polyatomic ions remain intact in solution. (b) Count the number of separate particles in the relevant beaker, then multiply by 0.1 mol and by Avogadro’s number. SOLUTION: (a) The formula is K2SO4, so the equation for dissociation is: K2SO4 (s) → 2K+ (aq) + SO42- (aq) 4-9 Sample Problem 4.1 There should be 2 cations for every 1 anion; beaker C represents this correctly. (b) Beaker C contains 9 particles, 6 K+ ions and 3 SO42- ions. 6.022x1023 particles 9 x 0.1 mol x 1 mol 4-10 = 5.420x1023 particles Sample Problem 4.2 PROBLEM: (a) (b) (c) (d) PLAN: 4-11 Determining Amount (mol) of Ions in Solution What amount (mol) of each ion is in each solution? 5.0 mol of ammonium sulfate dissolved in water 78.5 g of cesium bromide dissolved in water 7.42x1022 formula units of copper(II) nitrate dissolved in water 35 mL of 0.84 M zinc chloride Write an equation for the dissociation of 1 mol of each compound. Use this information to calculate the actual number of moles represented by the given quantity of substance in each case. Sample Problem 4.2 SOLUTION: (a) The formula is (NH4)2SO4 so the equation for dissociation is: (NH4)2SO4 (s) → 2NH4+ (aq) + SO42- (aq) 5.0 mol (NH4)2SO4 x 2 mol NH4+ 1 mol (NH4)2SO4 5.0 mol (NH4)2SO4 x 1 mol SO42- 1 mol (NH4)2SO4 4-12 = 10. mol NH4+ = 5.0 mol NH4+ Sample Problem 4.2 SOLUTION: (b) The formula is CsBr so the equation for dissociation is: CsBr (s) → Cs+ (aq) + Br- (aq) + 78.5 g CsBr x 1 mol CsBr x 1 mol Cs 212.8 g CsBr 1 mol CsBr = 0.369 mol Cs+ There is one Cs+ ion for every Br- ion, so the number of moles of Br- is also equation to 0.369 mol. 4-13 Sample Problem 4.2 SOLUTION: (c) The formula is Cu(NO3)2 so the formula for dissociation is: Cu(NO3)2 (s) → Cu2+ (aq) + 2NO3- (aq) 7.42x1022 formula units Cu(NO3)2 x 1 mol 6.022x1023 formula units = 0.123 mol Cu(NO3)2 0.123 mol Cu(NO3)2 x 1 mol Cu2+ 1 mol Cu(NO3)2 = 0.123 mol Cu2+ ions There are 2 NO3- ions for every 1 Cu2+ ion, so there are 0.246 mol NO3- ions. 4-14 Sample Problem 4.2 SOLUTION: (d) The formula is ZnCl2 so the formula for dissociation is: ZnCl2 (s) → Zn2+ (aq) + 2Cl- (aq) 35 mL soln x 1 L x 103 mL 2.9x10-2 mol ZnCl2 x 0.84 mol ZnCl2 = 2.9x10-2 mol ZnCl2 1 L soln 2 mol Cl- = 5.8x10-2 mol Cl- 1 mol ZnCl2 There is 1 mol of Zn2+ ions for every 1 mol of ZnCl2, so there are 2.9 x 10-2 mol Zn2+ ions. 4-15 Writing Equations for Aqueous Ionic Reactions The molecular equation shows all reactants and products as if they were intact, undissociated compounds. This gives the least information about the species in solution. 2AgNO3 (aq) + Na2CrO4 (aq) → Ag2CrO4 (s) + 2NaNO3 (aq) When solutions of silver nitrate and sodium chromate mix, a brick-red precipitate of silver chromate forms. 4-16 The total ionic equation shows all soluble ionic substances dissociated into ions. This gives the most accurate information about species in solution. 2Ag+ (aq) + 2NO3- (aq) + 2Na+ (aq) + CrO42- (aq) → Ag2CrO4 (s) + 2Na+ (aq) + NO3- (aq) Spectator ions are ions that are not involved in the actual chemical change. Spectator ions appear unchanged on both sides of the total ionic equation. 2Ag+ (aq) + 2NO3- (aq) + 2Na+ (aq) + CrO42- (aq) 4-17 → Ag2CrO4 (s) + 2Na+ (aq) + 2NO3- (aq) The net ionic equation eliminates the spectator ions and shows only the actual chemical change. 2Ag+ (aq) + CrO42- (aq) 4-18 → Ag2CrO4 (s) Figure 4.4 4-19 An aqueous ionic reaction and the three types of equations. Precipitation Reactions • In a precipitation reaction two soluble ionic compounds react to give an insoluble products, called a precipitate. • The precipitate forms through the net removal of ions from solution. • It is possible for more than one precipitate to form in such a reaction. 4-20 Figure 4.5 The precipitation of calcium fluoride. 2NaF (aq) + CaCl2 (aq) → CaF2 (s) + 2NaCl (aq) 2 Na+ (aq) + 2 F- (aq) + Ca2+ (aq) + 2 Cl- (aq) 2 NaF(aq) + CaCl2 (aq) 4-21 → → CaF2(s) + 2 Na+ (aq) + 2 Cl- (aq) CaF2(s) + 2 NaCl (aq) Figure 4.6 The precipitation of PbI2, a metathesis reaction. 2NaI (aq) + Pb(NO3)2 (aq) → PbI2 (s) + NaNO3 (aq) 2Na+(aq) + 2I- (aq) + Pb2+ (aq) + 2NO3- (aq) → PbI2 (s) + 2Na+ (aq) + 2NO3-(aq) Pb2+ (aq) + 2I- (aq) → PbI2 (s) Precipitation reactions are also called double displacement reactions or metathesis. 2NaI (aq) + Pb (NO3)2 (aq) → PbI2 (s) + 2NaNO3 (aq) Ions exchange partners and a precipitate forms, so there is an exchange of bonds between reacting species. 4-22 Predicting Whether a Precipitate Will Form • Note the ions present in the reactants. • Consider all possible cation-anion combinations. • Use the solubility rules to decide whether any of the ion combinations is insoluble. – An insoluble combination identifies the precipitate that will form. 4-23 Table 4.1 Solubility Rules for Ionic Compounds in Water Soluble Ionic Compounds 1. All common compounds of Group 1A(1) ions (Li+, Na+, K+, etc.) and ammonium ion (NH4+) are soluble. 2. All common nitrates (NO3-), acetates (CH3COO- or C2H3O2-) and most perchlorates (ClO4-) are soluble. 3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble, except those of Ag+, Pb2+, Cu+, and Hg22+. All common fluorides (F-) are soluble except those of Pb2+ and Group 2A(2). 4. All common sulfates (SO22-) are soluble, except those of Ca2+, Sr2+, Ba2+, Ag+, and Pb2+. Insoluble Ionic Compounds 1. All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca2+). 2. All common carbonates (CO32-) and phosphates (PO43-) are insoluble, except those of Group 1A(1) and NH4+. 3. All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH4+. 4-24 Sample Problem 4.3 Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations PROBLEM: Predict whether or not a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) potassium fluoride (aq) + strontium nitrate (aq) → (b) ammonium perchlorate (aq) + sodium bromide (aq) → PLAN: 4-25 Note reactant ions, write the possible cation-anion combinations, and use Table 4.1 to decide if the combinations are insoluble. Write the appropriate equations for the process. Sample Problem 4.3 SOLUTION: (a) The reactants are KF and Sr(NO3)2. The possible products are KNO3 and SrF2. KNO3 is soluble, but SrF2 is an insoluble combination. Molecular equation: 2KF (aq) + Sr(NO3)2 (aq) → 2 KNO3 (aq) + SrF2 (s) Total ionic equation: 2K+ (aq) + 2F- (aq) + Sr2+ (aq) + 2NO3- (aq) → 2K+ (aq) + 2NO3- (aq) + SrF2 (s) K+ and NO3- are spectator ions Net ionic equation: Sr2+ (aq) + 2F- (aq) → SrF2 (s) 4-26 Sample Problem 4.3 SOLUTION: (b) The reactants are NH4ClO4 and NaBr. The possible products are NH4Br and NaClO4. Both are soluble, so no precipitate forms. Molecular equation: NH4ClO4 (aq) + NaBr (aq) → NH4Br (aq) + NaClO4 (aq) Total ionic equation: NH4+ (aq) + ClO4- (aq) + Na+ (aq) + Br- (aq) → NH4+ (aq) + Br- (aq) + Na+ (aq) + ClO4- (aq) All ions are spectator ions and there is no net ionic equation. 4-27 Sample Problem 4.4 Using Molecular Depictions in Precipitation Reactions PROBLEM: The following molecular views show reactant solutions for a precipitation reaction (with H2O molecules omitted for clarity). (a) Which compound is dissolved in beaker A: KCl, Na2SO4, MgBr2, or Ag2SO4? (b) Which compound is dissolved in beaker B: NH4NO3, MgSO4, Ba(NO3)2, or CaF2? 4-28 Sample Problem 4.4 PLAN: Note the number and charge of each kind of ion and use Table 4.1 to determine the ion combinations that are soluble. SOLUTION: (a) Beaker A contains two 1+ ion for each 2- ion. Of the choices given, only Na2SO4 and Ag2SO4 are possible. Na2SO4 is soluble while Ag2SO4 is not. Beaker A therefore contains Na2SO4. (b) Beaker B contains two 1- ions for each 2+ ion. Of the choices given, only CaF2 and Ba(NO3)2 match this description. CaF2 is not soluble while Ba(NO3)2 is soluble. Beaker B therefore contains Ba(NO3)2. 4-29 Sample Problem 4.4 PROBLEM: (c) Name the precipitate and spectator ions when solutions A and B are mixed, and write balanced molecular, total ionic, and net ionic equations for this process. (d) If each particle represents 0.010 mol of ions, what is the maximum mass (g) of precipitate that can form (assuming complete reaction)? PLAN: (c) Consider the cation-anion combinations from the two solutions and use Table 4.1 to decide if either of these is insoluble. SOLUTION: The reactants are Ba(NO3)2 and Na2SO4. The possible products are BaSO4 and NaNO3. BaSO4 is insoluble while NaNO3 is soluble. 4-30 Sample Problem 4.4 Molecular equation: Ba(NO3)2 (aq) + Na2SO4 (aq) → 2NaNO3 (aq) + BaSO4 (s) Total ionic equation: Ba2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + SO42- (aq) → 2Na+ (aq) + 2NO3- (aq) + BaSO4 (s) Na+ and NO3- are spectator ions Net ionic equation: Ba2+ (aq) + SO42- (aq) → BaSO4 (s) 4-31 Sample Problem 4.4 PLAN: (d) Count the number of each kind of ion that combines to form the solid. Multiply the number of each reactant ion by 0.010 mol and calculate the mol of product formed from each. Decide which ion is the limiting reactant and use this information to calculate the mass of product formed. SOLUTION: There are 4 Ba2+ particles and 5 SO42- particles depicted. 2+ 4 Ba2+ particles x 0.010 mol Ba x 1 mol BaSO4 = 0.040 mol BaSO 4 1 mol Ba2+ 1 particle 24 SO42- particles x 0.010 mol SO4 x 1 mol BaSO4 = 0.050 mol BaSO4 1 mol SO421 particle 4-32 Sample Problem 4.4 Ba2+ ion is the limiting reactant, since it yields less BaSO4. 0.040 mol BaSO4 x 233.4 g BaSO4 = 9.3 g BaSO 4 1 mol BaSO4 4-33 Acid-Base Reactions An acid is a substance that produces H+ ions when dissolved in H2O. HX H2O → H+ (aq) + X- (aq) A base is a substance that produces OH- ions when dissolved in H2O. MOH H2O → M+ (aq) + OH- (aq) An acid-base reaction is also called a neutralization reaction. 4-34 Figure 4.7 The H+ ion as a solvated hydronium ion. H+ interacts strongly with H2O, forming H3O+ in aqueous solution. 4-35 Table 4.2 Selected Acids and Bases Acids Bases Strong Strong hydrochloric acid, HCl sodium hydroxide, NaOH hydrobromic acid, HBr potassium hydroxide, KOH hydriodic acid, HI calcium hydroxide, Ca(OH)2 nitric acid, HNO3 strontium hydroxide, Sr(OH)2 sulfuric acid, H2SO4 barium hydroxide, Ba(OH)2 perchloric acid, HClO4 Weak hydrofluoric acid, HF phosphoric acid, H3PO4 acetic acid, CH3COOH (or HC2H3O2) 4-36 Weak ammonia, NH3 Figure 4.8 Acids and bases as electrolytes. Strong acids and strong bases dissociate completely into ions in aqueous solution. They are strong electrolytes and conduct well in solution. 4-37 Figure 4.8 Acids and bases as electrolytes. Weak acids and weak bases dissociate very little into ions in aqueous solution. They are weak electrolytes and conduct poorly in solution. 4-38 Determining the Number of H+ (or OH-) Ions in Solution Sample Problem 4.5 PROBLEM: How many H+(aq) ions are in 25.3 mL of 1.4 M nitric acid? PLAN: Use the volume and molarity to determine the mol of acid present. Since HNO3 is a strong acid, moles acid = moles H+. volume of HNO3 convert mL to L and multiply by M mol of HNO3 mole of H+ = mol of HNO3 mol of H+ multiply by Avogadro’s number number of H+ ions 4-39 Sample Problem 4.5 SOLUTION: 35.3 mL soln x 1 L x 1.4 mol HNO3 = 0.035 mol HNO3 103 mL 1 L soln One mole of H+(aq) is released per mole of nitric acid (HNO3). H2O HNO3 (aq) = 0.035 mol HNO3 x 4-40 → H+ (aq) + NO3- (aq) 1 mol H+ 23 x 6.022x10 ions 1 mol HNO3 1 mol = 2.1x1022 H+ ions Sample Problem 4.6 Writing Ionic Equations for Acid-Base Reactions PROBLEM: Write balanced molecular, total ionic, and net ionic equations for the following acid-base reactions and identify the spectator ions. (a) hydrochloric acid (aq) + potassium hydroxide (aq) → (b) strontium hydroxide (aq) + perchloric acid (aq) → (c) barium hydroxide (aq) + sulfuric acid (aq) → PLAN: All reactants are strong acids and bases (see Table 4.2). The product in each case is H2O and an ionic salt. Write the molecular reaction in each case and use the solubility rules to determine if the product is soluble or not. 4-41 Sample Problem 4.6 SOLUTION: (a) hydrochloric acid (aq) + potassium hydroxide (aq) → Molecular equation: HCl (aq) + KOH (aq) → KCl (aq) + H2O (l) Total ionic equation: H+ (aq) + Cl- (aq) + K+ (aq) + OH- (aq) → K+ (aq) + Cl- (aq) + H2O (l) Net ionic equation: H+ (aq) + OH- (aq) → H2O (l) Spectator ions are K+ and Cl- 4-42 Sample Problem 4.6 SOLUTION: (b) strontium hydroxide (aq) + perchloric acid (aq) → Molecular equation: Sr(OH)2 (aq) + 2HClO4 (aq) → Sr(ClO4)2 (aq) + 2H2O (l) Total ionic equation: Sr2+ (aq) + 2OH- (aq) + 2H+ (aq) + 2ClO4- (aq) → Sr2+ (aq) + 2ClO4- (aq) + 2H2O (l) Net ionic equation: 2H+ (aq) + 2OH- (aq) → 2H2O (l) or H+ (aq) + OH- (aq) → H2O (l) Spectator ions are Sr2+ and ClO4- 4-43 Sample Problem 4.6 SOLUTION: (c) barium hydroxide (aq) + sulfuric acid (aq) → Molecular equation: Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l) Total ionic equation: Ba2+ (aq) + 2OH- (aq) + 2H+ (aq) + SO42- (aq) → BaSO4 (s) + H2O (l) The net ionic equation is the same as the total ionic equation since there are no spectator ions. This reaction is both a neutralization reaction and a precipitation reaction. 4-44 Figure 4.9 4-45 An aqueous strong acid-strong base reaction as a proton-transfer process. Figure 4.11 A gas-forming reaction with a weak acid. Molecular equation NaHCO3 (aq) + CH3COOH(aq) → CH3COONa (aq) + CO2 (g) + H2O (l) Total ionic equation Na+ (aq)+ HCO3- (aq) + CH3COOH (aq) → CH3COO- (aq) + Na+ (aq) + CO2 (g) + H2O (l) Net ionic equation HCO3-(aq) + CH3COOH (aq) → CH3COO- (aq) + CO2 (g) + H2O (l) 4-46 Sample Problem 4.7 Writing Proton-Transfer Equations for Acid-Base Reactions PROBLEM: Write balanced total and net ionic equations for the following reactions and use curved arrows to show how the proton transfer occurs. (a) hydriodic acid (aq) + calcium hydroxide (aq) → Give the name and formula of the salt present when the water evaporates. (b) potassium hydroxide (aq) + propionic acid (aq) → Note that propionic acid is a weak acid. Be sure to identify the spectator ions in this reaction. 4-47 Sample Problem 4.7 PLAN: In (a) the reactants are a strong acid and a strong base. The acidic species is therefore H3O+, which transfers a proton to the OH- from the base. SOLUTION: Total Ionic Equation: H+ transferred to OH- 2H3O+ (aq) + 2I- (aq) + Ca2+ (aq) + 2OH- (aq) → 2I- (aq) + Ca2+ (aq) + 4H2O (l) Net Ionic Equation: H3O+ (aq) + OH- (aq) → + H2O (l) When the water evaporates, the salt remaining is CaI2, calcium iodide. 4-48 Sample Problem 4.7 PLAN: In (b) the acid is weak; therefore it does not dissociate much and largely exists as intact molecules in solution. SOLUTION: Total Ionic Equation: H+ transferred to OH- K+ (aq) + OH- (aq) + CH3CH2COOH (aq) → K+ (aq) + H2O (l) + CH3CH2COO- (aq) Net Ionic Equation: CH3CH2COOH (aq) + OH- (aq) → CH3CH2COO- (aq) + H2O (l) K+ is the only spectator ion in the reaction. 4-49 Acid-Base Titrations • In a titration, the concentration of one solution is used to determine the concentration of another. • In an acid-base titration, a standard solution of base is usually added to a sample of acid of unknown molarity. • An acid-base indicator has different colors in acid and base, and is used to monitor the reaction progress. • At the equivalence point, the mol of H+ from the acid equals the mol of OH- ion produced by the base. – Amount of H+ ion in flask = amount of OH- ion added • The end point occurs when there is a slight excess of base and the indicator changes color permanently. 4-50 Figure 4.11 4-51 An acid-base titration. Sample Problem 4.8 Finding the Concentration of Acid from a Titration PROBLEM: A 50.00 mL sample of HCl is titrated with 0.1524 M NaOH. The buret reads 0.55 mL at the start and 33.87 mL at the end point. Find the molarity of the HCl solution. PLAN: Write a balanced equation for the reaction. Use the volume of base to find mol OH-, then mol H+ and finally M for the acid. volume of base (difference in buret readings) multiply by M of base mol of OHuse mole ratio as conversion factor mol of H+ and acid divide by volume (L) of acid molarity (M) of acid 4-52 Sample Problem 4.8 SOLUTION: NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) volume of base = 33.87 mL – 0.55 mL = 33.32 mL 33.32 mL soln x 1L x 0.1524 mol NaOH = 5.078x10-3 mol NaOH 103 mL 1 L soln Since 1 mol of HCl reacts with 1 mol NaOH, the amount of HCl = 5.078x10-3 mol. 5.078x10-3 mol HCl x 103 mL 1L 50.00 mL 4-53 = 0.1016 M HCl Oxidation-Reduction (Redox) Reactions Oxidation is the loss of electrons. The reducing agent loses electrons and is oxidized. Reduction is the gain of electrons. The oxidizing agent gains electrons and is reduced. A redox reaction involves electron transfer Oxidation and reduction occur together. 4-54 Figure 4.12 4-55 The redox process in compound formation. Table 4.3 Rules for Assigning an Oxidation Number (O.N.) General rules 1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 0 2. For a monoatomic ion: O.N. = ion charge 3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge. Rules for specific atoms or periodic table groups 1. For Group 1A(1): O.N. = +1 in all compounds 2. For Group 2A(2): O.N. = +2 in all compounds 3. For hydrogen: O.N. = +1 in combination with nonmetals 4. For fluorine: O.N. = -1 in combination with metals and boron 5. For oxygen: O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F) O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group 6. For Group 7A(17): ...
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