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CHAPTER 5 INTRODUCTION TO REACTIONS IN AQUEOUS SOLUTIONS PRACTICE EXAMPLES 1A In determining total Cl , we recall the definition of molarity: moles of solute per liter of solution. 0.438 mol NaCl 1 mol Cl from NaCl, Cl = = 0.438 M Cl 1 L soln 1 mol NaCl 0.0512 mol MgCl2 2 mol Cl from MgCl2 , Cl = = 0.102 M Cl 1 L soln 1 mol MgCl2 Cl total = Cl from NaCl + Cl from MgCl2 = 0.438 M + 0.102 M = 0.540 M Cl 1B 2A (a) 1.5 mg F1 g F 1 mol F = 7.9 10-5 M F L 1000 mg F 18.998 g F (b) 1.00 106 L 7.9 105 mol F1 mol CaF2 78.075 g CaF2 1 kg = 3.1 kg CaF2 1 mol CaF2 1000 g 1L 2 mol F In each case, we use the solubility rules to determine whether either product is insoluble. The ions in each product compound are determined by simply “switching the partners” of the reactant compounds. The designation “(aq)” on each reactant indicates that it is soluble. (a) Possible products are potassium chloride, KCl, which is soluble, and aluminum hydroxide, Al OH 3 , which is not. Net ionic equation: Al3+ aq + 3 OH aq Al OH 3 s (b) Possible products are iron(III) sulfate, Fe 2 SO 4 3 , and potassium bromide, KBr, both of which are soluble. No reaction occurs. 2B (c) Possible products are calcium nitrate, Ca(NO3)2, which is soluble, and lead(II) iodide, PbI 2 , which is insoluble. The net ionic equation is: Pb 2+ aq + 2 I aq PbI 2 s (a) Possible products are sodium chloride, NaCl, which is soluble, and aluminum phosphate, 3 AlPO 4 , which is insoluble. Net ionic equation: Al3+ aq + PO 4 aq AlPO 4 s (b) Possible products are aluminum chloride, AlCl 3 , which is soluble, and barium sulfate, BaSO 4 , which is insoluble. Net ionic equation: Ba 2+ aq + SO 4 97 2 aq BaSO 4 s Chapter 5: Introduction to Reactions in Aqueous Solutions (c) Possible products are ammonium nitrate, NH 4 NO 3 , which is soluble, and lead (II) carbonate, PbCO 3 , which is insoluble. Net ionic equation: Pb 2+ aq + CO3 2 aq PbCO3 s 3A Propionic acid is a weak acid, not dissociated completely in aqueous solution. Ammonia similarly is a weak base. The acid and base react to form a salt solution of ammonium propionate. NH 3 aq + HC3 H 5 O 2 aq NH 4 aq + C3 H 5 O 2 aq 3B Since acetic acid is a weak acid, it is not dissociated completely in aqueous solution (except at infinite dilution); it is misleading to write it in ionic form. The products of this reaction are the gas carbon dioxide, the covalent compound water, and the ionic solute calcium acetate. Only the latter exists as ions in aqueous solution. CaCO3 s + 2 HC 2 H 3 O 2 aq CO 2 g + H 2 O l + Ca 2+ aq + 2 C 2 H 3 O 2 aq 4A (a) This is a metathesis or double displacement reaction. Elements do not change oxidation states during this reaction. It is not an oxidation–reduction reaction. (b) The presence of O2(g) as a product indicates that this is an oxidation–reduction reaction. Oxygen is oxidized from O.S. = -2 in NO3- to O.S. = 0 in O2(g). Nitrogen is reduced from O.S. = +5 in NO3- to O.S. = +4 in NO2. 4B Vanadium is oxidized from O.S. = +4 in VO2+ to an O.S. = +5 in VO2+ while manganese is reduced from O.S. = +7 in MnO4- to O.S. = +2 in Mn2+. 5A Aluminum is oxidized (from an O.S. of 0 to an O.S. of +3 ), while hydrogen is reduced (from an O.S. of +1 to an O.S. of 0). Oxidation : Al s Al3+ aq + 3 e 2 Reduction: 2 H aq + 2 e + H 2 g 3 Net equation : 2 Al s + 6 H + aq 2 Al3+ aq + 3 H 2 g 5B Bromide is oxidized (from 1 to 0), while chlorine is reduced (from 0 to 1 ). Oxidation : 2 Br aq Br2 l + 2 e Reduction: Cl 2 g + 2 e 2 Cl aq Net equation : 2 Br aq + Cl2 g Br2 l + 2 Cl aq 6A Step 1: Write the two skeleton half reactions. MnO 4 aq Mn 2+ aq and Fe 2+ aq Fe3+ aq Step 2: Balance each skeleton half reaction for O (with H 2 O ) and for H atoms (with H + ). MnO 4 aq 8 H + aq Mn 2+ aq 4 H 2 O(l) 98 and Fe 2+ aq Fe3+ aq Chapter 5: Introduction to Reactions in Aqueous Solutions Step 3: Balance electric charge by adding electrons. MnO 4 aq 8 H + aq 5 e Mn 2+ aq 4 H 2 O(l) and Fe 2+ aq Fe3+ aq e Step 4: Combine the two half reactions Fe2+ aq Fe3+ aq + e 5 MnO 4 aq + 8 H + aq + 5 e Mn 2+ aq + 4 H 2O(l) MnO 4 aq + 8 H + aq + 5 Fe 2+ aq Mn 2+ aq + 4 H 2 O(l) + 5 Fe3+ aq 6B Step 1: Uranium is oxidized and chromium is reduced in this reaction. The “skeleton” 2+ 2 UO 2+ aq UO 2 aq and Cr2 O7 (aq) Cr 3 (aq) half-equations are: Step 2: First, balance the chromium skeleton half-equation for chromium atoms: 2 Cr2 O7 aq 2 Cr 3+ aq Next, balance oxygen atoms with water molecules in each half-equation: 2 2 UO 2+ aq + H 2O(l) UO 2 aq and Cr2 O7 (aq) 2Cr 3 (aq) 7H 2 O(l) Then, balance hydrogen atoms with hydrogen ions in each half-equation: 2 UO 2+ aq + H 2 O(l) UO 2 aq + 2 H + aq 2 Cr2 O7 (aq) 14H (aq) 2Cr 3 (aq) 7H 2 O(l) Step 3: Balance the charge of each half-equation with electrons. 2 UO 2+ aq + H 2 O(l) UO 2 aq + 2 H + aq + 2 e Cr2 O7 2 aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2O(l) Step 4: Multiply the uranium half-equation by 3 and add the chromium half-equation to it. 2 UO 2+ aq + H 2 O(l) UO 2 aq + 2 H + aq + 2 e 3 Cr2 O7 2 aq +14 H + aq + 6 e 2 Cr 3+ aq + 7 H 2O(l) 3 UO 2+ (aq)+Cr2 O 7 2- (aq)+14 H + (aq)+3 H 2 O(l) 3 UO 2 2+ (aq)+2 Cr 3+ (aq)+7 H 2 O(l)+6 H + (aq) Step 5: Simplify. Subtract 3 H 2 O (l) and 6 H+ (aq) from each side of the equation. 3 UO 2+ aq + Cr2 O7 7A 2 aq + 8 H + aq 3 UO2 2 aq + 2 Cr 3+ aq + 4 H 2O(l) Step 1: Write the two skeleton half-equations. 2 S(s) SO 3 (aq ) and OCl (aq ) Cl (aq ) Step 2: Balance each skeleton half-equation for O (with H 2 O ) and for H atoms (with H + ). 3 H 2O(l) + S s SO3 2 aq + 6 H + OCl (aq) 2H Cl (aq) H 2 O(l) Step 3: Balance electric charge by adding electrons. 3 H 2 O(l) + S s SO3 2 aq + 6 H + (aq) + 4 e OCl (aq) 2H (aq) 2e Cl (aq) H 2 O(l) 99 Chapter 5: Introduction to Reactions in Aqueous Solutions Step 4: Change from an acidic medium to a basic one by adding OH to eliminate H + . 2 3H 2 O(l) + S s + 6 OH (aq) SO3 aq + 6 H + (aq) + 6 OH (aq) + 4 e OCl aq + 2 H + (aq) + 2 OH (aq) + 2 e Cl aq + H 2 O(l) + 2 OH (aq) Step 5: Simplify by removing the items present on both sides of each half-equation, and combine the half-equations to obtain the net redox equation. 2 {S s + 6 OH (aq) SO3 aq + 3 H 2 O(l) + 4 e } 1 {OCl aq + H 2 O(l) + 2 e Cl aq + 2 OH (aq)} 2 S s + 6 OH (aq) + 2 OCl aq 2H 2 O(l) SO 3 2 aq + 3 H 2O(l) + 2 Cl aq + 4OH - Simplify by removing the species present on both sides. 2 Net ionic equation: S s + 2 OH aq + 2 OCl aq SO3 aq + H 2 O(l) + 2 Cl aq 7B Step 1: Write the two skeleton half-equations. MnO 4 aq MnO 2 s and SO32 (aq) SO 4 2 (aq) Step 2: Balance each skeleton half-equation for O (with H 2 O ) and for H atoms (with H + ). MnO 4 aq + 4 H + aq MnO 2 s + 2 H 2 O(l) 2 2 SO3 (aq) H 2 O(l) SO 4 (aq) 2H (aq) Step 3: Balance electric charge by adding electrons. MnO 4 aq + 4 H + aq + 3 e MnO 2 s + 2 H 2 O(l) SO3 2 aq + H 2 O(l) SO 4 2 aq + 2 H + aq + 2 e Step 4: Change from an acidic medium to a basic one by adding OH to eliminate H + . MnO 4 SO3 2 aq + 4 H + aq + 4 OH aq + 3 e MnO 2 s + H 2 O(l) + 4 OH aq aq + H 2 O(l) + 2 OH aq SO4 2 aq + 2 H + aq + 2 OH aq + 2 e Step 5: Simplify by removing species present on both sides of each half-equation, and combine the half-equations to obtain the net redox equation. {MnO 4 aq + 2 H 2 O(l) + 3 e MnO 2 s + 4 OH aq } 2 aq + 2 OH aq SO4 2 aq + H 2O(l) + 2 e } 3 2 2 MnO 4 aq + 3SO3 aq + 6 OH - (aq) + 4 H 2 O(l) 2 2 MnO 2 s + 3SO 4 aq + 3H 2 O(l) + 8 OH aq {SO3 2 Simplify by removing species present on both sides. Net ionic equation: 2 2 2 MnO 4 aq + 3SO3 aq + H 2 O(l) 2 MnO 2 s + 3SO 4 aq + 2 OH aq 100 Chapter 5: Introduction to Reactions in Aqueous Solutions 8A 8B Since the oxidation state of H is 0 in H2 (g) and is +1 in both NH3(g) and H2O(g), hydrogen is oxidized. A substance that is oxidized is called a reducing agent. In addition, the oxidation state of N in NO2 (g) is +4 , while it is 3 in NH 3 ; the oxidation state of the element N decreases during this reaction, meaning that NO2 (g) is reduced. The substance that is reduced is called the oxidizing agent. In Au CN 2 aq , gold has an oxidation state of +1; Au has been oxidized and, thus, Au(s) (oxidization state = 0), is the reducing agent. In OH- (aq), oxygen has an oxidation state of -2; O has been reduced and thus, O2(g) (oxidation state = 0) is the oxidizing agent. 9A We first determine the amount of NaOH that reacts with 0.500 g KHP. 1 mol KHP 1 mol OH 1 mol NaOH n NaOH = 0.5000 g KHP = 0.002448 mol NaOH 204.22 g KHP 1 mol KHP 1 mol OH 0.002448 mol NaOH 1000 mL [NaOH] = = 0.1019 M 24.03 mL soln 1 L 9B The net ionic equation when solid hydroxides react with a strong acid is OH- + H+ H2O. There are two sources of OH-: NaOH and Ca(OH)2. We compute the amount of OH- from each source and add the results. moles of OH from NaOH: = 0.235 g sample 92.5 g NaOH 1 mol NaOH 1 mol OH = 0.00543 mol OH 100.0 g sample 39.997 g NaOH 1 mol NaOH moles of OH from Ca OH 2 : = 0.235 g sample 7.5 g Ca OH 2 100.0 g sample 1 mol Ca OH 2 74.093 g Ba OH 2 2 mol OH 1 mol Ca OH 2 = 0.00048 mol OH total amount OH - = 0.00543 mol from NaOH + 0.00048 mol from Ca OH 2 = 0.00591 mol OH [HCl] = 0.00591 mol OH 1 mol H + 1 mol HCl 1000 mL soln = 0.130 M 45.6 mL HCl soln 1 mol OH 1 mol H + 1 L soln 10A First, determine the mass of iron that has reacted as Fe 2+ with the titrant. The balanced chemical equation provides the essential conversion factor to answer this question. Namely: 5 Fe 2+ aq MnO 4 aq 8 H aq 5 Fe 3+ aq Mn 2+ aq 4 H 2 Ol mass Fe = 0.04125 L titrant 0.02140 mol MnO 4 1 L titrant Then determine the % Fe in the ore. % Fe = 5 mol Fe 2+ 1 mol MnO 4 55.847 g Fe 1 mol Fe 2+ = 0.246 g Fe 0.246 g Fe 100% = 65.4% Fe 0.376 g ore 10B The balanced equation provides us with the stoichiometric coefficients needed for the solution. 10 CO 2 g 2 Mn 2+ aq 8 H 2 O l Namely: 5 C2 O 4 2- aq 2 MnO 4 aq 16 H aq 101 Chapter 5: Introduction to Reactions in Aqueous Solutions 1 mol Na 2 C 2 O 4 amount MnO 4 = 0.2482 g Na 2 C 2 O 4 134.00 g Na 2 C 2 O 4 = 0.0007409 mol MnO 4 1 mol C 2 O 4 2 1 mol Na 2 C 2 O 4 2 mol MnO 4 5 mol C 2 O 4 2 0.0007409 mol MnO 4 1000 mL 1 mol KMnO 4 [KMnO 4 ] = = 0.03129 M KMnO4 23.68 mL soln 1 L 1 mol MnO 4 INTEGRATIVE EXAMPLE A. First, balance the equation. Break down the reaction of chlorate and ferrous ion as follows: ClO3 +6H +6e Cl +3H 2 O 6 Fe 2 Fe3 e Net reaction: ClO 3 6Fe 2 6H Cl 6Fe3 3H 2 O The reaction between Fe2+ and Ce4+ is already balanced. To calculate the moles of Fe2+ that remains after the reaction with ClO3-, determine the moles of Ce4+ that react with Fe2+: mol Ce4+ = 0.01259 L × 0.08362 M = 1.0527×10-3 mol = mol of excess Fe2+ total mol of Fe2+ = 0.0500 L × 0.09101 = 4.551×10-3 mol Therefore, the moles of Fe2+ reacted = 4.551×10-3 - 1.0527×10-3 = 3.498×10-3 mol. To determine the mass of KClO3, use the mole ratios in the balanced equation in conjunction with the molar mass of KClO3. 1 mol ClO3 1 mol KClO3 122.54 g KClO3 3 2 3.498 10 mol Fe 2 6 mol Fe 1 mol ClO3 1 mol KClO3 = 0.07144 g KClO3 %KClO3 = B. 0.07144 g 100% = 49.89% 0.1432 g First, balance the equation. Break down the reaction of arsenous acid and permanganate as follows: 5 H 3 AsO3 + H 2 O H3 AsO 4 + 2e- + 2H + 2 MnO 4 + 8H + 5e- Mn 2+ + 4H 2 O Net reaction: 5H 3AsO3 + 2MnO 4 6H + 5H 3AsO 4 + 2Mn 2+ + 3H 2O moles of MnO4- = 0.02377 L × 0.02144 M = 5.0963×10-4 mol 102 Chapter 5: Introduction to Reactions in Aqueous Solutions To calculate the mass of As, use the mole ratios in the balanced equation in conjunction with the molar mass of As: 5 mol H 3 AsO3 1 mol As 74.922 g As 5.0963 104 mol MnO 4 2 mol MnO 4 1 mol H 3AsO3 1 mol As = 0.095456 g As 0.095456 g mass% As = 100% = 1.32% 7.25 g EXERCISES Strong Electrolytes, Weak Electrolytes, and Nonelectrolytes 1. (a) Because its formula begins with hydrogen, HC6 H 5O is an acid. It is not listed in Table 5-1, so it is a weak acid. A weak acid is a weak electrolyte. (b) Li 2SO 4 is an ionic compound, that is, a salt. A salt is a strong electrolyte. (c) MgI 2 also is a salt, a strong electrolyte. (d) CH3 CH 2 2 O is a covalent compound whose formula does not begin with H. Thus, it is neither an acid nor a salt. It also is not built around nitrogen, and thus it does not behave as a weak base. This is a nonelectrolyte. (e) Sr OH 2 is a strong electrolyte, one of the strong bases listed in Table 5-2. 3. HCl is practically 100% dissociated into ions. The apparatus should light up brightly. A solution of both HCl and HC2H3O2 will yield similar results. In strongly acidic solutions, the weak acid HC2H3O2 is molecular and does not contribute to the conductivity of the solution. However, the strong acid HCl is practically dissociated into ions and is unaffected by the presence of the weak acid HC2H3O2. The apparatus should light up brightly. 5. (a) (c) Barium bromide: strong electrolyte Ammonia: weak electrolyte (b) Propionic acid: weak electrolyte Ion Concentrations 7. 0.238 mol KNO 3 1 mol K + = 0.238 M K + 1 L soln 1 mol KNO 3 (a) K+ = (b) NO3 = 0.167 mol Ca NO3 2 1 L soln 2 mol NO3 = 0.334 M NO3 1 mol Ca NO3 2 103 Chapter 5: Introduction to Reactions in Aqueous Solutions (c) (d) Al3+ = Na + 0.083 mol Al2 SO 4 3 1 L soln 2 mol Al3+ = 0.166 M Al3+ 1 mol Al2 SO 4 3 0.209 mol Na 3 PO 4 3 mol Na + = = 0.627 M Na + 1 L soln 1 mol Na 3 PO 4 9. Conversion pathway approach: OH = 0.132 g Ba OH 2 8H 2 O 1000 mL 1 mol Ba OH 2 8H 2 O 2 mol OH 275 mL soln 1 L 315.5 g Ba OH 2 8H 2 O 1 mol Ba OH 2 8 H 2 O = 3.04 103 M OH Stepwise approach: 0.132 g Ba OH 2 8H 2 O 1000 mL = 0.480 g/L 275 mL soln 1 L 0.00152 mol Ba OH 2 ×8H 2O 0.480 g 1 mol Ba OH 2 8H 2 O = 315.5 g Ba OH 2 8H 2 O L L 0.00152 mol Ba OH 2 ×8H 2 O L 11. (a) [Ca 2+ ] = (c) 13 1 L solution 32.8 mg K + + (b) [K ] = 14.2 mg Ca 2+ 100 mL solution [Zn 2+ ] = 225 g Zn 2+ 1 mL solution 2 mol OH 1 mol Ba OH 2 8 H 2 O 1 g Ca 2+ 1000 mg Ca 2+ 1 g K+ 1000 mg K 1 g Zn + 2+ 1 mol Ca 2+ 40.078 g Ca 2+ 3.54 104 M Ca 2+ 1000 mL solution 1 L solution 2+ 1 10 g Zn 6 = 3.04 10-3 M OH - 1000 mL solution 1 L solution 1 mol K + 39.0983 g K 8.39 103 M K + + 1 mol Zn 2+ 65.39 g Zn 2+ 3.44 103 M Zn 2+ In order to determine the solution with the largest concentration of K+, we begin by converting each concentration to a common concentration unit, namely, molarity of K+. 0.0850 M K 2 SO 4 2 mol K + 0.17 M K + 1 L solution 1 mol K 2 SO 4 1000 mL solution 1.25 g KBr 1 mol KBr 1 mol K + 0.105 M K + 100 mL solution 1 L solution 119.0023 g KBr 1 mol KBr 1000 mL solution 8.1 mg K + 1 g K+ 1 mol K + 0.207 M K + 1 mL solution 1 L solution 1000 mg K + 39.0983 g K + Clearly, the solution containing 8.1 mg K+ per mL gives the largest K+ of the three solutions. 104 Chapter 5: Introduction to Reactions in Aqueous Solutions 15. Determine the amount of I in the solution as it now exists, and the amount of I in the solution of the desired concentration. The difference in these two amounts is the amount of I that must be added. Convert this amount to a mass of MgI 2 in grams. 1 L moles of I in final solution = 250.0 mL 0.1000 mol I = 0.02500 mol I 1000 mL 1 L soln 1 L 0.0876 mol KI 1 mol I moles of I in KI solution = 250.0 mL = 0.0219 mol I 1000 mL 1 L soln 1 mol KI 1 mol MgI 2 278.11 g MgI 2 1000 mg mass MgI 2 required = 0.02500 0.0219 mol I 2 mol I 1 mol MgI 2 1 g = 4.3 102 mg MgI 2 17. moles of chloride ion 0.625 mol KCl 1 mol Cl 0.385 mol MgCl 2 2 mol Cl = 0.225 L + 0.615 L 1 L soln 1 mol KCl 1 L soln 1 mol MgCl 2 0.615 mol Cl = 0.732 M = 0.141 mol Cl + 0.474 mol Cl = 0.615 mol Cl- Cl = 0.225 L + 0.615 L Predicting Precipitation Reactions 19. In each case, each available cation is paired with the available anions, one at a time, to determine if a compound is produced that is insoluble, based on the solubility rules of Chapter 5. Then a net ionic equation is written to summarize this information. (a) Pb 2+ aq + 2 Br aq PbBr2 s (b) No reaction occurs (all are spectator ions). (c) Fe3+ aq + 3 OH aq Fe OH 3 s 21. (a) 23. Mixture Result (Net Ionic Equation) HI a + Zn NO3 2 (aq): No reaction occurs. (b) CuSO 4 aq + Na 2 CO3 aq : (c) Cu NO3 2 aq + (a) Cu 2+ aq + CO3 aq CuCO3 s 3 Na 3 PO 4 aq : 3Cu 2+ aq + 2 PO 4 aq Cu 3 PO 4 2 s 2 Add K 2 SO 4 aq ; BaSO 4 s will form and MgSO4 will not precipitate. BaCl2 s + K 2 SO 4 aq BaSO 4 s + 2 KCl aq 105 Chapter 5: Introduction to Reactions in Aqueous Solutions (b) Add H 2 O l ; Na 2 CO3 s dissolves, but MgCO3 (s) will not dissolve (appreciably). water Na 2 CO3 s 2 Na + aq + CO3 (c) 2 aq Add KCl(aq); AgCl(s) will form, while Cu(NO3)2 (s) will dissolve. AgNO3 s + KCl aq AgCl s + KNO3 aq 25. (a) Mixture Sr NO3 2 aq + K 2SO 4 aq : Net Ionic Equation 2 Sr 2+ aq + SO 4 aq SrSO 4 s (b) Mg NO3 2 aq + NaOH aq : Mg 2+ aq + 2 OH aq Mg OH 2 s (c) BaCl2 aq + K 2SO 4 aq : 2 Ba 2+ (aq) SO 4 (aq) BaSO4 (s) (upon filtering, KCl (aq) is obtained) Acid–Base Reactions 27. The type of reaction is given first, followed by the net ionic equation. (a) Neutralization: OH aq + HC2 H3 O2 aq H 2 O l + C2 H3 O 2 aq (b) No reaction occurs. This is the physical mixing of two acids. (c) Gas evolution: FeS s + 2 H + aq H 2 S g + Fe 2+ aq aq + H + aq "H 2 CO3 aq " H + aq Mg 2+ aq + H 2 g (d) Gas evolution: HCO3 (e) Redox: Mg s + 2 29. 31. As a salt: NaHSO4 aq Na + aq + HSO 4 As an acid: HSO 4 aq + OH aq H 2 O l + CO 2 g aq 2 H 2 O l + SO 4 aq Use (b) NH3(aq): NH3 affords the OH- ions necessary to form Mg(OH)2(s). Applicable reactions: {NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)} 2 MgCl2(aq) Mg2+(aq) + 2 Cl-(aq) Mg2+(aq) + 2 OH-(aq) Mg(OH)2(s) 106 Chapter 5: Introduction to Reactions in Aqueous Solutions Oxidation–Reduction (Redox) Equations 33. 35. (a) The O.S. of H is +1, that of O is 2 , that of C is +4 , and that of Mg is +2 on each side of this equation. This is not a redox equation. (b) The O.S. of Cl is 0 on the left and 1 on the right side of this equation. The O.S. of Br is 1 on the left and 0 on the right side of this equation. This is a redox reaction. (c) The O.S. of Ag is 0 on the left and +1 on the right side of this equation. The O.S. of N is +5 on the left and +4 on the right side of this equation. This is a redox reaction. (d) On both sides of the equation the O.S. of O is 2 , that of Ag is +1 , and that of Cr is +6 . Thus, this is not a redox equation. (a) Reduction: aq + 6 H + aq + 4 e S2O32 aq + 3 H 2O(l) 2 NO3 aq +10 H + aq + 8 e N 2 O g + 5 H 2 O l Al s + 4 OH aq Al OH 4 aq + 3 e 2SO3 (b) Reduction: 37. 2 (c) Oxidation: (a) Oxidation: { 2 I aq I 2 s + 2 e Reduction: { MnO 4 aq + 8 }5 H + aq + 5 e Mn 2+ aq + 4 H 2 O l }2 Net: 10 I aq + 2 MnO 4 aq +16 H + aq 5 I 2 s + 2 Mn 2+ aq + 8 H 2 O l (b) Oxidation: { N 2 H 4 l N 2 g + 4 H + aq + 4 e } 3 + Reduction: { BrO3 aq + 6 H aq + 6 e Br aq + 3 H 2 O l Net: (c) 3 N 2 H 4 l + 2 BrO3 aq 3 N 2 g + 2 Br aq + 6 H 2 O l Oxidation: Fe 2+ aq Fe3+ aq + e Reduction: VO 4 Net: 3 aq + 6 Fe 2+ aq + VO 4 3 H + aq + e VO 2+ aq + 3 H 2 O l aq + 6 H + aq Fe3+ aq + VO2+ aq + 3 H 2 O l (d) Oxidation: { UO 2+ aq + H 2 O l UO2 Reduction: { NO3 aq + 4 H + aq + 3 e Net: 39. (a) } 2 3 UO 2+ aq + 2 NO 3 aq + 2 aq + 2 H + aq + 2 e NO g + 2 H 2 O l 2 H + aq 3 UO 2 2 aq + 2 }3 }2 NO g + H 2 O l Oxidation: { MnO 2 s + 4 OH aq MnO 4 aq + 2 H 2 O(l) + 3 e } 2 Reduction: ClO3 aq + 3 H 2 O(l) + 6 e Cl aq + 6 OH aq Net: 2 MnO 2 s + ClO3 aq + 2 OH aq 2MnO 4 aq + Cl aq + H 2 O(l) 107 Chapter 5: Introduction to Reactions in Aqueous Solutions (b) Oxidation: { Fe OH 3 s + 5 OH aq FeO 4 2 aq + 4 H 2 O(l) + 3 e Reduction: { OCl aq + H 2 O(l) + 2 e Cl aq + 2OH - aq Net: 2 Fe OH 3 s + 3 OCl aq + 4 OH aq 2FeO 4 (c) 2 aq + 3 }2 }3 Cl aq + 5 H 2 O(l) Oxidation: { ClO 2 (aq) + 2 OH aq ClO3 aq + H 2 O l + e } 5 Reduction: ClO 2 (aq) + 2 H 2 O l + 5 e Cl (aq) + 4 OH aq 6 ClO 2 (aq) + 6 OH aq 5ClO3 aq + Cl aq + 3 H 2 O (l) Net: (d) 41. (a) Oxidation: (Ag (s) Ag+ (aq) + 1 e¯ ) 3 Reduction: 4 H2O(l) + CrO42- + 3 e¯ Cr(OH)3(s) + 5 OH Net: 3 Ag(s) + CrO42- + 4 H2O(l) 3 Ag+(aq) + Cr(OH)3(s) + 5 OHOxidation: Cl2 g +12 OH aq 2 ClO3 aq + 6 H 2 O(l) +10 e Reduction: { Cl 2 g + 2 e 2 Cl aq (b) 43. (a) }5 Net: 6 Cl2 g +12 OH aq 10 Cl aq + 2 ClO3 aq + 6 H 2 O(l) Or: 3 Cl2 g + 6 OH aq 5 Cl aq + ClO3 aq + 3 H 2 O(l) aq + 2 H 2O(l) 2 HSO3 aq + 2 H + aq + 2 e 2 2 Reduction: S2 O 4 aq + 2 H + aq + 2 e S2 O3 aq + H 2 O (l) 2 2 Net: 2 S2 O 4 aq + H 2 O(l) 2 HSO3 aq + S2 O3 aq Oxidation: S2 O 4 2 aq + H 2 O l NO3 aq + 2 H + aq + 2 e + 2+ Reduction: { MnO 4 aq + 8 H aq + 5 e Mn aq + 4 H 2 O l Oxidation: { NO 2 }5 }2 Net: 5 NO 2 aq + 2 MnO 4 aq + 6 H + aq 5 NO3 aq + 2 Mn 2+ aq + 3 H 2 O l (b) (c) Oxidation: { Mn2+ (aq) + 4 OH- (aq) MnO2 (s) + 2 H2O (l) + 2 e- } 3 Reduction: { MnO4- (aq) + 2 H2O (l) + 3 e- MnO2 (s) + 4 OH- (aq) } 2 Net: 3 Mn2+ (aq) + 2 MnO4- (aq) + 4 OH- (aq) 5 MnO2 (s) + 2 H2O (l) Oxidation: { C2 H 5 OH CH 3 CHO + 2 H + aq + 2 e } 3 aq +14 H aq + 6 e 2 Cr aq + 7 H 2 O l Net: Cr2 O 7 2 aq + 8 H + aq + 3 C2 H 5 OH 2 Cr 3+ aq + 7 H 2 O l + 3 CH 3 CHO Reduction: Cr2 O 7 2 + 108 3+ Chapter 5: Introduction to Reactions in Aqueous Solutions 45. For the purpose of balancing its redox equation, each of the reactions is treated as if it takes place in acidic aqueous solution. (a) 2 H2O(g) + CH4(g) CO2(g) + 8 H+(g) + 8 e{2 e- + 2 H+(g) + NO(g) ½ N2(g) + H2O(g) }4 CH4(g) + 4 NO(g) 2 N2(g) + CO2(g) + 2 H2O(g) (b) {H2S(g) 1/8 S8(s)+ 2 H+(g) + 2 e}2 4 e- + 4 H+(g) + SO2(g) 1/8 S8(s) + 2 H2O(g) 2 H2S(g) + SO2(g) 3/8 S8(s) + 2 H2O(g) or 16 H2S(g) + 8 SO2(g) 3 S8(s) + 16 H2O(g) (c) {Cl2O(g) + 2 NH4+(aq) + 2 H+(aq) + 4 e- 2 NH4Cl(s) + H2O(l) {2 NH3(g) N2(g) + 6 e- + 6 H+(aq) 6 NH3(g) + 6 H+(aq) 6 NH4+(aq) 10 NH3(g) +3 Cl2O(g) 6 NH4Cl(s) + 2 N2(g) + 3 H2O(l) }3 } 2 Oxidizing and Reducing Agents 47. The oxidizing agents experience a decrease in the oxidation state of one of their elements, while the reducing agents experience an increase in the oxidation state of one of their elements. aq is the reducing agent; the O.S. of S = +4 in SO32 and = +6 in SO 4 2 . MnO 4 aq is the oxidizing agent; the O.S. of Mn = +7 in MnO 4 and + 2 in Mn 2+ . 2 (a) SO3 (b) H 2 g is the reducing agent; the O.S. of H = 0 in H 2 g and = +1 in H 2 O g . NO 2 g is the oxidizing agent; the O.S. of N = +4 in NO 2 g and 3 in NH 3 (g). (c) Fe CN 6 4 aq is the reducing agent; the O.S. of Fe = +2 in Fe CN 6 3 4 and = +3 in Fe CN 6 . H 2O 2 aq is the oxidizing agent; the O.S. of O = 1 in H 2 O 2 and = 2 in H 2 O . Neutralization and Acid–Base Titrations 49. The problem is most easily solved with amounts in millimoles. VNaOH = 10.00 mL HCl aq 0.128 mmol HCl 1 mmol H + 1 mmol OH 1 mL HCl aq 1 mmol HCl 1 mmol H + 109 Chapter 5: Introduction to Reactions in Aqueous Solutions 51. 1 mL NaOH aq 1 mmol NaOH = 13.3 mL NaOH aq soln 0.0962 mmol NaOH 1 mmol OH The net reaction is OH aq + HC3 H5O 2 aq H 2 O(l) + C3 H 5O2 aq . Conversion pathway approach: Vbase = 25.00 mL acid 0.3057 mmol HC3 H 5 O 2 1 mL acid 1 mmol KOH 1 mmol HC3 H 5 O 2 1 mL base 2.155 mmol KOH = 3.546 mL KOH solution Stepwise approach: 25.00 mL acid 0.3057 mmol HC3 H 5 O 2 = 7.643 mmol HC3 H 5 O 2 1 mL acid 1 mmol KOH 7.643 mmol HC3H 5O 2 = 7.643 mmol KOH 1 mmol HC3 H 5 O 2 7.643 mmol KOH 53. 1 mL base 2.155 mmol KOH NaOH aq + HCl aq NaCl aq + H 2 O(l) is the titration reaction. 0.02834 L [NaOH] = 55. = 3.546 mL KOH solution 0.1085 mol HCl 1mol NaOH 1L soln 1 mol HCl = 0.1230 M NaOH 0.02500 L sample The mass of acetylsalicylic acid is converted to the amount of NaOH, in millimoles, that will react with it. 0.32 g HC9 H 7 O 4 1 mol HC9 H 7 O 4 1 mol NaOH 1000 mmol NaOH NaOH = 23 mL NaOH aq 180.2 g HC9 H 7 O 4 1 mol HC9 H 7 O 4 1 mol NaOH = 0.077 M NaOH 57. The equation for the reaction is HNO3 aq + KOH aq KNO3 aq + H 2 O 1 . This equation shows that equal numbers of moles are needed for a complete reaction. We compute the amount of each reactant. mmol HNO 3 = 25.00 mL acid 0.132 mmol HNO 3 = 3.30 mmol HNO 3 1 mL acid 0.318 mmol KOH = 3.18 mmol KOH 1 mL base There is more acid present than base. Thus, the resulting solution is acidic. mmol KOH = 10.00 mL acid 110 Chapter 5: Introduction to Reactions in Aqueous Solutions 59. Vbase = 5.00 mL vinegar 61. 1.01 g vinegar 4.0 g HC2 H 3O 2 1 mol HC2 H 3O 2 1 mL 100.0 g vinegar 60.0 g HC2 H3O2 1 mol NaOH 1 L base 1000 mL = 34 mL base 1 mol HC 2 H 3O 2 0.1000 mol NaOH 1 L Answer is (d): 120 % of necessary titrant added in titration of NH3 required for equivalence point 5 NH3 + 5 HCl + 1 HCl 5 NH4+ + 6 Cl- + H3O+ (depicted in question's drawing ) 20 % excess Stoichiometry of Oxidation–Reduction Reactions 63. Conversion pathway approach: 0.1078 g As 2 O 3 [ MnO 4 ]= 1mol As 2 O 3 197.84 g As 2 O 3 22.15 mL 4 mol MnO 4 5 mol As 2 O3 1L 1mol KMnO 4 1mol MnO 4 1000 mL Stepwise approach: mol KMnO 4 L solution 1mol As 2 O3 0.1078 g As 2 O3 = 5.449 10-4 mol As 2 O3 197.84 g As 2 O3 [ KMnO 4 ]= 5.449 10-4 mol As 2 O3 4.359 10-4 mol MnO 4 22.15 mL 1L 1000 mL [ KMnO 4 ]= 4 mol MnO 4 = 4.359 10-4 mol MnO 4 5 mol As 2 O3 1mol KMnO 4 1mol MnO 4 = 4.359 10-4 mol KMnO 4 0.02215 L solution mol KMnO 4 L solution = 4.359 10-4 mol KMnO 4 0.02215 L solution 111 = 1.968 10-2 M = 0.01968 M KMnO 4 Chapter 5: Introduction to Reactions in Aqueous Solutions 65. First, we will determine the mass of Fe, then the percentage of iron in the ore. 2 0.05051 mol Cr2 O7 1 L 6 mol Fe 2+ 55.85 g Fe mass Fe = 28.72 mL 2 1000 mL 1 L soln 1 mol Fe 2+ 1 mol Cr2 O7 0.4861g Fe mass Fe = 0.4861 g Fe % Fe = 100% 53.23% Fe 0.9132 g ore 67. First, balance the titration equation: }5 aq 2 CO2 g + 2 e Reduction: { MnO 4 aq + 8 H + aq + 5 e Mn 2+ aq + 4 H 2 O l } 2 2 Net: 5 C2 O 4 aq + 2 MnO 4 aq +16 H + aq 10 CO 2 g + 2 Mn 2+ aq + 8 H 2 O l Oxidation: { C2 O 4 2 mass Na 2C2O4 =1.00 L satd soln Na 2 C2 O 4 1000 mL 25.8 mL satd soln KMnO 4 0.02140 mol KMnO 4 1L 5.00 mL satd soln Na 2 C2 O 4 1000 mL KMnO4 1 mol MnO 4 5 mol C2 O4 2 1 mol Na 2 C2 O4 134.0 g Na 2 C2 O 4 1 mol KMnO4 2 mol MnO4 1 mol Na 2 C2 O 4 1 mol C2 O 4 2 mass Na 2C2O4 = 37.0 g Na 2 C2 O 4 Integrative and Advanced Exercises 71. A possible product, based on solubility rules, is Ca 3 (PO 4 ) 2 . We determine the % Ca in this compound. 3 40.078 g Ca 2 30.974 g P 8 15.999 g O 120.23 g Ca 61.948 g P 127.99 g O 310.17 g 120.23 g Ca % Ca 100% 38.763% 310.17 g Ca 3 (PO 4 ) 2 molar mass Thus, Ca 3 (PO 4 ) 2 is the predicted product. The net ionic equation follows. 2 3 Ca 2 (aq) 2 HPO 4 (aq) Ca 3 (PO 4 ) 2 (s) 2 H (aq) 74. Let us first determine the mass of Mg in the sample analyzed. Conversion pathway approach: 1 mol Mg 2 P 2 O7 2 mol Mg 24.305 g 0.0120 g Mg 222.55 g Mg 2 P 2 O7 1 mol Mg 2 P2 O7 1 mol Mg 0.0120 g Mg ppm Mg 106 g sample 108 ppm Mg 110.520 g sample mass Mg 0.0549 g Mg 2 P2 O7 112 Chapter 5: Introduction to Reactions in Aqueous Solutions Stepwise approach: 0.0549 g Mg 2 P2 O 7 1 mol Mg 2 P 2 O7 = 2.47 10-4 mol Mg 2 P 2 O7 222.55 g Mg 2 P 2 O 7 2.47 10-4 mol Mg 2 P 2 O7 4.93 10-4 mol Mg 2 mol Mg 4.93 10-4 mol Mg 1 mol Mg 2 P2 O 7 24.305 g 0.0120 g Mg 1 mol Mg ppm Mg 106 g sample 0.0120 g Mg 108 ppm Mg 110.520 g sample 75. Let V represent the volume of added 0.248 M CaCl2 that must be added. We know that [Cl ] = 0.250 M, but also, 0.248 mol CaCl 2 0.186 mol KCl 1 mol Cl 2 mol Cl V 0.335 L 1 L soln 1 mol KCl 1 L soln 1 mol CaCl 2 [Cl ] 0.335 L V 0.250 (0.335 V ) 0.0838 0.250 V 0.0623 0.496 V V 0.0838 0.0623 0.0874 L 0.496 0.250 80. (a) [FeS2 + 8 H2O → Fe3+ + 2 SO42− + 16 H+ + 15 e−] × 4 [O2 + 4 H+ + 4 e− → 2 H2O] × 15 overall: 4 FeS2(s) + 15 O2(g) + 2 H2O(l) → 4 Fe3+(aq) + 8 SO42−(aq) + 4 H+(aq) (b) One kilogram of tailings contains 0.03 kg (30 g) of S. We have moles of FeS2 = 30 g S 1 mol S 1 mol FeS2 0.468 mol FeS2 32.07 g S 2 mol S moles of H+ = 0.468 mol FeS2 4 mol H + 0.467 mol H + 4 mol FeS2 moles of CaCO3 = 0.467 mol H + 1 mol CaCO3 0.234 mol CaCO3 2 mol H + mass of CaCO3 = 0.234 mol CaCO3 100.09 g CaCO3 23.4 g CaCO3 1 mol CaCO3 113 Chapter 5: Introduction to Reactions in Aqueous Solutions 83. Oxidation : {2 Cl (aq) Cl 2 (g) 2 e } 3 2 Reduction : Cr2 O 7 (aq) 14 H (aq) 6 e 2 Cr 3 (aq) 7 H 2 O 2 Net : 6 Cl (aq) Cr2 O 7 (aq) 14 H (aq) 2 Cr 3 (aq) 7 H 2 O 3 Cl 2 (g) We need to determine the amount of Cl2(g) produced from each of the reactants. The limiting reactant is the one that produces the lesser amount of Cl2.. 1.15 g 30.1 g HCl 1 mol HCl 1 mol Cl 3 mol Cl 2 1 mL 100. g soln 36.46 g HCl 1 mol HCl 6 mol Cl 1.54 mol Cl 2 amount Cl 2 325 mL 2 98.5 g K 2 Cr2 O 7 1 mol K 2 Cr2 O 7 1 mol Cr2 O 7 3 mol Cl 2 amount Cl 2 62.6 g 100. g sample 294.2 g K 2 Cr2 O 7 1 mol K 2 Cr2 O 7 1 mol Cr2 O 7 2 0.629 mol Cl 2 , the amount produced from the limiting reactant Then we determine the mass of Cl2(g) produced. = 0.629 mol Cl2 × 85. Cl 2 (g) NaClO 2 (aq) NaCl(aq) ClO 2 (g) 70.91 g Cl2 = 44.6 g Cl2 1 mol Cl2 (not balanced) Cl 2 (g) 2 NaClO 2 (aq) 2 NaCl(aq) 2 ClO 2 (g) amount ClO 2 1 gal 88. 2 mol ClO2 67.45 g ClO 2 3.785 L 2.0 mol NaClO 2 1 gal 1 L soln 2 mol NaClO 2 1 mol ClO2 97 g ClO 2 produced 5.0 102 g ClO 2 (g) 100 g ClO 2 calculated (a) First, balance the redox equations needed for the calculation. Oxidation: {HSO3- (aq)+ H2O(l) SO42- (aq) + 3 H+ (aq) + 2 e- } × 3 Reduction: {IO3- (aq) + 6 H+ (aq) + 6 e- I-(aq) + 3 H2O(l) } × 1 Net: 3 HSO3- (aq) + IO3- (aq) 3 SO42- (aq) + 3 H+ (aq) + I- (aq) The solution volume of 5.00 L contains 29.0 g NaIO3. This represents 29.0 g/197.9g/mol NaIO3 = 0.147 mol NaIO3. (b) From the above equation, we need 3 times that molar amount of NaHSO3, which is 3(0.147 mol) = 0.441 mol NaHSO3; the molar mass of NaHSO3 is 104.06 g/mol. The required mass then is 0.441(104.06) = 45.9 g. 114 Chapter 5: Introduction to Reactions in Aqueous Solutions For the second process: Oxidation: {2 I-(aq) I2(aq) + 2 e- } × 5 Reduction: {2 IO3- (aq) + 12 H+ (aq) + 10 e- I2(aq) + 6 H2O(l) } × 1 Net: 5 I-(aq) + IO3- (aq) + 6 H+ (aq) 3 I2(aq) + 3 H2O(l) In Step 1, we produced 1 mol of I- for every mole of IO3- reactant; therefore we had 0.147 mol I-. In step 2, we require 1/5 mol IO3- for every mol of I-. We require only 1.00 L of the solution in the question instead of the 5.00 L in the first step. 89. Mg(OH)2(aq) + 2 HCl(aq) → MgCl2(aq) + 2 H2O(l) (1) Al(OH)3(aq) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l) (2) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) (3) initial moles of HCl = 0.0500 L 0.500 mol = 0.0250 mol 1L moles of HCl that reacted with NaOH = moles of HCl left over from reaction with active ingredients = 0.0165 L 0.377 mol NaOH 1 mol HCl = 6.22 10-3 mol 1L 1 mol NaOH moles of HCl that react with active ingredients = 0.0250 mol - 6.22 10-3 mol = 0.0188 mol # moles HCl that # moles HCl that react with Mg(OH) + react with Al(OH) = total moles of HCl reacted/used 3 2 # moles HCl that react with Mg(OH)2 = 1 mol Mg(OH) 2 2 mol HCl X grams Mg(OH) 2 58.32 g Mg(OH) 2 1 mol Mg(OH) 2 # moles HCl that react with Al(OH)3 = 1 mol Al(OH)3 3 mol HCl 0.500-X grams Al(OH)3 78.00 g Al(OH)3 1 mol Al(OH)3 115 Chapter 5: Introduction to Reactions in Aqueous Solutions 2X 3(0.500 X) 0.0188 58.32 78.00 X = 0.108, therefore the mass of Mg(OH)2 in the sample is 0.108 grams. % Mg(OH)2 = (0.108/0.500) × 100 = 21.6 %Al = 100 − %Mg(OH)2 = 78.4 91. 0.1386 g AgI 1 mol AgI 1 mol CHI3 1 mol C19 H16 O 4 308.33 g C19 H16O 4 234.77 g AgI 3 mol AgI 1 mol CHI3 1 mol C19 H16 O 4 0.06068 g C19 H16 O 4 % C19 H16 O 4 = 0.06068 g 100 0.4346 % 13.96 g 93. (a) CaO(s) + H2O(l) → Ca2+(aq) + 2 OH−(aq) H2PO4−(aq) + 2 OH−(aq) → PO43−(aq) + 2 H2O(l) HPO4−(aq) + OH−(aq) → PO43−(aq) + H2O(l) 5 Ca2+(aq) + 3 PO43−(aq) + OH−(aq) → Ca5(PO4)3OH(s) 10.0 103 g P 1 mol P 1 mol PO34 5 mol Ca 2 1.00 10 L L 30.97 g P 1 mol P 3 mol PO34 4 (b) 1 mol CaO 56.08 g CaO = 301.80 g CaO = 302 g = 0.302 kg 1 mol Ca 2 1 mol CaO FEATURE PROBLEMS 94. From the volume of titrant, we can calculate both the amount in moles of NaC5 H 5 and (through its molar mass of 88.08 g/mol) the mass of NaC5 H 5 in a sample. The remaining mass in a sample is that of C 4 H 8O (72.11 g/mol), whose amount in moles we calculate. The ratio of the molar amount of C 4 H 8O in the sample to the molar amount of NaC5 H 5 is the value of x. 116 Chapter 5: Introduction to Reactions in Aqueous Solutions Conversion pathway approach: 0.1001 mol HCl 1 mol NaOH 1 mol NaC5 H 5 1 L soln 1 mol HCl 1 mol NaOH = 0.001493 mol NaC5 H 5 moles of NaC5 H 5 = 0.01492 L 88.08 g NaC5 H 5 mass of C4 H8O = 0.242 g sample 0.001493 mol NaC5 H 5 1 mol NaC5 H 5 = 0.111 g C4 H8O 1mol C4 H8O 72.11g C4 H8O = 1.03 0.001493 mol NaC5 H 5 0.110 g C4 H8O x= Stepwise approach: 0.1001 mol HCl 0.01492 L = 1.493 10-3 mol HCl 1 L soln 1 mol NaOH 1.493 10-3 mol HCl = 1.493 10-3 mol NaOH 1 mol HCl 1 mol NaC5 H 5 1.493 10-3 mol NaOH 1.493 10-3 mol NaC5 H 5 1 mol NaOH 88.08 g NaC5 H 5 1.493 10-3 mol NaC5 H5 0.1315 g NaC5 H 5 1 mol NaC5 H 5 mass of C4 H8O = 0.242 g sample 0.1315 g NaC5 H5 = 0.111 g C4 H8O 0.111g C4 H8O 1mol C4 H8O = 1.54 10-3 mol C4 H8O 72.11g C4 H8O 1.54 10-3 mol C4 H8O = 1.03 0.001493 mol NaC5 H 5 For the second sample, parallel calculations give 0.001200 mol NaC5 H 5 , 0.093 g C 4 H 8 , x = 1.1. There is rounding error in this second calculation because it is limited to two significant figures. The best answer is from the first run x ~1.03 or 1. The formula is NaC5H5(THF)1. 95. First, we balance the two equations. Oxidation: H 2 C2 O 4 aq 2 CO 2 g + 2 H + aq + 2 e Reduction: MnO 2 s + 4 H + aq + 2 e Mn 2+ aq + 2 H 2 O l Net: H 2 C 2 O 4 aq + MnO 2 s + 2 H + aq 2 CO 2 g + Mn 2+ aq + 2 H 2O l 117 Chapter 5: Introduction to Reactions in Aqueous Solutions Oxidation: { H 2 C2 O 4 aq 2 CO 2 g + 2 H + aq + 2 e }5 Reduction: { MnO 4 aq + 8 H + aq + 5 e Mn 2+ aq + 4 H 2 O l }2 Net: 5 H 2 C2 O 4 aq + 2 MnO 4 aq + 6 H + aq 10 CO 2 g + 2 Mn 2+ aq + 8 H 2 O l Next, we determine the mass of the excess oxalic acid. 0.1000 mol KMnO 4 1mol MnO 4 5 mol H 2 C2 O 4 mass H 2C 2 O 4 2H 2 O 0.03006 L 1L 1mol KMnO 4 2 mol MnO 4 1 mol H 2 C 2 O 4 2H 2 O 126.07 g H 2 C 2 O 4 2H 2 O = 0.9474 g H 2C2O4 2H 2O 1 mol H 2 C 2 O 4 1 mol H 2 C 2 O 4 2H 2 O The mass of H 2 C 2 O 4 2 H 2 O that reacted with MnO2 = 1.651 g 0.9474 g = 0.704 g H 2C2O4 2H 2O mass MnO 2 = 0.704 g H 2 C 2 O 4 2H 2 O 1 mol H 2 C 2 O 4 1 mol MnO 2 126.07 g H 2 C 2 O 4 2H 2 O 1 mol H 2 C 2 O 4 86.9 g MnO 2 1 mol MnO 2 = 0.485 g MnO 2 % MnO 2 97. 0.485g MnO 2 100% 91.0% MnO 2 0.533g sample The molecular formula for CH3CH2OH is C2H6O and for CH3COOH is C2H4O2. The first step is to balance the oxidation–reduction reaction. Oxidation: [C2H6O + H2O → C2H4O2 + 4 H+ + 4 e−] × 3 Reduction: [Cr2O72− + 14 H+ + 6e− → 2 Cr3+ + 7 H2O] × 2 Overall: 3 C2H6O + 2 Cr2O72− + 16 H+ → 3 C2H4O2 + 4 Cr3+ + 11 H2O Before the breath test: 0.75 mg K 2 Cr2 O7 1g 1 mol 1000 mL = 8.498 10-4 M 3 mL 1000 mg 294.19 g 1L = 8×10-4 M (to 1 sig fig) For the breath sample: 0.05 g C2 H 6 O 2.38×107 g C2 H 6 O 1 mL blood = BrAC = 100 mL blood 2100 mL breath mL breath mass C2H6O = 2.38×107 g C2 H 6 O × 500. mL breath = 1.19 × 10−4 g C2H6O mL breath 118 Chapter 5: Introduction to Reactions in Aqueous Solutions Calculate the amount of K2Cr2O7 that reacts: – 1 mol C2 H 6 O 2 mol Cr2 O72 1 mol K 2 Cr2 O7 1.19 10 g C2 H 6 O – 46.068 g C2 H 6 O 3 mol C2 H 6 O 1 mol Cr2 O72 4 = 1.72 106 mol K 2 Cr2 O7 # mol K2Cr2O7 remaining = moles K2Cr2O7 before – moles K2Cr2O7 that reacts moles K2Cr2O7 before = 0.75 mg K 2 Cr2 O7 1g 1 mol = 2.5 10-6 mol 1000 mg 294.19 g # mol K2Cr2O7 remaining = 2.5 × 10−6 mol − 1.72 × 10−6 mol = 0.78 × 10−6 mol concentration of K2Cr2O7 after the breath test = 0.78 × 10−6 mol/0.003 L = 2.6 × 10−4 mol/L = 3 × 10−4 mol/L (to 1 sig fig) 102. The answer is (b). Conversion pathway approach: 0.0050 mol Ba(OH) 2 2 mol OH 0.300 L = 0.0030 mol 1L 1mol Ba(OH) 2 Stepwise approach: 0.0050 mol Ba(OH) 2 0.300 L = 1.5 10-3 mol Ba(OH) 2 1L 2 mol OH 1.5 10-3 mol Ba(OH) 2 = 0.0030 mol 1mol Ba(OH) 2 103. The answer is (d), because H2SO4 is a strong diprotic acid and theoretically yields 0.20 mol of H+ for every 0.10 mol of H2SO4. 104. The answer is (c). Based on the solubility guidelines in Table 5-1, carbonates (CO32-) are insoluble. 105. The answer is (a). Reaction with ZnO gives ZnCl2 (soluble) and H2O. There is no reaction with NaBr and Na2SO4, since all species are aqueous. By the process of elimination, (a) is the answer. 106. Balanced equation: 2 KI + Pb(NO 3 ) 2 2KNO3 + PbI 2 Net ionic equation: 2I- + Pb 2 PbI 2 (s) 119 Chapter 5: Introduction to Reactions in Aqueous Solutions 107. Balanced equation: Na 2 CO3 + 2HCl 2NaCl + H 2 O + CO 2 Net ionic equation: CO 23 + 2H H 2O (l) + CO 2 (g) 108. (a) (b) (c) Balanced equation: 2 Na 3PO 4 + 3 Zn(NO 3 ) 2 6NaNO3 + Zn 3 (PO 4 ) 2 Net ionic equation: 3 Zn 2+ + PO34 Zn 3 (PO 4 ) 2 (s) Balanced equation: 2 NaOH + Cu(NO 3 ) 2 Cu(OH) 2 + 2 NaNO3 Net ionic equation: Cu 2+ + 2 OH Cu(OH) 2 (s) Balanced equation: NiCl2 + Na 2 CO 3 NiCO3 + 2 NaCl Net ionic equation: Ni 2+ + CO32 NiCO3 (s) 109. (a) Species oxidized: N in NO (b) Species reduced: O2 (c) Oxidizing agent: O2 (d) Reducing agent: NO (e) Gains electrons: O2 (f) Loses electrons: NO 110. The answer is (b). The charges need to be balanced on both sides. Using a coefficient of 4, the charges on both sides of the reaction becomes +12. 111. The answer is (d), 5 ClO- to 1 I2. The work to balance the half-reactions is shown below: Reduction: 5ClO + 2H + + 2e Cl + H 2 O Oxidation: I 2 + 6H 2 O 2IO3 + 10e + 12H To combine the above reactions, the oxidation reaction should be multiplied by 5. The combined equation is: Combined: 5ClO + I 2 + H 2O 5Cl + 2IO3 + 2H 112. The answer is (a). The balanced half-reaction is as follows: NpO 2 + 4 H + + e- Np 4 + 2H 2 O 113. (a) False. Based on solubility rules, BaCl2 dissolves well in water. Therefore, it is a strong electrolyte. (b) True. Since H- is a base, H2O is by necessity an acid. It also reduces H- (-1) to H2 (0). (c) False. The product of such a reaction would be NaCl and H2CO3, neither of which precipitates out. (d) False. HF is among the strongest of weak acids. It is not a strong acid, because it doesn’t completely dissociate. 120 Chapter 5: Introduction to Reactions in Aqueous Solutions (e) True. For every mole of Mg(NO3)2, there are 3 moles of ions, in contrast to 2 moles of ions for NaNO3. 114. (a) No. Oxidation states of C, H or O do not change throughout the reaction. (b) Yes. Li is oxidized to Li+ and H in H2O is reduced from +1 to 0 in H2. (c) Yes. Ag is oxidized and Pt is reduced. (d) No. Oxidation states of Cl, Ca, H, and O remain unchanged. 121