CHAPTER 5 INTRODUCTION TO REACTIONS IN AQUEOUS SOLUTIONS PRACTICE EXAMPLES

Transcription

1 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. mol NaCl mol Cl from NaCl, Cl = =0.8M Cl 1 L soln 1 mol NaCl mol MgCl mol Cl from MgCl, Cl = = 0.10 M Cl 1 L soln 1 mol MgCl Cl total = Cl from NaCl Cl from MgCl = 0.8 M 0.10 M = 0.50 M Cl 1B (a) mg F L 1 g F 1000 mg F 1 mol F g F = M F - (b) L mol F 1 mol CaF 1L mol F g CaF 1 mol CaF 1 kg 1000 g =.1 kg CaF A 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 aq OH aq Al OH s Al OH, which is not. Net ionic equation: (b) Possible products are iron(iii) sulfate, Fe SO, and potassium bromide, KBr, both of which are soluble. No reaction occurs. (c) Possible products are calcium nitrate, Ca(NO ), which is soluble, and lead(ii) iodide, Pb aq I aq PbI s PbI, which is insoluble. The net ionic equation is: B (a) Possible products are sodium chloride, NaCl, which is soluble, and aluminum phosphate, Al aq PO aq AlPO s AlPO, which is insoluble. Net ionic equation: (b) Possible products are aluminum chloride, AlCl, which is soluble, and barium sulfate, BaSO, which is insoluble. Net ionic equation: Ba aq SO aq BaSO s 97

2 (c) Possible products are ammonium nitrate, NHNO, which is soluble, and lead (II) carbonate, Pb aq CO aq PbCO s PbCO, which is insoluble. Net ionic equation: A B A 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 aq HC H O aq NH aq C H O aq 5 5 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. CaCO s HC H O aq CO g H O l Ca aq C H O aq (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 O (g) as a product indicates that this is an oxidation reduction reaction. Oxygen is oxidized from O.S. = - in NO - to O.S. = 0 in O (g). Nitrogen is reduced from O.S. = 5 in NO - to O.S. = in NO. B 5A 5B 6A Vanadium is oxidized from O.S. = in VO to an O.S. = 5 in VO while manganese is reduced from O.S. = 7 in MnO - to O.S. = in Mn. Aluminum is oxidized (from an O.S. of 0 to an O.S. of ), while hydrogen is reduced (from an O.S. of 1 to an O.S. of 0). Oxidation : Al s Al aq e Reduction: H aq e H g Net equation : Als 6 H aq Al aq H g Bromide is oxidized (from 1 to 0), while chlorine is reduced (from 0 to 1). Oxidation : Br aq Br l e Cl g e Cl aq Reduction: Net equation : Br aq Cl g Br l Cl aq Step 1: Write the two skeleton half reactions. MnO aq Mn aq and Fe aq Fe aq Step : Balance each skeleton half reaction for O (with HO ) and for H atoms (with H ). MnO aq 8 H aq Mn aq H O(l) and Fe aq Fe aq 98

3 Step : Balance electric charge by adding electrons. and MnO aq 8 H aq 5 e Mn aq H O(l) Fe aq Fe aq e Step : Combine the two half reactions Fe aq Fe aq e 5 MnO aq 8 H aq 5 e Mn aq H O(l) MnO aq 8 H aq 5 Fe aq Mn aq H O(l) 5 Fe aq 6B 7A Step 1: Uranium is oxidized and chromium is reduced in this reaction. The skeleton half-equations are: UO aq UO aq and CrO 7 (aq) Cr (aq) Step : First, balance the chromium skeleton half-equation for chromium atoms: CrO 7 aq Cr aq Next, balance oxygen atoms with water molecules in each half-equation: UO aq HO(l) UO aq and CrO 7 (aq) Cr (aq) 7HO(l) Then, balance hydrogen atoms with hydrogen ions in each half-equation: UO aq H O(l) UO aq H aq 7 Cr O (aq) 1H (aq) Cr (aq) 7H O(l) Step : Balance the charge of each half-equation with electrons. UO aq H O(l) UO aq H aq e CrO7 aq 1 H aq 6 e Cr aq 7 HO(l) Step : Multiply the uranium half-equation by and add the chromium half-equation to it. UO aq H O(l) UO aq H aq e Step 5: 7 Cr O aq 1 H aq 6 e Cr aq 7 H O(l) - UO (aq)cr O (aq)1 H (aq) H O(l) UO (aq) Cr (aq)7 H O(l)6 H (aq) 7 Simplify. Subtract HO (l) and 6 H (aq) from each side of the equation. UO aq Cr O aq 8 H aq UO aq Cr aq H O(l) 7 Step 1: Write the two skeleton half-equations. S(s) SO ( aq ) and OCl ( aq ) Cl ( aq ) Step : Balance each skeleton half-equation for O (with H O ) and for H atoms (with H ). HO(l) Ss SO aq 6 H OCl (aq) H Cl (aq) HO(l) Step : Balance electric charge by adding electrons. H O(l) S s SO aq 6 H (aq) e OCl (aq) H (aq) e Cl (aq) H O(l) 99

4 Step : Change from an acidic medium to a basic one by adding OH to eliminate H. H O(l) S s 6 OH (aq) SO aq 6 H (aq) 6 OH (aq) e OCl aq H (aq) OH (aq) e Cl aq H O(l) 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. {S s 6 OH (aq) SO aq H O(l) e } 1 {OCl aq H O(l) e Cl aq OH (aq)} - S s 6 OH (aq) OCl aq H O(l) SO aq H O(l) Cl aq OH Simplify by removing the species present on both sides. S s OH aq OCl aq SO aq H O(l) Cl aq Net ionic equation: 7B Step 1: Write the two skeleton half-equations. MnO aq MnOs and SO (aq) SO (aq) Step : Balance each skeleton half-equation for O (with H O ) and for H atoms (with H ). MnO aq H aq MnO s H O(l) Step : SO (aq) H O(l) SO (aq) H (aq) Balance electric charge by adding electrons. MnO aq H aq e MnO s H O(l) SO aq H O(l) SO aq H aq e Step : Change from an acidic medium to a basic one by adding OH to eliminate H. MnO aq H aq OH aq e MnO s H O(l) OH aq SO aq H O(l) OH aq SO aq H aq OH aq 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 aq H O(l) e MnO s OH aq } - MnO s SO aq H O(l)8 OH aq {SO aq OH aq SO aq H O(l) e } MnO aq SO aq 6 OH (aq) H O(l) Simplify by removing species present on both sides. Net ionic equation: MnO aq SO aq H O(l) MnO s SO aq OH aq 100

5 8A Since the oxidation state of H is 0 in H (g) and is 1 in both NH (g) and H O(g), hydrogen is oxidized. A substance that is oxidized is called a reducing agent. In addition, the oxidation state of N in NO (g) is, while it is innh ; the oxidation state of the element N decreases during this reaction, meaning that NO (g) is reduced. The substance that is reduced is called the oxidizing agent. 8B In Au CN 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 -; O has been reduced and thus, O (g) (oxidation state = 0) is the oxidizing agent. 9A We first determine the amount of NaOH that reacts with g KHP. 1 mol KHP 1 mol OH 1 mol NaOH n NaOH = g KHP = mol NaOH 0. g KHP 1 mol KHP 1 mol OH mol NaOH 1000 ml [NaOH] = = M.0 ml soln 1 L 9B The net ionic equation when solid hydroxides react with a strong acid is OH - H H O. There are two sources of OH - : NaOH and Ca(OH). We compute the amount of OH - from each source and add the results. moles of OH from NaOH: 9.5 g NaOH 1 mol NaOH 1 mol OH = 0.5 g sample = mol OH g sample g NaOH 1 mol NaOH : moles of OH from Ca OH 7.5 g Ca OH 1 mol Ca OH mol OH = 0.5 g sample = mol OH g sample 7.09 g Ba OH 1 mol Ca OH - total amount OH = mol from NaOH mol from Ca OH = mol OH mol OH 1 mol H 1 mol HCl 1000 ml soln [HCl] = = 0.10 M 5.6 ml HCl soln 1 mol OH 1 mol H 1 L soln 10A First, determine the mass of iron that has reacted as Fe with the titrant. The balanced chemical equation provides the essential conversion factor to answer this question. Namely: 5 Fe aqmno aq 8 H aq 5 Fe aq Mn aq HOl mol MnO 5 mol Fe g Fe mass Fe = L titrant = 0.6 g Fe 1 L titrant 1 mol MnO 1 mol Fe Then determine the % Fe in the ore. % Fe = g Fe g ore 100% = 65.% Fe 10B The balanced equation provides us with the stoichiometric coefficients needed for the solution. - Namely: 5 C O aq MnO aq16 H aq10 CO g Mn aq8 H Ol 101

6 1 mol Na C O 1 mol C O mol MnO amount MnO = 0.8 g Na C O 1.00 g Na C O 1 mol Na C O 5 mol C O = mol MnO mol MnO 1000 ml 1 mol KMnO [KMnO ] = = M KMnO.68 ml soln 1 L 1 mol MnO INTEGRATIVE EXAMPLE A. First, balance the equation. Break down the reaction of chlorate and ferrous ion as follows: ClO 6H 6e Cl H O 6 Fe Fe e Net reaction: ClO 6Fe 6H Cl 6Fe H O The reaction between Fe and Ce is already balanced. To calculate the moles of Fe that remains after the reaction with ClO -, determine the moles of Ce that react with Fe : mol Ce = L M = mol = mol of excess Fe total mol of Fe = L = mol Therefore, the moles of Fe reacted = = mol. To determine the mass of KClO, use the mole ratios in the balanced equation in conjunction with the molar mass of KClO. 1 mol ClO 1 mol KClO 1.5 g KClO mol Fe 6 mol Fe 1 mol ClO 1 mol KClO = g KClO g %KClO = 100% = 9.89% 0.1 g B. First, balance the equation. Break down the reaction of arsenous acid and permanganate as follows: 5 H AsO H O H AsO e H - - MnO 8H 5e Mn HO Net reaction: 5H AsO MnO 6H 5H AsO Mn H O moles of MnO - = L 0.01 M = mol 10

7 To calculate the mass of As, use the mole ratios in the balanced equation in conjunction with the molar mass of As: 5 mol HAsO 1 mol As 7.9 g As mol MnO mol MnO 1 mol H AsO 1 mol As = g As g mass% As = 100% = 1.% 7.5 g EXERCISES Strong Electrolytes, Weak Electrolytes, and Nonelectrolytes 1. (a) Because its formula begins with hydrogen, HC H O 6 5 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) (c) LiSOis an ionic compound, that is, a salt. A salt is a strong electrolyte. MgI also is a salt, a strong electrolyte. (d) CHCH 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 is a strong electrolyte, one of the strong bases listed in Table 5-.. HCl is practically 100% dissociated into ions. The apparatus should light up brightly. A solution of both HCl and HC H O will yield similar results. In strongly acidic solutions, the weak acid HC H O 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 HC H O. The apparatus should light up brightly. 5. (a) Barium bromide: strong electrolyte (b) Propionic acid: weak electrolyte (c) Ammonia: weak electrolyte Ion Concentrations 7. (a) (b) K mol KNO = L soln 1 mol K 1 mol KNO = 0.8 M K mol Ca NO mol NO NO = = 0. M NO 1 L soln 1 mol Ca NO 10

8 (c) (d) 0.08 mol Al SO mol Al Al = = M Al 1 L soln 1 mol Al SO Na mol Na PO = L soln mol Na 1 mol Na PO = 0.67 M Na 9. Conversion pathway approach: 0.1 g Ba OH 8H O 1 mol Ba OH 8H O 1000 ml mol OH OH = 75 ml soln 1 L 15.5g Ba OH 8H O 1 mol Ba OH 8 H O =.010 M OH Stepwise approach: 0.1 g Ba OH 8H O 1000 ml = 0.80 g/l 75 ml soln 1 L 0.80 g 1 mol Ba OH 8H O mol Ba OH 8H O = L 15.5g Ba OH 8H O L mol Ba OH 8H O mol OH =.010 M OH L 1 mol Ba OH 8 H O (a) (b) (c) 1. mg Ca 1 g Ca 1 mol Ca [Ca ] =.5 10 M Ca 1 L solution 1000 mg Ca g Ca.8 mg K 1 g K 1000 ml solution 1 mol K [K ] = M K 100 ml solution 1000 mg K 1 L solution g K 5 g Zn 1 g Zn 1000 ml solution 1 mol Zn [Zn ] =.10 M Zn 6 1 ml solution 110 g Zn 1 L solution 65.9 g Zn 1 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 M KSO mol K 0.17 M K 1 L solution 1 mol K SO 1.5 g KBr 1000 ml solution 1 mol KBr 1 mol K M K 100 ml solution 1 L solution g KBr 1 mol KBr 8.1 mg K 1000 ml solution 1 g K 1 mol K 0.07 M K 1 ml solution 1 L solution 1000 mg K g K Clearly, the solution containing 8.1 mg K per ml gives the largest K of the three solutions. 10

9 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 in grams. 1 L mol I moles of I in final solution = 50.0 ml 1000 ml 1 L soln = mol I 1 L mol KI 1 mol I moles of I in KI solution = 50.0 ml = mol I 1000 ml 1 L soln 1 mol KI 1 mol MgI g MgI 1000 mg mass MgI required = mol I mol I 1 mol MgI 1 g =. 10 mg MgI 17. moles of chloride ion 0.65 mol KCl 1 mol Cl 0.85 mol MgCl mol Cl = 0.5 L L 1 L soln 1 mol KCl 1 L soln 1 mol MgCl = 0.11 mol Cl 0.7 mol Cl = mol Cl mol Cl Cl = = 0.7 M 0.5 L 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 aq Br aq PbBr s (b) No reaction occurs (all are spectator ions). Fe aq OH aq Fe OH s (c) 1. Mixture Result (Net Ionic Equation) (a) HIa Zn NO (aq): No reaction occurs. (b) CuSOaq Na COaq : Cu aq CO aq CuCOs (c) Cu NO aq Na PO aq : Cu aq PO aq Cu PO s. (a) Add KSO aq; BaSO s will form and MgSO will not precipitate. BaCl s K SO aq BaSO s KClaq 105

10 (b) Add HOl; NaCO water Na CO s Na aq CO aq s dissolves, but MgCO (s) will not dissolve (appreciably). (c) Add KCl(aq); AgCl(s) will form, while Cu(NO ) (s) will dissolve. AgNO s KCl aq AgCl s KNO aq 5. Mixture Net Ionic Equation Sr NO aq K SO aq : Sr aq SO aq SrSO s (a) (b) Mg NO aq NaOH aq : Mg aq OH aq Mg OH s BaCl aq K SO aq : Ba (aq) SO (aq) BaSO (s) (c) (upon filtering, KCl (aq) is obtained) Acid Base Reactions 7. The type of reaction is given first, followed by the net ionic equation. (a) Neutralization: OH aq HC H O aq H Ol C H O aq (b) No reaction occurs. This is the physical mixing of two acids. FeS s H aq H S g Fe aq (c) Gas evolution: (d) Gas evolution: HCO aq H aq "HCOaq " HOl CO g (e) Redox: Mg s H aq Mg aq H g 9. As a salt: NaHSO aq Na aq HSO aq As an acid: HSO aq OH aq H Ol SO aq 1. Use (b) NH (aq): NH affords the OH - ions necessary to form Mg(OH) (s). Applicable reactions: {NH (aq) H O(l) NH (aq) OH - (aq)} MgCl (aq) Mg (aq) Cl - (aq) Mg (aq) OH - (aq) Mg(OH) (s) 106

11 Oxidation Reduction (Redox) Equations. (a) The O.S. of H is 1, that of O is, that of C is, and that of Mg is on each side of this equation. This is not a redox equation. (b) (c) (d) 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. 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 on the right side of this equation. This is a redox reaction. On both sides of the equation the O.S. of O is, that of Ag is 1, and that of Cr is 6. Thus, this is not a redox equation. 5. (a) Reduction: (b) Reduction: NO aq 10 H aq 8 e NOg 5 HOl (c) Oxidation: Als OH aq AlOH aq e 7. (a) Oxidation: { SO aq 6 H aq e S O aq H O(l) I aq I s e } 5 Reduction: { MnO aq 8 H aq 5 e Mn aq HOl } Net: 10 I aq MnO aq 16 H aq 5 I s Mn aq 8 H O l (b) Oxidation: { Reduction: { BrO aq 6 H aq 6 e Br aq HOl N H l N g H aq e } } Net: N H l BrO aq N g Br aq 6 H Ol (c) Oxidation: Fe aq Fe aq e Reduction: VO aq 6 H aq e VO aq HOl Net: Fe aq VO aq 6 H aq Fe aq VO aq H Ol (d) Oxidation: { UO aq HO l UO aq H aq e } Reduction: { NO aq H aq e NOg HOl } Net: UO aq NO aq H aq UO aq NOg H Ol 9. (a) Oxidation: { Reduction: ClO aq H O(l) 6 e Cl aq 6 OH aq MnO s OH aq MnO aq H O(l) e } Net: MnO s ClO aq OH aq MnO aq Cl aq H O(l) 107

12 (b) Oxidation: { - Reduction: { OCl aq H O(l) e Cl aq OH aq Fe OH s 5 OH aq FeO aq HO(l) e } } Net: Fe OH s OCl aq OH aq FeO aq Cl aq 5 H O(l) (c) Oxidation: { ClO (aq) OH aq ClO aq HOl e Reduction: ClO (aq) HOl 5 e Cl (aq) OH aq Net: } 5 6 ClO (aq) 6 OH aq 5ClO aq Cl aq H O (l) (d) Oxidation: (Ag (s) Ag (aq) 1 e ) - Reduction: H O(l) CrO e Cr(OH) (s) 5 OH - - Net: Ag(s) CrO H O(l) Ag(aq) Cr(OH) (s) 5 OH - 1. (a) Oxidation: Reduction: { Cl g e Cl aq Cl g 1 OH aq ClO aq 6 HO(l) 10 e } 5 Net: Or: 6 Cl g 1 OH aq 10 Cl aq ClO aq 6 H O(l) Cl g 6 OH aq 5 Cl aq ClO aq H O(l) (b) Oxidation: Reduction: S O aq H O(l) HSO aq H aq e S O aq H aq e S O aq H O (l) Net: S O aq H O(l) HSO aq S O aq. (a) Oxidation: { NO aq HOl NO aq H aq e } 5 Reduction: { MnO aq 8 H aq 5 e Mn aq HOl } Net: 5 NO aq MnO aq 6 H aq 5 NO aq Mn aq H Ol (b) Oxidation: { Mn (aq) OH - (aq) MnO (s) H O (l) e - } Reduction: { MnO - (aq) H O (l) e - MnO (s) OH - (aq) } Net: Mn (aq) MnO - (aq) OH - (aq) 5 MnO (s) H O (l) (c) Oxidation: { CH5OH CHCHO H aq e Reduction: Cr O7 aq 1 H aq 6 e Cr aq 7 HOl Net: } Cr O aq 8 H aq C H OH Cr aq 7 H O l CH CHO

13 5. For the purpose of balancing its redox equation, each of the reactions is treated as if it takes place in acidic aqueous solution. (a) H O(g) CH (g) CO (g) 8 H (g) 8 e - { e - H (g) NO(g) ½ N (g) H O(g) } CH (g) NO(g) N (g) CO (g) H O(g) (b) {H S(g) 1/8 S 8 (s) H (g) e - } e - H (g) SO (g) 1/8 S 8 (s) H O(g) H S(g) SO (g) /8 S 8 (s) H O(g) or 16 H S(g) 8 SO (g) S 8 (s) 16 H O(g) (c) {Cl O(g) NH (aq) H (aq) e - NH Cl(s) H O(l) } { NH (g) N (g) 6 e - 6 H (aq) } 6 NH (g) 6 H (aq) 6 NH (aq) 10 NH (g) Cl O(g) 6 NH Cl(s) N (g) H O(l) Oxidizing and Reducing Agents 7. 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. (a) SO aq MnO aq is the reducing agent; the O.S. of S = in SO and =6 in SO. is the oxidizing agent; the O.S. of Mn =7 in MnO and in Mn. (b) H gis the reducing agent; the O.S. of H = 0 in H g and=1 in NO g is the oxidizing agent; the O.S. of N = in NO g and HOg. in NH (g). (c) Fe CN aq 6 is the reducing agent; the O.S. of Fe = in FeCN and = in FeCN. H O aq in HO and= 6 6 is the oxidizing agent; the O.S. of O = 1 in HO. Neutralization and Acid Base Titrations 9. The problem is most easily solved with amounts in millimoles. V 0.18 mmol HCl 1 mmol H 1 mmol OH NaOH = ml HClaq 1 ml HCl aq 1 mmol HCl 1 mmol H 109

14 1 mmol NaOH 1 ml NaOH aq 1 mmol OH mmol NaOH = 1. ml NaOH aq soln 51. The net reaction is OH aq HC H O aq H O(l) C H O aq. 5 5 Conversion pathway approach: mmol HC H O 1 mmol KOH 1 ml base 5 V base = 5.00 ml acid 1 ml acid 1 mmol HC H O.155 mmol KOH 5 =.56 ml KOH solution Stepwise approach: mmol HC H O ml acid = 7.6 mmol HC H O 1 ml acid 1 mmol KOH 7.6 mmol HCH5O = 7.6 mmol KOH 1 mmol HC H O 5 1 ml base 7.6 mmol KOH =.56 ml KOH solution.155 mmol KOH 5 5. NaOHaq HClaq NaClaq HO(l) is the titration reaction mol HCl 1mol NaOH 0.08 L 1Lsoln 1 molhcl [NaOH] = = 0.10 M NaOH L sample 55. The mass of acetylsalicylic acid is converted to the amount of NaOH, in millimoles, that will react with it. 0. g HC9H7O 1 mol HC9H7O 1 mol NaOH 1000 mmol NaOH NaOH = ml NaOH aq 180. g HC H O 1 mol HC H O 1 mol NaOH = M NaOH The equation for the reaction is HNO aq KOH aq KNO aq H 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 = 5.00 ml acid mmol HNO ml acid =.0 mmol HNO 0.18 mmol KOH mmol KOH = ml acid =.18 mmol KOH 1 ml base There is more acid present than base. Thus, the resulting solution is acidic. 110

15 V 1.01 g vinegar.0 g HC H O 1 mol HC H O = 5.00 ml vinegar 59. base 1 ml g vinegar 60.0 g HCHO 1 mol NaOH 1 L base 1 mol HC H O mol NaOH ml L = ml base 61. Answer is (d): 10 % of necessary titrant added in titration of NH 5 NH 5 HCl 1 HCl required for equivalence point 0 % excess 5 NH 6 Cl - H O (depicted in question's drawing ) Stoichiometry of Oxidation Reduction Reactions 6. Conversion pathway approach: [MnO ]= 1mol As O mol MnO 1mol KMnO g As O g As O 5 mol As O 1molMnO 1L.15 ml 1000 ml = M KMnO Stepwise approach: mol KMnO [KMnO ]= L solution 1mol As O g As O = mol As O g AsO mol MnO mol As O =.5910 mol MnO - - 5mol AsO 1mol KMnO mol MnO = molMnO 1L.15 ml L solution 1000 ml mol KMnO.5910 mol KMnO mol KMnO - - [ KMnO ]= = = M L solution L solution 111

16 65. First, we will determine the mass of Fe, then the percentage of iron in the ore. 1 L mol CrO 7 6 mol Fe g Fe mass Fe = 8.7 ml 1000 ml 1 L soln 1 mol Cr O 7 1 mol Fe 0.861g Fe mass Fe = g Fe % Fe = 100% 5. % Fe 0.91 g ore 67. First, balance the titration equation: Oxidation: { C O aq CO g e } 5 } Reduction: { MnO aq 8 H aq 5 e Mn aq H Ol Net: 5 C O aq MnO aq 16 H aq 10 CO g Mn aq 8 H Ol mass mass =1.00 L satd soln Na C O NaCO 1000 ml 5.8 ml satd soln KMnO mol KMnO 1 L 5.00 ml satd soln Na C O 1000 ml KMnO 1 mol MnO 5 molc O 1 mol Na C O 1.0 g Na C O 1 mol KMnO molmno 1 molc O 1 mol Na C O = 7.0 g Na C O NaCO Integrative and Advanced Exercises 71. A possible product, based on solubility rules, is Ca (PO ). We determine the % Ca in this compound. molar mass g Ca 0.97 g P g O 10. g Ca g P g O g 10. g Ca % Ca 100% 8.76% g Ca (PO ) Thus, Ca (PO ) is the predicted product. The net ionic equation follows. Ca (aq) HPO (aq) Ca (PO ) (s) H (aq) 7. Let us first determine the mass of Mg in the sample analyzed. Conversion pathway approach: 1 mol Mg P O mol Mg.05 g 7 mass Mg g MgPO g Mg.55 g MgP O7 1 mol MgPO 7 1 mol Mg g Mg ppm Mg 10 g sample 108 ppm Mg g sample 11

17 Stepwise approach: 1 mol Mg P O g Mg P O =.710 mol Mg P O g MgP O7 mol Mg.710 mol Mg P O.910 mol Mg mol Mg PO 7.05 g 1 mol Mg mol Mg g Mg g Mg ppm Mg 10 g sample 108 ppm Mg g sample 75. Let V represent the volume of added 0.8 M CaCl that must be added. We know that [Cl ] [Cl 0.5 L ] = 0.50 M, but also, mol KCl 1L soln 1mol Cl V 1mol KCl 0.5 L V 0.8 mol CaCl 1L soln mol Cl 1mol CaCl (0.5 V ) V V V L (a) [FeS 8 H O Fe SO 16 H 15 e ] [O H e H O] 15 overall: FeS (s) 15 O (g) H O(l) Fe (aq) 8 SO (aq) H (aq) (b) One kilogram of tailings contains 0.0 kg (0 g) of S. We have 1molS 1molFeS moles of FeS = 0 g S 0.68 mol FeS.07 g S mol S moles of H = moles of CaCO = molh 0.68 mol FeS 0.67 mol H molfes 1 mol CaCO 0.67 mol H 0. mol CaCO molh g CaCO mass of CaCO = 0. mol CaCO. g CaCO 1molCaCO 11

18 8. Oxidation :{ Cl (aq) Cl (g) e } Reduction :CrO 7 (aq) 1 H (aq) 6 e Cr (aq) 7 H O Net :6 Cl (aq) CrO 7 (aq) 1 H (aq) Cr (aq) 7 H O Cl (g) We need to determine the amount of Cl (g) produced from each of the reactants. The limiting reactant is the one that produces the lesser amount of Cl.. amount Cl amount Cl 1.15 g 0.1g HCl 5 ml 1mL 100. g soln 1.5 mol Cl 1mol HCl 6.6 g HCl 1mol Cl 1mol HCl mol Cl 6 mol Cl 98.5 g K CrO 7 1mol K CrO 7 1mol CrO 7 mol Cl 6.6 g 100. g sample 9. g K CrO 7 1mol K CrO 7 1mol CrO mol Cl, the amount produced from the limiting reactant g Cl Then we determine the mass of Cl (g) produced. = 0.69 mol Cl =.6 g Cl 1 mol Cl 85. Cl (g) NaClO (aq) NaCl(aq) ClO (g) (not balanced) Cl (g) NaClO (aq) NaCl(aq) ClO (g) amount ClO g ClO (g) 100 g ClO calculated.785 L.0 mol NaClO mol ClO 67.5 g ClO 1 gal 1 gal 1 L soln mol NaClO 1 mol ClO 97 g ClO produced 88. (a) First, balance the redox equations needed for the calculation. Oxidation: {HSO - (aq) H O(l) SO - (aq) H (aq) e - } Reduction: {IO - (aq) 6 H (aq) 6 e - I - (aq) H O(l) } 1 Net: HSO - (aq) IO - (aq) SO - (aq) H (aq) I - (aq) The solution volume of 5.00 L contains 9.0 g NaIO. This represents 9.0 g/197.9g/mol NaIO = 0.17 mol NaIO. (b) From the above equation, we need times that molar amount of NaHSO, which is (0.17 mol) = 0.1 mol NaHSO ; the molar mass of NaHSO is g/mol. The required mass then is 0.1(10.06) = 5.9 g. 11

19 For the second process: Oxidation: { I - (aq) I (aq) e - } 5 Reduction: { IO - (aq) 1 H (aq) 10 e - I (aq) 6 H O(l) } 1 Net: 5 I - (aq) IO - (aq) 6 H (aq) I (aq) H O(l) In Step 1, we produced 1 mol of I - for every mole of IO - reactant; therefore we had 0.17 mol I -. In step, we require 1/5 mol IO - 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) (aq) HCl(aq) MgCl (aq) H O(l) (1) Al(OH)(aq) HCl(aq) AlCl(aq) HO(l) () HCl(aq) NaOH(aq) NaCl(aq) HO(l) () mol L = mol initial moles of HCl = 1 L moles of HCl that reacted with NaOH = moles of HCl left over from reaction with active ingredients = 0.77 mol NaOH 1 mol HCl 1 L 1 mol NaOH L = 6.10 mol moles of HCl that react with active ingredients = mol mol = mol # moles HCl that react with Mg(OH) # moles HCl that = total moles of HCl reacted/used react with Al(OH) # moles HCl that react with Mg(OH) = X grams Mg(OH) 1 mol Mg(OH) mol HCl 58. g Mg(OH) 1 mol Mg(OH) # moles HCl that react with Al(OH) = X grams Al(OH) 1 mol Al(OH) mol HCl g Al(OH) 1 mol Al(OH) 115

20 X (0.500 X) X = 0.108, therefore the mass of Mg(OH) in the sample is grams. % Mg(OH) = (0.108/0.500) 100 = 1.6 %Al = 100 %Mg(OH) = mol AgI 1 mol CHI 1 mol C H O 08. g C H O g AgI.77 g AgI mol AgI 1 mol CHI 1 mol C H O g C H O % C19H16O = g % 1.96 g 9. (a) CaO(s) H O(l) Ca (aq) OH (aq) H PO (aq) OH (aq) PO (aq) H O(l) HPO (aq) OH (aq) PO (aq) H O(l) 5 Ca (aq) PO (aq) OH (aq) Ca 5 (PO ) OH(s) (b) g P 1 mol P 1molPO 5 mol Ca L L 0.97 g P 1 mol P mol PO 1 mol CaO g CaO = g CaO = 0 g = 0.0 kg 1molCa 1molCaO FEATURE PROBLEMS 9. From the volume of titrant, we can calculate both the amount in moles of NaC 5 H 5 and (through its molar mass of g/mol) the mass of NaC 5 H 5 in a sample. The remaining mass in a sample is that of CH8 O (7.11 g/mol), whose amount in moles we calculate. The ratio of the molar amount of CH8 O in the sample to the molar amount of NaC 5 H 5 is the value of x. 116

21 Conversion pathway approach: mol HCl 1 mol NaOH 1 mol NaC H moles of NaC5H 5 = L 1 L soln 1 mol HCl 1 mol NaOH = mol NaC H 5 5 mass of C H O = 0. g sample mol NaC H = g C H O molCH8O 0.110g CH8O 7.11g C H 8 O x = = mol NaC H g NaC H 1 mol NaC H Stepwise approach: mol HCl L = mol HCl 1 L soln - 1 mol NaOH mol HCl = mol NaOH 1 mol HCl - 1 mol NaC5H mol NaOH mol NaC H 1 mol NaOH g NaC5H mol NaC5H g NaC5H5 1 mol NaC H mass of C H O = 0. g sample g NaC H = g C H O molC H O 0.111g C H O = mol C H O g CH8O mol NaC H mol CH8O = For the second sample, parallel calculations give mol NaC 5 H 5, 0.09 g C H 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 ~1.0 or 1. The formula is NaC 5 H 5 (THF) First, we balance the two equations. H C O aq CO g H aq e Oxidation: Reduction: MnO s H aq e Mn aq H Ol Net: H C O aq MnO s H aq CO g Mn aq H Ol 8, 117

22 Oxidation: { HCO aq CO g H aq e } 5 Reduction: { MnO aq 8 H aq 5 e Mn aq HOl } Net: 5 H C O aq MnO aq 6 H aq 10 CO g Mn aq 8 H Ol Next, we determine the mass of the excess oxalic acid mol KMnO 1mol MnO 5mol HCO mass HCOHO L 1L 1mol KMnO mol MnO 1 mol HCO HO g HCO HO = 0.97 g HCO HO 1 mol HCO 1 mol HCO HO The mass of H C O H O that reacted with MnO = g 0.97 g = 0.70 g HCO HO 1 mol H C O 1 mol MnO 86.9 g MnO mass MnO = 0.70 g H C O H O g H C O H O 1 mol H C O 1 mol MnO %MnO = 0.85 g MnO 0.85g MnO 100% 91.0% MnO 0.5g sample 97. The molecular formula for CH CH OH is C H 6 O and for CH COOH is C H O. The first step is to balance the oxidation reduction reaction. Oxidation: [C H 6 O H O C H O H e ] Reduction: [Cr O 7 1 H 6e Cr 7 H O] Overall: C H 6 O Cr O 7 16 H C H O Cr 11 H O Before the breath test: 0.75 mg K Cr O 1 g 1 mol 1000 ml 7 = M ml 1000 mg 9.19 g 1 L = M (to 1 sig fig) For the breath sample: 0.05 g CH6O 1 ml blood BrAC = = 100 ml blood 100 ml breath g CH6O ml breath mass C H 6 O = g CH6O ml breath 500. ml breath = g C H 6 O 118

23 Calculate the amount of K Cr O 7 that reacts: 1 mol CH6O mol CrO 7 1 mol K CrO g CH6O mol CH6O 1molCrO g C H O 6 = mol KCrO 7 # mol K Cr O 7 remaining = moles K Cr O 7 before moles K Cr O 7 that reacts moles K Cr O 7 before = 1 g 1 mol 1000 mg 9.19 g mg KCrO 7 =.5 10 mol # mol K Cr O 7 remaining = mol mol = mol concentration of K Cr O 7 after the breath test = mol/0.00 L =.6 10 mol/l = 10 mol/l (to 1 sig fig) 10. The answer is (b). Conversion pathway approach: mol Ba(OH) mol OH 0.00 L = mol 1 L 1mol Ba(OH) Stepwise approach: mol Ba(OH) L = mol Ba(OH) 1 L - mol OH mol Ba(OH) = mol 1mol Ba(OH) 10. The answer is (d), because H SO is a strong diprotic acid and theoretically yields 0.0 mol of H for every 0.10 mol of H SO. 10. The answer is (c). Based on the solubility guidelines in Table 5-1, carbonates (CO - ) are insoluble The answer is (a). Reaction with ZnO gives ZnCl (soluble) and H O. There is no reaction with NaBr and Na SO, since all species are aqueous. By the process of elimination, (a) is the answer Balanced equation: KI Pb(NO ) - Net ionic equation: I Pb PbI (s) KNO PbI 119

24 107. Balanced equation: Na CO HCl NaCl H O CO Net ionic equation: CO H H O (l) CO (g) 108. (a) Balanced equation: Na PO Zn(NO ) 6NaNO Zn (PO ) Net ionic equation: Zn PO Zn (PO ) (s) (b) (c) Balanced equation: NaOH Cu(NO ) Cu(OH) NaNO Net ionic equation: Cu OH Cu(OH) (s) Balanced equation: NiCl Na CO NiCO NaCl Net ionic equation: Ni CO NiCO (s) 109. (a) Species oxidized: N in NO (b) Species reduced: O (c) Oxidizing agent: O (d) Reducing agent: NO (e) Gains electrons: O (f) Loses electrons: NO 110. The answer is (b). The charges need to be balanced on both sides. Using a coefficient of, the charges on both sides of the reaction becomes The answer is (d), 5 ClO - to 1 I. The work to balance the half-reactions is shown below: Reduction: 5ClO H e Cl HO Oxidation: I 6HO IO 10e 1H To combine the above reactions, the oxidation reaction should be multiplied by 5. The combined equation is: Combined: 5ClO I H O 5Cl IO H 11. The answer is (a). The balanced half-reaction is as follows: - NpO H e Np H O 11. (a) False. Based on solubility rules, BaCl dissolves well in water. Therefore, it is a strong electrolyte. (b) True. Since H - is a base, H O is by necessity an acid. It also reduces H - (-1) to H (0). (c) False. The product of such a reaction would be NaCl and H CO, 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. 10

25 (e) True. For every mole of Mg(NO ), there are moles of ions, in contrast to moles of ions for NaNO. 11. (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 H O is reduced from 1 to 0 in H. (c) Yes. Ag is oxidized and Pt is reduced. (d) No. Oxidation states of Cl, Ca, H, and O remain unchanged. 11