CHM 201 Problems Solution for Chap 4 Exercises, Fall 2013

56 CHAPTER 4 TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY Aqueous Solutions: Strong and Weak Electrolytes 10. Only statement b is true. A concentrated solution can also contain a nonelectrolyte dissolved in water, for example, concentrated sugar water. Acids are either strong or weak electrolytes. Some ionic compounds are not soluble in water, so they are not labeled as a specific type of electrolyte. 11. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that results in unequal charge distribution in the overall molecule. Polar molecules have a partial negative end and a partial positive end. These are not full charges as in ionic compounds but are charges much smaller in magnitude. Water is a polar molecule and dissolves other polar solutes readily. The oxygen end of water (the partial negative end of the polar water molecule) aligns with the partial positive end of the polar solute, whereas the hydrogens of water (the partial positive end of the polar water molecule) align with the partial negative end of the solute. These opposite charge attractions stabilize polar solutes in water. This process is called hydration. Nonpolar solutes do not have permanent partial negative and partial positive ends; nonpolar solutes are not stabilized in water and do not dissolve. b. KF is a soluble ionic compound, so it is a strong electrolyte. KF(aq) actually exists as separate hydrated K+ ions and hydrated F ions in solution. C6H12O6 is a polar covalent molecule that is a nonelectrolyte. C6H12O6 is hydrated as described in part a. c. RbCl is a soluble ionic compound, so it exists as separate hydrated Rb+ ions and hydrated Cl ions in solution. AgCl is an insoluble ionic compound so the ions stay together in solution and fall to the bottom of the container as a precipitate. d. HNO3 is a strong acid and exists as separate hydrated H+ ions and hydrated NO3  ions in solution. CO is a polar covalent molecule and is hydrated as explained in part a. 12. The electrolyte designation refers to how completely the dissolved solute breaks up into ions. Strong electrolytes completely break up into ions when in water, weak electrolytes only partially break up into ions (less than 5% usually), and nonelectrolytes do not break up into ions when they dissolve in water. The conductivity apparatus illustrated in Figure 4.4 of the text is one way to experimentally determine the type of electrolyte. As illustrated, a bright light indicates many charge carriers (ions) are present and the solute is a strong electrolyte. A dim light indicates few ions are present so the solute is a weak electrolyte, and no light indicates no ions are present so the solute is a nonelectrolyte. CHAPTER 4 SOLUTION STOICHIOMETRY 57 13. a. Ba(NO3)2(aq)  Ba2+(aq) + 2 NO3 (aq); picture iv represents the Ba2+ and NO3  ions present in Ba(NO3)2(aq). b. NaCl(aq)  Na+(aq) + Cl(aq); picture ii represents NaCl(aq). c. K2CO3(aq)  2 K+(aq) + CO3 2(aq); picture iii represents K2CO3(aq). d. MgSO4(aq)  Mg2+(aq) + SO4 2(aq); picture i represents MgSO4(aq). 14. MgSO4(s)  Mg2+(aq) + SO4 2(aq); NH4NO3(s)  NH4 +(aq) + NO3 (aq) 15. Solution A: L 0.1 4 molecules ; solution B: L 0.1 molecules 5.1 L 0.4 6 molecules  Solution C: L 0.1 molecules 2 L 0.2 4 molecules  ; solution D: L 0.1 molecules 3 L 0.2 6 molecules  Solution A has the most molecules per unit volume so solution A is most concentrated. This is followed by solution D, then solution C. Solution B has the fewest molecules per unit volume, so solution B is least concentrated. Solution Concentration: Molarity 16. a. 1.00 L solution × L .0 50 molH SO 4 2 = 0.50 mol H2SO4 0.50 mol H2SO4 × molH2SO 4 18 1L = 2.8 × 2 10 L concentrated H2SO4 or 28 mL Dilute 28 mL of concentrated H2SO4 to a total volume of 1.00 L with water. The resulting 1.00 L of solution will be a 0.50 M H2SO4 solution. b. We will need 0.50 mol HCl. 0.50 mol HCl × molHCl 12 1L = 4.2 × 2 10 L = 42 mL Dilute 42 mL of concentrated HCl to a final volume of 1.00 L. c. We need 0.50 mol NiCl2. 0.50 mol NiCl2 × 6H O NiCl mol 6H O .69g NiCl 237 NiCl mol 6H O molNiCl 1 2 2 2 2 2 2 2     = 118.8 g NiCl2•6H2O ≈ 120 g Dissolve 120 g NiCl2•6H2O in water, and add water until the total volume of the solution is 1.00 L. 58 CHAPTER 4 SOLUTION STOICHIOMETRY d. 1.00 L × L .0 50 molHNO 3 = 0.50 mol HNO3 0.50 mol HNO3 × molHNO3 16 1L = 0.031 L = 31 mL Dissolve 31 mL of concentrated reagent in water. Dilute to a total volume of 1.00 L. e. We need 0.50 mol Na2CO3. 0.50 mol Na2CO3 × mol .99g Na CO 105 3 2 = 53 g Na2CO3 Dissolve 53 g Na2CO3 in water, dilute to 1.00 L. 17. a. 2.00 L mol .00 g NaOH 40 L .0 250 mol NaOH   = 20.0 g NaOH Place 20.0 g NaOH in a 2-L volumetric flask; add water to dissolve the NaOH, and fill to the mark with water, mixing several times along the way. b. 2.00 L 00molNaOH .1 L stock 1 L .0 250molNaOH   = 0.500 L Add 500. mL of 1.00 M NaOH stock solution to a 2-L volumetric flask; fill to the mark with water, mixing several times along the way. c. 2.00 L 4 2 4 2 4 2 K CrO mol .20g K CrO 194 L .0 100molK CrO   = 38.8 g K2CrO4 Similar to the solution made in part a, instead using 38.8 g K2CrO4. d. 2.00 L 4 2 4 2 75molK CrO .1 L stock 1 L .0 100molK CrO   = 0.114 L Similar to the solution made in part b, instead using 114 mL of the 1.75 M K2CrO4 stock solution. 18. 75.0 mL × 1. g 46 mol 1 mL .0 79g  = 1.3 mol C2H5OH; molarity = 250L .0 mol 3.1 = 5.2 M C2H5OH 19. Mol Na2CO3 = 0.0700 L × L molNa CO 0.3 3 2 = 0.21 mol Na2CO3 Na2CO3(s) → 2 Na+(aq) + CO3 2(aq); mol Na+ = 2(0.21 mol) = 0.42 mol CHAPTER 4 SOLUTION STOICHIOMETRY 59 Mol NaHCO3 = 0.0300 L × L molNaHCO 0.1 3 = 0.030 mol NaHCO3 NaHCO3(s) → Na+(aq) + HCO3 (aq); mol Na+ = 0.030 mol MNa = volume total totalmolNa  = 1000L .0 45mol .0 .0 0300L 0700L .0 .0 030mol .0 42 mol    = 4.5 M Na+ 20. 25.0 g (NH4)2SO4 × .15 g 132 1 mol = 1.89 × 101 mol (NH4)2SO4 Molarity = L mL 1000 0. mL 100 mol 10 .1 89 1    = 1.89 M (NH4)2SO4 Moles of (NH4)2SO4 in final solution = 10.00 × 103 L × L .1 89 mol = 1.89 × 102 mol (NH4)2SO4 Molarity of final solution = L mL 1000 50.00) mL 10.00 ( mol 10 .1 89 2     = 0.315 M (NH4)2SO4 (NH4)2SO4(s) → 2 NH4 +(aq) + SO4 2(aq);  MNH4 = 2(0.315) = 0.630 M;  2 MSO4 = 0.315 M 21. Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 40.00 g/mol Mass NaOH = 0.2500 L × NaOH mol .00g NaOH 40 L .0 400molNaOH  = 4.00 g NaOH 22. Stock solution: 1.584 g Mn2+ ×   2 2 .94g Mn 54 1molMn = 2.883 × 102 mol Mn2+; M = 000L .1 mol 10 .2 883 2  = 2.883 × 102 M Mn2+ Solution A contains: 50.00 mL × L mol 10 883 .2 mL 1000 1L 2   = 1.442 × 103 mol Mn2+ Molarity = L mL 1000 0. mL 1000 mol 10 .1 442 3    = 1.442 × 10-3 M Mn2+ 60 CHAPTER 4 SOLUTION STOICHIOMETRY Solution B contains: 10.00 mL × L mol 10 442 .1 mL 1000 1L 3   = 1.442 × 105 mol Mn2+ Molarity = 2500L .0 mol 10 .1 442 5  = 5.768 × 105 M Mn2+ Solution C contains: 10.00 × 103 L × L mol 10 .5 768 5  = 5.768 × 107 mol Mn2+ Molarity = 5000L .0 mol 10 .5 768 7  = 1.154 × 106 M Mn2+ 23. Stock solution = mL g steroid 10 00 .2 0. mL 500 g 10 0. 10 0. mL 500 10 0. mg 5 3       100.0 × 10-6 L stock × mL 10 g steroid 00 .2 L 1000mL 5   = 2.00 × 106 g steroid This is diluted to a final volume of 100.0 mL. 4. g steroid 336 molsteroid 1 L mL 1000 0. mL 100 g steroid 10 .2 00 6     = 5.95 × 108 M steroid 24. a. MCa(NO3)2 = 100 L .0 .0 100 molCa(NO ) 3 2 = 1.00 M Ca(NO3)2(s)  Ca2+(aq) + 2 NO3 (aq); MCa2 = 1.00 M;  MNO3 = 2(1.00) = 2.00 M b. MNa2SO4 = 25L .1 molNa SO 5.2 4 2 = 2.0 M Na2SO4(s)  2 Na+(aq) + SO4 2(aq); MNa = 2(2.0) = 4.0 M;  2 MSO4 = 2.0 M c. 5.00 g NH4Cl × .49g NH Cl 53 molNH Cl 1 4 4 = 0.0935 mol NH4Cl MNH4Cl = 5000L .0 .0 0935molNH Cl 4 = 0.187 M NH4Cl(s)  NH4 +(aq) + Cl(aq);  MNH4 = MCl = 0.187 M CHAPTER 4 SOLUTION STOICHIOMETRY 61 d. 1.00 g K3PO4 × .27 g 212 1molK PO 4 3 = 4.71 × 103 mol K3PO4 MK3PO4 = 2500L .0 mol 10 .4 71 3  = 0.0188 M K3PO4(s)  3 K+(aq) + PO4 3(aq); MK = 3(0.0188) = 0.0564 M;  3 MPO4 = 0.0188 M 25. Mol solute = volume (L) × molarity       L mol ; AlCl3(s) → Al3+(aq) + 3 Cl(aq) Mol Cl = 0.1000 L × 3 3 AlCl mol molCl 3 L .0 30 molAlCl   = 9.0 × 2 10 mol Cl MgCl2(s)  Mg2+(aq) + 2 Cl(aq) Mol Cl = 0.0500 L × 2 2 MgCl mol molCl 2 L .0 60 molMgCl   = 6.0 × 2 10 mol Cl NaCl(s)  Na+(aq) + Cl(aq) Mol Cl = 0.2000 L × NaCl mol molCl 1 L .0 40 molNaCl   = 8.0 × 2 10 mol Cl 100.0 mL of 0.30 M AlCl3 contains the most moles of Cl ions. 26. 10. g AgNO3 × 3 3 25molAgNO .0 L 1 9. g 169 1molAgNO  = 0.24 L = 240 mL 27. a. 5.0 ppb Hg in water = H O mL g Hg 10 0.5 H O mL ng Hg 0.5 2 9 2    L mL 1000 6. g Hg 200 molHg 1 mL g Hg 10 0.5 9     = 2.5 × 108 M Hg b. L mL 1000 4. g CHCl 119 molCHCl 1 mL g CHCl 10 0.1 3 3 3 9     = 8.4 × 109 M CHCl3 c. 10.0 ppm As = mL g As 10 0. 10 mL 10 0. μg As 6   L mL 1000 .92g As 74 molAs 1 mL g As 10 10 0. 6     = 1.33 × 104 M As 62 CHAPTER 4 SOLUTION STOICHIOMETRY d. L mL 1000 5. g DDT 354 molDDT 1 mL g DDT 10 .0 10 6     = 2.8 × 107 M DDT 28. We want 100.0 mL of each standard. To make the 100. ppm standard: mL 100.μg Cu × 100.0 mL solution = 1.00 × 104 µg Cu needed 1.00 × 104 µg Cu × 0. μg Cu 1000 1mLstock = 10.0 mL of stock solution Therefore, to make 100.0 mL of 100. ppm solution, transfer 10.0 mL of the 1000.0 ppm stock solution to a 100-mL volumetric flask, and dilute to the mark. Similarly: 75.0 ppm standard, dilute 7.50 mL of the 1000.0 ppm stock to 100.0 mL. 50.0 ppm standard, dilute 5.00 mL of the 1000.0 ppm stock to 100.0 mL. 25.0 ppm standard, dilute 2.50 mL of the 1000.0 ppm stock to 100.0 mL. 10.0 ppm standard, dilute 1.00 mL of the 1000.0 ppm stock to 100.0 mL. Precipitation Reactions 29. For the following answers, the balanced molecular equation is first, followed by the complete ionic equation, and then the net ionic equation. a. (NH4)2SO4(aq) + Ba(NO3)2(aq)  2 NH4NO3(aq) + BaSO4(s) 2 NH4 +(aq) + SO4 2(aq) + Ba2+(aq) + 2 NO3 (aq)  2 NH4 +(aq) + 2 NO3 (aq) + BaSO4(s) Ba2+(aq) + SO4 2(aq)  BaSO4(s) is the net ionic equation (spectator ions omitted). b. Pb(NO3)2(aq) + 2 NaCl(aq)  PbCl2(s) + 2 NaNO3(aq) Pb2+(aq) + 2 NO3 (aq) + 2 Na+(aq) + 2 Cl(aq)  PbCl2(s) + 2 Na+(aq) + 2 NO3 (aq) Pb2+(aq) + 2 Cl(aq) → PbCl2(s) c. The possible products, potassium phosphate and sodium nitrate, are both soluble in water. Therefore, no reaction occurs. d. No reaction occurs because all possible products are soluble. CHAPTER 4 SOLUTION STOICHIOMETRY 63 e. CuCl2(aq) + 2 NaOH(aq)  Cu(OH)2(s) + 2 NaCl(aq) Cu2+(aq) + 2 Cl(aq) + 2 Na+(aq) + 2 OH(aq)  Cu(OH)2(s) + 2 Na+(aq) + 2 Cl(aq) Cu2+(aq) + 2 OH(aq)  Cu(OH)2(s) 30. The following schemes show reagents to add in order to precipitate one ion at a time. In each scheme, NaOH can be added to precipitate the last remaining ion. 31. Use Table 4.1 to predict the solubility of the possible products. a. Possible products = Hg2SO4 and Cu(NO3)2; precipitate = Hg2SO4 b. Possible products = NiCl2 and Ca(NO3)2; both salts are soluble so no precipitate forms. c. Possible products = KI and MgCO3; precipitate = MgCO3 d. Possible products = NaBr and Al2(CrO4)3; precipitate = Al2(CrO4)3 Na2SO4 Ni2+ Hg2 2+, Ni2+ Hg2Cl2(s) c. PbSO4(s) Cu2+ Ag+, Cu2+ Ag+, Pb2+, Cu2+ b. AgCl(s) NaCl NaCl BaSO4(s) a. AgCl(s) Cr3+ , Cr3+ Ag+ Ag+, Ba2+, Cr3+ Na2SO4 NaCl 64 CHAPTER 4 SOLUTION STOICHIOMETRY 32. a. Hg2(NO3)2(aq) + CuSO4(aq)  Hg2SO4(s) + Cu(NO3)2(aq) Hg2 2+(aq) + 2 NO3 (aq) + Cu2+(aq) + SO4 2(aq)  Hg2SO4(s) + Cu2+(aq) + 2 NO3 (aq) Hg2 2+(aq) + SO4 2(aq)  Hg2SO4(s) b. No reaction occurs since both possible products are soluble. c. K2CO3(aq) + MgI2(aq)  2 KI(aq) + MgCO3(s) 2 K+(aq) + CO3 2(aq) + Mg2+(aq) + 2I(aq)  2 K+(aq) + 2 I(aq) + MgCO3(s) Mg2+(aq) + CO3 2(aq)  MgCO3(s) d. 3 Na2CrO4(aq) + 2 Al(Br)3(aq) → 6 NaBr(aq) + Al2(CrO4)3(s) 6 Na+(aq) + 3 CrO4 2(aq) + 2 Al3+(aq) + 6 Br(aq)  6 Na+(aq) + 6 Br(aq) + Al2(CrO4)3(s) 2 Al3+(aq) + 3 CrO4 2(aq)  Al2(CrO4)3(s) 33. Use the solubility rules in Table 4.1. Some soluble bromides by Rule 2 would be NaBr, KBr, and NH4Br (there are others). The insoluble bromides by Rule 3 would be AgBr, PbBr2, and Hg2Br2. Similar reasoning is used for the other parts to this problem. Sulfates: Na2SO4, K2SO4, and (NH4)2SO4 (and others) would be soluble, and BaSO4, CaSO4, and PbSO4 (or Hg2SO4) would be insoluble. Hydroxides: NaOH, KOH, Ca(OH)2 (and others) would be soluble, and Al(OH)3, Fe(OH)3, and Cu(OH)2 (and others) would be insoluble. Phosphates: Na3PO4, K3PO4, (NH4)3PO4 (and others) would be soluble, and Ag3PO4, Ca3(PO4)2, and FePO4 (and others) would be insoluble. Lead: PbCl2, PbBr2, PbI2, Pb(OH)2, PbSO4, and PbS (and others) would be insoluble. Pb(NO3)2 would be a soluble Pb2+ salt. 34. Pb(NO3)2(aq) + 2 KI(aq)  PbI2(s) + 2 KNO3(aq) (molecular equation) Pb2+(aq) + 2 NO3 (aq) + 2 K+(aq) + 2 I(aq)  PbI2(s) + 2 K+(aq) + 2 NO3 (aq) (complete) The 1.0 mol of Pb2+ ions would react with the 2.0 mol of I ions to form 1.0 mol of the PbI2 precipitate. Even though the Pb2+ and I ions are removed, the spectator ions K+ and NO3  are still present. The solution above the precipitate will conduct electricity because there are plenty of charge carriers (K+ and NO3  ions) present in solution. 35. a. When CuSO4(aq) is added to Na2S(aq), the precipitate that forms is CuS(s). Therefore, Na+ (the gray spheres) and SO4 2 (the bluish green spheres) are the spectator ions. CuSO4(aq) + Na2S(aq)  CuS(s) + Na2SO4(aq); Cu2+(aq) + S2(aq)  CuS(s) CHAPTER 4 SOLUTION STOICHIOMETRY 65 b. When CoCl2(aq) is added to NaOH(aq), the precipitate that forms is Co(OH)2(s). Therefore, Na+ (the gray spheres) and Cl- (the green spheres) are the spectator ions. CoCl2(aq) + 2 NaOH(aq)  Co(OH)2(s) + 2 NaCl(aq) Co2+(aq) + 2 OH(aq)  Co(OH)2(s) c. When AgNO3(aq) is added to KI(aq), the precipitate that forms is AgI(s). Therefore, K+ (the red spheres) and NO3  (the blue spheres) are the spectator ions. AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq); Ag+(aq) + I(aq) → AgI(s) 36. XCl2(aq) + 2 AgNO3(aq)  2 AgCl(s) + X(NO3)2(aq) 1.38 g AgCl × molAgCl 2 molXCl 1 4. g 143 1molAgCl 2  = 4.81 × 3 10 mol XCl2 2 3 2 molXCl 10 4.91 g XCl 1.00   = 208 g/mol; x + 2(35.45) = 208, x = 137 g/mol The metal X is barium (Ba). 37. M2SO4(aq) + CaCl2(aq)  CaSO4(s) + 2 MCl(aq) 1.36 g CaSO4 × 4 4 2 4 4 CaSO mol molM SO 1 .15g CaSO 136 1molCaSO  = 9.99 × 103 mol M2SO4 From the problem, 1.42 g M2SO4 was reacted, so: molar mass = 4 2 3 4 2 molM SO 10 99 .9 42g M SO .1   = 142 g/mol 142 amu = 2(atomic mass M) + 32.07 + 4(16.00), atomic mass M = 23 amu From periodic table, M is Na (sodium). 38. a. Cu(NO3)2(aq) + 2 KOH(aq)  Cu(OH)2(s) + 2 KNO3(aq) Solution A contains 2.00 L × 2.00 mol/L = 4.00 mol Cu(NO3)2, and solution B contains 2.00 L × 3.00 mol/L = 6.00 mol KOH. Let’s assume in our picture that we have 4 formula units of Cu(NO3)2 (4 Cu2+ ions and 8 NO3  ions) and 6 formula units of KOH (6 K+ ions and 6 OH ions). With 4 Cu2+ ions and 6 OH ions present, then OH is limiting. One Cu2+ ion remains as 3 Cu(OH)2(s) formula units form as precipitate. The following illustration summarizes the ions that remain in solution and the relative amount of precipitate that forms. Note that K+ and NO3  ions are spectator ions. In the illustration, V1 is the volume of solution A or B and V2 is the volume of the combined solutions with V2 = 2V1. The illustration exaggerates the amount of precipitate that would actually form. 66 CHAPTER 4 SOLUTION STOICHIOMETRY b. The spectator ion concentrations will be one-half the original spectator ion concentrations in the individual beakers because the volume was doubled. Or using moles, MK = 00 L .4 .6 00 molK  = 1.50 M and  MNO3 = 00 L .4 .8 00 molNO 3  = 2.00 M. The concentration of OH ions will be zero because OH is the limiting reagent. From the drawing, the number of Cu2+ ions will decrease by a factor of four as the precipitate forms. Because the volume of solution doubled, the concentration of Cu2+ ions will decrease by a factor of eight after the two beakers are mixed: MCu = 2.00 M       8 1 = 0.250 M Alternately, one could certainly use moles to solve for MCu2 : Mol Cu2+ reacted = 2.00 L ×     molOH 2 molCu 1 L .3 00 molOH 2 = 3.00 mol Cu2+ reacted Mol Cu2+ present initially = 2.00 L × L .2 00 molCu 2 = 4.00 mol Cu2+ present initially Excess Cu2+ present after reaction = 4.00 mol  3.00 mol = 1.00 mol Cu2+ excess MCu2 = .2 00 L 00 L .2 .1 00 molCu 2   = 0.250 M Mass of precipitate = 6.00 mol KOH × 2 2 2 Cu(OH) mol .57 g Cu(OH) 97 molKOH 2 1molCu(OH)  Mass of precipitate = 293 g Cu(OH)2 39. 2 AgNO3(aq) + Na2CrO4(aq)  Ag2CrO4(s) + 2 NaNO3(aq) NO3 - K+ Cu(OH)2 Cu(OH)2 Cu(OH)2 NO3 - K+ NO3 - K+ NO3 - Cu2+ NO3 - K+ NO3 - K+ NO3 - K+ NO3 - V1 V2 CHAPTER 4 SOLUTION STOICHIOMETRY 67 0.0750 L × 4 2 4 2 3 4 2 3 Na CrO mol .98g Na CrO 161 molAgNO 2 molNa CrO 1 L .0 100molAgNO   = 0.607 g Na2CrO4 40. 2 Na3PO4(aq) + 3 Pb(NO3)2(aq)  Pb3(PO4)2(s) + 6 NaNO3(aq) 0.1500 L × 4 3 4 3 2 3 4 3 2 3 100mol Na PO .0 L Na PO 1 molPb(NO ) 3 mol Na PO 2 L .0 250molPb(NO )   = 0.250 L = 250. mL Na3PO4 41. 2 AgNO3(aq) + CaCl2(aq)  2 AgCl(s) + Ca(NO3)2(aq) Mol AgNO3 = 0.1000 L × L .0 20 molAgNO 3 = 0.020 mol AgNO3 Mol CaCl2 = 0.1000 L × L .0 15molCaCl 2 = 0.015 mol CaCl2 The required mol AgNO3 to mol CaCl2 ratio is 2 : 1 (from the balanced equation). The actual mole ratio present is 0.020/0.015 = 1.3 (1.3 : 1). Therefore, AgNO3 is the limiting reagent. Mass AgCl = 0.020 mol AgNO3 × AgCl mol 4. g AgCl 143 molAgNO 1 molAgCl 1 3  = 2.9 g AgCl The net ionic equation is Ag+(aq) + Cl(aq)  AgCl(s). The ions remaining in solution are the unreacted Cl ions and the spectator ions, NO3  and Ca2+ (all Ag+ is used up in forming AgCl). The moles of each ion present initially (before reaction) can be easily determined from the moles of each reactant. 0.020 mol AgNO3 dissolves to form 0.020 mol Ag+ and 0.020 mol NO3 . 0.015 mol CaCl2 dissolves to form 0.015 mol Ca2+ and 2(0.015) = 0.030 mol Cl. Mol unreacted Cl = 0.030 mol Cl initially  0.020 mol Cl reacted Mol unreacted Cl = 0.010 mol Cl MCl = .0 1000L 1000L .0 010molCl .0 volume total 010molCl .0     = 0.050 M Cl The molarity of the spectator ions are:  MNO3 = 2000L .0 .0 020molNO 3  = 0.10 M NO3 ;  MCa2 = 2000L .0 .0 015molCa 2 = 0.075 M Ca2+ 42. Use the silver nitrate data to calculate the mol Cl present, then use the formula of douglasite to convert from Cl to douglasite. The net ionic reaction is Ag+ + Cl  AgCl(s). 0.03720 L mol .88g douglasite 311 molCl 4 moldouglasite 1 Ag mol molCl 1 L .0 1000molAg         = 0.2900 g douglasite 68 CHAPTER 4 SOLUTION STOICHIOMETRY Mass % douglasite = 4550g .0 .0 2900g × 100 = 63.74% 43. All the sulfur in BaSO4 came from the saccharin. The conversion from BaSO4 to saccharin uses the molar masses and formulas of each compound. 0.5032 g BaSO4 g S 32.07 g saccharin 183.9 g BaSO 233.4 g S 32.07 4   = 0.3949 g saccharin tablet .49 mg 39 tablet g 10 949 .3 tablets 10 3949g .0 Tablet Averagemass 2      Average mass % = 5894g .0 .0 3949g saccharin × 100 = 67.00% saccharin by mass 44. All the Tl in TlI came from Tl in Tl2SO4. The conversion from TlI to Tl2SO4 uses the molar masses and formulas of each compound. 0.1824 g TlI 8. g Tl 408 9. g Tl SO 504 3. g TlI 331 2044. g Tl 4 2   = 0.1390 g Tl2SO4 Mass % Tl2SO4 = 486g pesticide .9 .0 1390g Tl SO 4 2 × 100 = 1.465% Tl2SO4 45. Use aluminum in the formulas to convert from mass of Al(OH)3 to mass of Al2(SO4)3 in the mixture. 0.107 g Al(OH)3 ×      3 4 3 2 3 3 3 molAl 2 molAl (SO ) 1 Al(OH) mol molAl 1 .00g 78 1molAl(OH) 4 3 2 4 3 2 Al (SO ) mol 342.17g Al (SO ) = 0.235 g Al2(SO4)3 Mass % Al2(SO4)3 = 45g .1 .0 235g × 100 = 16.2% 46. There are many acceptable choices for spectator ions. We will generally choose Na+ and NO3  as the spectator ions because sodium salts and nitrate salts are usually soluble in water. a. Fe(NO3)3(aq) + 3 NaOH(aq)  Fe(OH)3(s) + 3 NaNO3(aq) b. Hg2(NO3)2(aq) + 2 NaCl(aq)  Hg2Cl2(s) + 2 NaNO3(aq) c. Pb(NO3)2(aq) + Na2SO4(aq)  PbSO4(s) + 2 NaNO3(aq) d. BaCl2(aq) + Na2CrO4(aq)  BaCrO4(s) + 2 NaCl(aq) CHAPTER 4 SOLUTION STOICHIOMETRY 69 47. Because a precipitate formed with Na2SO4, the possible cations are Ba2+, Pb2+, Hg2 2+, and Ca2+ (from the solubility rules). Because no precipitate formed with KCl, Pb2+, and Hg2 2+ cannot be present. Because both Ba2+ and Ca2+ form soluble chlorides and soluble hydroxides, both these cations could be present. Therefore, the cations could be Ba2+ and Ca2+ (by the solubility rules in Table 4.1). For students who do a more rigorous study of solubility, Sr2+ could also be a possible cation (it forms an insoluble sulfate salt, whereas the chloride and hydroxide salts of strontium are soluble). Acid-Base Reactions 48. Strong bases contain the hydroxide ion, OH. The reaction that occurs is H+ + OH → H2O. 0.0120 L ×     H mol molOH 1 L .0 150molH = 1.80 × 3 10 mol OH The 30.0 mL of the unknown strong base contains 1.80 × 3 10 mol OH . 0300L .0 molOH 10 8.1 3    = 0.0600 M OH The unknown base concentration is one-half the concentration of OH ions produced from the base, so the base must contain 2 OH in each formula unit. The three soluble strong bases that have 2 OH ions in the formula are Ca(OH)2, Sr(OH)2, and Ba(OH)2. These are all possible identities for the strong base. 49. If we begin with 50.00 mL of 0.100 M NaOH, then: 50.00 × 103 L × L .0 100mol = 5.00 × 103 mol NaOH to be neutralized. a. NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) 5.00 × 103 mol NaOH × mol 0.100 L soln 1 NaOH mol 1molHCl  = 5.00 × 102 L or 50.0 mL b. 2 NaOH(aq) + H2SO3(aq)  2 H2O(l) + Na2SO3(aq) 5.00 × 103 mol NaOH × 3 2 3 2 molH SO 0.100 L soln 1 molNaOH 2 1molH SO  = 2.50 × 102 L or 25.0 mL c. 3 NaOH(aq) + H3PO4(aq)  Na3PO4(aq) + 3 H2O(l) 5.00 × 103 mol NaOH × 4 3 4 3 molH PO 0.200 L soln 1 molNaOH 3 1molH PO  = 8.33 × 103 L or 8.33 mL d. HNO3(aq) + NaOH(aq)  H2O(l) + NaNO3(aq) 70 CHAPTER 4 SOLUTION STOICHIOMETRY 5.00 × 103 mol NaOH × 3 3 molHNO 0.150 L soln 1 NaOH mol 1molHNO  = 3.33 × 102 L or 33.3 mL e. HC2H3O2(aq) + NaOH(aq)  H2O(l) + NaC2H3O2(aq) 5.00 × 103 mol NaOH × 2 3 2 2 3 2 molHC H O 0.200 L soln 1 NaOH mol 1molHC H O  = 2.50 × 102 L or 25.0 mL f. H2SO4(aq) + 2 NaOH(aq)  2 H2O(l) + Na2SO4(aq) 5.00 × 103 mol NaOH × 4 2 4 2 molH SO 0.300 L soln 1 molNaOH 2 1molH SO  = 8.33 × 103 L or 8.33 mL 50. a. NH3(aq) + HNO3(aq)  NH4NO3(aq) (molecular equation) NH3(aq) + H+(aq) + NO3 (aq)  NH4 +(aq) + NO3 (aq) (complete ionic equation) NH3(aq) + H+(aq)  NH4 +(aq) (net ionic equation) b. Ba(OH)2(aq) + 2 HCl(aq)  2 H2O(l) + BaCl2(aq) Ba2+(aq) + 2 OH(aq) + 2 H+(aq) + 2 Cl(aq)  Ba2+(aq) + 2 Cl(aq) + 2 H2O(l) OH(aq) + H+(aq)  H2O(l) c. 3 HClO4(aq) + Fe(OH)3(s)  3 H2O(l) + Fe(ClO4)3(aq) 3 H+(aq) + 3 ClO4 -(aq) + Fe(OH)3(s)  3 H2O(l) + Fe3+(aq) + 3 ClO4 (aq) 3 H+(aq) + Fe(OH)3(s)  3 H2O(l) + Fe3+(aq) d. AgOH(s) + HBr(aq)  AgBr(s) + H2O(l) AgOH(s) + H+(aq) + Br(aq)  AgBr(s) + H2O(l) AgOH(s) + H+(aq) + Br(aq)  AgBr(s) + H2O(l) 51. a. Perchloric acid reacted with potassium hydroxide is a possibility. HClO4(aq) + KOH(aq)  H2O(l) + KClO4(aq) b. Nitric acid reacted with cesium hydroxide is a possibility. HNO3(aq) + CsOH(aq)  H2O(l) + CsNO3(aq) c. Hydroiodic acid reacted with calcium hydroxide is a possibility. 2 HI(aq) + Ca(OH)2(aq)  2 H2O(l) + CaI2(aq) CHAPTER 4 SOLUTION STOICHIOMETRY 71 52. We get the empirical formula from the elemental analysis. Out of 100.00 g carminic acid there are: 53.66 g C × .011g C 12 1molC = 4.468 mol C; 4.09 g H × 008g H .1 1molH = 4.06 mol H 42.25 g O × .999g O 15 1molO = 2.641 mol O Dividing the moles by the smallest number gives: .1 54 .2 641 .4 06 .1 692; .2 641 .4 468   These numbers don’t give obvious mole ratios. Let’s determine the mol C to mol H ratio: 10 11 .1 10 .4 06 .4 468   So let's try 10 .4 06 = 0.406 as a common factor: 10 ;0. .0 406 .4 06 11 ;0. .0 406 .4 468   .6 50 .0 406 .2 641  Therefore, C22H20O13 is the empirical formula. We can get molar mass from the titration data. The balanced reaction is HA(aq) + OH(aq)  H2O(l) + A(aq) where HA is an abbreviation for carminic acid, an acid with one acidic proton (H+). 18.02 × 103 L soln × NaOH mol molcarminicacid 1 soln L 0.0406molNaOH  = 7.32 × 104 mol carminic acid Molar mass = mol g 492 mol 10 32 .7 3602g .0 4    The empirical formula mass of C22H20O13  22(12) + 20(1) + 13(16) = 492 g. Therefore, the molecular formula of carminic acid is also C22H20O13. 53. HCl and HNO3 are strong acids; Ca(OH)2 and RbOH are strong bases. The net ionic equation that occurs is H+(aq) + OH(aq)  H2O(l). Mol H+ = 0.0500 L × HCl mol molH 1 L .0 100molHCl   + 0.1000 L × 3 3 HNO mol molH 1 L .0 200molHNO   = 0.00500 + 0.0200 = 0.0250 mol H+ 72 CHAPTER 4 SOLUTION STOICHIOMETRY Mol OH = 0.5000 L × 2 2 Ca(OH) mol molOH 2 L .0 0100molCa(OH)   + 0.2000 L × RbOH mol molOH 1 L .0 100molRbOH   = 0.0100 + 0.0200 = 0.0300 mol OH We have an excess of OH, so the solution is basic (not neutral). The moles of excess OH = 0.0300 mol OH initially  0.0250 mol OH reacted (with H+) = 0.0050 mol OH excess. M M 3 OH 10 9.5 8500L .0 0050mol .0 .0 2000) L .0 5000 .0 1000 .0 05000 ( .0 0050molOH           54. Because KHP is a monoprotic acid, the reaction is (KHP is an abbreviation for potassium hydrogen phthalate): NaOH(aq) + KHP(aq)  NaKP(aq) + H2O(l) 0.1082 g KHP × KHP mol molNaOH 1 .22g KHP 204 1molKHP  = 5.298 × 104 mol NaOH There is 5.298 × 104 mol of sodium hydroxide in 34.67 mL of solution. Therefore, the concentration of sodium hydroxide is: L 10 .67 34 mol 10 298 .5 3 4     = 1.528 × 102 M NaOH 55. HC2H3O2(aq) + NaOH(aq)  H2O(l) + NaC2H3O2(aq) a. 16.58 × 103 L soln  NaOH mol molaceticacid 1 soln L .0 5062molNaOH  = 8.393 × 103 mol acetic acid Concentration of acetic acid = 01000L .0 mol 10 .8 393 3  = 0.8393 M HC2H3O2 b. If we have 1.000 L of solution: total mass = 1000. mL × mL .1 006g = 1006 g solution Mass of HC2H3O2 = 0.8393 mol × mol 60.052g = 50.40 g HC2H3O2 Mass % acetic acid = g 1006 50.40g × 100 = 5.010% 56. 39.47 × 103 L HCl × HCl mol molNH 1 L .0 0984molHCl 3  = 3.88 × 103 mol NH3 Molarity of NH3 = L 10 .00 50 mol 10 88 .3 3 3     = 0.0776 M NH3 CHAPTER 4 SOLUTION STOICHIOMETRY 73 57. Ba(OH)2(aq) + 2 HCl(aq)  BaCl2(aq) + 2 H2O(l); H+(aq) + OH(aq)  H2O(l) 75.0 × 103 L × L .0 250molHCl = 1.88 × 102 mol HCl = 1.88 × 102 mol H+ + 1.88 × 102 mol Cl 225.0 × 103 L × L .0 0550molBa(OH) 2 = 1.24 × 102 mol Ba(OH)2 = 1.24 × 102 mol Ba2+ + 2.48 × 102 mol OH The net ionic equation requires a 1 : 1 mol ratio between OH and H+. The actual mol OH to mol H+ ratio is greater than 1 : 1, so OH is in excess. Because 1.88 × 102 mol OH will be neutralized by the H+, we have (2.48  1.88) × 102 = 0.60 × 102 mol OH in excess. MOH = .0 2250L 0750L .0 molOH 10 0.6 volume total molOH excess 3       = 2.0 × 102 M OH 58. Because KHP is a monoprotic acid, the reaction is: NaOH(aq) + KHP(aq)  H2O(l) + NaKP(aq) Mass KHP = 0.02046 L NaOH × KHP mol .22g KHP 204 NaOH mol molKHP 1 NaOH L .0 1000molNaOH   = 0.4178 g KHP 59. The acid is a diprotic acid (H2A) meaning that it has two H+ ions in the formula to donate to a base. The reaction is H2A(aq) + 2 NaOH(aq)  2 H2O(l) + Na2A(aq), where A2 is what is left over from the acid formula when the two protons (H+ ions) are reacted. For the HCl reaction, the base has the ability to accept two protons. The most common examples are Ca(OH)2, Sr(OH)2, and Ba(OH)2. A possible reaction would be 2 HCl(aq) + Ca(OH)2(aq)  2 H2O(l) + CaCl2(aq). 60. Let HA = unknown acid; HA(aq) + NaOH(aq)  NaA(aq) + H2O(l) Mol HA present = 0.0250 L × molNaOH 1 molHA 1 L .0 500molNaOH  = 0.0125 mol HA HA mol x g HA = 0125molHA .0 .2 20g HA , x = molar mass of HA = 176 g/mol Empirical formula mass ≈ 3(12) + 4(1) + 3(16) = 88 g/mol. Because 176/88 = 2.0, the molecular formula is (C3H4O3)2 = C6H8O6. 61. The pertinent reactions are: 2 NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2 H2O(l) HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) 74 CHAPTER 4 SOLUTION STOICHIOMETRY Amount of NaOH added = 0.0500 L × L .0 213mol = 1.07 × 102 mol NaOH Amount of NaOH neutralized by HCl: 0.01321 L HCl × HCl mol molNaOH 1 HCl L .0 103molHCl  = 1.36 × 103 mol NaOH The difference, 9.3 × 103 mol, is the amount of NaOH neutralized by the sulfuric acid. 9.3 × 103 mol NaOH × molNaOH 2 1molH SO 4 2 = 4.7 × 103 mol H2SO4 Concentration of H2SO4 = 1000L .0 mol 10 7.4 3  = 4.7 × 102 M H2SO4 62. 2 H3PO4(aq) + 3 Ba(OH)2(aq)  6 H2O(l) + Ba3(PO4)2(s) 0.01420 L × 2 2 4 3 2 4 3 0521molBa(OH) .0 L Ba(OH) 1 molH PO 2 molBa(OH) 3 L .0 141molH PO   = 0.0576 L = 57.6 mL Ba(OH)2 Oxidation-Reduction Reactions 63. a. SrCr2O7: Composed of Sr2+ and Cr2O7 2 ions. Sr, +2; O, 2; Cr, 2x + 7(2) = 2, x = +6 b. Cu, +2; Cl, 1; c. O, 0; d. H, +1; O, 1 e. Mg2+ and CO3 2 ions present. Mg, +2; O, 2; C, +4; f. Ag, 0 g. Pb2+ and SO3 2 ions present. Pb, +2; O, 2; S, +4; h. O, 2; Pb, +4 i. Na+ and C2O4 2 ions present. Na, +1; O, 2; C, 2x + 4(2) = 2, x = +3 j. O, 2; C, +4 k. Ammonium ion has a 1+ charge (NH4 +), and sulfate ion has a 2 charge (SO4 2). Therefore, the oxidation state of cerium must be +4 (Ce4+). H, +1; N, 3; O, 2; S, +6 l. O, 2; Cr, +3 64. a. UO2 2+: O, 2; for U: x + 2(2) = +2, x = +6 b. As2O3: O, 2; for As: 2(x) + 3(2) = 0, x = +3 c. NaBiO3: Na, +1; O, 2; for Bi: +1 + x + 3(2) = 0, x = +5 d. As4: As, 0 e. HAsO2: Assign H = +1 and O = -2; for As: +1 + x + 2(2) = 0, x = +