Chapter 15 Solutions

Chapter 15 Solutions 1. A homogeneous mixture is a combination of two (or more) pure substances that is uniform in composition and appearance throughout. Examples of homogeneous mixtures in the real world include rubbing alcohol (70% isopropyl alcohol, 30% water) and gasoline (a mixture of hydrocarbons). 2. solvent, solute 3. When an ionic solute dissolves in water, a given ion is pulled into solution by the attractive ion dipole force exerted by several water molecules. For example, in dissolving a positive ion, the ion is approached by the negatively charged end of several water molecules: if the attraction of the water molecules for the positive ion is stronger than the attraction of the negative ions near it in the crystal, the ion leaves the crystal and enters solution. After entering solution, the dissolved ion is surrounded completely by water molecules, which tends to prevent the ion from reentering the crystal. 4. One substance will mix with and dissolve in another substance if the intermolecular forces are similar in the two substances, so that when the mixture forms, the forces between particles in the mixture will be similar to the forces present in the separate substances. Sugar and ethyl alcohol molecules both contain polar OH groups, which are comparable to the polar OH structure in water. Sugar or ethyl alcohol molecules can hydrogen-bond with water molecules and intermingle with them freely to form a solution. Substances like petroleum (whose molecules contain only carbon and hydrogen) are very nonpolar and cannot form interactions with polar water molecules. 5. saturated 6. unsaturated 7. variable 8. large 9. Increasing the surface area of a solid increases the amount of solid that comes in contact with the solvent. Typically, particles of a solid are broken into smaller pieces by grinding. This increases the surface-to-volume ratio of each particle and thus increases the amount of solid in contact with the solvent. 10. An increase in temperature means an increase in the average kinetic energy. Thus, in a warmer solution the particles of the liquid solvent are moving more rapidly. Because of this, there is an increased rate in solute solvent interaction, which increases the rate of dissolving. 11. An increase in temperature means an increase in average kinetic energy. In a warmer solution, the solvent and solute particles are moving more rapidly. If the solute particles are gaseous, faster moving particles are more likely to have enough energy to escape from the liquid. 154

Solutions 155 12. a. b. c. 5.00 g CaCl 2 (95.0 g H 2 O + 5.00 g CaCl 2 ) 1.00 g CaCl 2 (19.0 g H 2 O + 1.00 g CaCl 2 ) 15.00 g CaCl 2 (285 g H 2 O + 15.00 g CaCl 2 ) 100 = 5.00% CaCl 2 100 = 5.00% CaCl 2 100 = 5.00% CaCl 2 d. 0.00200 g CaCl 2 (0.0380 g H 2 O + 0.00200 g CaCl 2 ) 100 = 5.00% CaCl 2 13. To say that a solution is x% NaCl means that 100 g of the solution would contain x g of NaCl. a. 6.25 g NaCl 11.5 g solution 100 g solution = 0.719 g NaCl b. 6.25 g solution c. 54.3 g solution d. 452 g solution 11.5 g NaCl 100 g solution 0.91 g NaCl 100 g solution 12.3 g NaCl 100 g solution = 0.719 g NaCl = 0.49 g NaCl = 55.6 g NaCl 14. 5.34 g KCl (5.34 g KCl + 152 g H 2 O) 100 = 5.34 g 157.34 g 100 = 3.39% KCl 15. 67.1 g CaCl 2 (67.1 g CaCl 2 + 275 g H 2 O) 100 = 19.6% CaCl 2 16. 285 g solution 5.00 g NaCl 100 g solution = 14.3 g NaCl 285 g solution 7.50 g Na 2 CO 3 100.0 g solution = 21.4 g Na 2 CO 3 17. g heptane = 93 g solution g pentane = 93 g solution 5.2 g heptane 100. g solution 2.9 g pentane 100. g solution = 4.8 g heptane = 2.7 g pentane 18. 0.105 g hexane = 93 g solution 4.8 g heptane 2.7 g pentane = 86 g hexane 19. 0.221 mol Ca 2+ ; 0.442 mol Cl

156 Chapter 15 20. To say that a solution has a concentration of 5 M means that in 1 L of solution (not solvent) there would be 5 mol of solute: to prepare such a solution one would place 5 mol of NaCl in a 1-L flask, and then add whatever amount of water is necessary so that the total volume would be 1 L after mixing. The NaCl will occupy some space, so the amount of water to be added will be less than. 21. Molarity = moles of solute liters of solution a. 250 ml = 0.25 L 0.50 mol KBr M = = 2.0 M 0.25 L solution b. 500 ml = 0.500 L 0.50 mol KBr M = = 1.0 M 0.500 L solution c. 750 ml = 0.75 L M = d. M = 22. Molarity = 0.50 mol KBr 0.75 L solution = 0.67 M 0.50 mol KBr 1.0 L solution = 0.50 M moles of solute liters of solution a. Molar mass of CuCl 2 = 134.45 g 125 ml = 0.125 L 4.25 g CuCl 2 1 mol 134.45 g = 0.0316 mol CuCl 2 M = 0.0316 mol CuCl 2 = 0.253 M 0.125 L solution b. Molar mass of NaHCO 3 = 84.01 g 11.3 ml = 0.0113 L 0.101 g NaHCO 3 1 mol 84.01 g = 0.00120 mol NaHCO 3 M = 0.00120 mol NaHCO 3 0.0113 L solution = 0.106 M c. Molar mass of Na 2 CO 3 = 105.99 g 52.9 g Na 2 CO 3 1 mol 105.99 g = 0.499 mol Na 2CO 3 M = 0.499 mol Na 2 CO 3 1.15 L solution = 0.434 M

Solutions 157 d. Molar mass of KOH = 56.11 g 1.5 ml = 0.0015 L 0.14 mg KOH 1 g 10 3 mg 1 mol 56.11 g = 2.50 10 6 mol KOH M = 2.50 x 106 mol KOH 0.0015 L solution = 1.67 10 3 M = 1.7 10 3 M 23. molar mass of KNO 3 = 101.1 g 225 ml = 0.225 L 45.3 g KNO 3 1 mol 101.1 g = 0.448 mol KNO 3 M = 0.448 mol KNO 3 0.225 L solution = 1.99 M 24. molar mass of I 2 = 253.8 g 225 ml = 0.225 L 5.15 g I 2 1 mol 253.8 g = 0.0203 mol I 2 M = 0.0203 mol I 2 0.225 L solution = 0.0902 M 25. molar mass of FeCl 3 = 162.2 g 1.01 g FeCl 3 1 mol FeCl 3 162.2 g FeCl 3 = 0.00623 mol FeCl 3 10.0 ml = 0.0100 L M = 0.00623 mol FeCl 3 0.0100 L solution = 0.623 M Since one mole of FeCl 3 contains one mole of Fe 3+ and three moles of Cl, the solution is 0.623 M in Fe 3+ and 3(0.623) = 1.87 M in Cl 26. molar mass of NaOH = 40.00 g 495 g NaOH = 1 mol 40.00 g = 12.4 mol NaOH M = 12.4 mol NaOH 20.0 L solution = 0.619 M 27. a. molar mass of HNO 3 = 63.02 g 127 ml = 0.127 L 0.127 L solution 0.105 mol HNO 3 solution = 0.0133 mol HNO 3 0.0133 mol HNO 3 63.02 g HNO 3 1 mol HNO 3 = 0.838 g HNO 3

158 Chapter 15 b. molar mass of NH 3 = 17.03 g 155 ml = 0.155 L 0.155 L solution 15.1 mol NH 3 solution = 2.34 mol NH 3 2.34 mol NH 3 17.03 g NH 3 1 mol NH 3 = 39.9 g NH 3 c. molar mass KSCN = 97.19 g 2.51 L solution 2.01 x 103 mol KSCN solution = 5.05 10 3 mol KSCN 5.05 10 3 97.19 g KSCN mol KSCN 1 mol KSCN = 0.491 g KSCN d. molar mass of HCl = 36.46 g 12.2 ml = 0.0122 L 0.0122 L solution 2.45 mol HCl solution = 0.0299 mol HCl 0.0299 mol HCl 36.46 g HCl 1 mol HCl = 1.09 g HCl 28. molar mass of NH 4 Cl = 53.49 g 450 ml = 0.450 L 0.450 L solution 0.251 mol NH 4 Cl 1 L solution 0.113 mol NH 4 Cl 53.49 g NH 4 Cl 1 mol NH 4 Cl = 0.113 mol NH 4 Cl = 6.04 g NH 4 Cl 29. a. 10.2 ml = 0.0102 L 0.0102 L 0.451 mol AlCl 3 0.0102 L 0.451 mol AlCl 3 b. 5.51 L 0.103 mol Na 3 PO 4 5.51 L 0.103 mol Na 3 PO 4 c. 1.75 ml = 0.00175 L 0.00175 L 1.25 mol CuCl 2 0.00175 L 1.25 mol CuCl 2 1 mol Al3+ 1 mol AlCl 3 = 4.60 10 3 mol Al 3+ 3 mol Cl = 1.38 10 2 mol Cl 1 mol AlCl 3 3 mol Na + 1 mol Na 3 PO 4 = 1.70 mol Na + 1 mol PO 3 4 3 = 0.568 mol PO 4 1 mol Na 3 PO 4 1 mol Cu2+ 1 mol CuCl 2 = 2.19 10 3 mol Cu 2+ 2 mol Cl 1 mol CuCl 2 = 4.38 10 3 mol Cl

Solutions 159 d. 25.2 ml = 0.0252 L 30. 250. ml = 0.250 L 0.0252 L 0.00157 mol Ca(OH) 2 0.0252 L 0.00157 mol Ca(OH) 2 1 mol Ca 2+ 1 mol Ca(OH) 2 = 3.96 10 5 mol Ca 2+ 2 mol OH 1 mol Ca(OH) 2 = 7.91 10 5 mol OH 0.250 L solution 0.100 mol AgNO 3 solution molar mass AgNO 3 = 169.9 g = 0.0250 mol AgNO 3 0.0250 mol AgNO 3 169.9 g AgNO 3 1 mol AgNO 3 = 4.25 g AgNO 3 31. M 1 V 1 = M 2 V 2 a. M 1 = 0.251 M? V 1 = 125 ml V 2 = 250. + 125 = 375 ml (0.251 M ) (125 ml) (375 ml) = 0.0837 M b. M 1 = 0.499 M? V 1 = 445 ml V 2 = 445 + 250. = 695 ml (0.499 M ) (445 ml) (695 ml) = 0.320 M c. M 1 = 0.101 M? V 1 = 5.25 L V 2 = 5.25 + 0.250 = 5.50 L (0.101 M ) (5.25 L) (5.50 L) = 0.0964 M d. M 1 = 14.5 M? V 1 = 11.2 ml V 2 = 11.2 + 250. = 261.2 ml (14.5 M ) (11.2 ml) (261.2 ml) = 0.622 M 32. M 1 V 1 = M 2 V 2 HCl: M 1 = 3.0 M 12.1 M V 1 = 225 ml V 2 =? V 2 = (3.0 M ) (225 ml) = 55.8 ml = 56 ml (12.1 M )

160 Chapter 15 HNO 3 : M 1 = 3.0 M 15.9 M V 1 = 225 ml V 2 =? V 2 = (3.0 M ) (225 ml) (15.9 M ) = 42.45 ml = 42 ml H 2 SO 4 : M 1 = 3.0 M 18.0 M V 1 = 225 ml V 2 =? V 2 = (3.0 M ) (225 ml) (18.0 M ) = 37.5 ml = 38 ml HC 2 H 3 O 2 : M 1 = 3.0 M 17.5 M V 1 = 225 ml V 2 =? V 2 = (3.0 M ) (225 ml) (17.5 M ) = 38.6 ml = 39 ml H 3 PO 4 : M 1 = 3.0 M 14.9 M V 1 = 225 ml V 2 =? V 2 = (3.0 M ) (225 ml) (14.9 M ) = 45.3 ml = 45 ml 33. M 1 V 1 = M 2 V 2 M 1 = 3.02 M V 1 =? 0.150 M V 2 = 125 ml = 0.125 L V 1 = (0.150 M ) (0.125 L) (3.02 M ) = 0.00621 L = 6.21 ml The student could prepare her solution by transferring 6.21 ml of the 3.02 M NaOH solution from a pipet or buret to a 125-mL volumetric flask, and then adding distilled water to the calibration mark of the flask. 34. M 1 V 1 = M 2 V 2 M 1 = 0.200 M 0.150 M V 1 = 500. ml = 0.500 L V 2 =? V 2 = (0.200 M ) (0.500 L) (0.150 M ) = 0.667 L = 667 ml Therefore, 667 500. = 167 ml of water must be added. 35. 27.2 ml = 0.0272 L 25.0 ml = 0.0250 L mol AgNO 3 = 0.0272 L solution 0.104 mol AgNO 3 solution = 0.002829 mol AgNO 3

Solutions 161 0.002829 mol AgNO 3 1 mol Cl = 0.002829 mol Cl 1 mol AgNO 3 M = 0.002829 mol Cl 0.0250 L = 0.113 M 36. Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaNO 3 (aq) 12.5 ml = 0.0125 L moles Ba(NO 3 ) 2 = 0.0125 L 0.15 mol Ba(NO 3 ) 2 = 1.88 10 3 mol Ba(NO 3 ) 2 From the balanced chemical equation for the reaction, if 1.88 10 3 mol Ba(NO 3 ) 2 is to be precipitated, then 1.88 10 3 mol Na 2 SO 4 will be needed. 1.88 10 3 mol Na 2 SO 4 1.00 L 0.25 mol Na 2 SO 4 = 0.0075 L required = 7.5 ml 37. 36.2 ml = 0.0362 L 37.5 ml = 0.0375 L Since each formula unit of CaCO 3 contains one Ca 2+ ion, and since each Na 2 CO 3 formula unit contains one CO 2 3 ion, we can say that mol Ca 2+ = 0.0362 L CaCl 2 0.158 mol CaCl 2 1 L CaCl 2 = 0.00572 mol Ca 2+ mol CO 3 2 = 0.0375 L Na 2 CO 3 0.149 mol Na 2 CO 3 1 L Na 2 CO 3 = 0.00559 mol CO 3 2 Since one Ca 2+ reacts with one CO 2 3, Na 2 CO 3 is the limiting reactant. Since 0.00559 mol CO 2 3 reacts, 0.00559 mol of CaCO 3 will form. molar mass CaCO 3 = 100.1 g 0.00559 mol CaCO 3 100.1 g CaCO 3 1 mol CaCO 3 = 0.560 g CaCO 3 38. Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + 2KNO 3 (aq) molar masses: Pb(NO 3 ) 2, 331.2 g; PbCrO 4, 323.2 g 1.00 g Pb(NO 3 ) 2 1 mol Pb(NO 3 ) 2 331.2 g Pb(NO 3 ) 2 = 0.003019 mol Pb(NO 3 ) 2 25.0 ml = 0.0250 L 0.0250 L solution 1.00 mol K 2 CrO 4 solution = 0.0250 mol K 2 CrO 4 Pb(NO 3 ) 2 is the limiting reactant: 0.003019 mol PbCrO 4 will form. 0.003019 mol PbCrO 4 323.2 g PbCrO 4 1 mol PbCrO 4 = 0.976 g PbCrO 4

162 Chapter 15 39. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 25.0 ml = 0.0250 L 0.150 mol NaOH 0.0250 L = 0.00375 mol NaOH 1 mol HCl 0.00375 mol NaOH = 0.00375 mol HCl 1 mol NaOH 0.00375 mol HCl 1 L solution 0.200 mol HCl = 0.01875 L = 18.8 ml 40. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 50.0 ml = 0.0500 L 48.7 ml = 0.0487 L 0.104 mol HCl 0.0500 L = 0.00520 mol HCl solution 1 mol NaOH 0.00520 mol HCl = 0.00520 mol NaOH 1 mol HCl 0.00520 mol NaOH M = = 0.107 M 0.0487 L 41. a. NaOH(aq) + HC 2 H 3 O 2 (aq) NaC 2 H 3 O 2 (aq) + H 2 O(l) 25.0 ml = 0.0250 L 0.0250 L 0.154 mol HC 2 H 3 O 2 0.00385 mol HC 2 H 3 O 2 = 0.00385 mol HC 2 H 3 O 2 1 mol NaOH 1 mol HC 2 H 3 O 2 = 0.00385 mol NaOH 0.00385 mol NaOH 1.00 mol NaOH = 0.00385 L = 3.85 ml NaOH b. HF(aq) + NaOH(aq) NaF(aq) + H 2 O(l) 35.0 ml = 0.0350 L 0.102 mol HF 0.0350 L = 0.00357 mol HF 1 mol HF 0.00357 mol HF = 0.00357 mol NaOH 1 mol NaOH 0.00357 mol NaOH = 0.00357 L = 3.57 ml 1.00 mol NaOH c. H 3 PO 4 (aq) + 3NaOH(aq) Na 3 PO 4 (aq) + 3H 2 O(l) 10.0 ml = 0.0100 L

Solutions 163 0.0100 L 0.143 mol H 3 PO 4 = 0.00143 mol H 3 PO 4 0.00143 mol H 3 PO 4 3 mol NaOH 1 mol H 3 PO 4 = 0.00429 mol NaOH 0.00429 mol NaOH 1.00 mol NaOH = 0.00429 L = 4.29 ml d. H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) 35.0 ml = 0.0350 L 0.0350 L 0.220 mol H 2 SO 4 = 0.00770 mol H 2 SO 4 0.00770 mol H 2 SO 4 2 mol NaOH 1 mol H 2 SO 4 = 0.0154 mol NaOH 0.0154 mol NaOH 1.00 mol NaOH = 0.0154 L = 15.4 ml 42. When H 2 SO 4 reacts with OH, the reaction is H 2 SO 4 (aq) + 2OH (aq) 2H 2 O(l) + SO 2 4 (aq) Since each mol of H 2 SO 4 provides two moles of H + ion, it is only necessary to take half a mole of H 2 SO 4 to provide one mole of H + ion. The equivalent weight of H 2 SO 4 is thus half the molar mass. 43. 1.53 equivalents OH ion are needed to react with 1.53 equivalents of H + ion. By definition, one equivalent of OH ion exactly neutralizes one equivalent of H + ion. 44. N = number of equivalents of solute number of liters of solution a. equivalent weight NaOH = molar mass NaOH = 40.00 g 0.113 g NaOH 1 equiv NaOH 40.00 g = 2.83 10 3 equiv NaOH 10.2 ml = 0.0102 L N = 2.83 x 103 equiv 0.0102 L = 0.277 N b. equivalent weight Ca(OH) 2 = 1 g 12.5 mg 10 3 mg 100. ml = 0.100 L molar mass 2 = 74.10 g 2 = 37.05 g 1 equiv 37.05 g = 3.37 10 4 equiv Ca(OH) 2

164 Chapter 15 N = 3.37 x 103 equiv 0.100 L = 3.37 10 3 N c. equivalent weight H 2 SO 4 = molar mass 2 12.4 g 1 equiv 49.05 g = 0.253 equiv H 2SO 4 = 98.09 g 2 = 49.05 g 155 ml = 0.155 L N = 0.253 equiv 0.155 L = 1.63 N 45. a. 0.134 M NaOH 1 equiv NaOH 1 mol NaOH = 0.134 N NaOH b. 0.00521 M Ca(OH) 2 2 equiv Ca(OH) 2 1 mol Ca(OH) 2 = 0.0104 N Ca(OH) 2 c. 4.42 M H 3 PO 4 3 equiv H 3 PO 4 1 mol H 3 PO 4 = 13.3 N H 3 PO 4 46. molar mass H 3 PO 4 = 98.0 g 35.2 g H 3 PO 4 1 mol H 3 PO 4 98.0 g H 3 PO 4 = 0.3592 mol H 3 PO 4 M = 0.3592 H 3 PO 4 = 0.3592 M = 0.359 M 0.3592 M H 3 PO 4 3 equiv H 3 PO 4 1 mol H 3 PO 4 = 1.08 N 47. H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) 0.145 M NaOH = 0.145 N NaOH 56.2 ml = 0.0562 L 0.145 equiv 0.0562 L NaOH = 0.00815 equiv NaOH 0.00815 equiv NaOH requires 0.00815 equiv H 2 SO 4 to react. 0.00815 equiv H 2 SO 4 0.172 equiv = 0.0474 L = 47.4 ml H 2SO 4 solution 48. 2NaOH(aq) + H 2 SO 4 (aq) Na 2 SO 4 (aq) + 2H 2 O(l) For the 0.125 N H 2 SO 4 : N acid V acid = N base V base (0.125 N) (24.2 ml) = (0.151 N) (V base )

Solutions 165 V base = 20.0 ml of the 0.151 N NaOH solution needed For the 0.125 M H 2 SO 4 : Since each H 2 SO 4 formula unit produces two H + ions, the normality of this solution will be twice its molarity. 0.125 M H 2 SO 4 = 0.250 N H 2 SO 4 N acid V acid = N base V base (0.250 N) (24.1 ml) = (0.151 N) (V base ) V base = 39.9 ml of the 0.151 N NaOH solution needed 49. Colligative properties are properties of a solution that depend only on the number, not the identity, of the solute particles. 50. For a solution to boil, bubbles must form in the solution. Solute particles block water from entering these bubbles. It is not the nature of these particles that matters, but the number of the particles; thus, it is a colligative property. 51. Antifreeze solution is a concentrated aqueous solution that has a lower freezing point than water. It will also have a higher boiling point than water (a solute in water both lowers the freezing point and raises the boiling point). 52. millimol CoCl 2 = 50.0 ml 0.250 M CoCl 2 = 12.5 millimol CoCl 2 This contains 12.5 millimol Co 2+ and 25.0 millimol Cl millimol NiCl 2 = 25.0 ml 0.350 M NiCl 2 = 8.75 millimol NiCl 2 This contains 8.75 millimol Ni 2+ and 17.5 millimol Cl Total millimol Cl after mixing = 25.0 + 17.5 = 42.5 millimol Cl Total volume after mixing = 50.0 ml + 25.0 ml = 75.0 ml M cobalt(ii) ion = M nickel(ii) ion = 12.5 millimol Co2+ 75.0 ml 8.75 millimol Ni2+ 75.0 ml = 0.167 M = 0.117 M M chloride ion = 42.5 millimol Cl 75.0 ml = 0.567 M 53. AgNO 3 (s) + NaCl(aq) AgCl(s) + NaNO 3 (aq) molar masses: AgNO 3, 169.9 g; AgCl, 143.4 g 10.0 g AgNO 3 1 mol AgNO 3 169.9 g AgNO 3 = 0.05886 mol AgNO 3 50. ml = 0.050 L

166 Chapter 15 0.050 L 1.0 x 102 mol NaCl = 0.00050 mol NaCl NaCl is the limiting reactant. 0.00050 mol AgCl form. 0.00050 mol AgCl 143.4 g AgNO 3 = 0.072 g AgCl (72 mg) 1 mol Since 1 mol AgNO 3 contains 1 mol Ag +, the mol Ag + remaining in solution = 0.05886 0.00050 = 0.05836 mol AgNO 3 0.05836 mol AgNO 3 = 0.05836 mol Ag + M silver ion = 0.05836 mol Ag+ 0.050 L = 1.167 M = 1.2 M 54. Ba(NO 3 ) 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + 2HNO 3 (aq) 37.5 ml = 0.0375 L 0.0375 L 0.221 mol H 2 SO 4 = 0.00829 mol H 2 SO 4 Since the coefficients of Ba(NO 3 ) 2 and H 2 SO 4 in the balanced chemical equation for the reaction are both one, then 0.00829 mol of Ba 2+ ion will be precipitated from the solution as BaSO 4. molar mass BaSO 4 = 233.4 g 0.00829 mol BaSO 4 233.4 g BaSO 4 1 mol BaSO 4 = 1.93 g BaSO 4 precipitate 55. molar mass H 2 O = 18.0 g 1.0 L water = 1.0 10 3 ml water 1.0 10 3 g water 1.0 10 3 g H 2 O 1 mol H 2 O 18.0 g H 2 O = 56 mol H 2O 56. molar mass CaCl 2 = 111.0 g 14.2 g CaCl 2 1 mol CaCl 2 111.0 g CaCl 2 = 0.128 mol CaCl 2 50.0 ml = 0.0500 L M = 0.128 mol CaCl 2 0.0500 L = 2.56 M

Solutions 167 57. M 1 V 1 = M 2 V 2 a. M 1 = 0.200 M? V 1 = 125 ml V 2 = 125 + 150. = 275 ml (0.200 M ) (125 ml) (275 ml) = 0.0909 M b. M 1 = 0.250 M? V 1 = 155 ml V 2 = 155 + 150. = 305 ml (0.250 M ) (155 ml) (305 ml) = 0.127 M c. M 1 = 0.250 M? V 1 = 0.500 L = 500. ml V 2 = 500. + 150. = 650. ml (0.250 M ) (500. ml) (650. ml) = 0.192 M d. M 1 = 18.0 M? V 1 = 15 ml V 2 = 15 + 150. = 165 ml (18.0 M ) (15 ml) (165 ml) = 1.6 M 58. a. 0.50 M HC 2 H 3 O 2 1 equiv HC 2 H 3 O 2 1 mol HC 2 H 3 O 2 = 0.50 N HC 2 H 3 O 2 b. 0.00250 M H 2 SO 4 2 equiv H 2 SO 4 1 mol H 2 SO 4 = 0.00500 N H 2 SO 4 c. 0.10 M KOH 1 equiv KOH 1 mol KOH = 0.10 N KOH 59. N acid V acid = N base V base N acid (10.0 ml) = (3.5 10 2 N)(27.5 ml) N acid = 9.6 10 2 N HNO 3 60. [A] = 4 mol 1.0 L [A] > [D] > [C] > [B] = 4.0 M, [B] = 6 mol 4.0 L = 1.5 M, [C] = 4 mol 2.0 L = 2.0 M, [D] = 6 mol 2.0 L = 3.0 M,