Chapter 15 Solutions

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1 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

2 Solutions a. b. c g CaCl 2 (95.0 g H 2 O g CaCl 2 ) 1.00 g CaCl 2 (19.0 g H 2 O g CaCl 2 ) g CaCl 2 (285 g H 2 O g CaCl 2 ) 100 = 5.00% CaCl = 5.00% CaCl = 5.00% CaCl 2 d g CaCl 2 ( g H 2 O g CaCl 2 ) 100 = 5.00% CaCl To say that a solution is x% NaCl means that 100 g of the solution would contain x g of NaCl. a g NaCl 11.5 g solution 100 g solution = g NaCl b g solution c 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 = g NaCl = 0.49 g NaCl = 55.6 g NaCl g KCl (5.34 g KCl g H 2 O) 100 = 5.34 g g 100 = 3.39% KCl g CaCl 2 (67.1 g CaCl g H 2 O) 100 = 19.6% CaCl g solution 5.00 g NaCl 100 g solution = 14.3 g NaCl 285 g solution 7.50 g Na 2 CO g solution = 21.4 g Na 2 CO 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 g hexane = 93 g solution 4.8 g heptane 2.7 g pentane = 86 g hexane mol Ca 2+ ; mol Cl

3 156 Chapter 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 = L 0.50 mol KBr M = = 1.0 M 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 = g 125 ml = L 4.25 g CuCl 2 1 mol g = mol CuCl 2 M = mol CuCl 2 = M L solution b. Molar mass of NaHCO 3 = g 11.3 ml = L g NaHCO 3 1 mol g = mol NaHCO 3 M = mol NaHCO L solution = M c. Molar mass of Na 2 CO 3 = g 52.9 g Na 2 CO 3 1 mol g = mol Na 2CO 3 M = mol Na 2 CO L solution = M

4 Solutions 157 d. Molar mass of KOH = g 1.5 ml = L 0.14 mg KOH 1 g 10 3 mg 1 mol g = mol KOH M = 2.50 x 106 mol KOH L solution = M = M 23. molar mass of KNO 3 = g 225 ml = L 45.3 g KNO 3 1 mol g = mol KNO 3 M = mol KNO L solution = 1.99 M 24. molar mass of I 2 = g 225 ml = L 5.15 g I 2 1 mol g = mol I 2 M = mol I L solution = M 25. molar mass of FeCl 3 = g 1.01 g FeCl 3 1 mol FeCl g FeCl 3 = mol FeCl ml = L M = mol FeCl L solution = M Since one mole of FeCl 3 contains one mole of Fe 3+ and three moles of Cl, the solution is M in Fe 3+ and 3(0.623) = 1.87 M in Cl 26. molar mass of NaOH = g 495 g NaOH = 1 mol g = 12.4 mol NaOH M = 12.4 mol NaOH 20.0 L solution = M 27. a. molar mass of HNO 3 = g 127 ml = L L solution mol HNO 3 solution = mol HNO mol HNO g HNO 3 1 mol HNO 3 = g HNO 3

5 158 Chapter 15 b. molar mass of NH 3 = g 155 ml = L L solution 15.1 mol NH 3 solution = 2.34 mol NH mol NH g NH 3 1 mol NH 3 = 39.9 g NH 3 c. molar mass KSCN = g 2.51 L solution 2.01 x 103 mol KSCN solution = mol KSCN g KSCN mol KSCN 1 mol KSCN = g KSCN d. molar mass of HCl = g 12.2 ml = L L solution 2.45 mol HCl solution = mol HCl mol HCl g HCl 1 mol HCl = 1.09 g HCl 28. molar mass of NH 4 Cl = g 450 ml = L L solution mol NH 4 Cl 1 L solution mol NH 4 Cl g NH 4 Cl 1 mol NH 4 Cl = mol NH 4 Cl = 6.04 g NH 4 Cl 29. a ml = L L mol AlCl L mol AlCl 3 b L mol Na 3 PO L mol Na 3 PO 4 c ml = L L 1.25 mol CuCl L 1.25 mol CuCl 2 1 mol Al3+ 1 mol AlCl 3 = mol Al 3+ 3 mol Cl = mol Cl 1 mol AlCl 3 3 mol Na + 1 mol Na 3 PO 4 = 1.70 mol Na + 1 mol PO = mol PO 4 1 mol Na 3 PO 4 1 mol Cu2+ 1 mol CuCl 2 = mol Cu 2+ 2 mol Cl 1 mol CuCl 2 = mol Cl

6 Solutions 159 d ml = L ml = L L mol Ca(OH) L mol Ca(OH) 2 1 mol Ca 2+ 1 mol Ca(OH) 2 = mol Ca 2+ 2 mol OH 1 mol Ca(OH) 2 = mol OH L solution mol AgNO 3 solution molar mass AgNO 3 = g = mol AgNO mol AgNO g AgNO 3 1 mol AgNO 3 = 4.25 g AgNO M 1 V 1 = M 2 V 2 a. M 1 = M? V 1 = 125 ml V 2 = = 375 ml (0.251 M ) (125 ml) (375 ml) = M b. M 1 = M? V 1 = 445 ml V 2 = = 695 ml (0.499 M ) (445 ml) (695 ml) = M c. M 1 = M? V 1 = 5.25 L V 2 = = 5.50 L (0.101 M ) (5.25 L) (5.50 L) = M d. M 1 = 14.5 M? V 1 = 11.2 ml V 2 = = ml (14.5 M ) (11.2 ml) (261.2 ml) = 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 )

7 160 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 ) = 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 =? M V 2 = 125 ml = L V 1 = (0.150 M ) (0.125 L) (3.02 M ) = 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 = M M V 1 = 500. ml = L V 2 =? V 2 = (0.200 M ) (0.500 L) (0.150 M ) = L = 667 ml Therefore, = 167 ml of water must be added ml = L 25.0 ml = L mol AgNO 3 = L solution mol AgNO 3 solution = mol AgNO 3

8 Solutions mol AgNO 3 1 mol Cl = mol Cl 1 mol AgNO 3 M = mol Cl L = M 36. Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaNO 3 (aq) 12.5 ml = L moles Ba(NO 3 ) 2 = L 0.15 mol Ba(NO 3 ) 2 = mol Ba(NO 3 ) 2 From the balanced chemical equation for the reaction, if mol Ba(NO 3 ) 2 is to be precipitated, then mol Na 2 SO 4 will be needed mol Na 2 SO L 0.25 mol Na 2 SO 4 = L required = 7.5 ml ml = L 37.5 ml = 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+ = L CaCl mol CaCl 2 1 L CaCl 2 = mol Ca 2+ mol CO 3 2 = L Na 2 CO mol Na 2 CO 3 1 L Na 2 CO 3 = mol CO 3 2 Since one Ca 2+ reacts with one CO 2 3, Na 2 CO 3 is the limiting reactant. Since mol CO 2 3 reacts, mol of CaCO 3 will form. molar mass CaCO 3 = g mol CaCO g CaCO 3 1 mol CaCO 3 = g CaCO Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + 2KNO 3 (aq) molar masses: Pb(NO 3 ) 2, g; PbCrO 4, g 1.00 g Pb(NO 3 ) 2 1 mol Pb(NO 3 ) g Pb(NO 3 ) 2 = mol Pb(NO 3 ) ml = L L solution 1.00 mol K 2 CrO 4 solution = mol K 2 CrO 4 Pb(NO 3 ) 2 is the limiting reactant: mol PbCrO 4 will form mol PbCrO g PbCrO 4 1 mol PbCrO 4 = g PbCrO 4

9 162 Chapter HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 25.0 ml = L mol NaOH L = mol NaOH 1 mol HCl mol NaOH = mol HCl 1 mol NaOH mol HCl 1 L solution mol HCl = L = 18.8 ml 40. HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 50.0 ml = L 48.7 ml = L mol HCl L = mol HCl solution 1 mol NaOH mol HCl = mol NaOH 1 mol HCl mol NaOH M = = M 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 = L L mol HC 2 H 3 O mol HC 2 H 3 O 2 = mol HC 2 H 3 O 2 1 mol NaOH 1 mol HC 2 H 3 O 2 = mol NaOH mol NaOH 1.00 mol NaOH = L = 3.85 ml NaOH b. HF(aq) + NaOH(aq) NaF(aq) + H 2 O(l) 35.0 ml = L mol HF L = mol HF 1 mol HF mol HF = mol NaOH 1 mol NaOH mol NaOH = 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 = L

10 Solutions L mol H 3 PO 4 = mol H 3 PO mol H 3 PO 4 3 mol NaOH 1 mol H 3 PO 4 = mol NaOH mol NaOH 1.00 mol NaOH = L = 4.29 ml d. H 2 SO 4 (aq) + 2NaOH(aq) Na 2 SO 4 (aq) + 2H 2 O(l) 35.0 ml = L L mol H 2 SO 4 = mol H 2 SO mol H 2 SO 4 2 mol NaOH 1 mol H 2 SO 4 = mol NaOH mol NaOH 1.00 mol NaOH = 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 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 = g g NaOH 1 equiv NaOH g = equiv NaOH 10.2 ml = L N = 2.83 x 103 equiv L = N b. equivalent weight Ca(OH) 2 = 1 g 12.5 mg 10 3 mg 100. ml = L molar mass 2 = g 2 = g 1 equiv g = equiv Ca(OH) 2

11 164 Chapter 15 N = 3.37 x 103 equiv L = N c. equivalent weight H 2 SO 4 = molar mass g 1 equiv g = equiv H 2SO 4 = g 2 = g 155 ml = L N = equiv L = 1.63 N 45. a M NaOH 1 equiv NaOH 1 mol NaOH = N NaOH b M Ca(OH) 2 2 equiv Ca(OH) 2 1 mol Ca(OH) 2 = N Ca(OH) 2 c M H 3 PO 4 3 equiv H 3 PO 4 1 mol H 3 PO 4 = 13.3 N H 3 PO molar mass H 3 PO 4 = 98.0 g 35.2 g H 3 PO 4 1 mol H 3 PO g H 3 PO 4 = mol H 3 PO 4 M = H 3 PO 4 = M = M 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) M NaOH = N NaOH 56.2 ml = L equiv L NaOH = equiv NaOH equiv NaOH requires equiv H 2 SO 4 to react equiv H 2 SO equiv = 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 N H 2 SO 4 : N acid V acid = N base V base (0.125 N) (24.2 ml) = (0.151 N) (V base )

12 Solutions 165 V base = 20.0 ml of the N NaOH solution needed For the 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 M H 2 SO 4 = 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 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 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 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 = = 42.5 millimol Cl Total volume after mixing = 50.0 ml ml = 75.0 ml M cobalt(ii) ion = M nickel(ii) ion = 12.5 millimol Co ml 8.75 millimol Ni ml = M = M M chloride ion = 42.5 millimol Cl 75.0 ml = M 53. AgNO 3 (s) + NaCl(aq) AgCl(s) + NaNO 3 (aq) molar masses: AgNO 3, g; AgCl, g 10.0 g AgNO 3 1 mol AgNO g AgNO 3 = mol AgNO ml = L

13 166 Chapter L 1.0 x 102 mol NaCl = mol NaCl NaCl is the limiting reactant mol AgCl form mol AgCl g AgNO 3 = g AgCl (72 mg) 1 mol Since 1 mol AgNO 3 contains 1 mol Ag +, the mol Ag + remaining in solution = = mol AgNO mol AgNO 3 = mol Ag + M silver ion = mol Ag L = M = 1.2 M 54. Ba(NO 3 ) 2 (aq) + H 2 SO 4 (aq) BaSO 4 (s) + 2HNO 3 (aq) 37.5 ml = L L mol H 2 SO 4 = 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 mol of Ba 2+ ion will be precipitated from the solution as BaSO 4. molar mass BaSO 4 = g mol BaSO 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 = ml water g water g H 2 O 1 mol H 2 O 18.0 g H 2 O = 56 mol H 2O 56. molar mass CaCl 2 = g 14.2 g CaCl 2 1 mol CaCl g CaCl 2 = mol CaCl ml = L M = mol CaCl L = 2.56 M

14 Solutions M 1 V 1 = M 2 V 2 a. M 1 = M? V 1 = 125 ml V 2 = = 275 ml (0.200 M ) (125 ml) (275 ml) = M b. M 1 = M? V 1 = 155 ml V 2 = = 305 ml (0.250 M ) (155 ml) (305 ml) = M c. M 1 = M? V 1 = L = 500. ml V 2 = = 650. ml (0.250 M ) (500. ml) (650. ml) = M d. M 1 = 18.0 M? V 1 = 15 ml V 2 = = 165 ml (18.0 M ) (15 ml) (165 ml) = 1.6 M 58. a 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 M H 2 SO 4 2 equiv H 2 SO 4 1 mol H 2 SO 4 = N H 2 SO 4 c 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) = ( N)(27.5 ml) N acid = N HNO [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,

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