a4s - Chemistry 2000B Spring 2002 Assignment 4 Solutions 1....

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Unformatted text preview: Chemistry 2000B Spring 2002 Assignment 4 Solutions 1. The required data are in Tables 11 and 12 of Appendix D (p A-18). We can break the calculation up into several pieces: Heat to melt solid: Heat to warm liquid from melting ( 38 8 C) to boiling point (357 C): ¢ ¨¢ ¡ 38 8 K § 357 ¥  £ ¢ £ § ¢ § 1 1 56 kJ ¢ ¥ ¦£ ¤ ¡  £ © ¡ ¡ 28 53 g 0 138 J K 1g 0 31 kJ ¨ ¡ 28 53 g 11 J g © ¢ £ ¢ Heat to vaporize liquid: The specific heat capacity of the vapor is given on a molar basis. The number of moles of mercury is 28 53 g nHg 0 1422 mol 200 59 g mol heat required to warm vapor to final temperature: 8 39 kJ ¢ ¢ § ¥ 500 357 K ¥ £ ¥ ¦£ ¤ ¡ £© £ ¢ ¢ ¤ 1 0 422 kJ ¢ ¡ ¥ ¢ ¡ 0 1422 mol 20 77 J K 1mol ¢ ¡ 28 53 g 294 J g © ¢ £ ¢ ¡  The total heat required is therefore 0 422 kJ 10 68 kJ ¥ ¢  8 39 ¢ ¢  1 56 ¢ ¢  0 31 ¢ 2. The solubility equilibrium for aluminium phosphate is PO3 aq 4 ¢  © Al3aq   ! AlPO4 s  2 The pKa of HPO4 , the conjugate acid of phosphate, is 12.44 (calculated from data in table 17.4, p 799). If the pH is well above 12.44, phosphate is the dominant species and the solubility should be low (but see below). As the pH approaches the pKa , there will be less 2 free phosphate in solution and more HPO4 . This will tend to increase the solubility. Well below the pKa , there will be very little free phosphate in solution and aluminium phosphate should be very soluble. © © There’s are additional complicating factors: Aluminium hydroxide isn’t very soluble either so at high pH, precipitation of this salt might remove aluminium from solution and therefore increase the amount of aluminium phosphate which will dissolve. Whether this will occur will depend on whether the reaction ¡   © 1 Al OH  £ 3OH aq 3s PO3 aq 4  © AlPO4 s  is spontaneous or not under the given conditions. This reaction can be broken down into 10 ¥ 1 1 9 10 % 13 "¢ ¥ $#  ©  £   ! ¡   ©   Al3aq K K 20 © Al3aq PO3 aq 4 Al OH 3 s AlPO4 s 3OH aq 33   ! so that the equilibrium constant for the overall reaction is 1 3 10 20 1 9 10 33 6 8 1012 . This is huge, so this reaction will be spontaneous under any reasonable conditions of high pH (high hydroxide ion concentration). In other words, we can “dissolve” aluminium phosphate at high pH, too, at least in the limited sense that AlPO 4 s will disappear to be replaced by Al OH 3 s , leaving phosphate in solution. "¢ ¥ "¢¤ © "¢ ©  ¡  £ I will admit that I had not thought of this second possibility when I framed this question. It just goes to show that you have to think about the chemistry of these things all the time. 3. The solubility equilibrium is Au3aq   !  3aAu3 , 3aAu3 3 27 aAu3 ¥ £& 4 © ¢ 10 © 80 10 gL ¢¤ "¢ ¥ '£ £& aAu3 ¢ £& ¥ £% ¡ £& ¤ "¢ ¡ ¢ © ¥ ¢ £¤ © © ¥% & ¡ ¥ "¢ ¥ "¢ " ¢¡ ¥ ¥ £& ¡  aAu3 solubility 3 10 46 3 7 10 48 27 1 4 10 12 1 4 10 12 mol L 577 6800 g mol 10 4 aI ¡ &   (a) The solubility equilibrium is Ca2aq  (  £ 2s 2CH3 COO aq ¢  © ¡ Ca CH3 COO   ! 4. aAu3 aAu3 ¡ Ksp ¡ Since aI 3I aq ¢  © AuI3 s The concentration of calcium acetate is 2 36 mol L ¢¤ ¢ ¥ 374 g L 158 167 g mol ¤ ¥ ) £ 2 ¤ ¢ ¡ Ca CH3 COO  The activity of calcium ions in solution is therefore1 aCa2 2 36 while the activity of acetate ions is aCH3 COO 2 2 36 4 73. The Ksp is therefore 2 36 4 73 2 ¥£ ¢ £ ¢ ¡ 2 ¡ ¢ ¥ £% ¥ £ aCH3 COO 52 9 ¢¢ ¥& ¢ ¡ ¡ £& ¡ ¥% aCa2 ¢ ¥ Ksp (b) The solubility of calcium acetate is so high that under normal conditions we would not expect to exceed its solubility. 1 Neglecting nonideal behavior. 2 5. This is based on the solubility equilibrium  ¥ 6 10 "¢ © We want aF 10 calcium must be 39 11 © Ksp ¥ ¡   ! For this reaction, 2F aq aCa2 ¢ £% £& ¡ ¢  © Ca2aq CaF2 s 2 aF at the end of the reaction. This means that the activity due to the excess ¥% 3 9 10 11 39 10 6 2 This is an excessively high (almost certainly unrealizable) concentration of calcium in solution. In other words, you probably can’t remove all of that fluoride by this reaction. However, assuming that you can actually get this much calcium dissolved in water, the number of moles of calcium ions in 50 L would be 1950 mol £© ¡ ¥& ¥ '£ £¤ ¡ 0 ¥ 1& The number of moles of fluoride removed from solution is 6 mol L 50 L £¤ © 10 ¡ 0 025 1 25 mol ¥ '£ § ¢ ¡ ¥ precip ¢ 0% nF ¢ " ¡¢ 39 mol L 50 L ¢ © aq ¥ nCa2 ¢ aCa2 By stoichiometry, the number of moles of calcium ions required to precipitate this much fluoride is 1 nCa2 precip 1 25 mol 0 625 mol 2 The total number of moles of calcium ions which we must add is therefore 1951 mol ¥ ¡ ¢ 1950 mol ¢ ¥ 3£  0 2& ¥ 0 625 ¢ total ¢ ¢ ¥ 0 )& nCa2 The calcium ions are added in the form of a nitrate with a molar mass of 164.0878 g/mol. The mass of calcium nitrate required is therefore ¤ ¢ £ ¥  again, clearly an absurd number. 320 kg ¥ 3£ 1951 mol 164 0878 g mol 4 ¡ 2 ¡ mCa NO3 6. The Henry’s law constant for oxygen in water is 1 66 10 6 mol L 1 mmHg 1 (table 14.2, p 653). The partial pressure of oxygen in air at sea level is 0.2095 atm (table 12.2, p 559). The concentration of oxygen in water which is in equilibrium with the atmosphere at sea level is therefore 2 64 10 © " ¢ ¥ (£ ¡ 0 2095 atm 760 mmHg atm 4 mol L ¢¤ ¢ ¤ £ ¢ ¡ 1 " £© mol L 1mmHg © © 6 © © 1 66 10 © " ¢ ¡ O2 aq ¥ 5    (The conversion factor from atm to mmHg came from the back cover of the book.) 1 L of water therefore contains 2 64 10 4 mol of oxygen. If we transfer this many moles of oxygen into a 1 L (0 001 m3) container, it would exert a pressure of 298 15 K £ ¢ £© © ¢ ¢ 3 1 ¥ 8 314 510 J K 1mol 0 001 m3 655 Pa ¢ £ 4 mol ¡ © 10 ¡ ¢ " 2 64 " ¢ ¡¢ ¥ nRT V © ¥ P ...
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This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.

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