Chapter 16 TAVSS

Chapter 16 TAVSS - CHAPTER 16 HALF EQUILIBRIA Problems to...

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CHAPTER 16 HALF EQUILIBRIA Problems to prepare students for Chem 162 hourly exam III. “Equilibria” covers Zumdahl 2 nd half of chapter 15. SOLUBILITY PRODUCT COMPLEX ION EQUILIBRIA E. Tavss 4/06 Chapter 15B practice problems 1
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SOLUBILITY PRODUCT 2 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 15-2 ND HALF EQUILIBRIA SOLUBILITY PRODUCT A solution is 0.0100M in both CrO 4 2- and SO 4 2- . Slowly, Pb(NO 3 ) is added to this solution. What is the concentration of CrO 4 2- when PbSO 4 first begins to precipitate? K sp of PbCrO 4 = 2.8 × 10 -13 ; K sp of PbSO 4 = 1.6 × 10 -8 . A. 2.8 × 10 -11 M B. 1.6 × 10 -18 M C. 0.01M. D . 1.75 × 10 -7 M E. 1.6 × 10 -4 M PbSO 4 Pb 2+ + SO 4 2- Initial Y 0 0.0100 Change -X +X +X Equilibrium Y-X +X 0.0100 + X As Pb + is added to the solution, PbCrO 4 precipitates. When the Pb + concentration increases, PbSO 4 would be about to precipitate. The titration is stopped at that point. PbSO 4 precipitation: [Pb 2+ ][SO 4 2- ] = 1.6 x 10 -8 [X][0.0100 + X] = 1.6 x 10 -8 [X][0.0100] = 1.6 x 10 -8 X = 1.6 x 10 -6 [Pb 2+ ] = 1.6 x 10 -6 M when PbSO 4 first begins to precipitate. PbCrO 4 precipitation: (1.6 x 10 -6 ) x [CrO 4 2- ] = 2.8 x 10 -13 [CrO 4 2- ] = 1.75 x 10 -7 when PbSO 4 first begins to precipitate Chapter 15B practice problems 2
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3 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA SOLUBILITY PRODUCT Compound K sp Bi 2 S 3 1.0 ×Y 10 -72 Ag 2 S 1.6 × 10 -49 HgS 1.6 × 10 -54 Mg(OH) 2 8.9 × 10 -12 MnS 2.3 × 10 -13 Which one of the following compounds is the least soluble (in moles/liter)? A. Bi2S3, B.Ag2S, C. HgS, D. Mg(OH)2, E. MnS HgS and MnS can be directly compared to each other since they have the same valences. HgS, with a smaller Ksp, is less soluble than MnS. Since the valences are different for HgS and the remaining three compounds, they all have to be compared via calculations: Bi 2 S 3 2Bi 3+ + 3S 2- Initial Y 0 0 Change -X +2X +3X Equilibrium Y-X +2X +3X [Bi3+]2[S2-]3 = Ksp [2X]2[3X]3 = 1x10-72 108X5 = 1x10-72 X = 1.56 x 10-15 = solubility of Bi2S3 Ag 2 S 2Ag + + S 2- Initial Y 0 0 Change -X +2X +X Equilibrium Y-X +2X +X [Ag+]2[S2-] = Ksp [2X]2[X] = 1.6x10-49 4X3 = 1.6x10-49 X = 3.43 x 10-17 = solubility of Ag2S HgS Hg 2+ + S 2- Initial Y 0 0 Change -X +X +X Equilibrium Y-X +X +X [Hg2+][S2-] = Ksp [X][X] = 1.6x10-54 X2 = 1.6x10-54 X = 1.26 x 10-27 = solubility of HgS Mg(OH) 2 Mg 2+ + 2OH - Initial Y 0 0 Change -X +X +2X Equilibrium Y-X +X +2X [Mg2+][2OH-] = Ksp [X][2X]2 = 1.6x10-54 X2 = 8.9x10-12 X = 2.98 x 10-6 = solubility of Mg(OH)2 Chapter 15B practice problems 3
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8 CHEM 162-2006 HOURLY EXAM III + ANSWERS CHAPTER 15 – APPLICATIONS OF AQUEOUS EQUILIBRIA SOLUBILITY PRODUCT The Ksp for BaF2 is 2.4 × 10-5. When 10 mL of 0.01 M NaF is mixed with 10 mL of 0.01 M Ba(NO3)2, will a precipitate form? A.
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Chapter 16 TAVSS - CHAPTER 16 HALF EQUILIBRIA Problems to...

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