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Solubility Lesson 8 Titrations & Max Ion Concentration Review Questions 1. Mg(OH)2 will have the greatest solubility in: A. NaOH B. Mg(NO3)2 C. H2O D. AgNO3 Review Questions 1. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ A. NaOH B. Mg(NO3)2 C. H2O D. AgNO3 + 2OH- Review Questions 1. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. NaOH Na+ lowers solubility B. Mg(NO3)2 Mg2+ lowers solubility C. H2O No effect solubility D. AgNO3 Ag+ increases solubility by reacting with OH- Review Questions 1. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. NaOH Na+ lowers solubility B. Mg(NO3)2 Mg2+ lowers solubility C. H2O No effect solubility D. AgNO3 Ag+ increases solubility by reacting with OH- Review Questions 2. Mg(OH)2 will have the greatest solubility in: A. 1.0 M NaNO3 B. NaOH C. Sr(OH)2 Review Questions 2. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ A. 1.0 M NaNO3 B. 1.0 M NaOH C. Sr(OH)2 + 2OH- Review Questions 2. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. 1.0 M NaNO3 No effect B. 1.0 M NaOH 1.0 M OH- lowers solubility C. Sr(OH)2 2.0 M OH- lowers solubility more Review Questions 2. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. 1.0 M NaNO3 No effect B. 1.0 M NaOH 1.0 M OH- lowers solubility C. Sr(OH)2 2.0 M OH- lowers solubility more remember: Sr(OH)2 1.0 M Sr2+ + 1.0 M 2OH2.0 M Review Questions 3. PbCl2 will have the greatest solubility in: PbCl2 ⇌ Pb2+ A. 1.0 M NaCl B. 1.0 M MgCl2 C. 1.0 M AlCl3 D. 2.0 M CaCl2 + 2Cl- Review Questions 3. PbCl2 will have the greatest solubility in: PbCl2 ⇌ Pb2+ A. 1.0 M NaCl B. 1.0 M MgCl2 C. 1.0 M AlCl3 D. 2.0 M CaCl2 + 2Cl1.0 M Cl- Review Questions 3. PbCl2 will have the greatest solubility in: PbCl2 ⇌ Pb2+ + 2Cl- A. 1.0 M NaCl 1.0 M Cl- B. 1.0 M MgCl2 2.0 M Cl- C. 1.0 M AlCl3 D. 2.0 M CaCl2 Review Questions 3. PbCl2 will have the greatest solubility in: PbCl2 ⇌ Pb2+ + 2Cl- A. 1.0 M NaCl 1.0 M Cl- B. 1.0 M MgCl2 2.0 M Cl- C. 1.0 M AlCl3 3.0 M Cl- D. 2.0 M CaCl2 Review Questions 3. PbCl2 will have the greatest solubility in: PbCl2 ⇌ Pb2+ + 2Cl- A. 1.0 M NaCl 1.0 M Cl- B. 1.0 M MgCl2 2.0 M Cl- C. 1.0 M AlCl3 3.0 M Cl- D. 2.0 M CaCl2 4.0 M Cl- 1. In a titration 3.61 mL of 0.0200M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl2 solution. Calculate the [Pb2+]. Pb2+ + 0.0100 L ?M [Pb2+] = 2I- 0.00361 L 0.0200 M PbI2(s) Equation Data 1. In a titration 3.61 mL of 0.0200M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl2 solution. Calculate the [Pb2+]. Pb2+ + 0.0100 L ?M [Pb2+] = 2I- 0.00361 L 0.0200 M 0.00361 L I- PbI2(s) Equation Data 1. In a titration 3.61 mL of 0.0200M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl2 solution. Calculate the [Pb2+]. Pb2+ + 0.0100 L ?M [Pb2+] = 2I- PbI2(s) 0.00361 L 0.0200 M 0.00361 L I- x 0.0200 mol 1L Equation Data 1. In a titration 3.61 mL of 0.0200M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl2 solution. Calculate the [Pb2+]. Pb2+ + 0.0100 L ?M [Pb2+] = 2I- 0.00361 L 0.0200 M PbI2(s) Equation Data 0.00361 L I- x 0.0200 mol x 1L 1 mol Pb2+ 2 mol I- 1. In a titration 3.61 mL of 0.0200M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl2 solution. Calculate the [Pb2+]. Pb2+ + 0.0100 L ?M [Pb2+] = 2I- 0.00361 L 0.0200 M PbI2(s) Equation Data 0.00361 L I- x 0.0200 mol x 1L 0.0100 L 1 mol Pb2+ 2 mol I- 1. In a titration 3.61 mL of 0.0200M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl2 solution. Calculate the [Pb2+]. Pb2+ + 0.0100 L ?M [Pb2+] 2I- 0.00361 L 0.0200 M PbI2(s) Equation Data = 0.00361 L I- x 0.0200 mol x 1L 0.0100 L = 0.00361 M 1 mol Pb2+ 2 mol I- 2. Determine the Ksp for PbCl2 from the results of the last question. ⇌ PbCl2(s) s Pb2+ s Ksp = [Pb2+][Cl-]2 Ksp = [s][2s] 2 Ksp = 4s3 Ksp = 4(0.00361)3 Ksp = 1.88 x 10-7 + 2Cl2s Maximum Ion Concentration 1. The maximum concentration of [AgBrO3] is lower in a solution of NaBrO3 than it would be in pure water. This is because the solution already has BrO3- present. What is the maximum [Ag+] possible in a 0.100M NaBrO3 solution? [AgBrO3] What molarity of [AgBrO3] is possible before it precipitates? 0.100 M BrO3AgBrO3(s) ⇌ Ksp = 5.3 x 10-5 = [Ag+] = Ag+ + BrO30.100 M [Ag+][BrO3-] [Ag+][0.100] [AgBrO3] = 5.3 x 10-4 M 2. Calculate the maximum number of grams of AgNO3 that will dissolve 100.0 mL of 0.200M AlCl3 without forming a precipitate. AgCl(s) ⇌ Ag+ + Cl0.600 M Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+][0.600] [Ag+] = 3.0 x 10-10 M 2. Calculate the maximum number of grams of AgNO3 that will dissolve 100.0 mL of 0.200M AlCl3 without forming a precipitate. AgCl(s) ⇌ Ag+ + Cl0.600 M Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+][0.600] [Ag+] = 0.1000 L AgNO3 3.0 x 10-10 M 2. Calculate the maximum number of grams of AgNO3 that will dissolve 100.0 mL of 0.200M AlCl3 without forming a precipitate. AgCl(s) ⇌ Ag+ + Cl0.600 M Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+][0.600] [Ag+] = 3.0 x 10-10 M 0.1000 L AgNO3 x 3.0 x 10-10 moles 1L 2. Calculate the maximum number of grams of AgNO3 that will dissolve 100.0 mL of 0.200M AlCl3 without forming a precipitate. AgCl(s) ⇌ Ag+ + Cl0.600 M Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+][0.600] [Ag+] = 3.0 x 10-10 M 0.1000 L AgNO3 x 3.0 x 10-10 moles x 1L 169.9 g 1 mole 2. Calculate the maximum number of grams of AgNO3 that will dissolve 100.0 mL of 0.200M AlCl3 without forming a precipitate. AgCl(s) ⇌ Ag+ + Cl0.600 M Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+][0.600] [Ag+] = 3.0 x 10-10 M 0.1000 L AgNO3 x 3.0 x 10-10 moles x 1L 169.9 g = 5.1 x 10-9 g 1 mole