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Solubility Lesson 8 Titrations & Max Ion Concentration

Solubility Lesson 8 Titrations & Max Ion Concentration. Review Questions 1. Mg(OH) 2 will have the greatest solubility in: A. NaOH B. Mg(NO 3 ) 2 C. H 2 O D. AgNO 3. Review Questions 1. Mg(OH) 2 will have the greatest solubility in: Mg(OH) 2 ⇌ Mg 2+ + 2OH - A. NaOH

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Solubility Lesson 8 Titrations & Max Ion Concentration

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  1. Solubility Lesson 8 Titrations & Max Ion Concentration

  2. Review Questions 1. Mg(OH)2 will have the greatest solubility in: A. NaOH B. Mg(NO3)2 C. H2O D. AgNO3

  3. Review Questions 1. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. NaOH B. Mg(NO3)2 C. H2O D. AgNO3

  4. 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-

  5. 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. AgNO3Ag+ increases solubility by reacting with OH-

  6. Review Questions 2. Mg(OH)2 will have the greatest solubility in: A. 1.0 M NaNO3 B. NaOH C. Sr(OH)2

  7. Review Questions 2. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. 1.0 M NaNO3 B. 1.0 M NaOH C. Sr(OH)2

  8. Review Questions 2. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. 1.0 M NaNO3No effect B. 1.0 M NaOH1.0 M OH- lowers solubility C. Sr(OH)22.0 M OH- lowers solubility more

  9. Review Questions 2. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. 1.0 M NaNO3No effect B. 1.0 M NaOH1.0 M OH- lowers solubility C. Sr(OH)22.0 M OH- lowers solubility more remember: Sr(OH)2  Sr2+ + 2OH- 1.0 M 1.0 M 2.0 M

  10. Review Questions 3. PbCl2 will have the greatest solubility in: PbCl2 ⇌ Pb2+ + 2Cl- A. 1.0 M NaCl B. 1.0 M MgCl2 C. 1.0 M AlCl3 D. 2.0 M CaCl2

  11. 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 C. 1.0 M AlCl3 D. 2.0 M CaCl2

  12. 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

  13. 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 AlCl33.0 M Cl- D. 2.0 M CaCl2

  14. 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 AlCl33.0 M Cl- D. 2.0 M CaCl24.0 M Cl-

  15. 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+ + 2I- PbI2(s)Equation 0.0100 L 0.00361 L Data ? M 0.0200 M [Pb2+] =

  16. 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+ + 2I- PbI2(s)Equation 0.0100 L 0.00361 L Data ? M 0.0200 M [Pb2+] = 0.00361 L I-

  17. 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+ + 2I- PbI2(s)Equation 0.0100 L 0.00361 L Data ? M 0.0200 M [Pb2+] = 0.00361 L I- x 0.0200 mol 1L

  18. 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+ + 2I- PbI2(s)Equation 0.0100 L 0.00361 L Data ? M 0.0200 M [Pb2+] = 0.00361 L I- x 0.0200 mol x 1 mol Pb2+ 1L 2 mol I-

  19. 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+ + 2I- PbI2(s)Equation 0.0100 L 0.00361 L Data ? M 0.0200 M [Pb2+] = 0.00361 L I- x 0.0200 mol x 1 mol Pb2+ 1L 2 mol I- 0.0100 L

  20. 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+ + 2I- PbI2(s)Equation 0.0100 L 0.00361 L Data ? M 0.0200 M [Pb2+] = 0.00361 L I- x 0.0200 mol x 1 mol Pb2+ 1L 2 mol I- 0.0100 L = 0.00361 M

  21. 2. Determine the Ksp for PbCl2 from the results of the last question. PbCl2(s)⇌ Pb2+ + 2Cl- s s 2s Ksp = [Pb2+][Cl-]2 Ksp = [s][2s] 2 Ksp= 4s3 Ksp= 4(0.00361)3 Ksp = 1.88 x 10-7

  22. 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? 0.100 M BrO3- [AgBrO3] What molarity of[AgBrO3] is possible before it precipitates? AgBrO3(s)⇌ Ag+ + BrO3- 0.100M Ksp = [Ag+][BrO3-] 5.3 x 10-5 = [Ag+][0.100] [Ag+] = [AgBrO3] = 5.3 x 10-4 M

  23. 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+ + Cl- 0.600M Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+][0.600] [Ag+] = 3.0 x 10-10 M

  24. 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+ + Cl- 0.600M Ksp = [Ag+][Cl-] 1.8 x 10-10 = [Ag+][0.600] [Ag+] = 3.0 x 10-10 M 0.1000 L AgNO3

  25. 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+ + Cl- 0.600M 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 1 L

  26. 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+ + Cl- 0.600M 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 169.9 g 1 L 1 mole

  27. 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+ + Cl- 0.600M 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 169.9 g = 5.1 x 10-9 g 1 L 1 mole

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