Solubility Lesson 8 Titrations & Max Ion Concentration

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 OH- lowers solubility Mg2+ lowers solubility No effect solubility D. AgNO3 Ag+ increases solubility by reacting with OH-

Review Questions 2. Mg(OH)2 will have the lowest solubility in: Mg(OH)2 ⇌ Mg2+ + 2OH- A. 1.0 M NaNO3 B. 1.0 M NaOH No effect 1.0 M OH- lowers solubility C. 1.0 M Sr(OH)2 2.0 M OH- lowers solubility more remember: Sr(OH)2  Sr2+ + 2OH- 1.0 M 1.0 M 2.0 M

Review Questions 3. PbCl2 will have the lowest solubility in: PbCl2 ⇌ Pb2+ + 2Cl- A. 1.0 M NaCl B. 1.0 M MgCl2 C. 1.0 M AlCl3 1.0 M Cl- 2.0 M Cl- 3.0 M Cl- D. 2.0 M CaCl2 4.0 M Cl-

Maximum Ion Concentration 4. What is the maximum [Ag+] possible in a 0.100M NaBrO3 without forming a precipitate @ 25 0C. [Ag+] 0.100 M BrO3- AgBrO3(s) ⇌ Ag+ + BrO3- 0.100 M Ksp = [Ag+][BrO3-] What is the molarity of [Ag+] just before it precipitates? 5.3 x 10-5 = [Ag+][0.100] [Ag+] = 5.3 x 10-4 M

5. Calculate the maximum number of grams of AgNO3 that will 5. Calculate the maximum number of grams of AgNO3 that will dissolve 100.0 mL of 0.200M AlCl3 without forming a precipitate @ 25 0C. AgCl(s) ⇌ Ag+ + Cl- 0.600 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 169.9 g = 5.1 x 10-9 g 1 L 1 mole

6. 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 0.0100 L 0.00361 L ? M 0.0200 M 0.00361 L I- x 0.0200 mol x 1 mol Pb2+ 1L 2 mol I- [Pb2+] = 0.0100 L = 0.00361 M