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Solubility Lesson 8 Titrations & Max Ion Concentration.

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Presentation on theme: "Solubility Lesson 8 Titrations & Max Ion Concentration."— Presentation transcript:

1 Solubility Lesson 8 Titrations & Max Ion Concentration

2 Review Questions 1. Mg(OH)2 will have the greatest solubility in: Mg(OH)2 ⇌ Mg OH- 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-

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

4 Review Questions 3. PbCl2 will have the lowest solubility in:
PbCl2 ⇌ Pb Cl- 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-

5 Maximum Ion Concentration
4. What is the maximum [Ag+] possible in a 0.100M NaBrO3 without forming a 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 = [Ag+][0.100] [Ag+] = x M

6 5. Calculate the maximum number of grams of AgNO3 that will
5. Calculate the maximum number of grams of AgNO3 that will dissolve mL of 0.200M AlCl3 without forming a 25 0C. AgCl(s) ⇌ Ag+ + Cl- 0.600 M Ksp = [Ag+][Cl-] 1.8 x = [Ag+][0.600] [Ag+] = 3.0 x M L AgNO3 x x moles x g = x g 1 L 1 mole

7 6. In a titration 3.61 mL of M NaI is required to completely precipitate all of the lead II ions in 10.0 mL of saturated PbCl2 solution. Calculate the [Pb2+]. Pb I-  PbI2 L L ? M M L I- x mol x 1 mol Pb2+ 1L 2 mol I- [Pb2+] = L = M


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