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Solubility Lesson 8 Review Notes

Solubility Lesson 8 Review Notes. Adding a Solid to a Saturated Solution. Consider the saturated solution - the rate of crystallizing equals the rate of dissolving. More solid AgCl is added. The solution is full so the ion concentrations remain constant. Ag +. Cl -.

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Solubility Lesson 8 Review Notes

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  1. Solubility Lesson 8 Review Notes

  2. Adding a Solid to a Saturated Solution Consider the saturated solution-the rateofcrystallizingequalsthe rate of dissolving More solid AgClis added The solution is full so the ionconcentrations remain constant Ag+ Cl- The rate of dissolving and rate of crystallizing both increase by the same amount.

  3. Describe the change in each of the following when more solid AgCl is added to a saturated solution. Constant 1. [Ag+] 2. [Cl-] Constant Ag+ Cl- 3. Rate of dissolving Increases 4. Rate of crystallizing Increases

  4. 5. The ions in hard water areMg2+andCa2+ Na2CO3can be added to water to remove these ions

  5. 6.Calculate the total ion concentration of 1.0 M AlCl3 AlCl3 Al3+ + 3Cl- 1.0 M 1.0 M 3.0 M Total4.0 M

  6. 7. What is the molar solubility of CaC2O4? Ksp = s2 2.3 x 10-9 = s2 4.8 x 10-5 M

  7. 8. The solubility is 7.1 x 10-5 M. The compound is: A. CaSO4 Ksp = s2 = 5.0 x 10-9 B. CaCO3

  8. 9. A solution of AgNO3 is added slowly to each of the following 0.10 M solutions. Which forms a precipitate first? A. NaCl ksp = 1.8 x 10-10 B. NaIO3 ksp = 3.2 x 10-8

  9. 10. Small amounts of AgNO3 are added to three solutions. One solution • a precipitate, which one is it? does not form A. NaCl ksp = 1.8 x 10-10 B. NaIO3 ksp = 3.2 x 10-8 C. NaBr ksp = 5.4 x 10-13

  10. 11. Which solution has the greatest conductivity? A. 1.0 M NaCl B. 1.0 M CaCO3 D. 1.0 M CaCl2 1.0 M3.0 M= 4.0 M C. 1.0 M AlCl3  Al3+ + 3Cl-  Ca2+ + 2Cl- 1.0 M2.0 M = 3.0 M

  11. 12. Calculate the maximum number of grams BaCl2that will dissolve in • 0.50 L of 0.20 M AgNO3solution. AgCl(s)⇄ Ag+ + Cl- 0.20 M Ksp = [Ag+][Cl-] 1.8 x 10-10 = [0.20][Cl-] [Cl-] = 9.0 x 10-10 M BaCl2(s)⇄ Ba2+ + 2Cl- 4.5 x 10-10 M 9.0 x 10-10 M x 4.5 x 10-10mole x 208.3 g 0.50 L = 4.7 x 10-8 g 1 mole 1 L

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