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Solubility Lesson 8 Review Notes

Solubility Lesson 8 Review Notes

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Solubility Lesson 8 Review Notes. Adding a Crystal to a Saturated Solution  Consider the saturated solution. Ag +. Cl -. Adding a Crystal to a Saturated Solution Note that the solution is saturated -. Ag +. Cl -. Adding a Crystal to a Saturated Solution - PowerPoint PPT Presentation

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Page 1: Solubility Lesson 8 Review Notes

SolubilityLesson 8

Review Notes

Page 2: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution  Consider the saturated solution     Cl-

 

Ag+

Page 3: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution      Note that the solution is

saturated-  Cl-

 

Ag+

Page 4: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution      Note that the solution is

saturated- filled to the max

 

Cl-

 

Ag+

Page 5: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution      Note that the solution is

saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

 

Cl-

 

Ag+

Page 6: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution      Note that the solution is

saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

 

More solid AgCl is added to the saturated solution above.

Cl-

 

Ag+

Page 7: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution      Note that the solution is

saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

 

More solid AgCl is added to the saturated solution above.

Cl-

 

Ag+

Page 8: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution      Note that the solution is

saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

 

More solid AgCl is added to the saturated solution above.

Cl-

 

Ag+

Page 9: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution      Note that the solution is

saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

 

More solid AgCl is added to the saturated solution above.The solution is full already so the ion concentrations remain constant!

Cl-

 

Ag+

Page 10: Solubility Lesson 8 Review Notes

Adding a Crystal to a Saturated Solution      Note that the solution is

saturated- filled to the max and the rate of crystallizing equals the rate of dissolving.

 

More solid AgCl is added to the saturated solution above.The solution is full already so the ion concentrations remain constant!The new solid dissolves at a greater rate, however crystallizes at an equally greater rate so there is no net change in ion concentration!

Cl-

 

Ag+

Page 11: Solubility Lesson 8 Review Notes

Describe the change in each of the following when more solid AgCl is added to a saturated solution.  1. [Ag+]

Page 12: Solubility Lesson 8 Review Notes

Describe the change in each of the following when more solid AgCl is added to a saturated solution.  1. [Ag+] Constant

2. [Cl-]

Page 13: Solubility Lesson 8 Review Notes

Describe the change in each of the following when more solid AgCl is added to a saturated solution.  1. [Ag+] Constant

2. [Cl-] Constant 3. Rate of dissolving

Page 14: Solubility Lesson 8 Review Notes

Describe the change in each of the following when more solid AgCl is added to a saturated solution.  1. [Ag+] Constant

2. [Cl-] Constant 3. Rate of dissolving Increases 4. Rate of crystallizing

Page 15: Solubility Lesson 8 Review Notes

Describe the change in each of the following when more solid AgCl is added to a saturated solution.  1. [Ag+] Constant

2. [Cl-] Constant 3. Rate of dissolving Increases 4. Rate of crystallizing Increases

Replay the last frame if you don't get this!

Page 16: Solubility Lesson 8 Review Notes

5. The ions in hard water are Mg2+ and Ca2+

Na2CO3 can be added to water to remove these ions

6. Calculate the total ion concentration of 1.0 M AlCl3

AlCl3 Al3+ + 3Cl-

1.0 M 1.0 M 3.0 M

Total 4.0 M

7. What is the solubility of CaC2O4?

Not on page 4On page 5 Must be low! They all are!

Page 17: Solubility Lesson 8 Review Notes

8. The solubility is 7.1 x 10-5 M. The compound is

A. CaSO4

B. CaCO3

Ksp = s2 = 5.0 x 10-9 B

9. A solution of AgNO3 is added slowly to each of the following 0.10 M solutions. Which forms a precipitate first?

A. NaCl

B. NaIO3

Page 18: Solubility Lesson 8 Review Notes

10. The solubility is 7.1 x 10-5 M. The compound is

A. CaSO4

B. CaCO3

Ksp = s2 = 5.0 x 10-9 B

11. 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

Page 19: Solubility Lesson 8 Review Notes

12. The solubility is 7.1 x 10-5 M. The compound is

A. CaSO4

B. CaCO3

Ksp = s2 = 5.0 x 10-9 B

13. 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 smaller

B. NaIO3 ksp = 3.2 x 10-8

Page 20: Solubility Lesson 8 Review Notes

14. Small amounts of AgNO3 are added to three solutions that have the same concentration. If only one solution does not form a precipitate, which one is it ?

A. NaCl

B. NaIO3

C. NaBr

Page 21: Solubility Lesson 8 Review Notes

14. Small amounts of AgNO3 are added to three solutions that have the same concentration. If only one solution does not form a precipitate, which one is it ?

A. NaCl ksp = 1.8 x 10-10

B. NaIO3 ksp = 3.2 x 10-8

C. NaBr ksp = 5.4 x 10-8

Page 22: Solubility Lesson 8 Review Notes

14. Small amounts of AgNO3 are added to three solutions that have the same concentration. If only one solution does not form a precipitate, which one is it ?

A. NaCl ksp = 1.8 x 10-10

B. NaIO3 ksp = 3.2 x 10-8

C. NaBr ksp = 5.4 x 10-8

Page 23: Solubility Lesson 8 Review Notes

14. Which solution has the greatest conductivity?

A. 1.0 M AgCl

B. 1.0 M CaCO3

C. 1.0 M AlCl3

D. 1.0 M CaCl2

Page 24: Solubility Lesson 8 Review Notes

14. Which solution has the greatest conductivity?

A. 1.0 M AgCl low

B. 1.0 M CaCO3 low

C. 1.0 M AlCl3 high

D. 1.0 M CaCl2 high

Page 25: Solubility Lesson 8 Review Notes

14. Which solution has the greatest conductivity?

A. 1.0 M AgCl low

B. 1.0 M CaCO3 low

C. 1.0 M AlCl3 high AlCl3 Al3+ + 3Cl-

D. 1.0 M CaCl2 high CaCl2 Al3+ + 2Cl-

Page 26: Solubility Lesson 8 Review Notes

15. Which solution has the greatest conductivity?

A. 1.0 M AgCl low

B. 1.0 M CaCO3 low

C. 1.0 M AlCl3 high AlCl3 Al3+ + 3Cl-

D. 1.0 M CaCl2 high CaCl2 Ca2+ + 2Cl-

4 ions versus 3

Page 27: Solubility Lesson 8 Review Notes

16. Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

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-

9.0 x 10-10 M

Page 28: Solubility Lesson 8 Review Notes

15. Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

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

Page 29: Solubility Lesson 8 Review Notes

15. Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution. .

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

0.50 L

Page 30: Solubility Lesson 8 Review Notes

15. Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

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

0.50 L x 4.5 x 10-10 mole1 L

Page 31: Solubility Lesson 8 Review Notes

15. Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

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

0.50 L x 4.5 x 10-10 mole x 208.3 g 1 L 1 mole

Page 32: Solubility Lesson 8 Review Notes

15. Calculate the maximum number of grams BaCl2 that will dissolve in 0.50 L of 0.20 M AgNO3 solution.

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

0.50 L x 4.5 x 10-10 mole x 208.3 g = 4.7 x 10-8 g1 L 1 mole

Page 33: Solubility Lesson 8 Review Notes

 

16. In an experiment to determine the solubility of BaF2, 500 mL

of the saturated solution was heated in an evaporating dish to remove the water. The evaporating dish and residue were heated two more times to ensure all of the water had been driven off.

Volume of saturated solution 500.0 mL Mass of evaporating dish 72.540 g Mass of evaporating dish & BaF2 after first heating 73.500 g

 Mass of evaporating dish & BaF2 after second heating 72.855 g

 Mass of evaporating dish & BaF2 after third heating 72.855 g

 Use this data to calculate the Ksp for BaF2.

Page 34: Solubility Lesson 8 Review Notes

[BaF2] = s = (72.855 - 72.540)g x 1 mole175.3 g

0.500 L

s = 0.0035938 M

BaF2(s) ⇄ Ba2+ + 2F-

s s 2s

Ksp = [Ba2+][F-]2

Ksp = [s][2s]2

Ksp = 4s3

Ksp = 4(0.0035938)3

Ksp = 1.86 x 10-7

Homework web review 1practice test 1 and 2