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Lesson 4 Calculating Molar Solubility From Ksp

Lesson 4 Calculating Molar Solubility From Ksp. 1. Calculate the molar solubility @ 25 o C for BaCrO 4 in units of g/L BaCrO 4(s) ⇌ Ba 2+ + CrO 4 2- s s s Ksp = [Ba 2+ ][CrO 4 2- ] Ksp = s 2 from page 5 1.2 x 10 -10 = s 2 s = 1.1 x 10 -5 M

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Lesson 4 Calculating Molar Solubility From Ksp

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  1. Lesson 4 Calculating Molar Solubility From Ksp

  2. 1. Calculate the molar solubility @ 25oC for BaCrO4 in units of g/L BaCrO4(s)⇌ Ba2+ + CrO42- s s s Ksp = [Ba2+][CrO42-] Ksp = s2 from page 5 1.2 x 10-10 = s2 s = 1.1 x 10-5 M note that solubility units are M! 1.1 x 10-5 mole x 253.3 g = 0.0028 g/L L 1 mole

  3. 2. Calculate the molar solubility @ 25oC for Cu(IO3)2 in units of g/L Cu(IO3)2 ⇌ Cu2+ + 2IO3- s s 2s Ksp = [Cu2+][IO3-]2 Ksp = [s][2s]2 Ksp = 4s3 6.9 x 10-8 = 4s3 s = 2.6 x 10-3 M note sig figs are 2 like the Ksp!

  4. 2.584 x 10-3moles L

  5. 2.584 x 10-3moles x 413.2 g L 1 mole

  6. 2.584 x 10-3moles x 413.2 g = 1.1 g/L L 1 mole

  7. 3. Calculate the molar solubility @ 25oC for Fe(OH)3. Calculate the mass required to prepare 2.0 L of the above saturated solution. Fe(OH)3⇌ Fe3+ + 3OH- s s 3s Ksp = [Fe3+][OH-]3 Ksp = [s][3s]3 2.6 x 10-39 = 27s4 s = 9.906 x 10-11 M 2.0 L

  8. 3. Calculate the molar solubility @ 25oC for Fe(OH)3. Calculate the mass required to prepare 2.0 L of the above solution. Fe(OH)3⇌ Fe3+ + 3OH- s s 3s Ksp = [Fe3+][OH-]3 Ksp = [s][3s]3 2.6 x 10-39 = 27s4 s = 9.906 x 10-11 M 2.0 L x 9.906 x 10-11moles 1 L

  9. 3. Calculate the molar solubility @ 25oC for Fe(OH)3. Calculate the mass required to prepare 2.0 L of the above solution. Fe(OH)3⇌ Fe3+ + 3OH- s s 3s Ksp = [Fe3+][OH-]3 Ksp = [s][3s]3 2.6 x 10-39 = 27s4 s = 9.906 x 10-11 M 2.0 L x 9.906 x 10-11moles x 106.8 g 1 L mole

  10. 3. Calculate the molar solubility @ 25oC for Fe(OH)3. Calculate the mass required to prepare 2.0 L of the above solution. Fe(OH)3⇌ Fe3+ + 3OH- s s 3s Ksp = [Fe3+][OH-]3 Ksp = [s][3s]3 2.6 x 10-39 = 27s4 s = 9.906 x 10-11 M 2.0 L x 9.906 x 10-11moles x 106.8 g = 2.1 x 10-8 g 1 L mole

  11. The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility. 4. Indicate the solid with the greatest solubility. PbSO4 ZnS AgCl BeS

  12. The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility. 4. Indicate the solid with the greatest solubility. PbSO4 start on page 4 ZnS AgCl BeS

  13. The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility. 4. Indicate the solid with the greatest solubility. PbSO4 low start on page 4 ZnS AgCl BeS

  14. The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility. 4. Indicate the solid with the greatest solubility. PbSO4 low start on page 4 ZnS low AgCl BeS

  15. The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility. 4. Indicate the solid with the greatest solubility. PbSO4 low start on page 4 ZnS low AgCl low BeS

  16. The size of the Ksp is related to the solubility of the ionic compound. For salts that are comparable (AB versus AB), the larger the Ksp of the salt, the greater is its solubility. 4. Indicate the solid with the greatest solubility. PbSO4 low start on page 4 ZnS low AgCl low BeShigh

  17. 5. Indicate the solid with the least solubility. PbSO4 ZnS AgCl BeS

  18. 5. Indicate the solid with the least solubility. PbSO4low start on page 4 ZnS low AgCl low BeShigh

  19. 5. Indicate the solid with the least solubility. PbSO4low start on page 4 ZnS low use page 5 if required AgCl low BeShigh

  20. 5. Indicate the solid with the least solubility. PbSO4low 1.8 x 10-8 start on page 4 ZnS low use page 5 if required AgCl low BeShigh

  21. 5. Indicate the solid with the least solubility. PbSO4low 1.8 x 10-8 start on page 4 ZnS low 2.0 x 10-25use page 5 if required AgCl low BeShigh

  22. 5. Indicate the solid with the least solubility. PbSO4low 1.8 x 10-8 start on page 4 ZnS low 2.0 x 10-25use page 5 if required AgCl low 1.8 x 10-10 BeShigh

  23. 5. Indicate the solid with the least solubility. PbSO4low 1.8 x 10-8 start on page 4 ZnS low 2.0 x 10-25use page 5 if required AgCl low 1.8 x 10-10 BeShigh

  24. 6. How many of the following salts could produce a solution with a concentration more than 0.10 M? Greater than 0.10 M means high solubility! FeSO4 ZnSO4 Al2(SO4)3 Na2SO4

  25. 6. How many of the following salts could produce a solution with a concentration more than 0.10 M? Greater than 0.10 M means high solubility! FeSO4 high ZnSO4 Al2(SO4)3 Na2SO4

  26. 6. How many of the following salts could produce a solution with a concentration more than 0.10 M? Greater than 0.10 M means high solubility! FeSO4 high ZnSO4 high Al2(SO4)3 Na2SO4

  27. 6. How many of the following salts could produce a solution with a concentration more than 0.10 M? Greater than 0.10 M means high solubility! FeSO4 high ZnSO4 high Al2(SO4)3 high Na2SO4

  28. 6. How many of the following salts could produce a solution with a concentration more than 0.10 M? Greater than 0.10 M means high solubility! FeSO4 high ZnSO4 high Al2(SO4)3 high Na2SO4high

  29. 6. How many of the following salts could produce a solution with a concentration more than 0.10 M? Greater than 0.10 M means high solubility! FeSO4 high ZnSO4 high Al2(SO4)3 high Na2SO4high four!

  30. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2.

  31. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2

  32. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M

  33. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] =

  34. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl

  35. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles 1 L

  36. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl

  37. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L

  38. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + 1Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M

  39. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M B. Ksp

  40. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M B. Ksp Ba(OH)2⇌ Ba2+ + 2OH-

  41. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M B. Ksp Ba(OH)2⇌ Ba2+ + 2OH- s s 2s

  42. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M B. Ksp Ba(OH)2⇌ Ba2+ + 2OH- s s 2s Ksp = [Ba2+][OH-]2

  43. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M B. Ksp Ba(OH)2⇌ Ba2+ + 2OH- s s 2s Ksp = [Ba2+][OH-]2 = [s][2s]2

  44. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M B. Ksp Ba(OH)2⇌ Ba2+ + 2OH- s s 2s Ksp = [Ba2+][OH-]2 = [s][2s]2 = 4s3

  45. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M B. Ksp Ba(OH)2⇌ Ba2+ + 2OH- s s 2s Ksp = [Ba2+][OH-]2 = [s][2s]2 = 4s3 = 4(0.1091)3

  46. 7. 40.00 mL of a saturated Ba(OH)2 solution is neutralized by adding 29.10 mL of 0.300 M HCl. Calculate the Ksp for Ba(OH)2. A. Titration 2HCl + Ba(OH)2 0.02910 L 0.04000 L 0.300 M? M [Ba(OH)2] = 0.02910 L HCl x 0.300 moles x 1 mole Ba(OH)2 1 L 2 moles HCl 0.0400 L s= 0.1091 M B. Ksp Ba(OH)2⇌ Ba2+ + 2OH- s s 2s Ksp = [Ba2+][OH-]2 = [s][2s]2 = 4s3 = 4(0.1091)3 = 5.20 x 10-3

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