1 / 15

Solubility Lesson 3 Calculating Ksp

Solubility Lesson 3 Calculating Ksp. The Molar Solubility is the molarity required to saturate or fill the solution at any given temperature. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature.

marva
Download Presentation

Solubility Lesson 3 Calculating Ksp

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solubility Lesson 3 Calculating Ksp

  2. The Molar Solubility is the molarity required to saturate or fill the solution at any given temperature.

  3. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp.

  4. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. Ba2+ CO32- BaCO3(s)

  5. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. BaCO3(s)⇌ Ba2+ + CO32- Ba2+ CO32- BaCO3(s)

  6. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. BaCO3(s)⇌ Ba2+ + CO32- s s s Ba2+ CO32- BaCO3(s)

  7. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. BaCO3(s)⇌ Ba2+ + CO32- s s s Ksp = [Ba2+][CO32-] Ba2+ CO32- BaCO3(s)

  8. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. BaCO3(s)⇌ Ba2+ + CO32- s s s Ksp = [Ba2+][CO32-] Ksp = [s][s] Ba2+ CO32- BaCO3(s)

  9. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. BaCO3(s)⇌ Ba2+ + CO32- s s s Ksp = [Ba2+][CO32-] Ksp = [s][s] Ksp = s2 Ba2+ CO32- BaCO3(s)

  10. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. BaCO3(s)⇌ Ba2+ + CO32- s s s Ksp = [Ba2+][CO32-] Ksp = [s][s] Ksp = s2 Ksp = (5.1 x 10-5)2 Ba2+ CO32- BaCO3(s)

  11. The Molar Solubility is the molarity required to saturate of fill the solution at any given temperature. 1. The solubility (s) of BaCO3 is 5.1 x 10-5 M @ 250 C. Calculate the solubility product or Ksp. BaCO3(s)⇌ Ba2+ + CO32- s s s Ksp = [Ba2+][CO32-] Ksp = [s][s] Ksp = s2 Ksp = (5.1 x 10-5)2 Ksp = 2.6 x 10-9 Ba2+ CO32- BaCO3(s)

  12. Ksp Solubility Product Special Keq Saturated solutions No Units Only Changes with Temperature

  13. 2. The solubility of PbBr2 is 0.012 M @ 25 0C. Calculate the Ksp. dissociation equation PbBr2(s)⇌ Pb2+ + 2Br- solubility s s 2s equilibriumexpression Ksp = [Pb2+][Br-]2 substitute & solve Ksp = [s][2s]2 Ksp = 4s3 Ksp = 4(0.012)3 Ksp = 6.9 x 10-6 Note that the Br- is doubled and then squared!

  14. 3. If 0.00243 g of Fe2(CO3)3 is required to saturate 100.0 mL of solution. What is the solubility product? Fe2(CO3)3⇌ 2Fe3+ + 3CO32- s 2s 3s s = 0.00243 g x 1 mole 291.6 g 0.100 L = 8.333 x 10-5 M Ksp = [Fe3+]2[CO32-]3 Ksp = [2s]2[3s]3 Ksp = 108s5 Ksp = 108(8.333 x 10-5)5 Ksp = 4.34 x 10-19

  15. 4. A 200.0 mL sample of a saturated solution of Mg(OH)2 weighs 222.1210 g. When the beaker containing the solution is evaporated to dryness it weighs 22.1213 g. The mass of the empty beaker is 22.1200 g. Calculate the Ksp. Mass of Beaker + Mg(OH)2 22.1213 g - Mass of Beaker - 22.1200 g Mass of Mg(OH)2 0.0013 g note sig figs s = 0.0013 g x 1 mole 58.3 g 0.2000 L = 1.1149 x 10-4 M Mg(OH)2⇌ Mg2+ + 2OH- s s 2s Ksp = [Mg2+][OH-]2 = [s][2s]2 = 4s3=4(1.1149 x 10-4)3 = 5.5 x 10-12

More Related