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Solubility Equilibrium

Solubility Equilibrium. Chapter 10. What happens to an insoluble salt when it is added to some water? What would you see? What does it mean to be “insoluble”?. “Insoluble” is a lie. All solids dissolve, but some dissolve a very, very tiny amount.

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Solubility Equilibrium

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  1. Solubility Equilibrium Chapter 10

  2. What happens to an insoluble salt when it is added to some water? What would you see? What does it mean to be “insoluble”?

  3. “Insoluble” is a lie. All solids dissolve, but some dissolve a very, very tiny amount. What would this look like if you could see it? Why are the arrows wrong? Pb2+(aq) I-(aq) I-aq PbI2(s)

  4. Because we have no change in the amount of solid sitting at the bottom of the water, yet ions are still forming, it has to be concluded that ions are also precipitating out at an equal rate. What do we call this? Pb2+(aq) I-(aq) I-aq PbI2(s)

  5. Equilibrium of an insoluble, or not very soluble, salt is called Solubility Product, so the constant is the Solubility Product Equilibrium Constant, or Ksp. Pb2+(aq) I-(aq) I-aq PbI2(s)

  6. For the reaction PbI2(s)  Pb2+(aq) + 2I-(aq) Ksp = [Pb2+][I-]2 Why is [I-] squared? What is missing from this K? Why?

  7. Where there’s a K, there’s a Q. When would you use Q? What would happen if Q is less than K? What would happen if Q is greater than K? What would happen if Q is equal to K?

  8. For the reaction PbI2(s)  Pb2+(aq) + 2I-(aq) According to LeChatelier’s Principle, what would happen if you added NaI? What is the name for this situation? What would happen to the value of K? What would happen to the solubility of PbI2?

  9. Let’s Prove it: For the reaction PbI2(s)  Pb2+(aq) + 2I-(aq) Ksp= 9.8 x 10-9 What is the [Pb2+]?

  10. Let’s Prove it: For the reaction PbI2(s)  Pb2+(aq) + 2I-(aq) Ksp= 9.8 x 10-9 If 0.20 M NaI is added, what is the new [Pb2+]?

  11. Ksp will not only tell us how much of an insoluble salt would dissolve (ionize), but how much of an ion would need to be added to get a precipitate. This could even be used to precipitate out one ion from a mixture of many for isolation of that ion - often called selective precipitate.

  12. Quick example of selective precipitation: A beaker has mixture of Na+ ions and Ag+ ions. How can I- ions be used to separate the Na+ from the Ag+?

  13. Not quick example of selective precipitation: A beaker has mixture of Cl- ions, Br- ions, and I- ions. If [Ag+] was added, in what order would the ions precipitate? KspAgCl = 1.8 x 10-10 KspAgBr = 3.3 x 10-13 KspAgI = 1.5 x 10-16

  14. Not quick example of selective precipitation: A beaker has mixture of Cl- ions, Br- ions, and I- ions, each present at 0.0010 M. How much [Ag+] would be needed to precipitate each ion? KspAgCl = 1.8 x 10-10 KspAgBr = 3.3 x 10-13 KspAgI = 1.5 x 10-16

  15. KspAgCl = 1.8 x 10-10 KspAgBr = 3.3 x 10-13 KspAgI = 1.5 x 10-16

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