1 / 18

The Solubility Product Principle

The Solubility Product Principle. Chapter 20. Solubility: the amount of compound that dissolves in a specified volume. Usually expressed as grams per Liter or grams per 100 mL Molar solubility : number of moles of a compound that dissolve to give 1 Liter of saturated solution.

tahir
Download Presentation

The Solubility Product Principle

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. The Solubility Product Principle Chapter 20

  2. Solubility: the amount of compound that dissolves in a specified volume. Usually expressed as grams per Liter or grams per 100 mL • Molar solubility: number of moles of a compound that dissolve to give 1 Liter of saturated solution.

  3. Solubility Product Expression: for a compound it is the product of the concentrations of its ions each raised to a power that corresponds to the number of ions formed. • Common ion effect: the solubility of a compound is less in a solution that contains an ion common to the compound than it is in pure water.

  4. Ex. 1) a. What is the solubility product expression for the dissociation of calcium carbonate? (Ksp = solubility product constant.)

  5. b. For aluminum fluoride?

  6. c. For barium nitride?

  7. Ex. 2) Calculate the molar solubility, concentrations of the constituent ions, and solubility in grams per liter of copper(II) hydroxide. Ksp = 1.6 x 10-19

  8. Ex. 3) Find [S2-] in saturated iron(III) sulfide Ksp = 1.40x10-88

  9. Ex. 4) 0.750 M FeI3 is added to Ex. 3. What is the final concentration of iron and sulfur ions?

  10. Ex. 5) Calculate the molar solubility and ion concentrations of aluminum fluoride in pure water and its solubility in grams per 100 mL. Ksp = 6.4 x 10-19. Then using the common ion effect, find the molar solubility of AlF3 in 0.20 M potassium fluoride solution along with the ion concentrations and compare the molar solubilities.

  11. Ex. 6) What is the solubility product constant if 0.477 g of silver dichromate is in a one liter solution?

  12. Fractional precipitation: a separation process that removes some ions from solution while leaving other ions with similar properties in the solution.

  13. Ex. 7) Solid silver nitrate is slowly added to a solution that is 0.010 M each in sodium bromide and sodium iodide. Calculate the [Ag+ ] required to initiate the precipitation of each silver halide. Which precipitates first? For AgBr: Ksp = 3.3 x 10-13 for AgI: Ksp = 1.5 x 10-16 For AgCl the Ksp= 1.8 x 10-10. Compare that to AgBr and AgI to determine which would precipitate last.

  14. Ex. 8) Using your answer from #7, find the percentage of Iodine ions that precipitated before silver bromide precipitates.

  15. Ex. 9) How many moles of bismuth(III) hydroxide will dissolve in 1.00 L of a solution with a pH=4.77 Ksp = 3.2x10-40

  16. If Qsp < Ksp Forward process is favored • No precipitation occurs; if solid is present, more solid can dissolve • If Qsp = Ksp Solution is just saturated • Solid and solution are in equilibrium; neither forward nor reverse process is favored. • If Qsp > Ksp Reverse process is favored • Precipitation occurs to form more solid Reaction Quotient in Precipitation Reactions

  17. Extra Credit Opportunity Many industries require extremely large amounts of water as a coolant in heat exchange processes. Muddy or cloudy water usually doesn’t work b/c the solids may clog filters or deposit sediment in the pipes and pumps. Murky water can be clarified by adding agents that form colloids which settle out in holding tanks or ponds before the clarified water is moved to the plant intakes. If 56 g of Ca(OH)2 and 75 g of MgSO4 were added to 520 liters of water, would any of the compounds form precipitates?

  18. Ksp Ca(OH)2 = 7.9 x 10-6 • Ksp MgSO4 = 4.8 x 10-5 • Ksp CaSO4 = 2.4 x 10-5 • Ksp Mg(OH)2 = 1.5 x 10-11

More Related