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Ch. 13 Precipitation Reactions

Ch. 13 Precipitation Reactions. Reactions in Solutions. Precipitation Reactions. How do you know if a precipitate will form when you mix two solutions? You look at the solubility of the products. Certain ionic compounds dissolve in water, other ones do not.

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Ch. 13 Precipitation Reactions

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  1. Ch. 13 Precipitation Reactions • Reactions in Solutions

  2. Precipitation Reactions • How do you know if a precipitate will form when you mix two solutions? • You look at the solubility of the products. • Certain ionic compounds dissolve in water, other ones do not. • Soluble: means it dissolves in water • Insoluble: means it makes a precipitate (a solid) and the solution will get cloudy.

  3. Soluble or Insoluble?

  4. The 3 Most Important Solubility Rules: • 1. All Nitrates, NO3- , are soluble (always true). • ie: KNO3, Ca(NO3)2, etc. • 2. Na, K, and NH4+ compounds are soluble • ie: NH4Cl, K2SO4, etc. • 3. Compounds with Pb, Ag, and Hg tend not to be soluble. (PbSO4, Ag3PO4, etc) • But: these are soluble with nitrates, for example, PbNO3is soluble. • This is just a broad overview. See Pg. 437 for specifics.

  5. Solubility Rules

  6. Example: • You mix silver nitrate and calcium chloride, what happens? • 2AgNO3(aq) + CaCl2(aq)  2AgCl(s) + Ca(NO3)2(aq) • But only include the ions which “participate.” • Ag+ (aq) + Cl-(aq)  AgCl(s) • This is the “Net ionic equation.” • The other ions are called “spectator ions.” • They are present before and after the reaction.

  7. Sample Problem: • Write the net ionic equation for the reaction which occurs when solutions of lead(II) nitrate and sodium bromide are mixed. • Pb(NO3)2(aq) + 2NaBr(aq)  PbBr2(s) + 2NaNO3(aq) • And the “net ionic” equation: • Pb2+ + 2Br1- PbBr2(s)

  8. The hazards of Ice Fishing….

  9. Try this one: • Copper(II)sulfate solution is mixed with sodium phosphate. • Write the overall reaction first, then write the “net ionic” equation. 3CuSO4 + 2Na3PO4 Cu3(PO4)2(s) + 3Na2SO4 • Net ionic: 3Cu2+ + 2PO43- Cu3(PO4)2(s)

  10. Diluting already made solutions: • So far, we just talked about making solutions from crystals and water. • How do you dilute an already made solution to the molarity you want? • Use: MdVd = McVc • “d” means “dilute • “c” means concentrate.

  11. Here’s a problem: • How would you make 500 mL of 3.0 M sulfuric acid from the stock solution, which is 17.8 M? • Md = 3.0 M Mc = 17.8 M • Vd = 500 mL Vc = ? • To make it, get 84.2 mL of the concentrated acid from the bottle, add it to a volumetric flask, and add water until you reach the 500 mL line.

  12. Try this one: • What would be the Molarity of a solution you made by diluting 12.5 mL of 11.7 Molar HCl to make 500. mL of solution? • Md = ? Mc = 11.7 M • Vd = 500 mL Vc = 12.5 mL

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