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Solubility Equilibria in Inorganic Chemistry

Explore the concepts of solubility-product constants and precipitation reactions involving "insoluble" salts. Learn how to write Ksp expressions, calculate solubility, and apply principles like the common-ion effect. Discover selective precipitation methods and factors affecting solubility. Dive into qualitative analysis methods to identify metallic elements in solutions.

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Solubility Equilibria in Inorganic Chemistry

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  1. MN(s) Mx+(aq) + Nx–(aq) etc. MN2(s) M2x+(aq) + 2 Nx–(aq) Solubility Equilibria -- involve the dissolution or precipitation of “insoluble” salts Consider a saturated solution of a typical salt: For these cases, the solubility-product constant is equal to (respectively): Ksp expressions NEVER have denominators because… Ksp = [Mx+] [Nx–] the “reactant” is always a solid, and we never write solids in equil. eqs. Ksp = [M2x+] [Nx–]2

  2. Nx–(aq) Nx–(aq) Mx+(aq) Mx+(aq) Mx+(aq) Nx–(aq) MN(s) PbCl2(s) Pb2+(aq) + 2 Cl–(aq) -- Ksp is the equilibrium constant between undissolved and dissolved ionic solute in a saturated aqueous solution. small Ksp… Not much dissolves. large Ksp… “Quite a bit” dissolves. Write the solubility-product constant expression for lead(II) chloride. Ksp = [Pb2+] [Cl–]2

  3. Cu(N3)2(s) ( ) mol Cu(N3)2 147.5 g g 5.40 x 10–4 = 8.0 x 10–2 L 1 mol L Copper(II) azide has Ksp = 6.3 x 10–10. Find the solubility of Cu(N3)2 in water, in g/L. Cu2+(aq) + 2 N3–(aq) Ksp = 6.3 x 10–10 = [Cu2+] [N3–]2 (Let [Cu2+] = x) ** In “plain-old” Cu(N3)2,: [N3–] = 2 [Cu2+] So…6.3 x 10–10 = x (2x)2 = 4x3 x = 5.40 x 10–4 M = [Cu2+] From eq. at top, 5.40 x 10–4 Mis also the [ ] of Cu(N3)2 that dissolves.

  4. Zn(OH)2(s) Find the solubility of zinc hydroxide (Ksp = 3.0 x 10–16) in a soln buffered at pH = 11.43. Zn2+(aq) + 2 OH–(aq) Ksp = 3.0 x 10–16 = [Zn2+] [OH–]2 from given pH… 3.0 x 10–16 = [Zn2+] (2.69 x 10–3)2 [Zn2+] = 4.1 x 10–11 M From eq. at top, 4.1 x 10–11 Mis also the [ ] of Zn(OH)2 that dissolves.

  5. For BaSO4(s) Ba2+(aq) + SO42–(aq)... Precipitation of Ions Ksp = [Ba2+] [SO42–] We could reach equilibrium from the left... i.e., start with BaSO4(s) or from the right... i.e., start w/Ba2+ (say, from Ba(NO3)2(aq)) and SO42– (say, from Na2SO4(aq)) At any given time, the ion product Q = [Ba2+] [SO42–] If Q > Ksp... more ppt will form. If Q < Ksp... more ions will dissolve. If Q = Ksp... ions and ppt are in eq.

  6. 0.008(0.10) 0.005(0.40) 0.50 0.50 Will a precipitate form from mixing 0.10 L of 8.0 x 10–3 M lead(II) nitrate and 0.40 L of 5.0 x 10–3 M sodium sulfate? Pb(NO3)2 and Na2SO4 (If any ppt forms, it will be of PbSO4. Look up the Ksp of PbSO4: 6.3 x 10–7.) [Pb2+]init = = 0.0016 M [SO42–]init = = 0.0040 M Q = [Pb2+] [SO42–] = 0.0016 (0.0040) = 6.4 x 10–6 Q > Ksp; ppt WILL form.

  7. silver chloride selective precipitation: using the different solubilities of ions to separate them Say we want to separate the Ag+ and Cu2+ in a soln that contains both. -- Add HCl. CuCl2 is… soluble, which means it won’t ppt out. AgCl is… insoluble (its Ksp is 1.8 x 10–10) so MOST of it will ppt out when we add HCl. long-stem funnel filtering out the AgCl(s). -- Separate by... filter paper AgCl(s) Erlenmeyer flask Cu2+ ions still in soln (along w/a “wee-little” Ag+)

  8. For example, with... CaF2(s) Ca2+(aq) + 2 F–(aq) (Sol. ) If you add Ca2+ or F–, shift . Factors That Affect Solubility 1. For solids, as temperature increases, solubility... 2. common-ion effect Use Le Chatelier’s principle. 3. pH and solubility Compounds with anions exhibiting basic properties (e.g., Mg(OH)2 / OH–, CaCO3 / CO32–, CaF2 / F–) ____ in solubility as solution becomes more acidic.

  9. AgCl(s) Ag+(aq) + Cl–(aq) complex ion If ammonia is added, we get… Ag+(aq) + 2 NH3(aq) Ag(NH3)2+(aq) 4. presence of complex ions: metal ions and the Lewis bases bonded to them e.g., In a soln of AgCl, we have the equilibrium: As Ag+ forms the complex ion w/NH3, Le Chatelier dictates that more AgCl will dissolve into the soln to replenish the decreased [ ] of Ag+. In general, the solubility of metal salts __ in the presence of suitable Lewis bases (e.g., NH3, CN–, OH–) if the metal forms a complex ion with the bases.

  10. Qualitative Analysis for Metallic Elements determines the presence or absence of a particular ion (as opposed to quantitative analysis) Steps: (1) separate ions into broad groups by solubility specific ions are then separated from the group and identified by specific tests (2)

  11. Add HCl. aqueous solution containing “god-knows-what” metal ions (but not much Ag+, Hg22+, or Pb2+) Bubble H2S(g) in. aqueous solution containing “god-knows-what” metal ions insoluble chlorides (AgCl, Hg2Cl2, PbCl2) ppt out. Filter and test for a specific ion. acid-insoluble sulfides (CuS, Bi2S3, CdS, PbS, HgS, As2S3, Sb2S3, SnS2) ppt out. Filter and test for a specific ion.

  12. Add (NH4)2S. Add (NH4)2PO4. aqueous solution w/even fewer metal ions base-insoluble sulfides and hydroxides (Al(OH)3, Fe(OH)3, Cr(OH)3, ZnS, NiS, CoS, MnS) ppt out. Filter and test for a specific ion. aqueous solution w/even FEWER metal ions insoluble phosphates (Ba3(PO4)2, Ca3(PO4)2, MgNH4PO4) ppt out. Filter and test for a specific ion. alkali metal ions still remain in soln.

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