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Harris Chapter 6

Harris Chapter 6. Supplements information In Zumdahl’s Chapter 13. Standard States Gibbs Free Energy Solubility Product Common Ion Effect. Complex Formations Protic Acids and Bases Autoprotolysis Polyprotic conjugates. Supplemental Content. Standard States.

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Harris Chapter 6

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  1. Harris Chapter 6 Supplements information In Zumdahl’s Chapter 13

  2. Standard States Gibbs Free Energy Solubility Product Common Ion Effect Complex Formations Protic Acids and Bases Autoprotolysis Polyprotic conjugates Supplemental Content

  3. Standard States • In (dimensionless!) mass action expressions, • [A]ameans ( [A] / 1 M )a • PAameans ( PA / 1 bar )a • 1 M is the standard reference for solute concentration • 1 bar is the standard reference for gas pressure ( 1 atm = 1.01325 bar)

  4. Gibbs Free Energy • Both K and G point to equilibrium. • Gequil = Gmin or for standard states, G • G° uses atmospheres instead of bar. • AT equilibrium G = 0; no tendency to go elsewhere! • For reactions not at standard concentration, • G = G + RT lnQ but for equilibrium, • 0 = G + RT lnK or K = e–G°/RT

  5. Solubility Product • An equilibrium between solids and their solutes in solution (aqueous if unstated). • AaBb(s)  a A(aq) + b B(aq) • Ksp = [A]a [B]b(pure solid set to 1) • Pb3(PO4)2, Ksp = 3.010–44 = (3x)3 (2x)2 = 108x5 • x = 7.710–10 moles/L of Pb3(PO4)2 dissolve. • MW = 811.54 g/mol means 6.310–7 g/L in soln.

  6. Common Ion Effect • Le Châtlier insists that adding more of a component motivates equilibrium to shift to consume it … to bring Q back to K. E.g., • Excess Pb2+(aq) reduces Pb3(PO4)2 solubility. • Swamp with [Pb2+] = 0.1M What effect? • Ksp = (0.1)3 (2x)2 = 0.004x2 • x = 2.710–21 or only 2.210–18 g/L solubility!

  7. Not only that, but Common Ion Effect works in all equilibria since Le Châtlier was right. Excess free hydronium ion reduces the % dissociation of a weak acid.

  8. Complex Formations • Complex: a Lewis Acid-Base association of an atom with coordinated ligands • E.g., Al(OH)2(H2O)4+ or PbI3(H2O)– • One result is that added species can dissolve a precipitate by forming a complex in accordance with Le Châtlier if the species is a reactant: • Pb2+(aq) + 3 I–(aq)  PbI3–(aq) instead of PbI2(s)

  9. Autoprotolysis • 2 H2O  H3O+ + OH– implies that water autoionizes to yield (damn few) protons • All weak acids can do likewise when they are the solvent: • 2 CH3CO2H(l)  CH3CO2H2+ + CH3CO2– • Kglacial = 3.510–15 at 25°C

  10. Polyprotic Conjugates • H2PO4– (acid) has HPO42– (conjugate base) • Ka2 = 6.3210–8 and Kb2 = 1.5810–7 since • HPO42–+ H+ H2PO4–Ka2–1 plus • H2O  H+ + OH– Kw yields • HPO42–+ H2O  H2PO4–+ OH– Kb2 = Kw/Ka2 • So for n-protic acid, Kai Kb(n+1–i) = Kw

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