Harris Chapter 6 Supplements information In Zumdahl’s Chapter 13

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

Standard States  In (dimensionless!) mass action expressions, [A] a means ( [A] / 1 M ) a P A a means ( P A / 1 bar ) a  1 M is the standard reference for solute concentration  1 bar is the standard reference for gas pressure ( 1 atm = bar)

Gibbs Free Energy  Both K and G point to equilibrium.  G equil = G min 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

Solubility Product  An equilibrium between solids and their solutes in solution (aqueous if unstated).  A a B b (s)  a A(aq) + b B(aq)  K sp = [A] a [B] b (pure solid set to 1) Pb 3 (PO 4 ) 2, K sp = 3.0  10 –44 = (3x) 3 (2x) 2 = 108x 5 x = 7.7  10 –10 moles/L of Pb 3 (PO 4 ) 2 dissolve. MW = g/mol means 6.3  10 –7 g/L in soln.

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 Pb 2+ (aq) reduces Pb 3 (PO 4 ) 2 solubility. Swamp with [Pb 2+ ] = 0.1M What effect? K sp = (0.1) 3 (2x) 2 = 0.004x 2 x = 2.7  10 –21 or only 2.2  10 –18 g/L solubility!

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.

Complex Formations  Complex: a Lewis Acid-Base association of an atom with coordinated ligands E.g., Al(OH) 2 (H 2 O) 4 + or PbI 3 (H 2 O) – 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: Pb 2+ (aq) + 3 I – (aq)  PbI 3 – (aq) instead of PbI 2 (s)

Autoprotolysis  2 H 2 O  H 3 O + + OH – implies that water autoionizes to yield (damn few) protons  All weak acids can do likewise when they are the solvent: 2 CH 3 CO 2 H( l )  CH 3 CO 2 H CH 3 CO 2 – K glacial = 3.5  10 –15 at 25°C

Polyprotic Conjugates  H 2 PO 4 – (acid) has HPO 4 2– (conjugate base)  K a2 = 6.32  10 –8 and K b2 = 1.58  10 –7 since HPO 4 2– + H +  H 2 PO 4 – K a2 –1 plus H 2 O  H + + OH – K w yields HPO 4 2– + H 2 O  H 2 PO 4 – + OH – K b2 = K w /K a2  So for n-protic acid, K ai  K b(n+1–i) = K w