Exam 3 Solutions

9 November 1998

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  1. O2 -(aq) doesn't exist because OH -(aq) is a weaker acid than H2O! With that in mind, find the (25°C) solubility of Ag2O in (a) neutral water and (b) pH 3 from 10-3 M HNO3, if you like, but the source is irrelevant as long as the counter ion doesn't precipitate Ag+(aq). g± = 1 at these concentrations. You'll need to know that KW=10 -14 and DfGq [kJ/mol] for Ag2O(s) [-11.20], Ag+(aq) [+77.11], H2O(liq) [-237.13], and OH -(aq) [-157.24]. Think KSP.

    1. Since
      O2 -(aq) + H2O(aq) arrow right 2 OH -(aq)
      we can safely conclude that
      Ag2O(s) + H2O(aq) arrow double 2 Ag+(aq) + 2 OH -(aq)
      DrGq = 2(+77.11) + 2(-157.24) - (-11.20) - (-237.13) kJ/mol
          = + 88.07 kJ/mol
      KSP = e-DrGq / RT = e-88,070 J/mol / [(8.315 J/mol K)(298 K)] = e- 35.5
          = 3.66×10 -16 = [Ag+]2 [OH - ]2 = (2x)4
      x = 6.92×10 -5     is the # of moles of Ag2O(s) which have dissolved to ions, giving a solubility of
      (231.8 g Ag2O / mol Ag2O)×(6.92×10 -5 M) = 16 mg/L

      Just to be on the safe side, we should check [OH - ] = 2x = 1.38×10 -4 to make sure that it hasn't inadvertantly fallen below 10 -7 where water's autoionization would have held it! (That can happen with less soluble oxides, but we dodged that bullet here.) If that had happened here, we would have used 10 -7 instead of 2x for [OH - ] in the equilibrium calculation.

    2. If [H+ ] = 10 -3 M, then [OH - ] is held to 10 -14 / 10 -3
      or [OH - ] = 10 -11 instead of 2x.

      [Ag+ ]2 [OH - ]2 = (2x)2 (10 -11 )2 = 3.66×10 -16
      (2x)2 = 3.66×10+6 or x = 960 (!)

      Now we don't imagine that 960 moles of Ag2O dissolve in a liter of pH 3 nitric acid. What we can say, however, is that silver oxide is quite soluble in acid solutions!

  2. Show that no matter how steep DrGq might be, equilibrium never lies at the reactant or product side. If the extent of reaction is x, this is because dDmixSq/dx is not finite at either x=0 or x=1 even though DmixSq=0 there! Use the A arrow double B equilibrium as your example since it's the simplest.

    XB=x and XA=(1-x) so
    DmixS = - R [ (1-x)ln(1-x) + xlnx ]
    dDmixS/dx = - R [ - (1-x)/(1-x) - ln(1-x) + x/x + lnx ]
        = + R [ ln(1-x) - lnx ]

    In the limit as x approaches 0, ln(1-x) is just 0 but lnx goes to minus infinity! Therefore DmixS has and infinitely positive slope at x=0 meaning that -TDmixS has an infinitely negative slope, pushing dDG/dx=0 away from x=0 regardless of the sign or magnitude of dDH/dx!

    The opposite happens as x approaches 1. lnx goes to 0 but ln(1-x) becomes infinitely negative, making -TDmixS infinitely positive and sucking dDG/dx=0 away from x=1 regardless of the sign or (finite) magnitude of dDH/dx!

    So no chemical reaction is ever truly quantitative.

  3. After hydrogen and oxygen, the most abundant atoms (as ions) in seawater [g/L] are Cl [19.4] and Na [10.8]. It can safely be assumed that the requirements for electrical neutrality are made up by other monocations and other anions are negligible. What pressure must a reverse osmosis purification plant provide? (Much more is actually used to improve the kinetics.)

    [Na+] = 10.8 g/L / 22.99 g/mol = 0.470 M
    [Cl - ] = 19.4 g/L / 35.45 g/mol = 0.547 M
    So there must be other (presumed mono) cations balancing off the Cl -, hence total concentration of ions of all sorts ought to be roughly 2[Cl - ] = 1.095 M.

    P = cRT = 1.095 M (0.08206 atm L/mol K)(298 K) = 26.76 atm

  4. Dry cells, NiCad rechargeables, and silver oxide (watch) batteries utilize solid reactant and products exclusively. Since they're usually alkaline or acid, OH - or H+ do have an influence, but they're usually neither produced nor consumed. (Otherwise how would any solutions maintain electrical neutrality?) So the question is "Why aren't solvated reactants or products used in ordinary batteries?" Not surprisingly, the answer involves thermodynamics.

    True, solutions would leak and present safety hazards, but the real reason is more fundamental.

    Since the voltage E = Eq - ( RT / nF ) lnQ, the potential fall from Eq to zero as equilibrium approaches and Q becomes K. But solids don't appear in Q! Or K. So batteries with solid reactants and products don't change their potential (a really useful property) until reactants are used up (or products fail to diffuse away from the electrodes).

  5. The lead storage (automobile) battery does consume H+ from H2SO4, but maintains electrical neutrality by converting both Pb and PbO2 to PbSO4. Estimate the voltage of a single cell of such a battery using 1 M acid (and ignoring g±) from DfGq [kJ/mol] of PbO2(s) [-217.33], PbSO4(s) [-811.3], and SO42 -(aq) [-744.53].

    Ignore any HSO4- that must also be around!

    Pb(s) + SO42 -(aq) arrow right PbSO4(s) + 2 e-
    2 e- + 4 H+(aq) + PbO2(s) + SO42 -(aq) arrow right PbSO4(s) + 2 H2O
    yields
    Pb(s) + PbO2(s) + 4H+(aq) + 2SO42 -(aq) arrow right 2PbSO4(s) + 2H2O
    where n = 2

    DrGq = 2(-811.3) + 2(-273.13) - (0) - (-217.33) - 4(0) - 2(-744.53)
        = - 336.47 kJ/mol

    But that presumes all of the species are present at unit activity. Even if all activity coefficients are one and molalities are well approximated by molarities, [H+] = 2 M not 1 M. So we have to correct DG.
    DrG = DrGq + RT ln(Q=1/[H+]4)
        = DrGq - 4 RT ln[H+] = - 343.34 kJ/mol
    non-negligibly different. Now DG = - nFE so
    E = - DG/nF = + 343,340 kJ/mol / [2(96,485 J/V mol)] = + 1.78 V

    Of course, automobile batteries have 6 of those cells in series giving a total voltage of 10.68 V to a 12 V battery. :-) So it must be more concentrated than 1 M acid.

Last modified 15 November 1998