Keys to Problem Set 1 - CHM 614/714 Answer Key to Problem...

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1 CHM 614/714 Answer Key to Problem Set #1 1. (a) At low concentrations of HA, the organic acid will dissociate, leaving less HA available to partition into the organic phase. (b) The buffered aqueous phase will prevent ionization of the organic acid and HA will distribute as a neutral molecule. (c) The dimerization (or polymerization) in the organic phase will tend to pull more and more HA into the organic phase. 2. D A = 30, D B = 15, V u.p. =V l.p. (a) p = D/(D + 1), q = 1/(D + 1) p A = 30/(30 + 1) = 96.77%, p B = 15/(15 + 1) = 93.75% q A = 1/(30 + 1) = 3.23%, q B = 1/(15 + 1) = 6.25% (b) To ensure the maximums are 10 tubes apart np A =np B + 10 n(0.9677) = n(0.9375) + 10 0.0302n = 10 n = 331 steps (c) Location of maximum concentration: r max = np r A = (331)(0.9677) = 320 r B = (331)(0.9375) = 310 3. D = 3.5, Vr = V 2 /V 1 = 1, n =? q n = [Vr/(Vr + D)] n 0.01 = [1/(1 + 3.5)] n 0.01 = (0.222) n log(0.01) = nlog(0.222) n = 3.06 Therefore, four extractions are necessary.
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2 4. K = D for those species which do not undergo secondary chemical equilibria. 5. -( G o i - G o j ) = RTln = (1.99 cal/K . mol)(298 K)ln(100/35)= 622 cal/mol Therefore, this separation is unfavorable from thermodynamics point of view. 6. HA ------> H + + A - K a = [H + ][A - ]/[HA] At pH = 3.0: D K 1 K a [ H ] 3 . 0 1 1 . 0 10 5 1 . 0 10 3 2 . 97 q 1 1 D V y V x 1 1 ( 2 . 97) ( 100 / 50 ) 1 6 . 94 0 . 1441 14. 41% At pH = 10.0: D K 1 K
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