University:
Massachusetts Institute of Technology
Course:
1.725J | Chemicals in the Environment: Fate and Transport
Academic year:

2022
Views:

332

Pages:

4
Author:

Sadivt3v

1-725 Problem Set #2 Solutions 1 - 5, 12, 15, 25, 27 5. Find maximum conc. of sulfide species Start with pHS = 1 atm and work through the K values. for CH2S: CH2S(aq) ⇌ PH2S K = 1 atm. mol/L = 0.103 mol/L for [HS-]: [H+][HS-] = 10^-7.02 [HS-] = 10^-7.02 [CH2S] / [H+] = 9.8 x 10^-3 mol/L for [S^2-]: [H+][S^2-] = 10^-13.9 [S^2-] = 10^-13.9 [HS-] / [H+] = 1.2 x 10^-10 mol/L or we can tell without calculation that [S^2-] will be negligible, because only above pH 13.9 will there be more S^2- than HS- (defn. of pKa) total: [H2S] + [HS-] + [S^2-] = 0.11 M b) reaction: CH2Br2 + HS- → products d[CH2Br2] / dt = -k [CH2Br2][HS-] assume [CH2Br2] << [HS-] then pseudo-first-order kinetics apply, with k' = k[HS-] So now we need to find [HS-], which is done in the same way as part a) [HS-] = 10^-7.02 (0.0102) / 10^-7 = 9.8 x 10^-4 mol/L 12. a) 0.5 = Co e^-kt ln 2 = kt1/2 k = ln 2 / t1/2 = 0.058 yr^-1 b) C = Coe^-kt = e^(-0.058 yr^-1)(25 yr) C/Co = 0.236 3/4 remaining = 0.764 3/4te = 1 - 0.236 = 0.764 te = 0.31 yr 15. Find Cwater, then use Kow and H to figure out partitioning Cw = 5 mg / 250 mL * 1000 mg/L = 1.88 x 10^-2 mol/L Kow = Coctanol / Cwater Coctanol = Kow * Cwater = 10^3.12 * (1.88 x 10^-2 mol/L) = 0.248 mol/L Convert to mass: 0.248 mol * 100.11 g/mol = 24.8 g in octanol H = Cair / Cwater Cair = H * Cw = (value for 20°C, assume same value at 25°C) * 1.88 x 10^-2 mol/L = 4.14 x 10^-5 mol/L Convert to mass: 4.14 x 10^-5 mol * 100.11 g/mol = 4.14 x 10^-3 g in air total: 0.005 + 24.8 + 2 x 10^-4 = 24.8 g 25. a) H = Cair / Cwater = pressure / solubility * 1 / RT = 8 x 10^-4 atm / 2 x 10^-4 mol/L * 1 / (0.08206 L atm/mol K * 298 K) = 0.048 (at 20°C) b) % of naphthalene in air = Cair * Va / (Cwater * Vw + Cair * Va) plug in: Va = 19 L, Vw = 1L = Cw(0.048)(19L) / (Cw(1L) + Cw(0.048)(19L)) = 0.471 = 47.1% 21. a) Species: H+, OH-, CH3COOH, CH3COO-, NH3, NH4+ H2O doesn't have to be included because its concentration is constant (we're looking for unknowns here) b) [CH3COO-][H+] / [CH3COOH] = 10^-4.8 [NH4+][OH-] / [NH3] = 10^-4.8 [H+][OH-] = 10^-14 c) [CH3COOH] + [CH3COO-] = 0.01 [NH3] + [NH4+] = 0.01 d) electroneutrality: # of + charges = # of - charges [H+] + [NH4+] = [OH-] + [CH3COO-] We have 5 equations and 5 unknowns, so we can solve for the concentrations and determine the pH.

1.725J Problem Set #2 Solutions

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