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16 Acid-Base Equilibria Solutions to Exercises (aq) 1L (aq) + Initial 0.10 M 0 0 equil. M X M 0.071 M 0.029 M 0.029 M = 1.2 X 10⁻² = [HSO₄⁻] = 0.10 ; Kₐ is relatively large, so use the quadratic. + 0.012 0.0012 = 0; = - 4(1)(-0.0012) 2 Total ion concentration = 0.10 M + 0.071 M + 0.029 M + 0.029 M = 0.229 = 0.23 M. Assume 100.0 mL of solution. 1.002 g/mL X 100.0 mL = 100.2 g solution. 120.1 g 0.10 M X 0.1000 = 0.010 mol NaHSO₄ X mol = 1.201 = 1.2 g 100.2 g soln - 1.201 g = 99.0 g = 0.099 kg H₂O m = mol kg H₂O ions = 0.229 0.0990 M L = 0.231 = 0.23 m ions = = = +0.12°C: Tb = 100.0 + 0.12 = 100.1°C 16.121 Rx 1: AH = D(H-F) + 2D(H-O) 3D(H-O) = D(H-F) - D(H-O) = 567 - 463 = 104 kJ Rx 2: = + 2D(H-O) 3D(H-O) = - D(H-O) = 431 kJ - 463 kJ = kJ The reaction involving is exothermic, while the reaction involving HF is endothermic, owing to the smaller bond dissociation enthalpy of H-Cl. is a stronger acid than HF, and the enthalpy of ionization for is exothermic, while that of HF is endothermic. This is consistent with the trend in acid strength for binary acids with heavy atoms (X) in the same family. That is, the longer and weaker the H-X bond, the stronger the acid (and the more exothermic the ionization reaction). 16.122 Calculate M of the solution from osmotic pressure, and Kb using the equilibrium expression for the hydrolysis of cocaine. Let Coc = cocaine and CocH+ be the conjugate acid of cocaine. П = MRT; M = П/RT = 52.7 288 torr K 760 1 atm torr 0.08206 mol L K atm = 0.002934 = 2.93 10⁻³ M Coc pH = 8.53; pOH = 14 pH = 5.47; [OH⁻] = = 3.39 X 10⁻⁶ = 3.4 M 500