Chem 142 Quiz 4 Spring 2009Name __________________Key___________ ___________________ Instructor: Martin Larter1. The weak base, trimethylamine [(CH3)3N] has a Kb = 6.4 × 10-5. Suppose 50.0 mL of 0.10 M trimethylamine is titrated by 0.20 M HCl.a) How much 0.20 M HCl must be added to reach the equivalence point? Answer: ____D___A) 100.0 mL B) 75.0 mL C) 50.0 mL D) 25.0 mL E) none0.10 M (CH3)3N * 50.0 mL = 0.0050 mol(CH3)3N1000 mL0.0050 mol(CH3)3N * 1mol HCl/ 1 mol(CH3)3N * 1000 mL/0.20 mol HCl = 25.0 mL HClb) Calculate the pH of the titrated solution at the equivalence point. Show your work. At equiv. point all (CH3)3N is converted to (CH3)3NH[(CH3)3NH] = 0.0050 mol (CH3)3NH /0.075 L = 6.67 x 10-2 M (CH3)3NH Ka = 1 x 10 -14= 1.6 x 10-10 6.4 × 10-5(CH3)3NH + H2O  (CH3)3N + H3O+Initial 6.67 x 10-2 M 0 0Change -x +x +xEquilibrium 6.67 x 10-2 M -x x xKa = [(CH3)3N][ H3O + ] 1.6 x 10-10 = x 2 Assume x <<< 6.67 x 10-2 M [(CH3)3NH] 6.67 x 10-2 M -x.6 x 10-10 = x 2 x = 3.2 x 10-6 M = [H3O+] pH = 5.49 6.67 x 10-2 MCheck: (3.2 x 10-6 M /6.67 x 10-2 M) *100 = 4.8 x10-3 % assumption ok2. A 25.0 mL sample of 0.130 M HN3 (hydrazoic acid; pKa = 4.74) is mixed (and diluted) with 10.0 mL of 0.130 M of NaOH. Based on your knowledge of acid-base titration curves, which of these is the most likelypH of the resulting solution? (Must do by ICE Tables)[HN3]I = 25.0 mL*(0.130 M HN3) = 9.29 x10-2 M Ka = 1.82 x10-5 35 mL[NaOH]I = 10.0 mL*(0.130 M NaOH = 3.71 x10-2 M 35 mLHN3 + NaOH.  N3- + H2OInitial 9.29 x10-2 M 3.71 x10-2 M 0Change -x -x +xFinal 5.58 x10-2 M 0 3.71 x10-2 MHN3 + H2O.  N3- + H3O+Initial 5.58 x10-2 M 3.71 x10-2 M 0Change -x +x +xEquilibrium 5.58 x10-2 M-x 3.71 x10-2 M+x xKa = [N3- ][ H3O + ] 1.82 x 10-5 = x*(3.71 x10 -2 M+x) Assume x <<< 3.71 x10-2 [HN3] 5.58 x10-2 M-x1.82 x 10-5 = x*(3.71 x10 -2 M) x = 2.74 x 10-5 M pH= 4.56 5.58 x10-2 MAssumption Check: (2.74 x 10-5 M / 3.71 x10-2 M) *100 = 7.39 x 10-2 % ASSUMPTION OKCheck: pH = 4.74 + log (3.71 x10-2 M / 5.58 x10-2 M) = 4.563. Determine the pH and the equilibrium concentrations of all species in a solution that contains 0.500 M NH3, 4.00 g NH4NO3, and 0.100 M (NH4)2SO4. Kb = 1.8 × 10-54.00 g NH4NO3 * (1mol NH4NO3 / 80.52 g NH4NO3) *(1mol NH4+)/ (1mol NH4NO3) =0.0500 M NH4+0.100 M (NH4)2SO4 * (1.00 L) * (2mol NH4+)/ (1mol (NH4)2SO4) = 0.200 M NH4+NH3 + H2O  NH4+ + OH-Initial 0.500 M 0..250 M 0Change -x +x +xEquilibrium 0.500 M-x 0.250 M+x x1.8 × 10-5= [0.250 M+x]*[x] Assume x <<< 0.500 M [0.500 M-x]1.8 × 10-5= [0.250 M]*[x] x = [OH-] = 3.6 x 10-5 M pOH = -[3.6 x 10-5 M] =4.44 [0.500 M]pH = 9.56Check: (3.6 x 10-5 M/0.500 M) *100 = 7.2 x 10-3% assumption