CH302 Spring 2009 Practice Quiz 3 Answer Key—The TA Version 1. Which of the following pairs of solutions would result in a buffer upon mixing? 1. 25 mL of 4 M HCl & 15 mL of 4 M HNO2 2. 200 mL of 0.5 M LiOH & 100 mL of 0.5 M H2SO4 3. 100 mL of 1 M NH3 & 10 mL of 10 M HNO3 4. 150 mL of 3 M Ba(OH)2 & 200 mL of 2 M HClO 5. 100 mL of 1 M CH3COOH & 50 mL of 1 M NaOH Correct Explanation: 100 mL of 1 M CH3COOH & 50 mL of NaOH, would equate to 0.1 mol of a weak acid and 0.05 mol of a strong base. Upon neutralization, the resulting solution would contain 0.05 mol each of acetic acid and its conjugate base, acetate - resulting in a buffered system. 2. What would be the pH of a solution prepared from 2 L of H2O, 85 g of NH3 and 98 g of NH4Br? Assume the Kb of ammonia is 2x10-5. 1. 4 2. 5.4 3. 10 Correct 4. 8.6 5. 7 Explanation: 85 g of NH3 x 1 mol/17 g = 5 mol NH3 5 mol NH3 / 2 L H2O = 2.5 M NH3 98 g of NH4Br x 1 mol/98 g = 1 mol NH4Br 1 mol NH4Br / 2 L H2O = 0.5 M NH4Br [OH-] = Kb(Cb\Ca) = 2x10-5(2.5 M/0.5 M) = 10-4 pH = 10 3. Two liters of a buffer containing 0.6 M CH3NH2 and 0.8 M CH3NH3Cl has 102.4 g of HI added to it. What is the new pH? Assume the Kb of CH3NH3 is 6x10-4. 1. 6 2. 3 3. 11 4. 4 5. 10 Correct 6. 8 Explanation: 102.4 g of HI x 1 mol/128 g = 0.8 mol HI [OH-] = Kb(Cb\Ca) = 6x10-4(0.2 M/1.2 M) = 10-4 pH = 10 4. A 0.08 M CH3NH2 solution is titrated against a 0.08 M HCl solution. Assuming the Kb of CH3NH2 is 4x10-10, what is the pH at the equivalence point? 1. 3 Correct 2. 7 3. 9 4. 5 5. not enough information Explanation: Because the titrant and analyte are equimolar, the volume of the system at the equivalance point will be double its initial value and the concentration of the conjugate acid will be half the initial concentration of the base. Vtotal = 2 x Vinitial Ca = 0.04 M CH3NH3+ Ka = Kw / Kb = 10-14 / 4x10-10 = 2.5x10-5 [H+] = (Ka·Ca)1/2 = (2.5x10-5·0.04)1/2 = (10-6)1/2 = 10-3 pH = 35. Consider the molecule ethylenediaminetetraacetic acid (EDTA): As drawn above, how many Ka would be needed to describe the complete deprotonation of EDTA? 1. 4 Correct 2. 6 3. 3 4. 5 Explanation: As drawn, EDTA has 4 ionizable protons and would thus require 4 Ka to express each deprotonation. In actuality, the nitrogen moieites are also ionizable, but as drawn, are already deprotonated. 6. What would be the difference in pH of a 1 M solution of NaH2AsO4 and a 1 M solution of Na2HAsO4? Assume H3AsO4 has a pKa1 of 2 and a pKa2 of 7 and a pKa3 of 12. 1. 7 2. 4.5 3. 9.5 4. 5 Correct 5. 2.5 6. 1.5 Explanation: For a solution composed of a single amphoteric species (H2AsO4-), pH = 0.5(pKa1 +pKa2) = 0.5(2 + 7) = 4.5 For a solution composed of a single amphoteric species (HAsO42-), pH = 0.5(pKa2 +pKa3) = 0.5(7 + 12) = 9.5 7. A student erroneously calculated that a solution consisting solely of a weak base dissolved in water had a pH of 6. Which two of the following might have been true? I. Kb < 10-11 II. Kb > 10-3 III. Cb > 10-1 IV. Cb < 10-4 1. I and IV only Correct 2. II and III only 3. I and III only 4. II and IV only Explanation: Having erroneously calculated a pH of 6 (pOH of 8) for a weak base solution suggests that student probably used the equation [OH-] = (Kb·Cb)1/2 and failed to notice that both the value of Kb and Cb were too small to satisfy the assumptions made when using [OH-] = (Kb·Cb)1/2. 8. An aqueous system with Na2CO3, NaCl and NH4Cl dissolved in it would require how many equations to find all the unknown equilibrium concentrations? 1. 3 2. 7 Correct 3. 4 4. 6 5. 9Explanation: An aqueous system with Na2CO3, NaCl and NH4Cl dissolved in it would have an unknown value for [H+], [OH-], [CO32-], [HCO3-], [H2CO3], [NH4+] and
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