CHEM 102 Winter 12 Make-Up Exam (A) Questions and Answers,100% CORRECT

CHEM 102 Winter 12 Make-Up Exam (A) Questions and Answers Potentially useful information: x  pH  pKa 2 a  log base acid  ΔG° = ΔH° - T ΔS° ΔG = ΔG° + RT ln Q ΔG° = - RT ln K R = 8.314 J mol-1 K-1 1. Please choose the letter “a” as your answer for this question. 2. A buffer solution a. contains more than the expected amount of solute for a particular temperature and is therefore unstable. b. contains the maximum amount of solute possible for a particular temperature. c. changes color upon addition of strong base. d. contains an equal number of hydronium and hydroxide ions. e. resists changes in pH upon addition of acid or base. 3. A buffer solution is 0.500 M in acetic acid and 0.500 M in sodium acetate. The pH is 4.74. What is the pH after the solution is diluted by a factor of 2? a. 4.44 b. 4.49 c. 4.74 d. 4.99 e. 5.04 4. What is the pH of a 500 mL solution containing 0.28 mol HClO (Ka = 3.5 × 10-8) and 0.31 mol ClO ? a. 7.50 b. 7.25 c. 7.65 d. 7.41 e. 7.20 5. Which acid, in combination with its conjugate base, would be the best choice to make a buffer of pH = 3.70? a. dihydrogen phosphate (Ka = 2.9  107) b. acetic acid (Ka = 1.8  105) c. benzoic acid (Ka = 6.3  105) d. formic acid (Ka = 1.8  104) e. hydrofluoric acid (Ka = 6.8  104) 6. In a buffer solution, if [A] [HA], which of the following is true? a. pH pKa b. pH = pKa c. pH pKa d. pH 7.00 e. pH 7.00 7. Which statement about the titration of 0.10 M HNO3 (strong acid) with 0.10 M KOH (strong base) is not correct? a. The endpoint always comes before the equivalence point. b. The pH at the equivalence point is 7.00. c. The initial pH is 1.00. d. At the equivalence point the pH increases sharply. e. The net ionic equation is H3O+ + OH  2H2O. 8. What volume of 0.106 M NaOH is needed to neutralize a 50.0 mL sample of 0.095 M HNO3? a. 55.8 mL b. 50.0 mL c. 44.8 mL d. 10.1 mL e. 5.19 mL 9. Which acid-base titration would yield a titration curve of the general form shown? a. H2CO3 titrated with NaOH b. NaOH titrated with H3PO4 c. Na3PO4 titrated with HCl d. H2SO4 titrated with NaOH e. H3PO4 titrated with NaOH 10. Which equation could be used to calculate the molar solubility (S) of Fe(OH)2? a. Ksp = 2 S2 b. Ksp = S2 c. Ksp = 4 S3 d. Ksp = S3 e. Ksp = 2 S3 11. The solubility of manganese(II) hydroxide (Mn(OH)2) is 2.2 × 10-5 M. What is the Ksp of Mn(OH)2? a. 1.1 × 10-14 b. 4.3 × 10-14 c. 2.1 × 10-14 d. 4.8 × 10-10 e. 2.2 × 10-5 12. What is the solubility of PbSO4 (Ksp = 1.8 × 10-8) in 0.1 M Na2SO4? a. 1.8 × 10-6 M b. 1.34 × 10-4 M c. 1.8 × 10-7 M d. 1.8 × 10-9 M e. 9.0 × 10-8 M 13. The Common Ion Effect predicts that the solubility of a salt in a solution already containing one of the ions in the salt will be than the solubility in water. This is predictable, because . a. more; Le Chatelier's Principle predicts that the common ion will suppress precipitation b. more; the presence of the common ion will increase that concentration, hence Ksp will fall c. less; Le Chatelier's Principle predicts that the common ion will suppress dissolving d. less; the presence of the common ion will increase that concentration, hence Ksp will rise e. less; at equilibrium, Q will be less than Ksp 14. Of the following, the entropy of gaseous is the largest at 25°C and 1 atm. a. H2 b. C2H6 c. C2H2 d. CH4 e. C2H4 15. Which reaction below would you expect to have an increase in entropy at 25°C? a. Ca(s) + Cl2(g) CaCl2(s) b. 2 H2O(l) 2 H2(g) + O2(g) c. CO2(g) C(s,graphite) + O2(g) d. N2(g) + 2 O2(g) 2 NO2(g) e. N2(g) + 3 H2(g) 2 NH3(g) 16. Calculate S of the following reaction. H2(g) + Cl2(g)  2 HCl(g) At 25°C, the values of S in J mol-1 K-1 are 130.7 (H2), 223.1 (Cl2), and 186.9 (HCl). a. –166.9 J mol-1 K-1 b. +166.9 J mol-1 K-1 c. +20.0 J mol-1 K-1 d. –20.0 J mol-1 K-1 e. +156.0 J mol-1 K-1 17. If a chemical reaction is non-spontaneous, it must be true that a. G° 0. b. G° 0. c. G° = 1. d. G° = 0. e. G° = K. 18. For a particular reaction, the value of H = –98.8 kJ and S = –141.5 J/K. This reaction is a. product-favored at any temperature, because S is negative. b. product-favored at any temperature, because H is negative. c. reactant-favored at any temperature, because H is negative. d. reactant-favored at any temperature, because S is negative. e. either reactant- or product-favored; cannot determine without the temperature. 19. Use the values given to calculate the value of Grxn for the reaction at 25C. 2 C(graphite) + H2(g)  C2H2(g) ΔS° = 58.84 J/mol-K, ΔH° = –226.8 kJ/mol a. –291.4 kJ b. –244.3 kJ c. –226.8 kJ d. –207.6 kJ e. –64.6 kJ 20. Calculate the value of Gf for NO(g) at 25C if Grxn = –173.1 kJ for the following reaction at the same temperature. 2NO(g)  N2(g) + O2(g) a. – 86.55 kJ/mol b. 86.55 kJ/mol c. – 173.1 kJ/mol d. 173.1 kJ/mol e. 346.2 kJ/mol 21. A certain reaction has Hrxn = +177.8 kJ, and Srxn = +160.5 J/K. Above what temperature does it become product-favored? a. 384C b. 630C c. 835C d. 1108C e. 1381C 22. If ΔG°f (C2H6) = –32.8 kJ/mol, ΔG°f (CO2) = –394.4 kJ/mol, and ΔG°f (H2O) = –237.1 kJ/mol, calculate the value of Grxn at 25C for the reaction given below. C2H6(g) + 7/2 O2(g)  2 CO2(g) + 3 H2O(l) a. –1467 kJ b. –598.7 kJ c. –1442 kJ d. 1615 kJ e. –756 kJ 23. Phosphorus and chlorine gases combine to produce phosphorus trichloride: P2 (g) + 3Cl2 (g) 2PCl3 (g) ΔG° at 298 K for this reaction is –642.9 kJ/mol. The value of ΔG at 298 K for a reaction mixture that consists of 1.5 atm P2, 1.6 atm Cl2, and 0.65 atm PCl3 is . a. –44.2 kJ/mol b. –3.88 × 103 kJ/mol c. –7.28 × 103 kJ/mol d. –708.4 kJ/mol e. –649.5 kJ/mol 24. The value of the equilibrium constant for a reaction is 2.65 × 10-6 at 45C. Calculate the value of Grxn at this temperature. a. +47.4 kJ b. +34.0 kJ c. +14.8 kJ d. +335 kJ e. -4.80 kJ 25. The value of ΔG° for the following reaction is 49.6 kJ/mol at 25°C. N2 (g) + 3H2 (g) 2NH3 (g) What is the equilibrium constant for this reaction? a. 2.0 × 10-9 b. 1.02 c. 1.05 d. 5.0 × 108 e. 9.67 × 103 chem102w12exammakeupa Answer Section MULTIPLE CHOICE 1. ANS: A PTS: 0 2. ANS: E PTS: 1 3. ANS: C PTS: 1 4. ANS: A PTS: 1 5. ANS: D PTS: 1 6. ANS: A PTS: 1 7. ANS: A PTS: 1 8. ANS: C PTS: 1 9. ANS: E PTS: 1 10. ANS: C PTS: 1 11. ANS: B PTS: 1 12. ANS: C PTS: 1 13. ANS: C PTS: 1 14. ANS: B PTS: 1 15. ANS: B PTS: 1 16. ANS: C PTS: 1 17. ANS: A PTS: 1 18. ANS: E PTS: 1 19. ANS: B PTS: 1 20. ANS: B PTS: 1 21. ANS: C PTS: 1 22. ANS: A PTS: 1 23. ANS: E PTS: 1 24. ANS: B PTS: 1 25. ANS: A PTS: 1