Part A. [2 points each] For each question, circle the letter of the one correct answer and enter the answer on the TEST SCORING SHEET in pencil only. The TEST SCORING ANSWER SHEET will be considered final. There is no penalty for incorrect answers. 1. The formula for the illegal drug cocaine is C 17 H 21 NO 4. What is the percentage of oxygen in the compound? A) 67.3% Answer: B) 21.1% MM: (17 12.01) (21 1.01) (1 14.01) (4 16.00) = 303.39 g/mol C) 6.93% %O = (4 16.00)/303.39 100% = 21.09% D) 4.62% 2. How many moles of K ions are present in 3.00L of a 0.141M solution of K 3 PO 4? A) 1.69 mol B) 1.27 mol Answer: C) 0.423 mol # moles K : 3.00L 0.141 mol/l 3 moles K /1 mole K 3 PO 4 = 1.27 mol D) 0.141 mol 3. Calculate the volume of 1.00 M Ca(OH) 2 (aq) needed to react completely with 50.0 ml of 2.24 M HNO 3 (aq). A) 224 ml Rxn: Ca(OH) 2 (aq) 2HNO 3 (aq) Ca(NO 3 ) 2 (aq) 2H 2 O(l) B) 112 ml # moles HNO 3 : 0.0500L 2.24 mol/l = 0.112 mol C) 56.0 ml # moles Ca(OH) 2 : 0.112 mol 1 mole Ca(OH) 2 /2 moles HNO 3 = 0.0560 moles D) 28.0 ml volume of Ca(OH) 2 : 0.0560 moles/1.00 mol/l = 0.0560 L or 56.0 ml 4. Consider the reaction: 4NH 3 (g) 5O 2 (g) 4NO(g) 6H 2 O(g) What are the limiting reactant and the number of moles of the other reactant remaining when 20 moles of ammonia gas and 20 moles of oxygen gas react? A) NH 3 (g) and 5 moles of O 2 (g) remaining NH 3 : 20 moles/4 moles per rxn = 5 times B) O 2 (g) and 1 moles of NH 3 (g) remaining O 2 : 20 moles/5 moles per rxn = 4 times (LR) C) NH 3 (g) and 1 moles of O 2 (g) remaining If rxn goes 4 times, 4 4 moles NH 3 of reacted D) O 2 (g) and 4 moles of NH 3 (g) remaining 20 moles 16 moles = 4 moles of NH 3 remain 5. Acid rain is produced by the following sequence of reactions: 4FeS 2 (s) 11O 2 (g) 2Fe 2 O 3 (s) 8SO 2 (g) 2SO 2 (g) O 2 (g) 2SO 3 (g) SO 3 (g) H 2 O(l) H 2 SO 4 (l) Calculate the moles of H 2 SO 4 that would be produced from 0.8347 moles of FeS 2 and excess oxygen. A) 3.339 mol Answer: 0.8347 moles FeS 2 8 mol SO 2 /4 mol FeS 2 = 1.669 mol SO 2 B) 1.669 mol 1.669 mol SO 2 2 mol SO 3 /2 mol SO 2 = 1.669 mol SO 3 C) 0.4174 mol 1.669 mol SO 3 1 mol H 2 SO 4 /1 mol SO 3 = 1.669 mol H 2 SO 4 D) 0.2087 mol Page 1
6. Which of the following pairs of solutions will produce a precipitate when mixed? A) LiClO 4 (aq) and CaBr 2 (aq) LiBr(aq) Ca(ClO 4 ) 2 (aq) B) K 2 CO 3 (aq) and NH 4 Cl(aq) (NH 4 ) 2 CO 3 (aq) KCl(aq) C) CuSO 4 (aq) and NaCH 3 COO(aq) Na 2 SO 4 (aq) Cu(CH 3 COO) 2 (aq) D) AgNO 3 (aq) and KI(aq) AgI(s) KNO 3 (aq) 7. Consider the following reaction: WO 3 (s) 3H 2 (g) W(s) 3H 2 O(g) When 0.495 moles of hydrogen gas react with excess WO 3 (s), 15.5 grams of W(s) are produced. What is the percent yield? (The molar mass of W is 183.84 g/mol.) A) 8.52% B) 17.0% Answer: # moles W: 0.495 moles H 2 1 mole W/3 moles H 2 = 0.165 mol C) 25.5% mass W: 0.165 mol W 183.84 g W/1 mole W = 30.33 g D) 51.1% % yield: 15.5 grams of W/30.33 g W 100% = 51.1% 8. At a certain temperature, K c = 9.0 for the equilibrium reaction N 2 O 4 (g) 2NO 2 (g) What is K c at the same temperature for NO 2 (g) ½N 2 O 4 (g)? A) 0.33 B) 0.22 Answer: Rxn is flipped and multiplied by ½ C) 0.11 K eq = (K 1 c ) ½ = (9 1 ) = 0.33 D) 0.012 9. At 24 C, the equilibrium constant, K c, for the reaction: 2NOBr(g) 2NO(g) Br 2 (g) is 3.07 10 4. If 1.2 moles of NOBr, NO and Br 2 are mixed in a 1.00 L container at 24 C, which of the following is true? A) [NOBr] = [NO] = [Br 2 ] at equilibrium conc = 1.2M B) NO(g) and Br 2 (g) will be formed until equilibrium is reached Q = (NO) 2 ( Br 2 )/(NOBr) 2 = 1.2 C) NOBr(g) will be formed until equilibrium is reached Q > K; too many products D) The system is at equilibrium and therefore no net change occurs. Rxn moves from right to left. 10. Which of the following has the strongest conjugate acid? A) aniline (pk b = 9.13) Answer: Want conjugate acid with smallest pk a value. B) nicotine (pk b = 5.98) pk a pk b = 14.00 C) ammonia.(pk b = 4.74) If pk b large, pk a will be small. D) methylamine (pk b = 3.34) Page 2
11. The pk b for dimethylamine, (CH 3 ) 2 NH is 3.27. This expression refers to which of the following reactions? A) (CH 3 ) 2 NH 2 (aq) H2 O(l) (CH 3 ) 2 NH(aq) H 3 O (aq) B) (CH 3 ) 2 NH(aq) H 3 O (aq) (CH 3 ) 2 NH 2 (aq) H2 O(l) C) (CH 3 ) 2 NH(aq) H 2 O(l) (CH 3 ) 2 NH 2 (aq) OH (aq) D) (CH 3 ) 2 NH 2 (aq) OH (aq) (CH3 ) 2 NH(aq) H 2 O(l) 12. What is the ph of 0.010 M HCN(aq).(K a = 4.9 10 10 )? A) 2.00 Rxn: HCN(aq) H 2 O(l) CN (aq) H 3 O (aq); K a B) 4.65 0.010 x x C) 5.65 [H 3 O ] = x = (K a 0.010)= 2.2 10 6 D) 8.35 ph = log [H 3 O ] = 5.65 13. A weak acid with K a = 1.8 10 4 is 4.2% ionized in a 0.10 M solution. In a 1.0 M solution, the percent ionization would be, A) less than 4.2%. B) 4.2%. More ionization occurs with dilution; opposite is true in this case. C) greater than 4.2%. 14. A water soluble salt composed of the anion of a weak acid and the cation of a strong base is dissolved in pure water. What is the ph of the resulting solution? A) The ph is greater than 7. anion of a weak acid = A (basic) B) The ph is 7.00 cation of strong base = group I and some group II metals (neutral) C) The ph is less than 7. 15. Which of the following properly arranges 0.10 M solutions of KNO 2, HClO, NH 3 and C 6 H 5 NH 3 I in order of increasing ph? A) HClO(aq) < C 6 H 5 NH 3 I(aq) < KNO 2 (aq) < NH 3 (aq) KNO 2 basic salt (K b =K w /K a = 2.3 10 11 ) B) KNO 2 (aq) < C 6 H 5 NH 3 I(aq) < HClO(aq) < NH 3 (aq) HClO weak acid (K a = 3.5 10 8 ) C) NH 3 (aq) < HClO(aq) < C 6 H 5 NH 3 I(aq) < KNO 2 (aq) NH 3 weak base (K b = 1.8 10 5 ) D) C 6 H 5 NH 3 I(aq) < HClO(aq) < KNO 2 (aq) < NH 3 (aq) C 6 H 5 NH 3 I acidic salt (K a =K w /K b = 2.3 10 5 ) 16. Which of the following pairs of ions cannot exist in large concentrations simultaneously in aqueous solution? A) NH 4 (aq) and OH (aq) weak acid strong base weak base water (reaction!) B) H 3 O (aq) and NO 3 (aq) strong acid neutral no reaction C) Ba 2 (aq) and OH (aq) neutral strong base no reaction D) Na (aq) and CH 3 COO (aq) neutral weak base no reaction Page 3
17. Solutions are made by combining equal volumes of the following solutions. Which is a buffer? A) 0.10 M HClO 4 and 0.10 M NaClO 4 strong acid neutral salt not a buffer solution. B) 0.10 M NaF and 0.050 M HF weak base weak acid conjugates in a ratio of 2:1. C) 0.10 M HF and 0.10 M NaOH weak acid strong base (1:1) stoichiometric point! D) 0.10 M NaF and 0.050 M NaOH weak base strong base no weak acid present 18. Choose the effective ph range of a HNO 2 NaNO 2 buffer. A) 2.4 4.4 B) 9.6 11.6 Effective ph range = pk a 1 C) 0.7 2.7 K a = 4.3 10 4 ; pk a = log K a = 3.37 D) 11.3 13.3 19. Which of the following indicators would be most suitable for the titration of 0.10 M (CH 3 ) 3 N(aq) with 0.10 M HClO 4 (aq)? A) thymol blue (pk IN = 1.7) Rxn: (CH 3 ) 3 N(aq) H 3 O (aq) (CH 3 ) 3 NH (aq) H 2 O(l) B) bromothymol blue (pk IN = 7.1) At the stoichiometric point, the ph would be acidic. C) phenolphthalein (pk IN = 9.4) K a (CH 3 ) 3 NH = K w /K b (CH 3 ) 3 N = 1.5 10 10 D) bromocresol green (pk IN = 4.7) Because the acid formed is so weak, the ph would not be 1.7. 20. What is the solubility product expression for Zn 3 (PO 4 ) 2 (s)? A) [3Zn 2 ] 3 [2PO 3 4 ] 2 B) [Zn 2 ] 3 [PO 3 4 ] 2 Rxn: Zn 3 (PO 4 ) 2 (s) 3Zn 2 3 (aq) 2PO 4 (aq) C) [Zn 3 ] 2 [PO 2 4 ] 3 K sp = [Zn 2 ] 3 [PO 3 4 ] 2 D) [Zn 2 3 ][(PO 3 4 ) 2 ] Page 4
Part B. Short Answers and Problems [Total of 30 points] [6 marks] 1. Hydrochloric acid reacts with aluminium metal to form aluminium chloride and hydrogen gas. (i) Write a balanced equation for this reaction. 2 marks 6HCl(aq) 2Al(s) fi 2AlCl 3 (aq) 3H 2 (g) ( 1 for each mistake) (ii) If excess hydrochloric acid reacts with 0.0556 moles of aluminium, how many litres of hydrogen gas are produced at 100.0 C and 760 Torr. (ratio used must match above equation) # moles of H 2 produced: 0.0556 moles Al 3 moles H 2 /2 moles Al = 0.0834 mol (1 point) PV = nrt 100.0 C = 373.15 K (1 point) 760 Torr = 1 atm = 101.3 kpa (1 point) V = (0.0834 mol)(0.0821 atml/molk)(373.15k)/1 atm = 2.56 L (1 point) OR V = (0.0834 mol)(8.314 kpal/molk)(373.15k)/101.3 kpa = 2.55 L (V = # moles 30.64) [8 marks] 2. A buffer was prepared by adding 500.0 ml of 0.255 M HBr(aq) to 500.0 ml of 0.565 M NH 3 (aq). Write the net ionic equation for the reaction and calculate the ph of this solution. NH 3 (aq) H 3 O (aq) «NH 4 (aq) H 2 O(l) (1 point) 0.5000 L 0.5000 L 0.565 M 0.255 M (1 pt) I: 0.2825 mol 0.1275 mol 0 K b = 1.8 10 5 (pk b = 4.74) C: 0.1275 0.1275 0.1275 K a = K w /K b = 5.6 10 10 E: 0.1550 (1 pt) 0 0.1275 (1 pt) pk a = 9.26 (2 point) ph = pk a log (NH 3 /NH 4 ) = 9.26 log (0.1550/0.1275) (1 point) = 9.34 (1 point) [ 4 marks] 3. i) Write the net ionic equation for the reaction that occurs when a small amount of KOH is added to a buffer solution containing CH 3 NH 2 and CH 3 NH 3 Cl. 2 marks CH 3 NH 3 (aq) OH (aq) «CH 3 NH 2 (aq) H 2 O(l) ( 1 for each mistake) ii) Calculate the equilibrium constant for the above reaction. K eq = 1/K b (CH 3 NH 2 ) = 1/3.6 10 4 = 2.8 10 3 OR 2800 2 marks PAGE TOTAL /18 Page 5
4. [12 points] The curve for the titration of 0.125 M Benzoic acid, C 6 H 5 COOH(aq), with 0.250 M NaOH(aq) is given below: 14.000 12.000 10.000 Stoichiometric Point ph 8.000 6.000 ph = pk a 1 point for each = 2 marks (must be clearly labelled) 4.000 2.000 0.000 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 Volume of base (ml) a) Write the net ionic equation for the titration reaction. C 6 H 5 COOH(aq) OH (aq) «C 6 H 5 COO (aq) H 2 O(l) (2 point) OR HA(aq) OH (aq) «A (aq) H 2 O(l) ( 1 for each individual mistake) b) The value of the equilibrium constant for this titration is (circle the correct answer). A) very large B) 1.0 C) very small (2 point) c) Clearly label on the curve above exactly where the stoichiometric point and the pk a of C 6 H 5 COOH fall on this titration curve. d) Write the formula(s) of the major species in the solution after 12 ml of base is added. C 6 H 5 COOH(aq), C 6 H 5 COO (aq), (not necessary to include Na (aq)) (1 point for each = 2 marks) e) Calculate the concentration of the benzoate ion, C 6 H 5 COO (aq), at the stoichiometric point. # moles of NaOH: 0.250 mol/l 0.0200 L = 0.00500 mol (1 point) # moles of NaOH = # moles of C 6 H 5 COOH 0.00500 mol C 6 H 5 COOH = 0.125 mol/l Volume at stoichiometric point Volume = 0.0400 L or 40.0 ml (1 point) Volume Total = 20.0 40.0 = 60.0 ml (1 point) [C 6 H 5 COO ] = 0.00500 mol/ 0.0600 L = 0.0833 M (1 point) Page 6 Could have also set one volume to 1L and then solved for other. Final answer will be the same. PAGE TOTAL /12
DATA SHEET Gas constant, R = 8.314 J mol 1 K 1 Avogadro s Number = 6.022 10 23 mol 1 R = 8.314 kpa L mol 1 K 1 1 atm = 101.3 kpa = 760 Torr R = 0.0821 atm L mol 1 K 1 0 C = 273.15 K Acidity and Basicity Constants Acid K a Base K b HNO 2 4.3 10 4 (C 2 H 5 ) 3N 1.0 10 3 HF 3.5 10 4 (CH 3 ) 2 NH 5.4 10 4 HCOOH 1.8 10 4 CH 3 NH 2 3.6 10 4 CH 3 COOH 1.8 10 5 (CH 3 ) 3 N 6.5 10 5 H 2 CO 3 4.3 10 7 (K a1 ) NH 3 1.8 10 5 HClO 3.5 10 8 NH 2 OH 1.1 10 8 HBrO 2.0 10 9 C 5 H 5 N (pyridine) 1.8 10 9 HCN 4.9 10 10 C 6 H 5 NH 2 (aniline) 4.3 10 10 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 1 H 2 He 2 3 Li 4 Be 5 B 6 C 7 N 8 O 9 F 10 Ne 3 11 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 4 19 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr 5 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 6 55 Cs 56 Ba 57 La 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 7 87 Fr 88 Ra 89 Ac 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds Molar Masses (g/mol) H 1.01 Al 26.98 C 12.01 Ar 39.95 N 14.01 P 30.97 O 16.00 S 32.06 F 19.00 Cl 35.45 Na 22.99 Cu 63.54 Ca 40.08 Ba 137.33 Page 7