Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 1

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before making your selection. The due time is Central time. PLEASE remember to bubble in your name, student ID number, and version number on the scantron! Msci 17 0631 17:06, general, multiple choice, > 1 min, fixed. 001 Which of the following equilibria is unaffected by a pressure change? 1. 2 NaCl(s) 2 Na(s) + Cl 2 (g) 2. H 2 (g) + I 2 (g) 2 HI(g) correct 3. 2 NO 2 (g) N 2 O 4 (g) 4. O 2 (g) 2 O(g) Equilibria is unaffected by pressure changes when the number of moles of gas on the product side is equal to the number of moles of gas on the reactant side. Msci 17 0615 17:06, general, multiple choice, > 1 min, fixed. 002 Suppose we let the exothermic reaction 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) come to equilibrium. Then we decrease the temperature of the reaction mixture, while holding the volume constant. What will be the effect on the net amount of SO 3 (g) present? 1. It increases. correct 2. It decreases. 3. It does not change. 4. The question cannot be answered without knowledge of the numerical value of K c or K p. 5. The question cannot be answered without knowledge of the numerical value of H. An exothermic reaction gives off heat and so decreasing temperature will favor the forward reaction, increasing [SO 3 ]. Holt da 18 1 practice 1 17:02, highschool, numeric, < 1 min, wordingvariable. 003 At equilibrium a mixture of N 2, H 2, NH 3 gas at 500 C is determined to consist of 0.602 mol/l of N 2, 0.420 mol/l of H 2, and 0.113 mol/l of NH 3. What is the equilibrium constant for the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g) at this temperature? Correct answer: 0.286294. [N 2 ] = 0.602 mol/l [NH 3 ] = 0.113 mol/l [H 2 ] = 0.42 mol/l K =? N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K = [NH 3] 2 [N 2 ][H 2 ] 3 (0.113 mol/l) 2 = (0.602 mol/l) (0.42 mol/l) 3 = 0.286294 Msci 17 0210x 17:02, general, multiple choice, < 1 min, fixed. 004 Consider the reaction A+2 B C + 2 D This reaction has an equilibrium constant of 3.7 10 3. Consider a reaction mixture with [A] = 2.0 10 2 M [C] = 2.4 10 6 M [B] = 1.7 10 4 M [D] = 3.5 10 3 M

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 2 Which of the following statements is definitely true? 1. The forward reaction can occur to a greater extent than the reverse reaction until equilibrium is established. 2. The system is at equilibrium. 3. The reverse reaction can occur to a greater extent than the forward reaction until equilibrium is established. correct 4. Heat will be evolved. 5. No conclusions about the system can be made without additional information. Q = [C] [D]2 [A] [B] 2 = (2.4 10 6 ) (0.0035) 2 (0.02) (0.00017) 2 = 0.0508651 > K Therefore the reaction will shift back to the left (reverse). Msci 17 0510 nowarning 17:05, general, multiple choice, < 1 min, normal. 005 For the reaction POCl 3 (g) POCl(g) + Cl 2 (g) K c = 0.30. An initial 0.3 moles of POCl 3 are placed in a 3 L container with initial concentrations of POCl and Cl 2 equal to zero. What is the final concentration of POCl 3? 1. final concentration = 0.079 M 2. final concentration = 0.021 M correct 3. final concentration = 0.185 M 4. final concentration = 0.285 M 5. final concentration = 0.042 M [POCl 3 ] initial = 0.3 mol 3 L = 0.1 M POCl 3(g) POCl (g) + Cl 2(g) ini, M 0.1, M x x x eq, M 0.1 x x x K c = [Cl 2] [POCl] [POCl 3 ] x 2 + 0.3 x 0.03 = 0 = x2 0.1 x = 0.3 x 2 = 0.03 0.3x x = 0.3 ± (0.3) 2 4 (1) ( 0.03) 2 (1) = 0.079 or 0.38 [POCl 3 ] = 0.1 x = 0.021 M Msci 17 0207 17:10, general, multiple choice, > 1 min, fixed. 006 Given the reversible reaction equation 2 CO(g) + O 2 (g) 2 CO 2 (g) which is the relationship between K c and K p? 1. K p = K c (R T ) 2 2. K p = K c R T 3. K p = K c 4. K p = K c (R T ) 1 correct 5. K p = K c (R T ) 2 The ideal gas law can be used to derive the link between K c and K p : P = n V R T = M R T M = P R T

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 3 or K c = [CO 2] 2 [CO] 2 [O 2 ] PCO 2 2 (R T ) 2 = PCO 2 (R T ) 2 PO 2 R T = P 2 CO 2 P 2 CO P O 2 R T = K p R T K p = K c (R T ) 1 Msci 17 1100 17:12, general, multiple choice, > 1 min,. 007 What is G for the reaction N 2 O 4 (g) + 1 2 O 2(g) N 2 O 5 (g) if each gas is present at a partial pressure of 0.1 atm and 25 C? 1. 20.03 kj/mol correct 2. 16.52 kj/mol 3. 17.46 kj/mol 4. 17.61 kj/mol 5. 17.4 kj/mol 6. 16.53 kj/mol 7. None of these Q = (P N2O 5 ) (P N2O 4 )(P O2 ) 1/2 = (0.1) (0.1) 0.1 = 3.16228 G 0 = G 0 f (N 2) 5 (g)) G 0 f (N 2O 4 (g)) 1 2 G0 f (O 2) = 115 kj/mol 97.82 kj/mol 0 = 17.18 kj/mol G = G 0 + R T ln K = 17.18 kj/mol ( ) 8.314 J + (298.15 K) ln(3.16228) mol K 1 kj 1000 J = 20.0338 kj/mol Msci 18 0313 18:03, general, multiple choice, > 1 min, fixed. 008 At 100 C, the ionization constant of water is K w = 1.0 10 13. What is the ph of pure water at 100 C? 1. 1.0 2. 6.5 correct 3. 7.0 4. 7.5 5. 13.0 2 H 2 O H 3 O + + OH K w = [H 3 O + ][OH ] = 1.0 10 13 In pure water, [H 3 O] = [OH ] = 1.0 10 13 M = 3.16228 10 7 M ph = log 3.16228 10 7 = 6.5 Mlib 66 0067 18:03, basic, multiple choice, > 1 min, fixed. 009 The solution which has the highest ph is the one with

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 4 1. 5 10 2 mol HCl in 1.0 L of solution. 2. 5 10 2 mol H 2 SO 4 in 0.5 L of solution. 3. 5 10 2 mol NaOH in 1.0 L of solution. 4. 5 10 2 mol Ba(OH) 2 in 0.5 L of solution. correct 5. All of these have the same ph. The basic solutions NaOH and Ba(OH) 2 will have ph above 7. The other solutions are acidic and will have ph lower than 7. Both NaOH and Ba(OH) 2 are strong bases which completely dissociate to give OH ions. NaOH Na + + OH [OH ] = 5 10 2 mol 1 L soln = 5 10 2 M Ba(OH) 2 Ba 2+ + 2 OH [OH ] = 2 (5 10 2 ) mol 0.5 L soln = 20 M Msci 19 5007 18:03, general, multiple choice, > 1 min, fixed. 010 What is the concentration of OH ions in a 0.40 M solution of KCN? The ionization constant of HCN is 4.0 10 10. 1. 1.60 10 10 2. 1.26 10 5 3. 6.25 10 5 4. 3.16 10 3 correct 5. 4.08 10 12 Msci 18 0418 18:04, general, multiple choice, > 1 min, fixed. 011 Given the following hypothetical acids, with pk a values: Acid pk a HP 1.92 HQ 10.33 H 2 R 7.21 H 4 Z + 9.25 H 2 X 11.62 Which of these is the strongest acid? 1. HP correct 2. HQ 3. H 2 R 4. H 4 Z 5. H 2 X Mlib 76 1460 18:99, basic, multiple choice, > 1 min, fixed. 012 Hard water deposits (calcium carbonate) have built up around your bathroom sink. Which one of the following would be best to dissolve the deposit? 1. vinegar correct 2. ammonia 3. bleach 4. lye Msci 18 0720 19:02, general, multiple choice, > 1 min, fixed. 013 Which one of the following mixtures will be a buffer when dissolved in a liter of water? 1. 0.2 mole of NaOH and 0.2 mole of HBr 2. 0.2 mole of NaCl and 0.3 mole of HCl

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 5 3. 0.4 mole of CH 3 COOH and 0.2 mole of NaOH correct 4. 0.5 mole of NH 3 and 0.5 mole of HCl 5. 0.3 mole of KOH and 0.2 mole of HBr A solution can only be buffer if an acid and conjugate base are both present or if a base and conjugate acid are both present at equilibirum. In the correct choice referring to 0.4 mole of CH 3 COOH, at first glance, we only have NaCH 3 COO and water. But there is twice as much acetic acid as there is sodium hydroxide, which means that there will also be leftover CH 3 COOH. So we do have an acid/conjugate base pair of acetic acid and sodium acetate. In the choice referring to 0.2 mole of NaOH, we only have water and NaBr at eq. In the choice referring to 0.2 mole of NaCl, we only have water, Na +, Cl and H +. In the choice referring to 0.5 mole of NH 3, we only have water, NH + 4 and Cl. In the choice referring to 0.3 mole of KOH, we only have water, KBr, and some extra K +, and OH. Msci 18 0872 19:03, general, multiple choice, > 1 min, normal. 014 What is the ph of a solution that is 0.01 M in NH 3 and 0.05 M in NH 4 Cl? K b = 1.8 10 5 for NH 3. 1. 8.56 correct 2. 9.56 3. 10.93 4. 5.05 5. 5.44 [ BH + ] poh = pk b + log B = log ( 1.8 10 5) + log 0.05 0.01 = 5.44 Thus ph = 14.0 poh = 14.0 5.44 = 8.56 Msci 18 0383x 18:03, general, multiple choice, < 1 min, fixed. 015 What is the ph of a 1.35 10 8 M HNO 3 aqueous solution? 1. 7.05 2. 6.95 correct 3. 7.00 4. 6.13 5. 7.87 6. 6.92 The H 3 O + present in solution have the following sources: Strong acid hydrolysis: HNO 3 + H 2 O H 3 O + + NO 3 [H 3 O + ] = [HNO 3 ] = 1.35 10 8 M Autoionization of water: 2 H 2 O H 3 O + + OH K w = 1.0 10 14 [H 3 O + ] = [OH ] = 1 10 7 M Total[H 3 O + ] = 1.135 10 7 M ph = log [H 3 O + ] = 6.945 DAL 19 003 19:99, general, multiple choice, < 1 min, fixed. 016

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 6 Derivation of the weak monoprotic acid equation yields a quadratic solution of the form [H + ] 2 K a = C HA [H + ]. With two assumptions this equation can be further simplified to a first order equation of the form K a = [H+ ] 2. C HA Under what conditions is this simplification most valid? 1. large C HA and large K a 2. small C HA and small K a 3. large C HA and small K a correct 4. small C HA and large K a When K a is small, the acid is very weak and dissociates very little at equilibrium, resulting in a fairly small H + concentration. When C HA is large, C HA H + C HA. DAL 19 004 19:99, general, multiple choice, < 1 min, normal. 017 A buffer is made by mixing 100 ml of 0.2 M NH 4 Cl with 200 ml of 0.6 M NH 3. How much NaOH can be added before the capacity of this buffer is reached? 1. 0.02 mol correct 2. 0.12 mol 3. 0.1 mol 4. 0.14 mol 5. This is a basic buffer so NaOH cannot be neutralized. NaOH (a strong base) will react with NH 4 Cl (a weak acid) control the ph. Only when all NH 4 Cl is used up is the capacity of the buffer reached. This occurs when (0.2 M)(0.1 L) = 0.02 mol NaOH has been added. Msci 18 0421 18:04, general, multiple choice, > 1 min, fixed. 018 We make a 1.0 M solution of an unknown acid HX. With a ph meter, we determine that the ph of the solution is 4.00. Which of the following statements about the acid HX is true? 1. HX is a strong acid. 2. HX is a weak acid with a K a value of about 10 4. 3. HX is a weak acid with a K a value of about 10 8. correct 4. HX is a weak acid with a K a value of about 10 10. 5. HX would probably be a good acid-base indicator. Msci 19 0618 19:08, general, multiple choice, > 1 min, fixed. 019 Calculate the ph of the solution resulting from the addition of 30.0 ml of 0.200 M HClO 4 to 60.0 ml of 0.150 M NaOH. 1. 12.52 correct 2. 11.88 3. 7.00 4. 1.48 5. 13.06 Here it s important to find out which of these two species (HClO 4 and NaOH) is in excess. The one that is in excess will determine the ph of this solution. From the formulas of

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 7 the two compounds, you can expect that they will react in a one-to-one fashion. So our first order of business will be to determine how many moles of each compound we have. For HClO 4, we have ( 1 L )( 0.200 mol ) 30.0 ml 1000 ml 1 L = 0.006 mol HClO 4 Likewise, for NaOH, we have ( 1 L )( 0.150 mol ) 60.0 ml 1000 ml 1 L = 0.009 mol NaOH So when HClO 4 and NaOH react, all of the HClO 4 will be consumed (it s the limiting reagent) and 0.009 mol 0.006 mol = 0.003 mol will remain. This 0.00300 mol excess of NaOH will determine the ph of this solution. The solution now is 30.0 ml + 60.0 ml = 90 ml and, since NaOH is a strong base (i.e., it s completely dissociated), it contains 0.003 mol OH. [OH ] is then [OH 0.003 mol ] = 0.09 L = 0.0333333 M which means that the poh of this solution is poh = log[oh ] = log(0.0333333) = 1.47712 However, we wanted ph. We can use the equation that relates ph to poh to get ph ph + poh = 14 ph + 1.47712 = 14 ph = 12.5229 Msci 19 0004 19:08, general, multiple choice, > 1 min, fixed. 020 A 100 ml sample of 0.100 M NH 3 solution is titrated to the equivalence point with 50 ml of 0.200 M HCl. What is the final [H 3 O + ]? The ionization constant of NH 3 is 1.8 10 5. 1. 6.09 10 6 M correct 2. 8.61 10 6 M 3. 1.10 10 3 M 4. 3.70 10 11 M 5. 1.00 10 7 M Neutralization: NH 3 0.1 M + HCl 0.2 M NH 4 Cl 100 ml 50 ml 10 mmol 10 mmol 10 mmol 10 mmol 10 mmol 0 mmol 0 mmol 10 mmol 10 mmol 150 ml = 0.067 M Equilibria re-established: K a NH 4 NH 3 + H+ 0.067 M 0.067 x x x K a = K w K b = K a = x2 C a x Assumption: 10 14 = 5.56 10 10 1.8 10 5 x = [H + ] = K a C a = (5.56 10 10 )(0.067) = 6.09 10 6 Msci 19 0734 19:08, general, multiple choice, > 1 min, fixed. 021 A 100 ml portion of 0.300 M acetic acid is being titrated with 0.200 M NaOH solution.

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 8 What is the [H + ] of the solution after 50.0 ml of the NaOH solution has been added? The ionization constant of acetic acid is 1.8 10 5. 1. 3.63 10 5 correct 2. 8.95 10 6 3. 1.21 10 5 4. 9.94 10 6 5. 6.01 10 4 Msci 19 0713 19:08, general, multiple choice, > 1 min, fixed. 022 The ph at the equivalence point of a weak acid-strong base titration is? because?. 1. greater than 7; of the hydrolysis of the conjugate base of the weak acid correct 2. greater than 7; of the presence of excess strong base 3. greater than 7; the strong base produces a greater ph change above 7 than a weak acid does below 7 4. equal to 7; the ph is 7 for all acid-base titrations 5. less than 7; of the use of the wrong indicator At the equivalence point the acid and base are used up and the conjugate base of the weak base that is produced hydrolyzes, producing OH, reusing the ph. titration: hydrolysis: HA + OH A + H 2 O A + H 2 O HA + OH Msci 19 0012 19:99, general, multiple choice, > 1 min, fixed. 023 Methyl orange is an indicator with a K a of 10 4. Its acid form, HIn, is red, while its base form, In-, is yellow. At ph 6.0, the indicator will be: 1. yellow correct 2. red 3. orange 4. blue 5. colorless Msci 18 0897 19:03, general, multiple choice, > 1 min, fixed. 024 A buffer was prepared by mixing 0.200 mole of ammonia (K b = 1.8 10 5 ) and 0.200 mole of ammonium chloride to form an aqueous solution with a total volume of. To 250 ml of this solution was added 50.0 ml of 1.00 M HCl. What is the ph of this solution? 1. 9.35 2. 9.73 3. 8.78 correct 4. 8.53 5. 8.38 6. 8.18 7. 7.87

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 9 0.2 mol [NH 3 ] = [NH + 0.2 mol 4 ] = 1.0 mol [HCl] = 1000 ml [Cl 0.2 mol ] = K b = 1.8 10 5 ph =? Initiall condition (ini): 0.2 mol [NH 3 ] = 250 ml = 100 mmol 1.0 mol [HCl] = 50 ml = 50 mmol 1000 ml [NH + 0.2 mol 4 ] = 250 ml = 100 mmol [Cl 0.2 mol ] = 250 ml = 100 mmol NH 3 + HCl NH + 4 + Cl ini, mmol 100 50 100 100, mmol 50 50 50 50 fin, mmol 50 0 150 150 Cl is a spectator ion. NH + 4 /NH 3 is a buffer system. ph = pk a + log [NH 3] [ NH + 4 ] ( 1.0 10 14 ) = log 1.8 10 5 = 8.77815 ( ) 50 + log 150 Msci 19 0722 19:08, general, multiple choice, > 1 min, fixed. 025 How many endpoints would be observed in a titration of the triprotic acid H 3 A? 1. 3 correct 2. 2 3. 1 4. None of these 5. 4 H 3 A H + + H 2 A H 2 A H + + HA 2 HA H + + A 3 These three dissociation equations show that three endpoints will be seen. Msci 20 0004 20:02, general, multiple choice, > 1 min, fixed. 026 Which of the following compounds has the LOWEST molar solubility in water? 1. CdS, K sp = 1.0 10 28 correct 2. Al(OH) 3, K sp = 2 10 32 3. PbSO 4, K sp = 1.3 10 8 4. Sn(OH) 2, K sp = 3 10 27 5. MgC 2 O 4, K sp = 8.6 10 5 Ksp from x 20:02, general, multiple choice, > 1 min, wording-variable. 027 The compound MX 3 has a molar solubility of 0.00255 M. What is the value of K sp for MX 3? 1. 1.14 10 9 correct 2. 3.81 10 10 3. 4.48 10 7 4. 4.23 10 11 5. 6.63 10 8 MX 3 M 3+ + 3 X 1 K sp = (x) (3x) 3 = 27x 4 = 3 (0.00255) 4 = 1.14 10 9.

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 10 Msci 20 0002 20:02, general, multiple choice, > 1 min, fixed. 028 Assume the molar solubility of silver chromate, Ag 2 CrO 4, is represented as x. Which of the following expressions correctly expresses the relationship between the molar solubility of silver chromate and the solubility product constant (K sp ) for this compound? 1. K sp = 4 x 3 correct 2. K sp = 2 x 3 3. K sp = x 2 4. K sp = 4 x 2 5. K sp = 2 x 2 The equation for the dissociation of Ag 2 CrO 4 is written as Ag 2 CrO 4 2 Ag + + CrO 4. The expression for K sp = [ Ag +] 2 [ CrO 4 ]. Since the molar solubility is x, K sp can be rewritten (in terms of x) as K sp = [2 x] 2 [x] = 4 x 3. [ Ag + ] = 2 x because for every salt molecule that dissociates, 2 Ag + ions are produced. Msci 20 0604 20:99, general, multiple choice, > 1 min, fixed. 029 AgCl would be least soluble in 1. pure water 2. 0.1 M CaCl 2 correct 3. 0.1 M HCl 4. 0.1 M HNO 3 5. 0.1 M NH 3 Msci 20 0403 20:04, general, multiple choice, > 1 min, fixed. 030 A solution is 0.01 M BaCl 2 and 0.02 M SrCl 2. Which cation can be selectively precipitated first with a concentrated Na 2 SO 4 solution? Compound K sp BaSO 4 1.5 10 9 SrSO 4 7.6 10 7. 1. Ba +2 correct 2. Sr +2 3. Both will precipitate at the same time. Before addition of Na 2 SO 4... [Ba 2+ ] = 0.01 M [Sr 2+ ] = 0.02 M [SO 2+ 4 ] = 0.02 M [Cl ] = not important There is a certain magic concentration of SO 2+ 4 that will cause each of these cations to form insoluble compounds (or precipitates). We just have to figure out which magic concentration is the smaller (since we ll get to that one first if we re increaing SO 4 concentration from zero). SrSO 4 Sr 2+ + SO 2 4 K sp = 7.6 10 7 = [Sr 2+ ][SO 2 4 ] = (0.02 M)[SO 2 4 ] Solve for magic [SO 2 4 ] = 3.8 10 5 BaSO 4 Ba 2+ + SO 2 4 K sp = 1.5 10 9 = [Ba 2+ ][SO 2 4 ] = (0.01 M)[SO 2 4 ] Solve for magic [SO 2 4 ] = 1.5 10 8 The concentration of sulfate ion that will cause BaSO 4 to precipitate is smaller than

Create assignment, 48975, Exam 2, Apr 05 at 9:07 am 11 the concentration that will cause SrSO 4 to precipitate. Therefore, Ba +2 will be precipitated first.