CHEM 10123/10125, Exam 2

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1 CHEM 10123/10125, Exam 2 March 7, 2012 (50 minutes) Name (please print) Please box your answers, and remember that significant figures, phases (for chemical equations), and units do count! 1. (13 points) SHOW ALL WORK. Ascorbic acid, H 2 A, has K a1 = 8.0 x 10 5 and K a2 = 1.6 x Determine the following values for a 2.5 M solution of this acid. Briefly show your work in the space below. (Reactions need not be shown here) a. [HA ] 1.4 x 10 2 (x 2 ) / (2.5) = 8.0 x10 5 b. ph 1.85 = log (1.4 x10 2 ) c. [A 2 ] 1.6 x10 12 M 2. (16 points) Briefly define the following terms: Lewis base electron pair donor Indicator a substance whose color depends on the ph Arrhenius base a substance that produces OH ions in aqueous solution Buffer range-- the ph values for which a buffer system is the most effective. (usually ±1 ph unit of either side of pk a ) 3. (8 points) For the amphoteric molecule HCO 3, write one equation showing it acting as an acid, and another equation showing it acting as a base (label which is which). Acid: HCO 3 (aq) H + (aq) + CO 3 2 (aq) Base: HCO 3 (aq) + H 2 O(l ) H 2 CO 3(aq) + OH (aq) 4. (5 points) Among the following, circle the weakest Bronsted-Lowry acid. HClO HClO 2 HClO 3 HBrO HBrO 2 HBrO 3 HBrO 5. (5 points) Among the following, circle the strongest Bronsted-Lowry base. HS HO HSe Br F Cl HO 1

2 6. (15 points) Arrange the following M solutions in order of increasing ph: NH 3, HNO 3, NaNO 2, HC 2 H 3 O 2, NaOH, NH 4 C 2 H 3 O 2, NH 4 ClO 4 HNO 3 <HC 2 H 3 O 2 <NH 4 ClO 4 <NH 4 C 2 H 3 O 2 <NH 3 <NaNO 2 <NaOH 7. (13 points) For each reaction, identify the Lewis acid and Lewis base, then predict the reaction product. Draw the complete and correct Lewis structure of the product for reaction (a) only. OH (aq) + CO 2(aq) HCO 3 (aq) base acid Cr 3+ (aq) + 6 H 2 O (l) [Cr(H 2 O) 6 ] 3+ (aq) acid base Al(OH) 3(aq) + OH (aq) Al(OH) 4 (aq) acid.base 8. SHOW ALL WORK. A ml sample of M HF is titrated with M NaOH. The initial ph is Calculate the ph when the neutralization is a) (12 points) 25% complete [note we used K a = 3.0 x 10 8 instead of HF K a!] ml x (0.150 mol HF)/(1000 ml HF soln) = mol HF When it is 25% complete, that means that (0.25)( mol) = mol of NaOH have been added, and = mol of HF remain. It is a buffer system, with K a = 3.0 x x 10 8 = [H + ][F ]/[HF] = [H + ]( )/( ) [H + ] = 9.0 x 10 8 M ph = 7.05 b) (5 points) 50% complete Buffer system, still, and if it s 50% complete then [HF] = [F ], so K a = [H + ] = 3.0 x 10 8 ph = 7.52 c) (20 points) 100% complete At this point, mol NaOH have been added and all of the acid has been neutralized to form mol F-. It is a solution of a weak base. The chemical equation is F (aq) + H 2 O (l) HF (aq) + OH (aq) To get the total volume, mol OH x (1000 ml NaOH soln)/(0.250 mol NaOH) = 12.0 ml NaOH soln 12.0 ml ml = L soln ( mol F )/( L soln) = M F 2

3 For the solution of a weak base we will need K b = (1.0 x )/(3.0 x 10 8 ) = x 10 7 F HF OH- I M 0 0 C -x +x +x E x M x x x 10 7 = [HF][OH ]/[F ] (x 2 )/( ) (assume x is small; clearly safe in this case) x = 1.77 x 10 4 M OH-, so poh = 3.75, and ph = (12 points) Fill in the blanks. [H + ] [OH ] ph poh 4.5 x 10 5 M 2.2 x x x (6 points) Coal can be used to generate hydrogen gas ( a potential fuel) by the endothermic reaction: C (s) + H 2 O (g) CO (g) + H 2(g) If this reaction is at equilibrium, predict whether each disturbance will result in the formation of additional hydrogen gas, the formation of less hydrogen gas, or have no effect on the quantity of hydrogen gas. a.) adding more C to the reaction no change b.) adding more H 2 O to the reaction c.) raising the temperature of the reaction d.) increasing the volume of the reaction e.) adding a catalyst to the reaction f.) adding an inert gas to the reaction favors H 2 favors H 2 favors H 2 no change except speed no change 11. (20 points) A solution is formed by mixing ml of M KOH with ml of a solution that is both M in CH 3 COOH and M in HI. Identify in the final solution a) all the solute species present b) the two most abundant solute species, and c) the two least abundant solute species ml of M KOH = mol KOH 100 ml of M HC 2 H 3 O 2 = mol HC 2 H 3 O ml of mol HI = mol HI Take the solute ions one at a time: The K + is a spectator ion and will not react. There are mol K +. 3

4 The OH starts at mol, but reacts with the mol H + formed from the HI. This leaves mol OH. The mol OH then reacts to neutralize the HC 2 H 3 O 2, converting mol of HC 2 H 3 O2 to C 2 H 3 O 2. Therefore, we now have mol HC 2 H 3 O 2 left, and the mol C2H 3 O 2 that was formed. The HI dissociated when dissolved into H + and I. The H + was neutralized by the OH as described above, leaving mol I. a) K +, OH, H +, HC 2 H 3 O 2, C 2 H 3 O 2, I b) K +, C 2 H 3 O 2 - c) H +, OH CHEM 10123/10125, Exam 2 March 7, 2012 (50 minutes) Name (please print) Please box your answers, and remember that significant figures, phases (for chemical equations), and units do count! 1. (13 points) SHOW ALL WORK. Carbonic acid, H 2 A, has K a1 = 4.4 x 10 7 and K a2 = 4.7 x Determine the following values for a 2.5 M solution of this acid. Briefly show your work in the space below. (Reactions need not be shown here) a. [HA ] 1.0 x10 3 M (x 2 ) / (2.5) = 4.4 x 10 7 b. ph 2.98 = log (1.0 x10 3 ) c. [C 6 H 6 O 6 2 ] 4.7 x10 11 M 2. (16 points) Briefly define the following terms: Bronsted acid proton donor Equivalence point In a titration, when number of moles of base is stoichiometrically equal to the number of moles of acid Buffer capacity is the amount of acid or base you can add without causing a large change in ph Lewis acid electron pair acceptor 3. (8 points) For the amphoteric molecule HS, write one equation showing it acting as an acid, and another equation showing it acting as a base (label which is which). 4

5 Acid: HS (aq) H + (aq) + S 2 (aq) Base: HS (aq) + H 2 O (l) H 2 S (g) + OH (aq) 4. (5 points) Among the following, circle the strongest Bronsted-Lowry base. ClO 3 ClO 4 ClO 2 - BrO 2 BrO 4 BrO 3 BrO 2 5. (5 points) Among the following, circle the weakest Bronsted-Lowry acid. CH 4 H 2 S SiH 4 PH 3 H 2 O NH 3 CH 4 6. (15 points) Arrange the following M solutions in order of decreasing ph: NH 3, HNO 3, NaNO 2, HC 2 H 3 O 2, NaOH, NH 4 C 2 H 3 O 2, NH 4 ClO 4 NaOH>NaNO 2 >NH 3 >NH 4 C 2 H 3 O 2 >NH 4 ClO 4 >HC 2 H 3 O 2 >HNO 3 7. (13 points) For each reaction, identify the Lewis acid and Lewis base, then predict the reaction product. Draw the complete and correct Lewis structure of the product for reaction (a) only. SO 2(aq) + OH (aq) HSO 3 (aq) acid.base B(OH) 3(aq) + OH (aq) B(OH) 4 (aq) acid base Al 3+ (aq) + 6 H 2 O (l) [Al(H 2 O) 6 ] 3+ (aq) acid base 8. SHOW ALL WORK. A ml sample of M propanoic acid (HC 3 H 5 O 2 ) is titrated with M KOH. The initial ph is Calculate the ph when the neutralization is a) (12 points) 25% complete ml x (0.165 mol HC 3 H 5 O 2 )/(1000 ml HC 3 H 5 O 2 soln) = mol HC 3 H 5 O 2 When it is 25% complete, that means that (0.25)( mol) = mol of KOH have been added, and = mol of HC 3 H 5 O 2 remain. It is a buffer system, with K a = 1.3 x x 10 5 = [H + ][C 3 H 5 O 2 ]/[HC 3 H 5 O 2 ] = [H + ]( )/( ) [H + ] = M ph =

6 b) (5 points) 50% complete Buffer system, still, and if it s 50% complete then [HC 3 H 5 O 2 ] = [C 3 H 5 O 2 ], so K a = [H + ] = 1.3 x 10 5 ph = 4.89 c) (20 points) 100% complete At this point, mol KOH have been added and all of the acid has been neutralized to form mol C 3 H 5 O 2 -. It is a solution of a weak base. The chemical equation is C 3 H 5 O 2 (aq) + H 2 O (l) HC 3 H 5 O 2 (aq) + OH (aq) To get the total volume, mol OH x (1000 ml KOH soln)/(0.300 mol KOH) = 16.5 ml KOH soln 16.5 ml ml = L soln ( mol C 3 H 5 O 2 )/( L soln) = M C 3 H 5 O 2 For the solution of a weak base we need K b = (1.0 x )/(1.3 x 10 5 ) = x C 3 H 5 O 2 HC 3 H 5 O 2 OH- I M 0 0 C -x +x +x E x M x x x = [HC 3 H 5 O 2 ][OH ]/[C 3 H 5 O 2 ] (x 2 )/(0.106) (assume x is small; clearly safe in this case) x = 9.03 x 10 6 M OH-, so poh = 5.04, and ph = (12 points) Fill in the blanks. [H + ] [OH ] ph poh M 6.7 x x x (6 points) Coal can be used to generate hydrogen gas ( a potential fuel) by the endothermic reaction: C (s) + H 2 O (g) CO (g) + H 2(g) If this reaction is at equilibrium, predict whether each disturbance will result in the formation of additional hydrogen gas, the formation of less hydrogen gas, or have no effect on the quantity of hydrogen gas. a.) raising the temperature of the reaction favors H 2 b.) removing some H 2 O from the reaction less H 2 c.) removing some of the C from the no change reaction d.) adding an inert gas to the reaction no change e.) adding a catalyst to the reaction no change except rate 6

7 f.) decreasing the volume of the reaction less H (20 points) A solution is formed by mixing ml of M KOH with ml of a solution that is both M in CH 3 COOH and M in HI. Identify in the final solution a) all the solute species present b) the two most abundant solute species, and c) the two least abundant solute species ml of M KOH = mol KOH 100 ml of M HC 2 H 3 O 2 = mol HC 2 H 3 O ml of mol HI = mol HI Take the solute ions one at a time: The K + is a spectator ion and will not react. There are mol K +. The OH starts at mol, but reacts with the mol H + formed from the HI. This leaves mol OH. The mol OH then reacts to neutralize the HC 2 H 3 O 2, converting mol of HC 2 H 3 O2 to C 2 H 3 O 2. Therefore, we now have mol HC 2 H 3 O 2 left, and the mol C2H 3 O 2 that was formed. The HI dissociated when dissolved into H + and I. The H + was neutralized by the OH as described above, leaving mol I. Finally, the autoionization of water happens in all aqueous solutions, though to a small extent only. Therefore both H + and OH are present in very small concentrations. (you don t need to calculate them, but if you want to, you can use the buffer equation to get the [H + ], followed by K w to get [OH ].) This means the answers are: a) K +, OH, H +, HC 2 H 3 O 2, C 2 H 3 O 2, I b) K +, C 2 H 3 O 2 - c) H +, OH 7

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