Exam 2 Answer Key - CHEM 3600 EXAM 2 Name Read through the...

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Unformatted text preview: CHEM 3600 EXAM 2 11/10/2003 Name:_____________________________ Read through the entire exam before starting. Make sure to get the easy ones done first. Remember to show all your work for full credit; there is extra room on p.4 if you need it. Possibly useful equations and numbers A- G = -nFE pH = 1 / 2(pK 1 + pK 2 ) pH = pK a + log [HA] [B] E = E - E E = 0.05916 V log K F = 9.649 10 4 C/mol e 0.05916 V log E = E + - n [A] n [ ] E 0.799 Ag+ + e- Ag(s) 0.222 AgCl(s) + e Ag(s) + Cl 3+ + 1.41 Au + 2e Au + 1.69 Au + e Au(s) -0.828 H2O + e H2(g) + OH 0.269 IO3 + 3H2O + 6e I + 6OH + 2 1.589 IO4 +2H + 2e IO3 + H O + 1.210 IO3 + 6H + 5e I2(s) + 3H2O 0.620 I2(aq) + 2e- 2I0.535 I3 + 2e 3I 1) (20 points) A 500 mg supplement tablet containing only Vitamin A (C18H30O, MW=262.43), Vitamin C (C6H8O6, MW=176.124), and Zinc (AW=65.39) in the form of sulfite (ZnSO3, MW=145.45) was analyzed by combustion for carbon and precipitation for zinc. Upon combustion, the tablet produced 1.194 g of CO2 (MW=44.01). The residue left from the combustion was dissolved and zinc was precipitated as zinc sulfide (ZnS, MW=97.45), and weighed 16.48 mg. What is the wt % of Vitamin A, Vitamin C, and Zinc in the tablet? Reaction 16.48mg ZnS 97.45 g/mol ZnS = 0.169 mol ZnS = mol Zn = mol ZnSO 3 145.45 g/mol = 24.60 mg ZnSO 3 500mg tablet - 24.60 mg ZnSO3 = 475.40 mg (Vitamin A + Vitamin C) x = mass Vitamin A; mass Vitamin C = 0.47540 - x 1.194 g CO 2 x .4754 - x 18 + 6 = 262.43 176.124 44.01 g/mol CO 2 3170.23x 748.555 - 1574.58 x + = 0.02713 mol CO 2 = 0.02713 mol C 46220.22 46220.22 1595.65 x + 748.555 = 1253.964 x = .3167 g = mass Vitamin A mass Vitamin C = .4754 - x = .1587 g wt % Vitamin A = 0.3167/0.500 *100 = 63.3% wt % Vitamin C = 0.1587 /0.500 *100 = 31.7% 65.39 wt % Zn = 0.02460 / 0.500 * 100 = 2.21% 145.45 2) (20 points total) A solution at equilibrium contains 0.100 M IO3-, 1x10-4 M I-, 1x10-5 M OH-, 1x10-3 M Au3+, and 0.100 M Au+. a) Write a balanced net reaction that can occur between the species in this solution (5) Au3+ + 2e- Au+ 1.41 V 0.269 V IO3 + 3H2O + 6e I + 6OH 3Au3+ + I- + 6OH- 3Au+ + IO3- + 3H2O 1.14 V b) Calculate K for the reaction (5) G = -6 FE = -660kJ K = 10 6 E / .05196 = 4 10131 Note: because I said "at equilibrium" in the problem, this is an acceptable answer: K= [Au ] [IO ] [Au ] [I ][OH ] + 3 - 3 3+ 3 - - 6 = (0.100) 3 (0.100) = 1 10 39 -3 3 -4 -5 6 (1 10 ) (110 )(1 10 ) c) Calculate E for the conditions given (5) Because I made the mistake of saying "at equilibrium" without giving you any numbers for temperature, etc., E would be equal to E If I had not said that, the answer would be this: + 3 IO -3 1.41 - 0.05916 log Au - 0.269 - 0.05196 log E= 3 6 6 Au 3+ I - OH - [ [ ] [ ] [ ][ ] = 1.37 V 6 d) Write the line notation for the cell that can be made with two Pt electrodes and the two half-reactions here (5) Pt(s) | Au 3+ , Au + || I - , OH - , IO3 | Pt(s) 3) (20 points total) a) Find the pH of a solution prepared by dissolving 27.75 g of ethylamine (CH3CH2NH2, MW=45.0837) and 16.73 g of ethylammonium chloride (CH3CH2NH3+Cl-,MW=81.5444) in 1.50 L of water. The pKa for ethylammonium is 10.636. (10) 27.75 g EtAm 45.0837 g/mol EtAm = 0.6155 mol EtAm 16.73 g EtAm + 81.5444 g/mol EtAm + = 0.2052 mol EtAm + 0.2052 b) If you added 15.72 mL of 1.000 M HClO4 is added to the solution, what is the pH? (10) 0.01572 L HClO 4 1.00 M = 0.01572 mol HClO 4 = 0.01572 mol H + mol EtAm = 0.6155 mol - 0.01572 mol = 0.5998 mol EtAm mol EtAm + = 0.2052 + 0.01572 mol = 0.2209 mol EtAm + pH = pK a [A ] = 10.636 + log 0.6155 = 11.11 + log - [HA] pH = pK a + log [HA] [A ] = 10.636 + log 0.5998 = 11.07 - 0.2209 4) (30 points total) A chemist dissolves barium oxalate (BaC2O4) in water. In solution, oxalate acts as a base and can accept two protons to form oxalic acid. In addition, barium can form a complex neutral species in solution with oxalate as well as one complex ion in solution with hydroxide. Use the systematic approach to setup equations to evaluate this system. a) Write the chemical reactions involved in this system (10) BaC2 O 4 ( s ) Ba 2+ (aq) + C 2 O 2- (aq) 4 2- - C 2 O 4 (aq) + H 2 O HC 2 O 4 (aq) + OH - (aq) HC 2 O - (aq) + H 2 O H 2 C 2 O 4 (aq) + OH - (aq) 4 Ba 2+ (aq) + C 2 O 2- (aq) BaC2 O 4 (aq) 4 2+ Ba (aq) + OH - (aq) BaOH + (aq) H 2 O H +(aq) + OH -(aq) b) Write the charge balance equation for this system (5) 2 Ba 2+ + BaOH + + H + = 2 C 2 O 2- + HC 2 O - + OH - 4 4 c) Write the mass balance equation(s) for this system (5) [ [ [ ] [ + 2- 4 [ [ [Ba ] + [BaOH ] = [C O ] + [HC O ] + [H C O ] 2+ 2 2 - 4 2 2 4 ([BaC2O4](aq) cancels as it's on both sides of the equation) d) Setup the equilibrium expressions for this solution (5) K sp = Ba 2+ C 2 O 2- 4 [ [ K b1 = [HC O ] [OH ] [C O ] 2 - 4 2 2- 4 + K b2 = [H 2 C 2 O 4 ] [OH - ] [HC O ] 2 - 4 1(BaC O ) = 2 4 [Ba ][C O ] 2+ 2 2- 4 [BaC 2 O 4 ] 1(BaOH ) = [BaOH ] [Ba ][OH ] + 2+ - K W = H + OH - [ ][ e) Count the equations and species. Can we solve for the concentrations of all species? (5) We have 8 unknowns ([Ba2+], [C2O42-], [HC2O4-], [H2C2O4], [BaC2O4](aq), [BaOH+], [H+], and [OH-]) and 8 equations (6 equilibria, charge balance, mass balance) so yes we can solve it! 5) (10 points total) Define/explain the following terms: a) Chelating ligand A ligand that can bind to an atom through more than atom (or a multidentate ligand) b) Buffer A solution made from weak acid and conjugate weak base that is resistant to change in pH c) Galvanic cell A voltaic cell (redox reaction) that uses a spontaneous chemical reaction to generate electricity d) Ion-selective electrode Electrodes which respond selectively to one ion by using a thin membrane which allows only that ion to pass between two reference electrodes based on the same reaction e) S.C.E. Saturated calomel electrode; A reference electrode based on the mercury-mercury chloride redox reaction ...
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This note was uploaded on 12/10/2008 for the course CHEMISTRY CHEM 3600 taught by Professor Philipj.silva during the Fall '07 term at Uni. Tartu.

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