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3 Pages

### Chapter 14

Course: CHM 435, Fall 2009
School: UNC Wilmington
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Word Count: 696

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14 Chapter 1. This is a two point standard addition, see p. 17 or p. 376 for formula and p. 17 for derivation of formula Cunk = S1C sVs 0.656 25.7 ppm 10.00 mL 168.592 = = ppm = 21.0 7 = 21.1 ppm ( S 2 - S1 )Vx ( 0.976 - 0.656) 25.0 mL 8.00 (lacking other error analysis information, answer should contain 3 sf's using general chemistry level sf rules) 2. This is another 2 point standard addition example with...

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14 Chapter 1. This is a two point standard addition, see p. 17 or p. 376 for formula and p. 17 for derivation of formula Cunk = S1C sVs 0.656 25.7 ppm 10.00 mL 168.592 = = ppm = 21.0 7 = 21.1 ppm ( S 2 - S1 )Vx ( 0.976 - 0.656) 25.0 mL 8.00 (lacking other error analysis information, answer should contain 3 sf's using general chemistry level sf rules) 2. This is another 2 point standard addition example with added step to determine concentration of analyte in a solid sample after determining its concentration in a solution containing the sample: C unk = S 1 C sV s 0.723 2.75 ppm 1.00 mL 1.988 = = ppm = 2.04 9 ppm ( S 2 - S1 )Vx ( 0.917 - 0.723) 5.00 mL 0.970 2.04 9 mg Cu 0.200 L 1g L 1000 mg 4.098 10 - 4 % Cu in sample = 100 = 100 = 0.0684 3 % 0.5990 g 0.5990 5. At 465 nm Bi(tu)63+ complex absorbs but the BiY- and Bi(III) do not (typo in line 3 of question reads "the Bi(III) or thiourea complex" but it should read "the Bi(III) thiourea complex") A decreases as Bi(tu)63+ is used up and BiY- is formed A levels off when titration is complete because there is no more Bi(tu)63+ mL titrant 8. a) Using equations from p. 179 of lab text (which are derived from simultaneous equations on p.377 of lecture text and p. 178 of lab text): C Co 2 + = (10.2 0.426) - ( 3228 0.026) = 4.3452 - 83.928 ( 3529 10.2) - ( 3228 428.9) 3.5995 10 4 - 1.3844 10 6 = - 79.582 = 5.90 10 -5 6 - 1.3484 10 C Ni 2 + = ( 3529 0.026) - ( 428.9 0.426) = 91.754 - 182.711 ( 3529 10.2) - ( 3228 428.9) 3.5995 10 4 - 1.3844 10 6 = - 90.957 = 6.74 10 -5 6 - 1.3484 10 10. a) A = bC , = A / bC = A / (1.00 cm)(5.00 x 10-4 M) (assuming no appreciable background signal but would have been better to determine from a calibration curve)) In HIn 485 nm 150 974 625 nm 1,808 362 Reminder: In 0.1 M NaOH, all HIn would be in In- form and in 0.1 M HCl all HIn would be in the HIn form. b) C HIn = (1808 0.567 ) - (150 0.395) ( 974 1808) - (150 362) = 1.025 10 3 - 59.25 965.88 = = 5.659 10 - 4 M 6 4 1.76099 10 - 5.430 10 1.70669 10 6 C In - = ( 974 0.395) - ( 362 0.567) ( 974 1808 ) - (150 362) = 384.73 - 205.25 179.47 = = 1.051 10 - 4 M 6 4 6 1.76099 10 - 5.430 10 1.70669 10 Ka expression is HIn + H2O H3O+ + In- Ka = [H3O+][In-] / [HIn] CIn- and CHIn solved for above are equilibrium concentrations needed for In- and HIn. The pH gives equilibrium M of H3O+. If pH = 5.00 then [H3O+] = x 1.0 10-5. Ka = (1.051 x 10-4)(1.0 x 10-5) / (5.659 x 10-4) = 1.85 x10-6 c) C HIn = C In - = (1808 0.492) - (150 0.245 ) ( 974 1808 ) - (150 362) = 889.53 - 36.75 852.786 = = 4.996 10 - 4 M 6 4 6 1.76099 10 - 5.430 10 1.70669 10 = 60.52 = 3.546 10 -5 M 6 1.70669 10 ( 974 0.245 ) - ( 362 0.492) = 238.63 - 178.10 ( 974 1808 ) - (150 362) 1.76099 10 6 - 5.430 10 4 CIn- and CHIn solved for above are equilibrium concentrations needed for In- and HIn. Because we have the Ka from part b we can now solve for pH using the equilibrium expression. Ka=(3.546 x 10-5)[H+])/(4.996 x 10-4)=1.85 x10-6; [H+]=(1.85 x10-6 x 4.996 x 10-4)/3.546 x 10-5 = 2.606 x10-5 pH = -log(2.606 x10-5) = 4.584 Note: (There is a typo in part c. It should have 625 nm instead of 635 nm). Parts b and c were assigned to show you neat (and useful) applications of simultaneous analysis. 11. (b) See spreadsheet in excel answer section. Note: Dilution from 25 to 50 mL is taken into account because all standards and unknowns treated same way so concentrations can be listed as those in 25 mL. 12. (b) For M = 1: C = 17.117 0.092 (or 17.12 0.09) %RSD = 0.53 or 0.5 For M = 3: C = 17.117 0.061 (or 17.12 0.06) %RSD = 0.35 or 0.4 Note: Assigned to remind you that measuring the unknown more than once reduces sc 20. The Kf for FeY- is 1.6 x 106 times larger than Kf for CuY2-. If a solution which contained Fe3+ and Cu2+ was titrated with Y4-, Cu2+ would not react until after Fe3+...

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