MT1-Sp11-key

MT1-Sp11-key - Name________________________...

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Unformatted text preview: Name________________________ GSI________________ Chemistry 4B, Spring 2011 February 7, 2011 (50 min., closed book) Name: KEY GSI: SID: Section: Write your name and that of your GSI on all 7 pages of the exam. Equations, constants, and data are provided below. In order to maximize your score on the exam: Do the questions you know how to do first. Circle your final answer. Go back and spend more time on the questions you find more challenging. Budget your time carefully ‐ don't spend too much time on one problem. Show all work for which you want credit and don't forget to include units. For the explanation sections, try keep your answers brief as well as complete. Topic Question Page Points Score Information 1‐2 IEC of Cr species Q1‐7 4 24 EDTA of Cr(III) Q8‐13 5 26 AAS of Cr Q14‐16 6 16 UV/Vis Q17‐19 7 14 Total 80 Useful Equations and Constants: pX = ‐ log X N0 = 6.02214 x 1023 mol‐1 kb = 1.380 × 10‐23 J K‐1 % % % h = 6.62607 × 10‐34 J s % me = 9.101939 × 10‐31 kg c = 2.99792 × 108 m s‐1 E = h 1 nm = 10‐9 m = c ∆ = 10‐6 Page 1 of 7 Name________________________ GSI________________ Hazards of Chromium Ions Two common oxidation states of chromium are Cr3+ and Cr6+. Safety information for chromium compounds are shown below. Cr2O3 Formula H2CrO4 Name chromic acid Chromium (III) oxide Health 4 2 Flammability 0 0 Reactivity 3 1 Contact 4 2 LD50* (rats) 80 mg/kg None measured Chromium (III) and (VI) compounds are used in many manufacturing process including chrome plating, dyes for paint, leather tanning and wood preservation. If groundwater become contaminated it is useful to determine how much of each species is present. Chromium (VI) causes cancer in humans. *Lethal Dose 50%, LD50 is the amount of material that will kill half the tested population. Ion Exchange Chromatography of Chromium Ions In an experiment analogous to the one you performed in lab, a mixture of Cr3+ and Cr6+ was separated using ion exchange chromatography. The ion exchange resin pictured in the schematic below is a cation exchange resin. Metal cations compete with H+ ions on the resin. Page 2 of 7 Name________________________ GSI________________ The fractional composition of Cr3+ and Cr6+ as a function of pH are shown below. Page 3 of 7 Name________________________ GSI________________ Ion Exchange Chromatography of Chromium Ions For the questions that follow, assume that all the species present are represented in the graphs above. 1. If Cr3+ binds to the resin, how many H+ ions will it displace? __3, 2, or 1______ 2. What is the pKa2 for H2CrO4? ___~6.5_____ 3. What is the pH of a 3M HCl solution? (Show your work) ____−0.5______ pH = ‐ log [H+] = ‐log 3 = −0.5 4. What is the primary species in solution for Cr3+ if the eluent is 3M HCl? ___ Cr3+_____ 5. What is the primary species in solution for Cr6+ if the eluent is 3M HCl? _____H2CrO4___ 6. Which species to you expect to elute first from the column, Cr3+ or Cr6+? ____Cr6+_____ 7. Explain your reasoning. The stationary phase is a cation exchange resin. Cations will bind but neither anions nor neutral (uncharged) species will bind. The primary species in solution at pH −0.5 are Cr3+ and H2CrO4. Cr3+ will be retained (displacing H+) and the Cr6+ existing as H2CrO4 will elute first. Page 4 of 7 Name________________________ GSI________________ EDTA titrations of Cr3+ The Cr3+ ion can be titrated quantitatively with EDTA. The Kf for the complexation reaction is 2.51×1023. 8. Write the chemical reaction for the titration of Cr3+ with EDTA (Y4−). Cr3+ (aq) + Y4− (aq) [CrY (aq)]1− 9. Calculate the conditional formation constant for the reaction if the solution is buffered to pH 8. Kf’ = ∙Kf =(4.2×10−3)(2.51×1023) = 1.05×1021 The concentration of Cr(III) was determined by back titration. For the reference titration a 5.00 mL aliquot of 0.2000 M Cr(III) solution was mixed with 10.00 mL of 0.2500 M EDTA. 10. What is the remaining volume of Cr(III) required to reach equivalence? Moles EDTA = total moles of Cr3+ at equivalence Moles EDTA = moles from aliquot + moles titrant (10.00 mL)(0.2500M EDTA) = (5.00 mL)(0.2000M) + moles titrant Moles titrant = 1.50×10−3 Volume titrant = moles/molarity = 1.50×10−3 /0.2000M = 7.50 mL For the samples of unknown, the same procedure was followed. For three trials, an average of 6.78 mL of standard Cr(III) was required. 11. What is the typical uncertainty in a 50 mL buret reading? ± 1.0mL ± 0.1mL ± 0.01mL 12. What is the average amount of Cr(III) unknown in the flasks? (report in moles) Moles EDTA = moles from aliquot of unknown + moles titrant (10.00 mL)(0.2500M EDTA) = moles unknown+ (6.78 mL)(0.2000M) Moles unknown = 1.14×10‐3 13. The Cr6+ ion does not bind as well as Cr3+ to EDTA. Explain why based on the ionic radii. Cr6+ is a smaller ion than Cr3+ To bind to metal ions EDTA must wrap around the ions in a 5 or 6 coordinate geometry Cr6+ must be too small for the EDTA to be able to bind effectively If the EDTA attempts to wrap tightly around the Cr6+ the functional groups in the EDTA will be crowded and repel each other (aka steric hindrance) Page 5 of 7 Name________________________ GSI________________ Flame AAS of Cr(VI) samples The Cr(VI) extracts of the contaminated water samples were analyzed by Atomic Absorption Spectroscopy (AAS). The calibration curve for a set of standard K2Cr2O7 solutions is shown below. Flame AAS of Cr(VI) Samples 0.70 0.60 Absorbance 0.50 0.40 y = 0.0657x ‐ 0.065 R² = 0.9911 0.30 0.20 0.10 0.00 0.00 2.00 4.00 6.00 8.00 10.00 12.00 Concentration (g/mL) 14. One unknown sample had an absorbance of 0.553. What is the concentration of Cr(VI) in the unknown? 0.553 = 0.657x – 0.065 X = 9.40 g/mL 15. A second unknown sample had an absorbance of 0.011. Can you determine an accurate concentration for this sample? Why or why not? No, an absorbance measurement that low would be unreliable because: 1) It is near the lower limit of detection for the instrument 2) It is near the lowest linear part of the calibration curve (the data does look a noisy at low concentrations) 16. The chromium standard solutions were prepared from a concentrated stock solution. The stock solution was prepared by transferring 0.2829 g (±0.0002 g) of K2Cr2O7 into a 1000.00 mL (±0.50 mL) volumetric flask. The sample was diluted to the mark and mixed thoroughly. Calculate the final concentration in g/mL. Report the absolute and percent error. (0.2829 g/1000.00 mL)×(1×106g/1 g) = 282.9 g/mL Balance error, (0.0002/0.2829)×100 = 0.07% Flask error, (0.50/1000.00)×100 = 0.05% % % % = 0.09% 282.9 g/mL ± 0.09% or 282.9 g/mL ± 0.2 g/mL Page 6 of 7 Name________________________ GSI________________ Absorbance UV/Vis The concentration of ions in the unknown can also be determined by UV/Vis spectroscopy. This experiment began with a separation of Cr3+ and Cr6+ using ion exchange chromatography. A 1.00mL sample of unknown was passed through the ion‐exchange column. The volume of the fractions collected was 10 mL. The UV/Vis spectrum for one of the fractions of Cr(VI) unknown is shown below. UV/Vis Spectrum of Cr(VI) fraction of an unknown 1.0 H2CrO4 0.75 0.50 0.25 0.00 200 400 600 800 Wavelength (nm) The extinction coefficient for Cr(VI) at 371 nm is 4730 L ∙mol−1 ∙ cm−1. 17. Using the data provided, determing the concentration of Cr(VI) in the original unknown. A = E b C C = (0.4)/(4730)(1) = 8×10−5 M in cuvette Original sample was 1 mL, diluted to ~10 mL fraction exiting the column Unknown is ten times more concentrated,s 8×10−4M 18. If Dr. Douskey’s pet rat Splinter drank 5.0 mL of the contaminated water that was the unknown in this analysis, would the rat die? Assume Splinter is a big rat with a mass of 1 kg. (8×10−4 mol/L)(0.005L)(118.01 g H2CrO4/mol) = 5×10−4 g = 0.5 mg H2CrO4 According ot the LD50 at the front of the exam it takes 80 mg/ 1 kg rat to kill half the population Splinter lives! (probably) 19. Sketch the basic schematics of a UV/Vis instrument below showing the few main components required for absorbance analysis. light source monochromator sample detector Page 7 of 7 ...
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