Quiz9 - (2*(116.12 1)+58.69)/58.69*0.02g = 98.5 mg. (one...

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Answer Key Chem 223 Post lab Quiz October 29, 2010 1. If a 10 mL aliquot of Ni 2+ solution is expected to contain 2 mg of nickel, which method for determining nickel would you choose, and why? 2 points for the explanation. 0.002 g Ni/58.69 g mol 1 = 0.034 millimoles. If precipitated with DMG, the dry mass would be (2*(116.12 1)+58.69)/58.69*0.002g = 9.9 mg. (one proton is removed from DMG in complexing the Ni 2+ ion). This is too little to weigh precisely. For EDTA titration, if the EDTA is ~ 0.02 M, titration volume would be 3.4×10 5 mol/2×10 2 mol L 1 = 1.7×10 3 L = 1.7 mL, too little for precise measurement. Further, Mg 2+ would interfere with the EDTA titration. Thus, the high dilution, low detection limit approach using spectrophotometry is preferred. 2. If a 10 mL aliquot of Ni 2+ solution is expected to contain 20 mg of nickel, which method for determining nickel would you choose, and why? 2 points for the explanation. 0.02 g Ni/58.69 g mol 1 = 0.34 millimoles. If precipitated with DMG, the dry mass would be
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Unformatted text preview: (2*(116.12 1)+58.69)/58.69*0.02g = 98.5 mg. (one proton is removed from DMG in complexing the Ni 2+ ion). This is enough that weighing precision should be adequate at the 1% level, but questionable at the 0.1% level. For EDTA titration, if the EDTA is ~ 0.02 M, titration volume would be 3.410 4 mol/210 2 mol L 1 = 1.710 2 L = 17 mL, about what one would desire. Further, Mg 2+ would interfere with the EDTA titration. Thus, gravimetry is the preferred method. 3. If you had a solution that was to be used to generate the unknown for this lab, why would you NOT use the spectrophotometric method to assay the solution? 1 point. The spectrophotometric method requires a standard solution for calibration. One can't standardize with the solution that is being prepared to be the standard! Useful data: atomic weight of Ni: 58.69. Molecular weight of DMG (neutral molecule): 116.12....
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