Chemistry lab 16 -postlab

Chemistry lab 16 -postlab - Chemistry Post-lab 16...

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Chemistry Post-lab 16 Calculations 1) a) In order to calculate the concentrations of FeSCN 2+ of the standard solutions, the molarity of HSCN of each dilution was obtained by using the formula Ci x Vi=Cf x Vf. Since the equilibrium shifted to the right due to the high concentration of Fe 3+ in the solution, the moles of FeSCN 2+ equaled to the moles of HSCN. Example: Dilution of solution A Ci x Vi = Cf x Vf Cf = 0.002 M x 1.00 mL / 100 mL = 2.00E-5.0 M HSCN The values of the standard solution are shown in Appendix A. b) In order to calculate the concentration of FeSCN 2+ of the equilibrium solutions, the εl, calculated by taking the slope of the straight line passing through the values of the standard solutions, was used. The slope of the straight line is m = (Af – Ai)/(Cf- Ci) Since A = Cεl, (Af – Ci) can be expressed as Cfεl-Ciεl or εl(Cf – Ci) Therefore, the slope of the line is m= εl(Cf – Ci)/(Cf – Ci) = εl The concentration of FeSCN 2+ can be calculated by diluting the absorbance by the slope of the straight line of the standard solution graph (εl ).
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C = A/εl [FeSCN 2+ ] = A/εl For instance: [FeSCN 2+ ] of tube 2 first reading = 2.33E-01 / 3.97E+03 = 5.87E-05 M FeSCN 2+ Values are shown in Appendix A. c) The number of moles of FeSCN 2+ at equilibrium in each mixture was determined by expressing the molarity in terms of (moles/mL), and multiplying this value by 10mL of total solution. For example: For tube 2 first reading 5.87E-5 moles/ 1000 mL x 10 mL = 5.87E-7 moles of FeSCN 2+ Values are shown in Appendix A. 2)
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This note was uploaded on 02/25/2012 for the course CHEM 4385 taught by Professor Staff during the Fall '11 term at Texas State.

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Chemistry lab 16 -postlab - Chemistry Post-lab 16...

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