# lab - 200 M.003#3 Determination of an equilibrium Constant...

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. . 00200 M 002 mL0 05 mL 5M concentrations of FeSCN 2+ are . × - 4 00 10 5M , . × - 8 00 10 5M , and . × - 1 20 10 4M respectively these values can be used to create the cali . . 0200 M 001 mL0 05 mL 5M . . 0200 M 003 mL0 05 mL M ituted into the equation = [ ] A y conc as the y value, in order to find the concentration of FeSCN 2+ . Exp. #3: Determination of an equilibrium Constant November, 15, 2009 Lauren Thomson Lab Partners: Jeffrey Leon & Nicole Thompson CHEM 1A03 Section 23 Pre-lab Questions: 1. Why do you calibrate the spectrophotometer? What will the calibration line be used for? The spectrophotometer must be calibrated as different molecules will absorb different wavelengths of light and the spectrophotometer cannot distinguish which molecule you are working with. The calibration line is obtained by plotting the various absorptions along with their respective concentrations. This line can be used to distinguish an unknown concentration with regards to any absorbency. 2. In an experiment, equal volumes of 0.00200 M FeCl 3 and 0.00300 M NaSCN were mixed together according to the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) The equilibrium concentration of FeSCN 2+ (aq) was 2.10 x 10 -4 M. Calculate the equilibrium constant for the reaction. Initial Concentrations: C 1 V 1 =C 2 V 2 [Fe 3+ ] = (0.00200M)(1)=C 2 (2) [SCN - ] = (0.00300M)(1)=C 2 (2) C 2 = 1.00 x 10 -3 M C 2 = 1.50 x 10 -3 M Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) I 1.00 x 10 -3 M 1.50 x 10 -3 M 0M C -X -X +X E 1.00 x 10 -3 M – X 1.00 x 10 -3 M – X X kc = ÷ . × - - . × - - x 1 00 10 3 x1 50 10 3 x = . × - x 2 10 10 4 = kc 206 Purpose:

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