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Determination of an Equilibrium Constant for a Chemical Rxn written lab

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McMaster University
Adam Hitchcock

Determination of an Equilibrium  Constant Purpose Aspectrophotometer will be used to attain an absorbance values for partA and B of this lab. In partAstudents 2+ are to calculate the concentration of FeSCN ions that formed.Agraph will also be made from absorbance 2+ values and FeSCN concentrations. For the second part of the lab students are given the concentrations of 3+ - 2+ KSCN and Fe(NO ) an3 3s expected find the unknown equilibrium concentrations of Fe ,SCN, and FeSCN 3+ - 2+ and the K value for equilibrium of Rxn: Fe (aq)SCN (aq)FeSCN (aq) Procedure Please refer to: Chemistry 1E03 Laboratory Manual; McMaster University: Hamilton, ON, 2013; p 43-47. Given Info & useful equations for calculations: FeSCN2+¿ ¿ eq ¿ [ConcentrationProducts]eq C [KSCN] 0.002 mol/L K eq Fe3+¿ = [ConcentrationReactants]eq ¿ SCN−¿eq ¿ ¿ C =0.200mol/L(partA)&0.002mol/L(partB) C V =C V [Fe(NO3)3] 1 1 2 2 A=y[conc] (where ‘y’is K & ‘A’is absorbance value) V Total.050L Rxn: Fe 3+(aq)CN (aq)FeSCN 2(aq) Observations and Sample Calculation PartA: Table 1:(absorp.Values) Flask# Vol. KSCN Vol.Fe(NO )3 3L) Absorp.Value(at wavelength 447 nm) (mL) 1 1.00 49.0 0.247 2 2.00 48.0 0.445 3 3.00 47.0 0.659 Sample Calc. for Flask #1 Using: C 1 =1 V 2 2 V [KSCN]0.001L (Flask#1) (0.002mol/L)(0.001L) C = 2 (0.05L) C =4.00∙10 mol/L is the concentration of FeSCN ions formed(also the initial concentration of SCN ions) for 2 flask # 1 There is less moles of KSCN than Fe(NO ) ,3 3 students can assume that all the SCN ions in the KSCN solution will be completely used up to make FeSCN ions, thus students may conclude that the KSCN initial concentration is equal to the concentration of FeSCN of ions formed. Table 2:(values of calculations in PartA) - Flask # Concent. Initial SCN (in mol/L) 1 4.00e-5 2 8.00e-5 3 1.20e-4 Graph: (using values from Table 1 & 2) Observation and Sample Calculation Part B: Table 3:(absorp.values) Test Tube # 1 2 3 4 5 KSCN(mL) 1.00 2.00 3.00 4.00 5.00 Fe(NO )3 3L) 5.00 5.00 5.00 5.00 5.00 H 2 (mL) 4.00 3.00 2.00 1.00 ---- Absorbance 0.116 0.229 0.35 0.45 0.561 Value (nm) 7 1 For Test Tube#1 (sample calculation) a) Using: C 1 1C V2 2 [C1 )V 1 ] C = 2 V2 (0.002mol/L)(1mL)=C (10mL)2 C 22.00e-4(mol/L) is the initial concentration of SCN - Convert to mol  b) Using: C V 1C1V 2 2 (0.002mol/L)(5mL)= C (10mL)2 3+ C 21.00e-3(mol/L) is the initial concentration of Fe c) Using:A=y[conc.] A=0.116(absorption value for test tube# 1 (at 447 nm)) y=5560.7 (from slope of calibration curve partA‘y=5560.7x’from the equation y=mx+b) 0.116=5560.7[conc. EQ. of FeSCN ] 2+ 2+ [Conc.EQ. FeSCN ]=2.086e-5 ≈ 2.07e-5 (mol/L) Table 5:Ice table general format for Determination of an EQ Constant Lab Fe 3+(aq) + SCN (aq) ⇌               FeSCN 2+(aq) 3+ - I(nitial) Fe initial SCN initial ---- C(oncentration) -x -x +x E(quilibrium) Fe3+initial SCN -initial x=calculated from absorption equation Table 6: Sample Ice table Test Tube #1 3+ - 2+ Fe (aq) + SCN (aq) ⇌               FeSCN (aq) I(nitial) 1e-3 2e-4 ---- C(oncentration) -x -x +x E(qu
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