C Final Exam CHEM 114 A student did an experiment to find the equilibrium consta

Question

C Final Exam CHEM 114 A student did an experiment to find the equilibrium constant K. of the following reaction: Fes (aq) +scN (aq) FescNR (aq) iron thiocyanate thiocyanoiron(ll) He prepared the following solutions using 0.0020 M FedNOob AND o.0020 KscN and measured their absorbance: (mL) He also prepared a standard solution of FescN by 18 mL of0200 MFe(Noob into a 20 x 150 mm test tube labeled 5'. Pipetted 2 mL of 0.0020 M KSCN into the same test tube and measured its absorbance. He/She collected the following data for trials 1 and 2 TEST TUBE 2 Absorbance TUBE 1 of standard calculate the initial concentration of Fe?", based on the dilution that results from adding KscN solution and water to the original 0.0020 M Fe(NOs3 solution calculate lFe3 l using the equation: This should be the same for all two test tubes Test Tube FeNonji

Explanation / Answer

For the given data,

reaction,

Fe3+ + SCN- <==> [FeSCN]2+

In test tube 1,

initial [Fe3+] = 0.002 M x 5 ml/10 ml = 0.001 M

In test tube 2,

initial [Fe3+] = 0.002 M x 5 ml/10 ml = 0.001 M

In test tube 1,

initial [SCN-] = 0.002 M x 2 ml/10 ml = 0.0004 M

In test tube 2,

initial [SCN-] = 0.002 M x 3 ml/10 ml = 0.0006 M

For standard solution,

[SCN-] is the lower concentration and hence the limiting reagent in the reaction

[SCN-] present in std = [FeSCN]2+ formed in std

[FeSCN]2+std = 0.002 M x 2 ml/20 ml = 0.0002 M

So taking absorbance values from above,

Test tube 1,

[FeSCN]2+eq = (0.153/0.549) x 0.0002 = 5.57 x 10^-5 M

Test tube 2,

[FeSCN]2+eq = (0.214/0.549) x 0.0002 = 7.79 x 10^-5 M

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