Final Garm CHEM114 22 A student did an experiment to find the equilibrium consta

Question

Final Garm CHEM114 22 A student did an experiment to find the equilibrium constant Ke of the following reaction: Fe3 (aq) scN (aq) FescN2 (aq) of 1.00 thiocyanoiron(Ill) iron(l) thiocyanate He prepared the following solutions using 0.0020 M Fe(NO) AND 0.0020 KscN and measured their absorbance He also prepared a standard solution of FescN by pipetting 18 mL of 0.200 M Fe(NO3)3 into a 20 x 150 mm test tub "5". Pipetted 2 ml of 0.0020 M KSCN into the same test tube and measured its absorbance HelShe collected the following data for trials 1 and 2 TEST TUBE 2 TEST Absorbance TUBE 1 Test tube 5) Calculate the initial concentration of Fe based on the dilution that results from adding KscN solution and water to original 0.0020 M Fe(NO, solution. Calculate Fe3 using the equation: [Fe' (Fe(Noss mL/total mLI x (0.0020 M) This should be the same for all two test tubes uestion 23 Number (mL)

Explanation / Answer

Fe3+ + SCN- <==> FeSCN2+

[Fe3+],

test tube 1 : initial [Fe3+] concentration = 0.002 M x 5 ml/10 ml = 0.001 M

test tube 2 : initial [Fe3+] concentration = 0.002 M x 5 ml/10 ml = 0.001 M

[SCN-],

test tube 1 : initial [SCN-] concentration = 0.002 M x 2 ml/10 ml = 0.0004 M

test tube 2 : initial [SCN-] concentration = 0.002 M x 3 ml/10 ml = 0.0006 M

[FeSCN]2+std = [SCN-] in standard solution

                        = 0.002 M x 2 ml/20 ml

                        = 0.0002 M

[FeSCN]2+eq = (Aeq/Astd) x [FeSCN]2+std

For Test tube 1,

[FeSCN]2+eq = (0.153/0.549) x 0.0002

                       = 5.57 x 10^-5 M

For Test tube 2,

[FeSCN]2+eq = (0.214/0.549) x 0.0002

                       = 7.80 x 10^-5 M

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