h. Calculate the molarity of complex ion, Fe(SCN)+, that was present at equilibr
ID: 1046519 • Letter: H
Question
h. Calculate the molarity of complex ion, Fe(SCN)+, that was present at equilibrium for tubes 1 through 12. Note that the concentration of KSCN was very large compared to the concentration of iron(III) nitrate. This resulted in the reaction for the formation of the complex ion being driven essentially to completion and you can assume that all the iron(III) ion was converted to the complex ion. However, to calculate the concentrations you must take dilution into account. (Note: the final volume for all 12 tubes was 8.00 mL.)Explanation / Answer
Concentration of Fe(SCN)2+ = concentration of [Fe3+] in solution
test tube 1, [Fe(SCN)2+ = 7.5 x 10^-5 M x 4 ml/8 ml = 3.75 x 10^-5 M
test tube 2, [Fe(SCN)2+ = 7.5 x 10^-5 M x 3.5 ml/8 ml = 3.30 x 10^-5 M
test tube 3, [Fe(SCN)2+ = 7.5 x 10^-5 M x 3.0 ml/8 ml = 2.8125 x 10^-5 M
test tube 4, [Fe(SCN)2+ = 7.5 x 10^-5 M x 2.5 ml/8 ml = 2.344 x 10^-5 M
test tube 5, [Fe(SCN)2+ = 7.5 x 10^-5 M x 2.0 ml/8 ml = 1.875 x 10^-5 M
test tube 6, [Fe(SCN)2+ = 7.5 x 10^-5 M x 1.5 ml/8 ml = 1.41 x 10^-5 M
test tube 7, [Fe(SCN)2+ = 7.5 x 10^-5 M x 1.0 ml/8 ml = 9.375 x 10^-6 M
test tube 8, [Fe(SCN)2+ = 7.5 x 10^-5 M x 0.8 ml/8 ml = 7.5 x 10^-6 M
test tube 9, [Fe(SCN)2+ = 7.5 x 10^-5 M x 0.6 ml/8 ml = 5.625 x 10^-6 M
test tube 10, [Fe(SCN)2+ = 7.5 x 10^-5 M x 0.4 ml/8 ml = 3.75 x 10^-6 M
test tube11, [Fe(SCN)2+ = 7.5 x 10^-5 M x 0.3 ml/8 ml = 2.8125 x 10^-6 M
test tube 12, [Fe(SCN)2+ = 7.5 x 10^-5 M x 0.2 ml/8 ml = 1.875 x 10^-6 M
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