A student prepared flask 3 at some non-standard temperature. She found the absor

Question

A student prepared flask 3 at some non-standard temperature. She found the absorbance of her solution and then determined that the equilibrium concentration of FeSCN^2+ was 0.00031 M. Determine the initial concentrations of both Fe^3- and SCN^- in the solution that would exist if no reaction happened (these would be the "initial" in your ICE chart). Determine the equilibrium concentration of Fe^3+ and SCN using the results from part [a] and the fact that the equilibrium concentrations of FeSCN^2+ is known. An ICE chart may be helpful here. Determine the equilibrium constant. Determine the initial concentrations of Fe^3+ and SCN^- in the solutions you will prepare for lab assuming that no reaction happens (again, these will be the "initial" in your ICE chart). What is the color of the starch-iodine complex that will be formed in Part V of the procedure? You will probably need to look this up somewhere, city your source!

Explanation / Answer

For the reaction of fe3+ with SCN- to form [FeSCN]2+ complex

[Fe3+]o or [SCN-]o = molarity of stock solution x volume of stock solution/total volume

[Fe3+]eq = [Fe3+]o - [FeSCN]2+eq

[SCN-]eq = [SCN-]o = [FeSCN]2+eq

Keq = [FeSCN]2+eq/[Fe3+]eq.[SCN-]eq

So,

Flask#                       [Fe3+]o                                                [SCN-]o                          

   1          0.002 M x 3 ml/10 ml = 6 x 10^-4                               0

   2          0.002 M x 3 ml/10 ml = 6 x 10^-4          0.002 M x 2 ml/10 ml = 4 x 10^-4

   3          0.002 M x 3 ml/10 ml = 6 x 10^-4          0.002 M x 3 ml/10 ml = 6 x 10^-4

   4          0.002 M x 3 ml/10 ml = 6 x 10^-4          0.002 M x 4 ml/10 ml = 8 x 10^-4

   5          0.002 M x 3 ml/10 ml = 6 x 10^-4          0.002 M x 5 ml/10 ml = 1 x 10^-3

Equilibrium concentrations

Flask#                        [Fe3+]eq                                                      [SCN-]eq                                        Keq

    1          6 x 10^-4 - 3.1 x 10^-4 = 2.9 x 10^-4                                   0                                                 0

    2          6 x 10^-4 - 3.1 x 10^-4 = 2.9 x 10^-4              4 x 10^-4 - 3.1 x 10^-4 = 9 x 10^-5            11877.40

    3          6 x 10^-4 - 3.1 x 10^-4 = 2.9 x 10^-4              6 x 10^-4 - 3.1 x 10^-4 = 2.9 x 10^-4          3686.10

    4          6 x 10^-4 - 3.1 x 10^-4 = 2.9 x 10^-4              8 x 10^-4 - 3.1 x 10^-4 = 4.9 x 10^-4          2181.56

    5          6 x 10^-4 - 3.1 x 10^-4 = 2.9 x 10^-4              1 x 10^-3 - 3.1 x 10^-4 = 6.9 x 10^-4          1549.22

mean Keq = 3858.856

3. Color of starc0iodine complex is deep purple or blue-black.

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